[
  {
    "id": 25001,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove, that for any $p, q \\in \\mathbb{N}$, such that $\\sqrt{11} > \\frac{p}{q}$, following inequality holds:\n$$\n\\sqrt{11} - \\frac{p}{q} > \\frac{1}{2pq}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We can assume that $p$ and $q$ are coprime, and since both sides of first inequality are positive, we can change it to $11q^2 > p^2$. The same way we can change second inequality:\n$$\n11p^2q^2 > p^4 + p^2 + \\frac{1}{4}.\n$$\nTo see this one holds, we will prove stronger one:\n$$\n11p^2q^2 \\geq p^4 + 2p^2.\n$$\nIndeed, dividing this inequality by $p^2$ we get $11q^2 \\geq p^2+2$, and since we already know that $11q^2 > p^2$ we only have to see, that $11q^2$ can't be equal to $p^2+1$. Since we know that the only reminders of squares (mod 11) are 0, 1, 3, 4, 5 and 9, $p^2+1$ can't be divisible by 11, and therefore $11q^2 \\neq p^2+1$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 25002,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 3$ be an integer, such that $4n + 1$ is a prime number. Prove that $4n + 1$ divides $n^{2n} - 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $p := 4n + 1$ is a prime number, each non-zero remainder modulo $p$ possesses a unique multiplicative inverse. Since $-4 \\cdot n \\equiv 1 \\pmod{p}$, we have $n \\equiv (-4)^{-1} \\pmod{p}$, from which we deduce that $n \\equiv -(2^{-1})^2$. Consequently,\n$$\nn^{2n} - 1 \\equiv (-(2^{-1})^2)^{2n} - 1 \\equiv (2^{-1})^{4n} - 1 \\equiv 0 \\pmod{p},\n$$\nby Fermat's Little Theorem.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25003,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\mathbb{T} = \\{1, 3, 6, 10, 15, \\dots\\}$ be the set of triangular numbers, i.e. numbers of the form $T_n = \\frac{n(n+1)}{2}$. Let $f$ be a function defined on the set of positive integers such that\n1) $f(n)$ is a positive integer for each $n$;\n2) $f(uv) = f(u)f(v)$ for any pair $(u, v)$ of coprime numbers;\n3) $f(a + b + c) = f(a) + f(b) + f(c)$ for $a, b, c \\in \\mathbb{T}$.\nProve that $f(n) = n$ for all $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "It is not difficult to find $f(n)$ for small $n$:\n$$\nf(1 \\cdot 1) = f(1)f(1) \\text{ therefore } f(1) = 1;\n$$\n$$\nf(3) = f(1) + f(1) + f(1) = 3;\n$$\n$$\nf(5) = f(1 + 1 + 3) = 5;\n$$\n$$\nf(10) = f(1 + 3 + 6) = 4 + 3f(2) \\text{ and } f(10) = f(2 \\cdot 5) = f(2)f(5) = 5f(2) \\text{ therefore } f(2) = 2.\n$$\nNow we use induction. Suppose that $f(n) = n$ for all $n < N$. Let us show that $f(N) = N$. Since $f$ is multiplicative we may assume that $N = p^r$ for some prime $p$. Consider several similar cases.\n\n1) $N = 3^r$. Then\n$$\nf(3T_{3^{r-1}}) = 3f(T_{3^{r-1}}) = 3f\\left(\\frac{3^{r-1}(3^{r-1}+1)}{2}\\right) = 3f(3^{r-1})f\\left(\\frac{3^{r-1}+1}{2}\\right)\n$$\nAnd from the other hand\n$$\nf(3T_{3^{r-1}}) = f\\left(\\frac{3^r(3^{r-1}+1)}{2}\\right) = f(3^r)f\\left(\\frac{3^{r-1}+1}{2}\\right)\n$$\nSo we conclude that $f(3^r) = 3^r$ since $f(3^{r-1}) = 3^{r-1}$ by induction hypothesis.\n\nand\n$$\nf(T_{s-1} + T_{s-1} + T_s) = f\\left(\\frac{s(3s-1)}{2}\\right) = f\\left(\\frac{s}{2}\\right)f(3s-1) = \\frac{s}{2}f(p^r).\n$$\nHence $f(p^r) = p^r$.\n\n3) $N = p^r$, where $p$ is an odd prime and $p^r = 3s + 1$. Similarly we have\n$$\n\\begin{aligned}\nf(T_{s-1} + T_s + T_s) &= \\frac{s(s-1)}{2} + \\frac{s(s+1)}{2} + \\frac{s(s+1)}{2} = \\frac{s(3s+1)}{2} = \\frac{sp^r}{2} \\\\\n&= f\\left(\\frac{s(3s+1)}{2}\\right) = f\\left(\\frac{s}{2}\\right)f(3s+1) = \\frac{s}{2}f(p^r).\n\\end{aligned}\n$$\nHence $f(p^r) = p^r$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25004,
    "subject": "Mathematics (Multi-modal)",
    "question": "An infinite set $B$ consisting of non-negative integers has the following property. For each $a, b \\in B$ ($a > b$) the number $\\frac{a-b}{(a,b)} \\in B$. Prove that $B$ contains all non-negative integers. Here $(a, b)$ is the greatest common divisor of numbers $a$ and $b$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $d$ is the greatest common divisor of all the numbers in set $B$, let $A = \\{b/d : b \\in B\\}$. Then for each $a, b \\in A$ ($a > b$) we have\n$$\n\\frac{a-b}{d(a,b)} \\in A. \\quad (*)\n$$\nObserve that the greatest common divisor of the set $A$ equals $1$, therefore we can find a finite subset $A_1 \\subset A$ for which $\\gcd A_1 = 1$. We may think that the sum of elements of $A_1$ is minimal possible. Choose numbers $a, b \\in A_1$ ($a > b$) and replace $a$ in the set $A_1$ with $\\frac{a-b}{d(a,b)}$. The greatest common divisor of the obtained set equals $1$. But the sum of numbers decreases by this operation, which contradicts the minimality of $A_1$.\n\nThus, $A_1 = \\{1\\}$. Therefore all the numbers in the set $A$ have residue $1$ modulo $d$. Take an arbitrary $a = kd + 1 \\in A$ and $b = 1$. Then $k \\in A$ by $(*)$ and hence $k = ds + 1$. But $(k, kd + 1) = 1$, therefore $\\frac{kd+1-ds-1}{d} = k - s = (d-1)s + 1 \\in A$, so $s$ is divisible by $d$. But $s \\in A$, therefore $s-1$ is also divisible by $d$, hence $d = 1$ (that means that $B = A$). Thus we have checked that if $a = kd + 1 = k + 1 \\in A$ then $a-1 = k \\in A$. Then all non-negative integers belong to $A$ because it is infinite.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25005,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that there exist only finitely many triples of positive integers $(n, a, b)$ such that:\n$$\nn! = 2^a - 2^b.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $n!$ is divisible by $3^{\\lfloor n/3 \\rfloor}$, $(2^{a-b}-1)2^b = 2^a - 2^b$ is divisible by $3^{\\lfloor n/3 \\rfloor}$. From the Lifting the Exponent Lemma we obtain that $a-b$ is divisible by $3^{\\lfloor n/3 \\rfloor-1}$. So $a-b \\ge 3^{\\lfloor n/3 \\rfloor-1} \\ge 3^{n/3-2}$. Hence, the right hand side of our prior equality is greater than\n$$\n2^{a-b} - 1 \\ge 2^{a-b-1} \\ge 2^{3^{n/3-2}-1}\n$$\nwhich for sufficiently large $n$ is obviously greater than $2^{n^2} > n!$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25006,
    "subject": "Mathematics (Multi-modal)",
    "question": "Does there exist a polynomial $f(x)$ with integer coefficients of degree four such that for any integer $k$, the polynomial $f(x) + k$ is either irreducible or has a rational root?",
    "options": [],
    "answer": "Yes; for example f(x) = x^4 + 2x.",
    "solution": "The answer is yes, and our example will be the polynomial $f(x) = x^4 + 2x$. Assume that there exist a constant $k$ such that $f(x) + k$ is reducible but has no rational zeros. This means that $f(x) + k$ factors as the product of two irreducible quadratic polynomials. Since $f(x) + k$ is monic so must the two factors be. Assume that $f(x) + k = (x^2 + a x + b)(x^2 + c x + d)$. By expanding, we find that\n$$\nx^4 + 2x + k = x^4 + (a + c)x^3 + (b + d + a c)x^2 + (a d + b c)x + b d\n$$\nBy comparing coefficients we see that $a + c = 0$, $b + d + a c = 0$ and $a d + b c = 2$. From the first equality we get $c = -a$, and the two others become $b + d = a^2$ and $(b - d)a = 2$. From $b + d = a^2$ it follows that $b + d$ and $a$ has the same parity. Since $b + d$ and $b - d$ has the same parity, it follows that $2 = (b - d)a$ is either odd or divisible by 4 which is a contradiction. Hence for any integer $k$ $f(x) + k$ is either irreducible or has a rational root, and the answer is therefore yes.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 25007,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for arbitrary positive integer $n$ the following inequality holds\n$$\n\\frac{1}{1^2 + 2019} + \\frac{1}{2^2 + 2019} + \\dots + \\frac{1}{n^2 + 2019} < \\frac{1}{22}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 25008,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all integers which can be expressed in the form\n$$\nx_1x_2 + x_2x_3 + \\dots + x_{k-1}x_k,\n$$\nwhere $x_1, x_2, \\dots, x_k$ are positive integers whose sum equals $2019$.",
    "options": [],
    "answer": "All integers from 2018 to 1019090 inclusive.",
    "solution": "First, easy induction shows that $\\sum_{i=1}^{k-1} x_i x_{i+1} \\ge (\\sum_{i=1}^k x_i) - 1$.\n\nBase case $k=2$: $x_1x_2 - (x_1+x_2-1) = (x_1-1)(x_2-1) \\ge 0$. Assuming the inequality holds for a certain $k$ and adding the obvious $x_kx_{k+1} \\ge x_{k+1}$, we get the claim for $k+1$.\n\nGiven the conditions of the problem, this shows that (for a given $n$) the quadratic form in question takes values $\\ge n-1$; and the minimum $n-1$ is attained e.g. for $k=n$ and all $x_i=1$.\n\nNow to the upper bound. The fine point is that $k$ is variable. So, let $x_1, \\dots, x_k$ be a $k$-string of positive integers with $\\sum x_i = n$, and with $k \\ge 4$; and let $V$ be the generated value $V = \\sum_{i=1}^{k-1} x_i x_{i+1}$. If $x_2 \\le x_3$, we merge $x_1$ with $x_2$; and if $x_2 > x_3$, we merge $x_3$ with $x_4$, thus creating the following $(k-1)$-string (with entries summing to $n$):\n$(x_1+x_2), x_3, \\dots, x_k, \\text{ resp. } x_1, x_2, (x_3+x_4), x_5, \\dots, x_k \\quad (\\text{if } k=4, x_5=0)$.\nIf $\\tilde{V}$ is the new value of the quantity under consideration then, in the first case $\\tilde{V} - V = x_1x_3 - x_1x_2 \\ge 0$; and in the second case\n$$\n\\tilde{V} - V = x_2x_4 - x_3x_4 + x_3x_5 \\ge 0\n$$\nAfter several steps $k$ comes down to $3$ and we arrive at a $3$-string $z_1, z_2, z_3$ (with $z_1 + z_2 + z_3 = n$) producing the value\n$$\nW = z_1z_2 + z_2z_3 = z_2(z_1 + z_3) \\ge V\n$$\nThe product of two integers with a given sum $n$ has a maximum\n$$\nM_n = \\lfloor n/2 \\rfloor \\cdot \\lceil n/2 \\rceil = \\lfloor (n^2 + 1)/4 \\rfloor\n$$\nTo show that all integer values between $n-1$ and $M_n$ are attained, we focus on strings $x_1, \\dots, x_k$ ending in $x_k=1$. We claim that these alone are enough to generate all those values. Induction again. Base $n=2$: obvious. Fix $n>2$ and assume that positive-integer strings with sum $n-1$, ending in a $1$, yield all values from $n-2$ to $M_{n-1}$. At the end of each of these strings (next to the terminal $1$) we attach another $1$; the value of the quadratic form grows by $1$. So we already have strings with sum $n$ and with last entry $1$, producing all values from $n-1$ to $M_{n-1}+1$.\n\nNow, if $n$ is even, $n=2m$, the triples (3-strings) $m-1, m, 1$ and $m, m-1, 1$ produce the values $M_n = m^2$ and $M_n-1 = m^2-1$; and the quadruples (4-strings) $j, m-2, m+1-j, 1$ with $j=1, \\dots, m$ give values from $m^2-2$ down to $m^2-m-1$ (which is below $M_{n-1}$). If $n$ is odd, $n=2m+1$, the value $M_n = m^2+m$ comes from the triples $m, m, 1$; and now the quadruples $j, m-1, m+1-j, 1$ with $j=1, \\dots, m$ yield the values from $m^2+m-1$ down to $m^2$ (below $M_{n-1}$). In each case, as $j$ ranges from $1$ to $m$, the generated values of the quadratic form sweep (with slight excess) the entire missing interval. Induction is completed and the claim results.\n\nThe answer follows: the values of $\\sum_{i=1}^{k-1} x_i x_{i+1}$ are all integers from $n-1$ to $M_n$ (inclusive).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25009,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m \\ge 2$ be a positive integer. Consider the $2m$ numbers\n$$\n1 \\cdot 2, 2 \\cdot 3, 3 \\cdot 4, \\dots, 2m(2m + 1).\n$$\nA move consists of choosing three numbers $a, b, c$, and replacing them with the single number\n$$\n\\frac{abc}{ab + bc + ca}.\n$$\nAfter $m-1$ such moves, only two numbers will remain. Supposing one of these is $\\frac{4}{3}$, show that the other exceeds 4.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denoting the new number\n$$\ng = \\frac{abc}{ab + bc + ca},\n$$\nits reciprocal is\n$$\n\\frac{1}{g} = \\frac{ab + bc + ca}{abc} = \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c}.\n$$\nThe sum of all reciprocals is therefore invariantly\n$$\n\\sum_{k=1}^{2m} \\frac{1}{k(k+1)} = \\sum_{k=1}^{2m} \\left( \\frac{1}{k} - \\frac{1}{k+1} \\right) = 1 - \\frac{1}{2m+1} = \\frac{2m}{2m+1}.\n$$\nThe other remaining number $x$ can then be calculated from the equation\n$$\n\\frac{1}{x} + \\frac{3}{4} = \\frac{2m}{2m+1},\n$$\nleading to\n$$\nx = \\frac{4(2m+1)}{2m-3} = 4 + \\frac{16}{2m-3} > 4.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25010,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all positive integers $n$ with the following property: For any integer $k$ the polynomial $x^n + k$ is either irreducible or has an integer root.",
    "options": [],
    "answer": "n = 1 or n is prime",
    "solution": "The answer is $n = 1$ or $n$ prime. It is clear that $n = 1$ has the desired property.\n\nAssume that $n > 1$. To see that $n$ must be a prime, assume that $n$ is composite, and let $d$ be a non-trivial divisor of $n$.\nWe consider $k = -2^d$. Since $2$ is a prime number, and $d < n$, $2^d$ is not a perfect $n$th-power so $x^n - 2^d = 0$ has no integer solutions. On the other hand, $x^n - 2^d = (x^{n/d})^d - 2^d$ is divisible by $x^{n/d} - 2$ which has degree less than $n$ since $d > 1$. Therefore $x^n + k$ has no integer root and is not irreducible, and $n$ does not have the desired property.\n\nAssume that $n$ is a prime. If $n = 2$, $n$ has the desired property, so assume that $n$ is odd. Let $k$ be given, and define $a = -k^{1/n}$, i.e. $a$ is the real root of $x^n + k$. If $k$ is a perfect $n$th-power, $a$ is an integer, and $x^n + k$ has an integer root. Assume that $k$ is not a perfect $n$th-power. Assume that $f(x)$ is a non-constant polynomial with integer coefficients dividing $x^n + k$. Any root $r \\in \\mathbb{C}$ of $f$ satisfy $r^n = -k$, and hence $|r|^n = k = |a|^n$ since also $a^n = -k$. Therefore $|r| = |a|$.\n\nAssume that $\\deg f = m$, and $f$ has roots $r_1, \\dots, r_m$. The constant term of $f$ is an integer since $f(x)$ divides $x^n + k$ which has integer coefficients. But the constant term is also given by $(-1)^m r_1 \\cdots r_m$, so $|(-1)^m r_1 \\cdots r_m| = |a^m|$ is an integer, and hence $a^m$ is an integer. But $a$ was given by $a = -k^{1/n}$ so since $n$ is a prime, and $k$ is not a perfect $n$th-power, $a^m$ is an integer if and only if $n \\mid m$. Therefore $n \\mid m$, and $\\deg f = m \\ge n$, so we must have $f(x) = x^n + k$ which shows that $x^n + k$ is irreducible, and $n$ has the desired property.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 25011,
    "subject": "Mathematics (Multi-modal)",
    "question": "For all non-negative real numbers $x$, $y$, $z$ with $x \\ge y$, prove inequality\n$$\n\\frac{x^3 - y^3 + z^3 + 1}{6} \\ge (x - y) \\sqrt{xyz}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "From AM-GM inequality we have $\\sqrt{xyz} \\le \\frac{xy+z}{2}$. Hence it suffices to prove that\n$$\n\\frac{x^3 - y^3 + z^3 + 1}{3} \\ge (x - y)(xy + z),\n$$\nwhich is equivalent to\n$$\n\\frac{(x - y)^3 + z^3 + 1}{3} \\ge z(x - y).\n$$\nThis inequality follows directly from AM-GM inequality applied to the numbers $(x-y)^3$, $z^3$ and $1$.\nNotice that\n$$\n\\frac{x^3 - y^3 + z^3 + 1}{6} = \\frac{(x - y)^3 + xy(x - y) + xy(x - y) + xy(x - y) + z^3 + 1}{6}.\n$$\nApplying AM-GM inequality this yields\n$$\n\\frac{x^3 - y^3 + z^3 + 1}{6} \\ge \\sqrt[6]{(x-y)^3 \\cdot (xy(x-y))^3} \\cdot z^3 \\cdot 1 = (x-y)\\sqrt{xyz}.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 25012,
    "subject": "Mathematics (Multi-modal)",
    "question": "Does there exist a positive integer $n$ such that the first and the second digit of $2^n$ (in decimal notation) are 3 and 9, respectively?",
    "options": [],
    "answer": "Yes",
    "solution": "Yes. Let us observe powers of two of the form $2^{10k} = (2^{10})^k = 1024^k = (1,024)^k \\cdot 10^{3k}$. Let $m$ be the smallest integer such that $(1,024)^m \\ge 3,9$ ($m$ must exist since $(1,024)^k$ is an exponential function and is unbounded from above). From the definition of $m$ we know that $(1,024)^{m-1} < 3,9$, therefore $(1,024)^m = (1,024)^{m-1} \\cdot 1,024 < 3,9 \\cdot 1,024 = 3,9+3,9 \\cdot 0,024 < 3,9+4 \\cdot 0,025 = 4,0$. Thus $3,9 \\le (1,024)^m < 4,0$ and the first digits of $2^{10m} = (1,024)^m \\cdot 10^{3m}$ are indeed 3 and 9.\n\n**Remark 1.** The value of $m$ found above is 58 and the respective power of two is $2^{580} = 3,957286... \\cdot 10^{174}$.\n\n**Remark 2.** The smallest power of two satisfying the requirements is $2^{95} = 39614081257132168796771975168$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25013,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that\n$$\nf(x^3 + y^3) = f(x^3) + 3x^3f(x)f(y) + 3f(x)(f(y))^2 + y^6f(y)\n$$\nfor every two real numbers $x$ and $y$.",
    "options": [],
    "answer": "f(x) = 0 for all real x; f(x) = x^3 for all real x",
    "solution": "Answer: $f(x) = 0$ and $f(x) = x^3$.\n\nSubstituting $x = y = 0$ into the original equation, we obtain $3(f(0))^3 = 0$ which implies\n$$\nf(0) = 0. \\qquad (\\text{Eq-1})\n$$\nSubstituting $x = 0$ into the original equation and applying (Eq-1), we get\n$$\nf(y^3) = y^6 f(y). \\qquad (\\text{Eq-2})\n$$\nReplacing the first term of the r.h.s. of the original equation by (Eq-2), we find\n$$\nf(x^3 + y^3) = x^6 f(x) + 3x^3 f(x)f(y) + 3f(x)(f(y))^2 + y^6 f(y). \\qquad (\\text{Eq-3})\n$$\nSwapping the variables $x$ and $y$ in (Eq-3), we find\n$$\n3x^3 f(x)f(y) + 3f(x)(f(y))^2 = 3y^3 f(y)f(x) + 3f(y)(f(x))^2. \\qquad (\\text{Eq-4})\n$$\nBy collecting similar terms in (Eq-4) and factorizing, we get\n$$\nf(x)f(y)(f(x) - x^3 - f(y) + y^3) = 0. \\qquad (\\text{Eq-5})\n$$\nSuppose that $f(x) = f(y)$ for some $x \\neq y$. Then $x^3 \\neq y^3$ as cubing is injective, whence $f(x) - x^3 - f(y) + y^3 \\neq 0$. By (Eq-5), we get $f(x)f(y) = 0$, which implies $f(x) = f(y) = 0$ by the choice of $x$ and $y$. Consequently, $f$ cannot take non-zero values more than once.\n\nSuppose now that $f(b) = 0$ for some $b \\neq 0$. Substituting $y = b$ into the original equation, we obtain $f(x^3 + b^3) = f(x^3)$ for all $x$. Now if $f(a) \\neq 0$ for some $a$ then taking $x = \\sqrt[3]{a}$ in the last equality contradicts the previous paragraph since $b^3 \\neq 0$. Hence $f(x) = 0$ for all $x$. This function satisfies the original equation.\n\nIt remains to consider the case where $f(x) = 0$ implies $x = 0$. Substituting $y = 1$ into (Eq-5), we obtain $f(x) - x^3 - f(1) + 1 = 0$ for all non-zero $x$. In other words, $f(x) = x^3 + c$ for all $x \\neq 0$ where $c = f(1) - 1$. After substituting $y = x$ into (Eq-2), simplifying now gives $c = x^6c$ which is possible only if $c = 0$. Hence $f(x) = x^3$ for all $x$. This function also satisfies the original equation.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25014,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $F_n$ be a sequence defined recursively by $F_1 = F_2 = 1$ and $F_{n+1} = F_n + F_{n-1}$ for $n \\ge 2$. Find all pairs of positive integers $(x, y)$ such that\n$$\n5F_x - 3F_y = 1\n$$",
    "options": [],
    "answer": "(3,4), (5,6), (6,7)",
    "solution": "From the equation $5F_x = 3F_y + 1$ we have\n$$\n3F_y + 1 = 5F_x > 3F_x + 1 \\implies y > x\n$$\nOn the other hand, if $y \\ge x + 2$ and $x > 1$ then\n$$\n3F_y + 1 \\ge 3F_{x+2} + 1 = 3(F_{x+1} + 3F_x) + 1 = 6F_x + 3F_{x-1} + 1 > 5F_x\n$$\nwe have a contradiction. Therefore $y = x + 1$ and we have to solve the equation becomes\n$3F_{x+1} + 1 = 5F_x$\nWe will show by induction that $3F_{x+1} + 1 < 5F_x$ for any $x \\ge 7$.\nIndeed, for $x = 7$ we have $F_7 = 13$, $F_8 = 21$. Therefore, $3F_8 + 1 = 3 \\cdot 21 + 1 = 64 < 5F_7 = 65$.\nFor $x = 8$ we have $3F_9 + 1 = 103 < 5F_8 = 105$.\nAssume $3F_{k+1} + 1 < 5F_k$ and $3F_{k+2} + 1 < 5F_{k+1}$ for some $k \\ge 7$. We have then $3(F_{k+1} + F_{k+2}) + 2 < 5(F_k + F_{k+1})$. Thus\n$$\n3F_{k+3} + 1 < 3F_{k+3} + 2 < 5F_{k+2}\n$$\nFor $x < 7$ we can check and see that $x \\in \\{3, 5, 6\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25015,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{R} \\to [-2019, 2019]$, for which\n$$\n2f(f(x)) + f(f(-x)) + 8f(x) + 4f(-x) + 4x = 0\n$$\nfor all real numbers $x$.",
    "options": [],
    "answer": "No such function exists.",
    "solution": "Let us define an auxiliary function $g: \\mathbb{R} \\to \\mathbb{R}$ by setting\n$$\ng(x) = f(f(x)) + 4f(x) + 4x\n$$\nfor all real numbers $x$. The given functional equation may be written as\n$$\n2g(x) = -g(-x),\n$$\nfor all real numbers $x$, so that\n$$\n4g(x) = -2g(-x) = -(-g(-(-x))) = g(x),\n$$\nfor all real numbers $x$. We obtain from this that\n$$\nf(f(x)) + 4f(x) + 4x = g(x) = 0,\n$$\nfor all real numbers $x$.\nLet now $x$ be an arbitrary nonzero real number, and let us define a sequence $a_0, a_1, a_2, \\dots$ by setting\n$$\na_0 = x, \\quad a_1 = f(x), \\quad a_2 = f(f(x)), \\quad a_3 = f(f(f(x))), \\quad \\dots\n$$\nso that\n$$\na_{n+1} = f(a_n) = f(f(a_{n-1})) = -4f(a_{n-1}) - 4a_{n-1} = -4a_n - 4a_{n-1},\n$$\nfor each positive integer $n$. The general term of the sequence may be written as a function of $n$ through\n$$\na_n = (A + Bn)(-2)^n,\n$$\nwhich holds for all nonnegative integers $n$, and where\n$$\nA = a_0 = x \\quad \\text{and} \\quad B = -\\frac{a_1}{2} - a_0 = -\\frac{f(x)}{2} - x.\n$$\nIf $B \\neq 0$, then we may choose a positive integer $n$ so that $|A + Bn| > 1$ and $2^n > 2019$, and thus\n$$\n|f(a_{n-1})| = |a_n| = |A + Bn| \\cdot |(-2)^n| > 2019,\n$$\nagainst the fact that $f(a_{n-1}) \\in [-2019, 2019]$. Therefore we must have $B = 0$. But now we may choose a positive integer $n$ so that $|x| 2^n > 2019$, leading to\n$$\n|f(a_{n-1})| = |a_n| = |a_0| \\cdot |(-2)^n| = |x| 2^n > 2019,\n$$\nwhich again produces a contradiction. We conclude that no function with the desired properties exists.\n\n\nIf $f$ satisfies the problem condition, then for all $x \\in \\mathbb{R}$ we have\n$$\n\\begin{aligned}\n4|x| &= |2f(f(x)) + f(f(-x)) + 8f(x) + 4f(-x)| \\\\\n&\\leq 2|f(f(x))| + |f(f(-x))| + 8|f(x)| + 4|f(-x)| \\\\\n&\\leq 2 \\cdot 2019 + 2019 + 8 \\cdot 2019 + 4 \\cdot 2019 = 30285.\n\\end{aligned}\n$$\nThis obviously does not hold for the big enough value of $x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25016,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that\n$$\nf(xf(y) - y^2) = (y + 1)f(x - y)\n$$\nholds for all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "f(x) = 0 for all x; f(x) = x + 1 for all x",
    "solution": "Putting $y = -1$ and $x = 0$, we obtain $f(-1) = 0$.\nSuppose that $f$ has another zero, i.e., there is an $a \\neq -1$ with $f(a) = 0$. Putting $x = 0$ and $y = -a$ gives $f(-a^2) = 0$. Now choosing $y = a$, we obtain $f(x - a) = 0$. Hence, $f(x) = 0$ for all $x$ which clearly is a solution.\n\nLet us assume now that $x = -1$ is the only zero of $f$. Then $x = y - 1$ gives $f((y - 1)f(y) - y^2) = 0$. Consequently, we have $(y - 1)f(y) = y^2 - 1$ for all $y$ which implies $f(x) = x + 1$ for all $x \\neq 1$. Setting $y = 2, x = 3$ gives $f(1) = 2$. Hence, $f(x) = x + 1$ for all $x$ which is another solution.\n\nAltogether, the only solutions are $f(x) = 0$ and $f(x) = x + 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25017,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest positive integer that cannot be written in the form $\\binom{a}{2} + \\binom{b}{2} + c$ with nonnegative integers $a, b, c$ satisfying $a \\ge b \\ge c$ and $a + b \\le 2019$.",
    "options": [],
    "answer": "1915900",
    "solution": "The number is $m = \\binom{1957}{2} + \\binom{63}{2} + 1 = 1,915,900$.\nAssume that $m$ has a representation as above. Then $a \\le 1957$ as $\\binom{1958}{2} > m$. On the other hand, by $\\binom{a}{2} + \\binom{b}{2} + c \\le \\binom{a+1}{2} + \\binom{b-1}{2} + (b-1)$ it follows that the largest number that can be represented as above with $a \\le 1957$ is $\\binom{1957}{2} + \\binom{62}{2} + 62 = m - 1$, a contradiction.\n\nIt remains to show that all natural numbers smaller than $m$ have a representation in the form $\\binom{a}{2} + \\binom{b}{2} + c$ with $a \\ge b \\ge c$ and $a+b \\le 2019$. If some number $k$ has such a representation and $c > 0$ or $c = 0, b \\ge 2$, then $k-1$ can be represented as $\\binom{a}{2} + \\binom{b}{2} + (c-1)$ or $\\binom{a}{2} + \\binom{b-1}{2} + (b-2)$, respectively. Hence, we can represent all integers between $\\binom{a}{2}$ and $k$ in the desired form. Therefore and because we have a representation for $m-1$ already, it suffices to show that the numbers $\\ell_s = \\binom{1957-s}{2} - 1$ with $s = 0, 1, \\dots, 1954$ can be represented. Now the claim follows by $\\ell_0 = \\binom{1956}{2} + \\binom{63}{2} + 2$ and $\\binom{1956-s}{2} < \\ell_s \\le \\binom{1956-s}{2} + \\binom{b}{2}$ with $b = \\min\\{63, 1956-s\\}$ for $s = 1, 2, \\dots, 1954$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25018,
    "subject": "Mathematics (Multi-modal)",
    "question": "A fair $k$-die is a die with $k$ sides and the numbers $1, 2, \\dots, k$ written one on each side such that the probability of getting one of these numbers when rolling the die is exactly $\\frac{1}{k}$ for each number. Let $n$ be a positive integer and $a_1 \\ge a_2 \\ge \\dots \\ge a_n > 0$ be integers. Georg has $n$ fair dice with $a_1, a_2, \\dots, a_n$ sides, respectively. When he rolls the $n$ dice, he sums the $n$ numbers he gets. Prove that the probability of getting a specific sum is at most $\\frac{1}{a_1}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25019,
    "subject": "Mathematics (Multi-modal)",
    "question": "Magician puts on the $3 \\times 100$ board cards with numbers from $1$ to $300$ in a \"snake-like\" line so that consecutive numbers are side-to-side adjacent (either horizontally or vertically, not diagonally). The numbers are written on the bottom sides of the cards, the upper sides are empty. After that the magician turns $k$ cards by his choice. For what minimum $k$ can it happen that open cards determine uniquely the whole snake?",
    "options": [],
    "answer": "2",
    "solution": "**Answer:** $k = 2$.\n\nExample. Put $300$ in the low left corner and $101$ in the cell above it.\n\nEstimation. One open card does not determine the snake uniquely, because all the cells of $3 \\times 100$ board can be considered as one cyclic path: let the rows of the board be denoted by letters *a*, *b*, *c* and columns be numbered from $1$ to $300$, then that path is\n$$\na_1-b_1-c_1-c_2-b_2-b_3-c_3-c_4-b_4-b_5-\\dots-c_{300}-b_{300}-a_{300}-a_{299}-a_{298}-\\dots-a_2-a_1\n$$\nIf we open only one card in this path we can orient this path in two ways and put cards such that they increase in the direction of the chosen orientation. So we have at least two configurations here.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25020,
    "subject": "Mathematics (Multi-modal)",
    "question": "A hacker is locked into an underground industrial complex. She is presented with a computer screen, on which appears a long message of length $72$, consisting of the symbols $E$, $X$, $I$, $T$, exactly $18$ letters of each kind in some seemingly random order. The message may be manipulated by inserting any one of the combinations\n$EX$, $XE$, $IT$, $TI$, $IXIXI$\nat an arbitrary place in the message. Such a combination may also be erased, wherever it may occur in the message.\nThe hacker may escape when the system is cracked, which happens when only the word `EXIT` is printed on the screen. Show that she may escape using less than $2019$ operations.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $18 = n$, so that the initial message has length $4n$, with exactly $n$ symbols of each kind.\nWe first establish an invariant. Assign\n$E = 3$, $X = -3$, $I = 2$, $T = -2$,\nand let $S$ denote the sum of the values of all symbols appearing in the message. Initially, $S = 0$, and the sum stays invariant under all the legal transformations.\n\nOur next observation is that we may always insert or delete the combination TETET, for\n$$\n\\emptyset \\mapsto \\{TI\\} \\mapsto \\{TEXI\\} \\mapsto \\{TETIXI\\} \\mapsto \\{TETEXIXI\\} \\mapsto \\{TETETIXIXI\\} \\mapsto \\{TETET\\},\n$$\nand this works also in reverse. Required are six operations.\n\nTo crack the system, first insert XE behind every I, and XE in front of every T:\n$I \\mapsto IXE$, \\quad $T \\mapsto XET$.\nThis requires at most $p_1 = 4n$ operations, after which the message has length at most $12n$. It may now be considered a sequence of the four possible strings\n$E$, $X$, $IX$, $ET$.\n\nSecond, expand any single $X$ (not preceded by an $I$) into $IXIXIX$, and any single $E$ (not succeeded by a $T$) into $ETETET$:\n$X \\mapsto IXIXIX$ (one operation), $E \\mapsto ETETET$ (six operations).\nThis requires at most $p_2 = 12n \\cdot 6 = 72n$ operations, and the message now has length at most $72n$.\nIt is at present reduced to some binary combination of the two strings\n$IX$, $ET$.\n\nThird, effectuate all possible reductions\n$$\n\\{ETIX\\} \\mapsto \\{EX\\} \\mapsto \\emptyset \\quad \\text{and} \\quad \\{IXET\\} \\mapsto \\{IT\\} \\mapsto \\emptyset.\n$$\nThere can be at most $18n$ such reductions, totalling $p_3 = 18n \\cdot 2 = 36n$ operations. The hacker will be left with a message of the types\n$ETET \\dots ET$ or $IXIX \\dots IX$\nor an empty screen. But since the invariant $S = 0$, the screen must now, in fact, be empty.\n\nFourth, insert **EXIT**, using $p_4 = 2$ more operations. The number of operations was at most\n$$\np_1 + p_2 + p_3 + p_4 = 4n + 72n + 36n + 2 = 112n + 2 = 112 \\cdot 18 + 2 = 2018 < 2019.\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 25021,
    "subject": "Mathematics (Multi-modal)",
    "question": "The two women Inger and Ellen play the following game. On a large piece of paper are $2019$ six-pointed snowflakes $*$, placed in general position (and not touching). Each of the six arms of the snowflake has an extreme end, called its *apex*.\n\nA move consists of drawing a curve connecting two apices, possibly belonging to the same snowflake. The curve may not intersect itself and may not touch any previously drawn curves. Also, it may not touch any of the snowflakes (except at the two apices forming its endpoints). Each apex may be used only once.\n\nInger makes the first move, and then the players alternate. The first player who cannot make a move has lost the game. Which player has a winning strategy?",
    "options": [],
    "answer": "Inger has a winning strategy.",
    "solution": "**Inger has a winning strategy.**\n\nLet Inger select an arbitrary snowflake $S$ and connect a pair of opposite apices. By the Jordan Curve Theorem, this curve (along with $S$) will divide the plane into two domains, the curve (along with $S$) being their common boundary. It is obvious that Inger may draw her curve in such a way that exactly $1009$ snowflakes fall in each domain. Each successive move must then be made entirely within one of these two domains. Since the two domains are combinatorially identical (though possibly not geometrically), Inger may respond to each move of Ellen in one domain by the corresponding move in the other, and will win the game.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25022,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a positive integer $n$, consider all nonincreasing functions $f: \\{1, \\dots, n\\} \\to \\{1, \\dots, n\\}$. Some of them have a fixed-point ($c = f(c)$), some don't have. Which are more numerous? Evaluate the difference between the sizes of the two sets of functions.",
    "options": [],
    "answer": "Functions with a fixed point are more numerous; the difference equals binom(2n−2, n−1) − binom(2n−2, n) = (1/n)·binom(2n−2, n−1), i.e., the Catalan number C_{n−1}.",
    "solution": "Lemma. Let $A$ be a set of $a$ consecutive integers and let $B$ be a set of $b$ consecutive integers.\nThere are exactly $\\binom{a+b-1}{a}$ nonincreasing functions $f: A \\to B$.\nProof. Let (wlog) $A = \\{1, \\dots, a\\}$, $B = \\{1, \\dots, b\\}$. A function $f: A \\to B$ is nonincreasing if and only if the function $g: A \\to C = \\{1, \\dots, a+b-1\\}$ given by $g(x) = f(x) + (a-x)$ is strictly decreasing. Any such function $g$ is uniquely identified with the set of its values $g(A)$; and the latter has size $a$. So there are as many functions $f$ in question as there are $a$-element subsets of $C$; the lemma is ready.\n\nNow we count the functions of the two types under consideration. A function $f$ that has a fixed-point $c \\in \\{1, \\dots, n\\}$ maps $\\{1, \\dots, c-1\\}$ (nonincreasingly) into $\\{c, \\dots, n\\}$, and it maps $\\{c+1, \\dots, n\\}$ (nonincreasingly) into $\\{1, \\dots, c\\}$; and there are no further restrictions. So, on the left side of $c$, we have (by the lemma with $a=c-1, b=n-c+1$) $\\binom{n-1}{c-1}$ choices; and right to $c$ (again by the lemma, with $a=n-c, b=c$) we have also $\\binom{n-1}{c-1}$ choices. The number of nonincreasing functions with a fixed-point is therefore equal to\n$$\nF_n = \\sum_{c=1}^{n} \\binom{n-1}{c-1} \\binom{n-1}{c-1}.\n$$\nIf $f$ has no fixed-point then there exists $c \\in \\{1, \\dots, n-1\\}$ such that $f(x) > x$ for $x \\le c$ and $f(x) < x$ for $x > c$. A similar reasoning applies: $f$ maps $\\{1, \\dots, c\\}$ into $\\{c+1, \\dots, n\\}$ and it maps $\\{c+1, \\dots, n\\}$ into $\\{1, \\dots, c\\}$ (arbitrarily nonincreasingly). On the left piece (lemma with $a=c, b=n-c$) there are $\\binom{n-1}{c}$ choices; on the right piece (lemma, $a=n-c, b=c$) $\\binom{n-1}{c-1}$ choices. So the number of nonincreasing functions without a fixed-point equals\n$$\nG_n = \\sum_{c=1}^{n-1} \\binom{n-1}{c} \\binom{n-1}{c-1}.\n$$\nConsider the polynomial $P_n(x) = (1+x)^{n-1}$. We recognize $F_n$ and $G_n$ as the coefficients of $x^{n-1}$ and $x^n$ in the product $P_n(x)P_n(x) = (1+x)^{2n-2}$. Thus $F_n = \\binom{2n-2}{n-1}$ and $G_n = \\binom{2n-2}{n}$, with the difference $F_n - G_n = \\binom{2n-2}{n-1} - \\binom{2n-2}{n}$. (This gets us another characterization of Catalan numbers.)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25023,
    "subject": "Mathematics (Multi-modal)",
    "question": "Several points are given in the plane. A child wants to draw $k$ (closed) discs in such a manner, that for any two points $A, B$ ($A \\neq B$) there exists a disc that contains only one of these points. What is the minimum $k$, such that for any initial configuration of 2019 points it is possible to draw the $k$ discs with the above property?",
    "options": [],
    "answer": "1010",
    "solution": "Answer: $k = 1010 = \\lfloor n/2 \\rfloor$, where $n$ is the number of points.\n\nWe say that discs separates two points if it contains only one of them.\n\nEstimation. Consider a set of $n$ points on a circle. Then any disc can separate at most two consecutive pairs, therefore we need at least $\\lfloor n/2 \\rfloor$ discs.\n\nExample. Let $S_1$ be a set of $n$ points, and $H_1$ be the convex hull of $S$. Let $D_1$ be a disc that contains exactly $\\lfloor n/2 \\rfloor$ points. There are points $A_1, B_1 \\in S_1$ in the boundary of $H_1$, such that $A_1 \\in D_1, B_1 \\notin D_1$ and the (open) interval $A_1B_1$ does not contain other points of $S_1$. Let $D_2$ be a disc that separates points $A_1, B_1$ from the other points of $S_1$.\n\nDenote the set $S_1 \\setminus \\{A_1, B_1\\}$ by $S_2$ and its convex hull by $H_2$. There are points $A_2, B_2 \\in S_2$ in the boundary of $H_2$, such that $A_2 \\in D_2, B_2 \\notin D_2$ and the (open) interval $A_2B_2$ does not contain other points of $S_2$. Let $D_3$ be a disc that separates points $A_2, B_2$ from the other points of $S_2$.\n\nBy continuing this procedure we get a set of $\\lfloor n/2 \\rfloor$ discs. (The description of the final step of the process we left to the reader. The last point that has no pair is $A_{\\lfloor n/2 \\rfloor}$.) This set of discs has the desired separating property. Indeed, let $P, Q \\in S_1$. If $P = A_i, Q = B_j$ for some $i, j$, then $D_1$ separates them. If $P = A_i$ and $Q = A_j$, $i < j$, (or if $P = B_i$ and $Q = B_j$, $i < j$), then $D_{i+1}$ separates them.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25024,
    "subject": "Mathematics (Multi-modal)",
    "question": "On a board the numbers $1, 2, 3, \\ldots, 500$ are written. In a game two players $A$ and $B$ take turns alternately and $A$ starts. In each turn they have to erase two numbers $n$ and $2n$ on the board. If a player can not take a turn, the player loses. Determine who has a winning strategy.",
    "options": [],
    "answer": "A",
    "solution": "First we notice that for each odd number $u$, $1 \\le u < 500$, the numbers $u, 2u, 2^2u, \\dots, 2^{k_u}u$, where $k_u$ is the largest integer such that $u2^{k_u} \\le 500$, form a string of length $k_u + 1$ such that it is only possible to erase each number in the string if you also erase one of its neighbours. We now divide the numbers on the board into strings and count the lengths of these strings. It is not difficult to see that the numbers of strings of different lengths are\n\n| length              | 1   | 2  | 3  | 4  | 5 | 6 | 7 | 8 | 9 |\n|---------------------|-----|----|----|----|---|---|---|---|---|\n| number of strings   | 125 | 62 | 32 | 15 | 8 | 4 | 2 | 1 | 1 |\n\nIt is easy to see that a configuration with an even number of strings of each length is a losing position: Player $B$ pairs a string with one of the same length, and whenever $A$ makes a turn on one string, $B$ copy that turn using the paired string. This type of configuration we call a symmetric configuration.\n\nNow if we have a winning configuration and add a symmetric configuration it is still a winning configuration: If $A$ is going to play starting from a winning configuration $W$ and we add a symmetric configuration $S$, then $A$ starts by following her winning strategy on $W$. Every time $B$ replies by changing a string in $W$, $A$ continues using her winning strategy on $W$, and every time $B$ replies by using a string in $S$, $A$ continues by a symmetric move on $S$. It is easy to see that this is a winning strategy.\n\nNow we prove that the above configuration is a winning one, and hence that $A$ has a winning strategy. You can not make a turn on a string of length $1$. Hence we just have to prove that the configuration with three strings of length $4$, $8$ and $9$ is a winning strategy. First $A$ erase the two middle numbers in the string of length $9$, and hence reduces it to two strings of lengths $3$ and $4$, leaving a configuration with two strings of length $3$ and $8$ plus a symmetric configuration. Now player $B$ can either play on the string of length $3$, leaving a string of length $8$, or $B$ can play on the string on length $8$ leaving either: one string of length $6$, one string of length $5$, two strings of length $2$ and $4$, two strings of length $3$. This shows how $A$ replies:\n\n| $B$ leaves      | $8$   | $3+6$ | $3+5$ | $3+2+4$ | $3+3+3$ |\n|-----------------|-------|-------|-------|---------|---------|\n| $A$ leaves      | $3+3$ | $3+3$ | $3+3$ | $3+2$   | $3+3$   |\n\nNotice that $A$ can destroy a string of length $3$ or $4$ in one turn. The situation $3+2$ is essentially symmetric since you can only make one move on each string. Hence $A$ leaves a symmetric configuration to $B$ in all the cases, and hence $A$ has a winning strategy.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25025,
    "subject": "Mathematics (Multi-modal)",
    "question": "The expressions $x+y$, $x-y$, $x^2+xy+y^2$ and $x^2-xy+y^2$ are written on the two sides of two cards in such a way that each side of each card contains exactly one of these expressions. The cards are laid on the table on top of each other in such a way that only the top side of the uppermost card is visible. Alice and Bob who know the expressions but not how they distribute on the invisible sides of the cards play the following game. Without inspecting the invisible sides of the cards, Alice picks one card according to her preference, the other card is left to Bob. Now both players may examine both sides of their card. Alice chooses a real value to either $x$ or $y$ according to her preference and tells her choice to Bob; then Bob chooses a real value to the other variable according to his preference. The player with larger product of the values of the expressions on two sides of their card wins. Does either of the players have a winning strategy and if yes then who does?",
    "options": [],
    "answer": "Alice",
    "solution": "**Answer:** Yes, Alice.\n\nLet Alice choose the card where at least one of the two expressions is a trinomial. She can do it as follows: if the visible side of the topmost card contains a trinomial then she can pick that card, otherwise the bottommost card definitely contains a trinomial and she can pick that one. By case study, we can show that Alice always can choose a value to $y$ in such a way that the product of the expressions on the sides of her card is larger than that the product of the expressions on the sides of Bob's card in the case of any value of $x$.\n\n* If one card contains expressions $x^2+xy+y^2$ and $x-y$ and the other card contains $x^2-xy+y^2$ and $x+y$ then the product of the expressions on the first card is $x^3-y^3$ and the product of the expressions on the second card is $x^3+y^3$. If Alice has the first card then she can choose a negative value to $y$, in which case $x^3-y^3 > x^3+y^3$ for any value of $x$. If Alice has the other card then she can choose a positive value to $y$, in which case $x^3+y^3 > x^3-y^3$ for any value of $x$.\n\n* If one card contains expressions $x^2+xy+y^2$ and $x+y$ and the other card contains $x^2-xy+y^2$ and $x-y$ then the product of the expressions on the first card is $(x^3+2xy^2)+(2x^2y+y^3)$ and the product of the expressions on the second card is $(x^3+2xy^2)-(2x^2y+y^3)$. Similarly to the previous case, the ordering between the products depends on the value of $y$ solely, because $x$ occurs with even exponent in all terms that have different signs in the expressions. Alice wins by choosing a positive value to $y$ if she has expressions with only plus signs and a negative value to $y$ if she has expressions with minus signs.\n\n* If one card contains expressions $x^2+xy+y^2$ and $x^2-xy+y^2$ and the other card contains $x+y$ and $x-y$ then the product of the expressions on the first card is $x^4+x^2y^2+y^4$ and the product of the expressions on the second card is $x^2-y^2$. According to the Alice's choice, she has the first card. She can win by assigning a real number whose absolute value is greater than 1 to $y$. Indeed, if Bob assigns a real number whose absolute value is not greater than 1 to $x$ then his product is negative and her product is positive, and if Bob assigns a real number whose absolute value is greater than 1 to $x$ then his product is less than $x^2$ whereas her product is larger than $x^4$ which is larger than $x^2$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 25026,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine if there is an integer $k \\ge 2$ such that if we partition the set $\\{2, 3, \\dots, k\\}$ in two parts, then at least one of the parts contains numbers $a, b$ and $c$ with $ab = c$? (We allow $a = b$.) If such a number $k$ exist, find the least $k$ with this property.",
    "options": [],
    "answer": "32",
    "solution": "We show first that $k = 32$ is such a number. Consider a partition $\\{U, V\\}$ of the set $\\{2, 3, \\dots, 32\\}$ where we may assume that $2 \\in U$. Towards contradiction, suppose that none of the parts contains numbers $a, b$ and $c$ with the desired property. As $2 \\in U$ and $2 \\cdot 2 = 4$, we have $4 \\in V$. Similarly, $4 \\cdot 4 = 16$ implies $16 \\in U$. Hence $2, 16 \\in U$, but $2 \\cdot 8 = 16$, so $8 \\in V$. We have concluded that $2, 16 \\in U$ and $4, 8 \\in V$, but $2 \\cdot 16 = 32 = 4 \\cdot 8$, which implies that the number $32$ cannot be in any of the parts, which is a contradiction. Therefore, $k = 32$ is a desired number.\n\nWe now prove that $k = 32$ is actually the least number with the desired property. We form the partition $\\{U, V\\}$ of the set $K = \\{2, 3, \\dots, 31\\}$ in the following way. For any number $n \\in \\mathbb{Z}_+$, consider its prime factorization representation $n = \\prod_{i=0}^{k-1} p_i$, and put $\\Omega(n) = k$. If $n < 32 = 2^5$, then $n$ is necessarily a product of at most four primes (with repetitions counted) or $\\Omega(n) \\ge 4$. Put\n$$\nU = \\{n \\in K \\mid \\Omega(n) = 1 \\text{ tai } \\Omega(n) = 4\\}\n$$\nand\n$$\nV = \\{n \\in K \\mid \\Omega(n) = 2 \\text{ tai } \\Omega(n) = 3\\}.\n$$\nLet $a, b, c \\in K$ be numbers with $ab = c$. We observe that $\\Omega(c) = \\Omega(a) + \\Omega(b)$. On the other hand, we have $\\Omega(c) \\le 4$, as $c \\in K$. If now $a, b \\in U$, then necessarily $\\Omega(a) = \\Omega(b) = 1$, which implies $\\Omega(c) = 2$ and $c \\in V$. If instead of that $a, b \\in V$, then $\\Omega(a) = \\Omega(b) = 2$ and $\\Omega(c) = 4$, so $c \\in U$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25027,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $S$ be a finite set with $n \\ge 2$ elements. Two players, $A$ and $B$, alternately choose nonempty proper subsets of $S$, where\n\n(1) it is not allowed to choose a set that contains a set previously chosen by any player,\n\n(2) it is not allowed to choose a set that is contained in any previously chosen set,\n\n(3) it is not allowed to choose a set whose union with any previously chosen set is $S$.\n\n$A$ begins. The player who first cannot choose a set anymore loses. Which player has a winning strategy?",
    "options": [],
    "answer": "Player A has a winning strategy.",
    "solution": "It is straightforward to verify that the following is a winning strategy for $A$. Player $A$ first chooses a singleton subset $\\{x\\} \\subset S$. After that, he responds to $B$ choosing a set $T$ by choosing $(S \\setminus \\{x\\}) \\setminus T$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25028,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $M(n)$ be the minimum number of cells that must be coloured in the $6 \\times n$ square grid to ensure that every $2 \\times 3$ rectangle (in any orientation) contains a coloured cell. Is it true that $M(n) = p_n + k_n^3$ for all $n \\ge 2$ where $p_n$ is a prime and $k_n$ is a nonnegative integer?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "No, it is not true. At first we note that $M(n)$ is a nondecreasing function and $M(n) \\ge n$ for all $n \\ge 2$ what follows from the fact that $6 \\times n$ rectangle can be cut into $n$ rectangles of size $2 \\times 3$, each of whom should contain a coloured cell.\n\nNext we show that $M(n+1) \\le M(n)+2$ for all $n \\ge 2$. Assume that we have the minimal colouring for $6 \\times n$ grid (blue rectangle in the following figure) and we add one more column to the right.\n![](attached_image_1.png)\nAt least one of the red or green cells are coloured. If a red cell is coloured then colouring cells marked A and C will finish the minimal colouring for the $6 \\times (n+1)$ grid. If a green cell is coloured then colouring B and C is enough.\n\nNeither 35 nor 36 can be expressed in the form $p + k^3$, where $p$ is a prime and $k$ is a nonnegative integer (just check $k \\in \\{0, 1, 2, 3\\}$). As $M(n)$ is an unbounded function which increases in each step by at most two, we will have that either $M(n) = 35$ or $M(n) = 36$ for some $n$.\n\nAlternatively one can show, that $M(20) = 22$, a simple check shows that 22 cannot be expressed in the form $p + k^3$. The following figure shows that 22 coloured cells are enough to cover a $6 \\times 20$ rectangle. To show that 21 coloured cell is insufficient we will show that $M(10) > 10$.\n![](attached_image_2.png)\nAssume, that $M(10) = 10$ and notice that $6 \\times 10$ rectangle can be cut into four $3 \\times 5$ rectangles. Each $3 \\times 5$ rectangle should contain at least two coloured cells therefore there will be at least two $3 \\times 5$ rectangles containing exactly 2 coloured cells. The only way to colour two cells in a $3 \\times 5$ rectangle is as in the next figure (left), because both red and green squares can contain at most one coloured cell and the only way to colour a single cell in a $3 \\times 3$ square is to put it in the centre.\n\n![](attached_image_3.png)\n![](attached_image_4.png)\nThere is only one way how two cells can be colored in a $3 \\times 5$ rectangle (left); two $3 \\times 5$ rectangles with two coloured cells cannot be placed side by side (right).\nNow let's see how these two $3 \\times 5$ rectangles containing two coloured cells can be placed. If they are placed side by side then the red rectangle remains empty (the right figure above), similarly one can show that they cannot be placed one above another. But if they are placed diagonally then the following figure shows how seven $2 \\times 3$ rectangles can be placed in the remaining space each requiring at least one coloured cell.\n\n![](attached_image_5.png)",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25029,
    "subject": "Mathematics (Multi-modal)",
    "question": "A *palindrome* is a word which is build using 2 letters and is equal to its reverse, for example: ABBA and ABABABABA are palindromes. Prove that any 2019-letter word (which uses 2 letters) is *build* by at most 808 palindromes.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note that any 5-letter word may be build with at most two palindromes (case-by-case checking). Any 2019-letter word can be divided into 404 5-letter words, hence it can be divided into at most 808 palindromes.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25030,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a directed graph $G$ outgoing degree of each vertex equals $3$ (loops and bidirectional edges are allowed and considered as cycles). Prove that the graph contains $2$ non-intersecting directed cycles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We proceed by induction on $n$, the number of vertices.\n\nFor the base case we consider $n = 4$ where the only possible digraph is the complete digraph which contains $2$ disjoint cycles of length $2$.\n\nWe now assume that any $3$-out digraph, with $n - 1 \\ge 4$ vertices, contains $2$ disjoint cycles.\n\nLet $G$ be a digraph on $n$ vertices failing our assumption.\n\nThen $G$ has no bidirectional edges. If $uw$ is a bidirectional edge then $G \\setminus \\{u, w\\}$ is still $1$-out so it contains a cycle, which paired with the $2$-cycle made by the bidirectional edge $uw$ gives the desired disjoint cycles.\n\nThe main idea allowing us to prove the result is using edge contractions. If there is an edge $uv$ such that $u, v$ have no common parent, we can modify $G$ to a new digraph $G'$ by removing $u$ and $v$ and adding a vertex $w$ for which its outgoing edges come to all children of $v$ and its ingoing edges come from all parents of $u$ and $v$. $G'$ has $n-1$ nodes and is still $3$-out, as $u, v$ had no common parent, so by the inductive assumption $G'$ contains $2$ disjoint cycles. If $w$ is not in any of the cycles they were contained in $G$ and we are done. The other option is that $w$ is contained in one of the cycles, this implies the other cycle is in $G$. We distinguish the cases when the cycle in-edge of $w$ comes from $u$'s in-edge or from $v$'s in-edge. Replacing $w$ in the first case by $uv$ and in the second by $v$ we get the other cycle in $G$ and we are done.\n\nThe only remaining option is when every edge of $G$ has a «witness» (defined as common parent to both its end-vertices). Let $v$ be a vertex having the smallest in-degree (we denote in-degree as $d^-$). As there are $3n$ edges in total we have $d^-(v) \\le 3$.\n\nCase 1: $d^-(v) = 0$. Then the edge starting in $v$ has no witnesses.\n\nCase 2: $d^-(v) = 1$. Then the edge ending in $v$ has no witnesses.\n\nCase 3: $d^-(v) = 2$. Let $u, w$ be two parents of $v$. Then $u$ has to be the witness to $wv$ and $w$ to $uv$ implying $uw$ is a bidirectional edge which is a contradiction.\n\nCase 4: $d^-(v) = 3$. Since our graph is exactly $3$-out and the minimal in-degree is $3$, then all the vertices have in-degree $3$. Given a vertex $x$ and its parents $u, v, w$ we show that $u, v, w$ make a $3$-cycle. Indeed, we notice that each of the witnesses of $ux, vx, wx$ must be among $u, v, w$ implying that the $u, v, w$ induced subgraph is $1$-in. This combined with the fact bidirectional edges do not exist, we conclude that $u, v, w$ make a $3$-cycle as claimed. If we reverse the edges of $G$ we notice that all the above arguments still apply, as $G$ is both exactly $3$-out and exactly $3$-in, so children of $x$ also form a $3$-cycle. As there are no bidirectional edges, the children and parents of $x$ are disjoint and give $2$ disjoint $3$-cycles.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 25031,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $G$ be a graph of diameter $2$ on $2019$ vertices. $62$ cops and $1$ robber are placed on the vertices of graph $G$. Everybody sees all others, they move in turn. Each move each person can stay in his vertex or go to any of the adjacent vertices (each cop stay or go independently of the others and they coordinate their actions during a game). The robber makes the first move, then the cops move (simultaneously), then the robber moves again, then cops, etc. The robber is caught if he is in the vertex occupied by cop. Prove that the cops can catch the robber.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The numbers $62$ and $2019$ have the following secret property:\n$$\n64 + 63 + \\dots + 3 = \\frac{64 \\cdot 65}{2} - 3 = 2077 > 2019.\n$$\nLet before the cops move the robber be in the vertex $A$.\nIf $\\deg A \\le 62$, the cops can catch the robber in one, two or three moves (due to diameter $2$ property): denote $B_1, B_2, \\dots$ the vertices adjacent to $A$, let in the first move the $i$-th cop go “towards” the $B_i$ (i.e. he either goes directly to $B_i$ if the corresponding edge exists or goes to the vertex adjacent to $B_i$ and in the next move to $B_i$); if the robber will leave the vertex $A$ and will go to some vertex $B_k$ he will be caught immediately or at the next move, otherwise the cops will occupy all vertices of $B_i$ in their second move and after that will catch the robber.\nIf $\\deg A \\ge 63$, let one of the cops go to $A$ and after that always stay there guarding the vertex $A$ and its neighbouring vertices. It means that for the robber the safe part of the graph is decreased by at least $64$ vertices.\nNow, if before the cops move the “safe degree” of the robber’s vertex is at most $61$, the remaining “active” cops again can catch the robber in one-two-three moves. Otherwise, let one of the active cops go to the vertex of degree at least $62$ and after that always guard this vertex and its neighbours. The safe part of the graph for the robber is decreased once again by at least $63$ vertices.\nContinuing this process we will find after the $62^{\\text{nd}}$ step that the safe area is empty.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 25032,
    "subject": "Mathematics (Multi-modal)",
    "question": "A circle with center $O$ passes through the vertices $B$ and $C$ of the triangle $ABC$ and intersects for the second time segments $AB$ and $AC$ in points $C_1$ and $B_1$ correspondingly. The lines $AO$, $BB_1$ and $CC_1$ are concurrent. Prove that triangle $ABC$ is isosceles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Segments $BB_1$ and $CC_1$ intersect inside triangle $ABC$, point $A$ is outside the circle, therefore ray $AO$ lies inside angle $BAC$. Assume that triangle $ABC$ is not isosceles. Then $\\angle BAO \\neq \\angle OAC$. Without loss of generality $\\angle BAO < \\angle OAC$. Let $B'_1$ and $C'$ be points symmetrical with respect to line $AO$ to points $C_1$ and $B$ correspondingly. Then triangle $ABC'$ is isosceles, and lines $C'C_1$ and $BB'_1$ intersect in some point $K$ on the line $AO$ (see fig.). Now point $B_1$ lies on the short arc $B'_1C'$, therefore the ray $BB_1$ is obtained from the ray $BB'_1$ by rotation in clockwise direction, and therefore the intersection point of $BB_1$ and $AO$ lies below point $K$. By the analogous reasons the intersection point of $C_1C$ and $AO$ lies above point $K$. So for non-isosceles triangle the concurrence from the problem statement is impossible.\n\n![](attached_image_1.png)\nSuppose that $B_1C_1$ intersects $BC$ at $X$. The polar line of $X$ is the line passing through $BC_1 \\cap B_1C = A$ and $BB_1 \\cap CC_1$. By assumption given in the problem statement, this line is $AO$. This is a contradiction since the polar line cannot pass through the center of the circle.\nIt follows that $B_1C_1 \\parallel BC$ and it is clear now that $ABC$ is isosceles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25033,
    "subject": "Mathematics (Multi-modal)",
    "question": "$AC$ is hypotenuse of right triangle $ABC$, $BH$ is its altitude. Points $M$ and $N$ are the midpoints of segments $AH$ and $CH$ correspondingly. Lines $BM$ and $BN$ intersect for second time the circumscribed circle of triangle $ABC$ in points $P$ and $Q$ correspondingly. Segments $AQ$ and $CP$ intersect in point $R$. Prove that line $BR$ passes through the midpoint of segment $MN$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $K$ be the midpoint of segment $BH$, $S$ be the intersection point of $AK$ and $BP$, $T$ be the intersection point of $CK$ and $BQ$. Then $SK$ and $KT$ is one third of the corresponding medians and $ST$ is parallel to $AC$.\n\nThe triangles $ABH$ and $BCH$ are similar. From this similarity and properties of inscribed angles we have\n$$\n\\angle KAB = \\angle NBC = \\angle QAC.\n$$\nHence $\\angle BAC = \\angle KAR$. But also $\\angle BAC = \\angle BPC = \\angle BPR$ as inscribed angles. Therefore quadrilateral $SAPR$ is cyclic and $\\angle ARS = \\angle APS = \\angle APB = \\angle AQB$. So, $RS$ is parallel to $BT$. By analogous reasoning $RT$ is parallel to $BS$. Hence $BSRT$ is parallelogram.\n\nDiagonal $BR$ of this parallelogram splits diagonal $ST$ on 2 equal parts, therefore it also splits the segment $MN$ which is parallel to $ST$ on 2 equal parts, QED.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25034,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 4$, and consider a non-intersecting $n$-gon $P_1P_2...P_n$ in the plane. Suppose that, to each $P_k$, there is a unique other vertex $Q_k$ among $P_1, ..., P_n$ that lies closest to it. The polygon is said to be *hostile* if $Q_k \\ne P_{k \\pm 1}$ for all $k$ (counting cyclically).\n\na. Prove that there exist no convex hostile polygons.\n\nb. Determine all $n$ for which there exists a concave hostile $n$-gon.",
    "options": [],
    "answer": "a) No convex hostile polygons exist. b) Hostile concave polygons exist for all integers n ≥ 4.",
    "solution": "(a) As an auxiliary result, we prove the following. There is no convex quadrilateral $ABCD$ in which $C$ is the closest neighbour of $A$, and $D$ is the closest neighbour of $B$. See Figure below:\n![](attached_image_1.png)\nConvex quadrilateral.\nIndeed, the diagonals $AC$ and $BD$ cross at a point $P$ by convexity. The Triangle Inequality yields\n$$\nAD + BC < (AP + PD) + (BP + PC) = (AP + PC) + (BP + PD) = AC + BD.\n$$\nSince $AC < AD$ by the first assumption, we must have $BC < BD$, contradicting the second assumption.\n\nNow, consider a convex hostile polygon $P_1P_2...P_n$. We say a vertex $P_k$ has *separation $h \\ge 2$* when its closest neighbour is $Q_k = P_{k+h}$. Among all vertices, select one with minimal separation $h$; without loss of generality, we may take it to be $P_1$, thus its closest neighbour is $Q_1 = P_{h+1}$. Since $h \\ge 2$, the vertex $P_2$ does not belong to the line $P_1P_{h+1}$. The closest neighbour of $P_2$ cannot lie on the opposite side of this line according to the auxiliary result proved above, i.e. $Q_2$ is to be found among the vertices $P_1, ..., P_{h+1}$. But then $P_2$ will have separation at most $h-1$, which contradicts the minimality of $h$.\n\n\n(b) *Answer:* Hostile concave polygons exist for all $n \\ge 4$.\nStart from the type of zig-zag construction given in Figure below, and dislocate the vertices slightly, so as to make all distances unequal (to make each vertex have a unique closest neighbour).\n![](attached_image_2.png)",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 25035,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a convex quadrilateral such that $|AD| < |AB|$ and $|CD| < |CB|$. Prove that $\\angle ABC < \\angle ADC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $D'$ be the point symmetric to point $D$ with respect to line $AC$. As triangles $ADC$ and $AD'C$ are similar, it suffices to prove that the inequalities $\\angle CAB + \\angle CAD' < 180^\\circ$ and $\\angle ACB + \\angle ACD' < 180^\\circ$ together imply $\\angle ABC < \\angle AD'C$.\n\nSuppose the contrary, i.e., $\\angle AD'C \\le \\angle ABC$. Then point $D'$ is located on the circumcircle of triangle $ABC$ or outside it. Let $E$ be the second intersection point of line $AD'$ with the circumcircle of triangle $ABC$ ($E = A$ if $AD'$ is tangent). Similarly, let $F$ be the second intersection point of line $CD'$ with the circumcircle of triangle $ABC$ ($F = C$ if $CD'$ is tangent). Consider three cases that may occur.\n\n* If $\\angle ACD' \\ge \\angle ACB$ and $\\angle CAD' \\ge \\angle CAB$ then point $B$ is located inside triangle $AD'C$ or on its side. Applying the triangle inequality twice now gives $|AB| + |BC| \\le |AD'| + |D'C|$, whereas adding the two inequalities given in the problem results in $|AD| + |DC| < |AB| + |AC|$. This leads to contradiction as $|AD'| = |AD|$ and $|D'C| = |DC|$.\n\n* If $\\angle ACD' < \\angle ACB$ then points $A, F, B, C$ occur on the circumcircle of triangle $ABC$ in this order whereas points $E$ and $B$ lie on different sides of line $CD'$. Consequently,\n$$\n\\angle FBC > \\angle EBC = 180^\\circ - \\angle D'AC > \\angle CAB = \\angle CFB,\n$$\nwhence in triangle $FBC$ we get $|CF| > |CB|$ implying $|CD'| > |CB|$, contradiction.\n\n* The case $\\angle CAD' < \\angle CAB$ is symmetric to the previous case.\nSince $AD < AB$, we have $\\angle ABD < \\angle ADB$. Similarly, $\\angle CBD < \\angle CDB$ because $CD < CB$. Hence\n$$\n\\angle ABC = \\angle ABD + \\angle CBD < \\angle ADB + \\angle CDB = \\angle ADC.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25036,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two points $M$ and $N$ have been selected from the edge $BC$ of a tetrahedron $ABCD$ so that the point $M$ lies between $B$ and $N$, the point $N$ lies between $M$ and $C$, and the angle between the planes $AND$ and $ACD$ is equal to the angle between the planes $AMD$ and $ABD$. Prove that\n$$\n\\frac{MB}{MC} + \\frac{NB}{NC} \\ge 2 \\cdot \\frac{|\\triangle ABD|}{|\\triangle ACD|},\n$$\nwhere $|\\triangle ABD|$ is the area of the triangle $\\triangle ABD$, and similarly $|\\triangle ACD|$ is the area of $\\triangle ACD$.\n\n![](attached_image_1.png)\n",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We denote the angle between the planes $AND$ and $ACD$ by $\\alpha$, and the angle between the planes $AMD$ and $AND$ by $\\beta$. Furthermore, when $P, Q, R$ and $S$ are four points in space which do not all lie on a plane, we denote the volume of the tetrahedron $PQRS$ by $|PQRS|$.\nThe triangles $\\triangle MBD$, $\\triangle MCD$, $\\triangle NBD$ and $\\triangle NCD$ have a common height, which is the distance of the point $D$ from the line $BC$. Therefore\n$$\n\\frac{MB}{MC} + \\frac{NB}{NC} = \\frac{|\\triangle MBD|}{|\\triangle MCD|} + \\frac{|\\triangle NBD|}{|\\triangle NCD|}.\n$$\nIn the same vein, since the points $M, B, C$ and $D$ lie on a common plane, which does not contain the point $A$, we must have\n$$\n\\frac{|\\triangle MBD|}{|\\triangle MCD|} = \\frac{|MBDA|}{|MCDA|}.\n$$\nSimilarly,\n$$\n\\frac{|\\triangle NBD|}{|\\triangle NCD|} = \\frac{|NBDA|}{|NCDA|}.\n$$\nLet us consider the volume of the tetrahedron $MBDA$. Naturally\n$$\n|MBDA| = \\frac{1}{3} \\cdot |\\triangle ABD| \\cdot h,\n$$\nwhere $h$ is the distance of $M$ from the plane $ABD$. Let $P$ be the projection of $M$ to the line $AD$ and let $Q$ be the projection of $M$ to the plane $ABD$. Then\n$$\nh = MQ, \\quad \\angle QPM = \\alpha, \\quad \\text{and} \\quad |\\triangle AMD| = \\frac{1}{2} \\cdot AD \\cdot PM.\n$$\nBecause in addition $h = PM \\cdot \\sin \\alpha$, we reach the conclusion\n$$\n|MBDA| = \\frac{2 \\cdot |\\triangle ABD| \\cdot |\\triangle AMD| \\sin \\alpha}{3 \\cdot AD}.\n$$\n\n$\\frac{MB}{MC} + \\frac{NB}{NC} = \\frac{|MBDA|}{|MCDA|} + \\frac{|NBDA|}{|NCDA|} \\ge 2 \\sqrt{\\frac{|MBDA| \\cdot |NBDA|}{|MCDA| \\cdot |NCDA|}} = 2 \\cdot \\sqrt{\\frac{\\frac{2 \\cdot |\\triangle ABD| \\cdot |\\triangle AMD| \\sin \\alpha}{3 \\cdot AD} \\cdot \\frac{2 \\cdot |\\triangle ABD| \\cdot |\\triangle AND| \\sin (\\alpha + \\beta)}{3 \\cdot AD}}{\\frac{2 \\cdot |\\triangle ACD| \\cdot |\\triangle AMD| \\sin (\\alpha + \\beta)}{3 \\cdot AD} \\cdot \\frac{2 \\cdot |\\triangle ACD| \\cdot |\\triangle AND| \\sin \\alpha}{3 \\cdot AD}}} = 2 \\cdot \\frac{|\\triangle ABD|}{|\\triangle ACD|}$\n(prepared by the Organizing Committee)\nLet $B'$ be a point so that $BB' \\parallel AD$ and $B'C \\perp AD$. Let $M', N'$ be points on $B'C$ so that $MM' \\parallel NN' \\parallel BB'$. Let $E$ be a point on $AD$ such that the plane $B'CE$ is perpendicular to $AD$. Then $\\frac{MB}{MC} = \\frac{M'B'}{M'C}$, $\\frac{NB}{NC} = \\frac{N'B'}{N'C}$, and $\\frac{|\\Delta ABD|}{|\\Delta ACD|} = \\frac{|\\Delta AB'D|}{|\\Delta ACD|} = \\frac{B'E}{CE}$.\nSince the angle between planes $AND$, $ACD$ is equal to the angle between planes $AMD$, $ABD$, it follows that lines $EM'$, $EN'$ are isogonal in the angle $B'EC$. It is well-known that $\\frac{M'B'}{M'C} \\cdot \\frac{N'B'}{N'C} = \\frac{B'E^2}{CE^2}$. Hence by AM-GM we obtain\n$$\n\\frac{MB}{MC} + \\frac{NB}{NC} = \\frac{M'B'}{M'C} + \\frac{N'B'}{N'C} \\ge 2 \\sqrt{\\frac{M'B'}{M'C} \\cdot \\frac{N'B'}{N'C}} = 2 \\cdot \\frac{B'E}{CE} = 2 \\cdot \\frac{|\\Delta ABD|}{|\\Delta ACD|}\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 25037,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with $AB = AC$. Let $M$ be the midpoint of $BC$. Let circles with diameters $AC$, $BM$ intersect at points $M$, $P$. Let $MP$ intersect $AB$ at $Q$. Let $R$ be a point on $AP$ such that $QR \\parallel BP$. Prove that $CP$ bisects $\\angle RCB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $\\{ACMP\\}$ is cyclic, we have $\\angle RPQ = \\angle ACM$. Moreover, $\\angle PQR = 90^\\circ = \\angle CMA$. Hence $\\triangle PQR \\sim \\triangle CMA$. It follows that\n$$\n\\frac{PR}{PQ} = \\frac{CA}{CM}. \\qquad (1)\n$$\nSimilarly, $\\angle PMB = \\angle PAC$ and $\\angle BPM = 90^\\circ = \\angle CPA$. Hence $\\triangle BPM \\sim \\triangle CPA$. It follows that\n$$\n\\frac{CA}{BM} = \\frac{CP}{BP}. \\qquad (2)\n$$\nUsing (1), (2), and the equality $BM = CM$ we obtain\n$$\n\\frac{PR}{PQ} = \\frac{CA}{CM} = \\frac{CA}{BM} = \\frac{CP}{BP}. \\qquad (3)\n$$\nUsing (3) and $\\angle QPB = 90^\\circ = \\angle RPC$ we obtain $\\triangle QPB \\sim \\triangle RPC$. In particular, $\\angle RCP = \\angle PBQ$. Hence\n$$\n\\angle RCP = \\angle PBQ = \\angle CBA - \\angle MBP = \\angle ACB - \\angle ACP = \\angle PCB.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25038,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a scalene triangle. Let $P$ be an interior point of $ABC$ such that $AP \\perp BC$. Assume that $BP$ and $CP$ intersect $AC$ and $AB$ at $X$ and $Y$, respectively. Prove that $AX = AY$ iff there exists a circle with centre lying on $BC$ and tangent to $AB$ and $AC$ at points $Y$ and $X$, respectively.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "$(\\Rightarrow)$ is trivial. For $(\\Leftarrow)$ let $AP \\cap BC = D$. Now **AD** is a bisector of angle **XDY**, for this we can argue in two ways\n\n* Let $XY \\cap BC = S$. Then $D(S, D; B, C) = 1$ and $\\angle ADB = 90^\\circ$, so statement.\n\n* Let $\\ell$ be a parallel line to **BC** passing through **A**. Let $DY \\cap \\ell = K$ and $DX \\cap \\ell = L$. Then using Ceva and Tales theorem we see that **A** is a midpoint of **KL**.\n\nSince **ABC** is a scalene we see that in triangle **XDY** bisector **DA** and perpendicular bisector of **XY** intersects at **A**, so **AXDY** is inscribed in circle **\\omega**. Now let **Z** be the second intersection of **\\omega** with **BC**. Then **AXZY** is a kite, so circle with centre **Z** and radius **ZX** satisfies problem statement.\nThe implication $\\Leftarrow$ is trivial. For the proof of $\\Rightarrow$, let $AP \\cap BC = D$. We prove that **DA** bisects angle **XDY**. We can argue in two ways:\n\n* Let $S = XY \\cap BC$ and $XY \\cap AD = T$. It is well-known that the quadruple $S, T, X, Y$ is harmonic. Since $\\angle SDT = \\frac{\\pi}{2}$, it follows that **DT** bisects angle **YDX**.\n\n* Let $\\ell$ be the line parallel to **BC** passing through **A**. Let $DY \\cap \\ell = K$ and $DX \\cap \\ell = L$. By Ceva and similarities we obtain $1 = \\frac{BD}{DC} \\cdot \\frac{CX}{XA} \\cdot \\frac{AY}{YB} = \\frac{BD}{DC} \\cdot \\frac{CD}{AL} \\cdot \\frac{AK}{DB} = \\frac{AK}{AL}$. Hence **A** is the midpoint of **KL** and since $\\ell \\perp AD$, it follows that **DA** bisects the angle **XDY**.\n\nSince **ABC** is scalene, **DA** is not the perpendicular bisector of **XY**. So, since $AX = AY$ and **DA** bisects **XDY**, it follows that **AXDY** is cyclic. Let **\\omega** be the circumcircle of **AXDY**. Let $Z \\neq D$ be the second intersection of **\\omega** with **BC**. Then **AXZY** is a kite, so the circle with centre **Z** and radius **ZX** satisfies the problem condition.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25039,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle and $H$ its orthocenter. Point $D$ lies on segment $AC$ and $E$ is its foot onto the line $BC$. Prove that $EH \\perp BD$ iff $BD$ bisects $AE$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $BD \\cap AH = X$. Then $XH \\perp BE$, so $EH \\perp BD \\iff H$ is orthocenter $BXE \\iff BH \\perp EX \\iff EX \\parallel AC \\iff AXED$ is parallelogram $\\iff BD$ bisects $AE$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25040,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle and $H$ its orthocenter. Let $P$ be an arbitrary point on segment $BC$. Prove that $H$ lies on the line passing through reflections of $P$ with respect to $AB$ and $AC$ iff triangle $ABC$ is right angled.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote by $P'$ and $P''$ reflections of $P$ with respect to $AB$ and $AC$. Easy to see that $P$ is an anti-Steiner point of the line $P'P''$, so $P \\equiv B$ or $P \\equiv C$— therefore $\\angle BAC = 90^\\circ$.\nThe implication $\\iff$ is very easy. We shall prove $\\implies$.\nSuppose that triangle $ABC$ is acute. Let $P'$ and $P''$ be the reflections of $P$ with respect to $AB$ and $AC$, respectively. Then $\\angle PAP'' = 2\\angle BAC$.\nLet $H'$, $H''$ be the reflections of $H$ with respect to $AB$ and $AC$, respectively. Since $H'$ lies on the circumcircle of $ABC$ and $P$ lies inside this circle, we have $\\angle APH' > \\angle ACH'$. Hence $\\angle AP'H = \\angle APH' > \\angle ACH' = \\frac{\\pi}{2} - \\angle BAC$. Similarly, $\\angle AP''H > \\frac{\\pi}{2} - \\angle BAC$. Since $H$ lies on $P'P''$, we get\n$$\n2\\angle BAC = \\angle P'AP'' = \\pi - \\angle APH' - \\angle AP''H < \\pi - (\\frac{\\pi}{2} - \\angle BAC) - (\\frac{\\pi}{2} - \\angle BAC) = 2\\angle BAC\n$$\na contradiction.\nWe obtain similar contradiction when triangle $ABC$ is obtuse.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25041,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCDEF$ be a convex hexagon in which $AB = AF$, $BC = CD$, $DE = EF$ and $\\angle ABC = \\angle EFA = 90^\\circ$. Prove that $AD \\perp CE$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider circle $\\omega$ with centre $A$ and radius $AB$. Note that $BC$ and $EF$ are tangent to $\\omega$ and so from the problem condition, the line $CE$ is a radical axis of $\\omega$ and $D$. Therefore $CE \\perp AD$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25042,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $X$, $Y$ be points on $AB$, $AC$ of triangle $ABC$, respectively, such that $B$, $C$, $X$, $Y$ lie on one circle.\nThe median of triangle $ABC$ from $A$ intersects the perpendicular bisector of $XY$ at $P$. Find $\\angle BAC$, if $PXY$ is equilateral.",
    "options": [],
    "answer": "60 or 120 degrees",
    "solution": "Since $AP$ is the symmedian of $AXY$ and $P$ lies on the perpendicular bisector of $XY$, then $PX$ and $PY$ are tangent to the circumcircle of triangle $AXY$. Therefore, we can easily find that $\\angle YAX = 60^\\circ$ or $\\angle YAX = 120^\\circ$.\n\nSince $B$, $C$, $X$, $Y$ lie on a circle, $XY$ is antiparallel to $BC$ in the angle $BAC$. So, the $A$-median of $ABC$ is the $A$-symmedian of $AXY$. Hence $PX$, $PY$ are tangent to the circumcircle of $AXY$. Let $O$ be the circumcenter of $AXY$.\n\nIf $A$ and $P$ lie on different sides of the line $XY$ then we obtain\n$$\n\\angle XAY = \\frac{1}{2} \\angle XOY = \\frac{1}{2} (180^\\circ - \\angle YPX) = \\frac{1}{2} (180^\\circ - 60^\\circ) = 60^\\circ.\n$$\nIf $A$ and $P$ lie on the same side of the line $XY$ then $\\angle XAY = 180^\\circ - \\frac{1}{2} \\angle XOY = 180^\\circ - \\frac{1}{2} \\cdot 120^\\circ = 120^\\circ$.\n\nHence, there are only two possible values of angle $XAY$: $60^\\circ$ and $120^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25043,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $N$ be a positive integer. Determine the number of pairs, $(a, b)$, of positive integers $a$ and $b$ such that the number\n$$\n\\frac{ab}{a+b}\n$$\nis a divisor of $N$.",
    "options": [],
    "answer": "d(N)^2",
    "solution": "In what follows $d(n)$ denotes the number of positive divisors of a positive integer $n$. We show that the number of pairs is $d(N)^2$. Let $m$ be a fixed positive divisor of $N$. We claim that the number of pairs, $(a, b)$, such that $\\frac{ab}{a+b} = m$ is $d(m)^2$. This becomes clear if we rewrite the equation as\n$$\n(a - m)(b - m) = m^2.\n$$\nSince both $a - m$ and $b - m$ are greater than $-m$, it follows that $a - m$ and $b - m$ are both positive divisors of $m^2$, so the number of pairs $(a, b)$ for a fixed $m \\mid N$ is $d(m^2)$, since we have $d(m^2)$ choices of $a$, and the choice of $a$ determines $b$ uniquely. The number of pairs is therefore\n$$\n\\sum_{m|N} d(m^2)\n$$\nAssume that $N$ has prime factorisation $p_1^{s_1} \\cdots p_t^{s_t}$. Since $d(kl) = d(k)d(l)$ when $\\text{gcd}(k, l) = 1$, it follows that\n$$\n\\begin{aligned}\n\\sum_{m|N} d(m^2) &= \\prod_{i=1}^{t} (d(1) + d(p_i^2) + \\cdots + d(p_i^{2s_i})) \\\\\n&= \\prod_{i=1}^{t} (1 + 3 + \\cdots + (2s_i + 1)) \\\\\n&= \\prod_{i=1}^{t} (s_i + 1)^2 \\\\\n&= d(N)^2\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25044,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let us consider a polynomial $P(x)$ with integer coefficients satisfying\n$$\nP(-1) = -4, \\quad P(-3) = -40, \\quad \\text{and} \\quad P(-5) = -156.\n$$\nWhat is the largest possible number of integers $x$ satisfying\n$$\nP(P(x)) = x^2?\n$$",
    "options": [],
    "answer": "0",
    "solution": "No such numbers $x$ can ever exist. To see this, let us recall that when $x$ is an integer, the right-hand side of the equation is always $\\equiv 0 \\pmod{3}$ or $1 \\pmod{3}$. We recall also that when $x$ and $y$ are integers, we have\n$$\nP(x) \\equiv P(y) \\pmod{3}, \\quad \\text{whenever} \\quad x \\equiv y \\pmod{3}.\n$$\nTherefore\n$$\nP(x) \\equiv P(-3) \\equiv -40 \\equiv 2 \\pmod{3} \\quad \\text{when} \\quad x \\equiv 0 \\pmod{3},\n$$\nand similarly\n$$\nP(x) \\equiv P(-5) \\equiv -156 \\equiv 0 \\pmod{3} \\quad \\text{when} \\quad x \\equiv 1 \\pmod{3},\n$$\nand\n$$\nP(x) \\equiv P(-1) \\equiv -4 \\equiv 2 \\pmod{3}, \\quad \\text{when} \\quad x \\equiv 2 \\pmod{3}.\n$$\nIn particular, $P(x)$ is always $\\equiv 0 \\pmod{3}$ or $\\equiv 2 \\pmod{3}$, and $P(P(x))$ is always $\\equiv 2 \\pmod{3}$. Therefore we always have $P(P(x)) \\not\\equiv x^2 \\pmod{3}$ for every integer $x$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 25045,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a positive integer $n$ it is known that the number $\\sqrt{12n^2 + 1}$ also is a positive integer. Prove that the number\n$$\n\\sqrt{\\frac{\\sqrt{12n^2 + 1} + 1}{2}}\n$$\nalso is a positive integer.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let's denote\n$$\n\\sqrt{12n^2 + 1} = 2k + 1 \\qquad (\\text{Eq-6})\n$$\nbecause if the number $\\sqrt{12n^2 + 1}$ is integer then it is odd. We have to prove that\n$$\n\\sqrt{\\frac{\\sqrt{12n^2 + 1} + 1}{2}} = \\sqrt{\\frac{(2k + 1) + 1}{2}} = \\sqrt{k + 1}\n$$\nis an integer, that is, that $k + 1$ is a perfect square.\n\nBy squaring both sides of (Eq-6) we get that $12n^2 + 1 = 4k^2 + 4k + 1$ or $k(k + 1) = 3n^2$. As $k$ and $k + 1$ are coprime then we have two options: $k = 3a^2$ and $k + 1 = b^2$ or $k = a^2$ and $k + 1 = 3b^2$ for some positive integers $a$ and $b$. But the second option is not possible as that would mean that $a^2 = 3b^2 - 1 = 2 \\pmod 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25046,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that there are infinitely many different triangles in the coordinate plane whose vertices have integer coordinates and whose side lengths are consecutive integers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "At first we will prove that there are infinitely many triangles whose side lengths are consecutive integers and whose area also is an integer. Let the side lengths of the triangle be $2m-1$, $2m$ and $2m+1$, then by Heron's formula its area is\n$$\nS = \\sqrt{\\frac{6m(2m+2)2m(2m-2)}{16}} = m\\sqrt{3(m^2-1)} \\quad (\\text{Eq-7})\n$$\nNow we show that the equation $3(m^2 - 1) = q^2$ has infinitely many integer solutions. We use the standard method for solving Pell's equations. As $q$ should be divisible by $3$ then we denote $q = 3l$ and our equation turns into $m^2 - 1 = 3l^2$. It has a solution $m = 2$ and $l = 1$ and if $(m, l)$ is a solution then $(2m + 3l, m + 2l)$ also is a solution.\nIndeed, we can check, that\n$$\n(2m + 3l)^2 - 1 = 3(m + 2l)^2\n$$\nis equivalent to\n$$\n4m + 12ml + 9l^2 - 1 = 3m^2 + 12ml + 12l^2\n$$\nwhat is equivalent to $m^2 - 1 = 3l^2$.\nWe have proved that there are infinitely many triangles whose side lengths are $2m-1$, $2m$ and $2m+1$ and whose area is $S = m\\sqrt{3(m^2-1)}$ where $\\sqrt{3(m^2-1)}$ is an integer. Let $ABC$ be such a triangle with $AB = 2m$, $BC = 2m-1$ and $AC = 2m+1$ and let $CH$ be the altitude drawn from $C$ to the side $AB$. As $S = \\frac{CH \\cdot AB}{2}$ we get that $CH = \\sqrt{3(m^2-1)}$ what is an integer. And\n$$\nAH = \\sqrt{AC^2 - CH^2} = \\sqrt{(2m+1)^2 - (3m^2-3)} = m + 2\n$$\nis an integer, too.\nNow we can put our triangle $ABC$ into a coordinate plain. Let $A$ be the origin point $(0,0)$ and let $B$ be the point with coordinates $(2m, 0)$. Then the point $C$ also has integer coordinates $(m+2, \\sqrt{3m^2-3})$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25047,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let odd positive integer $a$ be not a perfect square, $b$, and $c$ be odd primes and\n$$\na^2 + a + 1 = 3(b^2 + b + 1)(c^2 + c + 1).\n$$\nProve that at least one of $b^2 + b + 1$ and $c^2 + c + 1$ is composite.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume that $b^2 + b + 1$ and $c^2 + c + 1$ are primes. Then\n$$\na^2 + a + 1 \\equiv 0 \\pmod{b^2 + b + 1}.\n$$\nThis quadratic equation modulo prime has at most 2 roots. We must have\neither $a \\equiv b \\pmod{b^2 + b + 1}$ or $a \\equiv b^2 \\pmod{b^2 + b + 1}$\n(in the latter case $a^2 + a + 1 \\equiv b^4 + b^2 + 1 = (b^2 + b + 1)(b^2 - b + 1) \\equiv 0$). We claim that we can rule out the smallest few such numbers of either of these forms. We have $a > b$, and $a \\neq b^2$ since $a$ is not a perfect square. Our next possible value for $a$ is $b^2 + b + 1 + b = (b + 1)^2$ which is even, as is the value after that, $b^2 + b + 1 + b^2$. Thus we have $a \\ge 2(b^2 + b + 1) + b > 2(b^2 + b + 1)$. By the same logic, $a > 2(c^2 + c + 1)$. But then we have\n$$\n3(b^2 + b + 1)(c^2 + c + 1) = a^2 + a + 1 > a^2 > 4(b^2 + b + 1)(c^2 + c + 1),\n$$\nand this is a contradiction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25048,
    "subject": "Mathematics (Multi-modal)",
    "question": "Chessboard has a form of $n \\times n$ torus, where $n = 2^{2k} + 1$. Prove that $n$ queens in the cells $(i, 2^k \\cdot i)$ ($1 \\le i \\le n$, multiplication modulo $n$) do not attack each other.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "It is clear that each vertical line contains 1 queen. Since $2^k$ is invertible modulo $2^{2k} + 1$, each horizontal line contains 1 queen. Each diagonal line also contains 1 queen because $2^k \\pm 1$ is invertible modulo $2^{2k} + 1$.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 25049,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ a prime number. Prove that $p^2 + p + 1$ is not a perfect cube.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume that $p^2 + p + 1 = k^3$. Then we have that $p(p+1) = k^3 - 1 = (k-1)(k^2 + k + 1)$. Since $p$ is prime, either $p \\mid k-1$ or $p \\mid k^2 + k + 1$. If $p \\mid k-1$, then $p \\le k-1$ and $k^3 \\ge (p+1)^3 > p^2 + p + 1 = k^3$, which is impossible. Now consider the possibility that $p \\mid k^2 + k + 1$. Since $p$ is prime, we have one of 3 cases.\n\n1) $p=3$, which does not give a solution;\n\n2) $p \\equiv 1 \\pmod 3$. Thus, $p^2 + p + 1 \\equiv 3 \\pmod 9$, but no perfect cube is congruent to 3 (mod 9) and so we again reach a contradiction.\n\n3) $p \\equiv 2 \\pmod 3$. Then $p=2$ is not a solution. And for odd $p$ we have $4(k^2 + k + 1)^2 \\equiv 0 \\pmod p$, or, equivalently, $(2k+1)^2 \\equiv -3 \\pmod p$. This is impossible because $-3$ is not a square residue modulo $p = 3\\ell + 2$. This detail is not quite elementary and can be checked via the properties of the Legendre symbol:\n$$\n\\left(\\frac{-3}{p}\\right) = \\left(\\frac{-1}{p}\\right) \\left(\\frac{3}{p}\\right) = (-1)^{\\frac{p-1}{2}} \\cdot \\left((-1)^{\\frac{3-1}{2}} \\cdot \\frac{p-1}{2} \\left(\\frac{p}{3}\\right)\\right) = \\left(\\frac{2}{3}\\right) = -1.\n$$\n\n\nAssume $p^2 + p + 1 = (x+1)^3$ for a certain $x \\ge 1$. This yields $p(p+1) = x(x^2 + 3x + 3)$. Clearly $p > x$, so $p$ (prime) must divide $x^2 + 3x + 3$; hence\n$$\nx^2 + 3x + 3 = ap \\quad \\text{and} \\quad p + 1 = ax\n$$\nfor a certain $a \\ge 1$. Elimination of $p$ leads to\n$$\nx^2 - (a^2 - 3)x + (a + 3) = 0,\n$$\na quadratic equation whose discriminant\n$$\nD = (a^2 - 3)^2 - 4(a + 3)\n$$\nmust be an even square. Thus $D \\le (a^2 - 5)^2$, recasting into $a^2 \\le a + 7$; and $a$ must be odd. This leaves 1 and 3 as admissible values of $a$; but neither of them yields an integer solution to the quadratic equation (displayed).",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25050,
    "subject": "Mathematics (Multi-modal)",
    "question": "A positive integer $N$ is called a *lighthouse number* if it has the following property. Any positive integer not exceeding $N$ which is relatively prime to it has at most two prime factors (possibly equal). Determine whether the lighthouse numbers are infinitely or finitely many and, in the latter case, find the greatest among them.",
    "options": [],
    "answer": "Finitely many; the greatest lighthouse number is 1260.",
    "solution": "**Answer:** The greatest lighthouse number is 1260.\n\nDenote by $p_n$ the $n$-th prime number. We first set out to prove that, for $n \\ge 5$, the inequality\n$$\np_{n+1}^3 \\le p_1 \\cdots p_n, \\quad n \\ge 5, \\qquad (\\text{Eq-8})\n$$\nholds. As is readily verified by computation, the inequality holds for $n = 5, 6$. Consider now the case $n \\ge 7$. By Bertrand's Postulate, $p_k < 2p_{k-1}$ for each $k$. Consequently,\n$$\n\\begin{aligned}\np_{n+1}^3 &< 2^3 p_n^3 = 2^3 p_n \\cdot p_n^2 \\\\\n&< 2^3 p_n \\cdot 2^2 p_{n-1}^2 = 2^3 p_n \\cdot 2^2 p_{n-1} \\cdot p_{n-1} \\\\\n&< 2^3 p_n \\cdot 2^2 p_{n-1} \\cdot 2 p_{n-2} = 64 p_n p_{n-1} p_{n-2} \\\\\n&\\le p_n p_{n-1} \\cdots p_1, \\end{aligned}\n$$\nin so far as $64 \\le p_1 p_2 \\cdots p_{n-3}$, which does indeed hold true when $n \\ge 7$.\n\nConsider now a number $N$ divisible by the primes $p_1, \\dots, p_n$, but not by $p_{n+1}$. Then $N$ is a lighthouse number precisely when $N < p_{n+1}^3$, for $p_{n+1}^3$ is the least number relatively prime to $N$ with more than two prime factors. But $p_1 \\cdots p_n \\le N < p_{n+1}^3$, which, in conjunction with (Eq-8), enforces $1 \\le n \\le 4$, so that the only possible prime divisors of $N$ are $2, 3, 5, 7$. There remains to examine these cases in detail.\n\nSuppose $n \\le 3$. Then\n$$\nN < p_4^3 = 7^3 = 343.\n$$\nSuppose now $n = 4$. Writing $N = p_1 p_2 p_3 p_4 M$, we find\n$$\n210M = p_1 p_2 p_3 p_4 M = N < p_5^3 = 11^3 = 1331,\n$$\nimplying $M \\le 6$, and\n$$\nN \\le 210 \\cdot 6 = 1260.\n$$\nConsequently, the lighthouse numbers are bounded by 1260, hence finite in number. Further, the number 1260 has the lighthouse property, for\n$$\np_5^3 = 11^3 = 1331 > 1260.\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25051,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that the equation $7^x = 1 + y^2 + z^2$ has no positive integral solution.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume that there exist positive integers $x$, $y$, $z$ satisfying the given equation. Let's consider two cases:\n\nIf $x$ is odd then $7^x \\equiv 7 \\pmod{8}$. Since each of $y^2$ and $z^2$ can have $0$, $1$, $4$ as the remainders when divided by $8$. We have contradiction because $1 + y^2 + z^2 \\equiv 1, 2, 3, 5, 6 \\pmod{8}$.\n\nIf $x$ is even then $x = 2n$. We have\n$$\ny^2 + z^2 = 7^{2n} - 1 = (7^n + 1)(7^n - 1)\n$$\nWe can show that the right hand side has at least one prime factor $p \\equiv 3 \\pmod{4}$ of odd order. Thus $p | (y^2 + z^2)$. It implies that $p | y$ and $p | z$. We have a contradiction because $p$ is a prime factor of even order of the left hand side.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 25052,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_0 > 0$ be a real number, and let\n$$\na_n = \\frac{a_{n-1}}{\\sqrt{1 + 2020 \\cdot a_{n-1}^2}}, \\quad \\text{for } n = 1, 2, \\dots, 2020.\n$$\nShow that $a_{2020} < \\frac{1}{2020}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $b_n = \\frac{1}{a_n^2}$. Then $b_0 = \\frac{1}{a_0^2}$ and\n$$\nb_n = \\frac{1 + 2020 \\cdot a_{n-1}^2}{a_{n-1}^2} = b_{n-1} \\left( 1 + 2020 \\cdot \\frac{1}{b_{n-1}} \\right) = b_{n-1} + 2020.\n$$\nHence $b_{2020} = b_0 + 2020^2 = \\frac{1}{a_0^2} + 2020^2$ and $a_{2020}^2 = \\frac{1}{\\frac{1}{a_0^2} + 2020^2} < \\frac{1}{2020^2}$ which shows that $a_{2020} < \\frac{1}{2020}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25053,
    "subject": "Mathematics (Multi-modal)",
    "question": "We call a positive integer $N$ splendid, if\n$$\nN = (a - b)^2 + (b - c)^2 + (c - a)^2\n$$\nfor some integers $a$, $b$ and $c$. If $M$ and $N$ are splendid positive integers, is the sum $M + N$ or the product $MN$ also necessarily splendid? How about the product $2MN$?",
    "options": [],
    "answer": "Sum M+N: no; Product MN: no; Product 2MN: yes.",
    "solution": "Answer: no to the first question, yes to the second.\nThe number $2$ is splendid since\n$$\n(1 - 0)^2 + (0 - 1)^2 + (1 - 1)^2 = 1^2 + 1^2 + 0^2 = 2.\n$$\nThe number $4$ is not splendid. Namely, if we had $4 = (a-b)^2 + (b-c)^2 + (c-a)^2$, at least one of these squares would have to be $2^2$ and the other two would then have to be $0^2$. But if two of the differences $a-b$, $b-c$ and $c-a$ vanish, then the third must vanish as well. Therefore the number $4$ is not splendid, and since $2+2$ and $2 \\cdot 2$ are not splendid even though $2$ is, it follows that the sum and the product of two splendid numbers need not be splendid.\nLet us then assume that $M$ and $N$ are splendid numbers and that $a$, $b$, $c$, $a'$, $b'$ and $c'$ are integers such that\n$$\nM = (a - b)^2 + (b - c)^2 + (c - a)^2\n$$\nand\n$$\nN = (a' - b')^2 + (b' - c')^2 + (c' - a')^2.\n$$\nThen the product $2MN$ is necessarily splendid for choosing\n$$\n\\begin{cases} A = 2(a(a' - c') + b(c' - b') + c(b' - a')) \\\\ B = 2(a'(b-c) + b'(a-b) + c'(c-a)) \\\\ C = 0, \\end{cases}\n$$\ngives\n$$\n2MN = (A - B)^2 + (B - C)^2 + (C - A)^2.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25054,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f : \\mathbb{R} \\setminus \\{0\\} \\to \\mathbb{R}$ so that\n$$\nf\\left(\\frac{1}{x}\\right) \\ge 1 - f(x) \\ge x^2 f(x)\n$$\nfor all $x \\in \\mathbb{R} \\setminus \\{0\\}$.",
    "options": [],
    "answer": "f(x) = 1/(x^2 + 1)",
    "solution": "Answer: $f(x) = \\frac{1}{x^2 + 1}$.\nWe have\n$$\n1 - f(x) \\ge x^2 f(x) \\iff \\frac{1}{x^2 + 1} \\ge f(x).\n$$\n\nIf we substitute $z = \\frac{1}{x}$ into $f(\\frac{1}{x}) \\ge 1 - f(x)$ we get this:\n$$\nf(z) \\ge 1 - f\\left(\\frac{1}{z}\\right) \\quad \\forall z \\in \\mathbb{R} \\setminus \\{0\\}.\n$$\nWe combine these results:\n$$\n\\begin{gather*}\n\\frac{1}{x^2 + 1} \\ge f(x) \\ge 1 - f\\left(\\frac{1}{x}\\right) \\quad \\forall x \\in \\mathbb{R} \\setminus \\{0\\} \\\\\n\\implies f\\left(\\frac{1}{x}\\right) \\ge 1 - \\frac{1}{x^2 + 1} = \\frac{x^2}{1+x^2} = \\frac{1}{\\frac{1}{x^2} + 1} \\quad \\forall x \\in \\mathbb{R} \\setminus \\{0\\}\n\\end{gather*}\n$$\nWe make the same substitution $z = \\frac{1}{x}$ and get that\n$$\nf(z) \\ge \\frac{1}{z^2 + 1} \\quad \\forall z \\in \\mathbb{R} \\setminus \\{0\\}.\n$$\n$$\n\\text{Hence } \\frac{1}{x^2 + 1} \\ge f(x) \\ge \\frac{1}{x^2 + 1} \\quad \\forall z \\in \\mathbb{R} \\setminus \\{0\\} \\implies\n$$\n$$\nf(x) = \\frac{1}{x^2 + 1} \\quad \\forall x \\in \\mathbb{R} \\setminus \\{0\\}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25055,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all real $x, y, z$ so that\n$$\n\\begin{aligned}\n& x^2 y + y^2 z + z^2 = 0 \\\\\n& z^3 + z^2 y + z y^3 + x^2 y = \\frac{1}{4}(x^4 + y^4)\n\\end{aligned}\n$$",
    "options": [],
    "answer": "x = y = z = 0",
    "solution": "Answer: $x = y = z = 0$.\n\n$y = 0 \\implies z^2 = 0 \\implies z = 0 \\implies \\frac{1}{4}x^4 = 0 \\implies x = 0$. $x = y = z = 0$ is a solution, so assume that $y \\neq 0$. Then $z = 0 \\implies x^2y = 0 \\implies x = 0 \\implies \\frac{1}{4}y^4 = 0$, which is a contradiction. Hence $z \\neq 0$. Now we solve the quadratic (first) equation w.r.t. $x, y$, and $z$.\n$$\n\\begin{aligned}\nx &= \\pm \\frac{\\sqrt{-4y^3z - 4yz^2}}{2y} \\\\\ny &= \\frac{-x^2 \\pm \\sqrt{x^4 - 4z^3}}{2z} \\\\\nz &= \\frac{-y^2 \\pm \\sqrt{y^4 - 4x^2y}}{2}\n\\end{aligned}\n$$\nThe discriminants must be non-negative.\n$$\n\\begin{aligned}\n-4y^3z - 4yz^2 &\\ge 0 \\\\\nx^4 - 4z^3 &\\ge 0 \\\\\ny^4 - 4x^2y &\\ge 0\n\\end{aligned}\n$$\nAdding the inequalities we get\n$$\n\\begin{aligned}\ny^4 - 4x^2y + x^4 - 4z^3 - 4y^3z - 4yz^2 &\\ge 0 \\\\\n\\frac{1}{4}(x^4 + y^4) &\\ge z^3 + z^2y + zy^3 + x^2y\n\\end{aligned}\n$$\nBut equation 2 says that $\\frac{1}{4}(x^4 + y^4) = z^3 + z^2y + zy^3 + x^2y$. This is only possible if all of the inequalities are in fact equalities, so we have\n$$\n\\begin{aligned}\n-4y^3z - 4yz^2 &= 0 \\\\\nx^4 - 4z^3 &= 0 \\\\\ny^4 - 4x^2y &= 0\n\\end{aligned}\n$$\nThis means that $x = \\pm \\frac{\\sqrt{0}}{2y} = 0 \\implies y = \\frac{-0^2 \\pm \\sqrt{0}}{2z} = 0$, which is a contradiction. Hence the only solution is $x = y = z = 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25056,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$ and $c$ be positive real numbers such that $a > b + c$. Prove that\n$$\n(a^2 - b^2 - c^2)(a^5 - b^5 - c^5) \\le (a^3 - b^3 - c^3)(a^4 - b^4 - c^4).\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We shall apply the following inequality due to Aczél:\n**Lemma.** Let $n \\in \\mathbb{Z}_+$, let $t, u \\in \\mathbb{R}$ and let $x, y \\in \\mathbb{R}^n$ so that $t^2 > |x|^2$ and $u^2 > |y|^2$. Then\n$$\n(tu - x \\cdot y)^2 \\ge (t^2 - |x|^2)(u^2 - |y|^2).\n$$\nLet us momentarily assume that this lemma is true. Since $a > b + c$, we clearly have $a^k > (b+c)^k > b^k + c^k$ for each $k \\in \\{2, 3, 4, 5\\}$, and we may apply the lemma twice to estimate\n$$\n\\begin{aligned}\n(a^2 - b^2 - c^2)(a^5 - b^5 - c^5) &\\le \\frac{(a^2 - b^2 - c^2)(a^4 - b^4 - c^4)^2}{(a^3 - b^3 - c^3)} \\\\\n&\\le \\frac{(a^2 - b^2 - c^2)(a^4 - b^4 - c^4)}{(a^3 - b^3 - c^3)} \\cdot \\frac{(a^3 - b^3 - c^3)^2}{a^2 - b^2 - c^2} \\\\\n&= (a^3 - b^3 - c^3)(a^4 - b^4 - c^4).\n\\end{aligned}\n$$\nThus, it only remains to prove the lemma.\n*Proof of the Lemma.* By homogeneity, it is enough to prove the lemma in the case in which $t = u = 1$, $|x| < 1$ and $|y| < 1$. Let $\\varphi \\in \\mathbb{R}$ so that $x \\cdot y = |x| \\cdot |y| \\cos \\varphi$. We may rewrite the claim of the lemma as\n$$\n|x|^2 + |y|^2 + |x|^2 |y|^2 \\cos^2 \\varphi \\ge 2 |x| \\cdot |y| \\cos \\varphi + |x|^2 |y|^2.\n$$\nSince $|x|^2 \\ge |x|^2 |y|^2 |\\cos \\varphi|$ and $1 \\ge |\\cos \\varphi|$, we may estimate by the rearrangement inequality that\n$$\n|x|^2 + |x|^2 |y|^2 \\cos^2 \\varphi \\ge |x|^2 |\\cos \\varphi| + |x|^2 |y|^2 |\\cos \\varphi|.\n$$\nSimilarly, since $|y|^2 \\ge |x|^2 |y|^2$ and $1 \\ge |\\cos \\varphi|$, we may estimate\n$$\n|y|^2 + |x|^2 |y|^2 |\\cos \\varphi| \\ge |y|^2 \\cos \\varphi + |x|^2 |y|^2.\n$$\nCombining the above estimates gives\n$$\n\\begin{aligned}\n|x|^2 + |y|^2 + |x|^2 |y|^2 \\cos^2 \\varphi &\\ge |x|^2 |\\cos \\varphi| + |y|^2 + |x|^2 |y|^2 |\\cos \\varphi| \\\\\n&\\ge |x|^2 |\\cos \\varphi| + |y|^2 |\\cos \\varphi| + |x|^2 |y|^2 \\\\\n&\\ge 2 |x| \\cdot |y| \\cos \\varphi + |x|^2 |y|^2,\n\\end{aligned}\n$$\nas required.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25057,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$ be positive real numbers with $abc = 1$. Prove that\n$$\n\\frac{a}{b(c+1)} + \\frac{b}{c(a+1)} + \\frac{c}{a(b+1)} \\ge \\frac{3}{2}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By Cauchy-Schwarz we have\n$$\n\\left( \\frac{a}{b(c+1)} + \\frac{b}{c(a+1)} + \\frac{c}{a(b+1)} \\right) \\left( (c+1) + (a+1) + (b+1) \\right) \\ge \\left( \\sqrt{\\frac{a}{b}} + \\sqrt{\\frac{b}{c}} + \\sqrt{\\frac{c}{a}} \\right)^2.\n$$\nTherefore it suffices to prove that\n$$\n\\begin{aligned}\n& \\frac{a}{b} + \\frac{b}{c} + \\frac{c}{a} + 2 \\left( \\sqrt{\\frac{a}{c}} + \\sqrt{\\frac{b}{a}} + \\sqrt{\\frac{c}{b}} \\right) = \\left( \\sqrt{\\frac{a}{b}} + \\sqrt{\\frac{b}{c}} + \\sqrt{\\frac{c}{a}} \\right)^2 \\\\\n& \\ge \\frac{3}{2}(a + b + c + 3).\n\\end{aligned}\n$$\nNow AM-GM and the condition $abc = 1$ imply that\n$$\n\\frac{a}{b} + 2\\sqrt{\\frac{a}{c}} \\ge 3\\sqrt[3]{\\frac{a}{b} \\cdot \\sqrt{\\frac{a}{c}} \\cdot \\sqrt{\\frac{a}{c}}} = 3\\sqrt[3]{\\frac{a^2}{bc}} = 3\\sqrt[3]{a^3} = a.\n$$\nAnalogously, we have $\\frac{b}{c} + 2\\sqrt{\\frac{b}{a}} \\ge 3b$ and $\\frac{c}{a} + 2\\sqrt{\\frac{c}{b}} \\ge 3c$. These inequalities together yield\n$$\n\\frac{a}{b} + \\frac{b}{c} + \\frac{c}{a} + 2 \\left( \\sqrt{\\frac{a}{c}} + \\sqrt{\\frac{b}{a}} + \\sqrt{\\frac{c}{b}} \\right) \\ge 3(a + b + c).\n$$\nFinally, we have by AM-GM $a+b+c \\ge 3\\sqrt[3]{abc} = 3$. This implies that\n$$\n3(a + b + c) \\ge \\frac{3}{2}(a + b + c + 3),\n$$\nwhich completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25058,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest positive integer $k$ with a property: for any three points with integer coordinates on a plane\n$$\nL_{\\max} - L_{\\min} > \\frac{1}{\\sqrt{k} \\cdot L_{\\max}},\n$$\nwhere $L_{\\max}$ and $L_{\\min}$ are respectively maximum and minimum distances among them.",
    "options": [],
    "answer": "4",
    "solution": "Answer: $k = 4$.\nAt first we note that $L_{\\max} > L_{\\min}$ because the triangle made by these three points is not equilateral. Assume the contrary: then the area of a triangle with vertices in integral points is a rational number, but the area of an equilateral triangle can be computed by formula $S = \\frac{a^2\\sqrt{3}}{4}$ which is irrational if $a^2$ is rational.\nAs $L_{\\max}^2$ and $L_{\\min}^2$ are integers we have $L_{\\max}^2 - L_{\\min}^2 \\ge 1$ from where we get\n$$\nL_{\\max} - L_{\\min} \\ge \\frac{1}{L_{\\max} + L_{\\min}} > \\frac{1}{2L_{\\max}} = \\frac{1}{\\sqrt{4}L_{\\max}}.\n$$\nIt remains to show that for $k \\le 3$ one can find three points for which $L_{\\max} - L_{\\min} \\le \\frac{1}{\\sqrt{k \\cdot L_{\\max}}}$. Consider an isosceles triangle with base and height against base both equal 2. Then the side length is $\\sqrt{5}$ and\n$$\nL_{\\max} - L_{\\min} = \\sqrt{5} - 2 < \\frac{1}{\\sqrt{3}\\sqrt{5}}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25059,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all triples $x, y, z$ of real numbers such that the following system holds:\n$$\n\\begin{cases} x^3 + y = z^2 \\\\ y^3 + z = x^2 \\\\ z^3 + x = y^2 \\end{cases}\n$$",
    "options": [],
    "answer": "(0, 0, 0)",
    "solution": "Answer. The only triple satisfying given system of equations is $(x, y, z) = (0, 0, 0)$.\n\nFirst, consider the case $x, y, z \\ge 0$. Note that for all $t \\ge 0$ we have $t^3 + t \\ge 2t^2$. Summing up the three equations we obtain $x^3 + x + y^3 + y + z^3 + z = x^2 + y^2 + z^2$. By our observation, the lefthandside is greater than or equal to $2(x^2 + y^2 + z^2)$. This yields $x^2 + y^2 + z^2 \\le 0$, which holds only for $x = y = z = 0$. The triple $(x, y, z) = (0, 0, 0)$ clearly satisfies the system.\n\nSecond, consider the case when at least two of $x, y, z$ are negative. Because of cyclicity, we may assume that $x, y < 0$. Then $z^2 = x^3 + y < 0$, which is impossible.\n\nWe are left with the case when exactly one of $x, y, z$ is negative, say $x$. Then $y, z \\ge 0$. Note that $y > y + x^3 = z^2$ and $z^3 > z^3 + x = y^2$, in particular both $y, z$ are positive. We have $z^3 > y^2 > yz^2$ which gives $z > y$. Further, $y > z^2 > y^2$, hence $1 > y$.\n\nOn the other hand, the first equation multiplied by $x$ gives $x^4 + xy = xz^2$, and the third equation multiplied by $y$ gives $yz^3 + xy = y^3$. Subtracting obtained equalities yields $x^4 - yz^3 = xz^2 - y^3$. Rearranging terms leads to $x^4 + y^3 = xz^2 + yz^3$. Hence $0 < x^4 + y^3 = z^2(x + yz)$ which implies $x + yz > 0$.\n\nFinally, note that $x^2 = y^3 + z^2 > x^3 + y^2 = z^2$. Keeping in mind that $z > 0 > x$, we obtain $-x > z$. Therefore $0 > x + z > x + yz > 0$ a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25060,
    "subject": "Mathematics (Multi-modal)",
    "question": "A real sequence $(a_n)_{n=0}^{\\infty}$ is defined recursively by $a_0 = 2$ and the recursion formula\n$$\na_n = \\begin{cases} a_{n-1}^2 & \\text{if } a_{n-1} < \\sqrt{3} \\\\ \\frac{a_{n-1}^2}{3} & \\text{if } a_{n-1} \\ge \\sqrt{3}. \\end{cases}\n$$\nAnother real sequence $(b_n)_{n=1}^{\\infty}$ is defined in terms of the first by the formula\n$$\nb_n = \\begin{cases} 0 & \\text{if } a_{n-1} < \\sqrt{3} \\\\ \\frac{1}{2^n} & \\text{if } a_{n-1} \\ge \\sqrt{3}, \\end{cases}\n$$\nvalid for each $n \\ge 1$. Prove that\n$$\nb_1 + b_2 + \\dots + b_{2020} < \\frac{2}{3}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The first step is to prove, using induction, the formula\n$$\na_n = \\frac{2^{2^n}}{3^{2^n}(b_1+b_2+\\cdots+b_n)}.\n$$\nThe base case $n = 0$ is trivial. Assume the formula is valid for $a_{n-1}$, that is,\n$$\na_{n-1} = \\frac{2^{2^{n-1}}}{3^{2^{n-1}}(b_1+b_2+\\cdots+b_{n-1})}.\n$$\nIf now $a_{n-1} < \\sqrt{3}$, then $b_n = 0$, and so\n$$\na_n = a_{n-1}^2 = \\frac{2^{2^n}}{3^{2^n}(b_1+b_2+\\cdots+b_{n-1})} = \\frac{2^{2^n}}{3^{2^n}(b_1+b_2+\\cdots+b_{n-1}+b_n)},\n$$\nwhereas if $a_{n-1} \\ge \\sqrt{3}$, then $b_n = \\frac{1}{2^n}$, and so\n$$\na_n = \\frac{a_{n-1}^2}{3} = \\frac{2^{2^n}}{3^{2^n}(b_1+b_2+\\cdots+b_{n-1})+1} = \\frac{2^{2^n}}{3^{2^n}(b_1+b_2+\\cdots+b_{n-1}+b_n)}.\n$$\nThis completes the induction.\nNext, we inductively establish the inequality $a_n \\ge 1$. The base case $n = 0$ is again trivial. Suppose $a_{n-1} \\ge 1$. If $a_{n-1} < \\sqrt{3}$, then\n$$\na_n = a_{n-1}^2 \\ge 1^2 = 1,\n$$\nwhereas if $a_{n-1} \\ge \\sqrt{3}$, then\n$$\na_n = \\frac{a_{n-1}^2}{3} \\ge \\frac{(\\sqrt{3})^2}{3} = 1,\n$$\nand the induction is complete.\nFrom\n$$\n1 \\le a_n = \\frac{2^{2^n}}{3^{2^n}(b_1+b_2+\\cdots+b_n)} = \\left( \\frac{2}{3^{b_1+b_2+\\cdots+b_n}} \\right)^{2^n},\n$$\nwe may then draw the conclusion\n$$\n3^{b_1+b_2+\\cdots+b_n} \\le 2.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25061,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all polynomial functions $P(x, y)$ with real coefficients which for all real numbers $x$ and $y$ satisfy\n$$\nP(x + y, x - y) = 2P(x, y).\n$$",
    "options": [],
    "answer": "P(x, y) = (a + b)x^2 + axy + by^2 for arbitrary real a, b.",
    "solution": "Answer: $P(x, y) = (a + b)x^2 + axy + by^2$, where $a$ and $b$ are real numbers.\nFrom the given equation,\n$$\nP(2x, 2y) = P((x+y) + (x-y), (x+y) - (x-y)) = 2P(x+y, x-y) = 4P(x, y). \\quad (1)\n$$\nLet $m, n$ be arbitrary non-negative integers. Consider term $c_{m,n}x^m y^n$ of the polynomial $P(x, y)$. Two polynomials are equal if and only if coefficients of all their respective terms are equal. On the left of Eq. (1), we have $c_{m,n}(2x)^m(2y)^n = 2^{m+n}c_{m,n}x^m y^n$. On the right, we have $4c_{m,n}x^m y^n$. Hence $c_{m,n} = 0$ unless $m + n = 2$.\nTherefore, the only terms which appear in the polynomial $P(x, y)$ are the square terms, i.e. $P(x, y) = cx^2 + axy + by^2$ for some coefficients $a, b, c$. By substituting to the given equation and cancelling the terms, we get\n$$\n(c + a + b)x^2 + 2(c - b)xy + (c - a + b)y^2 = 2cx^2 + 2axy + 2by^2.\n$$\nThe respective coefficients of the r.h.s. and the l.h.s. must be equal. Therefore, $c + a + b = 2c$, from which $c = a + b$. This makes the other respective coefficients equal as well. Therefore, all polynomials in the form $P(x, y) = (a + b)x^2 + axy + by^2$ ($a, b$ are arbitrary real numbers) satisfy the given equality.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 25062,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $n > 1$ guests at Georg's birthday party and each guest is friends with at least one other guest. Georg organizes a party game among the guests. Each guest receives a jug of water such that there are no two guests with the same amount of water in their jugs. Every guest now proceeds as follows. The guest takes one cup for each of their friends at the party and distributes all the water from their jug evenly in the cups. They then pass a cup to each of their friends. Each guest having received a cup of water from each of their friends pours the water they have received into their jug. What is the smallest possible number of guests that do not have the same amount of water as they started with?",
    "options": [],
    "answer": "2",
    "solution": "Answer: 2.\nIf there are guests $1$, $2$, $\\ldots$, $n$ and guest $i$ is friends with guest $i-1$ and $i+1$ modulo $n$ (e.g. guest $1$ and guest $n$ are friends). Then if guest $i$ has $i$ amount of water in their jug at the start of the game, then only guest $1$ and $n$ end up with a different amount of water than they started with.\n\nTo show that there always will be at least two guests with a different amount of water at the end of the game than they started with, let $x_i$ and $d_i$ be the amount of water and number of friends, respectively, that guest $i$ has. Define $z_v = x_v/d_v$ and assume without loss of generality that the friendship graph of the party is connected. Since every friend has at least one friend, there must exist two guests $a$ and $b$ at the party with the same number of friends by the pigeonhole principle. They must satisfy $z_a \\neq z_b$. Thus, the sets\n$$\nS = \\{c \\mid z_c = \\min_d z_d\\} \\text{ and } T = \\{c \\mid z_c = \\max_d z_d\\}\n$$\nare non-empty and disjoint. Since we assumed the friendship graph to be connected, there exists a guest $c \\in S$ that has a friend $d$ not in $S$. Let $F$ be the friends of $c$ at the party. Then the amount of water in $c$'s cup at the end of the game is\n$$\n\\sum_{f \\in F} z_f \\geq z_d + (d_c - 1)z_c > d_c \\cdot z_c = x_c.\n$$\nThus, $c$ ends up with a different amount of water at the end of the game. Similarly, there is a guest in $T$ that ends up with a different amount of water at the end of the game than what they started with.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25063,
    "subject": "Mathematics (Multi-modal)",
    "question": "Alice and Bob are playing hide and seek. Initially, Bob chooses a secret fixed point $B$ in the unit square. Then Alice chooses a sequence of points $P_0, P_1, \\ldots, P_N$ in the plane. After choosing $P_k$ (but before choosing $P_{k+1}$) for $k \\ge 1$, Bob tells \"warmer\" if $P_k$ is closer to $B$ than $P_{k-1}$, otherwise he says \"colder\". After Alice has chosen $P_N$ and heard Bob’s answer, Alice chooses a final point $A$. Alice wins if the distance $AB$ is at most $1/2020$, otherwise Bob wins. Show that if $N = 18$, Alice cannot guarantee a win.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $S_0$ be the set of all points in the square, and for each $1 \\le k \\le N$, let $S_k$ be the set of possible points $B$ consistent with everything Bob has said. For each $k$, we then have that $S_k$ is the disjoint union of the two possible values $S_{k+1}$ can take for each of Bob's possible answers. Hence one of these must have area $\\le \\frac{|S_{k+1}|}{2}$, and the other must have area $\\ge \\frac{|S_{k+1}|}{2}$. Suppose now that Alice always receives the answer resulting in the greater half. After receiving $N$ answers, then, $|S_N| \\ge \\frac{1}{2^N}$. If Alice has a winning strategy, there must be a point $A$ in $S_N$ so that the circle of radius $\\frac{1}{2020}$ centered at $A$ contains $S_N$. Hence $\\frac{\\pi}{2020^2} \\ge \\frac{1}{2^N}$. It therefore suffices to show that this inequality does not hold for $N=18$. This follows from the estimates $\\pi \\le 2^2$ and $2020 > 1024 = 2^{10}$, meaning that $\\frac{\\pi}{2020^2} > \\frac{2^2}{2^{20}} = \\frac{1}{2^{18}}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25064,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a school with $n$ students, a team of students must be chosen to clean the school. A team of students is called *good* if the school principal believes that they can manage to clean the school. It is known that at least one team is good, and that if a team is good, then it is also good after adding any other student. Show that the average number of students in good teams is at least $\\frac{n}{2}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $S$ be the set students and let $G$ be the set of good teams. For any student $s$ consider the trivial injection from the set of good teams without $s$ to the set of good teams containing $s$ by taking a team $T$ and mapping it to $T \\cup \\{s\\}$. Since this is an injection we conclude that there are at least as many good teams with $s$ than without $s$. This means that\n$$\n\\sum_{T \\in G} 1_{(s \\in T)} \\geq \\frac{|G|}{2}.\n$$\nThe average number of students in good teams is then\n$$\n\\frac{1}{|G|} \\sum_{T \\in G} |T| = \\frac{1}{|G|} \\sum_{T \\in G} \\sum_{s \\in S} 1_{(s \\in T)} = \\frac{1}{|G|} \\sum_{s \\in S} \\sum_{T \\in G} 1_{(s \\in T)} \\geq \\frac{1}{|G|} \\sum_{s \\in S} \\frac{|G|}{2} = \\frac{n}{2}.\n$$\nNote: The last part is just linearity of expectation in disguise.\nLet $\\mathcal{F}$ be the set of good teams of students and denote by $\\mathcal{F}_k$ the set of good teams of students with exactly $k$ students. Now, for some $k < n$ construct the following bipartite graph. The vertices on the left side are the teams of $\\mathcal{F}_k$ and on the right side the teams of $\\mathcal{F}_{k+1}$. For every team $A \\in \\mathcal{F}_k$ let there be an edge from $A$ to $A \\cup \\{i\\} \\in \\mathcal{F}_{k+1}$ for every $i$ a student not in $A$. Clearly, the degree of every vertex on the left is $n-k$ and the degree of every vertex on the right is $\\leq k+1$. Hence,\n$$\n\\frac{|\\mathcal{F}_{k+1}|}{|\\mathcal{F}_k|} \\geq \\frac{n-k}{k+1} = \\frac{\\binom{n}{k+1}}{\\binom{n}{k}}. \\qquad (4)\n$$\nThis implies that the function $k \\mapsto |\\mathcal{F}_k|/\\binom{n}{k}$ is weakly increasing. Thus, for every $k < \\frac{n}{2}$, $|\\mathcal{F}_k| \\leq |\\mathcal{F}_{n-k}|$, by symmetry of the binomial coefficient, such that $\\sum_{k<n/2} |\\mathcal{F}_k| \\leq \\sum_{k \\geq n/2} |\\mathcal{F}_k|$. The conclusion follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25065,
    "subject": "Mathematics (Multi-modal)",
    "question": "6 lines are chosen in 3-dimensional space. Find the largest number of points at which at least 3 of the chosen lines intersect.",
    "options": [],
    "answer": "4",
    "solution": "Answer: 4.\nLet us call a point *rich* if at least 3 of the chosen lines intersect there. Let the number of rich points be $n$. Let us count pairs $(P, s)$, in which $P$ is a rich point and $s$ is a chosen line passing through that point. There are at least $3n$ such pairs because each of the $n$ rich points appears in at least 3 pairs. On the other hand, for every chosen line all lines passing through the rich points on that line are different (two lines cannot intersect at more than one point). Therefore, on each of the chosen lines there are at most $\\lfloor \\frac{6-1}{2} \\rfloor = 2$ rich points and the number of pairs counted can be at most $2 \\cdot 6 = 12$. Consequently, $3n \\leq 12$, i.e., $n \\leq 4$.\nA construction for 4 rich points is provided by the lines defined by the 6 sides of a tetrahedron. 3 of the lines intersect at each of the 4 vertices of the tetrahedron.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25066,
    "subject": "Mathematics (Multi-modal)",
    "question": "A mason has bricks with dimensions $2 \\times 5 \\times 8$ and other bricks with dimensions $2 \\times 3 \\times 7$. She also has a box whose interior has dimensions $10 \\times 11 \\times 14$. The bricks and the interior of the box are all rectangular parallelepipeds. The mason packs bricks into the box filling the entire volume of its interior. How many bricks does she pack in the box?",
    "options": [],
    "answer": "24",
    "solution": "Answer: 24.\nLet the number of $2 \\times 5 \\times 8$ bricks in the box be $x$, and the number of $2 \\times 3 \\times 7$ bricks $y$. We must figure out the sum $x + y$. The volume of the box is divisible by $7$, and so is the volume of any $2 \\times 3 \\times 7$ brick. The volume of a $2 \\times 5 \\times 8$ brick is not divisible by $7$, which means that $x$ must be divisible by $7$.\n\nThe volume of the box is $10 \\cdot 11 \\cdot 14$. The volume of the $2 \\times 5 \\times 8$ bricks in the box is $x \\cdot 2 \\cdot 5 \\cdot 8 = 80x$. Since this volume cannot exceed the volume of the box, we must have\n$$\nx \\le \\frac{10 \\cdot 11 \\cdot 14}{80} = \\frac{11 \\cdot 7}{4} = \\frac{77}{4} < 20.\n$$\nSince $x$ was divisible by $7$, and certainly nonnegative, we conclude that $x$ must be $0$, $7$ or $14$. Let us explore each of these possibilities separately.\n\nIf we had $x = 0$, then the volume of the $2 \\times 3 \\times 7$ bricks, which is $y \\cdot 2 \\cdot 3 \\cdot 7$, would be equal to the volume of the box, which is $10 \\cdot 11 \\cdot 14$. However, this is not possible since the volume of the $2 \\times 3 \\times 7$ bricks is divisible by three whereas the volume of the box is not. Thus $x$ must be $7$ or $14$.\n\nIf we had $x = 7$, then equating the total volume of the bricks with the volume of the box would give\n$$\n7 \\cdot 2 \\cdot 5 \\cdot 8 + y \\cdot 2 \\cdot 3 \\cdot 7 = 10 \\cdot 11 \\cdot 14,\n$$\nso that\n$$\ny \\cdot 2 \\cdot 3 \\cdot 7 = 10 \\cdot 11 \\cdot 14 - 7 \\cdot 2 \\cdot 5 \\cdot 8 = 1540 - 560 = 980.\n$$\nHowever, again the left-hand side, the volume of the $2 \\times 3 \\times 7$ bricks, is divisible by three, whereas the right-hand side, $980$, is not. Thus we cannot have $x = 7$ either, and the only possibility is $x = 14$.\n\nSince $x = 14$, equating the volumes of the bricks and the box gives\n$$\n14 \\cdot 2 \\cdot 5 \\cdot 8 + y \\cdot 2 \\cdot 3 \\cdot 7 = 10 \\cdot 11 \\cdot 14,\n$$\nwhich in turn leads to\n$$\ny \\cdot 2 \\cdot 3 \\cdot 7 = 10 \\cdot 11 \\cdot 14 - 14 \\cdot 2 \\cdot 5 \\cdot 8 = 1540 - 1120 = 420,\n$$\nso that\n$$\ny = \\frac{420}{2 \\cdot 3 \\cdot 7} = \\frac{420}{42} = 10.\n$$\nThus the number of bricks in the box can only be $14 + 10 = 24$. Finally, for completeness, let us observe that $14$ bricks with dimensions $2 \\times 5 \\times 8$ can be used to fill a volume with dimensions $10 \\times 8 \\times 14$, and $10$ bricks with dimensions $2 \\times 3 \\times 7$ can be used to fill a volume with dimensions $10 \\times 3 \\times 14$, so that these $24$ bricks can indeed be packed in the box.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 25067,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that there exists a polygon on a square grid that can be tiled with dominoes ($1 \\times 2$ or $2 \\times 1$ figures) in exactly 2020 ways.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $A_n$ be a figure that consists of a $2 \\times 2$ square combined with $n$ L-shapes ($A_3$ is shown in Figure ?? (a)). We will prove by induction that $A_n$ can be tiled with dominoes in exactly $2n + 2$ ways.\n\nThe base case for $A_0$ is evident. Now assume that we have proved it for $A_k$ and consider the figure $A_{k+1}$. Its bottom rightmost square can be tiled in two ways. If it is tiled with a horizontal domino (Figure ?? (b)) then this leads to a unique tiling for all $A_{k+1}$.\n\nIf it is tiled with a vertical domino, then consider the bottom rightmost square of the remaining figure. If it is tiled with a vertical domino (Figure ?? (c)), then this again leads to a unique tiling for all $A_{k+1}$. But if it is tiled with a horizontal domino (Figure ?? (d)), then the remaining figure is $A_k$ that can be tiled in $2k + 2$ ways.\n\nTherefore $A_{k+1}$ can be covered with dominoes in exactly $1 + 1 + (2k + 2) = 2k + 4$ ways, which finishes the induction step.\n\nNow we see that $A_{1009}$ can be covered in exactly 2020 ways.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25068,
    "subject": "Mathematics (Multi-modal)",
    "question": "Each vertex of complete bipartite graph $K_{128,128}$ is occupied by a person. In the beginning all the persons are unfamiliar. If the persons are in the adjacent vertices, they immediately get to know each other. It is allowed to take several edges without common vertices and swap the persons in the endpoints of each edge. What is the minimum number of such operations we need to get all the persons acquainted?",
    "options": [],
    "answer": "6",
    "solution": "Answer: 6.\nLet $r = 6$. Assign each person a string $x = (x_0, x_1, \\dots, x_r) \\in \\{0, 1\\}^{r+1}$ such that all persons who started on the same side have the same first bit $x_0$. We now describe an $r$-rounds strategy for acquaintance. In the $i$'th round move all persons with $x_i = 0$ to A and all agents with $x_i = 1$ to B. Now if two persons are assigned strings $x$ and $x'$ such that $x_i \\neq x'_i$, then in the $i$'th round they will be on different sides of the graph, and hence will be acquainted.\n\nWe now claim that $r$ rounds are also necessary. Indeed, suppose we have a $t$-rounds strategy for acquaintance. Assign each person a string $x = (x_0, x_1, \\dots, x_t) \\in \\{0, 1\\}^{t+1}$, where $x_i = 0$ for $i \\le t$ if and only if in the $i$'th round the person was in A. Note that two agents met during the $t$ rounds if and only if their strings are different. This implies $2^{t+1} \\ge 2n$, and thus $t \\ge r$, as required.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 25069,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. A restaurant offers a choice of $n$ starters, $n$ main dishes, $n$ desserts and $n$ wines. A merry company dines at the restaurant, with each guest choosing a starter, main dish, dessert and wine. No two people place exactly the same order. It so happens that there is no collection of $n$ guests such that their orders coincide in three of these aspects, but in the fourth one they all differ. (For example, there are no $n$ people that order exactly the same three courses of food, but $n$ different wines.) What is the maximal number of guests?",
    "options": [],
    "answer": "n^4 - n^3",
    "solution": "Answer: The maximal number of guests is $n^4 - n^3$.\nThe possible menus are represented by quadruples\n$$\n(a, b, c, d), \\quad 1 \\le a, b, c, d \\le n.\n$$\nLet us count those menus satisfying\n$$\na + b + c + d \\not\\equiv 0 \\pmod{n}.\n$$\nThe numbers $a, b, c$ may be chosen arbitrarily ($n$ choices for each), and then $d$ is required to satisfy only $d \\not\\equiv -a - b - c$. Hence there are\n$$\nn^3(n - 1) = n^4 - n^3\n$$\nsuch menus.\nIf there are $n^4 - n^3$ guests, and they have chosen precisely the $n^4 - n^3$ menus satisfying $a+b+c+d \\not\\equiv 0 \\pmod{n}$, we claim that the condition of the problem is fulfilled. So suppose there is a collection of $n$ people whose orders coincide in three aspects, but differ in the fourth. With no loss of generality, we may assume they have ordered exactly the same food, but $n$ different wines. This means they all have the same value of $a, b$ and $c$, but their values of $d$ are distinct. A contradiction arises since, given $a, b$ and $c$, there are only $n-1$ values available for $d$.\nWe now show that for $n^4 - n^3 + 1$ guests (or more), it is impossible to obtain the situation stipulated in the problem. The $n^3$ sets\n$$\nM_{a,b,c} = \\{(a, b, c, d) \\mid 1 \\le a, b, c, d \\le n\\}, \\quad 1 \\le a, b, c \\le n,\n$$\nform a partition of the set of possible menus, totalling $n^4$. When the number of guests is at least $n^4 - n^3 + 1$, there are at most $n^3 - 1$ unselected menus. Therefore, there exists a set $M_{a,b,c}$ which contains no unselected menus. That is, all the $n$ menus in $M_{a,b,c}$ have been selected, and the condition of the problem is violated.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25070,
    "subject": "Mathematics (Multi-modal)",
    "question": "Each vertex $v$ and each edge $e$ of a graph $G$ are assigned numbers $f(v) \\in \\{1, 2\\}$ and $f(e) \\in \\{1, 2, 3\\}$, respectively. Let $S(v)$ be the sum of numbers assigned to the edges incident to $v$ plus the number $f(v)$. We say that assignment $f$ is cool if $S(u) \\neq S(v)$, for every pair of adjacent vertices in $G$. Prove that every graph has a cool assignment.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $v_1, v_2, \\dots, v_n$ be any ordering of the vertices of $G$. Initially each vertex assigned number $1$, and each edge assigned number $2$. One may imagine that there is a chip lying on each vertex, while two chips are lying on each edge. We are going to refine this assignment so as to get a cool one by performing the following greedy procedure.\n\nTo explain what we do in the $i$th step, denote by $x_1, x_2, \\dots, x_k$ all backward neighbors of $v_i$, and let $e_j = v_i x_j$, with $j = 1, 2, \\dots, k$, denote the corresponding backward edges. For each edge $e_j$ we have two possibilities: (1) if there is only one chip on $x_j$, then we may move one chip from $e_j$ to $x_j$ or do nothing, (2) if there are two chips on $x_j$ we may move one chip from $x_j$ to $e_j$ or do nothing. Notice that none of the sums $S(x_j)$ may change as a result of such action. Also, any action on each edge may change the total sum for $v_i$ just by one. Hence there are $k+1$ possible values for $S(v_i)$. So, at least one combination of chips gives a sum which is different from each of $S(x_j)$. We fix this combination and go to the next step. The proof is complete.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 25071,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider $2n$ rays (half-lines) in the plane such that no two rays are parallel (the endpoints of the rays may coincide). Prove that there exists a line in the plane that does not pass through any of the endpoints and intersects with exactly $n$ rays.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us choose any circle such that all of the endpoints of the rays are inside the circle. Let us also choose a tangent line on the circle that is not parallel to any of the rays. Let this tangent line be $l_0$ and let $l_\\alpha$ be a tangent line we get after rotating $l_0$ counterclockwise by angle $\\alpha$ with respect to the center of the circle. Let $f(\\alpha)$ be the number of rays that intersect with $l_\\alpha$.\n\nSince $l_0$ and $l_\\pi$ are two parallel lines and all of the endpoints of the rays are between them, we know from the definition of $l_0$ that $f(0) + f(\\pi) = 2n$. Without loss of generality, let $f(0) \\le n$ and $f(\\pi) \\ge n$. Because no two rays are parallel, when we continuously increase $\\alpha$ (i.e. rotate the tangent line), $f(\\alpha)$ can at any time only change by at most one. Therefore, $f(\\alpha)$ ranges over all integral values between $f(0)$ and $f(\\pi)$ and there exists a value of $\\alpha$ for which $f(\\alpha) = n$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25072,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with $AB > AC$. The bisector of $\\angle BAC$ intersects the side $BC$ at $D$. The circles with diameters $BD$ and $CD$ intersect the circumcircle of $\\triangle ABC$ a second time at $P \\neq B$ and $Q \\neq C$, respectively. The lines $PQ$ and $BC$ intersect at $X$. Prove that $AX$ is tangent to the circumcircle of $\\triangle ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The key observation is that the circumcircle of $\\triangle DPQ$ is tangent to $BC$. This can be proved by angle chasing:\n$$\n\\begin{aligned}\n\\angle BDP &= 90^\\circ - \\angle PBD = 90^\\circ - \\angle PBC = 90^\\circ - (180^\\circ - \\angle CQP) \\\\\n&= \\angle CQP - 90^\\circ = \\angle DQP.\n\\end{aligned}\n$$\nNow let the tangent to the circumcircle of $\\triangle ABC$ at $A$ intersect $BC$ at $Y$. It is well-known (and easy to show) that $YA = YD$. This implies that $Y$ lies on the radical axis of the circumcircles of $\\triangle ABC$ and $\\triangle PDQ$, which is the line $PQ$. Thus $Y \\equiv X$, and the claim follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25073,
    "subject": "Mathematics (Multi-modal)",
    "question": "On a plane, there are given points $A_0, B_0, C_0$ (not necessarily distinct) such that $A_0B_0 + B_0C_0 + C_0A_0 = 1$. Points $A_1, B_1, C_1$ (not necessarily distinct) are chosen in such a way that $A_1B_1 = A_0B_0$ and $B_1C_1 = B_0C_0$. Points $A_2, B_2, C_2$ are chosen as a permutation of points $A_1, B_1, C_1$. Finally, points $A_3, B_3, C_3$ (not necessarily distinct) are chosen in such a way that $A_3B_3 = A_2B_2$ and $B_3C_3 = B_2C_2$. Find the least and the greatest possible value of $A_3B_3 + B_3C_3 + C_3A_3$.",
    "options": [],
    "answer": "1/3 and 3",
    "solution": "Answer: $\\frac{1}{3}$ and $3$.\n\nDenote the lengths $A_0B_0, B_0C_0, C_0A_0$ by $x, y, z$ in non-increasing order. Similarly, denote the lengths $A_1B_1, B_1C_1, C_1A_1$ by $x', y', z'$ in non-increasing order, and the lengths $A_3B_3, B_3C_3, C_3A_3$ by $x'', y'', z''$ in non-increasing order. (As permuting the points does not change the distances, we do not need a separate vector for $A_2B_2, B_2C_2, C_2A_2$.) Then we have $x + y + z = 1$, $y + z \\ge x$, $y' + z' \\ge x'$, $y'' + z'' \\ge x''$. By construction, triples $(x, y, z)$ and $(x', y', z')$ have two values in common (but not necessarily at corresponding places), similarly $(x', y', z')$ and $(x'', y'', z'')$ have two values in common.\n\nUsing these observations, calculate:\n$$\n\\begin{aligned}\nx'' + y'' + z'' &\\le 2(y'' + z'') \\le 2(x' + y') \\le 2(y' + y' + z') \\\\\n&\\le 2(x + x + y) \\le 6x \\le 3(x + y + z) = 3.\n\\end{aligned}\n$$\n\nWe can achieve the value $3$ as follows. Let $A_0B_0 = \\frac{1}{2}$ and $C_0 = A_0$. Let $A_1 = A_0$, $B_1 = B_0$ and $\\overrightarrow{B_1C_1} = -\\overrightarrow{B_0C_0}$. Let $A_2 = A_1$ and $B_2 = C_1$, $C_2 = B_1$. Finally, let $A_3 = A_2$, $B_3 = B_2$ and $\\overrightarrow{B_3C_3} = -\\overrightarrow{B_2C_2}$. By construction, $A_3B_3 = 1$, $B_3C_3 = \\frac{1}{2}$ and $C_3A_3 = \\frac{3}{2}$, so $A_3B_3 + B_3C_3 + C_3A_3 = 3$.\n\nThis establishes the upper bound. For the lower bound, note that all steps are reversible and the 3-step process itself is symmetric. By scaling, we can also make the initial configuration to satisfy the conditions of the problem. Hence all processes satisfying the conditions of the problem and achieving a final value $t$ are in one-to-one correspondence with processes satisfying the conditions of the problem and achieving the final value $\\frac{1}{t}$. This shows that the lower bound is $\\frac{1}{3}$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 25074,
    "subject": "Mathematics (Multi-modal)",
    "question": "Figure shows two non-intersecting circles $\\alpha$ and $\\beta$ in space. We say that circle $\\alpha$ *devours* circle $\\beta$ since one chord of $\\beta$ (solid) is strictly contained in a chord of $\\alpha$ (dashed).\nThe question is whether it is possible to place three circles $\\alpha, \\beta$ and $\\gamma$ in space so as to have $\\alpha$ devouring $\\beta$ devouring $\\gamma$ devouring $\\alpha$. The radii of the circles need not be equal.\n![](attached_image_1.png)\n",
    "options": [],
    "answer": "No, it is impossible.",
    "solution": "Answer: No, it is impossible.\nConsider a point $X$ on a chord drawn in circle $\\alpha$, as in Figure ???. The power of $X$ with respect to $\\alpha$ is given by the familiar expression $p_{\\alpha}(X) = -xy$.\n\nLet us examine the case of two circles. The situation when $\\alpha$ devours $\\beta$ is represented in Figure ???. The point $X$, lying on the common chord, is coplanar with both circles, and so we may compute its power with respect to both of them. Evidently $p_{\\alpha}(X) < p_{\\beta}(X)$, for the chord in $\\beta$ is contained within the chord in $\\alpha$, and the distances to the periphery are correspondingly smaller.\n![](attached_image_2.png)\nFigure 2: One circle.\n![](attached_image_3.png)\nFigure 3: Two circles.\n\nSuppose finally we have three circles $\\alpha, \\beta, \\gamma$, somehow cyclically devouring each other, as requested per the problem. Each of the three circles determines a plane; their common point $X$ will be coplanar with all the circles. Using the above reasoning, we arrive at the contradiction\n$$\np_{\\alpha}(X) < p_{\\beta}(X) < p_{\\gamma}(X) < p_{\\alpha}(X).\n$$",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 25075,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCDE$ be a convex pentagon inscribed in a circle $\\omega$ such that $CD \\parallel BE$. The line tangent to $\\omega$ at $B$ intersects the line $AC$ at a point $F$ such that $A$ lies between the points $C$ and $F$. The lines $BD$ and $AE$ intersect at $G$. Prove that the line $FG$ is tangent to the circumcircle of $ADG$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $CD \\parallel BE$, we have $BC = DE$. Note that\n$$\n\\angle BFC = \\angle BAC - \\angle ABF = \\angle DBE - \\angle AEB = \\angle BGA,\n$$\nhence $ABGF$ is cyclic. It follows that $\\angle AGF = \\angle ABF = \\angle ADG$, hence the circumcircle of $ADG$ is tangent to $FG$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25076,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with the circumcircle $\\omega$. The internal angle bisectors of $\\angle ABC$ and $\\angle ACB$ intersect $\\omega$ at $X$ and $Y$, respectively. Let $K$ be a point on $CX$ such that $\\angle KAC = 90^\\circ$. Similarly, let $L$ be a point on $BY$ such that $\\angle LAB = 90^\\circ$. Let $S$ be the midpoint of arc $CAB$ of $\\omega$. Prove that $SK = SL$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "W.l.o.g. let $AB < AC$. We will prove that triangles $KXS$ and $SYL$ are congruent by SAS, which will finish the proof.\nAs $BX$ and $CY$ are angle bisectors, we obtain:\n$$\n\\frac{1}{2} C\\hat{A}B = C\\hat{X}S = C\\hat{X} + \\hat{X}S = \\frac{1}{2} C\\hat{X}A + \\hat{X}S.\n$$\nThis implies $\\hat{X}S = \\frac{1}{2} A\\hat{Y}B = \\hat{Y}B$ and therefore $SX = YB$. Note that $BY = YA$, hence $Y$ is the midpoint of the hypotenuse $BL$ in $\\triangle ABL$. Thus $SX = YB = YL$. Similarly, we get $SY = XK$.\nFinally, as $S$ is the midpoint of arc $C\\hat{A}B$, we obtain $\\angle SXC = \\angle BYS$, thus $\\angle KXS = \\angle SYL$, finishing the proof of congruency.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25077,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\omega$ be a semicircle with diameter $XY$. Let $M$ be the midpoint of $XY$. Let $A$ be an arbitrary point on $\\omega$ such that $AX < AY$. Let $B$ and $C$ be points lying on the segments $XM$ and $YM$, respectively, such that $BM = CM$. The line through $C$ parallel to $AB$ intersects $\\omega$ at $P$. The line through $B$ parallel to $AC$ intersects $\\omega$ at $Q$. The line $PQ$ intersects the line $XY$ at a point $S$. Prove that the line $AS$ is tangent to $\\omega$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $D$ be the point symmetric to $A$ with respect to $M$. Then $ABDC$ is a parallelogram as $M$ bisects both $BC$ and $AD$. It follows that $AB \\parallel CD$ and $AC \\parallel BD$. This means that $P$ lies on $CD$ and $Q$ lies on $BD$.\nWe shall prove that the tangent line at $A$ to $\\omega$ intersects $PQ$ at a point $S'$ such that $\\frac{PS'}{S'Q} = \\frac{PS}{SQ}$. This in turn will imply that $S = S'$ giving the thesis.\nSince $AS'$ is tangent to $\\omega$, we have $\\angle APS' = \\angle S'AQ$. Hence triangles $APS'$, $QAS'$ are similar as they have equal angles. Hence\n$$\n\\frac{PS'}{S'Q} = \\frac{PS'}{AS'} \\cdot \\frac{AS'}{S'Q} = \\frac{AP}{AQ} \\cdot \\frac{AP}{AQ} = \\frac{AP^2}{AQ^2}.\n$$\nWe now calculate the ratio $\\frac{PS}{QS}$. By Menelaos' theorem for triangle $PQD$ and transversal $XY$ we have\n$$\n\\frac{PS}{SQ} \\cdot \\frac{QB}{BD} \\cdot \\frac{DC}{CP} = 1.\n$$\nSince $ABDC$ is a parallelogram, we have $BD = AC$ and $DC = AB$. This yields\n$$\n\\frac{PS}{SQ} = \\frac{BD}{QB} \\cdot \\frac{CP}{DC} = \\frac{AC \\cdot CP}{AB \\cdot BQ}.\n$$\nSince $AM = MP = MD = MQ$, we have $\\angle APC = 90^\\circ = \\angle AQB$. Moreover, since $ABDC$ is a parallelogram, $\\angle ABQ = \\angle ACP$. This shows that triangles $ABQ$, $ACP$ are similar. In particular $\\frac{AC}{AB} = \\frac{AP}{AQ} = \\frac{CP}{BQ}$. Therefore\n$$\n\\frac{PS}{SQ} = \\frac{AC \\cdot CP}{AB \\cdot BQ} = \\frac{AP^2}{AQ^2} = \\frac{PS'}{S'Q},\n$$\nwhich, as was observed before, finishes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25078,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle with a circumcircle $\\omega$. Let $\\ell$ be the tangent line to $\\omega$ at $A$. Let $X$ and $Y$ be the projections of $B$ onto lines $\\ell$ and $AC$, respectively. Let $H$ be the orthocenter of $BXY$. Let $CH$ intersect $\\ell$ at $D$. Prove that $BA$ bisects angle $CBD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\nNote that $XH \\perp BY \\perp AC$ and $YH \\perp BX \\perp AD$. Therefore $XH \\parallel AC$ and $YH \\parallel AD$. It follows that\n$$\n\\frac{AD}{AX} = \\frac{CD}{CH} = \\frac{CA}{CY} \\implies \\frac{AD}{CA} = \\frac{AX}{CY} = \\frac{AB \\sin \\angle XAB}{CB \\sin \\angle YCB}.\n$$\nSince $\\ell$ is tangent to $\\omega$, we have $\\angle XAB = \\angle YCB$. Thus the sines in the equality above cancel out and we obtain\n$$\n\\frac{AD}{CA} = \\frac{AB}{CB}.\n$$\nThis, along with $\\angle DAB = \\angle ACB$, proves that $\\triangle DAB \\sim \\triangle ACB$ by SAS. Therefore $\\angle CBA = \\angle ABD$. This shows that $BA$ bisects angle $CBD$.\nLet $D'$ be a point on $\\ell$ such that $\\angle CBA = \\angle ABD'$. Let $Z$ and $T$ be projections of $A$ onto $BC$ and $BD'$, respectively. Note that the circle with diameter $AB$ passes through $X, Y, Z, T$. By Pascal's theorem for hexagon $AXZBTY$, points $D', C$, and $H' := XZ \\cap TY$ are collinear.\n![](attached_image_2.png)\nWe have\n$$\n\\angle XZB = \\angle XAB = \\angle CAB = 90^\\circ - \\angle ZBY\n$$\nwhich shows that $XZ \\perp BY$.\nBy definition of $D'$, \n$$\n\\angle BD'A = 180^\\circ - \\angle D'AB - \\angle ABD' = 180^\\circ - \\angle ACB - \\angle CBA = \\angle BAC.\n$$\nTherefore\n$$\n\\angle D'AT = 90^\\circ - \\angle TD'A = 90^\\circ - \\angle BAC,\n$$\nhence\n$$\n\\angle XYT + \\angle BXY = \\angle XAT + \\angle BAY = 90^\\circ - \\angle BAC + \\angle BAC = 90^\\circ.\n$$\nThis shows that $YT \\perp BX$.\nSince $XZ \\perp BY$ and $YT \\perp BX$, it follows that $H'$ is the orthocenter of $BXY$, i.e. $H' = H$. Since $D', C, H'$ are collinear and $D'$ lies on $\\ell$, it follows that $D' = D$. Therefore $\\angle ABD = \\angle CBA$ and we are done.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25079,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\omega$ and $I$ be the incircle and the incentre of a triangle $ABC$, respectively. Let $E$ and $F$ be the tangency points of $\\omega$ with the sides $AC$ and $AB$, respectively. The perpendicular bisector of $AI$ intersects $AC$ at a point $P$. Point $Q$ lies on $AB$ and satisfies $QI \\perp FP$. Prove that $EQ \\perp AB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $PF$ intersect $QI$ at $X$. Let $Y$ be the point symmetric to $E$ with respect to $PI$.\nLet $\\omega_1$ be the circle with diameter $PI$. Let $\\omega_2$ be the circle with diameter $FI$.\nNote that $E, X$, and $Y$ all lie on $\\omega_1$, and that $X$ lies on $\\omega_2$. The line $XI$ is the radical axis of $\\omega_1$ and $\\omega_2$ because $X$ and $I$ lie on both circles. The circles $\\omega$ and $\\omega_2$ are tangent at $F$ and therefore their common tangent $AB$ is the radical axis of $\\omega$ and $\\omega_2$. Finally, $EY$ is the radical axis of $\\omega$ and $\\omega_1$ as both circles pass through $E$ and $Y$. It follows that $AB, XI, $ and $EY$ concur at the radical centre of $\\omega, \\omega_1$, and $\\omega_2$, which is $Q$. The conclusion is that $Q$ lies on $EY$.\nSince $P$ lies on the perpendicular bisector of $AI$, we have $\\angle PIA = \\angle IAP = \\frac{1}{2}\\angle BAC = \\angle BAI$. It follows that $PI \\parallel AB$. On the other hand, $PI \\perp EY$ as $E$ and $Y$ are symmetric with respect to $PI$. The thesis follows.\nLet $Q'$ be the projection of $E$ into $AB$. Let $\\xi$ be the spiral similarity which maps $A$ to $E$ and $I$ to $F$.\nNote that $\\angle IAP = \\angle PIA = \\frac{1}{2}\\angle BAC = \\angle IFE = \\angle FEI$. This shows that triangle $AIP$ is similar to triangle $EFI$ and hence $\\xi$ maps $P$ to $I$. On the other hand, $\\angle IFA = \\frac{\\pi}{2} = \\angle FQ'E$ and $\\angle AIF = \\frac{\\pi}{2} - \\frac{1}{2}\\angle BAC = \\angle EFQ'$. This shows that triangles $AIF$ and $EFQ'$ are similar and therefore $\\xi$ maps $F$ to $Q'$.\nNow observe that $\\angle (PF, IQ') = \\angle (AI, FE) = \\frac{\\pi}{2}$ and therefore $PF \\perp IQ'$. This means that $Q'$ coincides with $Q$. As a consequence, $Q$ is the projection of $E$ into $AB$ and therefore $EQ \\perp AB$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25080,
    "subject": "Mathematics (Multi-modal)",
    "question": "A triangle $ABC$ is given. Call its orthocenter $H$. Define $\\omega$ as the circle through $B$, $C$, and $H$, and define $\\Gamma$ as the circle with diameter $AH$. Let $X$ be the other intersection of $\\omega$ and $\\Gamma$, and let the reflection of $\\Gamma$ over $AX$ be $\\gamma$.\nSuppose $\\gamma$ and $\\omega$ intersect again at $Y \\neq X$, and line $AH$ and $\\omega$ intersect again at $Z \\neq A$. Show that the circle through $A$, $Y$, $Z$ passes through the midpoint of segment $BC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $M$ be the midpoint of $BC$. We first show that $X$ lies on $AM$. Consider $A'$, the reflection of $A$ across $M$. As $ABA'C$ is a parallelogram, we have that $\\angle BA'C = \\angle BAC = 180^\\circ - \\angle BHC$, which in turn gives us that $A'$ lies on $\\omega$. Now $\\angle HBA' = \\angle HBC + \\angle CBA' = \\angle HBC + \\angle ACB = 90^\\circ$. Hence $HA$ is a diameter of $\\omega$. In particular we must have $\\angle HXA' = 90^\\circ$. Consequently $\\angle AXA' = \\angle AXH + \\angle HXA' = 90^\\circ + 90^\\circ = 180^\\circ$, i.e. $A, X, A'$ are collinear. But $A, M, A'$ collinear by definition, hence $X$ lies on the $A$-median.\n\nNow it suffices to show that $\\angle AYZ = \\angle AMZ$. We note the two following facts:\n* $\\angle AHX = \\angle AYX$, since $\\omega$ and $\\Gamma$ have the same radius and the two angles span the same chord $AX$.\n* $\\omega$ is the reflection of the circumcircle of $ABC$ across $BC$. That gives us that $Z$ is the reflection of $A$ across $D$, the feet of the $A$-altitude to $BC$.\n\nHence we can write: $\\angle AYZ = \\angle AYX + \\angle XYZ = \\angle AHX + (180^\\circ - \\angle XHZ) = 2\\angle AHX = 2\\angle AMD = \\angle AMZ$, which is what we wanted.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25081,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the Euclidean plane, the points $A = (0,0)$ and $B = (1,0)$ and the open half-strip\n$$\nS = \\{(x, y) : 0 < x < 1, y > 0\\}\n$$\nwith width 1 and vertices $A$ and $B$.\nFind all functions $f: S \\to S$ satisfying the following conditions for all $P, Q \\in S$:\n(i) $f(f(P)) = P$\n(ii) If $P, Q, A$ are collinear, then $f(P), f(Q)$ and $B$ are collinear.\n(iii) If $f(P) = P$ and $f(Q) = Q$, then there is a circle containing $A, B, P$ and $Q$.",
    "options": [],
    "answer": "The unique function maps each point to the orthocenter of the triangle formed by that point and the two endpoints of the base segment.",
    "solution": "Fix any $0 < \\alpha < 90^\\circ$ and consider the ray of points $P \\in S$ with $\\angle BAP = \\alpha$. This ray (or the part of it which is in $S$) is mapped by (ii) under $f$ to a ray starting from $B$ with a certain angle $\\beta = \\beta(\\alpha)$, i.e., $\\angle f(P)BA = \\beta(\\alpha)$ for all such $P$. By (i), this second ray is mapped to the first ray.\nNow there is a unique point $P \\in M$ such that $\\angle BAP = \\alpha$ and $\\angle PBA = \\beta$. Since the rays are mapped to each other, $f(P)$ must also lie on both rays and hence $f(P) = P$ is a fixed point. But then (iii) implies that $\\angle APB + \\angle PBA = \\alpha + \\beta(\\alpha)$ is constant.\nConsidering $\\alpha \\to 90^\\circ$ and $\\alpha \\to 0^\\circ$ we see that the only possible value of this constant is $90^\\circ$ and hence $\\beta(\\alpha) = 90^\\circ - \\alpha$.\nSo we always have $\\angle f(P)BA = 90^\\circ - \\angle PBA$ and by (i) also $\\angle PBA = 90^\\circ - \\angle f(P)BA$ and so $f(P)$ is the orthocenter $H(ABP)$ of $ABP$. This function $P \\mapsto H(ABP)$ is indeed a solution.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25082,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$, $B$, $C$, $P$ and $Q$ be five pairwise different points in the plane. Suppose that $A$, $B$ and $C$ are not collinear and that\n$$\n\\frac{AP}{BP} = \\frac{AQ}{BQ} = \\frac{21}{20},\n$$\n$$\n\\frac{BP}{CP} = \\frac{BQ}{CQ} = \\frac{20}{19}.\n$$\nProve that the line $PQ$ contains the circumcentre of the triangle $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We choose a coordinate system such that the circumcentre $O$ of $\\triangle ABC$ lies in the origin and the circumradius of $\\triangle ABC$ is equal to $1$. Let us write $A = (a_1, a_2)$, $B = (b_1, b_2)$ and $C = (c_1, c_2)$. The following statements about a point $X = (x_1, x_2)$ are equivalent:\n$$\n\\frac{AX}{BX} = \\frac{21}{20} \\Leftrightarrow 400 \\left( (x_1 - a_1)^2 + (x_2 - a_2)^2 \\right) = 441 \\left( (x_1 - b_1)^2 + (x_2 - b_2)^2 \\right).\n$$\nWe simplify the equation using the relations $a_1^2 + a_2^2 = 1$, etc. We see that if $X \\in \\{P, Q\\}$, then\n$$\n0 = 41(x_1^2 + x_2^2) - (441b_1 - 400a_1)x_1 - (441b_2 - 400a_2)x_2 + 41,\n$$\nand similarly\n$$\n[39(441b_1 - 400a_1) - 41(400c_1 - 361b_1)]x_1 + [39(441b_2 - 400a_2) - 41(400c_2 - 361b_2)]x_2 = 0,\n$$\nwhich simplifies to\n$$\nx_1(-39a_1 + 80b_1 - 41c_1) + x_2(-39a_2 + 80b_2 - 41c_2) = 0.\n$$\nThe numbers $-39a_1 + 80b_1 - 41c_1$ and $-39a_2 + 80b_2 - 41c_2$ cannot vanish at the same time, because otherwise $A$, $B$ and $C$ would be collinear. Hence the previous equation defines a line through $O$ which contains $P$ and $Q$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25083,
    "subject": "Mathematics (Multi-modal)",
    "question": "For $n = \\prod_i p_i^{e_i}$ ($p_i$ primes, $e_i$ positive integers), define $\\psi(n) = \\prod_i (p_i + 1)^{e_i - 1}$. Show that by repeated application of $\\psi$, we may transform any starting number into a number of the form $2^m$ for $m \\ge 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "As $p + 1$ is even when $p$ is odd, we have that the greatest prime divisor of $p + 1$ is strictly less than $p$ for all primes $p \\neq 2$. Therefore, the greatest prime divisor of the number on the blackboard decreases at every step, unless it is $2$, in which case the number is evidently of the desired form.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25084,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$ and $B$ be sets of positive integers with $|A| \\ge 2$ and $|B| \\ge 2$. Let $S$ be a set consisting of $|A| + |B| - 1$ numbers of the form $ab$ where $a \\in A$ and $b \\in B$. Prove that there exist pairwise distinct $x, y, z \\in S$ such that $x$ is a divisor of $yz$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We use induction on $k = |A| + |B| - 1$.\nFor $k = 3$ we have $|A| = |B| = 2$. Let $A = \\{x, y\\}$, $B = \\{z, t\\}$. Then $S$ consists of three numbers from the set $\\{xz, yz, xt, yt\\}$. Relabelling the elements of $A$ and $B$ if necessary, we can assume without loss of generality that the missing number is $yt$. Then $xz, xt, yz \\in S$ and $xz \\mid xt \\cdot yz$ which concludes the base case of induction.\n\nFor the inductive step, suppose the thesis holds for some $k-1 \\ge 3$. Since $k = |A|+|B|-1 \\ge 4$, we have that $\\max(|A|, |B|) \\ge 3$, WLOG assume $|A| \\ge 3$. Since the set $S$ consists of $k = |A|+|B|-1 > |A|$ elements, by pigeonhole principle there exists a number $x \\in A$ which appears as the first of the two factors of at least two elements of $S$. So, there exist $y, z \\in B$ with $xy, xz \\in S$. If there exists $t \\in A \\setminus \\{x\\}$ such that $ty \\in S$ then we are done because $xy \\mid xz \\cdot ty$. If there exists no such $t$ then apply the inductive hypothesis to the sets $A \\setminus \\{x\\}$, $B$ and $S \\setminus \\{xy\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25085,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a positive integer $n$, we let $s(n)$ denote the sum of the proper divisors of $n$ (excluding $n$ itself). Does there exist a positive integer $a$ for which the equation\n$$\ns(n) = a + n\n$$\nhas infinitely many solutions?",
    "options": [],
    "answer": "Yes; a = 12",
    "solution": "Answer: Yes, for example $a = 12$.\nWhenever $p$ is prime,\n$$\ns(6p) = 1 + 2 + 3 + 6 + p + 2p + 3p = 12 + 6p.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 25086,
    "subject": "Mathematics (Multi-modal)",
    "question": "An integer valued function $f(n)$ is multiplicative (that means $f(ab) = f(a)f(b)$ for all coprime $a$ and $b$) and satisfies the equation\n$$\nf(m + k - 3) = f(m) + f(k) - f(3) \\quad \\text{for all primes } m \\text{ and } k.\n$$\nFind $f(11)$.",
    "options": [],
    "answer": "1 or 11",
    "solution": "Answer: 1 or 11.\nNote that $f(1) = 1$ and the following equalities hold:\n$$\n\\begin{aligned}\nf(1) &= f(2 + 2 - 3) = f(2) + f(2) - f(3), \\\\\nf(7) &= f(5 + 5 - 3) = f(5) + f(5) - f(3), \\\\\nf(10) &= f(2)f(5) = f(11 + 2 - 3) = f(11) + f(2) - f(3), \\\\\nf(11) &= f(7 + 7 - 3) = f(7) + f(7) - f(3), \\\\\nf(15) &= f(3)f(5) = f(11 + 7 - 3) = f(11) + f(7) - f(3).\n\\end{aligned}\n$$\nWhen we solve these equations with respect to $a = f(2)$, $b = f(3)$, $c = f(5)$, $d = f(7)$, $e = f(11)$, we obtain quadratic equation $a^2 - 3a + 2 = 0$. And obtain two solutions: $f(k) = 1$ for $k = 2, 3, 5, 7, 11$ and $f(k) = k$ for $k = 2, 3, 5, 7, 11$. These solutions correspond to the functions $f(x) \\equiv 1$ and $f(x) = x$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25087,
    "subject": "Mathematics (Multi-modal)",
    "question": "Richard and Kaarel are taking turns to choose numbers from the set $\\{1, \\dots, p-1\\}$ where $p > 3$ is a prime. Richard is the first one to choose. Any number which has been chosen by one of the players can be chosen again be neither of the players. Every number chosen by Richard is multiplied with the very next number chosen by Kaarel. Kaarel wins the game if at some moment during the game the sum of all the found products is divisible by $p$. Richard wins if this does not happen i.e. the players run out of numbers before any of the sums is divisible by $p$. Can either of the players guarantee their victory regardless of their opponent's moves and if so, which one?",
    "options": [],
    "answer": "Yes, Kaarel.",
    "solution": "Answer: Yes, Kaarel.\n\nLet us split the numbers in the set to the following pairs: $(1, p-1)$, $(2, p-2)$, $\\dots$, $\\left(\\frac{p-1}{2}, \\frac{p+1}{2}\\right)$. If Richard chooses some number $a$, then let Kaarel choose the other number from the pair i.e. $p-a$. This forces Richard to choose a number from a pair in which both of the numbers have not been chosen yet and hence Kaarel can make his desired move. The residues modulo $p$ of the products are of the form $-a^2$. The residue of the sum of all the products is congruent to $-(1^2 + 2^2 + \\dots + (\\frac{p-1}{2})^2)$. For every natural number $n$, we have $1^2 + 2^2 + \\dots + n^2 = \\frac{n(n+1)(2n+1)}{6}$, therefore $1^2 + 2^2 + \\dots + (\\frac{p-1}{2})^2 = \\frac{(p-1)p(p+1)}{24}$. This must be an integer and as $p$ and $24$ are coprime, $\\frac{(p-1)p(p+1)}{24}$ must be divisible by $p$. Therefore, when the last number is chosen from the set, the sum of the products is divisible by $p$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25088,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $N$ be a positive integer. Determine the number of pairs, $(a, b)$, of positive integers $a$ and $b$ such that the number\n$$\n\\frac{ab}{a+b}\n$$\nis a divisor of $N$.",
    "options": [],
    "answer": "d(N)^2",
    "solution": "In what follows $d(n)$ denotes the number of positive divisors of a positive integer $n$. We show that the number of pairs is $d(N)^2$. Let $m$ be a fixed positive divisor of $N$. We claim that the number of pairs, $(a, b)$, such that $ab/(a+b) = m$ is $d(m)^2$. This becomes clear if we rewrite the equation as\n$$\n(a - m)(b - m) = m^2.\n$$\nSince both $a - m$ and $b - m$ are greater than $-m$, it follows that $a - m$ and $b - m$ are both positive divisors of $m^2$, so the number of pairs $(a, b)$ for a fixed $m \\mid N$ is $d(m^2)$, since we have $d(m^2)$ choices of $a$, and the choice of $a$ determines $b$ uniquely. The number of pairs is therefore\n$$\n\\sum_{m \\mid N} d(m^2)\n$$\nAssume that $N$ has prime factorization $p_1^{s_1} \\cdots p_t^{s_t}$. Since $d(kl) = d(k)d(l)$ when $\\gcd(k, l) = 1$, it follows that\n$$\n\\begin{aligned}\n\\sum_{m \\mid N} d(m^2) &= \\prod_{i=1}^{t} (d(1) + d(p_i^2) + \\cdots + d(p_i^{2s_i})) \\\\\n&= \\prod_{i=1}^{t} (1 + 3 + \\cdots + (2s_i + 1)) \\\\\n&= \\prod_{i=1}^{t} (s_i + 1)^2 \\\\\n&= d(N)^2\n\\end{aligned}\n$$\nas desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25089,
    "subject": "Mathematics (Multi-modal)",
    "question": "(a) Is there a positive integer $a$ such that $((a^2 - 2)^3 + 1)^a - 1$ is a perfect square?\n(b) Is there a positive integer $a$ such that $((a^2 - 2)^3 + 1)^{a+1} - 1$ is a perfect square?",
    "options": [],
    "answer": "(a) No; (b) No",
    "solution": "Answer: (a) No; (b) No.\n\na. If $a$ is even then $((a^2 - 2)^3 + 1)^a$ is clearly a perfect square. An integer that differs from it by 1 cannot be a square of a positive integer.\nIf $a$ is odd then $a^2 \\equiv 1 \\pmod 4$, implying that $a^2 - 2 \\equiv -1 \\pmod 4$ and $(a^2 - 2)^3 \\equiv -1 \\pmod 4$. Then $4 \\mid (a^2 - 2)^3 + 1$ and also $4 \\mid ((a^2 - 2)^3 + 1)^a$. Thus $((a^2 - 2)^3 + 1)^a - 1 \\equiv 3 \\pmod 4$, implying that $((a^2 - 2)^3 + 1)^a - 1$ is not a perfect square.\n\nb. Denote $x = (a^2 - 2)^3 + 1$ and $k = a + 1$. The number $x^k - 1$ can be represented in the form $(x - 1)(x^{k-1} + x^{k-2} + \\dots + x + 1)$. Note that\n$$\n\\begin{aligned}\nx^{k-1} + x^{k-2} + \\dots + x + 1 &= (x^{k-1} + 2x^{k-2} + \\dots + (k-1)x) \\\n& \\quad - (x^{k-2} + \\dots + (k-2)x + (k-1)) + k \\\\\n&= (x - 1)(x^{k-2} + \\dots + (k-2)x + (k-1)) + k,\n\\end{aligned}\n$$\nimplying that\n$$\n\\gcd(x - 1, x^{k-1} + x^{k-2} + \\dots + x + 1) = \\gcd(x - 1, k).\n$$\nAs $k = a+1 \\mid a^2-1 = (a^2-2)+1 \\mid (a^2-2)^3+1 = x$, we have $\\gcd(x-1,k) = \\gcd(k-1,k) = 1$. Thus if $x^k - 1$ were a perfect square, the factor $x-1$ would be a perfect square. But $x-1 = (a^2-2)^3$; this could be a perfect square only if $a^2 - 2$ were a perfect square, which is impossible. Consequently, the given number is not a perfect square.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25090,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a prime number $p$ and a positive integer $n$, denote by $f(p, n)$ the largest integer $k$ such that $p^k \\mid n!$. Let $p$ be a given prime number and let $m$ and $c$ be given positive integers. Prove that there exists infinitely many positive integers $n$ such that $f(p, n) - c$ is divisible by $m$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We denote $v_p(n)$ for the largest power of $p$ dividing $n$.\nWe start with a lemma.\n**Lemma.** For any prime $q$ and modulus $m'$ not divisible by $q$, there exists infinitely many powers $q^n$ of $q$ such that $v_p(q^n!) \\equiv 1 \\pmod{m'}$.\n*Proof.* Define $a_k = v_q(q^k!)$. We then have $a_{k+1} = q a_k + 1$. This sequence is eventually periodic modulo $m'$. It must actually be periodic starting from $0$, as $a_i \\equiv a_{i+T} \\pmod{m'}$ implies $q a_{i-1} \\equiv q a_{i+T-1} \\pmod{m'}$ and therefore $a_{i-1} \\equiv a_{i+T-1} \\pmod{m'}$, since $q \\nmid m'$. Thus, for infinitely many $n$ we have $a_n \\equiv a_1 = 1 \\pmod{m'}$.\n\nWe now turn to solving the problem. Write $m = p^t m'$, where $p \\nmid m'$. The sequence $v_p(p!), v_p(p^2!), v_p(p^3!), \\dots$ is eventually constant modulo $p^t$. Denote this constant by $C$. Since $p \\nmid C$, by the Chinese remainder theorem there exists a positive integer $s$ such that $C s \\equiv c \\pmod{p^t}$ and $s \\equiv c \\pmod{m'}$. Now, choose\n$$\nn = p^{b_1} + p^{b_2} + \\dots + p^{b_s},\n$$\nwhere $b_i$ are distinct positive integers such that $v_p(p_i^b!) \\equiv 1 \\pmod{m'}$ (possible by the lemma) and large enough such that $v_p(p_i^b!) \\equiv C \\pmod{p^t}$. We have\n$$\nv_p(n!) = v_p(p_1^b!) + \\dots + v_p(p_s^b!) \\equiv C s \\equiv c \\pmod{p^t}\n$$\nand\n$$\nv_p(n!) = v_p(p_1^b!) + \\dots + v_p(p_s^b!) \\equiv s \\equiv c \\pmod{m'},\n$$\nwhich proves $v_p(n!) \\equiv c \\pmod{m}$. Since there are infinitely many possible choices $n$, we are done.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25091,
    "subject": "Mathematics (Multi-modal)",
    "question": "Denote by $d(n)$ the number of (positive) divisors of a positive integer $n$. Prove that there are infinitely many positive integers $n$ such that $\\lceil \\sqrt{3} \\cdot d(n) \\rceil$ divides $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note that $\\lfloor \\sqrt{3} \\cdot 8 \\rfloor = 13$. Therefore all numbers with $8$ divisors that are divisible by $13$ satisfy the condition. There are infinitely many of those, for example, all numbers in the form $13p^3$, where $p$ is a prime different from $13$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25092,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that there are infinitely many mutually coprime positive integers $a$, $b$ and $c$ such that\n$$\n\\left\\lfloor \\frac{a^2}{2020} \\right\\rfloor + \\left\\lfloor \\frac{b^2}{2020} \\right\\rfloor = \\left\\lfloor \\frac{c^2}{2020} \\right\\rfloor.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "At first we note that if $b$ is divisible by $2020$ and $a^2 + b^2 = c^2$ then also\n$$\n\\lfloor \\frac{a^2}{2020} \\rfloor + \\lfloor \\frac{b^2}{2020} \\rfloor = \\lfloor \\frac{c^2}{2020} \\rfloor.\n$$\nConsider the equality $(m^2 - n^2)^2 + (2mn)^2 = (m^2 + n^2)^2$ and take $n = 2020$ and $m > 2020$ to be any number coprime to $n$. Then all three numbers $a = m^2 - n^2$, $b = 2mn$ and $c = m^2 + n^2$ are mutually coprime and thus satisfy the given equation.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25093,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ be a real number. Prove that there exist real numbers $b$ and $c$ such that the inequalities\n$$ \\min \\{\\sin(x), \\sin(a+x)\\} \\le b \\sin(x+c) \\le \\max \\{\\sin(x), \\sin(a+x)\\} $$\nhold for all real numbers $x$ and the equalities hold only if $\\sin(x) = \\sin(a+x)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We recall that two real numbers $r$ and $s$ always satisfy the trigonometric identity $\\frac{1}{2}[\\sin(r+s) + \\sin(r-s)] = \\cos(s) \\sin(r)$.\nSubstituting $r = x + \\frac{a}{2}$ and $s = \\frac{a}{2}$, we see that $\\cos(\\frac{a}{2}) \\sin(x + \\frac{a}{2})$ is for all real numbers $x$ the arithmetic mean of the numbers $\\sin(x + a)$ and $\\sin(x)$. Since the arithmetic mean of two numbers lies in the closed interval bounded by the two numbers, $b = \\cos(\\frac{a}{2})$ and $c = \\frac{a}{2}$ is a suitable choice for $b$ and $c$. Furthermore, the arithmetic mean of two numbers differs from the two numbers if they are not equal. $\\square$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 25094,
    "subject": "Mathematics (Multi-modal)",
    "question": "We call an ordered pair $(x, y)$ of real numbers *auroral* if the equations $x = y(3 - y)^2$ and $y = x(3 - x)^2$ hold simultaneously.\nFind all integers $k$ for which there exists an auroral pair $(x, y)$ of real numbers with $x + y = k$.",
    "options": [],
    "answer": "{0, 3, 4, 5, 8}",
    "solution": "The numbers we are looking for are $k \\in \\{0, 3, 4, 5, 8\\}$.\nA pair $(x, x) \\in \\mathbb{R}^2$ is auroral if and only if $x = x(3 - x)^2$, and it is easy to see that this cubic equation has the solution set $\\{0, 2, 4\\}$. These pairs give us 0, 4 and 8 as possible values for $k$.\n\nNext we investigate the case $k = 3$. Here we can simplify the given equations as $x = yx^2$ and $y = xy^2$. The number $x$ cannot be zero in this case, since otherwise $y$ and $k$ would also be zero. We can conclude that $xy = 1$. The equations $x + y = 3$ and $xy = 1$, according to Vieta's Theorem, imply that $x$ and $y$ are the solutions of the equation $\\lambda^2 - 3\\lambda + 1 = 0$. Hence\n$$\n(x, y) = \\left( \\frac{3 + \\sqrt{5}}{2}, \\frac{3 - \\sqrt{5}}{2} \\right) \\quad \\text{or} \\quad (x, y) = \\left( \\frac{3 - \\sqrt{5}}{2}, \\frac{3 + \\sqrt{5}}{2} \\right)\n$$\nand it is easy to verify that both pairs are actually auroral. These pairs give us 3 as a possible value for $k$.\n\nA simple calculation shows that if a pair $(x, y) \\in \\mathbb{R}^2$ is auroral, then the pair $(4 - x, 4 - y)$ is also auroral. From the pairs we have just found, we can therefore construct auroral pairs\n$$\n(x, y) = \\left( \\frac{5 - \\sqrt{5}}{2}, \\frac{5 + \\sqrt{5}}{2} \\right) \\quad \\text{and} \\quad (x, y) = \\left( \\frac{5 + \\sqrt{5}}{2}, \\frac{5 - \\sqrt{5}}{2} \\right),\n$$\nwhich give us 5 as a possible value for $k$.\n\nNext we investigate the case $k = 4$, where we can simplify the first given equation as $x = (4 - x)(x - 1)^2$. We already know $x = 2$ must be a solution of this cubic equation, since the pair $(2, 2)$ is auroral. By polynomial division we calculate the quotient of $x - (4 - x)(x - 1)^2$ by $x - 2$ as $x^2 - 4x + 2$. The roots of this quadratic polynomial yield pairs\n$$\n(x, y) = (2 + \\sqrt{2}, 2 - \\sqrt{2}) \\quad \\text{and} \\quad (x, y) = (2 - \\sqrt{2}, 2 + \\sqrt{2})\n$$\nand it is easy to verify that both pairs are actually auroral.\n\nIf $(x, y) \\in \\mathbb{R}^2$ is auroral, then $x$ satisfies the polynomial equation\n$$\nx - x(3 - x)^2 [3 - x(3 - x)^2]^2 = 0\n$$\nwhich is of degree 9. Since we have already found nine values for $x$ that must satisfy the equation, there are no more auroral pairs other than those already found. $\\square$\nLet $f(x) = x(3 - x)^2$. It is easy to check that if $x < 0$ then $f(x) < x$. In particular $f(f(x)) < f(x) < x$ in this case, so that the pair $(x, f(x))$ cannot be auroral. Similarly, $f(x) > x$ if $x > 4$, so the pair $(x, f(x))$ cannot be auroral in this case either.\n\nSuppose that $(x, y) \\in \\mathbb{R}^2$ is auroral. According to the previous remark $x \\in [0, 4]$, and similarly, $y \\in [0, 4]$. Hence we may write $x = 2 + 2r$ and $y = 2 + 2s$ with $r, s \\in [-1, 1]$. After substitution and simplification, the equation $x = y(3 - y)^2$ transforms into the equation $r = 4s^3 - 3s$. Recall the trigonometric identities for threefold angles. If $s = \\cos(\\alpha)$ for some $\\alpha \\in \\mathbb{R}$, then $r = 4 \\cos^3(\\alpha) - 3 \\cos(\\alpha) = \\cos(3\\alpha)$. In the same way $s = 4r^3 - 3r = \\cos(9\\alpha)$.\n\nWe can deduce that $9\\alpha = 2\\pi m + \\alpha$ or $9\\alpha = 2\\pi l - \\alpha$ for some integers $m$ and $l$. In the former case we have $8\\alpha = 2\\pi m$, so that $m \\in \\{0, 1, 2, 3, 4\\}$, and the corresponding possible auroral pairs can be found in Figure 3. In the former case we have $10\\alpha = 2\\pi l$, so that $l \\in \\{0, 1, 2, 3, 4, 5\\}$, where $l = 0$ and $l = 5$ result in angles that we have already considered in the first case. We consider the other options in Figure 4 taking into account the well-known identities $\\cos(\\pi/5) = (1 + \\sqrt{5})/4$ and $\\cos(3\\pi/5) = (1 - \\sqrt{5})/4$. $\\square$\n\n| m | $8\\alpha$ | $\\alpha$ | $r$ | $s$ | $x$ | $y$ | $x + y$ |\n|---|----------|---------|-----|-----|-----|-----|---------|\n| 0 | 0        | 0       | 1   | 1   | 4   | 4   | 8       |\n| 1 | $2\\pi$   | $\\pi/4$ | $\\sqrt{2}/2$ | $-\\sqrt{2}/2$ | $2 + \\sqrt{2}$ | $2 - \\sqrt{2}$ | 4 |\n| 2 | $4\\pi$   | $\\pi/2$ | 0   | 0   | 2   | 2   | 4       |\n| 3 | $6\\pi$   | $3\\pi/4$| $-\\sqrt{2}/2$ | $\\sqrt{2}/2$ | $2 - \\sqrt{2}$ | $2 + \\sqrt{2}$ | 4 |\n| 4 | $8\\pi$   | $\\pi$   | -1  | -1  | 0   | 0   | 0       |\n\nFigure 3: Auroral pairs and their sums\n\n| l | $10\\alpha$ | $\\alpha$ | $r$ | $s$ | $x$ | $y$ | $x + y$ |\n|---|------------|----------|-----|-----|-----|-----|---------|\n| 1 | $2\\pi$     | $\\pi/5$  | $(1 + \\sqrt{5})/4$ | $(1 - \\sqrt{5})/4$ | $(5 + \\sqrt{5})/2$ | $(5 - \\sqrt{5})/2$ | 5 |\n| 2 | $4\\pi$     | $2\\pi/5$ | $(-1 + \\sqrt{5})/4$ | $(-1 - \\sqrt{5})/4$ | $(3 + \\sqrt{5})/2$ | $(3 - \\sqrt{5})/2$ | 3 |\n| 3 | $6\\pi$     | $3\\pi/5$ | $(1 - \\sqrt{5})/4$ | $(1 + \\sqrt{5})/4$ | $(5 - \\sqrt{5})/2$ | $(5 + \\sqrt{5})/2$ | 5 |\n| 4 | $8\\pi$     | $4\\pi/5$ | $(-1 - \\sqrt{5})/4$ | $(-1 + \\sqrt{5})/4$ | $(3 - \\sqrt{5})/2$ | $(3 + \\sqrt{5})/2$ | 3 |\n\nFigure 4: Auroral pairs and their sums",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 25095,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ that satisfy the equation\n$$ f(x)^n f(x+y) = f(x)^{n+1} + x^n f(y) $$\nfor all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "All solutions are: the zero function; the identity function; and, additionally when the parameter is even, the negation function.",
    "solution": "The functions we are looking for are $f : \\mathbb{R} \\to \\mathbb{R}$, $f(x) = 0$ and $f : \\mathbb{R} \\to \\mathbb{R}$, $f(x) = x$. For $n$ even $f : \\mathbb{R} \\to \\mathbb{R}$, $f(x) = -x$ is also a solution.\n\nThroughout the solution, $P(x_0, y_0)$ will denote the substitution of $x_0$ and $y_0$ for $x$ and $y$, respectively, in the given equation.\n\n$P(x, 0)$ for $x \\neq 0$ gives\n$$\nf(x)^{n+1} = f(x)^{n+1} + x^n f(0)\n$$\nand therefore\n$$\nf(0) = \\frac{f(x)^{n+1} - f(x)^{n+1}}{x^n} = 0.\n$$\n\n$P(x, -x)$ for $x \\neq 0$ gives\n$$\n0 = f(x)^n f(0) = f(x)^{n+1} + x^n f(-x),\n$$\nand therefore\n$$\nf(-x) = -\\frac{f(x)^{n+1}}{x^n}.\n$$\n\n$$\nf(x)(x^{n^2+2n} - f(x)^{n^2+2n}) = 0.\n$$\nIf there exists an $a \\neq 0$ for which $f(a) = 0$, then $P(a, y)$ yields\n$$\n0 = a^n f(y),\n$$\nwhich means that $f(y) = 0$ for all $y \\in \\mathbb{R}$. This is a solution to the equation for all $n$.\n\nIf instead $f(x) \\neq 0$ for all $x \\neq 0$, then we have\n$$\nx^{n^2+2n} = f(x)^{n^2+2n}.\n$$\nIf $n$ is odd, then so is $n(n + 2) = (n^2 + 2n)$, meaning $f(x) = x$ for all $x \\in \\mathbb{R}$. This is a solution to the equation.\n\nIf $n$ is even, then so is $n(n + 2) = (n^2 + 2n)$, meaning $f(x) = \\pm x$ for all $x \\in \\mathbb{R}$. Both $f(x) = x$ and $f(x) = -x$ are solutions to the equation. In all other cases there must exist $x, y \\neq 0$ such that $f(x) = x$ and $f(y) = -y$. Then $P(x, y)$ yields\n$$\nx^n f(x + y) = x^{n+1} - x^n y,\n$$\nwhich after dividing by $x^n \\neq 0$ yields\n$$\nf(x + y) = x - y.\n$$\nSince $(f(x))^2 = x^2$ for all $x \\in \\mathbb{R}$, we have $(x + y)^2 = (x - y)^2$. That is $4xy = 0$ which is impossible as $x, y \\neq 0$.\n\nThere are therefore no more solutions to the equation. $\\square$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 25096,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b, c$ be real numbers, representing the side lengths of a triangle. Prove that\n$$ 4(a + b)(a + c)(b + c) \\ge (a + b + c)^3. $$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Given are three real numbers $a, b, c$ such that $0 < a, b, c < \\frac{a+b+c}{2}$ (the inequality $a, b, c < \\frac{a+b+c}{2}$ is equivalent to the triangle inequality). Introduce the variables\n$$\nx = \\frac{-a + b + c}{2(a + b + c)}, \\quad y = \\frac{a - b + c}{2(a + b + c)} \\quad \\text{and} \\quad z = \\frac{a + b - c}{2(a + b + c)}\n$$\nThe conditions imposed on $a, b$ and $c$ are equivalent to\n$$\n0 < x, y, z, \\quad \\text{and} \\quad x + y + z = \\frac{1}{2}\n$$\nClearly\n$$\n\\frac{b+c}{a+b+c} = \\frac{1}{2} + x, \\quad \\frac{c+a}{a+b+c} = \\frac{1}{2} + y \\quad \\text{and} \\quad \\frac{a+b}{a+b+c} = \\frac{1}{2} + z\n$$\nTherefore we have:\n$$\n\\begin{align*} \\frac{(a+b)(a+c)(b+c)}{(a+b+c)^3} &= \\frac{b+c}{a+b+c} \\cdot \\frac{c+a}{a+b+c} \\cdot \\frac{a+b}{a+b+c} \\\\ &= \\left(\\frac{1}{2} + x\\right) \\cdot \\left(\\frac{1}{2} + y\\right) \\cdot \\left(\\frac{1}{2} + z\\right) \\\\ &= \\frac{1}{8} + \\frac{x+y+z}{4} + \\frac{yz+zx+xy}{2} + xyz \\\\ &= \\frac{1}{8} + \\frac{1/2}{4} + \\frac{yz+zx+xy}{2} + xyz \\\\ &> \\frac{1}{4} \\end{align*}\n$$\nWithout loss of generality it may be assumed that $a \\le b \\le c$. Observe that\n$$\n((a-t)+b)(b+(c+t)) = (a+b)(b+c) - t(c-a) - t^2 < (a+b)(b+c).\n$$\nfor $t > 0$.\nThe initial numbers $a, b, c$ satisfy the triangle inequality $a + b > c$. Choose $t$ such that\n$$\n(a-t)+b=c+t,\n$$\nit is clear that $0 < t < a$. If we replace in the inequality under consideration $a$ and $c$ with $a-t, c+t$, the sums $a+c$ and $a+b+c$ remain unchanged, and the product $(a+b)(b+c)$ decreases. Hence it is sufficient to prove the given inequality for positive $a, b, c$ under assumption $a+b=c$. In this case\n$$\n4 \\cdot \\frac{a+b}{a+b+c} \\cdot \\frac{a+c}{2c} \\cdot \\frac{b+c}{2c} = 4 \\cdot \\frac{1}{2} \\cdot \\frac{a+c}{2c} \\cdot \\frac{b+c}{2c} = \\frac{ab+c(a+b)+c^2}{2c^2} = \\frac{ab+2c^2}{2c^2} > 1.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25097,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b, c$ be real numbers, representing the side lengths of a triangle. Prove that\n$$\n\\sqrt[3]{(a^2 + bc)(b^2 + ca)(c^2 + ab)} > \\frac{a^2 + b^2 + c^2}{2}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We claim that\n$$\na^2 + bc > \\frac{a^2 + b^2 + c^2}{2},\n$$\nwhich will finish the proof. Note that the claimed inequality is equivalent to\n$$\n\\begin{aligned}\na^2 + bc > \\frac{a^2 + b^2 + c^2}{2} &\\iff 2a^2 + 2bc > a^2 + b^2 + c^2 \\\\\n&\\iff a^2 > (b-c)^2 \\iff a > |b-c|,\n\\end{aligned}\n$$\nwhich holds due to the assumption of $a, b, c$ being side lengths of a triangle.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25098,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given is a triangle with real side lengths $a$, $b$, $c$ and real area $A$. Prove that\n$$ 3a^2 + 3b^2 - c^2 \\ge 4\\sqrt{3}A. $$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\gamma$ be the angle between the sides $a$ and $b$. We have\n$$\n\\begin{aligned}\na^2 + b^2 &\\ge 2ab \\\\\n&\\ge 2ab \\sin(\\gamma - 30^\\circ) \\\\\n&= 2ab [\\sin(\\gamma) \\cos(30^\\circ) - \\cos(\\gamma) \\sin(30^\\circ)] \\\\\n&= ab \\sin(\\gamma)\\sqrt{3} - ab \\cos(\\gamma) \\\\\n&= 2\\sqrt{3}A - \\frac{1}{2}(a^2 + b^2 - c^2),\n\\end{aligned}\n$$\nwhich implies the desired inequality.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25099,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all triples $(a, b, c)$ of real numbers such that\n$$ \\cos(ax) + \\cos(bx) = 2 \\cos(cx) $$\nholds for all $x \\in \\mathbb{R}$.",
    "options": [],
    "answer": "All real triples with |a| = |b| = |c|; equivalently, all triples of the forms (t, t, t), (−t, t, t), (t, −t, t), (t, t, −t) for real t.",
    "solution": "The triples we are looking for have the forms $(t, t, t)$, $(-t, t, t)$, $(t, -t, t)$, $(t, t, -t)$ where $t \\in \\mathbb{R}$.\nIf $c = 0$ then $\\cos(ax) + \\cos(bx) = 2$ for all $x \\in \\mathbb{R}$. Since $\\cos(ax) \\le 1$ and $\\cos(bx) \\le 1$, we must have $\\cos(ax) = 1$ and $\\cos(bx) = 1$ for any $x$. Thus $a = b = 0$, which clearly works.\n\nNow, suppose that $c \\neq 0$. Plugging in $x = \\frac{2\\pi}{c}$ we find $\\cos(2\\pi \\cdot \\frac{a}{c}) + \\cos(2\\pi \\cdot \\frac{b}{c}) = 2$. Therefore $\\cos(2\\pi \\cdot \\frac{a}{c}) = 1$ and $\\cos(2\\pi \\cdot \\frac{b}{c}) = 1$. This means that $2\\pi \\frac{a}{c} = 2k\\pi$ and $2\\pi \\frac{b}{c} = 2l\\pi$ for some integers $k$ and $l$, i.e. $a = ck$ and $b = cl$.\n\nNote that $k \\neq 0$. Indeed, otherwise $LHS \\ge 0$ for all $x$, but $RHS$ has negative values. For similar reason $l \\neq 0$. We shall prove that $|k| = |l| = 1$. Suppose otherwise. Assume without loss of generality that $|k| \\ge |l|$. Plug in $x = \\frac{\\pi}{a}$. We obtain $-1 + \\cos(\\frac{l}{k}\\pi) = 2\\cos(\\frac{1}{k}\\pi)$. Note that $LHS < -1 + 1 = 0$. Moreover, if $|k| \\ge 2$ then $RHS$ is nonnegative, yielding a contradiction. Therefore we have $|k| = 1$. This leads to $|l| = 1$. This means: $|a| = |b| = |c|$. Clearly, such triples work. $\\square$\nApplying the operator $\\frac{d}{dx}$ two times and four times we obtain $-a^2 \\cos(ax) - b^2 \\cos(bx) = -2c^2 \\cos(cx)$ and $a^4 \\cos(ax) + b^4 \\cos(bx) = 2c^4 \\cos(cx)$, respectively. Plugging in $x = 0$ we obtain $a^2 + b^2 = 2c^2$ and $a^4 + b^4 = 2c^4$. Therefore\n$$\n2(a^4 + b^4) = 4c^4 = (2c^2)^2 = (a^2 + b^2)^2 = a^4 + 2a^2b^2 + b^4,\n$$\nwhich yields $0 = \\text{LHS} - \\text{RHS} = a^4 + b^4 - 2a^2b^2 = (a^2 - b^2)^2$. Therefore $a^2 = b^2$. From $2c^2 = a^2 + b^2 = 2a^2$ we obtain $c^2 = a^2$. Therefore $|a| = |b| = |c|$ and we check directly that all such triples work. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25100,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all sequences $(a_1, a_2, ...)$ of positive integers satisfying\n$$ a_{n+1}^2 = 1 + (n + 2021)a_n $$\nfor all $n \\ge 1$.",
    "options": [],
    "answer": "a_n = n + 2019",
    "solution": "Clearly for $C = 1$ we have the solution $(a_n)_{n=1}^{\\infty} = (n + 2019)_{n=1}^{\\infty}$. Let's prove that this is the only value for $C$ that works.\nAssume $(a_n)_{n=1}^{\\infty}$ is a solution and let $(b_n)_{n=1}^{\\infty} = (a_n - n)_{n=1}^{\\infty}$. We claim that for $n > |C| + 2021^2$:\n(i) If $b_n < 2019$, then $b_n < b_{n+1} < 2019$.\n(ii) If $b_n > 2019$, then $b_n > b_{n+1} > 2019$.\nIt is clear that these two claims implies that $b_n = 2019$ for all large $n$ and hence that $C = 1$.\nLet us prove the claims:\n(i) First of all, $b_n \\le 2018$ implies that\n$$\n\\begin{aligned}\na_{n+1}^2 &\\le C + (n + 2021)(n + 2018) \\\\\n&= (n + 2020)^2 - n + C + 2018 \\cdot 2021 - 2020^2 \\\\\n&< (n + 2020)^2\n\\end{aligned}\n$$\nand hence $a_{n+1} < n + 2020$ so that indeed $b_{n+1} < 2019$.\n\nMoreover, we have\n$$\n\\begin{align*}\na_{n+1}^2 &= C + (n + 2021)(n + b_n) \\\\\n&= (n + 1 + b_n)^2 + (2019 - b_n)n + 2021b_n + C - (b_n + 1)^2 \\\\\n&\\ge (n + 1 + b_n)^2 + n + C - 2019^2 \\\\\n&> (n + 1 + b_n)^2\n\\end{align*}\n$$\nand hence $a_{n+1} > n + 1 + b_n$ so that indeed $b_{n+1} > b_n$.\n(ii) First of all, $b_n \\ge 2020$ implies that\n$$\na_{n+1}^2 \\geq C + (n + 2021)(n + 2020) = (n + 2020)^2 + n + C + 2021 > (n + 2020)^2\n$$\nand hence $a_{n+1} > n + 2020$ so that indeed $b_{n+1} > 2019$.\nMoreover, we have\n$$\n\\begin{align*}\na_{n+1}^2 &= C + (n + 2021)(n + b_n) \\\\\n&= (n + 1 + b_n)^2 + (2019 - b_n)n + 2021b_n + C - (b_n + 1)^2 \\\\\n&\\le (n + 1 + b_n)^2 - n + C \\\\\n&< (n + 1 + b_n)^2\n\\end{align*}\n$$\nand hence $a_{n+1} < n + 1 + b_n$ so that indeed $b_{n+1} < b_n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25101,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all integers $C$ for which there exists a sequence $(a_1, a_2, ...)$ of positive integers satisfying\n$$\na_{n+1}^2 = C + (n + 2021)a_n\n$$\nfor all $n \\ge 1$.",
    "options": [],
    "answer": "1",
    "solution": "Clearly for $C = 1$ we have the solution $(a_n)_{n=1}^\\infty = (n + 2019)_{n=1}^\\infty$. Let's prove that this is the only value for $C$ that works.\nAssume $(a_n)_{n=1}^\\infty$ is a solution and let $(b_n)_{n=1}^\\infty = (a_n - n)_{n=1}^\\infty$. We claim that for $n > |C| + 2021^2$:\n(i) If $b_n < 2019$, then $b_n < b_{n+1} < 2019$.\n(ii) If $b_n > 2019$, then $b_n > b_{n+1} > 2019$.\nIt is clear that these two claims implies that $b_n = 2019$ for all large $n$ and hence that $C = 1$.\nLet us prove the claims:\n(i) First of all, $b_n \\le 2018$ implies that\n$$\n\\begin{align*}\na_{n+1}^2 &\\le C + (n + 2021)(n + 2018) \\\\\n&= (n + 2020)^2 - n + C + 2018 \\cdot 2021 - 2020^2 \\\\\n&< (n + 2020)^2\n\\end{align*}\n$$\nand hence $a_{n+1} < n + 2020$ so that indeed $b_{n+1} < 2019$.\n\n$$\n\\begin{align*}\na_{n+1}^2 &= C + (n + 2021)(n + b_n) \\\\\n&= (n + 1 + b_n)^2 + (2019 - b_n)n + 2021b_n + C - (b_n + 1)^2 \\\\\n&\\ge (n + 1 + b_n)^2 + n + C - 2019^2 \\\\\n&> (n + 1 + b_n)^2\n\\end{align*}\n$$\nand hence $a_{n+1} > n + 1 + b_n$ so that indeed $b_{n+1} > b_n$.\n(ii) First of all, $b_n \\ge 2020$ implies that\n$$\na_{n+1}^2 \\geq C + (n + 2021)(n + 2020) = (n + 2020)^2 + n + C + 2021 > (n + 2020)^2\n$$\nand hence $a_{n+1} > n + 2020$ so that indeed $b_{n+1} > 2019$.\nMoreover, we have\n$$\n\\begin{align*}\na_{n+1}^2 &= C + (n + 2021)(n + b_n) \\\\\n&= (n + 1 + b_n)^2 + (2019 - b_n)n + 2021b_n + C - (b_n + 1)^2 \\\\\n&\\le (n + 1 + b_n)^2 - n + C \\\\\n&< (n + 1 + b_n)^2\n\\end{align*}\n$$\nand hence $a_{n+1} < n + 1 + b_n$ so that indeed $b_{n+1} < b_n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25102,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\Gamma$ be a circle in the plane and $S$ be a point on $\\Gamma$. Two brothers, Mario and Luigi, drive around the circle $\\Gamma$ with their go-karts. They both start at $S$ at the same time, they both drive for exactly 6 minutes at constant speed counterclockwise around the track. During these 6 minutes, Luigi makes exactly one lap around $\\Gamma$ while Mario, who is three times as fast, accomplishes three laps.\nWhile Mario and Luigi drive their go-karts, Princess Daisy positions herself such that she is always exactly in the middle of the two brothers. When she reaches a point she has already visited, she marks it with a banana.\nHow many points in the plane, apart from $S$, are marked with a banana after the race?",
    "options": [],
    "answer": "3",
    "solution": "Without loss of generality, we assume that $\\Gamma$ is the unit circle and $S = (1, 0)$. Three points are marked with bananas:\n(i) After 45 seconds, Luigi has passed through an arc with a subtended angle of $45^\\circ$ and is at the point $(\\sqrt{2}/2, \\sqrt{2}/2)$, whereas Mario has passed through an arc with a subtended angle of $135^\\circ$ and is at the point $(-\\sqrt{2}/2, \\sqrt{2}/2)$. Therefore Daisy is at the point $(0, \\sqrt{2}/2)$ after 45 seconds. After 135 seconds, Mario and Luigi's positions are exactly the other way round, so the princess is again at the point $(0, \\sqrt{2}/2)$ and puts a banana there.\n(ii) Similarly, after 225 seconds and after 315 seconds, Princess Daisy is at the point $(0, -\\sqrt{2}/2)$ and puts a banana there.\n(iii) After 90 seconds, Luigi is at $(0, 1)$ and Mario at $(0, -1)$, so that Daisy is at the origin of the plane. After 270 seconds, Mario and Luigi's positions are exactly the other way round, hence Princess Daisy drops a banana at the point $(0, 0)$.\nWe claim that no other point in the plane, apart from these three points and $S$, is marked with a banana. Let $t_1$ and $t_2$ be two different times when Daisy is at the same place. For $n \\in \\{1, 2\\}$ we write Luigi's position at time $t_n$ as a complex number $z_n = \\exp(ix_n)$ with $x_n \\in ]0, 2\\pi[$. At this time, Mario is located at $z_i^3$ and Daisy at $(z_i^3 + z_j)/2$.\n\nAccording to our assumption we have $(z_1^3 + z_1)/2 = (z_2^3 + z_2)/2$ or, equivalently, $(z_1 - z_2)(z_1^2 + z_1z_2 + z_2^2 + 1) = 0$. We have $z_1 \\neq z_2$, so that we must have $z_1^2 + z_1z_2 + z_2^2 = -1$.\nWe proceed with an observation of the structure of $\\Gamma$ as a set of complex numbers. Suppose that $z \\in \\Gamma \\setminus \\{S\\}$. Then $z+1+z^{-1} \\in \\Gamma$ if and only if $z \\in \\{i, -i, -i, i\\}$. For a proof of the observation note that $z+1+z^{-1} = z+1+\\bar{z}$ is a real number for every $z \\in \\mathbb{C}$ with norm $|z| = 1$. So it lies on the unit circle if and only if it is equal to 1, in which case the real part of $z$ is equal to 0, or it is equal to $-1$, in which case the real part of $z$ is equal to $-1$. We apply the observation to the number $z = z_1/z_2$, which satisfies the premise since $z+1+\\bar{z} = -\\bar{z_1} \\cdot \\bar{z_2} \\in \\Gamma$. Therefore, one of the following cases must occur.\n(i) We have $z = \\pm i$, that is, $z_1 = \\pm iz_2$. Without loss of generality we may assume $z_2 = iz_1$. It follows that $-1 = z_1^2 + z_1z_2 + z_2^2 = iz_1^2$, so that $z_1 = \\exp(i\\pi/4)$ or $z_1 = \\exp(5i\\pi/4)$. In the former case $(z_1, z_2) = (\\exp(i\\pi/4), \\exp(3i\\pi/4))$, which matches case (1) above. In the latter case $(z_1, z_2) = (\\exp(5i\\pi/4), \\exp(7i\\pi/4))$, which matches case (2) above.\n(ii) We have $z = -1$, that is, $z_2 = -z_1$. It follows that $-1 = z_1^2 + z_1z_2 + z_2^2 = z_1^2$, so that $z_1 = i$ or $z_1 = -i$. This matches case (3) above.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25103,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ be a polynomial with real coefficients and $n \\ge 1$ be an integer. Prove that there exists a non-zero polynomial $q$ such that the coefficients of $p \\cdot q$ vanish for each power that is not a multiple of $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $x$ be the variable. We want to find a real non-zero polynomial $q$ such that $p(x) \\cdot q(x) = s(x^n)$ for some real polynomial $s$. If $p$ is the zero polynomial then $p \\cdot q = 0$ for every polynomial $q$. It can therefore be assumed that $p \\neq 0$.\n\nLet $y = x^n$. Then $y^m$ is a real polynomial for each $m \\in \\mathbb{N}$. For each $m \\in \\mathbb{N}$ let $r_m(x)$ be the remainder of the polynomial division of $y^m = (x^n)^m$ by $p(x)$. Then each $r_m$ is of degree less than the $\\deg(p)$, the degree of $p$. Consider the polynomials $r_0, r_1, \\dots, r_{\\deg(p)}$. We want to find coefficients $s_0, s_1, \\dots, s_{\\deg(p)}$ such that $s_0 \\cdot r_0(x) + s_1 \\cdot r_1(x) + \\dots + s_{\\deg(p)} \\cdot r_{\\deg(p)}(x) = 0$. By considering the coefficients this is equivalent to a system with $\\deg(p)$ linear equations and $\\deg(p) + 1$ unknowns, $(s_0, s_1, \\dots, s_n$ are the unknowns). As there are more unknowns than equations it follows that there exists a solution $(s_0, s_1, \\dots, s_n) \\neq (0, 0, \\dots, 0)$.\n\nTake a solution $(s_0, s_1, \\dots, s_n)$ to the system of linear equations and let $s(x) = s_0 + s_1 \\cdot x + \\dots + s_{\\deg(p)} x^p$. This is non-zero polynomial. As $r_m(x)$ is the remainder of the polynomial division of $y^m$ by $p(x)$ it follows that $p(x)$ divides $y^m - r_m(x)$ for all $m \\in \\mathbb{N}$. Hence $p(x)$ divides\n$$\n\\begin{aligned}\n& s_0 \\cdot (y^0 - r_0(x)) + s_1 \\cdot (y^1 - r_1(x)) + \\dots + s_{\\deg(p)} \\cdot (y^{\\deg(p)} - r_{\\deg(p)}(x)) \\\\\n&= s(y) - (s_0 \\cdot r_0(x) + s_1 \\cdot r_1(x) + \\dots + s_{\\deg(p)} \\cdot r_{\\deg(p)}(x)) \\\\\n&= s(y)\n\\end{aligned}\n$$\nIt follows that $q(x) = s(y)/p(x)$ is a polynomial. As $s$ is non-zero it follows that $q$ is non-zero as well. We have therefore found a non-zero polynomial $q$ such that $p(x) \\cdot q(x) = s(x^n)$ as desired. $\\square$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 25104,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an equilateral triangle, find all positive integers $n$ such that it can be partitioned into $n$ equilateral triangles (not necessarily of the same size).",
    "options": [],
    "answer": "All positive integers except 2, 3, and 5",
    "solution": "An equilateral triangle can be partitioned into one piece, that is a partition with the triangle itself as the only piece.\nWe also note that the cases $n = 6$ and $n = 8$ are possible as figure 5 shows\n\n![](attached_image_1.png)\nFigure 5: The cases $n = 6$ and $n = 8$\n\nAssume that the original triangle can be partitioned into $n$ equilateral triangles. Partition the equilateral triangle into four parts as depicted in figure 6.\nThen partition one of the resulting equilateral triangles into $n$ parts. We have therefore partitioned the equilateral triangle into $n + 3$ parts.\nAs the equilateral triangle can be partitioned into 1 part, it follows that the triangle can be partitioned into $n$ equilateral triangles if $n = 3k + 1$ for some $k$. Similarly, the constructions for $n = 6$ and $n = 8$ show that the triangle can be partitioned into $n = 6 + 3k$ parts and $n = 8 + 3k$ parts for any $k$.\n\n![](attached_image_2.png)\nFigure 6: Partition of an equilateral triangle into four equilateral triangles\n\nWe have therefore shown that the equilateral triangle can be partitioned into $n$ equilateral triangles for all $n$ except $n = 2, 3, 5$. We proceed to show that the equilateral triangle can not be partitioned into $n$ equilateral triangles if $n \\in \\{2, 3, 5\\}$.\nConsider the cases:\n(i) Assume that $n = 2$. By the pigeonhole principle, one triangle shares two vertices with the original triangle, and will therefore be the entire triangle. This is absurd, so no partition for $n = 2$ is possible.\n(ii) Assume that $n = 3$. As in the case above, no triangle shares two vertices with the original triangle. So each triangle shares exactly one vertex with the original triangle. In each of the 3 smaller triangles, let $a_i$, $i = 1, 2, 3$ be the side opposing the vertex common with the original triangle. The side $a_1$ lies inside the triangle, so it must be a side of two smaller triangles. However, the only internal segments of the other triangles are $a_2$ and $a_3$, but $a_1$ can only coincide with either $a_2$ or $a_3$. We conclude that no partition is possible for $n = 3$.\n(iii) Assume that $n = 5$. As above, no triangle shares two vertices with the original triangle. Consider the three triangles that share a vertex with the original triangle, and the sides $a_i$ as above. As this is a partition, we know the $a_i$'s intersect either on the sides of the large triangle or outside it. We get four cases depending on how they intersect.\n* If all three pairs intersect outside of the original triangle, the remaining shape is a convex hexagon.\n* If two pairs intersect outside of the original triangle, and one pair on a side of the triangle, we get a convex pentagon.\n* If two pairs intersect on the sides of the triangle, and one pair on the side of the triangle, we get an isosceles trapezoid.\n* If all pairs intersect on the sides of the triangle, we get an equilateral triangle\nWe are to split the remaining convex shape, a hexagon, a pentagon, an isosceles trapezoid, or a triangle, into two triangles. The only convex shape split up into two equilateral triangles is the rhombus with one angle of 60°. We conclude that no partition is possible for $n = 5$.\nWe have shown that the equilateral triangle can be partitioned into $n$ parts for all positive integers except $n \\in \\{2, 3, 5\\}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25105,
    "subject": "Mathematics (Multi-modal)",
    "question": "John has a string of paper where $n$ real numbers $a_i \\in [0, 1]$, for all $i \\in \\{1, \\dots, n\\}$, are written in a row. Show that for any given $k < n$, he can cut the string of paper into $k$ pieces, between adjacent numbers, in such a way that sum of the numbers on each piece does not differ from another by more than 1.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote the sums on each piece by\n$$\n\\begin{align*}\nS_1 &= a_1 + a_2 + \\dots + a_{m_1}, \\\\\nS_2 &= a_{m_1+1} + a_{m_1+2} + \\dots + a_{m_2}, \\\\\n\\vdots \\\\\nS_k &= a_{m_{k-1}+1} + \\dots + a_{m_k}.\n\\end{align*}\n$$\nBy abuse of notation $S_i$ will both denote the set of numbers enclosed by cuts and its sum, the meaning of which must be determined by the context.\n\nWe will start the following algorithm. During this algorithm we will move some elements to the neighbouring piece and construct new sequence of pieces $S^* = (S_1^*, S_2^*, \\dots, S_k^*)$. Empty pieces may appear.\n\n(i) Find $p \\le k$ such that $S_p$ is the piece with the maximum sum of elements.\n\n(ii) If $S_p \\le \\min(S_1, \\dots, S_k) + 1$ we are done.\n\n(iii) If $S_p > \\min(S_1, \\dots, S_k) + 1$, let $S_q$ be the pieces with minimum sum of elements nearest to $S_p$ (ties broken arbitrarily) and let $S_h$ be the next pieces to $S_q$ between $S_p$ and $S_q$ (it is non empty by the choice of $S_q$). Then either $p < q$ and then $h = q - 1$ and we define $S^*$ by moving the last element from $S_h = S_{q-1}$ to $S_q$, or $q < p$, and then $h = q + 1$ and $S^*$ is obtained by moving the first element of $S_h = S_{q+1}$ to $S_q$. If $p = h$ then set $S = S^*$ and go to step (1). If $p \\ne h$ then set $S = S^*$ and proceed to step (2).\n\nNote that in step (3) each number $S_i^*$ is at most $S_p$ and no new pieces with sum $S_p$ is created. Indeed, $S_h^* < S_h \\le S_p$, and for some $j$ $S_q^* = S_q + a_j < S_p$ since $a_j \\in [0, 1]$ and $S_p > \\min(S_1, \\dots, S_k) + 1$. It is clear also that $\\max(S_1, \\dots, S_k)$ does not increase during the algorithm.\n\nNote also that in step (3) the pieces $S_h$ may become empty. Then, in the next iteration of the algorithm, $q = h$ will be chosen since $\\min(S_1, \\dots, S_k) = S_h = 0$ and in step (3) $S_h^*$ will become non empty (but one of its neighbours may become empty, etc.).\n\nClaim. Step (3) is repeated at most $kn$ times with $S_p$ being the same maximal pieces in $S^*$ and in $S$.\n\nProof. Let $s_i$ be the number of elements in $i$-th pieces. Then the number\n$$\n\\sum_{i=1}^{k} |i - p|s_i\n$$\ntakes positive integral values and is always less than $kn$. It is clear that this number decreases during the algorithm.\n\nThus after at most $kn$ iteration of (3), the algorithm decreases the value of $S_p$ and so goes to (1). Consequently it decreases either the number of pieces with maximal sums or $\\max(S_1, \\dots, S_k)$. As there are only finitely many ways to split the sum onto pieces, the algorithm eventually terminates at (2). $\\square$",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 25106,
    "subject": "Mathematics (Multi-modal)",
    "question": "For which values of $n$ does there exist a circumscribed $n$-gon with side lengths $1, 2, \\dots, n$ (in any order)?",
    "options": [],
    "answer": "n = 4k or n = 4k + 1 for positive integers k",
    "solution": "It exists if $n = 4k$ or $n = 4k + 1$ where $k$ is a positive integer.\n\nLet us consider $n$-gon $P_1P_2 \\dots P_n$. Tangent points of the inscribed circle divide each of its sides in two segments. Lengths of these segments that have a common vertex $P_i$ are equal. Denote the length of tangent segments that originate at point $P_i$ by $A_i$. It means that side lengths of the $n$-gon can be expressed as $P_iP_{i+1} = A_i + A_{i+1}$ for all $i = 1, 2, \\dots, n$ where we consider points cyclically ($P_{n+1} = P_1$ and $A_{n+1} = A_1$).\n\nWe can show that the converse is true as well. That is, if we can find $n$ positive real numbers $A_i$, $i = 1, 2, \\dots, n$ such that the sequence $(A_1 + A_2, A_2 + A_3, \\dots, A_n + A_1)$ is a permutation of $(1, 2, \\dots, n)$ then there is a circumscribed polygon $P_1P_2 \\dots P_n$ with side lengths $1, 2, \\dots, n$.\n\nTo show this we start with a circle of arbitrary radius $R$ and construct points $P_1, P_2, \\dots, P_n$ outside this circle so that the length of the tangent segments from $P_i$ to the circle are of length $A_i$ and the \"right\" tangent segment from $P_i$ touches the circle at the same point as the \"left\" tangent segment from $P_{i-1}$.\n\nNow we almost have the $n$-gon except that possibly the \"right\" tangent point of $P_1$ does not match the \"left\" touching point of $P_n$. This can be easily fixed by adjusting the radius $R$ of the circle, using continuity.\n\nNow we solve the problem by considering 4 cases:\n\ni. First let's consider the case when $n = 4k$. In this case such circumscribed $n$-gon exists. The $4k$ segments $A_i$ can be of lengths\n$$\n\\begin{aligned}\nA_1 &= \\frac{1}{2},\\ A_2 = \\frac{1}{2},\\ A_3 = \\frac{3}{2},\\ A_4 = \\frac{3}{2}, \\dots, A_{2k-1} = \\frac{2k-1}{2},\\ A_{2k} = \\frac{2k-1}{2}, \\\\\nA_{2k+1} &= \\frac{2k+1}{2},\\ A_{2k+2} = \\frac{6k-1}{2},\\ A_{2k+3} = \\frac{2k-1}{2},\\ A_{2k+4} = \\frac{6k-3}{2}, \\dots, \\\\\nA_{4k-1} &= \\frac{3}{2},\\ A_{4k} = \\frac{4k+1}{2}.\n\\end{aligned}\n$$\nOne can see that the values of the sums of the consecutive elements $A_1+A_2, A_2+A_3, \\dots, A_{4k-1}+A_{4k}, A_{4k}+A_{4k+1}$ are exactly $1, 2, \\dots, 2k, 4k, 4k-1, \\dots, 2k+1$, respectively.\n\nii. In the case $n = 4k + 1$ the construction is similar, we can choose $4k + 1$ segments of length\n$$\n\\begin{aligned}\nA_1 &= \\frac{1}{2}, & A_2 &= \\frac{1}{2}, & A_3 &= \\frac{5}{2}, & A_4 &= \\frac{5}{2}, \\dots, \\\\\nA_{2k+1} &= \\frac{4k+1}{2}, & A_{2k+2} &= \\frac{4k+1}{2}, & A_{2k+3} &= \\frac{4k-1}{2}, \\\\\nA_{2k+4} &= \\frac{4k-3}{2}, & A_{2k+5} &= \\frac{4k-5}{2}, & \\dots, & A_{4k+1} &= \\frac{3}{2}.\n\\end{aligned}\n$$\nIn this case the values of the sums of consecutive elements $A_1 + A_2, A_2 + A_3, \\dots, A_{4k-1} + A_{4k}, A_{4k} + A_{4k+1}$ are $1, 3, 5, \\dots, 4k+1, 4k, 4k-2, \\dots, 2$, respectively.\n\niii. In case when $n = 4k + 2$ such a polygon does not exist. To prove this we note that in case if the number of the sides of the circumscribed polygon is even then the sum of the odd numbered sides is equal to the sum of the even numbered sides. It is evident as two segments of equal length that originate from the same vertex contribute to different sums. But the total sum of the side lengths is an odd number what means that it is impossible to split the sides into two parts with equal sum of lengths.\n\niv. In case $n = 4k + 3$ such a polygon also does not exist. In this case we can express $A_1$ as\n$$\n\\begin{aligned}\nA_1 &= (A_1 + A_2 + \\dots + A_n) - (A_2 + A_3) - (A_4 + A_5) - \\dots - (A_{4k+2} + A_{4k+3}) = \\\\\n&= \\frac{P_1 P_2 + P_2 P_3 + \\dots + P_{4k+3} P_1}{2} - P_2 P_3 - P_4 P_5 - \\dots - P_{4k+2} P_{4k+3}\n\\end{aligned}\n$$\nAs the sum of the length of the sides is an even number then we conclude that $A_1$ is a positive integer. The same is true for all $A_2, A_3, \\dots$ as well. But now we have a contradiction as the side of length 1 cannot be split in two parts, each of which has positive integer length.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25107,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $t$ be a non-zero real number and $n$ be a positive integer. Let $a_1, a_2, \\dots, a_{2n-1}$ be real numbers (not necessarily distinct). Prove that there exist indices $i_1 < i_2 < \\dots < i_n$ such that $a_{i_k} - a_{i_{l}} \\neq t$ for any $1 \\le k, l \\le n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $G = (V, E)$ be a graph with vertex set $V = \\{1, 2, \\dots, 2n-1\\}$ and edge set $E = \\{\\{i, j\\} : |a_i - a_j| = t\\}$. Note that $G$ has no odd cycles. Indeed, if $j_1, \\dots, j_{2k+1}$ is a cycle, then for all $l = 1, 3, 5, \\dots, 2k-1$ the number $a_{jl}$ differs from $a_{j,l}$ by $2t$ or $0$. Hence $a_{j_1}$ differs from $a_{j_{2k+1}}$ by an even multiple of $t$. Therefore there is no edge between $j_1$ and $j_{2k+1}$ contradicting the assumption that $j_1, \\dots, j_{2k+1}$ is a cycle.\n\nSince $G$ has no odd cycles, it is bipartite. Therefore $V$ can be split into two disjoint sets $V_1, V_2$ such that there is no edge between any two vertices of $V_1$ and there are no edges between any two vertices in $V_2$. Since $V$ has $2n-1$ elements, one of the sets $V_1, V_2$ has at least $n$ elements. Without loss of generality assume that $V_1$ has at least $n$ elements. Then for $k = 1, 2, \\dots, n$ simply define $i_k$ to be the $k$-th least element of $V_1$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25108,
    "subject": "Mathematics (Multi-modal)",
    "question": "Anne, Ellie and Milo play a game on a hexagonal board tiled with hexagons, $n > 2$ hexagons on each side. The game begins with a token on a tile in one of corner of the board. Ellie and Milo are on the same team, playing against Anne, and they win if the token lands on center of the board. Anne, Ellie and Milo take turns moving the token, Anne begins, then Ellie and then Milo.\nThe turns proceed as follows:\n* Anne has to move the token to an adjacent hexagon, in any direction.\n* Ellie has to move the token by two hexagons in any of the 6 possible directions.\n* Milo has a choice of passing the turn, or moving the token by three hexagons in any of the 6 possible directions.\nFind all $n > 2$ for which Ellie and Milo have a winning strategy.\n\n![](attached_image_1.png)\nFigure 1: A board with 5 hexagons on each side.",
    "options": [],
    "answer": "n ≡ 1 (mod 3)",
    "solution": "We colour the board in three colours in such a way that no neighbouring tiles are of the same colour. We can give each hexagon a coordinate using $\\vec{e}_1 = (1, 0)$ and $\\vec{e}_2 = (\\cos(120^\\circ), \\sin(120^\\circ)) = (\\frac{-1}{2}, \\frac{\\sqrt{3}}{2})$ as basis. Let the center square be the origin. Then each hexagon has center at $a \\cdot \\vec{e}_1 + b \\cdot \\vec{e}_2$, $(a, b) \\in \\mathbb{Z}^2$. The tuple $(a, b)$ is the coordinate for a given hexagon; its neighbours are $(a + 1, b)$, $(a + 1, b + 1)$, $(a, b + 1)$, $(a - 1, b)$, $(a - 1, b - 1)$ and $(a, b - 1)$.\n\nWe colour the hexagon with coordinates $(a, b)$ with colour number $(a+b) \\pmod{3}$. It is clear that neighbouring hexagons do not share a colour. (In fact this is the only three colouring of a hexagonal tiling). See figure 8.\n\nWe see that if $n \\equiv 1 \\pmod{3}$, the token begins in a space in the same colour as the center hexagon, let that colour be grey. By regarding a few cases, we see that whatever Anne does, Ellie and Milo can end their turns by getting the token to a prescribed grey hexagon of the closest grey hexagons. Therefore they can get the token to the center.\n\nIf $n \\not\\equiv 1 \\pmod{3}$, the token does not begin on the same grey colour as the center. Say the stone begins on a white tile, and say the third colour is black. Anne can always move the stone to a grey hexagon that is not on the same horizontal/diagonal line as the center hexagon. Then Anne moves the stone to a white or black hexagon. After Milo moves the token is still again on a white/black hexagon. Anne can continue this indefinitely, with the token never reaching the center.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25109,
    "subject": "Mathematics (Multi-modal)",
    "question": "A directed graph $G$ has 2021 vertices located on a circle, 101 directed edges go from each vertex of $G$ to the next 101 vertices in the counter-clockwise direction. We wish to colour the edges in such a way that for any two vertices $u$ and $v$ one can choose a directed path from $u$ to $v$ in which no two edges have the same colour. What is the minimum number of colours needed?",
    "options": [],
    "answer": "21",
    "solution": "Answer: $\\lceil \\frac{2021}{101} \\rceil = 21$ colours.\n\nLabel the vertices along the cycle: $v_0, v_1, \\dots, v_{2020}$. It is clear that the distance from $v_0$ to any of $v_1, \\dots, v_{101}$ equals 1; the distance from $v_0$ to any of $v_{102}, \\dots, v_{202}$ equals 2, etc. Therefore, the distance from $v_0$ to $v_{2020}$ is the longest one, it equals 20.\n\nE.g., for each $i$, $1 \\le i \\le 20$, let\n$$\nV_i = \\{v_{101(i-1)+1}, v_{101(i-1)+2}, \\dots, v_{101i}\\}\n$$\nand we colour the edges that go from vertices of $V_i$ in $i$-th colour. The edges that starts in $v_0$ we colour in 21-st colour. It is evident that for any two vertices $v_i$ and $v_j$ we can choose a directed path from $v_i$ to $v_j$ that intersects each $V_\\ell$ in at most one vertex (with the only one exception: we allow $v_i$ and $v_j$ to belong to the same set $V_\\ell$). The edges of this path have pairwise different colours.\n\nAssume that we can colour edges in 20 colours. Then consider a path that in each step jumps from the current vertex $v_i$ to $v_{i+101}$. Here and henceforth we are considering addition modulo 2021. Since $\\gcd(101, 2021) = 1$, this path is a Hamiltonian cycle $C$.\n\nFor each $v_i$ the sub-path from $v_i$ to $v_{i+2020}$ in $C$ consist of 20 edges and this is the shortest path from $v_i$ to $v_{i+2020}$. Since we have a colouring in 20 colours only, the edges of this path have 20 different colours. Thus the edge between $v_i$ and $v_{(i+101)}$ and the edge from $v_{(i+2020)}$ to $v_{(i+2121)}$ are of the same colour. Since $\\gcd(2020, 2021) = 1$, this means that all edges in $C$ have the same colour, a contradiction.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 25110,
    "subject": "Mathematics (Multi-modal)",
    "question": "During one test a service dog smells a pile of coins and barks if there is a fake coin in it. If a dog is sick, whether it barks or not does not depend on the presence of fake coin, it happens randomly. Suppose $k \\le 2^s$ and that we have $2^k$ coins, exactly one of which is fake. Assume that we have an excess number of service dogs, one of them is sick but we do not know which one. Device a strategy to find the fake coin using at most $k+s+2$ tests.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Number the coins by $k$-digit binary numbers from $00\\ldots0$ to $11\\ldots1$. Let $A_i$ be the set of coins which have $0$ in $i$-th position of the binary number. The first $k$ tests we perform with the help of $k$ different dogs. In the $i$-th test we determine whether the set $A_i$ contains the fake coin. Without loss of generality we may assume that the dogs determined that all the digits in the number of the fake coin are $0$'s. Due to the possible presence of the sick dog in these tests, it means in fact that the binary number of the fake coin contains at most one $1$.\n\nIn the next test we let a new dog determine whether the coin $00\\ldots0$ is genuine. If the new dog barks then the coin is really fake, for otherwise two dogs had given us a false answer. If the new dog does not bark we find a dog we have not used before to test the suspected coin.\n\n(i) If the last two dogs disagree one of them must be sick and hence the first $k$ dogs must be healthy. In this case the coin $00\\ldots0$ is the fake one.\n\n(ii) If the last two dogs agree (by not barking) it follows that both of them are healthy. The reason is that if one of the last two dogs was sick and did not bark, it would mean that the first $k$ dogs were healthy, implying that the coin $00\\ldots0$ is fake, but then the other of the last two dogs is healthy and did not bark at the fake coin, a contradiction. Therefore one of the first $k$ dogs gave a wrong verdict. In this case we have $k$ possible candidates for the fake coin. We can find the fake coin using the last dog and $s$ tests using binary search.\n\nIt follows that no more than $k + s + 2$ tests are needed. $\\square$",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 25111,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let's call the intersection of two segments *almost perfect* if for each of them the length of the segment is at least $2021$ times the distance between its midpoint and the intersection point.\nProve that there exists a closed broken line that intersects each of its segments at least once and for which all its intersections are almost perfect.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider two equilateral triangles with common centre and parallel sides. The closed broken line $A_1B_2C_1A_2B_1C_2$ has three intersections, because of symmetry we will consider only one.\nAssume that $A_1B_2$ intersects $A_2B_1$ in point $X$. Triangles $A_1B_1X$ and $A_2B_2X$ are similar therefore $\\frac{A_1X}{XB_2} = \\frac{A_1B_1}{A_2B_2}$. It is enough to choose initial triangles of close enough size $1 < \\frac{A_1B_1}{A_2B_2} < \\frac{2022}{2021}$ to make the intersection $X$ almost perfect. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25112,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $2021$ points on a plane, no three of which are collinear. For every $5$ points there exists at least $4$ among them which are concyclic. Is it necessarily true that at least $2020$ of the points are concyclic?",
    "options": [],
    "answer": "Yes",
    "solution": "Answer: Yes.\n\nLet us first prove a lemma that if $4$ points $A$, $B$, $C$, $D$ all lie on circle $\\Gamma$ and some two points $X$, $Y$ do not lie on $\\Gamma$, then these $6$ points are pairs of intersections of three circles, circle $\\Gamma$ and two other circles. Indeed, according to the problem statement there are $4$ points among $A$, $B$, $C$, $X$, $Y$ which are concyclic. These $4$ points must include points $X$ and $Y$ because if one of them is not, then the other one must lie on $\\Gamma$. Without loss of generality, take $A$, $B$, $X$, $Y$ to lie on the same circle. Similarly for points $A$, $C$, $D$, $X$, $Y$ there must be $4$ points which are concyclic. Analogously, they must include points $X$ and $Y$. Point $A$ cannot be one of them because two circles cannot have more than two common points. Therefore, points $C$, $D$, $X$, $Y$ are concyclic which proves the lemma.\n\nLet us first solve the problem for the case for which there exist $5$ points which lie on one circle $\\Gamma$. Label these points $A$, $B$, $C$, $D$, $E$. Let us assume that there exists two points which do not lie on $\\Gamma$, label them $X$ and $Y$. According to the previously proven lemma, points $A$, $B$, $C$, $D$, $X$, $Y$ must be the pairwise intersections of $3$ circles. Without loss of generality, let the intersections of $\\Gamma$ with one of the other circles be $A$ and $B$ and with the other circle $C$ and $D$. Similarly, $A$, $B$, $C$, $E$, $X$, $Y$ must be the pairwise intersections of three circles one of which is $\\Gamma$. This is not possible as none of the points $A$, $B$, $C$ lies on the circumcircle of triangle $EXY$. This contradiction shows that at most $1$ point can lie outside circle $\\Gamma$, i.e. at least $2020$ points lie on circle $\\Gamma$.\n\nIt remains to look at the case for which no $5$ points lie on the same circle. Let $A$, $B$, $C$, $D$, $E$ be arbitrary $5$ points. Without loss of generality, let $A$, $B$, $C$, $D$ be concyclic and $E$ a point not on this circle. According to the lemma, for every other point $F$ and points $A$, $B$, $C$, $D$, $E$, the $6$ points are the intersections of circle $\\Gamma$ and some two other circles. But in total, there are $3$ such points because one of the two circles must go through $E$ and some $2$ points out of $A$, $B$, $C$, $D$, while the other circle must go through point $E$ and the other two points out of $A$, $B$, $C$, $D$. There are only three partitions of $A$, $B$, $C$, $D$ into two sets. There is a contradiction, as there are $2021 > 5 + 3$ points in total. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25113,
    "subject": "Mathematics (Multi-modal)",
    "question": "A convex $n$-gon is triangulated with triangulation $P$ of $n-3$ non-intersecting diagonals. An *independent set* of vertices is a set of vertices of $P$ where no two vertices are connected by a side or by a diagonal. Let $i(P)$ be the number of ways to choose an independent set of vertices with respect to the triangulation $P$. Prove that the minimum $i(P)$ is achieved for a zig-zag triangulation.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "1) Let $z_n$ be the number of ways to choose an independent set of vertices for a zig-zag triangulation of an $n$-gon. It is known that each triangulation has at least two vertices which are not the endpoints of any diagonal. Let $A$ be one of these vertices for the zig-zag triangulation (see figure above).\nIf the independent set does not contain $A$, then by removing the vertex $A$ we obtain a zig-zag triangulated $(n-1)$-gon. If the independent set contains $A$, then it does not contain its two neighbours, and by removing $A$ and its two neighbours we obtain a zig-zag triangulated $(n-3)$-gon.\nTherefore $z_n$ satisfies the recurrence relation\n$$\nz_n = z_{n-1} + z_{n-3}.\n$$\n\n![](attached_image_1.png)\nFigure 9: A zig-zag triangulation of a 10-gon.\n\n2) We will consider a triangulation of $n$-gon as a graph: the vertices of $n$-gon are vertices of the graph, the sides and the diagonals are edges.\n\n3) Now we will prove the problem statement $i(P) \\ge z_n$ by induction on $n$. Base $n \\le 6$ is trivial.\nLet the statement hold for all convex triangulations on less than $n$ vertices. Consider a triangulated $n$-gon $P$ as a graph. The triangulation has at least two vertices of degree 2. Let $A$ be one of these vertices, $u, v$ be its neighbours.\nAs in part 1), consider independent sets that do not contain $A$. The number of these sets equals the number of independent sets in graph $P \\setminus A$. By induction hypothesis it is at most $z_{n-1}$.\nNow consider independent sets that contain $A$ (and do not contain $u$ and $v$). The number of these sets equals the number of independent sets in the graph $H = P \\setminus \\{A, u, v\\}$ which has $n-3$ vertices. But generally speaking, graph $H$ is not a graph of a triangulation. Let us add some edges to $H$ in order to obtain graph $H^*$ of some triangulation. This operation of adding new diagonals creates new neighbours in the graph, and thus decreases the number of independent sets. Then $i(H) \\ge i(H^*) \\ge z_{n-3}$.\nThus,\n$$\ni(P) \\ge i(P \\setminus A) + i(H) \\ge z_{n-1} + z_{n-3} = z_n.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25114,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $I$ be the incenter of a triangle $ABC$. Let the incircle of $ABC$ be tangent to $CA$ and $AB$ at $E$ and $F$, respectively. Lines $BI$ and $CI$ intersect line $EF$ at $Y$ and $Z$, respectively. Denote by $M$, $N$ midpoints of segments $BC$ and $YZ$, respectively. Prove that $MN$ is parallel to $AI$.\n\n![](attached_image_1.png)\nFigure 22",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Refer to figure 22. Let us start with proving a known simple lemma stating that $\\angle BYC = 90^\\circ$. To that end, as $\\angle IEC = 90^\\circ$, it is enough to prove that $IEYC$ is cyclic. Indeed:\n$$\n\\begin{align*} \n\\angle CEY &= \\angle AEF \\\\ \n&= 90^\\circ - \\frac{1}{2} \\angle BAC \\\\ \n&= 90^\\circ - \\frac{1}{2} (180^\\circ - \\angle ABC - \\angle ACB) \\\\ \n&= \\frac{1}{2} \\angle ABC + \\frac{1}{2} \\angle ACB \\\\ \n&= \\angle IBC + \\angle ICB \\\\ \n&= \\angle YIC. \n\\end{align*}\n$$\nThis shows that $YIEC$ is cyclic. Since $AC$ is tangent to the incircle at $E$,\n$$\n\\angle BIC = \\angle CYI = \\angle CEI = 90^\\circ\n$$\nSimilarly, we may prove that $\\angle BZC = 90^\\circ$.\nAll this implies that $BYZC$ is cyclic with $M$ being its centre. Hence $N$ is the midpoint of its chord $YZ$ and therefore $MN \\perp EF$. However, $AI \\perp EF$, so $MN \\parallel AI$ as desired. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25115,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be an acute triangle. Denote by $E$ and $F$ the feet of the altitudes from $B$ and $C$, respectively. Let $H$ be the intersection of $BE$ and $CF$. Let $D$ be on the same side of line $BC$ as $A$ and satisfy:\n$$\n\\angle DBC = \\angle DCB = \\angle BAC.\n$$\nLet $N$ be the midpoint of $EF$. Prove that points $H$, $D$ and $N$ are collinear.\n\n![](attached_image_1.png)\nFigure 23",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Refer to figure 23. Notice that as triangles $HEF$ and $HCB$ are similar with different orientations and $C$, $H$, $F$ and $B$, $H$, $E$ collinear, then the statement is equivalent to $DH$ being a symmedian from $H$ in $BHC$.\n\nAs $\\angle CBD = \\angle BAC = 180^\\circ - \\angle BHC$, the line $DB$ is tangent to $o(BHC)$. Similarly, $DC$ is tangent to $o(BHC)$. The line $HD$ is the line connecting $H$ with the intersection of tangents to $o(BHC)$ at $B$ and $C$, and so $HD$ is indeed the symmedian in $BHC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25116,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $O$ be the circumcenter of triangle $ABC$. Let $D$, $E$ and $F$ be the reflection of $O$ over lines $BC$, $CA$ and $AB$, respectively. Show that lines $AD$, $BE$ and $CF$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Refer to figure 12.\n\nAs $O$ is the circumcenter of $\\triangle ABC$ it follows that line segments $OB$ and $OC$ are concurrent. As $D$ is the reflection of $O$ across $BC$ it follows that segments $OB$ and $DB$ are concurrent and segments $OC$ and $DC$ are concurrent. It follows that $OBDC$ is a rhombus and therefore also a parallelogram. Similarly $OCEA$ is also a parallelogram.\n\nAs lines $BD$ and $OC$ are parallel and lines $OC$ and $AE$ are parallel it follows from the transitivity of parallelism that lines $BD$ and $AE$ are parallel.\n\nAs lines $OB$ and $CD$ are parallel and lines $OA$ and $CE$ are parallel it follows from Desargues's theorem (the Euclidean plane is a translation plane) that lines $AB$ and $ED$ are parallel. We have hence shown that $ABDE$ is a parallelogram. Similarly it can be shown that $AFDC$ is also a parallelogram.\n\nLet $P$ be the intersection point of lines $AD$ and $BE$. As $ABDE$ is a parallelogram it follows that $P$ is the midpoint of segment $AD$. As $P$ is the midpoint of segment $AD$ and $AFDC$ is a parallelogram it follows that line $DF$ passes through $P$. We have therefore demonstrated that $P$ is a common point on $AD$, $BE$ and $CF$, that is lines $AD$, $BE$ and $CF$ are concurrent. $\\square$\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25117,
    "subject": "Mathematics (Multi-modal)",
    "question": "Point $P$ lies inside triangle $ABC$, $M$ is the midpoint of side $BC$ and $P'$ is symmetric to $P$ with respect to $M$. $K$ and $H$ are projections of $P$ on $AB$ and $AC$ respectively and $KM = HM$. Prove that $\\angle PAB = \\angle CAP'$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote by $A'$, $K'$, $H'$ the points symmetrical to $A$, $K$, $H$ with respect to $M$, as in figure 13.\nIt is clear that $KH'H'K$ is a rectangle and the quadrilateral $AKPH$ is inscribed, denote its circumcircle by $\\omega$. Then\n$$\n\\angle CHK' = 90^\\circ - \\angle AHK = \\angle PHK = \\angle PAK.\n$$\nLet $L$ be the second intersection point of line $HK'$ and $\\omega$. The above equality of angles means that $\\angle LHA = \\angle PAK$, therefore $AL = KP$, and then $AKPL$ is a rectangle. Hence $AL \\parallel KP \\parallel K'P'$ and $AP'K'L$ is parallelogram. Thus, $HK' \\parallel AP'$ and then $\\angle CHK' = \\angle CAP'$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25118,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let points $A$ and $B$ lie on circle $\\omega$ with center $O$. Assume that $O$ does not lie on line $AB$. Let point $C$ lie on segment $AB$ and denote by $M$ and $N$ the midpoints of segments $AC$ and $CB$, respectively. The circumcircle of $AON$ intersects $\\omega$ at $A$ and $K$ and the circumcircle of $BOM$ intersects $\\omega$ at $B$ and $L$. Moreover, circumcircles of $AON$ and $BOM$ intersect each other at $O$ and $X$. Prove that the quadrilateral $CKXL$ is cyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Solution.** Let $K'$ and $H'$ be the projections of $P'$ onto $AB$ and $AC$ respectively, as in figure 14. Now, $HPP'H'$ is a right angled trapezium, and $M$ is the midpoint of $PP'$. If $M'$ is the midpoint of $HH'$, then $M'M||HP$, so $M'M \\perp HH'$. Therefore, $MH = MH'$. Similarly, $MK = MK'$, and so as $MK = MH, K', K, H, H'$ lie on a circle with centre $M$.\nAs $KHH'K'$ is cyclic, we have $\\angle H'K'A = \\angle KHA$. By considering right angles, we see that $AH'P'K'$ and $AHPK$ are cyclic.\nWe get\n$$\n\\begin{align*} \n\\angle PAB &= \\angle PAK \\\\ \n&= \\angle PHK \\\\ \n&= 90^\\circ - \\angle KHA \\\\ \n&= 90^\\circ - \\angle H'K'A \\\\ \n&= \\angle P'K'H' \\\\ \n&= \\angle P'AH' \\\\ \n&= \\angle P'AC \n\\end{align*}\n$$\nas required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25119,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\omega_1$ and $\\omega_3$ be two circles, touching externally in a common point $P$. Let further $\\omega_2$ and $\\omega_4$ be two circles touching externally in $P$. Suppose that for $i \\in \\{1, 2, 3, 4\\}$ $\\omega_i$ intersect $\\omega_{(i \\pmod 4)+1}$ again in $A_i$. Let $\\ell_1$ be the common tangent of $\\omega_1$ and $\\omega_3$ and $\\ell_2$ be the common tangent of $\\omega_2$ and $\\omega_4$. Show that $A_1, A_2, A_3$ and $A_4$ are con-cyclic if and only if $\\ell_1$ and $\\ell_2$ are orthogonal.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Solution.** In figure 15, we have\n$$\n\\angle CNK = \\angle ANK = \\angle AOK = 2\\angle ABK = 2\\angle NBK.\n$$\nHence triangle $BNK$ is isosceles, so $NK = NB = NC$, and therefore $\\angle BKC$ is right.\nIf we let $KC$ intersect $\\omega$ at $K'$, we see that $K', O$ and $B$ are collinear, as $\\angle BKK'$ is right.\nLet us note that $C$ lies on $OX$. Indeed,\n$$\n\\mathrm{Pow}(C, AONKX) = AC \\cdot CN = \\frac{1}{2} AC \\cdot CB = CM \\cdot CB = \\mathrm{Pow}(C, BOMLX).\n$$\nNow,\n$$\n\\angle LXC = \\angle LXO = \\angle LBO = \\angle LBK' = \\angle LKK' = \\angle LKC\n$$\nShowing that $XLCK$ is cyclic, as required.\n\n![](attached_image_1.png)",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 25120,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $I$ be the incenter of a triangle $ABC$. Let $F$ and $G$ be the feet of the perpendiculars drawn from $A$ to the lines $BI$ and $CI$, respectively. Rays $AF$ and $AG$ intersect the circumcircles of the triangles $CFI$ and $BGI$ second times at points $K$ and $L$, respectively. Prove that line $AI$ bisects the segment $KL$.\n\n![](attached_image_1.png)\nFigure 16",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $\\angle IFK = 90^\\circ$, then $IK$ is the diameter of the circumcircle of $CFI$, hence also $\\angle ICK = 90^\\circ$. Similarly is $IL$ the diameter of the circumcircle of $BGI$ and $\\angle IBL = 90^\\circ$. Therefore are the lines $CK$ and $GL$ parallel, also $BL$ and $FK$ are parallel.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25121,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $AD$ be the $A$-altitude of an acute-angled triangle $ABC$. The internal bisector of angle $DAC$ intersects $BC$ at $K$. Let $L$ be the projection of $K$ onto $AC$. Let $M$ be the intersection point of $BL$ and $AD$. Let $P$ be the intersection point of $MC$ and $DL$. Prove that $PK \\perp AB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Solution.** Since $\\angle IFK = 90^\\circ$, then $IK$ is the diameter of the circumcircle of $CFI$, hence also $\\angle ICK = 90^\\circ$. Similarly is $IL$ the diameter of the circumcircle of $BGI$ and $\\angle IBL = 90^\\circ$. Therefore are the lines $CK$ and $GL$ parallel, also $BL$ and $FK$ are parallel.\nLet the lines $CK$ and $BL$ intersect at $D$, as seen in figure 17. From the above we get that $DKAL$ is a parallelogram. Note that $D$ is the excenter with respect to the vertex $A$ of the triangle $ABC$, since the lines $BL$ and $CK$ are perpendicular to the corresponding internal angle bisectors. The excenter lies on the internal angle bisector $AI$, hence $AI$ bisects the diagonal $KL$.\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25122,
    "subject": "Mathematics (Multi-modal)",
    "question": "Assume that $ABCD$ is a cyclic quadrilateral with circumcircle $\\Omega$. Assume lines $AB$ and $CD$ intersect at point $P$ and lines $AD$ and $BC$ intersect at $Q$. Let $\\Gamma$ be the circumcircle of triangle $APQ$. Then $\\Omega$ and $\\Gamma$ intersect in two points, $A$ is one of them and $R$ is the other. Assume $C \\neq R$. Prove that line $CR$ passes through $M$ where $M$ is the midpoint of linesegment $PQ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Solution 1.** Let $X$ be a point on $BC$ such that $LX \\perp AB$, as seen in figure 18. It is enough to prove that\n$$\n\\frac{DP}{PL} = \\frac{DK}{KX}\n$$\nbecause then $PK \\parallel LX$ and $LX \\perp AB$.\nApplying Menelaos for triangle *BDL* and transversal *MPC* we get\n$$\n\\frac{DP}{PL} \\cdot \\frac{LM}{MB} \\cdot \\frac{BC}{CD} = 1,\n$$\nand Menelaus for triangle *BLC* and transversal *AMD* gives\n$$\n\\frac{BM}{ML} \\cdot \\frac{LA}{AC} \\cdot \\frac{CD}{DB} = 1.\n$$\nMultiplying these two equalities yields\n$$\n\\frac{DP \\cdot BC \\cdot AL}{PL \\cdot BD \\cdot AC} = 1.\n$$\nNote, however, that $AL = AD = AC \\sin \\gamma$, $BD = AB \\cos \\beta$, and, by the sine rule, $\\frac{AB}{BC} = \\frac{\\sin \\gamma}{\\sin \\alpha}$, where $\\alpha = \\angle BAC$, $\\beta = \\angle CBA$ and $\\gamma = \\angle ACB$. Therefore\n$$\n\\frac{DP}{PL} = \\frac{BD \\cdot AC}{BC \\cdot AL} = \\frac{AB \\cos \\beta \\cdot AC}{BC \\cdot AC \\sin \\gamma} = \\frac{\\sin \\gamma \\cos \\beta}{\\sin \\alpha \\sin \\gamma} = \\frac{\\cos \\beta}{\\sin \\alpha}.\n$$\nOn the other hand, since $DK = KL$, $\\angle KLX = \\pi - \\alpha$, and $\\angle LXX = \\frac{\\pi}{2} - \\beta$, we have by the sine rule\n![](attached_image_1.png)\n\nTherefore\n$$\n\\frac{DP}{PL} = \\frac{\\cos \\beta}{\\sin \\alpha} = \\frac{DK}{KX}\n$$\n**Solution 2.** Let $\\omega$ be the circle with center $K$ an radius $KD$, as in figure 19. Then $\\omega$ is tangent to $AD$ and $AL$. Let $BC$ intersect $\\omega$ at $D$ and $Q$. Let $BM$ intersect $\\omega$ at $L$ and $R$. Let $QP$ intersect $BL$ at $S$.\nCross-ratio chasing gives, through the projections $BL \\to D$-pencil $\\to \\omega \\to L$-pencil $\\to BC \\to P$-pencil $\\to BL$,\n$$\n\\begin{aligned}\n(L, R; M, B) &= (DL, DR; DM, DB) = (L, R; D, Q) = (LC, LB; LD, LQ) \\\\\n&= (C, B; D, Q) = (PC, PB; PD, PQ) = (M, B; L, S) = (L, S; M, B),\n\\end{aligned}\n$$\ntherefore $R = S$.\nIt is clear now that $P$ lies on the polar lines of both $A$ and $B$ with respect to $\\omega$, therefore $AB$ is the polar line of $P$. This implies that $PK \\perp AB$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25123,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with circumcircle $\\Gamma$ and circumcenter $O$. Denote by $M$ the midpoint of $BC$. Point $D$ is the reflection of $A$ over $BC$, and $E$ is the intersection of $\\Gamma$ and ray $MD$. Let $S$ be the circumcenter of triangle $ADE$. Prove that $A, E, M, O$, and $S$ are concyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Solution.** Take $S$ to be the point such that $CPSQ$ is a parallelogram, as seen in figure 20. For points $X, Y, Z$ let $\\text{rot } XYZ$ denote the morphism on translations induced by the rotation that takes line $XY$ to line $XZ$, modulo half turn. As $ABCD$ is a cyclic quadrilateral it follows that $\\text{rot } BAD = \\text{rot } BCD$. It follows that $\\text{rot } PAD = \\text{rot } BCP$ and hence $\\text{rot } PAQ = \\text{rot } QCP = \\text{rot } PSQ$. It follows that $S$ lies on $\\Gamma$.\nUsing the cyclic quadrilateral *AQRS* it follows that $\\text{rot } RSQ = \\text{rot } RAQ$. Considering the cyclic quadrilateral *ABCD* it follows that $\\text{rot } RAQ = \\text{rot } RAD = \\text{rot } RCD = \\text{rot } RCP$. Hence, $\\text{rot } RSQ = \\text{rot } RCP$. As lines *CP* and *SQ* are parallel it follows that line *SR* is parallel to line *CR*. Now *R* is a common point so it follows that line *SR* = *CR* = *CS*.\nAs lines *CS* and *PQ* are diagonals in a parallelogram *CPSQ* it follows that line *CR* = *CS* passes through *M*, the midpoint of linesegment *PQ*.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25124,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that no non-zero integers $a$, $b$, $x$, $y$ satisfy\n$$\n\\begin{cases} ax - by = 256 \\\\ ay + bx = 1 \\end{cases}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If we use the Diophantine sum of squares equality\n$$\n(ax - by)^2 + (ay + bx)^2 = (a^2 + b^2)(x^2 + y^2)\n$$\nthen we can see that for a system\n$$\n\\begin{cases} ax - by = s \\\\ ay + bx = t \\end{cases}\n$$\nto have a solution in positive integers the number $s^2 + t^2$ must be a composite number.\nThe number corresponding to the equation, $256^2 + 1^2 = 2^{16} + 1$, is a prime number. (In fact, this is a rather well-known fact, as primes of the form $2^n + 1$ relate to constructible polygons. These primes are called Fermat primes, and $2^{16} + 1$ is the largest known to date.) This shows that no solution can exist in non-zero integers, as it would give a factorisation of the prime with each factor $> 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25125,
    "subject": "Mathematics (Multi-modal)",
    "question": "Distinct positive integers $a$, $b$, $c$, $d$ satisfy\n$$\n\\begin{cases}\na|b^2 + c^2 + d^2 \\\\\nb|a^2 + c^2 + d^2 \\\\\nc|a^2 + b^2 + d^2 \\\\\nd|a^2 + b^2 + c^2\n\\end{cases} .\n$$\nAlso it is known that none of them is larger than the product of three others. What is the largest possible number of primes among them?",
    "options": [],
    "answer": "3",
    "solution": "At first we note that the given condition is equivalent to $a$, $b$, $c$, $d$ dividing $a^2 + b^2 + c^2 + d^2$.\nIt is possible that three of the given numbers are primes, for example for $a = 2$, $b = 3$, $c = 13$ and $d = 26$. In this case $2^2 + 3^2 + 13^2 + 26^2 = 13 \\cdot 66$ which is divisible by all four given numbers. Furthermore we will show that it is impossible that all four of them are primes.\nLet us assume that $a$, $b$, $c$ and $d$ are primes. As the sum $a^2 + b^2 + c^2 + d^2$ is divisible by each of them then it is divisible also by their product $abcd$. If one of the primes is equal to $2$, then we obtain a contradiction: the sum of four squares is odd, but its divisor $abcd$ is even. Therefore all four primes are odd, and $a^2+b^2+c^2+d^2=0 \\pmod 4$. Hence $a^2+b^2+c^2+d^2$ is divisible by $4abcd$ which leads to a contradiction as it is easy to see that $a^2+b^2+c^2+d^2 < 4abcd$. Indeed, this is equivalent to\n$$\n\\frac{a}{bcd} + \\frac{b}{acd} + \\frac{c}{abd} + \\frac{d}{abc} < 4\n$$\nwhich is true as none of the numbers exceed the product of three other and equality can hold only for the largest of the four.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25126,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all integer triples $(a, b, c)$ satisfying the equation:\n$$\n5a^2 + 9b^2 = 13c^2\n$$",
    "options": [],
    "answer": "(0, 0, 0)",
    "solution": "Observe that $(a, b, c) = (0, 0, 0)$ is a solution. Assume that the equation has a solution $(a_0, b_0, c_0) \\neq (0, 0, 0)$. Let $d = \\gcd(a_0, b_0, c_0) > 0$. Let $(a, b, c) = (a_0/d, b_0/d, c_0/d)$. Then $\\gcd(a, b, c) = 1$. From $5a_0^2 + 9b_0^2 = 13c_0^2$ it follows that:\n$$\n5a^2 + 9b^2 = 5 \\left(\\frac{a_0}{d}\\right)^2 + 9 \\left(\\frac{b_0}{d}\\right)^2 = \\frac{5a_0^2 + 9b_0^2}{d^2} = \\frac{13c_0^2}{d^2} = 13 \\left(\\frac{c_0}{d}\\right)^2 = 13c^2\n$$\nhence $(a, b, c)$ is also a solution.\nAs $(a_0, b_0, c_0) \\neq (0, 0, 0)$ it follows that $(a, b, c) \\neq (0, 0, 0)$. Consider the equation modulo $13$. It follows that $5a^2 + 9b^2 = 13c^2 \\equiv 0 \\pmod{13}$, that is $5a^2 \\equiv -9b^2 \\equiv 4b^2 \\pmod{13}$. Multiplying by $8$ gives:\n$$\na^2 \\equiv 40a^2 = 8 \\cdot 5a^2 \\equiv 8 \\cdot 4b^2 = 32b^2 \\equiv 6 \\cdot b^2 \\pmod{13}\n$$\nIf $13\\mid b$ then $6b^2 \\equiv 6 \\cdot 0^2 = 0 \\pmod{13}$ and therefore $a^2 \\equiv 0 \\pmod{13}$, that is $13\\mid a^2$. As $13$ is prime it follows that $13\\mid a$. Hence $13$ divides $a$ and $b$. It follows that $13^2\\mid 5a^2 + 9b^2 = 13c^2$. Consequently $13$ divides $c^2$. As $13$ is prime, $13\\mid c$. This means that $13$ divides $a, b$ and $c$ contradicting the fact that $\\gcd(a, b, c) = 1$. We conclude that $13 \\nmid b$ does not hold.\nAs $13\\mid b$ does not hold and $13$ is a prime it follows that $b$ and $13$ are relatively prime. Therefore there exists $x \\in \\mathbb{Z}$ such that $b \\cdot x \\equiv 1 \\pmod{13}$. Multiplying by $x^2$ gives:\n$$\n(a \\cdot x)^2 = a^2 \\cdot x^2 \\equiv 6 \\cdot b^2 \\cdot x^2 = 6 \\cdot (b \\cdot x)^2 \\equiv 6 \\cdot 1^2 = 6 \\pmod{13}\n$$\nThat is $y^2 \\equiv 6 \\pmod{13}$ where $y = a \\cdot x$. As $y^2 \\equiv 6 \\pmod{13}$ it follows that $y$ and $13$ are relatively prime. By Fermat's little theorem it follows that $y^{12} \\equiv 1 \\pmod{13}$. Hence:\n$$\n\\begin{aligned}\n1 &\\equiv y^{12} = (y^2)^6 \\equiv 6^6 = (6^2)^3 \\equiv (36)^3 \\equiv 10^3 \\\\\n&= 10^2 \\cdot 10 = 100 \\cdot 10 \\equiv 9 \\cdot 10 = 90 \\equiv 12 \\pmod{13}\n\\end{aligned}\n$$\nbut $1 \\not\\equiv 12 \\pmod{13}$ so we have a contradiction. We conclude that the equation $5a^2 + 9b^2 = 13c^2$ has no solution besides the solution $(a, b, c) = (0, 0, 0)$.\n\n$\\boxed{(a, b, c) = (0, 0, 0)}$ is the only integer solution.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 25127,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all polynomials $P$ with integer coefficients such that the number $P(a) - P(b)$ is divisible by $a + b$ for all integers $a, b$, provided that $a + b \\neq 0$",
    "options": [],
    "answer": "All polynomials with integer coefficients whose odd-degree terms all have zero coefficients (i.e., even polynomials).",
    "solution": "Answer: all polynomials whose every odd-degree term has zero coefficient.\n\nLet $P(x) = P_0(x) + P_1(x)$, where $P_0$ and $P_1$ are polynomials whose all non-zero terms have either even or odd degree, respectively.\nThen we can write $P_0(x) = Q(x^2)$, where polynomial $Q$ is obtained from polynomial $P_0$ by dividing degrees of all non-zero terms by $2$. Now, for any integers $a, b$ the number $P_0(a) - P_0(b) = Q(a^2) - Q(b^2)$ is divisible by $a^2 - b^2$, and hence also by $a+b$. Thus, if every odd-degree term of $P$ has zero coefficient, then the condition of the problem is satisfied.\n\nOn the other hand, if polynomial $P$ satisfies the condition of the problem, then also $P - P_0 = P_1$ must satisfy it. Note that for every real $x$, $P_1(-x) = -P_1(x)$, i.e. $P_1$ is an odd function. By substituting $b$ by $-b$ in the condition of the problem we obtain that $a - b|P_1(a) + P_1(b)$ holds for any distinct integers $a$ and $b$. Since also $a - b|P_1(a) - P_1(b)$, then for any integers $a, b$ we have $a - b|2P_1(a)$. But for any $a$ there exists such $b$ that $|a - b| > 2P_1(a)$. From this we conclude that $P_1(a) = 0$ for any integer $a$. Altogether we have $P = P_0$, i.e. coefficients of all odd-degree terms are zero. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25128,
    "subject": "Mathematics (Multi-modal)",
    "question": "$A$ is a 20-element set of residue classes modulo $397$. Prove that for any non-negative integer $n$ there exist $x_1, x_2, x_3, x_4 \\in A$, such that $x_1 \\neq x_2$ and\n$$\n(x_1 - x_2)n \\equiv x_3 - x_4 \\pmod{397}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Observe that $397$ is a prime. If $n$ is divisible by $397$, the statement is trivial.\n\nFix an arbitrary positive integer $n$ not divisible by $397$. If for $x_1, x_2, x_3, x_4 \\in A$, where $(x_1, x_4)$ and $(x_2, x_3)$ are different ordered pairs, we have\n$$\nx_1 n + x_4 \\equiv x_2 n + x_3 \\pmod{397},\n$$\nthen we are done (if $x_1 = x_2$ then $x_3 = x_4$ due to this equivalence).\n\nOtherwise, this equivalence is impossible and therefore a map $(a, b) \\mapsto an + b$ is an injective map from $A \\times A$ to $\\mathbb{Z}_{397}$. But this is impossible since the number of elements in $A \\times A$ is $400 > 397$. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 25129,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all $k \\in \\mathbb{Z}$ such that there exists a function $f : \\mathbb{Z} \\to \\mathbb{Z}$ satisfying\n$$\nf(f(n)) = n + k\n$$\nfor all $n \\in \\mathbb{Z}$.",
    "options": [],
    "answer": "All even integers k",
    "solution": "If $k \\in \\mathbb{Z}$ is even then for $f : \\mathbb{Z} \\to \\mathbb{Z}$, $f(x) = x + \\frac{k}{2}$ we get:\n$$\nf(f(n)) = \\left(n + \\frac{k}{2}\\right) + \\frac{k}{2} = n + k\n$$\nFor $k$ even it is therefore possible to find function $f : \\mathbb{Z} \\to \\mathbb{Z}$ with the property that $f(f(n)) = n + k$ for all $n \\in \\mathbb{Z}$. It can therefore be assumed that $k$ is odd, in particular $k$ is non-zero.\nFor all $n \\in \\mathbb{Z}$ we get following:\n$$\nf(n) - n = (f(n) + k) - (n + k) = f(f(f(n))) - (n + k) = f(n + k) - (n + k)\n$$\nUsing induction it can be shown that $f(n + m \\cdot k) - (n + m \\cdot k) = f(n) - n$ for all $m \\in \\mathbb{N}$. If $p, q \\in \\mathbb{Z}$, and $p \\equiv q(\\text{mod } |k|)$ then there is a natural number $m$ such that $p = q + m \\cdot k$ or $q = p + m \\cdot k$. In either case $f(m) - m = f(n) - n$,\nequivalently $f(m) - f(n) = m - n$. As $m \\equiv n \\pmod{|k|}$, $f(m) - f(n) = m - n \\equiv 0 \\pmod{|k|}$, that is $f(m) \\equiv f(n) \\pmod{|k|}$.\nIf $m \\in \\mathbb{Z}$, $f(f(m-k)) = (m-k)+k = m$ so $m$ is in the image of $f$. As $m$ is arbitrary this means that $f$ is surjective. If $m, n \\in \\mathbb{Z}$ and $f(m) = f(n)$, we get:\n$$\nm = (m + k) - k = f(f(m)) - k = f(f(n)) - k = (n + k) - k = n,\n$$\nthat is $f$ is injective. As $f$ is both injective and surjective it is bijective. Assume $m, n \\in \\mathbb{Z}$ and $f(m) \\equiv f(n) \\pmod{|k|}$. Then\n$$\nm \\equiv m + k \\equiv f(f(m)) \\equiv f(f(n)) \\equiv n + k \\equiv n \\pmod{|k|}\n$$\nLet $h : \\{0, 1, \\dots, |k| - 1\\} \\to \\{0, 1, \\dots, |k| - 1\\} : x \\mapsto (f(x) \\pmod{|k|})$. From last equation we infer that $h$ is injective. As\n$$\nh(h(n)) = f(f(n) \\pmod{|k|}) \\pmod{|k|} = f(f(n)) \\pmod{|k|} = (n + k) \\pmod{|k|} = n\n$$\nfor all $n \\in \\{0, 1, \\dots, |k| - 1\\}$. That is $h$ is an involution and we see that $h$ is bijective. Assume $h$ has a fixed point $n_0$. As $n_0 = h(n_0) = f(n_0) \\pmod{|k|}$ we conclude that $f(n_0) - n_0 = m \\cdot |k|$ where $m \\in \\mathbb{Z}$.\nIt has already been shown that $f(n_0 + m \\cdot |k|) - (n_0 + m \\cdot |k|) = f(n_0) - n_0 = m \\cdot |k|$ so:\n$$\n\\begin{aligned}\nk &= f(f(n_0)) - n_0 \\\\\n  &= (f(f(n_0)) - f(n_0)) + (f(n_0) - n_0) \\\\\n  &= (f(n_0 + m \\cdot |k|) - (n_0 + m \\cdot |k|)) + m \\cdot |k| \\\\\n  &= m \\cdot |k| + m \\cdot |k| \\\\\n  &= 2 \\cdot m |k|\n\\end{aligned}\n$$\nThis implies $|k| = 2|m||k|$. As $k \\neq 0$ we get $2|m| = 1$ which is impossible as 1 is odd. The assumption that $n_0$ is a fixed point of $h$ must therefore be false.\nGiven $n \\in \\mathbb{Z}$ $h(n) \\neq n$ and $h(h(n)) = n$ so the sets $\\{n, h(n)\\}$ and $\\{h(n), h(h(n))\\}$ are equal and each contains two distinct elements. Now\n$$\n\\{0, 1, \\dots, |k| - 1\\} = \\bigcup \\{\\{n, h(n)\\} | n \\in \\{0, 1, \\dots, |k| - 1\\}\\}.\n$$\nAs each subset of $A := \\{\\{n, h(n)\\} | n \\in \\{0, 1, \\dots, |k| - 1\\}\\}$ contains two elements it follows that the union $\\bigcup A = \\{0, 1, \\dots, |k| - 1\\}$ contains an even number of elements. The cardinality of $\\{0, 1, \\dots, |k| - 1\\}$ is $|k|$ which is odd and we get a contradiction. This shows that if $k$ is odd there is no function $f : \\mathbb{Z} \\to \\mathbb{Z}$ such that $f(f(n)) = n + k$ for all $n \\in \\mathbb{Z}$.\nFunction $f : \\mathbb{Z} \\to \\mathbb{Z}$ satisfying $f(f(n)) = n + k$ for all $n \\in \\mathbb{Z}$ can therefore be found if and only if $k$ is even. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25130,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 2$ be an integer. Let $a_1, a_2, \\dots, a_n \\in \\{1, 2, 3, \\dots, 2n\\}$ be such that $\\text{lcm}(a_i, a_j) > 2n$ for any $1 \\le i < j \\le 2n$. Prove that\n$$\na_1 a_2 \\dots a_n \\mid (n+1)(n+2) \\dots (2n-1)(2n).\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For every $i = 1, 2, \\dots, n$ let $a_i = b_i \\cdot 2^{c_i}$ where $b_i$ is odd.\nNote that the $b_i$'s are pairwise distinct. Indeed, if $b_i = b_j$ for some $i < j$ then one of the numbers $a_i, a_j$ divides the other one, so $\\text{lcm}(a_i, a_j) = \\max(a_i, a_j) \\le 2n$ which is a contradiction.\nAlso, it is clear that each $b_i$ belongs to $\\{1, 3, 5, \\dots, 2n-1\\}$. Since there are exactly $n$ $b_i$'s and the set $\\{1, 3, 5, \\dots, 2n-1\\}$ has exactly $n$ elements, we have\n$$\n\\{b_1, b_2, \\dots, b_n\\} = \\{1, 3, 5, \\dots, 2n-1\\}.\n$$\nNow, for every $i$ let $d_i$ be the greatest $d$ such that $b_i \\cdot 2^d \\le 2n$. Note that $b_i \\cdot 2^{d_i} \\in \\{n+1, n+2, \\dots, 2n\\}$ as otherwise $b_i \\cdot 2^{d_i+1} \\le 2n$, contradicting maximality of $d_i$.\nNote that the numbers $b_i \\cdot 2^{d_i}$ are pairwise distinct (because $\\mathbb{Z}$ is a unique factorization domain). Again, we have $n$ pairwise distinct numbers $b_1 \\cdot 2^{d_1}, b_2 \\cdot 2^{d_2}, \\dots, b_n \\cdot 2^{d_n}$ belonging to the $n$-element set $\\{n+1, n+2, \\dots, 2n\\}$, hence\n$$\n\\{b_1 \\cdot 2^{d_1}, b_2 \\cdot 2^{d_2}, \\dots, b_n \\cdot 2^{d_n}\\} = \\{n+1, n+2, \\dots, 2n\\}.\n$$\nReindexing $a_i$'s if necessary, we can assume that $b_i \\cdot 2^{d_i} = n + i$ for every $i$. Clearly, $c_i \\le d_i$, so $a_i = b_i \\cdot 2^{c_i}$ for every $b_i \\cdot 2^{d_i} = n + i$. As a consequence,\n$$\na_1 a_2 \\dots a_n \\mid (n+1)(n+2)\\dots(2n-1)(2n),\n$$\n\nas desired. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25131,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer and let $a_1, a_2, \\dots, a_n$ and $b_1, b_2, \\dots, b_n$ be positive real numbers. Prove that\n$$\na_1 \\sqrt{\\frac{a_1}{b_1}} + a_2 \\sqrt{\\frac{a_2}{b_2}} + \\dots + a_n \\sqrt{\\frac{a_n}{b_n}} \\ge (a_1 + a_2 + \\dots + a_n) \\sqrt{\\frac{a_1 + a_2 + \\dots + a_n}{b_1 + b_2 + \\dots + b_n}}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For $a, b \\in \\mathbb{R}_{>0}$ we consider the function $f: \\mathbb{R}_{\\ge 0} \\to \\mathbb{R}$ with $f(x) = b x^3 - a x$ for all $x \\in \\mathbb{R}_{\\ge 0}$.\nFor $x \\ge 0$ we have $b(x - \\sqrt{\\frac{a}{3b}})^2 (x + 2\\sqrt{\\frac{a}{3b}}) \\ge 0$ which is equivalent to\n$$\nf(x) = b x^3 - a x \\geq -\\frac{2a}{3}\\sqrt{\\frac{a}{3b}} = f\\left(\\sqrt{\\frac{a}{3b}}\\right) = \\min(f).\n$$\nFor $i \\in \\{1, \\dots, n\\}$ consider the functions $f_i: \\mathbb{R}_{\\ge 0} \\to \\mathbb{R}$ with $f_i(x) = b_i x^3 - a_i x$ for all $x \\in \\mathbb{R}_{\\ge 0}$, and put $f = f_1 + f_2 + \\dots + f_n$. Note that\n$$\nf(x) = (b_1 + b_2 + \\dots + b_n)x^3 - (a_1 + a_2 + \\dots + a_n)x\n$$\nfor all $x \\in \\mathbb{R}_{\\ge 0}$. It follows that\n$$\n\\min(f_i) = -\\frac{2}{3\\sqrt{3}} \\cdot a_i \\sqrt{\\frac{a_i}{b_i}}\n$$\nfor all $i \\in \\{1, \\dots, n\\}$ and\n$$\n\\min(f) = -\\frac{2}{3\\sqrt{3}} \\cdot (a_1 + a_2 + \\dots + a_n) \\sqrt{\\frac{a_1 + a_2 + \\dots + a_n}{b_1 + b_2 + \\dots + b_n}}.\n$$\nThe desired inequality follows from the inequality\n$$\n\\min(f_1) + \\min(f_2) + \\dots + \\min(f_n) \\le \\min(f_1 + f_2 + \\dots + f_n)\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25132,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $D := \\mathbb{R} \\setminus \\{0,1\\}$. Find all functions $f:D \\rightarrow D$ which satisfy for any $x, y \\in \\mathbb{R}$ with $x, xy \\in D$ the equation\n$$\nf(f(xy)) = 1 - \\frac{1}{y f(f(f(x)))}\n$$",
    "options": [],
    "answer": "f(x) = 1/(1 - x) for all x in D",
    "solution": "Plugging in $y = \\frac{a}{x}$ for $a, x \\in D$ gives\n$$\nf(f(a)) = 1 - \\frac{x}{a f(f(f(x)))}.\n$$\nOn the other hand, $y = 1$ and $x = a$ gives for $a \\in D$\n$$\nf(f(a)) = 1 - \\frac{1}{f(f(f(a)))}.\n$$\nfrom which we conclude $\\frac{x}{f(f(f(x)))}$ is constant for all $x \\in D$. Thus it follows $f(f(f(x))) = Cx$ for some constant $C \\neq 0$. Plugging this into the second equation gives $f(f(x)) = 1 - \\frac{1}{Cx}$. Replace here $x$ with $f(x)$. This is allowed, since $f(x) \\in D$ follows from the definition of $f$. Thus\n$$\nCx = f(f(f(x))) = 1 - \\frac{1}{Cf(x)}, \\text{ i.e. } f(x) = \\frac{1}{C(1 - Cx)}.\n$$\n\nSo we get $f(x) = \\frac{1}{1-x}$ for all $x \\in D$, which clearly solves the given functional equation, so we are done.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 25133,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest possible value of\n$$\n\\frac{x^{2023} + 203}{17x^7 + 7x^{17}}\n$$\nover all positive real numbers $x$.",
    "options": [],
    "answer": "17/2",
    "solution": "Answer: $\\frac{17}{2}$.\n\nNote that $2023 = 7 \\cdot 17^2 = 7 \\cdot 289 = 17 \\cdot 119$. Using AGM on the numbers $x^{2023}, 1, 1, \\dots, 1$, where there are 288 ones, yields\n$$\nx^{2023} + 288 \\geq 289 \\sqrt[289]{x^{2023}} = 289x^7.\n$$\nAnalogously, using AGM on the numbers $x^{2023}, 1, 1, \\dots, 1$, where there are 118 ones, yields\n$$\nx^{2023} + 118 \\geq 119 \\sqrt[119]{x^{2023}} = 119x^{17}.\n$$\nIn both inequalities the equality holds if and only if $x = 1$. Adding them yields $2x^{2023} + 406 \\geq 289x^7 + 119x^{17}$, which yields $2(x^{2023} + 203) \\geq 17(17x^7 + 7x^{17})$. Therefore $\\frac{x^{2023}+203}{17x^7+7x^{17}} \\geq \\frac{17}{2}$, where $x = 1$ is still the equality case. Therefore the smallest possible value of $\\frac{x^{2023}+203}{17x^7+7x^{17}}$ is $\\frac{17}{2}$.\nDenote\n$$\nf(x) = \\frac{x^{2023} + 203}{17x^7 + 7x^{17}}.\n$$\nThen\n$$\nf'(x) = 119x^6 \\cdot \\frac{118x^{2023} + 288x^{2023} - 203x^{10} - 203}{(17x^7 + 7x^{17})^2}.\n$$\nTo show that $f$ obtains its minimal value at $x = 1$, we will show that $f'$ is negative for $0 < x < 1$, zero at $x = 1$ and positive for $x > 1$. This is equivalent to showing the same for $118x^{2023} + 288x^{2023} - 203x^{10} - 203$. We see that its value is indeed zero at $x = 1$ and that (e.g. by computing the geometric sums on the right-hand side)\n$$\n\\begin{aligned}\n& 118x^{2023} + 288x^{2023} - 203x^{10} - 203 \\\\\n&= (x-1) \\cdot (118(x^{2022} + \\dots + x^{2023})) \\\\\n& \\quad + 406(x^{2022} + \\dots + x^{10}) + 203(x^9 + \\dots + x + 1).\n\\end{aligned}\n$$\nThe second factor is clearly positive for all positive $x$. So $f'$ is negative for $0 < x < 1$, zero at $x = 1$ and positive for $x > 1$, as desired. Thus the minimal value of $f(x)$ is\n$$\nf(1) = \\frac{1 + 203}{17 \\cdot 1 + 7 \\cdot 1} = \\frac{204}{24} = \\frac{17}{2}.\n$$\nLike in Solution 2, reduce the problem to having to show that $118x^{2033} + 288x^{2023} - 203x^{10} - 203$ is negative for $0 < x < 1$ and positive for $x > 1$.\nFor $0 < x < 1$, we have $x^{2033} < x^{2023} < x^{10} < 1$, so $118x^{2033} + 85x^{2023} - 203x^{10} < 0$ and $203x^{2023} - 203 < 0$. Adding those yields the desired result.\nFor $x > 1$, we have $x^{2033} > x^{2023} > x^{10} > 1$, so $118x^{2033} + 85x^{2023} - 203x^{10} > 0$ and $203x^{2023} - 203 > 0$. Adding those yields the desired result.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25134,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\mathbb{Z}^+$ be the set of positive integers. Find all strictly increasing functions $\\mathbb{Z}^+ \\to \\mathbb{Z}^+$ with $f(1) = 1$ that satisfy the equation\n$$\n3 \\cdot (f(1) + f(2) + \\dots + f(n)) = f(n+1) + f(n+2) + \\dots + f(2n)\n$$\nfor all $n \\in \\mathbb{Z}^+$.",
    "options": [],
    "answer": "f(n) = 2n - 1",
    "solution": "The strictly increasing function $\\mathbb{Z}^+ \\to \\mathbb{Z}^+$ with $f(n) = 2n-1$ for all $n \\in \\mathbb{Z}^+$ satisfies $f(1) = 1$ and solves the functional equation, since $1+3+\\dots+(2n-1) = n^2$ and $(2n+1)+(2n+3)+\\dots+(4n-1) = (2n)^2-n^2 = 3n^2$ for all $n \\in \\mathbb{Z}^+$.\nWe claim that no other function is suitable. Let $f:\\mathbb{Z}^+ \\to \\mathbb{Z}^+$ be a function that meets all requirements of the problem.\nLet $k \\in \\mathbb{Z}^+$. Note that the given functional equation for $k$ and $k+1$ implies\n$$\n3 \\cdot \\sum_{l=1}^{k} f(l) = \\sum_{l=k+1}^{2k} f(l), \\\\\n3 \\cdot \\sum_{l=1}^{k+1} f(l) = \\sum_{l=k+2}^{2k+2} f(l);\n$$\nthe difference of the two equations yields $3f(k+1) = -f(k+1) + f(2k+1) + f(2k+2)$. In other words, the equation\n$$\n4f(k+1) = f(2k+1) + f(2k+2) \\quad (*)\n$$\nholds for all $k \\in \\mathbb{Z}^+$.\nEquation (*) implies that the numbers $f(2k+1)$ and $f(2k+2)$ have the same parity for every $k \\in \\mathbb{Z}^+$. Since $f$ is strictly increasing, we can deduce that $f(2k+2) \\ge f(2k+1)+2$. Shifting indices we also\n\nobtain $4f(k+2) = f(2k+3) + f(2k+4)$ from equation (*). Note that $f(2k+3) \\ge f(2k+2) + 1 \\ge f(2k+1) + 3$. Similarly, since $f(2k+3)$ and $f(2k+4)$ must have the same parity, $f(2k+4) \\ge f(2k+3) + 2 \\ge f(2k+2) + 3$, so that\n$$\n\\begin{aligned}\n4f(k+2) &= f(2k+3) + f(2k+4) \\\\\n&\\ge (f(2k+1) + 3) + (f(2k+2) + 3) \\\\\n&= 4f(k+1) + 6.\n\\end{aligned}\n$$\nWe can conclude that\n$$\nf(k+2) \\ge f(k+1) + 2 \\text{ for all } k \\in \\mathbb{Z}^{+}. \\qquad (**)\n$$\nNow we are ready to show that $f(n) = 2n - 1$ for all $n \\in \\mathbb{Z}^+$. More precisely, we use strong induction to show that $f(2k-1) = 4k-3$ and $f(2k) = 4k-1$ for all $k \\in \\mathbb{Z}^+$. The claim implies $f(n) = 2n-1$ for all $n \\in \\mathbb{Z}^+$.\nFor the start of the induction, note that we have $f(1) = 1$ by definition; the given condition for $n=1$ implies $f(2) = 3f(1) = 3$. Hence, the equations $f(2k-1) = 4k-3$ and $f(2k) = 4k-1$ are true for $k=1$.\nFor the induction step, let $k \\ge 1$ and assume that $f(2l-1) = 4l-3$ and $f(2l) = 4l-1$ for all $l \\in \\{1, \\dots, k\\}$. We want to show that $f(2k+1) = 4k+1$ and $f(2k+2) = 4k+3$.\nSince $k+1 \\le 2k$ the induction hypothesis implies $f(k+1) = 2k+1$. Equation (*) implies $f(2k+1)+f(2k+2) = 8k+4$. By induction hypothesis $f(2k) = 4k-1$, so that by virtue of inequality ($**$) we have $f(2k+1) \\ge 4k+1$ and $f(2k+2) \\ge 4k+3$. Since the sum of the two function values is $8k+4$, we must have $f(2k+1) = 4k+1$ and $f(2k+2) = 4k+3$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 25135,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_1, a_2, \\dots, a_{2023}$ be positive real numbers with\n$$\na_1 + a_2^2 + a_3^3 + \\dots + a_{2023}^{2023} = 2023.\n$$\nShow that\n$$\na_1^{2023} + a_2^{2022} + \\dots + a_{2022}^2 + a_{2023} > 1 + \\frac{1}{2023}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us prove that conversely, the condition\n$$\na_1^{2023} + a_2^{2022} + \\dots + a_{2023} \\le 1 + \\frac{1}{2023}\n$$\nimplies that\n$$\nS := a_1 + a_2^2 + \\dots + a_{2023}^{2023} < 2023.\n$$\nThis is trivial if all $a_i$ are less than $1$. So suppose that there is an $i$ with $a_i \\ge 1$, clearly it is unique and $a_i < 1 + \\frac{1}{2023}$. Then we have\n$$\n\\begin{aligned}\na_i^i &< \\left(1 + \\frac{1}{2023}\\right)^{2023} = 1 + \\sum_{k=1}^{2023} \\frac{1}{k!} \\cdot \\frac{2023}{2023} \\cdot \\frac{2022}{2023} \\dots \\cdot \\frac{2023-k+1}{2023} \\\\\n&< 1 + \\sum_{k=1}^{2023} \\frac{1}{k!} \\le 1 + \\sum_{k=0}^{2022} \\frac{1}{2^k} < 3,\n\\end{aligned}\n$$\n$$\n\\sum_{\\substack{k=1, \\\\ k \\ne i}}^{1011} a_k^k \\le 1011 \\quad \\text{and} \\quad \\sum_{\\substack{k=1012, \\\\ k \\ne i}}^{2023} a_k^k \\le \\sum_{\\substack{k=1012, \\\\ k \\ne i}}^{2023} a_k^{2024-k} < \\frac{1}{2023}.\n$$\nHence we have\n$$\nS = a_i^i + \\sum_{\\substack{k=1, \\\\ k \\ne i}}^{1011} a_k^k + \\sum_{\\substack{k=1012, \\\\ k \\ne i}}^{2023} a_k^k < 3 + 1011 + \\frac{1}{2023} < 2023.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25136,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all (not necessarily strictly) monotonic functions $f: \\mathbb{R} \\to \\mathbb{R}$ with\n$$\nf(x+y)^3 = f(x^3) + f(y^3) \\text{ for all } x, y \\in \\mathbb{R}.\n$$",
    "options": [],
    "answer": "All such functions are: f(x) = 0 for all x; f(x) = cube_root(x); f(x) = -cube_root(x); f(x) = sqrt(2) for all x; f(x) = -sqrt(2) for all x.",
    "solution": "By substituting $x = y = 0$ we see that $f(0)^3 = 2f(0)$ which implies $f(0)^3 - 2f(0) = 0$. This means that $f(0) = 0$, $f(0) = -\\sqrt{2}$ or $f(0) = \\sqrt{2}$.\n\nBy substituting $y = 0$ we get $f(x)^3 = f(x^3) + f(0^3) = f(x^3) + f(0)$.\n\n1) Let's first investigate the case $f(0) = 0$: in this case, $f(x)^3 = f(x^3)$, so $(f(x+y))^3 = f(x^3) + f(y^3) = f(x)^3 + f(y)^3$, in which case $g(x) = f(x)^3$ is a Cauchy functional equation and due to function being monotonic must be of the form $c'x$ for some $c' \\in \\mathbb{R}$. This means that the solution must be $f(x) = \\sqrt[3]{c'x} = c\\sqrt[3]{x}$. Here $\\sqrt[3]{x}$ is a function defined on the entire set of real numbers $\\mathbb{R}$. Substituting this to the original equation we get $(c\\sqrt[3]{x+y})^3 = c\\sqrt[3]{x^3} + c\\sqrt[3]{y^3}$ which implies $c^3(x+y) = c(x+y)$, so $c^3 - c = 0$ and $c$ must be $-1$, $0$ or $1$. From this we get solutions $f(x) = 0$, $f(x) = \\sqrt[3]{x}$ and $f(x) = -\\sqrt[3]{x}$.\n\n2) Let's then investigate the case $f(0) = \\sqrt{2}$: By substituting $x = 1$ and $y = 0$ we get $f(1)^3 = f(1^3) + f(0) = f(1) + \\sqrt{2}$ and $f(1)^3 - f(1) - \\sqrt{2} = 0$. $f(1) = \\sqrt{2}$ is a solution, so this can be modified to $(f(1) - \\sqrt{2})(f(1)^2 + \\sqrt{2}f(1) + 1) = 0$. The latter part doesn't have real roots, so the only option is $f(1) = \\sqrt{2}$. Now, substituting $y = 1$ in the original equation we get $f(x+1)^3 = f(x^3) + \\sqrt{2}$. For $x = 1$ we get $f(2)^3 = f(1) + \\sqrt{2} = 2\\sqrt{2}$. This means that also $f(2) = \\sqrt{2}$.\n\nSubstituting $x = 2$ and $y = 0$ we get $f(2)^3 = f(8) + f(0)$ and substituting $x = 2, y = 1$ we get $f(3)^3 = f(8) + f(1)$, from which we get $f(3) = \\sqrt[3]{f(8) + f(1)} = \\sqrt[3]{f(8) + \\sqrt{2}} = \\sqrt[3]{2\\sqrt{2} - \\sqrt{2} + \\sqrt{2}} = \\sqrt{2}$. Let us now use induction and assume that $f(x) = \\sqrt{2}$ for all $x = 1, 2, \\dots, n-1$. Now let's substitute $x = \\sqrt[3]{n}, y = 1$: then $f(\\sqrt[3]{n} + 1)^3 = f(\\sqrt[3]{n}^3) + f(1^3) = f(n) + \\sqrt{2}$, from which $f(n) = f(\\sqrt[3]{n} + 1)^3 - \\sqrt{2} = 2\\sqrt{2} - \\sqrt{2} = \\sqrt{2}$, since $\\sqrt[3]{n} + 1 \\le n - 1$ when $n \\le (n-2)^3$, which is true when $n \\ge 4$. Because $f$ is monotonic, also all real values between positive integer values must have $f(x) = \\sqrt{2}$.\n\nLet's then substitute $x = -1$ and $y = 0$ in the original equation: we get $f(-1)^3 = f(-1)+f(0) = f(-1)+\\sqrt{2}$. From here we can use the same deductions as before to prove that $f(x) = \\sqrt{2}$ also for all negative $x$.\n\n3) Let's then investigate the case $f(0) = -\\sqrt{2}$: By substituting $x = 1$ and $y = 0$ we get $f(1)^3 = f(1^3) + f(0) = f(1) - \\sqrt{2}$ which implies $f(1)^3 - f(1) + \\sqrt{2} = 0$. $f(1) = -\\sqrt{2}$ is a solution, so this can be modified to $(f(1) + \\sqrt{2})(f(1)^2 - \\sqrt{2}f(1) + 1) = 0$. The latter part doesn't have real roots, so the only option is $f(1) = -\\sqrt{2}$. Now, substituting $y = 1$ in the original equation we get $f(x+1)^3 = f(x^3) - \\sqrt{2}$. For $x = 1$ we get $f(2)^3 = f(1) - \\sqrt{2} = -2\\sqrt{2}$. This means that also $f(2) = -\\sqrt{2}$. Proving that $f(3) = -\\sqrt{2}$ goes similarly to 2). Similarly to case 2) we can also use induction to first prove that $f(x) = -\\sqrt{2}$ then use the similar deductions to 2) to prove it also for all negative $x$.\n\nIt's easy to see that constant functions $0$, $\\sqrt{2}$ and $-\\sqrt{2}$ fulfill the original functional equation. So in the end we get solutions $f(x) = \\sqrt[3]{x}$, $f(x) = -\\sqrt[3]{x}$, $f(x) = 0$, $f(x) = \\sqrt{2}$ and $f(x) = -\\sqrt{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25137,
    "subject": "Mathematics (Multi-modal)",
    "question": "For sets $S$ and $T$ consisting of positive real numbers, define $S + T = \\{s + t \\mid s \\in S, t \\in T\\}$ and $\\frac{1}{S} = \\{\\frac{1}{s} \\mid s \\in S\\}$. Define the sets $A_1, A_2, A_3, \\dots$ recursively by $A_1 = \\{1\\}$ and\n$$\nA_n = \\bigcup_{i=1}^{n-1} \\left( (A_i + A_{n-i}) \\cup \\left( \\frac{1}{\\frac{1}{A_i} + \\frac{1}{A_{n-i}}} \\right) \\right)\n$$\nfor all integers $n \\ge 1$. Prove that for all integers $n \\ge 1$ we have\n$$\n2^{n-1} \\le |A_n| \\le 8^n.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "*Solution:* We start with a lemma.\n**Lemma.** For any $n$, $x \\in A_n$ implies $\\frac{1}{x} \\in A_n$.\n*Proof.* Induction. Case $n = 1$ is clear. Now, if $x = a_i + a_{n-i} \\in A_i + A_{n-i} \\subset A_n$, then $\\frac{1}{a_i} \\in A_i$ and $\\frac{1}{a_{n-i}} \\in A_{n-i}$, and thus\n$$\n\\frac{1}{x} \\in \\frac{1}{\\frac{1}{A_i} + \\frac{1}{A_{n-i}}} \\subset A_n.\n$$\nSimilarly if $x = (\\frac{1}{a_i} + \\frac{1}{a_{n-i}})^{-1}$, then\n$$\n\\frac{1}{x} = \\frac{1}{a_i} + \\frac{1}{a_{n-i}} \\in A_i + A_{n-i} \\subset A_n. \\quad \\square\n$$\n**Lower bound.**\nWe now prove that $|A_{n+1}| \\ge 2|A_n|$ for all $n$, which proves the lower bound. Let $a$ denote the number of elements in $A_n$ which are larger than or equal to $1$ and let $b$ denote the number of elements in $A_n$ which are strictly larger than $1$. Clearly $a = b$ if $1 \\notin A_n$ and otherwise $a = b + 1$.\n\nFirst note that $A_1 + A_n = \\{1\\} + A_n \\subset A_{n+1}$, so $A_{n+1}$ contains at least $a$ elements which are $\\ge 2$. Also note that for any $a_n \\in A_n$ for which $a_n > 1$ we have\n$$\n\\frac{1}{2} < a_{n+1} := \\frac{1}{\\frac{1}{1} + \\frac{1}{a_n}} < 1\n$$\nand $a_{n+1} \\in A_n$. Thus, $A_{n+1}$ contains at least $b$ elements in $(\\frac{1}{2}, 1)$ and thus, by the lemma, at least $b$ elements in $(1, 2)$.\nTherefore, $A_{n+1}$ has at least $a+b$ elements greater than $1$. By the lemma $A_{n+1}$ thus has at least $2(a+b)$ elements. If $a=b$, then $2|A_n| = 4a = 2(a+b) \\le |A_{n+1}|$ and if $a=b+1$, then $2|A_n| = 2(2b+1) = 2(a+b) \\le |A_{n+1}|$.\n\n**Upper bound.**\nWe then prove the upper bound. Define $s_n := |A_n|$, let $c_n := \\frac{1}{n+1} \\binom{2n}{n}$ be the $n$-th Catalan number, and $b_n := 2^n c_{n-1}$. Then\n$$\n\\begin{aligned}\n\\sum_{i=1}^{n-1} b_i b_{n-i} &= \\sum_{i=1}^{n-1} 2^i \\cdot c_{i-1} \\cdot 2^{n-i} \\cdot c_{n-2-(i-1)} \\\\\n&= 2^n \\cdot \\sum_{i=0}^{n-2} c_i c_{(n-2)-i} = 2^n \\cdot c_{n-1} = b_n,\n\\end{aligned}\n$$\nwhere we used the well-known recursion formula for the Catalan numbers.\nWe prove that $s_n < b_n$ for all $n$, which certainly is enough since $b_n = \\frac{2^n}{n} \\binom{2(n-1)}{n-1} < 2^n \\cdot 2^{2n} = 8^n$. The proof is by induction. The cases $n \\le 4$ may be checked by hand, as we have\n$$\ns_1 = 1 < 2 = b_1, \\quad s_2 = 2 < 4 = b_2, \\quad s_3 = 4 < 16 = b_3, \\quad \\text{and} \\quad s_4 = 9 < 80 = b_4.\n$$\nWe now assume $n \\ge 5$.\nWe have the trivial bound\n$$\ns_n \\le \\sum_{i=1}^{\\lfloor \\frac{n}{2} \\rfloor} 2s_i s_{n-i}.\n$$\n\ns_n \\le \\sum_{i=1}^{n-1} s_i s_{n-i}.\n$$\nUse the induction hypothesis to get\n$$\ns_n \\le \\sum_{i=1}^{n-1} s_i s_{n-i} < \\sum_{i=1}^{n-1} b_i b_{n-i} = b_n.\n$$\nFor $n$ even we need more care. Note that $|A_{n/2} + A_{n/2}| \\le \\binom{s_{n/2}}{2} + s_{n/2}$ and\n$$\n\\left| \\frac{1}{\\frac{1}{A_{n/2}} + \\frac{1}{A_{n/2}}} \\right| \\le \\binom{s_{n/2}}{2} + s_{n/2}.\n$$\nTherefore, for $n$ even we have\n$$\ns_n \\le \\sum_{i=1}^{\\frac{n}{2}-1} 2s_i s_{n-i} + s_{n/2}^2 + s_{n/2} = \\sum_{i=1}^{n-1} s_i s_{n-i} + s_{n/2}.\n$$\nWe now note that for $n \\ge 6$ we have, by the induction hypothesis, $s_3 s_{n-3} < (b_3 - 1) b_{n-3}$, and hence\n$$\n\\sum_{i=1}^{n-1} s_i s_{n-i} + s_{n/2} < \\sum_{i=1}^{n-1} b_i b_{n-i} - b_{n-3} + b_{n/2} \\le \\sum_{i=1}^{n-1} b_i b_{n-i} = b_n,\n$$\nas Catalan numbers and also the $b_n$ are increasing, concluding the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25138,
    "subject": "Mathematics (Multi-modal)",
    "question": "Denote a set of equations in the real numbers with variables $x_1, \\dots, x_m \\in \\mathbb{R}$ Flensburgian if there exists an $i \\in \\{1, \\dots, m\\}$ such that every solution of the set of equations where all the variables are pairwise different, satisfies $x_i > x_j$ for all $j \\ne i$.\nDetermine for which positive integer $n \\ge 1$, the following set of two equations\n$$\na^n + b = a \\text{ and } c^{n+1} + b^2 = ab\n$$\nin the three real variables $a, b, c$ is Flensburgian.",
    "options": [],
    "answer": "n = 1 or n is even",
    "solution": "The set of equations given in the problem statement is Flensburgian precisely when $n$ is even or $n = 1$.\n\nTo see that it is not Flensburgian when $n \\ge 3$ is odd, notice that if $(a, b, c)$ satisfies the set of equations then so does $(-a, -b, -c)$. Hence, if there exists a single solution to the set of equations where all the variables are different then the set of equations cannot be Flensburgian. This is in fact the case, e.g., consider $(a, b, c) = \\left(\\frac{1}{2}, \\frac{2^{n-1}-1}{2^n}, \\left(\\frac{2^{n-1}-1}{2^{2n}}\\right)^{\\frac{1}{n+1}}\\right)$.\n\nIf $n = 1$ from the first equation we know $b = 0$. But then from the second equation it follows $c = 0$. Hence, there cannot be any solution with pairwise different variables $a, b, c$. Therefore, the set of equations is Flensburgian for $n = 1$.\n\nThe rest of the solution is dedicated to prove that the set of equations is indeed Flensburgian when $n$ is even.\n\nThe first equation yields $b = a - a^n \\le a$, since $a^n \\ge 0$ when $n$ is even. The inequality is strict whenever $a \\ne 0$ and the case $a = 0$ implies $b = 0$, i.e. $a = b$, which we can disregard. Substituting the relation $b = a - a^n$ into the second equation yields\n$$\n0 = c^{n+1} + (a - a^n)^2 - a(a - a^n) = c^{n+1} + a^{2n} - a^{n+1}, \\text{ i.e.\\ } \\\\ c^{n+1} = a^{n+1} - a^{2n} < a^{n+1}\n$$\nsince we can disregard $a = 0$ and $2n$ is even. Since $n + 1$ is odd, the polynomial $x^{n+1}$ is strictly increasing, implying that $c < a$. Hence, when $n$ is even, all solutions of the set of equations where $a, b, c$ are pairwise different satisfy $a > b$ and $a > c$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 25139,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer, $x_1, \\dots, x_n$ and $y_1, \\dots, y_n$ be positive real numbers satisfying $x_n = n$ and $y_i y_{i+1} \\ge 1$ for $1 \\le i < n$ and $y_n y_1 \\ge 1$. Moreover, let $x_0 = 0$.\nDetermine the minimal possible value of the expression\n$$\n\\sum_{i=1}^{n} \\sqrt{(x_i - x_{i-1})^2 + y_i}.\n$$",
    "options": [],
    "answer": "n*sqrt(2)",
    "solution": "We claim that the minimal value of the desired expression is $n\\sqrt{2}$ achieved by $x_i = i$ and $y_i = 1$ for all $i$. In what follows below, $Y$ denotes the sum $Y = \\sum_{i=1}^{n} \\sqrt{y_i}$.\nThe solution consists of proving two separate inequalities.\n\nBy taking the global product of all the inequalities of the form $y_i y_{i+1} \\ge 1$ and $y_n y_1 \\ge 1$ we obtain\n$$\n\\prod_{i=1}^{n} y_i^2 \\geq 1.\n$$\nThus from the AGM inequality we obtain\n$$\n\\sum_{i=1}^{n} \\sqrt{y_i} \\geq n \\prod_{i=1}^{n} y_i^{\\frac{1}{2n}} \\geq n\n$$\nwith equality in the left-hand inequality for equal $y_i$ and in the right-hand inequality for $y_i = 1$.\nHence, by combining the two inequalities we obtain\n$$\n\\sum_{i=1}^{n} \\sqrt{(x_i - x_{i-1})^2 + y_i} \\geq \\sqrt{n^2 + Y^2} \\geq \\sqrt{2n^2} = n\\sqrt{2}\n$$\nwith equality iff $y_i = 1$ for all $i$ and $x_i = x_{i-1} + 1$, i.e. $x_i = i$.\n\n\n2nd Solution:\nFor the two inequalities we give a second proof, each.\n**First Inequality:**\nAgain we consider the $y_i$ as constants, and rewrite\n$$\n\\sqrt{(x_i - x_{i-1})^2 + y_i} = \\sqrt{y_i} \\sqrt{1 + \\left(\\frac{x_i - x_{i-1}}{\\sqrt{y_i}}\\right)^2}.\n$$\nBy considering the second derivative $\\frac{\\partial^2}{\\partial x^2} \\sqrt{1+x^2} = \\frac{1}{(1+x^2)^{\\frac{3}{2}}} > 0$ for all $x$ we find that $f(x) = \\sqrt{1+x^2}$ is a convex function.\nHence, by applying the weighted Jensen's Inequality with weights $\\frac{\\sqrt{y_i}}{Y}$ we obtain\n\n\\begin{aligned}\n\\sum_{i=1}^{n} \\frac{\\sqrt{y_i}}{Y} f\\left(\\frac{x_i - x_{i-1}}{\\sqrt{y_i}}\\right) &\\ge f\\left(\\sum_{i=1}^{n} \\frac{x_i - x_{i-1}}{Y}\\right) = f\\left(\\frac{n}{Y}\\right), \\text{ i.e.} \\\\\n\\sum_{i=1}^{n} \\sqrt{(x_i - x_{i-1})^2 + y_i} &\\ge \\sqrt{n^2 + Y^2}\n\\end{aligned}\n$$\nwith equality if and only if $\\frac{x_i - x_{i-1}}{\\sqrt{y_i}}$ is independent of $i$. Hence the equality case may be established identically to the first solution.\n**Second Inequality:**\nBy rewriting the given inequalities as $\\sqrt[4]{y_i y_{i+1}} \\ge 1$, taking the global sum and using the Rearrangement Inequality we obtain\n$$\n\\sum_{i=1}^{n} \\sqrt{y_i} \\ge \\sum_{i=1}^{n} \\sqrt[4]{y_i y_{i+1}} \\ge n\n$$\nwith equality iff $y_i = y_{i+1}$ and $y_i y_{i+1} = 1$ for all $i$, i.e. $y_i = 1$ for all $i$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 25140,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest positive real number $\\alpha$, such that\n$$\n\\frac{x+y}{2} \\geq \\alpha\\sqrt{xy} + (1-\\alpha)\\sqrt{\\frac{x^2+y^2}{2}}\n$$\nfor all positive real numbers $x$ and $y$.",
    "options": [],
    "answer": "1/2",
    "solution": "Let's prove that $\\alpha = \\frac{1}{2}$ works. Then the following inequality should hold for all positive real numbers $x$ and $y$:\n$$\n\\begin{aligned}\n\\frac{x+y}{2} &\\ge \\frac{1}{2}\\sqrt{xy} + \\frac{1}{2}\\sqrt{\\frac{x^2+y^2}{2}} \\\\\n\\Leftrightarrow (x+y)^2 &\\ge xy + \\frac{x^2+y^2}{2} + 2\\sqrt{xy \\cdot \\frac{x^2+y^2}{2}} \\\\\n\\Leftrightarrow (x+y)^2 &\\ge 4\\sqrt{xy \\cdot \\frac{x^2+y^2}{2}} \\\\\n\\Leftrightarrow (x+y)^4 &\\ge 8xy(x^2+y^2) \\\\\n\\Leftrightarrow (x-y)^4 &\\ge 0\n\\end{aligned}\n$$\nwhich is true, so we showed that $\\alpha = \\frac{1}{2}$ actually works.\n\nNow it remains to show that $\\alpha \\ge \\frac{1}{2}$. Let's consider $x = 1 + \\varepsilon$ and $y = 1 - \\varepsilon$ where $\\varepsilon < 1$. Then the inequality becomes\n$$1 \\ge \\alpha\\sqrt{1-\\varepsilon^2} + (1-\\alpha)\\sqrt{1+\\varepsilon^2}, \\text{ i.e.}$$ $$\\alpha \\ge \\frac{\\sqrt{1+\\varepsilon^2}-1}{\\sqrt{1+\\varepsilon^2}-\\sqrt{1-\\varepsilon^2}}.$$\nNotice that\n$$\n\\begin{align*}\n\\frac{\\sqrt{1+\\varepsilon^2}-1}{\\sqrt{1+\\varepsilon^2}-\\sqrt{1-\\varepsilon^2}} &= \\frac{(\\sqrt{1+\\varepsilon^2}-1)(\\sqrt{1+\\varepsilon^2}+1)(\\sqrt{1+\\varepsilon^2}+\\sqrt{1-\\varepsilon^2})}{(\\sqrt{1+\\varepsilon^2}-\\sqrt{1-\\varepsilon^2})(\\sqrt{1+\\varepsilon^2}+\\sqrt{1-\\varepsilon^2})(\\sqrt{1+\\varepsilon^2}+1)} \\\\\n&= \\frac{\\varepsilon^2(\\sqrt{1+\\varepsilon^2}+\\sqrt{1-\\varepsilon^2})}{2\\varepsilon^2(\\sqrt{1+\\varepsilon^2}+1)} = \\frac{\\sqrt{1+\\varepsilon^2}+1-1+\\sqrt{1-\\varepsilon^2}}{2(\\sqrt{1+\\varepsilon^2}+1)} \\\\\n&= \\frac{1}{2} - \\frac{1-\\sqrt{1-\\varepsilon^2}}{2(\\sqrt{1+\\varepsilon^2}+1)} = \\frac{1}{2} - \\frac{(1-\\sqrt{1-\\varepsilon^2})(1+\\sqrt{1-\\varepsilon^2})}{2(\\sqrt{1+\\varepsilon^2}+1)(1+\\sqrt{1-\\varepsilon^2})} \\\\\n&= \\frac{1}{2} - \\frac{\\varepsilon^2}{2(\\sqrt{1+\\varepsilon^2}+1)(1+\\sqrt{1-\\varepsilon^2})} \\\\\n&> \\frac{1}{2} - \\frac{\\varepsilon^2}{4(1+\\sqrt{2})}.\n\\end{align*}\n$$\nAs $\\varepsilon$ can be arbitrarily small this expression can get arbitrarily close to $\\frac{1}{2}$. This means that $\\alpha < \\frac{1}{2}$ cannot hold, as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25141,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. Each cell of an $n \\times n$ table is coloured in one of $k$ colours where every colour is used at least once. Two colours $A$ and $B$ are said to touch each other, if there exists a cell coloured in $A$ sharing a side with a cell coloured in $B$. The table is coloured in such a way that each colour touches at most 2 other colours. What is the maximal value of $k$?",
    "options": [],
    "answer": "k = 2n − 1 for n ≠ 2, and k = 4 for n = 2",
    "solution": "$k = 2n-1$ when $n \\neq 2$ and $k = 4$ when $n = 2$.\n$k = 2n - 1$ is possible by colouring diagonally as shown in the figure below and when $n = 2$, $k = 4$ is possible by colouring each cell in a unique colour.\n![](attached_image_1.png)\n\nWe consider the graph, where each node represents a colour and two nodes are linked, if the colours they represent touch. This graph is connected and since each colour touches at most 2 colours every node has at most degree 2. This means that the graph is either one long chain or one big cycle.\n![](attached_image_2.png)\n\nWe now look at the case when $n$ is odd. Consider the cell in the center of the table. From this cell we can get to any other cell by passing through at most $n-1$ cells. Therefore from the node representing this cell, we can get to any node through at most $n-1$ edges. But if the graph has $2n$ or more nodes, then for every node there is a node which is more than $n-1$ edges away. So we must have $k \\le 2n-1$ for all odd $n$.\n\nWhen $n$ is even we consider the 4 center cells. If they all have a different colour, then they form a 4-cycle in the graph, meaning the graph has only 4 nodes. If two of the center cells have the same colour, then from this colour you will be able to get to all other cells passing through at most $n-1$ cells. By same the arguments as in the odd case, we get $k \\le \\max(2n-1, 4)$ for even $n$.\n\nSo overall we have $k \\le 2n-1$ for $n \\ne 2$ and $k \\le 4$ for $n = 2$ as desired.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 25142,
    "subject": "Mathematics (Multi-modal)",
    "question": "The Euclidean plane is dissected into several disjoint *bounded* regions (not further subdivided) by $n$ intersecting equilateral triangles of side length $1$. Determine the smallest value of $n$ such that the number of parts thus arising can be at least $2023$.\n\nThe picture shows four intersecting triangles.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "27",
    "solution": "We will prove the following claim:\n**Claim:** For $n \\in \\mathbb{Z}^+$, the maximal number of bounded regions of the plane generated by $n$ triangles (of whatever size and shape) is $1 + 3n(n-1)$.\n\nOnce the claim is established, we can easily see that the inequality $1 + 3n(n-1) \\ge 2023$ is equivalent to $n \\ge 27$, so that we need at least $27$ triangles.\n\nBelow, we will establish that $27$ equilateral triangles of side length $1$ are actually enough.\n\n*Proof.* By induction. The case $n=1$ is trivial.\n\nSuppose that the claim is true for $n$. Let $C = \\{T_1, \\dots, T_n\\}$ be a configuration consisting of triangles $T_1, \\dots, T_n$. Let us see what happens when a new triangle $T$ is added to the configuration. We enumerate the points of intersection of $T$ with the elements of $C$, starting at some point of $T$, in clockwise direction; let the set of these points be $\\mathcal{P} := \\{P_1, \\dots, P_k\\}$. Clearly, each region newly arising through the addition of $T$ will have some part of its border in common with $T$; since the regions are not supposed to be further subdivided, the part of the border of a new region will be the section between two successive elements of $\\mathcal{P}$, where we also count $P_1$ as a successor of $P_k$. Each such line segment can at best generate one new region by either cutting an already bounded into two parts or by splitting off some bounded part of the unbounded region around $C$. It follows that the number of newly generated regions is at most $k$. Now, two triangles can have at most $6$ points of intersection, and $C$ consists of $n$ triangles, so that $k \\le 6n$. Consequently, the addition of $T$ can yield at most $6n$ new regions. By assumption, the number of regions prior to the addition of $T$ was at most $1 + 3n(n-1)$. Thus, after adding $T$, the number of regions is at most $1 + 3n(n-1) + 6n = 1 + 3n(n+1)$.\n\n(To see that two triangles can intersect in at most $6$ points, note that a triangle, being convex, can be intersected by a line in at most $2$ points, and each triangle consists of three line segments.) $\\square$\n\n*Proof that $27$ triangles suffice.* We prove by induction the more general claim that the upper bound $1 + 3n(n-1)$ for the number of bounded regions generated by $n$ triangles can actually be attained. Again, for the case $n=1$, this is trivial.\n\nWe now assume inductively that we have produced a configuration $C = \\{T_1, \\dots, T_n\\}$ of $n$ triangles such that:\n1. All elements of $C$ have the same orthocenter.\n2. Each two distinct elements of $C$ intersect in six points.\n3. No point lies on three elements of $C$.\n4. The number of bounded regions is $1 + 3n(n-1)$.\n\nWe will show that $C$ can be extended to a configuration with one extra triangle $T$ such that (1)-(3) still hold for the new configuration and such that the number of bounded regions is increased by $6n$.\n\nTo this end, consider rotations of $T_1$ around its orthocenter. There are only finitely many angles for which the respective rotation will pass through one of the finitely many points of intersection in $C$ (or such that the result will be identical to an element of $C$); thus (because there are infinitely many angles of rotation), there is an angle $\\alpha$ such that the result $T$ of the rotation of $T_1$ around the orthocenter by $\\alpha$ will not pass through any prior point of intersection. Thus (1) and (3) are satisfied for $C \\cup \\{T\\}$. It is easy to see that, rotating an equilateral triangle around its orthocenter will either yield an identical triangle or exactly six points of intersection; the first case is excluded by construction, thus (2) is satisfied as well.\n\nThus, $T$ has $6n$ points of intersection with the elements of $C$, let these be $P_1, \\dots, P_{6n}$. We need to show each polygonal line $P_iP_{i+1}$ (which is either a line segment or comprised of two line segments meeting at a vertex of $T$) generates a new bounded region; let us enumerate these segments as $s_1, \\dots, s_{6n}$. To see this, note that, for each such segment $s_k = P_kP_{k+1}$ (with $s_{6n} = P_{6n}P_1$), one of the following cases holds:\n1. $s_k$ runs through a region already bounded by $C$ and $s_1, \\dots, s_{k-1}$, splitting into two parts.\n2. $s_k$ runs through the unbounded region of $C$, splitting off a bounded part of it.\n3. $s_k$ runs through the unbounded region of $C$ without splitting off a bounded part of it.\n\nIn the cases (1) and (2), a new region is generated, while in case (3), this is not the case. But note that case (3) cannot occur: For if $P_k$ belongs to $T_i$ and $P_{k+1}$ belongs to $T_j$ then, by the case assumption, we have $i \\neq j$ and, by assumption (2) about $C$, $T_i$ and $T_j$ intersect, so that $s_k$ must generate a new bounded region.\n\nIt follows that $6n$ new bounded regions are added, so that we now have $1+3n(n-1) = 1+3(n+1)n$ regions, which finishes the induction. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25143,
    "subject": "Mathematics (Multi-modal)",
    "question": "For positive integers $n$, $k$ let $S(n, k)$ be the number of ways to partition a set with $n$ elements into $k$ non-empty parts. For example $S(3, 2) = 3$, because there are three ways to partition the set $\\{1, 2, 3\\}$ into two parts. For another example, $S(4, 2) = 7$.\nLet $p$ be a prime number, and $m$, $n$ positive integers such that $p-1 \\mid m-n$. Prove that $S(m, i) \\equiv S(n, i) \\pmod{p}$ for all $i = 1, 2, \\dots, p-1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\mathrm{Epi}(n, i)$ be the number of surjective functions from a set with $n$ elements to a set with $i$ elements. It is not hard to see that $S(n, i) = \\frac{\\mathrm{Epi}(n, i)}{i!}$. Indeed, a surjective function from $\\{1, \\dots, n\\}$ to $\\{1, \\dots, i\\}$ is the same thing as a partition of $\\{1, \\dots, n\\}$ into $i$ parts, together with a numbering of the blocks of the partition, from $1$ to $i$. For every partition there exists $i!$ numberings, and therefore $S(n, i) \\cdot i! = \\mathrm{Epi}(n, i)$.\n\nWhen $1 \\le i \\le p-1$, $\\gcd(p, i!) = 1$. It follows that for $1 \\le i \\le p-1$, $S(n, i) \\equiv S(m, i) \\pmod{p}$ if and only if $\\mathrm{Epi}(n, i) \\equiv \\mathrm{Epi}(m, i) \\pmod{p}$. So it is enough to prove that if $p-1 \\mid m-n$ then $\\mathrm{Epi}(n, i) \\equiv \\mathrm{Epi}(m, i) \\pmod{p}$ for all $1 \\le i \\le p-1$.\n\nThere is a formula for $\\mathrm{Epi}(n, i)$:\n$$\n\\mathrm{Epi}(n, i) = \\sum_{j=0}^{i-1} (-1)^j (i-j)^n \\binom{i}{j}.\n$$\nThis formula is based on nothing more than the inclusion-exclusion principle. The number of surjective functions is the number of all functions $-i$ times the number of functions that miss one particular element $+ \\binom{i}{2}$ times the number of functions that miss two particular elements, etc. The number of functions that miss $j$ specified elements is $(i-j)^n$.\n\nNow suppose $m$, $n$ are positive integers satisfying $p-1 \\mid m-n$. Then $a^m \\equiv a^n \\pmod{p}$ for all integers $a$. When $a$ is not divisible by $p$ this follows from Fermat's little theorem. When $a$ is divisible by $p$, both sides are zero mod $p$. But we only need the cases when $a = 1, \\dots, p-1$ anyway.\n\nIt follows that if $p-1 \\mid m-n$ then $\\mathrm{Epi}(n,i) \\equiv \\mathrm{Epi}(m,i) \\pmod{p}$ for all $1 \\le i \\le p-1$, and we are done.\n\n\nLemma:\nFor every $i > 1$ and $n > i$:\n$$\nS(n,i) = i \\cdot S(n-1,i) + S(n-1,i-1)\n$$\n*Proof.* Select an element $x$. There are $S(n-1,i)$ ways to partition the remaining elements into $i$ different sets. By adding $x$ to any of them, we get a partition of the original set. All the partitions obtained this way are different and the only partitions that cannot be obtained this way are the ones where $x$ ends up alone. But in that case the $n-1$ other elements must form $i-1$ sets, yielding another $S(n-1,i-1)$ partitions. $\\square$\n\nLemma:\nFor every $p$ prime, $p > i \\ge 1$ and $\\lambda \\ge 0$:\n$$\nS(i + \\lambda(p-1), i) = 1 \\pmod{p}.\n$$\n*Proof.* As $S(n,1) = 1$ and $S(i,i) = 1$, this is obvious for $i = 1$ or $\\lambda = 0$. We proceed by induction over $i$ and $\\lambda$. Using the induction hypothesis and the previous lemma:\n$$\n\\begin{aligned}\nS(i + \\lambda(p-1), i) &= i \\cdot S(i-1 + \\lambda(p-1), i) + S(i-1 + \\lambda(p-1), i-1) \\\\\n&= i \\cdot S(i-1 + \\lambda(p-1), i) + 1 \\quad \\text{mod } p\n\\end{aligned}\n$$\nOtherwise applying this $p-1$ times gives\n$$\n\\begin{aligned}\nS(i + \\lambda(p-1), i) &= i^{p-1} \\cdot S(i + (\\lambda - 1)(p-1), i) + \\sum_{j=0}^{p-2} i^j \\quad \\text{mod } p \\\\\n&= i^{p-1} \\cdot 1 + \\frac{i^{p-1} - 1}{i-1} \\quad \\text{mod } p \\\\\n&= 1 \\cdot 1 + 0 \\quad \\text{mod } p.\n\\end{aligned}\n$$\nWhere the last equality holds by Fermat's little theorem as $p > i > 1$ implies $i \\neq 0, 1 \\pmod{p}$. $\\square$\n\nWe can now proceed to proving the problem statement by induction: If $i=1$ the statement is obvious. If $m=i$ or $n=i$, we have the statement of the second lemma. So assume $i>1$ and $n>i$, $m>i$ and that the problem statement holds for all smaller values. Then\n$$\n\\begin{aligned}\nS(n,i) - S(m,i) &= i \\cdot (S(n-1,i) - S(m-1,i)) + \\\\\n&\\quad + S(n-1,i-1) - S(m-1,i-1) \\\\\n&= i \\cdot 0 + 0 \\quad \\text{mod } p \\\\\n&= 0 \\quad \\text{mod } p.\n\\end{aligned}\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25144,
    "subject": "Mathematics (Multi-modal)",
    "question": "Fifty marbles are lying in a heap. Alfred and Bodil take turns in removing a positive number of marbles, Alfred starts and at each step the number of marbles removed needs to be a prime or a square. The winner is the one who empties the heap. Who wins if both play in an optimal way?",
    "options": [],
    "answer": "Bodil",
    "solution": "We make a table inductively that shows whether the player, whose turn it is if there are $n$ marbles left, wins (marked with a W) or loses (marked with an L).\n\n| $n$ | 1<br>W<br>-1 | 2<br>W<br>-2 | 3<br>W<br>-3 | 4<br>W<br>-4 | 5<br>W<br>-5 | 6<br>L | 7<br>W<br>-1 | 8<br>W<br>-2 | 9<br>W<br>-3 | 10<br>W<br>-4 |\n|---|---|---|---|---|---|---|---|---|---|---|\n| $n$ | 11<br>W<br>-5 | 12<br>L | 13<br>W<br>-1 | 14<br>W<br>-2 | 15<br>W<br>-3 | 16<br>W<br>-4 | 17<br>W<br>-5 | 18<br>L | 19<br>W<br>-1 | 20<br>W<br>-2 |\n| $n$ | 21<br>W<br>-3 | 22<br>W<br>-4 | 23<br>W<br>-5 | 24<br>L | 25<br>W<br>-1 | 26<br>W<br>-2 | 27<br>W<br>-3 | 28<br>W<br>-4 | 29<br>W<br>-5 | 30<br>L |\n| $n$ | 31<br>W<br>-1 | 32<br>W<br>-2 | 33<br>W<br>-3 | 34<br>W<br>-4 | 35<br>W<br>-5 | 36<br>W<br>-36 | 37<br>W<br>-7 | 38<br>L | 39<br>W<br>-1 | 40<br>W<br>-2 |\n| $n$ | 41<br>W<br>-3 | 42<br>W<br>-4 | 43<br>W<br>-5 | 44<br>L | 45<br>W<br>-1 | 46<br>W<br>-2 | 47<br>W<br>-3 | 48<br>W<br>-4 | 49<br>W<br>-5 | 50<br>L |\n\nThe proof of the correctness of the table is inductive. If there are at most 35 marbles left, a player cannot remove a number of marbles of the type $6k$ with $k \\in \\mathbb{Z}$, because the numbers 6, 12, 18, 24 and 30 are neither prime numbers nor squares. We can conclude inductively that, as long as $n \\le 35$, the number $n$ should be marked with an L if it is divisible by 6 (since a legal move will yield a number that is not divisible by 6), and it should be marked with a **W** if it is not divisible by 6 (since removing 1, 2, 3, 4 or 5 marbles yields to a number that is divisible by 6).\n\nThis argument is not valid for $n = 36$, as one can remove all 36 marbles in one step and thus win.\n\nThe number $n = 37$ is also winning, since one can remove 7 marbles to bring the opponent into a losing position with 30 marbles.\n\nWith $n = 38$ marbles left it is not possible to win, since one would need to remove $38 - 30 = 8$, $38 - 24 = 14$, $38 - 18 = 20$, $38 - 12 = 26$, $38 - 6 = 32$ or all 38 marbles to bring the opponent into a losing position, but the numbers 8, 14, 20, 26, 32 and 38 are neither prime numbers nor squares.\n\nThe next five numbers, 39, ..., 43 are winning positions, because a removal of 1, ..., 5 marbles leaves the opponent in the losing position with 38 marbles.\n\nHowever, $n = 44$ is again a losing position, because one would need to remove $44 - 38 = 6$, $44 - 30 = 14$, $44 - 24 = 20$, $44 - 18 = 26$, $44 - 12 = 32$, $44 - 6 = 38$ or all 44 marbles to force the opponent into a losing position, but these moves are not allowed.\n\nThe next five numbers, 45, ..., 49 are winning positions, because a removal of 1, ..., 5 marbles leaves the opponent in the losing position with 44 marbles.\n\nNow the number 50 is left and it is again a losing position, because Alfred would need to remove $50 - 44 = 6$, $50 - 38 = 12$, $50 - 30 = 20$, $50 - 24 = 26$, $50 - 18 = 32$, $50 - 12 = 38$, $50 - 6 = 44$ or all 50 marbles to bring Bodil into a losing position, but these moves are not allowed. Hence Bodil will win the game if she plays in an optimal way.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25145,
    "subject": "Mathematics (Multi-modal)",
    "question": "The Perseverance is NASA's Mars rover exploring this planet. Every day it starts in home base and then goes north, south, east or west. Every one kilometer the robot makes one $90^\\circ$ turn. Moreover, the robot doesn't want to check the same place twice during the day (except for the home base, which is always the starting and the ending point of its trip). What are the possible lengths of the robot's path?",
    "options": [],
    "answer": "All positive multiples of four",
    "solution": "Let's define the coordinate system with the origin point at the home base and vertical-horizontal axes. W.l.o.g. assume that the first move was east and the path had length of $n$. Then each odd move changed the $x$ coordinate of the robot by $1$ and each even move changed the $y$ coordinate by $1$.\n\nAt the end of the day both coordinates were equal to zero again, so there had to be an even number of odd and an even number of even moves. That implies that only $n$ divisible by $4$ can fulfill the conditions.\n\nFor $n = 4$ we have a square path. For $n = 8$ we had $4$ changes of $x$ coordinate and $4$ changes of $y$, so the whole path was inside some\n\nFor $n = 12$ there is a path in the shape of \"+\" with first $4$ moves like $(\\rightarrow, \\uparrow, \\rightarrow, \\uparrow)$. Now we can change the middle $(\\uparrow, \\rightarrow)$ sequence by $(\\downarrow, \\rightarrow, \\uparrow, \\rightarrow, \\uparrow, \\leftarrow)$. Thanks to this change the robot explored new territory south-east from the one before explored. We got $+4$ of length of the path. There we can do it again and again, reaching any length of $4k + 8$ for all $k \\in \\mathbb{Z}^+$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25146,
    "subject": "Mathematics (Multi-modal)",
    "question": "A robot moves along the lines of an infinitely large square grid. It starts at an intersection and turns $90^\\circ$ at each intersection it encounters. At some point it reaches its starting point without having visited any other point twice. What are the possible path lengths of the robot?",
    "options": [],
    "answer": "Exactly the integers divisible by four other than eight; that is, 4 and all integers at least 12 that are multiples of 4.",
    "solution": "*Solution:* See Solution of C2.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25147,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. In this problem, we consider labellings of the squares of a chessboard of size $n \\times n$ with the natural numbers from $1$ to $n^2$ such that every number is used exactly once. Given such a labelling, we say a positive integer is a *rook product* if it is the product of the labels of $n$ squares which have the property that if you place a rook on each of them, no two rooks will attack each other.\n(Two rooks are attacking each other, if and only if they are in the same row or column.)\n\na. Let $n = 8$. Determine whether there exists a labelling of an $8 \\times 8$ chessboard such that the following condition is fulfilled: The difference of any two rook products is always divisible by $65$.\n\nb. Let $n = 10$. Determine whether there exists a labelling of a $10 \\times 10$ chessboard such that the following condition is fulfilled: The difference of any two rook products is always divisible by $101$.",
    "options": [],
    "answer": "a. No. b. Yes.",
    "solution": "a. No, there is no such labelling.\nOn the contrary, we show that for every labelling there exist two rook products whose difference is not divisible by $65$. Suppose that an $8 \\times 8$ chessboard is labelled with the numbers $1, 2, \\ldots, 64$ such that no number is used twice.\nWe can construct a rook product that is divisible by $13$ by placing a rook on the square with the label $13$ and the other seven rooks non attackingly, but otherwise arbitrarily.\nWe can construct a rook product that is not divisible by $13$ as follows. Notice that only four labels are divisible by $13$, namely $13, 26, 39$, and $52$. These four labels are located in at most four rows; we denote the index set of these rows $R \\subseteq \\{1, 8\\}$. Similarly, there are at least four columns that do not contain any of these four labels; we denote the index set of these columns $C \\subseteq \\{1, 8\\}$. Since $|R| \\leq |C|$ it is possible to place non attacking rooks in rows $R$ using only columns from $C$. The remaining rooks are placed in the remaining rows non attackingly, but otherwise arbitrarily. The resulting rook product is not divisible by $13$ since the rooks avoid the squares whose labels are divisible by $13$.\n\nThe difference of the two rook products is not divisible by $13$, since one rook product is divisible by $13$ whereas the other one is not. Hence the difference is not divisible by $65$.\n\nb. Yes, there is such a labelling.\nFor $k \\in [0, 99]$ we define $a_k = 2^k \\pmod{101}$; in other words, $a_k$ is the remainder of $2^k$ when divided by $101$. Note that $a_k \\neq 0$ since no power of $2$ is divisible by $101$. Hence $1 \\leq a_k \\leq 100$ for all $k \\in [0, 99]$.\n\nSince $r < s$ and $s$ is the smallest positive integer with $2^s \\equiv 1 \\pmod{101}$, we must have $r = 0$. In other words, $100$ is divisible by $s$; in other words, $s$ is a divisor of $100$.\nWe claim that $s = 100$. If this was not the case, we would have $s|20$ or $s|50$, which implies that $2^{20} \\equiv 1 \\pmod{101}$ or $2^{50} \\equiv 1 \\pmod{101}$. However $2^{10} = 1024 \\equiv 14 \\pmod{101}$, so that $2^{20} \\equiv 14^2 \\equiv 196 \\equiv -6 \\not\\equiv 1 \\pmod{101}$ and $2^{50} \\equiv (2^{20})^2 \\cdot 2^{10} \\equiv (-6)^2 \\cdot 14 \\equiv 504 \\equiv -1 \\not\\equiv 1 \\pmod{101}$.\nNow assume that $k, l \\in [0, 99]$ are positive integers with $k > l$ and $a_k = a_l$. Then we have $2^k \\equiv 2^l \\pmod{101}$. We conclude that $0 \\equiv 2^k - 2^l \\equiv 2^l \\cdot (2^{k-l} - 1) \\pmod{101}$. Since $2^l$ and $101$ are coprime, it follows that $2^{k-l} - 1 \\equiv 0 \\pmod{101}$ and $2^{k-l} \\equiv 1 \\pmod{101}$. This cannot be true, since $k-l \\in [1, 99]$, but $s = 100$ is the smallest positive integer with $2^s \\equiv 1 \\pmod{101}$. Hence $a_k \\neq a_l$.\nWe conclude that the numbers $a_k$ with $k \\in [0, 99]$ are a hundred pairwise different numbers from the set $[1, 100]$, hence they are a permutation of the set $[1, 100]$ as it was required.\n**Definition:** Let $p$ be a prime. Consider an $n \\times n$-square of elements $a_{i,j} \\in \\mathbb{F}_p^*$ (for $i,j = 1, \\dots, n$), which are not necessarily distinct. We call it *rooky*, if all its rook-products are equal as elements in $\\mathbb{F}_p^*$.\nWe will provide a classification of all *rooky* squares. Of course, most of this is not necessary when writing down a solution to the given problem, but it may still be interesting...\n\n**Lemma:** A square is *rooky* if and only if for all $i$, $j$, $k$, $\\ell$:\n$$\na_{i,j} \\cdot a_{k,\\ell} = a_{i,\\ell} \\cdot a_{k,j} \\quad (1)\n$$\n*Proof.* If we swap the rows of two rooks and keep their columns, it turns one valid rook formation into another. When comparing their rook products, we can ignore all $n - 2$ values of rooks that were not moved. The remaining values are $a_{i,j} \\cdot a_{k,\\ell}$ resp. $a_{i,\\ell} \\cdot a_{k,j}$ for certain $i$, $j$, $k$, $\\ell$. This gives equality (1) for *rooky* squares.\nConversely assume that (1) holds. Then we have to compare two arbitrary rook products. But they can be transformed into each other by a sequence of several swaps of two rooks. Due to (1) the rook product does not change at any of these steps, so the rook products of the original configurations are the same as well. $\\Box$\n\n**Lemma:** A *rooky* square is uniquely determined by the elements of its first row and first column.\n*Proof.* Indeed the previous lemma implies that\n$$\na_{i,j} \\cdot a_{1,1} = a_{i,1} \\cdot a_{1,j}\n$$\nwhich determines $a_{i,j}$ uniquely because $a_{1,1}$ is a unit. $\\Box$\n\nOne can actually prove directly that the square obtained that way is *rooky*, but it is simpler to continue directly to\n\n**Proposition:** Let $\\lambda_i \\in \\mathbb{F}_p^*$ ($i = 1, \\dots, n$) and $\\mu_j \\in \\mathbb{F}_p^*$ ($j = 1, \\dots, n$) be arbitrary elements. Then the square with\n$$\na_{i,j} = \\lambda_i \\cdot \\mu_j\n$$\nis *rooky*. Moreover any *rooky* square can be obtained this way.\n\nLet us prove the converse: By the previous lemma, it suffices to find $\\lambda_i$s and $\\mu_j$s that recreate the values of the first row and column. For this simply set $\\lambda_i = a_{i,1}$ and $\\mu_j = \\frac{a_{1,j}}{a_{1,1}}$. $\\Box$\n\n**Proposition:** For any prime $p > n^2$, there exists a *rooky* square with only distinct elements.\n*Proof.* Choose any primitive root $\\alpha \\in \\mathbb{F}_p^*$. Then set $\\lambda_i = \\alpha^{i-1}$, $\\mu_j = \\alpha^{n \\cdot (j-1)}$ and $a_{i,j} = \\lambda_i \\cdot \\mu_j = \\alpha^{i-1+n \\cdot (j-1)}$. This provides indeed a *rooky* square. The values in the square are $\\alpha^0, \\alpha^1, \\dots, \\alpha^{n^2-1}$. As we have chosen a primitive root, these are all distinct. $\\Box$\n\nFor $n = 10$, $p = 101$ and $\\alpha = 2$, this reproduces exactly the construction given in the previous solution.\n\n*Proof.* The square with $a_{i,j} = \\lambda_i \\cdot \\mu_j$ is *rooky*, because any rook product has the value\n$$\n\\prod_i \\lambda_i \\cdot \\prod_j \\mu_j.\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25148,
    "subject": "Mathematics (Multi-modal)",
    "question": "(a) Let $n = 8$. Determine whether there exists a labelling of an $8 \\times 8$ grid such that the following condition is fulfilled: The difference of any two transversal products is always divisible by $65$.\n\n(b) Let $n = 10$. Determine whether there exists a labelling of a $10 \\times 10$ grid such that the following condition is fulfilled: The difference of any two transversal products is always divisible by $101$.",
    "options": [],
    "answer": "Yes for n = 8 (mod 65), and yes for n = 10 (mod 101).",
    "solution": "See solution of Problem C 3.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25149,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n > 1$ be an integer and let an $2 \\times n$ grid be given, where the cells are labelled as $(x, y)$ with $1 \\le x \\le 2$ and $1 \\le y \\le n$. A grasshopper lives on the grid and occasionally jumps from one cell to another. The *length* of a jump from a cell $(x_1, y_1)$ to $(x_2, y_2)$ is defined as $|x_1 - x_2| + |y_1 - y_2|$. Determine the largest possible sum of lengths of jumps the grasshopper can make when it starts from one cell and never jumps to a cell it has visited before.",
    "options": [],
    "answer": "n^2 + 2n - 3",
    "solution": "Answer: The solution is $n^2 + 2n - 3$.\n\nCombining the x-bound and the y-bound, we get in both cases the upper bound $n^2 + 2n - 2$. This can still be improved by observing that it is not possible to satisfy both bounds with equality. For $n = 2k + 1$, in order to get $n^2 - 1$ units in y-direction, the two end points have to be $(0, k+1)$ and $(1, k+1)$ (so that they are never in the smaller part). Without loss of generality, the first step goes from $(0, k+1)$ into the set $\\{(1, y) : y \\ge k+2\\}$. Then, if we always want to change the x-coordinate and also always move between $y \\ge k+2$ and $y \\le k+1$ (which we have to do in order to make the y-bound tight), we can only alternate between $\\{(1, y) : y \\ge k+2\\}$ and $\\{(0, y) : y \\le k\\}$. Similarly, for $n = 2k$, without loss of generality, the first step goes from $\\{(0, y) : y \\le k\\}$ to $\\{(0, y) : y \\ge k+1\\}$, and if we don't want to lose anything in x- or y-direction we can only alternate between these two sets.\n\n$$\n2k + 2 \\sum_{y=1}^{k} (n - 2y) - 1 + 2 \\sum_{y=k+2}^{n} (2y - n - 1) = 4k^2 + 4k - 1 = n^2 - 2,\n$$\nand for $n = 2k$,\n$$\nk + k + \\sum_{y=1}^{k-1} [(n - 2y) + (n - 2y + 1)] + \\sum_{y=k+2}^{n} [(2y - n - 2) + (2y - n - 1)] \\\\ = 4k^2 - 2 = n^2 - 2.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25150,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the land of Flensburg there is a single, infinitely long, street with houses numbered $2, 3, \\ldots$ The police in Flensburg is trying to catch a thief which every day moves from the house where he is currently hiding to one of its neighbouring houses.\nTo taunt the local law enforcement the thief reveals every day the highest prime divisor of the house he will move to.\nEvery Sunday the police are allowed to search a single house, and they catch the thief if they search the house he is currently occupying. Determine if the thief will be able to escape the police indefinitely or if the police has a strategy to catch the thief in finite time.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will prove that the police are always able to catch the thief in finite time.\nLet $h_i$ denote the house the thief stays at the $i$-th night and $p_i$ denote the greatest prime divisor of $h_i$.\nThe police know that he stays at different neighbouring houses every night, so $|h_{i+1} - h_i| = 1$ for all non-negative integers $i$. Let us assume that the police are given the address of the thief's first two hiding spots, then we will prove by induction that the police can determine $h_i$ precisely except being unable to distinguish between houses numbered $2$ and $4$.\n\nAssume the police knows $h_{i-2}$ and $h_{i-1}$, then they know that $h_i = h_{i-2}$ or $h_i = 2h_{i-1} - h_{i-2}$. In the first case they will receive $p_i = p_{i-2}$ and in the latter case they will receive $p_i$ as the biggest prime divisor of $2h_{i-1} - h_{i-2}$. Assume that they are unable to distinguish between these two cases, i.e., that $p_i = p_{i-2}$, which implies\n$$\np_{i-2} \\mid 2h_{i-1} - h_{i-2}, \\text{ i.e. } p_{i-2} \\mid 2h_{i-1}, \\text{ i.e. } p_{i-2} \\mid 2, \\text{ i.e. } p_{i-2} = 2\n$$\nsince $|h_{i-1} - h_{i-2}| = 1$ implies $\\gcd(h_{i-1}, h_{i-2}) = 1$. Moreover, since $p_i = p_{i-2} = 2$ are the biggest prime divisors of $h_i = 2h_{i-1} - h_{i-2}$ and $h_{i-2}$ they must both be powers of $2$. However, the only powers of two with a difference of exactly $2$ are $2$ and $4$. Hence $\\{h_{i-2}, 2h_{i-1} - h_{i-2}\\} = \\{2, 4\\}$, i.e. $h_{i-1} = \\frac{2+4}{2} = 3$.\n\nThus, either the police will with certainty be able to determine $h_i$ or $h_{i-1} = 3$, in which case $h_i$ may equal either $2$ or $4$. To complete the inductive step we observe that the police are always able to determine the parity of $h_j$, since it changes every day. Thus, in the future if the police know that $h_j \\in [2, 4]$, then they can either determine $h_j = 3$ or $h_j \\in \\{2, 4\\}$. However, the only way for the thief to leave the interval $[2, 4]$ is to go to house number $5$, in which case the police will be alerted by receiving $p_j = 5$, and they can again with certainty determine $h_j = 5$ and $h_{j-1} = 4$ preserving our inductive hypothesis.\n\nTo summarize, if the police knows both $h_0$ and $h_1$, then they can always determine $h_i$ with certainty until $h_{i-1} = 3$. After this point they will have known the two last hiding places of the thief if he leaves the interval $[2, 4]$, restoring the inductive hypothesis, or otherwise, if he never leaves $[2, 4]$ be able to determine his position, up to confusion about $2$ and $4$ using the parity of the day.\n\nNow, to catch the thief in finite time, they may methodically try to guess all viable pairs of $(h_0, h_1)$, i.e. $h_0, h_1 \\in \\mathbb{N}_{\\ge 2}$ and $|h_0 - h_1| = 1$, of which there are countably many.\nFor each viable starting position, let us consider either the immediate Sunday or the one after that, since each week has an odd amount of days, we are certain that exactly one of these days gives us that the thief is hiding in an odd house (given our assumption on his starting position). Thus, due to our inductive hypothesis, we can precisely determine where the thief will be, and search this house.\nIf the thief is hiding in that house, the police wins, and if not, they will with certainty know that their guess of starting positions was incorrect, and move onto the next guess. By the above argument, each guess of initial starting positions requires at most two weeks, meaning that the police will catch the thief in finite time.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25151,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the least positive integer $n$ for which it is possible to draw an $n$-gon along the lines of a square grid, whose perimeter is $n$ and whose area is $n$. Here the boundary of the $n$-gon may not visit any point more than once.",
    "options": [],
    "answer": "32",
    "solution": "Answer: $n = 32$. We start by showing that such a $32$-gon exists.\nIt remains to prove that this is indeed the smallest possible $n$. We will do that by proving that for all $n < 32$ the area $S$ of an $n$-gon with perimeter $P = n$ is smaller than $n$.\n\nAt first we note that $P = n$ if and only if all sides of the $n$-gon are of length $1$. From this it follows that $n$ should be an even number as there should be an even number of horizontal and an even number of vertical sides. Denote $n = 2m$ and the vertices of the $n$-gon as $A_1A_2\\dots A_{2m}$. We call a vertex $A_i$ odd (even) if its index $i$ is odd (even).\n\nNow we can consider the sides of the $n$-gon in pairs. Let vertex $A_i$ be located at the point $(x, y)$. Vertex $A_{i+2}$ is then located at $(x\\pm1, y\\pm1)$. It means that by moving from $A_i$ to $A_{i+2}$ (we count vertices modulo $2m$) we have moved by a unit vector in the “diagonal” coordinate system with unit vectors $(1, 1)$ and $(1, -1)$. We denote these steps $\\rightarrow$ and $\\leftarrow$, respectively.\n\nLet's start at the vertex $A_1$ and travel around the $n$-gon in $m$ steps by stopping at all odd vertices $A_1 \\to A_3 \\to \\dots \\to A_{2m-1} \\to A_1$. Denote the total number of $\\rightarrow$ steps as $2a$ and the total number of $\\leftarrow$ steps as $2b$ (both of them must be even numbers). Therefore $m = 2a + 2b$ and $n = 4a + 4b$, hence $n$ is divisible by $4$. It means that we have to prove that $S < n$ for all $P = n \\le 28$.\n\nLet's consider the most “northeastern”, “northwestern”, “southeastern” and “southwestern” vertices of our $n$-gon, denote them $NE, NW, SE, SW$ respectively (see figure). W.l.o.g. assume that $NE$ and at least one more of them are odd. Our $n$-gon is contained in a rectangle whose sides are parallel to vectors $(1, 1)$ and $(1, -1)$ and pass through these four vertices. Denote its width and height as $W$ and $H$. Our $n$-gon's area is no larger than $W \\cdot H - S_g$ where $S_g$ is the area of the “gray triangles”. It can be computed as half of the area of the “square ring” between outer and inner (gray) rectangles:\n$$\nS_g = \\frac{WH - (W - \\sqrt{2})(H - \\sqrt{2})}{2} = \\frac{\\sqrt{2}(W + H)}{2} - 1\n$$\ntherefore the area of the $n$-gon is no larger than $S \\le W H - \\frac{\\sqrt{2}(W+H)}{2} + 1$. By using the inequality $WH \\le \\frac{(W+H)^2}{4}$ we can transform it into\n$$\nS \\le \\frac{(W+H)^2}{4} - \\frac{\\sqrt{2}(W+H)}{2} + 1 = \\frac{(W+H-\\sqrt{2})^2}{4} + \\frac{1}{2}. \\quad (*)\n$$\nRecall that the $NE$ vertex is odd. Assume at first that $SW$ also is odd. Then as travelling from $NE$ to $SW$ uses at most $b$ $\\nearrow$ steps, then we conclude that $W \\le b\\sqrt{2}$. If the vertex $SW$ is even, then the estimation is slightly worse: we can travel in at most $b$ $\\nearrow$ steps from $NE$ to one of the neighbours of $SW$ and $SW$ itself is further $\\sqrt{2}/2$ away from it in $\\nearrow$ direction, so that $W \\le b\\sqrt{2} + \\sqrt{2}/2$.\n\nIf $SW$ is even then at least one of vertices $NW$ and $SE$ is odd. In that case for $H$ we get similar estimation $H \\le a\\sqrt{2} + \\sqrt{2}/2$. But if $SW$ is odd then it could happen that both $NW$ and $SE$ are even, in which case we have to do the $\\sqrt{2}/2$ correction in both ends what gives us the estimation $H \\le a\\sqrt{2} + \\sqrt{2}$. To summarize, we have two possible cases:\n$$\n\\begin{cases} W \\le b\\sqrt{2} \\\\ H \\le a\\sqrt{2} + \\sqrt{2} \\end{cases} \\quad \\text{or} \\quad \\begin{cases} W \\le b\\sqrt{2} + \\frac{\\sqrt{2}}{2} \\\\ H \\le a\\sqrt{2} + \\frac{\\sqrt{2}}{2} \\end{cases}\n$$\nWe see that in both cases $W + H \\le a\\sqrt{2} + b\\sqrt{2} + \\sqrt{2}$.\nNow we can plug that into $(*)$ to obtain that\n$$\nS \\le \\frac{(a+b)^2}{2} + \\frac{1}{2}.\n$$\nRecall that the perimeter of our $n$-gon is $4a+4b$. It remains to prove that if $4a+4b \\le 28$ then $S < P$ that is $\\frac{(a+b)^2}{2} + \\frac{1}{2} < 4a+4b$ if $a+b \\le 7$. But this inequality can be rewritten as $(a+b)(a+b-8) < -1$ which is obviously true if $a+b \\le 7$.\nThe parity arguments in the given solution can be combined: A grid-$n$-gon with perimeter $n$ alternates between horizontal and vertical unit segments, so the number of horizontal segments equals the number of vertical segments. As this common number must be even (#left steps = #right steps, #up steps = #down steps), we have that $n$ is divisible by $4$.\n\nGiven the construction for $n=32$, it remains to be shown that it can't be done for $n \\le 28$. Viewing the polygon as a collection of unit squares and turning this into a graph by joining two squares that share an edge, this is a variant of the isoperimetric problem in the grid graph with an additional constraint coming from the prescribed number of polygon vertices. Without this constraint we have the following bound.\n\nLemma 1: For any polygon $P$ with axis-parallel sides, vertices in $\\mathbb{Z}^2$, area $A$ and perimeter $u$, we have $u \\ge 2(a+b)$ where\n$$\na = \\max\\{x \\mid (x, y) \\in P\\} - \\min\\{x \\mid (x, y) \\in P\\}, \\text{ and}\n$$\n$$\nb = \\max\\{y \\mid (x, y) \\in P\\} - \\min\\{y \\mid (x, y) \\in P\\}.\n$$\nProof. Without loss of generality we have $\\min\\{x \\mid (x, y) \\in P\\} = \\min\\{y \\mid (x, y) \\in P\\} = 0$. The boundary contains at least $2a$ segments of the form $((x, y), (x + 1, y))$ and at least $2b$ segments of the form $((x, y), (x, y + 1))$. $\\square$\n\nThis bound itself is not good enough to rule out all relevant values for $n$, but we can add the following. In the $n$-gon with area $n$ and perimeter $n$, consider the unit squares that are furthest north, east, west and south. Since horizontal and vertical segments on the boundary of the polygon alternate, these are four distinct squares, and they correspond to vertices of degree $1$ in the corresponding subgraph of the grid. Deleting these four squares, we obtain an $n'$-gon with area $n-4$, perimeter $n-8$ and $n' \\ge n-16$. Now Lemma 1 rules out $n \\in \\{4, 8, 12, 16, 20, 24\\}$ immediately. For instance for $n=24$, we need area $20$ and perimeter $16$, but Lemma 1 provides a lower bound of $18$ for the perimeter. For $n=28$, the reduced polygon has area $24$ and perimeter $20$. This can be achieved, but only with a $6 \\times 4$-rectangle or a $5 \\times 5$-square with a corner unit square removed while we would need at least $12$ vertices.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25152,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine if there exists a triangle that can be cut into 101 congruent triangles.",
    "options": [],
    "answer": "Yes",
    "solution": "Answer: Yes, there is.\nChoose an arbitrary positive integer $m$ and draw a height in the right triangle with ratio of legs $1 : m$. This height cuts the triangle in two similar triangles with similarity coefficient $m$. The largest of them can further be cut into $m^2$ smaller equal triangles by splitting all sides in $m$ equal parts and connecting corresponding points with parallel lines. Thus a triangle can be split into $m^2 + 1$ equal triangles.\n\nThe figure shows this for $m = 4$, but in our problem we must take $m = 10$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25153,
    "subject": "Mathematics (Multi-modal)",
    "question": "Some points are marked on a circle. Each marked point is coloured red, green or blue. In one step, one can erase two marked points of different colours that have no marked points between them and mark a new point between the locations of the erased points with the third colour. In a final state, all marked points have the same colour which is called the colour of the final state. Find all positive integers $n$ for which there exists an initial state of $n$ marked points with one missing colour, from which one can reach a final state of any of the three colours by applying a suitable sequence of steps.",
    "options": [],
    "answer": "All even integers greater than 2",
    "solution": "Answer: All even numbers $n$ greater than 2.\n\nIf $n = 2$ then the colour of the final state is uniquely determined. We show now that required initial states are impossible for odd $n$. Note that if one colour is missing then the numbers of marked points of existing two colours have different parities, i.e., the difference of these numbers is odd. Each step keeps the parity of the difference of the numbers of marked points of these two colours unchanged. Hence in every intermediate state and also in the final state, one of these two colours is represented. Consequently, a final state of the third colour is impossible.\n\nFor every even number $n > 2$, an initial state with 2 consecutive points marked with one colour and $n-2$ points marked with another colour satisfies the conditions of the problem. Indeed, if $n > 4$ then with two symmetric steps, one can reach a similar state where the number of points marked with the more popular colour is 2 less. Hence it suffices to solve the case $n = 4$. In this case, making one step leads to a state with 3 marked points, all with different colours. In order to obtain a final state of any given colour, one can replace points of the other two colours with a new point of the given colour. This completes the solution.\nDefinition: Call a configuration *colourful*, if the final state may have any of the three colours.\n\nThe cases $n = 2$ and $n$ odd are excluded as in the first solution, so let $n > 2$ be even. To construct *colourful* configurations, we consider linear configurations, i.e. one where the points are placed on a line instead of a circle. There is only difference to the circular situation: We may not choose the two end points for the replacement step. So it suffices to construct linear *colourful* configurations.\n\nWe start by providing explicit examples for $n = 4$ and $n = 6$ (with the bold letters being replaced):\n$$\n\\begin{align*}\n\\mathbf{RGRG} &\\rightarrow \\mathbf{BRG} \\rightarrow \\mathbf{BB} \\\\\n\\mathbf{RGRG} &\\rightarrow \\mathbf{RBG} \\rightarrow \\mathbf{RR} \\\\\n\\mathbf{RGRG} &\\rightarrow \\mathbf{RBG} \\rightarrow \\mathbf{GG} \\\\\n\\mathbf{RGRRGR} &\\rightarrow \\mathbf{BRRGR} \\rightarrow \\mathbf{BRRB} \\rightarrow \\mathbf{BRG} \\rightarrow \\mathbf{BB} \\\\\n\\mathbf{RGRRGR} &\\rightarrow \\mathbf{RBRGR} \\rightarrow \\mathbf{RBBR} \\rightarrow \\mathbf{GBR} \\rightarrow \\mathbf{RR} \\\\\n\\mathbf{RGRRGR} &\\rightarrow \\mathbf{RBRGR} \\rightarrow \\mathbf{RBBR} \\rightarrow \\mathbf{GBR} \\rightarrow \\mathbf{GG}.\n\\end{align*}\n$$\n\nNext observe that the concatenation of several linear *colourful* configurations is again *colourful*: Indeed, each part can be transformed\n\n$$\n|R| \\equiv |G| \\equiv |B| \\pmod{2}\n$$\nand it contains at least two colours.\n\n*Proof.* We have already seen in the solution above that $|R| - |G| \\pmod 2$, $|G| - |B| \\pmod 2$ and $|B| - |R| \\pmod 2$ are invariants. Moreover it is obvious that we need at least two colours to be able to do anything. So the conditions are necessary.\n\nWe prove that they are sufficient: For $n = 1$ and $n = 2$ there is nothing to do. For $n = 3$ the conditions require $|R| = |G| = |B| = 1$ and the configuration indeed *colourful*. We continue by induction for $n > 3$: As $n > 3$, there is at least one colour with more than one point, so assume wlog. $|R| > 1$. Having at least two colours, we can find a pair of two different colours, one of which is red. Assume w.l.o.g. that the other is green. As a first step replace these two points. The resulting configuration has $|R| - 1$ red, $|G| - 1$ green and $|B| + 1$ blue points, so it satisfies $|R| - 1 = |G| - 1 = |B| + 1 \\pmod 2$. Moreover due to $|R| > 1$ is has at least one red and one blue point. So by induction the configuration is *colourful*, and hence so was our original state. □\n\n*Proof.* Immediate.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25154,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $k$ be a non-negative integer and let $A$ be a nonempty finite set. We define a word from the alphabet $A$ of length $k$ as a sequence of elements of $A$ of length $k$. (Notice that by definition the empty sequence is a word of length $0$.)\n\n* Let $B$ be the set of all words from the nine-element alphabet $\\{b, @, l, t, i, c, w, a, y\\}$, whose lengths are at most $2023$.\n* Moreover, let $M$ be the set of all words from the four-element alphabet $\\{m, a, t, h\\}$, whose lengths are at most $2025$ and which contain the element $t$ exactly twice.\n(For example, $tata \\in M$, because it has length $4$ and contains the letter $t$ twice, but $mahatma \\notin M$, because it contains the letter $t$ only once.)\n\nProve that the difference $|B| - |M|$ is divisible by $3^{2023}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For $k \\in \\mathbb{N}_0$ we denote by $b_k$ the number of words from the alphabet $\\{b, @, l, t, i, c, w, a, y\\}$ of maximum length $k$, so that $b_{2023} = |B|$. Furthermore, we denote by $m_k$ the number of words of maximum length $k+2$ from the alphabet $\\{m, a, t, h\\}$, that contain the letter $t$ exactly twice, so that $m_{2023} = |M|$.\nFor example, we have $b_0 = 1$ (coming from the empty word), $b_1 = b_0 + 9 = 10$ (adding $9$ one-letter words) and $b_2 = b_1 + 9^2 = 91$ (adding $9^2$ two-letter words). More generally, we have\n$$\nb_k = b_{k-1} + 9^k \\quad (*)\n$$\nfor all $k \\ge 1$, since we can form exactly $9^k$ words of length $k$ from the given alphabet. The recursion\n$$\nb_k = 1 + 9b_{k-1} \\quad (**)\n$$\nis also true for all $k \\ge 1$, since every word of maximum length $k$ from the given alphabet is either the empty word (1 case) or the concatenation of a word of maximum length $k-1$ and one of the nine letters from the given alphabet ($9b_{k-1}$ cases).\n\nOn the other hand, we have $m_0 = 1$ (coming from the word $tt$), $m_1 = m_0 + 9 = 10$ (adding $9$ words of the form $tx$, $txt$ and $xtt$ with $x \\in \\{h, a, m\\}$) and $m_2 = m_1 + 54 = 64$ (adding $6 \\cdot 9 = 54$ words of the form $ttxy$, $txty$, $txyt$, $xtty$, $xtyt$ and $xytt$ with $x, y \\in \\{h, a, m\\}$).\n\n**Lemma.** The recursion\n$$\nm_{k+2} = 9m_{k+1} - 27m_k + 27m_{k-1} + 1\n$$\nis true for $k \\ge 1$.\n\n*Proof of the lemma.* We will consider two further sequences, first the sequence $(p_k)_{k \\in \\mathbb{N}_0}$, where $p_k$ is the number of words of maximum length $k+2$ from the alphabet $\\{m, a, t, h\\}$, that contain the letter $t$ exactly once. We will also consider the sequence $(q_k)_{k \\in \\mathbb{N}_0}$, where $q_k$ is the number of words of maximum length $k+2$ from the alphabet $\\{m, a, t, h\\}$, that do not use the letter $t$ at all (i.e. words of maximum length $k+2$ from the alphabet $\\{h, a, m\\}$).\n\nLet $k \\ge 1$ for the moment. A word of maximum length $k+2$ from the alphabet $\\{m, a, t, h\\}$, that contains the letter $t$ exactly twice, is either\n* the concatenation of the letter $t$ with a word of maximum length $k+1$ with exactly one occurrence of the letter $t$ (of which there are $p_{k-1}$ cases) or\n* the concatenation of a letter $x \\in \\{h, a, m\\}$ with a word of maximum length $k+1$ with exactly two occurrences of the letter $t$ (of which there are $3 \\cdot m_{k-1}$ cases).\n\nWe conclude that\n$$\nm_k = p_{k-1} + 3m_{k-1}. \\qquad (***)\n$$\nIn a similar spirit, we obtain\n$$\np_k = q_{k-1} + 3p_{k-1} \\qquad (****)\n$$\nand\n$$\nq_k = 1 + 3q_{k-1}. \\qquad (†)\n$$\nEquation (***) tells us that $p_{k-1} = m_k - 3m_{k-1}$, from which we obtain $p_k = m_{k+1} - 3m_k$ by shifting indices. We plug both expressions into equation (****) and obtain\n$$\n\\begin{align*}\nm_{k+1} - 3m_k &= q_{k-1} + 3(m_k - 3m_{k-1}) && (††) \\\\\n\\Leftrightarrow q_{k-1} &= m_{k+1} - 6m_k + 9m_{k-1}.\n\\end{align*}\n$$\nPlugging equation (††) and its index shift into equation (†) yields the recursion from the lemma. $\\square$\n\nThanks to the lemma. By induction hypothesis, there exist natural numbers $r, s, t \\in \\mathbb{N}$ such that $m_{k-1} = b_{k-1} + 3^{k-1}r$, $m_{k-2} = b_{k-2} + 3^{k-2}s$ and $m_{k-3} = b_{k-3} + 3^{k-3}t$. It follows that\n$$\nm_k \\equiv 9b_{k-1} - 27b_{k-2} + 27b_{k-3} + 1 \\pmod{3^k},\n$$\nfrom which we conclude\n$$\n\\begin{align*}\nm_k - b_k &\\equiv -b_k + 9b_{k-1} - 27b_{k-2} + 27b_{k-3} + 1 \\\\\n&\\equiv -b_k + 9b_{k-1} - 27(b_{k-1} - 9^{k-1}) + 27(b_{k-1} - 9^{k-1} - 9^{k-2}) + 1 \\\\\n&\\equiv -b_k + 9b_{k-1} + 1 \\pmod{3^k}.\n\\end{align*}\n$$\nthanks to equation (*). By equation (**) we have $m_k - b_k \\equiv 0 \\pmod{3^k}$, which finishes the induction step.\n\nWe now show that $b_k \\equiv m_k \\pmod{3^k}$ for all $k \\in \\mathbb{N}_{\\ge 0}$, which yields the desired statement when we put $k = 2023$.\nWe are going to show our claim by strong mathematical induction on $k$. The claim is true for $k=0$, because the difference $b_0 - m_0 = 0 - 0 = 0$ is divisible by $3^0 = 1$. It is true for $k=1$, because the difference $b_1 - m_1 = 10 - 10 = 0$ is divisible by $3^1 = 3$. It is true for $k=2$, because the difference $b_2 - m_2 = 91 - 64 = 27$ is divisible by $3^2 = 9$.\nLet $k = 2023$. Since there are exactly $9^i$ words of length $i$ in the set $B$ for every $i \\in [0, k]$, we have\n$$\n|B| = \\sum_{i=0}^{k} 9^i = \\frac{9^{k+1} - 1}{9 - 1} = \\frac{1}{8} (9^{k+1} - 1)\n$$\nthanks to the formula for the geometric series. In particular, multiplication by $16$ brings us to the congruence $16|B| \\equiv -2 \\pmod{3^k}$, since $2 \\cdot 9^{k+1}$ is divisible by $3^k$.\n\nMoreover, every word in $M$ is bounded by length by $k+2$. If $i$ parametrises the number of occurrences of letters different from $t$ in a word in $M$, then the word's length is $i+2$ and there are $\\binom{i+2}{2}$ possibilities to choose two places for the two letters $t$. The other $i$ letters can be arbitrary elements from the set $\\{h, a, m\\}$. It follows that\n$$\n|M| = \\sum_{i=0}^{k} \\binom{i+2}{2} 3^i.\n$$\nDefine a function $f:\\mathbb{R} \\setminus \\{1\\} \\to \\mathbb{R}$ by\n$$\nf(x) = \\sum_{i=0}^{k} x^{i+2} = \\frac{x^{k+3} - 1}{x - 1} - x - 1\n$$\nfor all $x$. Its first derivative is given by\n$$\nf'(x) = \\sum_{i=0}^{k} (i+2)x^{i+1} = \\frac{(k+2)x^{k+3} - (k+3)x^{k+2} + 1}{(x-1)^2} - 1;\n$$\nits second derivative $f''(x) = \\sum_{i=0}^{k} (i+2)(i+1)x^i$ is given by\n$$\n\\frac{(k+2)(k+1)x^{k+3} - 2(k+3)(k+1)x^{k+2} + (k+3)(k+2)x^{k+1} - 2}{(x-1)^3}.\n$$\nIf we plug in $x = 3$ in the last equation, every summand in the numerator of the previous fraction is divisible by $3^k$ except for $-2$. It follows that $-2 \\equiv (3-1)^3 \\cdot f''(3) = 8 \\cdot 2|M| \\pmod{3^k}$.\nThe congruence $16|M| \\equiv -2 \\equiv 16|B| \\pmod{3^k}$ means that $16(|B| - |M|)$ is divisible by $3^k$. Since $16$ is coprime to $3^k$, the difference $|B| - |M|$ is also divisible by $3^k = 3^{2023}$.\nWe consider the 3-adic expansion of $x := -\\frac{1}{2} = \\frac{1}{1-3} = \\sum_{k=0}^{\\infty} 3^k$ and $y := -\\frac{1}{8} = \\frac{1}{1-9} = \\sum_{\\ell=0}^{\\infty} 3^{2\\ell}$. Observe that $y = x^3$ and, therefore,\n$$\n\\sum_{\\ell=0}^{\\infty} 3^{2\\ell} = \\left( \\sum_{k=0}^{\\infty} 1 \\cdot 3^k \\right)^3 = \\sum_{k=0}^{\\infty} a_k \\cdot 3^k,\n$$\nwhere $a_k$ is the number of presentations of $k$ as an ordered sum of three summands. Since we can place between $k$ objects two “plus” symbols in $\\binom{k+2}{2}$ ways and interpret the number of objects before the first plus as the first summand, between the two pluses as the second and after the second plus sign as the third summand, we have $a_k = \\binom{k+2}{2}$. Hence, we get\n$$\n\\sum_{\\ell=0}^{\\infty} 3^{2\\ell} = \\sum_{k=0}^{\\infty} \\binom{k+2}{2} \\cdot 3^k.\n$$\nReducing this 3-adic identity modulo $3^{2023}$ gives\n$$\n|B| \\equiv |M| \\pmod{3^{2023}}.\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25155,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be a triangle and let $J$ be the center of the excircle opposite to $A$. The reflection of $J$ in $BC$ is $K$. $E$ and $F$ are on $BJ$ and $CJ$, respectively, such that $\\angle EAB = \\angle CAF = 90^\\circ$. Prove that $\\angle FKE + \\angle FJE = 180^\\circ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\nLet $JK$ intersect $BC$ at $X$. We'll prove a key claim:\n\n**Claim:** $\\triangle BEK$ is similar to $\\triangle BAX$.\n\n*Proof.* Note that $\\angle EAB = 90^\\circ = \\angle KXB$. Also, since $BJ$ bisects $\\angle CBA$, we get $\\angle ABE = \\angle JBX = \\angle XBK$. Hence $\\triangle EBA \\sim \\triangle KBX$. From that, we see that the spiral similarity that sends the line segment $EA$ to $KX$ has center $B$. So the spiral similarity that sends the line segment $EK$ to $AX$ has center $B$. Thus $\\triangle BEK \\sim BAX$. $\\square$\n\nIn a similar manner, we get $\\triangle CFK$ is similar to $\\triangle CAX$.\n\n---\n\n$$\n\\begin{align*} \n\\angle FKE + \\angle FJE &= \\angle FKE + \\angle BKC \\\\ \n&= 360^\\circ - \\angle EKB - \\angle CKF \\\\ \n&= 360^\\circ - \\angle AXB - \\angle CXA \\\\ \n&= 360^\\circ - 180^\\circ \\\\ \n&= 180^\\circ \n\\end{align*}\n$$\nas desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25156,
    "subject": "Mathematics (Multi-modal)",
    "question": "In an acute triangle $\\triangle ABC$ with $|AB| \\neq |AC|$, the angle bisector of $\\angle BAC$ intersects side $BC$ and $\\odot(ABC)$ at the points $D$ and $M_A$, respectively. Let points $X$ and $Y$ be the feet of perpendiculars from $M_A$ to sides $AB$ and $AC$, respectively. The tangent of $\\odot(BXM_A)$ at the point $X$ and the tangent of $\\odot(CYM_A)$ at the point $Y$ intersect at the point $T$. Suppose that lines $AT$ and $BC$ intersect at the point $S$. Show that $\\odot(TSM_A)$ passes through the midpoint of segment $AD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Therefore, quadrilateral $M_AXTY$ is cyclic as well. Combining this with $AXM_AY$ being cyclic, gives us that $M_AXTAY$ is cyclic. Moreover, note that\n$$\n\\angle TAM_A = \\angle TYM_A = \\angle YCM_A = \\angle ACM_A.\n$$\nThis means that $AT$ is tangent to $\\odot(ABC)$. Also note that $\\angle STM_A = 180^\\circ - \\angle M_ATA = 90^\\circ$.\n\nNow let $Z$ be the midpoint of $AD$, then $\\angle SZM_A = 90^\\circ$, since $\\triangle SAD$ is an isosceles triangle. Combining this with $\\angle STM_A = \\angle SZM_A = 90^\\circ$, we have that $STZM_A$ is a cyclic quadrilateral, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25157,
    "subject": "Mathematics (Multi-modal)",
    "question": "In an acute scalene triangle $\\triangle ABC$ points $P, Q$ lie on the smaller arcs $\\widehat{AB}$ and $\\widehat{AC}$, respectively, and are the intersection of the midline parallel to $BC$ with the circumcircle of $\\triangle ABC$. Points $X$ and $Y$ are the intersection of the perpendicular bisectors of segments $AB$ and $AC$, respectively, with the tangent of $\\odot(ABC)$ at point $A$. Define point $T \\neq A$ as the intersection of $\\odot(PXA)$ and $\\odot(QYA)$. If $AD$ is an altitude in the triangle $\\triangle ABC$, prove that\n$$ \\frac{|TX|}{|TY|} = \\frac{|DB|}{|DC|} $$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\nShortlist: Keep confidential!\n\n**Step 1:** Points $X, P, O, Q, Y$ lie on a circle.\n*Proof.* Since point $X$ lies on the perpendicular bisector of $AB$, we have that $XA, XB$ are tangents to $\\odot(ABC)$. This means that $\\angle XAO = \\angle OBX = 90^{\\circ}$, therefore quadrilateral $OAXB$ is cyclic. From Power of the Point we have\n$$\n|XM| \\cdot |MO| = |AM| \\cdot |MB| = |PM| \\cdot |MQ|\n$$\nsince quadrilateral $APBQ$ is cyclic too. This implies that points $X, P, O, Q$ lie on the same circle. Similarly, we can prove that points $Y, P, O, Q$ lie on the same circle, so points $X, P, O, Q, Y$ lie on a circle as desired. $\\square$\n\n**Step 2:** $\\angle APX = \\angle YQA$\n*Proof.* Observe that $\\angle XAP = \\angle AQP$. Moreover, $\\angle AYP = \\angle XQP$, since quadrilateral $XPQY$ is cyclic. We deduce that\n$$\n\\angle AQX = \\angle AQP - \\angle XQP = \\angle XAP - \\angle AYP = \\angle APY.\n$$\nRemember that $\\angle XPY = \\angle XQY$ as again quadrilateral $XPQY$ is cyclic. Consequently, we have\n$$\n\\angle XPA = \\angle XPY - \\angle APY = \\angle XQY - \\angle AQX = \\angle AQY\n$$\nas desired. $\\square$\n\nNotice that $\\triangle XAO$ and $\\triangle YAO$ are right angle triangles, therefore\n$|XA| = |AO| \\cdot \\tan \\angle ACB$ and $|AY| = |AO| \\cdot \\tan \\angle CBA$.\nThis means that $\\frac{|AX|}{|AY|} = \\frac{\\tan \\angle ACB}{\\tan \\angle CBA}$. Similarly, $\\triangle ABD$ and $\\triangle ACD$ are right angle triangles, therefore\n$|AD| = |BD| \\cdot \\tan \\angle CBA$ and $|AD| = |CD| \\cdot \\tan \\angle ACB$.\nThis means that $\\frac{|BD|}{|DC|} = \\frac{\\tan \\angle ACB}{\\tan \\angle CBA}$. We deduce that\n$$\n\\frac{|TX|}{|TY|} = \\frac{|AX|}{|AY|} = \\frac{\\tan \\angle ACB}{\\tan \\angle CBA} = \\frac{|BD|}{|DC|}\n$$\nas desired. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25158,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be an acute triangle and $P$ be an inner point of $\\triangle ABC$. Let $K$, $L$ and $M$ be the reflections of $P$ across $BC$, $AC$ and $AB$, respectively. Let $D$ and $E$ be the second points of intersection of $\\odot(PBC)$ with lines $AB$ and $AC$, respectively. Let lines $MD$ and $LE$ intersect at $F$. Prove that $F$, $A$ and $K$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "$$\n\\angle BGF = \\angle BGK = \\angle BCK = \\angle PCB = \\angle PDB = \\angle BDM = \\angle BDF.\n$$\nSimilarly, $GFCE$ is cyclic.\n\nLastly, notice that $A$ is the radical centre of circumcircles of $GFBD$, $GFCE$ and $BDCE$. Thus, $F$, $A$ and $K$ are collinear.\nConstruct point $I$ as the second point of intersection of $MD$ with the circumcircle of $\\triangle BPC$. Then $IKPM$ is cyclic with $B$ being its circumcentre (using directed angles):\n$$\n\\begin{gather*}\n\\angle BIM = \\angle BID = \\angle BPD = \\angle DMB = \\angle IMB \\\\\n\\implies |BM| = |BP| = |BK| = |BI|.\n\\end{gather*}\n$$\nWe now show that $C$, $K$ and $I$ are collinear. $\\angle KCB = \\angle BCP$ by reflection and $\\angle BCP = \\angle ICB$, since they subtend equal chords $BI$ and $BP$. Thus, $\\angle KCB = \\angle ICB$ and thus $C$, $K$ and $I$ are collinear.\n\nThus, $K$, $F$ and $A$ are collinear.\nLet $\\bar{F}$ denote the isogonal conjugate of $F$ with respect to $\\triangle ADE$.\n\n**Claim.** $\\triangle DEF \\sim \\triangle CBK$ (with opposite orientation).\n\n*Proof.* Using directed angles mod $180^\\circ$, we have\n$$\n\\angle \\bar{F}DE = \\angle ADF = \\angle ADM = \\angle PDA = \\angle PDB = \\angle PCB = \\angle BCK,\n$$\nwhere we used that $BCPD$ is cyclic. Similarly, $\\angle DEF = \\angle KBC$. Hence $\\triangle DEF \\sim \\triangle CBK$. $\\Box$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25159,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\omega_1$ and $\\omega_2$ be circles with no common points. Points $M$ and $N$ are chosen on the circles $\\omega_1$ and $\\omega_2$, respectively, such that the tangent to the circle $\\omega_1$ at $M$ and the tangent to the circle $\\omega_2$ at $N$ intersect at $P$ and $\\triangle PMN$ is an isosceles triangle with apex $P$. The circles $\\omega_1$ and $\\omega_2$ meet the segment $MN$ again at $A$ and $B$, respectively. The line $PA$ meets the circle $\\omega_1$ again at $C$ and the line $PB$ meets the circle $\\omega_2$ again at $D$. Prove that $\\angle BCN = \\angle ADM$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $\\triangle MPN$ is an isosceles triangle, we have $\\angle PMA = \\angle PMN = \\angle MNP = \\angle BNP$. By tangent and chord theorem, $\\angle MCA = \\angle PMA = \\angle BNP = \\angle BDN$.\nSince $\\angle MCP = \\angle MNP$, the quadrilateral $CMPN$ is cyclic. Analogously, from $\\angle PDN = \\angle PMN$, we get that $NDMP$ is cyclic. Since $C$ and $D$ both lie on the circumcircle of $\\triangle NPM$, points $P, N, M, C$ and $D$ are concyclic.\nFrom inscribed angles subtending arcs with the same length, we get that $\\angle MDP = \\angle MCP = \\angle MNP = \\angle PDN = \\angle PMN = \\angle PCN$.\nThe power of $P$ with respect to $\\omega_1$ gives us that $|PM|^2 = |PA| \\cdot |PC|$. The power of $P$ with respect to $\\omega_2$ gives us that $|PN|^2 = |PB| \\cdot |PD|$. Since $|PM| = |PN|$, the powers of $P$ with respect to $\\omega_1$ and $\\omega_2$ are equal ($P$ lies on the radical axis). Hence, $|PA| \\cdot |PC| = |PB| \\cdot |PD|$, which implies that $ABDC$ is cyclic. From inscribed angles subtending the arc $AB$, we get that $\\angle ACB = \\angle ADB$.\nHence, $\\angle BCN = \\angle ACN - \\angle ACB = \\angle MDB - \\angle ADB = \\angle MDA$.\nSince $\\triangle MPN$ is an isosceles triangle, we have $\\angle PMA = \\angle PMN = \\angle MNP = \\angle BNP$. By tangent and chord theorem, $\\angle MCA = \\angle PMA = \\angle BNP = \\angle BDN$.\nSince $\\angle MCP = \\angle MNP$, the quadrilateral MPNC is cyclic, which means that $P$ lies on the circumcircle of $\\triangle MNC$. Since $\\triangle MPN$ is isosceles, the perpendicular bisector of $MN$ passes through $P$. Since the intersection point of the angle bisector and the perpendicular bisector of the opposite side of the triangle lies on the circumcircle, it follows that $CP$ bisects angle $\\angle MCN$. Hence, $\\angle MCP = \\angle PCN$. Analogously, since $\\angle PDN = \\angle PMN$, it follows that $NDMP$ is cyclic and the circumcircle of $\\triangle MND$, the perpendicular bisector of $MN$ and the angle bisector of $\\angle MDN$ meet at $P$. Hence, $\\angle MDP = \\angle PDN = \\angle MCP = \\angle PCN$.\nNow we continue as in the previous solution.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25160,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCDEF$ be a cyclic and convex hexagon. A point $P$ is called *admissible* if it does not lie on the circumcircle of $ABCDEF$ or on any of the lines $AD$, $BE$ and $CF$. An admissible point is called *fantastic* if $\\odot(ADP)$, $\\odot(BEP)$ and $\\odot(CFP)$ intersect in exactly two points. Prove that if there exists a fantastic point, then all admissible points are fantastic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The problem follows readily from the following claim:\n\n**Claim:** An admissible point $P$ is fantastic if and only if the lines $AD$, $BE$ and $CF$ all intersect in a point.\n\n*Proof.* Let $\\Gamma$ be the circumcircle of $ABCDEF$. Start by supposing that $P$ is fantastic. Let the other intersection of $\\odot(ADP)$, $\\odot(BEP)$ and $\\odot(CFP)$ be $Q$. Since the radical axis of $\\Gamma$ and of $\\odot(ADP)$ and $\\odot(BEP)$ intersect, we have that the lines $AD$, $BE$ and $PQ$ intersect. Likewise we also have that $AD$, $CF$ and $PQ$ intersect. Therefore the lines $AD$, $BE$ and $CF$ all intersect in a point.\n\nSuppose now that $AD$, $BE$ and $CF$ all intersect in a point $S$. Let $PS$ intersect $\\odot(ADP)$ again in $Q$. Now we have that\n$$\n|SC| \\cdot |SF| = |SB| \\cdot |SE| = |SA| \\cdot |SD| = |SP| \\cdot |SQ|\n$$\nHence $\\odot(ADP)$, $\\odot(BEP)$ and $\\odot(CFP)$ intersect in $P$ and $Q$.\n\n$\\Box$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25161,
    "subject": "Mathematics (Multi-modal)",
    "question": "Does there exist a scalene triangle whose incenter lies on its Euler line?",
    "options": [],
    "answer": "No",
    "solution": "The answer is **No**.\nLet $\\triangle ABC$ be a triangle with orthocenter $H$, circumcenter $O$ and incenter $I$. Assume $I$ lies on $OH$. We'll show that $\\triangle ABC$ must be isococles, contradicting the scalene property of the statement.\n\nFirst a well-known claim:\n**Claim.** $O$ and $H$ are isogonal conjugates in $\\triangle ABC$.\n*Proof.* Let $AH$ intersect $BC$ in $D$. Then\n$$\n\\angle BAH = \\angle BAD = 90^\\circ - \\angle CBA = 90^\\circ - \\frac{1}{2} \\angle AOC = \\angle OAC,\n$$\nso $AH$ and $AO$ are isogonal in $\\angle BAC$. Similarly, $BH$ and $AO$ are isogonal in $\\angle CBA$, so the conclusion follows. □\n\nNote that at most one of the vertices can lie on the Euler line because the centroid lies in the interior of $\\triangle ABC$. Therefore we may assume that $OH$ passes through neither $B$ or $C$.\n\nNow from the claim, $BI$ bisects $\\angle HBO$ and $CI$ bisects $\\angle HCO$. And since $I$ lies on $OH$, we get from the Angle Bisector Theorem, that\n$$\n\\frac{|BH|}{|BO|} = \\frac{|IH|}{|IO|} = \\frac{|CH|}{|CO|}\n$$\nso $|BH| = |CH|$ since $|BO| = |CO|$. Now this implies that $OH$ is the perpendicular bisector of $BC$. In particular, $OH \\perp BC$ so $A$ lies on $OH$. That is, $A$ lies on the perpendicular bisector of $BC$ so $|AB| = |AC|$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25162,
    "subject": "Mathematics (Multi-modal)",
    "question": "Does there exist a scalene triangle whose incenter, circumcenter and orthocenter are collinear?",
    "options": [],
    "answer": "No",
    "solution": "See Problem G 3.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25163,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be an acute triangle with $|AB| > |AC|$. The internal angle bisector of $\\angle BAC$ intersects $BC$ at $D$. Let $O$ be the circumcenter of $\\triangle ABC$. Let $AO$ intersect the segment $BC$ at $E$. Let $J$ be the incenter of $\\triangle AED$. Prove that if $\\angle ADO = 45^\\circ$ then $|OJ| = |JD|$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\alpha = \\angle BAC$, $\\beta = \\angle CBA$, $\\gamma = \\angle ACB$. We have\n$$\n\\begin{align*} \n\\angle DJA &= 90^\\circ + \\frac{1}{2} \\angle DEA = 90^\\circ + \\frac{1}{2} (\\angle EBA + \\angle BAE) \\\\ \n&= 90^\\circ + \\frac{1}{2} (\\beta + 90^\\circ - \\gamma) = 135^\\circ + \\frac{\\beta}{2} - \\frac{\\gamma}{2} \n\\end{align*}\n$$\nand\n$$\n\\begin{align*} \n\\angle DOA &= 180^\\circ - \\angle OAD - \\angle ADO = 180^\\circ - (\\angle OAC - \\angle DAC) - 45^\\circ \\\\ \n&= 135^\\circ - \\left(90^\\circ - \\beta - \\frac{\\alpha}{2}\\right) = 135^\\circ - \\left(\\frac{1}{2}(\\alpha + \\beta + \\gamma) - \\beta - \\frac{\\alpha}{2}\\right) \\\\ \n&= 135^\\circ + \\frac{\\beta}{2} - \\frac{\\gamma}{2} \n\\end{align*}\n$$\nTherefore, $\\angle DJA = \\angle DOA$, hence quadrilateral $ADJO$ is cyclic.\nSince $AJ$ is the bisector of $\\angle OAD$, the arcs $OJ$ and $JD$ are equal.\nHence $|OJ| = |JD|$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25164,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be an isosceles triangle with $|AB| = |AC|$. Let $D$ be an arbitrary point on segment $BC$. Let $X$ and $Y$ be points on $AB$ and $AC$, respectively, such that $XY \\parallel BC$ and $XY$ passes through the midpoint of $AD$. Prove that if the circumcenter of $\\triangle ABC$ lies on $\\odot(AXY)$ then the quadrilateral $AYDX$ is a parallelogram.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note that $\\angle XYO = \\angle XAO = \\angle OAY = \\angle OXY$, hence the triangle $\\triangle OXY$ is isosceles. Let $K, L, M, N$ be the midpoints of $AB, AC, AD, XY$ respectively. Note that $K, L, M$ are collinear because they lie on the midline of $\\triangle ABC$ parallel to $BC$. Moreover, $M$ lies on $XY$ by assumption. Since $XY \\parallel BC \\parallel KL$, the lines $KL$ and $XY$ do not coincide. This means that the lines $XY$ and $KL$ intersect at $M$.\n\nNote that $K, L, N$ are the projections of $O$ onto $AX, AY$ and $XY$, respectively. Therefore $K, L, N$ lie on the Simson line of $O$.\n\nThis means that $M = N$, i.e. the segments $AD$ and $XY$ share a common midpoint. Therefore $AXDY$ is a parallelogram.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25165,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be an acute triangle with $|AB| < |AC|$ and incenter $I$. Let $D$ be the projection of $I$ onto $BC$. Let $H$ be the orthocenter of $\\triangle ABC$. Prove that if $\\angle IDH = \\angle CBA - \\angle ACB$ then $|AH| = 2 \\cdot |ID|$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $H'$ be the reflection of $H$ in $BC$. It is well-known (and easy to prove) that $H'$ lies on the circumcircle of $\\triangle ABC$. Let $O$ be the circumcenter of $\\triangle ABC$. We have\n$$\n\\begin{align*}\n\\angle OH'A &= \\angle HAO = \\angle BAC - \\angle BAH - \\angle OAC \\\\\n&= \\angle BAC - 2(90^\\circ - \\angle CBA) = \\angle CBA - \\angle ACB \\\\\n&= \\angle IDH = \\angle H'HD = \\angle DH'A,\n\\end{align*}\n$$\nhence $O, D, H'$ are collinear. Also note that $\\angle HAO = \\angle H'HD$ implies that $AO \\parallel HD$.\n\nSince $OM \\parallel AH$, the above equality gives that $A, O, E$ are collinear.\nLet $D'$ be the reflection of $D$ in $I$. It is well-known (and easy to prove) that $D'$ lies on $AE$. Since $AH \\parallel OM$ and $AD' \\parallel HD$, quadrilateral $AHDD'$ is a parallelogram. Therefore $|AH| = |DD'| = 2 \\cdot |ID|$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25166,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a triangle $\\triangle ABC$ let the incircle be tangent to $BC$, $CA$, $AB$ at $D$, $E$, $F$, respectively, and let the excircle opposite to $A$ be tangent to $BC$, $CA$, $AB$ at $P$, $Q$, $R$, respectively. Let $EX$ and $FY$ be altitudes in triangle $\\triangle DEF$, and $QZ$ and $RW$ altitudes in triangle $\\triangle PQR$. Prove that the points $X$, $Y$, $Z$, $W$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $K$ be the orthogonal projection of $B$ onto the angle bisector of $\\angle BAC$. Now by the Iran lemma and its analogue for the excircle (both easy angle chases), $K$ lies on lines $DE$ and $PQ$. Now let $L$ be the orthogonal projection of $K$ onto line $AB$. Our main claim is that $L$ lies on both $XY$ and $WY$ and thus $X$, $Y$, $W$ are colinear. This also suffices since then a symmetric claim with respect to $C$ gives $X$, $Y$, $Z$ colinear. We use directed angles.\n\n**Claim:** $L$ lies on $XY$.\n**Proof:** Clearly $KLFY$ is cyclic. Also letting $I$ be the incenter of $\\triangle ABC$, we see that $BFIK$ is cyclic. Now\n$$\n\\angle XYD = \\angle XFE = \\angle CDE\n$$\nwhich is quite easy to chase to be\n$$\n\\angle BIK = \\angle BFK = \\angle LFK = \\angle LYK = \\angle LYD.\n$$\n\n**Claim:** $L$ lies on $WY$.\n**Proof:** First note that $\\angle QKE = 90^\\circ$ as\n$$\n\\angle KEQ + \\angle EQK = \\angle DEC + \\angle CQP = 90^\\circ\n$$\nand since $FR$ and $EQ$ are symmetric with respect to $AK$, we also have $\\angle FKR = 90^\\circ$. Now as $KLRW$ is cyclic, we have\n$$\n\\angle WLR = \\angle WKR = \\angle QKR\n$$\nand by the perpendicularities, we have that this is just\n$$\n\\angle EKF = \\angle YKF = \\angle YLF.\n$$\n\nSo $L$ lies on both $XY$ and $WY$, and thus $X$, $Y$, $W$ are collinear. By symmetry, $X$, $Y$, $Z$ are also collinear, so $X$, $Y$, $Z$, $W$ are collinear.\nFirst we show that $XY$ and $BC$ are parallel. Indeed, since $EFXY$ is cyclic and $BC$ is tangent to the incircle at $D$, we have\n$$\n\\angle DXY = \\angle FED = \\angle FDB = \\angle XDB.\n$$\nSimilarly, $ZW$ and $BC$ are parallel. Moreover, the points $X$, $Y$, $Z$ and $W$ all lie on the same side of $BC$ because, for example, $X$ lies in the interior of segment $DF$ (since $\\angle EDF < 90^\\circ$), and similarly for the other points. Therefore it suffices to show that the distance from $BC$ to $XY$ and $ZW$ is the same.\n\nWe will use the following Lemma:\n**Lemma.** Let $\\triangle ABC$ be a non-rectangular triangle with altitudes $BU$ and $CV$. Then $d(A, UV) = d(A, BC) \\cdot |\\cos \\angle A|$, where $d(P, \\ell)$ denotes the distance from point $P$ to line $\\ell$.\n*Proof.* Let $\\phi$ be the homothety with center $A$ and factor $\\cos \\angle A$, followed by reflection across the internal angle bisector of $\\angle A$. Then $\\phi$ maps $B$ to $U$ and $C$ to $V$. Hence $\\triangle ABC$ and $\\triangle AUV$ are similar with factor $|\\cos \\angle A|$, from which the Lemma follows.\n\nAfter applying the Lemma to $\\triangle DEF$ and $\\triangle PQR$ and noting that\n$$\n\\cos \\angle RPQ = \\cos(180^\\circ - \\angle EDF) = -\\cos \\angle EDF,\n$$\nit remains to show that $d(D, EF) = d(P, QR)$.\nHowever, since $EF$ and $RQ$ are parallel and $|BD| = |PC|$, this is equivalent to showing that $d(B, EF) = d(C, QR)$. But this can be done as follows:\n$$\n\\begin{align*}\nd(B, EF) &= |BF| \\cdot \\sin \\angle BFE = |BD| \\cdot \\sin \\angle EFA \\\\\n&= |PC| \\cdot \\sin \\angle AEF = |CQ| \\cdot \\sin \\angle AQR \\\\\n&= d(C, QR).\n\\end{align*}\n$$\n\nTherefore, the lines $XY$ and $ZW$ are parallel and equidistant from $BC$, so the points $X$, $Y$, $Z$, $W$ are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25167,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be a triangle with centroid $G$. Let $D$, $E$ and $F$ be the circumcenters of $\\triangle BCG$, $\\triangle CAG$ and $\\triangle ABG$, respectively. The point $X$ is defined as the intersection of the perpendiculars from $E$ to $AB$ and $F$ to $AC$. Prove that $DX$ bisects the segment $EF$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The two parts may be completed independently, and in the three solutions below we demonstrate different approaches to both parts, though one can create valid solutions combining either first part with either second part.\nLet $\\omega_B$, $\\omega_C$ denote the circumcircles of triangles $ABG$ and $ACG$ respectively, and the points $Y$ and $Z$ the second intersection of the line through $B$ parallel to $AC$ and $\\omega_B$ and the second intersection of the line through $C$ parallel to $AB$ and $\\omega_C$.\nThe lines $BY$ and $CZ$ thus intersect at $A'$, the reflection of $A$ across the midpoint of $BC$, and in particular on the $A$-median. Using Power of a Point from $A'$ with respect to the circles $\\omega_B$ and $\\omega_C$ we obtain\n\n$$\n|A'B| \\cdot |A'Y| = |A'A| \\cdot |A'G| = |A'C| \\cdot |A'E|\n$$\nimplying from the converse of Power of a Point that the quadrilateral $YBCZ$ is cyclic. The perpendicular bisector of $BY$ is orthogonal to $BY \\parallel AC$ and passes through $F$ and thus $X$ as well. Similarly, the perpendicular bisector of $CZ$ passes through $Z$. Hence $X$ is the center of circle ($YBCZ$) and thus on the perpendicular bisector of the line $BC$.\nLet $M$ and $N$ denote the midpoints of $BC$ and $EF$, respectively. To prove that $N$ lies on the perpendicular bisector of $BC$, let $V$ and $W$ denote the second intersections of $\\omega_B$ and $\\omega_C$ with the line $BC$, respectively.\nFrom Power of a Point from $M$ with respect to $\\omega_B$ and $\\omega_C$ we obtain\n$$\n|MV| \\cdot |MB| = |MG| \\cdot |MA| = |WM| \\cdot |CM| \\implies |MV| = |WM|\n$$\nso $M$ is the midpoint of the segment $VW$. Let $E'$, $N'$, $F'$ denote the projections of $E$, $N$ and $F$ onto $BC$ respectively. Since $N$ is the midpoint of $EF$, $N'$ will be the midpoint of $E'F'$.\nMoreover, from the fact that $E$ and $F$ are the centers of $\\omega_B$ and $\\omega_C$ we get that $E'$ and $F'$ are the midpoints of $BV$ and $WC$, and hence $M$ is the midpoint $E'F'$ as well, implying $N' = M$ and that $N$ is on the perpendicular bisector of $BC$.\nLet $G'$ denote the reflection of $G$ across the midpoint of $BC$. We begin by proving that triangles $ABC$ and $DFE$ are orthological, with orthology centers $G'$ and $X$.\nObserve that $G'$ is on the $A$-median and thus $AG' \\perp EF$. Furthermore, quadrilateral $BGCG'$ is a parallelogram and hence $BG' \\parallel CG \\perp DE$ and $CG' \\parallel BG \\perp DF$. Hence, $G'$ is the first orthology center of $\\triangle ABC$ and $\\triangle DFE$.\n\nThus, by the property of orthological triangle, the second orthology center must exist, which is defined as the common intersection of the normal from $D$ to $BC$, $E$ to $AB$ and $F$ to $AC$, i.e. the point $X$. Since $D$ is on the perpendicular bisector of $BC$, by virtue of being the circumcenter of triangle $BGC$, and $XD \\perp BC$ so must point $X$.\nMoreover, let $O$ denote the circumcenter of triangle $ABC$. Then $EO \\perp AC \\perp FX$ implies $EO \\parallel FX$ and $FO \\perp AB \\perp EX$ implies $FO \\parallel EX$, meaning that quadrilateral $FOEX$ is a parallelogram. Hence, the midpoint of $EF$ lies on the line $XOD$ i.e. the perpendicular bisector of segment $BC$.\nLet $M$ be the midpoint of $BC$. Let $N$ be the intersection of $EF$ and $DM$. We claim that $N$ is the midpoint of $EF$.\nNamely, we have $\\triangle DEN \\sim \\triangle CGM$ because corresponding pairs of sides are orthogonal. Similarly, $\\triangle DFN \\sim \\triangle BGM$. Hence proving that $|EN| = |FN|$, as desired.\n\nNext, let $X'$ resp. $X''$ denote the intersection of $DN$ with the perpendicular from $E$ to $AB$ resp. the perpendicular from $F$ to $AC$. Just as above we have $\\triangle ENX' \\sim \\triangle AMB$ and $\\triangle FNX'' \\sim AMC$, thus\nThis shows that $X = X' = X''$ lies on $DN$.\n*Remark:* That the medians of triangle $DEF$ coincide with the perpendicular bisectors of triangle $ABC$ implies that the centroid",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25168,
    "subject": "Mathematics (Multi-modal)",
    "question": "In an acute triangle $\\triangle ABC$ with $|AB| \\neq |AC|$, the perpendicular bisectors of sides $AC$ and $AB$ intersect segment $BC$ at points $D$ and $E$, respectively. The tangents to $\\odot(ABC)$ at the points $B$ and $C$ intersect $\\odot(ABD)$ and $\\odot(ACE)$ at points $Y$ and $Z$, respectively. Suppose that lines $YD$ and $ZE$ intersect at point $X$. Define points $U$ and $V$ to be the intersections of $\\odot(DEX)$ and the lines $AD$ and $AE$, respectively. Prove that the lines $UV, BC$ and $XA$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $O$ be the circumcentre of $\\odot(ABC)$. We proceed in several steps.\n\n**Step 1:** Point $O$ lies on $\\odot(ABD)$ and $\\odot(ACE)$.\n*Proof.* Note that since $|AD| = |DC|$, we have $\\angle CDA = 180^\\circ - 2\\angle ACB$, therefore $\\angle ADB = 2\\angle ACB = \\angle AOB$, which means that quadrilateral $AODB$ is cyclic. Similarly, we can prove that quadrilateral $AOEC$ is cyclic. $\\square$\n\n**Step 2:** Points $Y$, $A$, $Z$ are collinear. Moreover, $YZ$ is tangent to $\\odot(ABC)$ at point $A$.\n*Proof.* Note that since $YB$ is tangent to $\\odot(ABC)$, we have that:\n$$\n\\angle AOY = \\angle YBA = \\angle ACB.\n$$\nSince $\\angle AOB = 2\\angle ACB$, we get that:\n$$\n\\angle YAB = \\angle YOB = \\angle AOB - \\angle AOY = \\angle ACB.\n$$\nThis means that $YA$ is tangent to $\\odot(ABC)$. Consequently, $|YA| = |YB|$ implies that $Y$ lies on the perpendicular bisector of $AB$. Similarly, we can prove that $AZ$ is tangent to $\\odot(ABC)$ and $Z$ lies on the perpendicular bisector of $AC$. Since the tangent line at a fixed point is unique, we conclude that points $Y$, $A$, $Z$ all lie on the tangent to $\\odot(ABC)$ at the point $A$. $\\square$\n\n**Step 3:** Points $Y$, $O$, $E$ and $Z$, $O$, $D$ are collinear. Moreover, $O$ is the orthocentre of $\\triangle YZX$.\n*Proof.* The first part follows from the previous result that $Y$ lies on the perpendicular bisector of $AB$, which is $OE$. A similar argument applies to $Z$, $O$, $D$.\nSince the radius of a circle is perpendicular to the corresponding tangent, note that:\n$$\n\\angle OBY = \\angle ODY = \\angle YAO = 90^\\circ \\quad \\text{and} \\quad \\angle OCZ = \\angle OEZ = 90^\\circ.\n$$\nThis gives us that $ZD \\perp YX$ and $YE \\perp ZX$, implying that $O$ is the orthocentre of $\\triangle YZX$. Also note that this immediately gives us that points $A$, $O$, $X$ are collinear, too. $\\square$\n\nTo finish the problem, note that quadrilateral $YDEZ$ is cyclic from $\\angle YDZ = \\angle YEZ = 90^\\circ$. Therefore:\n$$\n\\angle DAO = \\angle DYO = \\angle OZE = \\angle OAE.\n$$\n\nSince $OX$ is the diameter of $\\odot(DEX)$ (note that $\\angle ODX = \\angle OEX = 90^\\circ$) and $OX$ is the bisector of $\\angle DAE$, by symmetry we have that $DV$ and $UE$ are parallel lines. It is well-known that the diagonals of an isosceles trapezoid intersect on the perpendicular bisector of their parallel sides. Therefore, lines $UV$, $DE$, $AX$ are concurrent, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25169,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\mathbb{Z}^+$ denote the set of positive integers, and $\\text{lcm}(n, m)$ the least positive integer that is divisible by both $n$ and $m$.\nFind all functions $f: \\mathbb{Z}^+ \\to \\mathbb{Z}^+$ such that\n$$\n\\text{lcm}(n \\cdot f(n^2), f(m^2)) = n \\cdot f(nm), \\text{ for all positive integers } n, m\n$$\nand such that $f(n)$ divides $n$ for all positive integers $n$.",
    "options": [],
    "answer": "All functions obtained by choosing an arbitrary subset S of primes and defining f(1)=1 and, for n=∏ p_i^{e_i}, f(n)=∏_{p_i | n, p_i ∈ S} p_i. Equivalently, for each prime p fix α_p ∈ {0,1} and set f(n)=∏_{p | n} p^{α_p}.",
    "solution": "First notice that $f(1)$ divides $1$, and hence $f(1) = 1$. For all primes $p$, we have\n$$\n\\begin{align*}\np \\cdot f(p^{2n}) &= \\text{lcm}(p \\cdot f(p^{2n}), 1) = \\text{lcm}(p \\cdot f(p^{2n}), f(1^2)) \\\\\n&= p \\cdot f(p^n \\cdot 1) = p \\cdot f(p^n),\n\\end{align*}\n$$\nwhich shows that $f(p^n) = f(p^{2n})$ for all $n$. Now we show by induction on $m$ that $f(p^m) = f(p)$. We know that $f(p^2) = f(p)$. Assume that $f(p^m) = f(p)$. Now\n$$\n\\begin{gather*}\np f(p \\cdot p^m) = \\text{lcm}(p \\cdot f(p^2), f(p^{2m})) = \\text{lcm}(p \\cdot f(p), f(p^m)) = p \\cdot f(p), \\\\\n\\text{and hence } f(p^{m+1}) = f(p).\n\\end{gather*}\n$$\nSince $f(p)$ divides $p$, we know that $f(p) = p^{\\alpha_p}$ for $\\alpha_p \\in \\{0, 1\\}$. For each prime $p_i$, let $f(p_i) = p_i^{\\alpha_{p_i}}$ with $\\alpha_{p_i} \\in \\{0, 1\\}$. For two primes $p_1 \\neq p_2$, we have\n$$\n\\begin{align*}\np_1^n f(p_1^n \\cdot p_2^m) &= \\text{lcm}(p_1^n \\cdot f(p_1^{2n}), f(p_2^{2m})) = \\text{lcm}(p_1^n \\cdot p_1^{\\alpha_{p_1}}, p_2^{\\alpha_{p_2}}) \\\\\n&= p_1^n \\cdot p_1^{\\alpha_{p_1}} \\cdot p_2^{\\alpha_{p_2}},\n\\end{align*}\n$$\nand hence $f(p_1^n \\cdot p_2^m) = p_1^{\\alpha_{p_1}} \\cdot p_2^{\\alpha_{p_2}}$ for all non-negative integers $n$ and $m$. By induction on $r$, it follows that\n$$\nf(p_1^{n_1} p_2^{n_2} \\cdots p_r^{n_r}) = p_1^{\\alpha_{p_1}} p_2^{\\alpha_{p_2}} \\cdots p_r^{\\alpha_{p_r}}.\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25170,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that the sum of the decimal digits of $2^{2^{2^{2023}}}$ is greater than $2023$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will prove the more general statement that, for every positive integer $n$, the sum of decimal digits of $2^{2^{2n}}$ is greater than $n$.\nLet $m = 2^{2n} = 4^n$, so that we need to consider the digits of $2^m$. It will suffice to prove that at least $n$ of these digits are different from $0$, since the last digit is at least $2$.\n\nLet $0 = e_0 < e_1 < \\dots < e_k$ be the positions of non-zero digits, so that $2^m = \\sum_{i=0}^k d_i \\cdot 10^{e_i}$ with $1 \\le d_i \\le 9$. Considering this number modulo $10^{e_j}$, for some $0 < j \\le k$, the residue $\\sum_{i=0}^{j-1} d_i \\cdot 10^{e_i}$ is a multiple of $2^{e_j}$, hence at least $2^{e_j}$, but on the other hand it is bounded by $10^{e_{j-1}+1}$.\nIt follows that $2^{e_j} < 10^{e_{j-1}+1} < 16^{e_{j-1}+1}$, and hence $e_j < 4(e_{j-1} + 1)$. With $e_0 = 4^0 - 1$ and $e_j \\le 4(e_{j-1} + 1) - 1$, it follows that $e_j \\le 4^j - 1$, for all $0 \\le j \\le k$. In particular, $e_k \\le 4^k - 1$ and hence\n$$\n2^m = \\sum_{i=0}^{k} d_i \\cdot 10^{e_i} < 10^{4^k} < 16^{4^k} = 2^{4 \\cdot 4^k} = 2^{4^{k+1}},\n$$\nwhich yields $4^n = m < 4^{k+1}$, i.e., $n - 1 < k$. In other words, $2^m$ has $k \\ge n$ non-zero decimal digits, as claimed.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25171,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a positive integer $m$, let $S(m)$ denote the positive integer whose decimal representation is equal to the octal representation of $m$. For example, $S(64) = 100$, $S(100) = 144$, $S(2023) = 3747$. Two positive integers $x, y$ satisfy the relation\n$$\n\\frac{S(x) + S(1)}{S(x + 1)} = \\frac{S(y) + S(2023)}{S(y + 2023)} < \\frac{8}{10}.\n$$\nWhat is the smallest possible value of $y$ if $x < 100$?",
    "options": [],
    "answer": "2137",
    "solution": "The answer is 2137.\n\nFirst of all, let's observe that $x = 63$, $y = 2137$ satisfies the given relation. In fact,\n$$\n\\frac{S(63) + S(1)}{S(64)} = \\frac{77 + 1}{100} = \\frac{78}{100} < \\frac{8}{10} \\quad \\text{and} \\\\\n\\frac{S(2137) + S(2023)}{S(4160)} = \\frac{4131 + 3747}{10100} = \\frac{7878}{10100} = \\frac{78}{100} < \\frac{8}{10}.\n$$\n\nNow let's assume that the octal representation of $x$ ends with exactly $k$ digits “7” (maybe $k = 0$). In other words, $x = a \\cdot 8^k + 8^k - 1$ where $a$ is a nonnegative integer satisfying $a \\bmod 8 < 7$. In particular, we know that $S(a + 1) = S(a) + 1$ because there is no carry in addition in the octal system, since $a \\bmod 8 < 7$. Therefore,\n$$\nS(x) = S(a \\cdot 8^k + 8^k - 1) = S(a) \\cdot 10^k + \\frac{7}{9}(10^k - 1) \\quad \\text{and} \\\\\nS(x + 1) = S((a + 1) \\cdot 8^k) = S(a + 1) \\cdot 10^k = (S(a) + 1) \\cdot 10^k.\n$$\n\nLet's observe that $k = 0$ would imply that $S(x) + S(1) = S(x + 1)$ and therefore $k \\neq 0$. $k \\ge 3$ would imply $x \\ge 8^3 - 1 > 100$ and therefore $k = 1$ or $k = 2$. If $k = 2$ then $a = 0$ (because $8^2 + 8^2 - 1 > 100$) and therefore $x = 8^2 - 1 = 63$ and $\\frac{S(x)+S(1)}{S(x+1)} = \\frac{78}{100}$. $k = 1$ is also impossible because:\n$$\n\\frac{10S(a) + 8}{10S(a) + 10} < \\frac{8}{10} \\implies 10S(a) + 8 < 8S(a) + 8 \\implies S(a) < 0.\n$$\n\nNow let's assume that we can find a solution with $y < 2137 = 4131_8$. Then $y$ has at most 4 digits in its octal representation. Let $d_i$ for $i = 0, 1, 2, 3$ be equal to 1 if there is a carry at the $i$-th position in addition of $y$ and $2023 = 3747_8$ in octal system. It is easy to conclude that\n$$\nS(y+2023)-S(y)-S(2023) = \\sum_{i=0}^{3} (-8) \\cdot 10^i \\cdot d_i + \\sum_{i=0}^{3} 10^{i+1} \\cdot d_i = 2 \\sum_{i=0}^{3} 10^i \\cdot d_i.\n$$\nWe have the implication:\n$$\n\\begin{aligned}\n\\frac{S(y) + S(2023)}{S(y + 2023)} &= \\frac{78}{100} & \\implies & \\frac{S(y) + S(2023)}{S(y + 2023) - S(y) - S(2023)} &= \\frac{78}{22} \\\\\n& \\implies & S(y) + S(2023) &= \\frac{78}{11} \\sum_{i=0}^{3} 10^i \\cdot d_i.\n\\end{aligned}\n$$\nWe know that $d_3 = 1$ because otherwise we would have:\n$$\nS(y) + S(2023) \\le \\frac{78}{11} \\cdot 111 < 3747 = S(2023).\n$$\nBecause 78 is coprime with 11, we need to have the divisibility $11|\\sum_{i=0}^{3} 10^i \\cdot d_i$. Since $d_3 = 1$, there are only three possible cases: $\\sum_{i=0}^{3} 10^i \\cdot d_i = 1001, 1100, 1111$. They correspond to $S(y) + S(2023) = 7098, 7800, 7878$, so $S(y) = 3351, 4053, 4131$. The first two cases are impossible because they assume $d_1 = 0$ but $5 + 4 > 7$, so there has to be a carry at the first position.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25172,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\mathbb{Z}^+$ be the set of positive integers and $(F_n)_{n \\in \\mathbb{Z}^+}$ be the Fibonacci sequence defined by $F_1 = F_2 = 1$ and $F_{n+1} = F_n + F_{n-1}$ for $n \\ge 2$. Consider the number\n$$\nN = 2^{2022}F_{2023} - 2^{2021}F_{2022} + 2^{2020}F_{2021} + \\dots + 2F_2 + F_1.\n$$\nProve that the binary expansion of $N$ contains more 1's than 0's.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $F_n$ is monotonically increasing, it is clear that\n$$\n0 < N < 2^{2022}F_{2023}.\n$$\nMoreover, since $F_{n+1} \\le 2F_n$, it is clear that $F_{2023} < 2^{2023}$ and hence $N$ has at most 4045 binary digits. It will thus suffice to prove that the last 2023 binary digits of $N$ are 1's, in other words, that $2^{2023} \\mid N + 1$.\nWe have\n\nAdding this three equations together, we get on the left hand side\n$$\n4N - 2N - N = N\n$$\nand on the right hand side\n$$\n\\begin{align*} \n&= 2^{2024} F_{2023} - 2^{2023} F_{2022} - 2^{2023} F_{2023} + \\dots \\\n& \\quad + \\sum_{i=2}^{2022} (-2)^i (F_{i-1} + F_i - F_{i+1}) - 2F_1 - 2(-F_2) - F_1 \\\n&= 2^{2024} F_{2023} - 2^{2023} (F_{2022} + F_{2023}) + 2(F_2 - F_1) - F_1 \\\n&= 2^{2024} F_{2023} - 2^{2023} F_{2024} - F_1. \n\\end{align*}\n$$\nHence\n$$\nN + 1 = 2^{2024} F_{2023} - 2^{2023} F_{2024}\n$$\nis indeed a multiple of $2^{2023}$ as desired.\nFor $n \\in \\mathbb{Z}_{\\ge 2}$, let $N_n := \\sum_{i=0}^{n} (-2)^i F_{i+1}$. As in the first proof, we want to show $2^{2023} \\mid N_{2022} + 1$, or more generally $2^{n+1} \\mid N_n + 1$. To do so, we prove the explicit representation of the sum as $N_n + 1 = (-1)^n \\cdot 2^{n+1} \\cdot F_{n-1}$.\nFor $n=2$, we have $N_2 + 1 = 2^2 F_3 - 2F_2 + F_1 + 1 = 8 = (-1)^2 \\cdot 2^3 \\cdot F_1$. The induction steps follows via\n$$\n\\begin{align*} \nN_{n+1} + 1 &= N_n + 1 + (-2)^{n+1} F_{n+2} \\\n&= -(-2)^{n+1} \\cdot F_{n-1} + (-2)^{n+1} \\cdot (F_{n+1} + F_n) \\\n&= (-2)^{n+1} \\cdot (-F_{n-1} + F_n + F_{n-1} + F_n) = -(-2)^{n+2} \\cdot F_n. \n\\end{align*}\n$$\nThus, $2^{n+1} \\mid N_n + 1$ and, in particular, $2^{2023} \\mid N_{2022}$, which proves the claim. The desired results follows from there as in the first proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25173,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a prime number $p$ and a polynomial $f$ with integer coefficients, define $\\text{Im}(p, f)$ be the set of integers $a \\in \\{0, 1, \\dots, p-1\\}$ such that there exists an integer $x$, for which $f(x) - a$ is divisible by $p$.\n\nProve that there exist nonconstant polynomials $f$ and $g$ such that, for infinitely many primes, the intersection of $\\text{Im}(p, f)$ and $\\text{Im}(p, g)$ is empty.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We take $f(x) = (x^2 + 1)^2$ and $g(y) = -(y^2 + 1)^2$ and prove that if $p \\equiv 3 \\pmod 4$ then the equation $f(x) \\equiv g(y) \\pmod p$ has no solution. Famously, there are infinitely many primes congruent to $3$ modulo $4$.\n\nRecall the fact that if $p \\equiv 3 \\pmod 4$ then the only solution to the equation $a^2 + b^2 \\equiv 0 \\pmod p$ is $a \\equiv b \\equiv 0 \\pmod p$. Hence, for $f(x) \\equiv g(y) \\pmod p$ to hold, we need\n$$\n(x^2 + 1)^2 + (y^2 + 1)^2 \\equiv 0 \\pmod{p}\n$$\nand thus\n$$\nx^2 + 1 \\equiv y^2 + 1 \\equiv 0 \\pmod{p},\n$$\nwhich is impossible for $p \\equiv 3 \\pmod 4$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25174,
    "subject": "Mathematics (Multi-modal)",
    "question": "For integers $a$ and $n > 0$, we say that $a$ is *n-expressible* if it can be written as a sum of distinct positive divisors of $n$. The natural number $n$ is *good* if $a$ being *n-expressible* implies $a-1$ being *n-expressible* for all $a \\ge 1$.\nDetermine for which $n$, both $n!$ and $n^n$ are good.",
    "options": [],
    "answer": "n! is good for all positive integers n. The number n^n is good exactly when n is even or n equals 1. Hence both n! and n^n are good precisely for n even or n = 1.",
    "solution": "For both versions of the problem, we begin with a lemma which classifies the good numbers in a more compact way.\n\nNote that the largest integer that can be written as a sum of distinct divisors of $n$ is the sum of all its positive divisors, denoted by $\\sigma(n)$. Hence, a positive integer $n$ is good iff all the integers in the interval $[0, \\sigma(n)]$ are $n$-expressible.\n\n**Lemma: Classification of Good Integers**\nLet the prime factorization of $n$ be $n = \\prod p_i^{a_i}$, where $a_i$ is positive and the primes are sorted in ascending order, i.e. $i > j \\implies p_i > p_j$. Then $n$ is good iff for all $i \\ge 0$, $p_{i+1} \\le 1 + \\sigma(\\prod_{j \\le i} p_j^{a_j})$.\n\n*Proof.* The only if direction may be seen from the fact that if we only consider sums of divisors containing the $i$ first prime divisors of $n$, $i \\ge 0$, then we can only create sums which are strictly smaller than $1 + \\sigma(\\prod_{j \\le i} p_j^{a_j})$ and if we use any divisor containing a prime divisor not among the $i$ first, then the smallest possible sum equals $p_{i+1}$, which yields the desired inequality.\n\nTo see that these inequalities are in fact sufficient, let $n_i = \\prod_{j \\le i} p_j^{a_j}$, where $n_0 = 0$, and let us prove by induction that, for all $i$, $n_i$ is a good number.\n\nThe base case $n_0 = 1$ is easily seen to be good, hence for the induction step, we are given that $n_i$ is good and that $p_{i+1} \\le 1 + \\sigma(n_i)$, and want to prove that $n_{i+1}$ is good, as well.\n\nThe sums of distinct positive divisors of $n_{i+1}$ may be broken up into $a_{i+1} + 1$ different separate sums, where each divisor $d$ is put into the group corresponding to how many times the prime $p_{i+1}$ divides the divisor $d$. Moreover, if we consider any sum on the form\n$$\nd = \\sum_{j=0}^{a_{i+1}} d_j p_{i+1}^j,\n$$\nwhere $d_j$ can be written as a sum of distinct positive divisors of $n_i$, i.e. $d_j \\in [0, \\sigma(n_i)]$, then $d$ may be written as a sum of distinct divisors of $n_{i+1}$, since we can split $d_j$ into its corresponding sum of distinct divisors of $n_i$, none of them being divisible by $p_{i+1}$.\n\nthis is essentially writing a number in base $p_{i+1}$, except that we may have more coefficients available than those in the interval $[0, p_{i+1}-1]$. However, the inequality $\\sigma(n_i) \\ge p_{i+1} - 1$ guarantees that we have at least these available.\n\nHence, to express any number $d$ in the interval $[0, \\sigma(n_{i+1})]$, we will start by determining the coefficient $d_{a_{i+1}}$, which we will put equal to\n$$\n\\left\\lfloor \\frac{d}{p_{i+1}} \\right\\rfloor .\n$$\nObserve that since $\\gcd(p_{i+1}^{a_{i+1}}, n_i) = 1$ and $\\sigma$ is multiplicative,\n$$\n\\sigma(n_{i+1}) = \\sigma(n_i) \\cdot (1 + p_{i+1} + \\cdots + p_{i+1}^{a_{i+1}}).\n$$\nHence, $d_{a_{i+1}} \\le \\sigma(n_i)$ and\n$$\n\\begin{align*}\nd - d_{a_{i+1}} p_{i+1}^{a_{i+1}} &\\le p_{i+1}^{a_{i+1}} - 1 = (p_{i+1} - 1) (1 + p_{i+1} + \\cdots + p_{i+1}^{a_{i+1}-1}) \\\\\n&\\le \\sigma(n_i) (1 + p_{i+1} + \\cdots + p_{i+1}^{a_{i+1}-1}).\n\\end{align*}\n$$\nThus, the remaining number to be expressed lies in the interval $[0, \\sigma(\\frac{n_{i+1}}{p_{i+1}})]$, which we know is expressible as as sum of the form\n$$\nd' = d - d_{a_{i+1}} p_{i+1}^{a_{i+1}} = \\sum_{j=0}^{j=a_{i+1}-1} d_j p_{i+1}^j\n$$\nby induction, since $d' \\in [0, \\sigma(\\frac{n_{i+1}}{p_{i+1}})]$.\n\nThus, any number in the interval $[0, \\sigma(n_{i+1})]$ is $n_{i+1}$ expressible, meaning that $n_{i+1}$ is good. $\\square$\n\nThe rest of the solution is dedicated to utilizing the **Classification of Good Integers**, to determine whether various integers are good.\n\n**For which $n$ is $n!$ good:**\nWe will prove that, for all positive integers $n, n!$ is good. Let us write $n! = \\prod p_i^{a_i}$. By virtue of being a factorial, the $p_i$-s in this product will be the first $I$ primes for some $I$. Indeed, if $p < q$ are primes and $q | n!$ then $q | m$ for some $m \\le n$, which implies $q \\le n, p \\le n$ and $p | (1 \\cdot 2 \\cdot \\dots \\cdot p - 1 \\cdot p + 1 \\cdot \\dots \\cdot n) \\cdot p = n!$.\n\nTo prove that $n!$ is good, we thus need to verify the inequalities $p_{i+1} \\le \\sigma(n_i) + 1 = (1+\\cdots+p_1^{a_1})\\cdots(1+\\cdots+p_i^{a_i}) + 1$. Note that the product $(1+\\cdots+p_1^{a_1})\\cdots(1+\\cdots+p_i^{a_i})$ is at least $p_1^{\\dots}p_i$. Thus, it is sufficient to verify that the next prime $p_{i+1}$ is in the interval $[p_i + 1, 1 + p_1 \\dots \\dots p_i]$. However, this can be done by the classic observation that $\\gcd(1 + p_1 \\dots \\dots p_i, p_j) = 1$ for all $1 \\le j \\le i$,\n\nTherefore, $n!$ is good for all positive integers $n$.\n\n**For which $n$ is $n^n$ good:**\nTo prove the if direction, we resort to the lemma **Classification of Good Integers**, and write $n^n$ in the form $n^n = \\prod p_i^{a_i}$. Observe that when $n$ is even, $p_1 = 2$.\n\nWe need to prove that, for all $i \\ge 0$, the inequality $p_{i+1} \\le 1 + \\sigma(n_i)$ holds. For $i = 0$, this is satisfied since $p_1 = 2$, and for $i \\ge 1$, we have $1 + \\sigma(n_i) \\ge 1 + \\sigma(n_1) = 2^{a_i+1} \\ge 2^{n+1}$, since $n | a_i$.\n\nTherefore, $p_{i+1} | n \\implies p_{i+1} \\le n \\le 2^{n+1} \\le 1 + \\sigma(n_i)$, meaning that $n^n$ is indeed good for even $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25175,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ be an odd prime. Let $a_1, a_2, \\dots, a_{p-1}$ be integers such that $i \\cdot a_i \\equiv 1 \\pmod{p}$, for all $i = 1, 2, \\dots, p-1$. Prove that\n$$\n2^p - 2 \\equiv p(a_1 - a_2 + \\dots + a_{p-2} - a_{p-1}) \\pmod{p^2}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "From the binomial formula, we have\n$$\n2^p = C_p^0 + \\dots + C_p^p.\n$$\nSince $C_p^0 = C_p^p = 1$, we also have\n$$\n2^p - 2 = C_p^1 + \\dots + C_p^{p-1}.\n$$\nFor $1 \\le k \\le p-1$, we will find the value of $C_p^k$ modulo $p^2$. We know that\n$$\nC_p^k = \\frac{p!}{k!(p-k)!} = p \\cdot \\frac{(p-1)!}{k!(p-k)!}.\n$$\nWrite $x_k = \\frac{(p-1)!}{k!(p-k)!}$. Note that $x_k$ is an integer, because $p x_k = \\frac{p!}{k!(p-k)!} = C_p^k$ is clearly an integer, but $p x_k = C_p^k$ is divisible by $p$, as there is no factor of $p$ in the denominator. Since\n$$\nk! = k \\cdot (k-1) \\cdot \\dots \\cdot 1 \\equiv (-1)^k (p-k) \\cdot (p-k+1) \\cdot \\dots \\cdot (p-1) \\pmod{p},\n$$\none has\n$$\n-k x_k \\equiv (p-k)x_k \\equiv \\frac{(p-k)(p-1)!}{(-1)^k (p-1)!(p-k)} \\equiv (-1)^k \\pmod{p}.\n$$\nTherefore, $(-1)^{k+1} k x_k \\equiv 1 \\pmod{p}$, which yields $(-1)^{k+1} x_k \\equiv a_k \\pmod{p}$ and $x_k \\equiv (-1)^{k+1} a_k \\pmod{p}$. Adding together these congruences, for $k=1, \\dots, p-1$, we get $x_1 + x_2 + \\dots + x_{p-2} + x_{p-1} \\equiv a_1 - a_2 - \\dots - a_{p-2} + a_{p-1} \\pmod{p}$. Therefore,\n$$\n\\begin{aligned}\n2^p - 2 &= C_p^1 + \\dots + C_p^{p-1} \\equiv p(x_1 + x_2 + \\dots + x_{p-2} + x_{p-1}) \\\\\n& \\equiv p(a_1 - a_2 + \\dots + a_{p-2} - a_{p-1}) \\pmod{p^2},\n\\end{aligned}\n$$\nas desired.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25176,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\delta(n)$ denote the number of positive divisors of $n$ and let $\\phi(n)$ denote the number of non-negative integers less than $n$ and relatively prime to $n$.\nFind all positive integers $n$ such that $\\delta(n) \\cdot \\phi(n) = n$.",
    "options": [],
    "answer": "1, 2",
    "solution": "Answer: $1$, $2$.\n\nLet $n = p_1^{\\alpha_1} \\cdots p_k^{\\alpha_k}$ be the canonical representation of $n$, where $\\alpha_1, \\dots, \\alpha_k$ are positive integers. It is known that\n$$\n\\delta(n) = (\\alpha_1 + 1) \\dots (\\alpha_k + 1)\n$$\nand\n$$\n\\phi(n) = (p_1 - 1) \\dots (p_k - 1) \\cdot p_1^{\\alpha_1 - 1} \\dots p_k^{\\alpha_k - 1}.\n$$\n\nHence, the given equation reduces to\n$$\n(\\alpha_1 + 1) \\dots (\\alpha_k + 1) = \\frac{p_1 \\dots p_k}{(p_1 - 1) \\dots (p_k - 1)}. \\quad (*)\n$$\nNote that $\\alpha_i + 1 \\ge 2 \\ge 1 + \\frac{1}{p_i-1} = \\frac{p_i}{p_i-1}$, for every $i=1, \\dots, k$, whereby equalities hold if and only if $\\alpha_i = 1$ and $p_i = 2$. Multiplying the inequalities $\\alpha_i + 1 \\ge \\frac{p_i}{p_{i-1}}$, for $i = 1, \\dots, k$, leads to the inequality obtained from $(*)$ by replacing equality with inequality. Hence equality $(*)$ holds if and only if every inequality that was multiplied holds as an equality. I.e., if $\\alpha_i = 1$ and $p_i = 2$, for every $i = 1, \\dots, k$. As primes in the canonical representation are distinct, this holds if and only if $k = 0$ or $k = 1$ and $p_1 = 2$, i.e., if $n = 1$ or $n = 2$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25177,
    "subject": "Mathematics (Multi-modal)",
    "question": "Call a pair of positive integers $(a, b)$ *carroty* if $S(a^{b+1}) = a^b$, where $S(m)$ is the digit sum of $m$. Find all *carroty* pairs $(a, b)$.",
    "options": [],
    "answer": "(1, b) for all positive integers b; (3, 2); (9, 1)",
    "solution": "*Answer:* $(a, b) \\in \\{(1, b) \\mid b \\in \\mathbb{Z}^+\\} \\cup \\{(3, 2), (9, 1)\\}$.\n\nLet us consider the case where $k \\ge 2$. Then $k \\le 2(k-1)$ and note that $b+1 \\le 2b$ as $b \\ge 1$. Put $(k-1)b =: x$, then $k(b+1) \\le 4(k-1)b = 4x$. So $10^x \\le 36x$. It is obvious that the only solutions in nonnegative integers to this inequality are $x = 0$ and $x = 1$. Indeed, for $x \\ge 2$, the left hand side grows faster. Therefore, either $k = 1$ or $k = 2$ and $b = 1$.\nNow we have only two cases left.\n\n**Case 1:** $b = 1$ and $k = 2$. We are left with the equation $S(a^2) = a$, for $10 \\le a < 100$. Then $a^2 < 10^4$, so $a = S(a^2) \\le 9 \\cdot 4 = 36$.\nMoreover, taking into account the fact that the sum of digits does not change the number modulo 9, $a^2 \\equiv a \\pmod{9}$, i.e., $a(a - 1) \\equiv 0 \\pmod{9}$, therefore $a \\equiv 0 \\pmod{9}$ or $a \\equiv 1 \\pmod{9}$. So now we are left only with numbers $a \\in \\{10, 18, 19, 27, 28, 36\\}$, which we can easily check by substitution and see that there are no solutions.\n\n**Case 2:** $k=1$. In the same way, by checking modulo 9, we get that $a^{b+1} \\equiv a^b \\pmod{9}$ implies $a^b(a-1) \\equiv 0 \\pmod{9}$. Therefore either $a=1$ or $a$ is divisible by 3. $a=1$ is an obvious solution with all $b \\in \\mathbb{Z}^+$.\nOtherwise, $a \\in \\{3, 6, 9\\}$. But then $a^{b+1} < 10^{b+1}$ and $S(a^{b+1}) \\le 9(b+1)$. Therefore, $3^b \\le a^b = S(a^{b+1}) \\le 9(b+1)$. But from $3^b \\le 9(b+1)$, we can conclude $b \\le 3$. Indeed, for $b \\ge 4$, the left hand side increases faster. So we are left with $a \\in \\{3, 6, 9\\}$ and $b \\le 3$. We check all these cases to determine that only $(a, b) = (3, 2)$ or $(a, b) = (9, 1)$ are solutions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25178,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p > 7$ be a prime number and let $A \\subseteq \\{0, 1, \\dots, p-1\\}$ consist of at least $\\frac{p-1}{2}$ elements. Show that for each integer $r$, there are elements $a, b, c, d \\in A$ such that\n$$\nab - cd \\equiv r \\pmod{p}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $P$ be the set of possible products $ab$, for $a, b \\in A$. Clearly, $|P| \\ge |aA| \\ge \\frac{p-1}{2}$, for any $a \\in A$. If $|P| \\ge \\frac{p+1}{2}$, then $|r + P| \\ge \\frac{p+1}{2}$, too. Hence, $|P| + |r + P| \\ge p + 1 > p$, so, by the Pigeonhole Principle, $P$ and $r + P$ must have an element in common. In other words, there are $p_1, p_2$ with $p_1 \\equiv r + p_2 \\pmod{p}$ and hence $p_1 - p_2 \\equiv r \\pmod{p}$, which gives a solution of the desired shape from the definition of $P$. So the only remaining case is that of $|P| = |A| = \\frac{p-1}{2}$.\n\nMultiplying all elements of $A$ with the same constant and reducing modulo $p$, if necessary, we may assume w.l.o.g. that $1 \\in A$. Then $A \\subseteq P$ and hence $A = P$. This means that the non-zero elements of $A$ form a group under multiplication.\n\nIf $0 \\in A$, then this group has size $\\frac{p-3}{2}$, which has to divide the group order $p-1$, and hence also has to divide $2 = p-1-2 \\cdot \\frac{p-3}{2}$. This is impossible for $p > 7$.\n\nConsequently, $0 \\notin A$ and the group has size $\\frac{p-1}{2}$ and hence is exactly the group of quadratic residues (here we use the existence of primitive roots implicitly).\n\nReplacing $r$ by $r+p$, if necessary, one may assume $r$ to be odd. Then put $b = d := 1 \\in A$, as well as\n$$\na \\equiv \\left( \\frac{r+1}{2} \\right)^2 \\pmod{p} \\quad \\text{and}\n$$\n$$\nc \\equiv \\left( \\frac{r-1}{2} \\right)^2 \\pmod{p}.\n$$\nThen $a, c \\in A$, too. This yields\n$$\nad - bc \\equiv a - c \\equiv \\left(\\frac{r+1}{2}\\right)^2 - \\left(\\frac{r-1}{2}\\right)^2 \\equiv r \\pmod{p},\n$$\nas required.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25179,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given real non-zero numbers $a$, $b$, $c$ ($b > 0$) so that two distinct roots of the equation $ax^2 + bx + c = 0$ are also the roots of the equation $x^3 + bx^2 + ax + c = 0$.\nProve that\n\na) $abc > 16$;\n\nb) $abc \\ge \\frac{3125}{108}$.\n\n(V. Karamzin)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $x_1$, $x_2$ be distinct roots of the equation $ax^2 + bx + c = 0$, i.e. be zeroes of the functions $f(x) = ax^2 + bx + c$ and $g(x) = x^3 + bx^2 + ax + c$. By condition, $f(0) = g(0) = c \\neq 0$. Let $F(x) = g(x) - f(x)$. Then $0$, $x_1$, $x_2$ are the distinct zeroes of the polynomial $F(x)$. So $F(x) = x(x - x_1)(x - x_2)$, hence\n$$\nx^3 + (b-a)x^2 + (a-b)x = x(x-x_1)(x-x_2) \\\\\n(b-a)x + (a-b) = -(x_1+x_2)x + x_1x_2 \\quad \\forall x \\in \\mathbb{R}.\n$$\nTherefore, $x_1+x_2 = a-b$ and $x_1x_2 = a-b$. By Vieta's theorem, $x_1+x_2 = -b/a$, $x_1x_2 = c/a$, which gives $c = b$. Further, $a-b = -b/a$ or $a^2-ab = -b$, i.e. $a^2 = b(a-1)$. Then $a \\neq 1$, and from $b > 0$ it follows that $a > 1$.\n\na) Since $b = a^2/(a-1)$, we have\n$$\nc = b = \\frac{a^2}{a-1} = \\frac{(a-1+1)^2}{a-1} = \\frac{(a-1)^2 + 2(a-1) + 1}{a-1} = (a-1) + 2 + \\frac{1}{a-1} \\ge 4.\n$$\nTherefore, $abc = ab^2 > 1 \\cdot 4^2 = 16$, as required.\n\n\nb) First solution. Since $ab = a^2 + b$, we see that\n$$\nabc = ab^2 = (a^2+b)b = \\left( \\frac{a^2b}{2} + \\frac{a^2b}{2} + \\frac{b^2}{3} + \\frac{b^2}{3} + \\frac{b^2}{3} \\right) \\ge 5\\sqrt[5]{\\frac{a^2b}{2} \\cdot \\frac{a^2b}{2} \\cdot \\frac{b^2}{3} \\cdot \\frac{b^2}{3} \\cdot \\frac{b^2}{3}} = 5\\sqrt[5]{\\frac{a^4b^8}{4 \\cdot 27}},\n$$\nhence\n$$\n(ab^2)^5 \\ge \\frac{5^5}{108}a^4b^8 \\implies abc = ab^2 \\ge \\frac{3125}{108},\n$$\nas required. Note that the equality is achieved only if $\\frac{a^2}{2} = \\frac{b}{3}$, then from $b = \\frac{a^2}{a-1}$ it follows that $a - 1 = 2/3$, i.e. $a = 5/3$.\n\n\n**Second solution.** We have $abc = ab^2 = \\frac{a^5}{(a-1)^2}$. We find the smallest value of $h(a) = \\frac{a^5}{(a-1)^2}$ on $(1; +\\infty)$. We see that\n$$\nh'(a) = 0 \\iff \\frac{5a^4(a-1)^2 - 2(a-1)a^5}{(a-1)^4} = \\frac{a^4(a-1)(3a-5)}{(a-1)^4} = 0,\n$$\nso $a = 5/3$. Since $h'(a) < 0$ on $(1; 5/3)$ and $h'(a) > 0$ on $(5/3; +\\infty)$, we see that the smallest value of $h$ on $(1; +\\infty)$ is equal to $h(5/3) = \\frac{5^5}{108}$. Therefore, $abc \\ge \\frac{5^5}{108}$, as required.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 25180,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all possible values of the expression\n$$\n\\frac{1}{a}\\left(\\frac{1}{b}+\\frac{1}{c}+\\frac{1}{b+c}\\right)+\\frac{1}{b}\\left(\\frac{1}{c}+\\frac{1}{a}+\\frac{1}{c+a}\\right)+\\frac{1}{c}\\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{a+b}\\right)-\\frac{1}{a+b+c}\\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}+\\frac{1}{a+b}+\\frac{1}{b+c}+\\frac{1}{c+a}\\right)+\\frac{1}{a^2}+\\frac{1}{b^2}+\\frac{1}{c^2}\n$$\nif positive numbers $a, b, c$ satisfy the condition $ab + bc + ca = abc$. (S. Chernov)",
    "options": [],
    "answer": "1",
    "solution": "Answer: $1$.\nBy condition, $1/a + 1/b + 1/c = 1$. Let $x = 1/a$, $y = 1/b$, $z = 1/c$. Then\n$$\nx + y + z = 1. \\quad (1)\n$$\nWe rewrite the initial expression as\n$$\n\\begin{align*}\n&+ \\left( \\frac{xyz}{x+y} - \\frac{xyz}{xy+yz+zx} \\cdot \\frac{xy}{x+y} \\right) - \\frac{xyz}{xy+yz+zx} = \\\\\n&= 1 + \\frac{xyz(xy + zx)}{(xy + yz + zx)(y + z)} + \\frac{xyz(xz + yz)}{(xy + yz + zx)(x + y)} + \\\\\n&\\qquad + \\frac{xyz(zy + xy)}{(xy + yz + zx)(z + x)} - \\frac{xyz}{xy + yz + zx} = \\\\\n&= 1 + \\frac{x^2yz}{xy + yz + zx} + \\frac{xyz^2}{xy + yz + zx} + \\frac{xy^2z}{xy + yz + zx} - \\frac{xyz}{xy + yz + zx} = \\\\\n&= \\therefore 1 + \\frac{xyz(x + y + z)}{xy + yz + zx} - \\frac{xyz}{xy + yz + zx} = 1.\n\\end{align*}\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 25181,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of positive integers $a$ and $b$ ($a \\le b$) such that\n$$\nab = 300 + 7[a, b] + 5(a, b),\n$$\nwhere $(a, b)$ is the greatest common divisor and $[a, b]$ is the least common multiple of $a$ and $b$.",
    "options": [],
    "answer": "[(12, 72), (24, 36)]",
    "solution": "Let $[a, b] = x$, and $(a, b) = y$. It is known (and easy to prove) that $ab = [a, b] \\cdot (a, b)$, so $ab = xy$. Then the initial equation can be rewritten in the form\n$$\nxy = 300 + 7x + 5y \\Leftrightarrow xy - 7x - 5y + 35 = 335 \\Leftrightarrow x(y - 7) - 5(y - 7) = 335 \\Leftrightarrow (x - 5)(y - 7) = 5 \\cdot 67.\n$$\nFactors on the left-hand side are nonnegative integers, and since $x = [a, b] \\ge (a, b) = y$, we have $(x - 5) > (y - 7)$. So we have two possibilities.\n\n1) $x - 5 = 67$ and $y - 7 = 5$. Then $x = 72$ and $y = 12$. It means that $a = 12n$, $b = 12m$ for some relatively prime $n, m$, and $ab = xy = 72 \\cdot 12$, i.e. $12n \\cdot 12m = 72 \\cdot 12$, thus $nm = 6$. Taking into account that $a \\le b$ and so $n \\le m$, we obtain $n = 1, m = 6$. Therefore $a = 12, b = 12 \\cdot 6 = 72$, or $n = 2, m = 3$, so $a = 12 \\cdot 2 = 24$, $b = 12 \\cdot 3 = 36$.\n\n2) $x - 5 = 335$ and $y - 7 = 1$. Then $x = 340$ and $y = 8$. But it is impossible since $[a, b] = x$ must be divisible by $y = (a, b)$, but $340 \\nmid 8$.\n\nTherefore, we have two required pairs $(12, 72)$, $(24, 36)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25182,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f(x)$ be a quadratic trinomial, and $a, b, c$ be pairwise distinct real numbers.\nGiven $f(a) = bc$, $f(b) = ac$, $f(c) = ab$, find $f(a + b + c)$.",
    "options": [],
    "answer": "ab + bc + ac",
    "solution": "Answer: $ab + bc + ac$.\nLet $f(x) = \\alpha x^2 + \\beta x + \\gamma$. Then\n$$\n\\alpha a^2 + \\beta a + \\gamma = bc, \\quad \\alpha b^2 + \\beta b + \\gamma = ac, \\quad \\alpha c^2 + \\beta c + \\gamma = ab. \\quad (1)\n$$\nSubtracting the second and the third equations from the first one, we have\n$$\n\\alpha(a^2 - b^2) + \\beta(a - b) = c(b - a), \\quad \\alpha(a^2 - c^2) + \\beta(a - c) = b(c - a).\n$$\nSince $a, b, c$ are pairwise distinct, we have\n$$\n\\alpha(a+b) + \\beta = -c, \\quad \\alpha(a+c) + \\beta = -b. \\quad (2)\n$$\nSubtracting the second equation from the first one, we have $\\alpha(b-c) = b-c$. Since $b-c \\ne 0$, we have $\\alpha = 1$. Then it follows from (2) that $\\beta = -(a+b+c)$, and it follows from (1) that $\\gamma = ab+bc+ac$.\nTherefore, $f(x) = x^2-(a+b+c)x+(ab+bc+ac)$, so $f(a+b+c) = ab+bc+ac$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25183,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of positive integers $a$ and $b$ such that\n$$\nab = 160 + 90(a, b),\n$$\nwhere $(a, b)$ is the greatest common divisor of $a$ and $b$.",
    "options": [],
    "answer": "(125, 2), (2, 125), (250, 1), (1, 250), (10, 34), (34, 10), (170, 2), (2, 170)",
    "solution": "By condition it follows that one of the numbers is divisible by $5$. Moreover, exactly one of the numbers is divisible by $5$, otherwise $(a, b)$ and $90(a, b)$ are divisible by $25$, and so $160$ is divisible by $25$, a contradiction. Let without loss of generality $a \\neq 5$, $b \\neq 5$. Then $a = 5c$, $(a, b) = (5c, b) = (c, b)$, and the equation can be rewritten as $bc = 32+18(c, b)$. Since $bc$ and $18(c, b)$ are divisible by $(c, b)$, it follows that $(c, b)$ is a factor of $32$, i.e. $(c, b) = 1, 2, 4, 8, 16, 32$.\n\nIf $(c, b) \\ge 4$, then $bc$ is divisible by $4^2 = 16$, so $18(c, b) = (bc - 32) \\neq 16$, and, therefore, $(c, b) \\neq 8$. Thus $bc \\neq 64$, i.e. $18(c, b) \\neq 32$. Therefore, $(c, b) \\neq 16$. If either $(c, b) = 16$ or $(c, b) = 32$, then $bc = 16^2$, but $bc = 32+18 \\cdot 16$ or $bc = 32+18 \\cdot 64$, which are impossible. Hence $(c, b) \\le 2$, i.e. is equal to either $1$ or $2$.\n\n1) Let $(c, b) = 1$. Then $bc = 50$, and at least one of the numbers is odd. Since $b \\neq 5$, we find that either $c = 25$ and $b = 2$ or $c = 50$ and $b = 1$. This gives the solutions of the initial equation: $(a, b) = (125, 2)$ and $(a, b) = (250, 1)$.\n\n2) Now let $(c, b) = 2$. Then $bc = 32+36=68$. Set $b=2b_1$, $c=2c_1$, where $b_1$ and $c_1$ are coprime. Then $b_1c_1 = 17$. Hence either $c_1 = 1$ and $b_1 = 17$ or $c_1 = 17$ and $b_1 = 1$. This gives the solutions of the initial equation: $(a, b) = (10, 34)$ and $(a, b) = (170, 2)$.\n\nSince the initial equation is symmetric with respect to $a$ and $b$, we have four more solutions: $(34; 10)$, $(2; 125)$, $(2; 170)$, $(1; 250)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25184,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a trapezium $ABCD$ with $BC \\parallel AD$, $AD > BC$. Let $M$ be the point of intersection of $AC$ and $BD$. A circle $\\Gamma_1$ passes through $M$ and touches $AD$ at $A$. A circle $\\Gamma_2$ passes through $M$ and touches $AD$ at $D$. Let $S$ be the point of intersection of the lines $AB$ and $DC$, $X$ be the point of intersection of the circle $\\Gamma_1$ and the line $AS$, $Y$ be the point of intersection of the circle $\\Gamma_2$ and the line $DS$, $O$ be the circumcenter of the triangle $ASD$.\nProve that $SO \\perp XY$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Draw the line $LM$, where $L$ is the second (different from $M$) point of intersection of $\\Gamma_1$ and $\\Gamma_2$. (If $\\Gamma_1$ touches $\\Gamma_2$ at $M$, then $LM$ is the common tangent of these circles at $M$.) Let $K$ be the point of intersection of the lines $LM$ and $AD$. By the tangent-secant theorem, we have $KA^2 = KM \\cdot KL = KD^2$ (or $KA^2 = KM^2 = KD^2$) $\\Rightarrow$ $KA = KD$.\n\n![](attached_image_1.png)\n\nIt is well-known fact that the midpoint of the base, of the trapezium, $K$, the point of intersection of the trapezium diagonals, $M$, and the point of intersection of the extensions of its lateral sides, $S$, lie on the same line. Then $SX \\cdot SA = SL \\cdot SM = SY \\cdot SD$. Hence\n$SX : SY = SD : SA$, which gives the similarity of the triangles $SXY$ and $SDA$. Thus $\\angle SYX = \\angle SAD$. The angles $SOD$ and $SAD$ are respectively a central and an inscribed angle of the same circle, and both of them are subtended by the same chord $SD$. Hence $\\angle SAD = 0.5\\angle SOD$. Therefore,\n$$\n\\begin{aligned}\n\\angle SYX + \\angle YSO &= \\angle SAD + \\angle YSO = 0.5\\angle SOD + \\angle YSO = \\\\\n&= 0.5(180^\\circ - \\angle OSD - \\angle ODS) + \\angle YSO = [\\angle OSD = \\angle ODS] = \\\\\n&= 0.5(180^\\circ - 2\\angle ODS) + \\angle YSO = 90^\\circ + (\\angle YSO - \\angle ODS) = 90^\\circ,\n\\end{aligned}\n$$\nwhich implies $SO \\perp XY$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25185,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all possible non-zero integers $a, b, c$, so that two distinct roots of the equation $ax^2+bx+c = 0$ are also the roots of the equation $x^3 + bx^2 + ax + c = 0$.",
    "options": [],
    "answer": "a = 2, b = 4, c = -4",
    "solution": "Answer: $a = 2$, $b = 4$, $c = -4$.\n\nLet $x_1, x_2$ be distinct roots of the equation $ax^2+bx+c=0$, i.e. be the zeroes of the function $f(x) = ax^2+bx+c$ and $g(x) = x^3+bx^2+ax+c$. By condition, $f(0) = g(0) = c \\ne 0$. Let $F(x) = g(x) - f(x)$. Then $0, x_1, x_2$ are the distinct zeroes of the polynomial $F(x)$. So, $F(x) = x(x - x_1)(x - x_2)$, then\n$$\nx^3 + (b-a)x^2 + (a-b)x = x(x-x_1)(x-x_2) \\Rightarrow \\\\\n(b-a)x + (a-b) = -(x_1+x_2)x + x_1x_2 \\quad \\forall x \\in \\mathbb{R}.\n$$\nTherefore, $x_1 + x_2 = a - b$ and $x_1x_2 = a - b$. By the Vieta theorem $x_1 + x_2 = -b/a$, $x_1x_2 = c/a$, hence $c = -b$. Further, $a - b = -b/a$ or $a^2 - ab = -b$, i.e. $a^2 = b(a - 1)$. Note that we have $a \\ne 1$, so\n$$\nb = \\frac{a^2}{a-1} = \\frac{(a^2-1)+1}{a-1} = a+1 + \\frac{1}{a-1}.\n$$\nSince $a$ and $b$ are integers, we see that $1/(a-1)$ is also integer, which is possible only if $a=0$ and $a=2$. But $a \\ne 0$, so $a=2$ and then $b=4$, $c=-4$.\n\nStraightforward calculation show that the obtained triple satisfy the problem condition. Indeed, the equation $2x^2+4x-4=0$ has two distinct roots $-1 \\pm \\sqrt{3}$. These numbers are the roots of the equation $x^3+4x^2-2x-4$ since\n$$\n(-1 \\pm \\sqrt{3})^3 + 4 \\cdot (-1 \\pm \\sqrt{3})^2 - 2 \\cdot (-1 \\pm \\sqrt{3}) - 4 = \\\\\n= -1 \\pm 3\\sqrt{3} - 3 \\cdot 3 \\pm 3\\sqrt{3} + 4 \\mp 8\\sqrt{3} + 12 - 2 \\pm 2\\sqrt{3} - 4 = 0.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 25186,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given real non-zero numbers $a$, $b$, $c$ ($b > 0$) so that two distinct roots of the equation $ax^2 + bx + c = 0$ are also the roots of the equation $x^3 + bx^2 + ax + c = 0$.\n\nProve that a) $abc > 16$; b) $abc \\ge \\frac{3125}{108}$. (V. Karamzin)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $x_1$, $x_2$ be distinct roots of the equation $ax^2 + bx + c = 0$, i.e. be zeroes of the functions $f(x) = ax^2 + bx + c$ and $g(x) = x^3 + bx^2 + ax + c$. By condition, $f(0) = g(0) = -c \\neq 0$. Let $F(x) = g(x) - f(x)$. Then $0$, $x_1$, $x_2$ are the distinct zeroes of the polynomial $F(x)$. So $F(x) = x(x - x_1)(x - x_2)$, hence\n$$\n\\begin{aligned}\nx^3 + (b - a)x^2 + (a - b)x &= x(x - x_1)(x - x_2) \\\\\n(b - a)x + (a - b) &= -(x_1 + x_2)x + x_1x_2 \\quad \\forall x \\in \\mathbb{R}.\n\\end{aligned}\n$$\nTherefore, $x_1 + x_2 = a - b$ and $x_1x_2 = a - b$. By Vieta theorem, $x_1 + x_2 = -b/a$, $x_1x_2 = -c/a$, which gives $c = b$. Further, $a - b = -b/a$ or $a^2 - ab = -b$, i.e. $a^2 = b(a - 1)$. Then $a \\neq 1$, and from $b > 0$ it follows that $a > 1$.\n\n**Second solution.** We have $abc = ab^2 = \\frac{a^5}{(a - 1)^2}$. We find the smallest value of $h(a) = \\frac{a^5}{(a - 1)^2}$ on $(1; +\\infty)$. We see that\n$$\nh'(a) = 0 \\iff \\frac{5a^4(a - 1)^2 - 2(a - 1)a^5}{(a - 1)^4} = \\frac{a^4(a - 1)(3a - 5)}{(a - 1)^4} = 0,\n$$\nso $a = 5/3$. Since $h'(a) < 0$ on $(1; 5/3)$ and $h'(a) > 0$ on $(5/3; +\\infty)$, we see that the smallest value of $h$ on $(1; +\\infty)$ is equal to $h(5/3) = \\frac{5^5}{108}$. Therefore, $abc \\ge \\frac{5^5}{108}$, as required.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 25187,
    "subject": "Mathematics (Multi-modal)",
    "question": "The side $AB$ is the least side in a triangle $ABC$. Points $M$ and $N$ are marked on the rays $CA$ and $CB$ respectively so that $CM = MB$, $CN = NA$. Let $O$ be a circumcenter of the triangle $ABC$.\nProve that $A$, $B$, $N$, $M$, and $O$ are concyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $P$ and $Q$ be the midpoints of $BC$ and $AC$ respectively. Since $CM = MB$ and $CN = NA$, we see that $M$ and $N$ lie on the perpendicular bisectors of the sides $BC$ and $AC$ respectively. Since $AB$ is the smallest side, the distance between $A$ and $B$ is less than the distance between $A$ and $C$, so $A$ and $C$ lie in the different half-planes with respect to the perpendicular bisector of the side $BC$. Thus $M$ lies on the side $AC$. Similarly, $N$ lies on the side $BC$. By condition, $CM = MB$, it follows that the triangle $BMC$ is isosceles and $\\angle BCM = \\angle MBC$. Similarly, $\\angle ACN = \\angle NAC$. Since $\\angle BCM = \\angle ACN$, we have\n\n![](attached_image_1.png)\n\n$$\n\\angle NAM = \\angle NAC = \\angle MBC = \\angle MBN.\n$$\n\nTherefore, $A$, $B$, $M$, $N$ lie on the same circle $\\Gamma$ (since the angles $NAM$ and $MBN$ are subtended by the same segment, $MN$). It remains to note that the angles $NOM$ and $ACB$ are the angles with mutually perpendicular sides, so either $\\angle NOM + \\angle ACB = \\angle NOM + \\angle MBN = 180^\\circ$ (see Fig. 1), or $\\angle NOM = \\angle ACB = \\angle MBN$ (see Fig. 2). It follows that $O$ lie on $\\Gamma$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25188,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the value of the expression\n$$(a+b+c) \\left( \\frac{1}{a+b-5c} + \\frac{1}{b+c-5a} + \\frac{1}{c+a-5b} \\right),$$\nif real $a$, $b$, $c$ satisfy the equality $$(a+b+c) \\left( \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\right) = \\frac{27}{2}$$ (all denominators are supposed to be different from zero).",
    "options": [],
    "answer": "9/5",
    "solution": "Set $a+b+c = x$ and $\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} = y$. Let $k = \\frac{x}{6}$, then\n$$\na+b-5c=6(k-c), \\quad b+c-5a=6(k-a), \\quad c+a-5b=6(k-b).\n$$\nIt follows that\n$$\n(a+b+c) \\left( \\frac{1}{a+b-5c} + \\frac{1}{b+c-5a} + \\frac{1}{c+a-5b} \\right) = \\frac{x}{6} \\left( \\frac{1}{k-a} + \\frac{1}{k-b} + \\frac{1}{k-c} \\right). \\quad (1)\n$$\nAdding the fractions in the right-hand side of (1), we obtain $\\frac{A}{B}$, where\n$$\nA = (k-a)(k-b) + (k-b)(k-c) + (k-c)(k-a) \\text{ and } B = (k-a)(k-b)(k-c). \\quad (2)\n$$\nLet $z = abc$, then $ab + bc + ca = yz$. From (2) it follows\n$$\nA = 3k^2 - 2zk + yz \\quad \\text{and}\\quad B = k^3 - zk^2 + yz - z. \\quad (3)\n$$\nSince $x = 6k$ and $xy = \\frac{27}{2}$, we have $y = \\frac{9}{4k}$. Substitute for $x$ and $y$ in (3) to obtain\n$$\nA = 3k^2 - 2 \\cdot 6k \\cdot k + \\frac{9}{4k}z = 9 \\left(-k^2 + \\frac{z}{4k}\\right), \\quad B = k^3 - 6k \\cdot k^2 + \\frac{9}{4k}zk - z = 5k \\left(-k^2 + \\frac{z}{4k}\\right).\n$$\nThus\n$$\n\\frac{A}{B} = \\frac{9}{5k}, \\quad \\text{and}\\quad \\frac{x}{6} \\cdot \\frac{A}{B} = k \\cdot \\frac{9}{5k} = \\frac{9}{5}.\n$$\nHence\n$$\n(a+b+c) \\left( \\frac{1}{a+b-5c} + \\frac{1}{b+c-5a} + \\frac{1}{c+a-5b} \\right) = \\frac{9}{5}.\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 25189,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all $x$ satisfying the equality\n$$\n[x^2] - [-x^2] - 8[x] + 2 = 0.\n$$\n(Here $[x]$ stands for the greatest integer not exceeding $x$.)",
    "options": [],
    "answer": "sqrt(3), sqrt(7), sqrt(11)",
    "solution": "Answer: $\\sqrt{3}$, $\\sqrt{7}$, $\\sqrt{11}$.\nSince $8[x]$ is an integer number, we can rewrite the initial equation as\n$[x^2 - 4[x] + 1] - [-x^2 - 4[x] + 1] = 0$.\nSet $y = x^2 - 4[x] + 1$, then $[y] = [-y]$.\nIt is easy to see that for $y > 0$ we have $0 \\le [y] = [-y] \\le -1$, which is impossible.\nSimilarly, for $y < 0$ we have $-1 \\ge [y] = [-y] \\ge 0$.\nSo $y = 0$, i.e. $x^2 - 4[x] + 1 = 0$.\nNote that for $x < 1$ we have $[x] \\le 0$, therefore $x^2 - 4[x] + 1 \\ge 1 > 0$.\nThus $x \\ge 1$.\nLet $x = n + a$, where $n \\in \\mathbb{N}$, $0 \\le a < 1$.\nThen $0 = x^2 - 4[x] + 1 = n^2 + 2na + a^2 - 4n + 1 = n(n - 4) + 2na + a^2 \\ge n(n - 4)$, so $n \\le 4$.\nTherefore, $1 \\le n \\le 4$.\n$$\n\\text{If } [x] = n = 1, \\text{ then } x^2 - 4[x] + 1 = x^2 - 4 + 1 = x^2 - 3 = 0, \\text{ so } x = \\sqrt{3}.\n$$\n$$\n\\text{If } [x] = n = 2, \\text{ then } x^2 - 4[x] + 1 = x^2 - 8 + 1 = x^2 - 7 = 0, \\text{ so } x = \\sqrt{7}.\n$$\n$$\n\\text{If } [x] = n = 3, \\text{ then } x^2 - 4[x] + 1 = x^2 - 12 + 1 = x^2 - 11 = 0, \\text{ so } x = \\sqrt{11}.\n$$\n$$\n\\text{If } [x] = n = 4, \\text{ then } x^2 - 4[x] + 1 = x^2 - 16 + 1 = x^2 - 15 = 0, \\text{ so } x = \\sqrt{15} < 4, \\text{ contrary to } [x] = 4.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25190,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that $|a\\sqrt{3} - b\\sqrt{5}| > \\frac{4}{4a + 5b}$ for any naturals $a$ and $b$.\n(V. Bernik)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Multiplying the left-hand side of the required inequality by $a\\sqrt{3} + b\\sqrt{5}$, we obtain $|a\\sqrt{3} - b\\sqrt{5}|(a\\sqrt{3} + b\\sqrt{5}) = |3a^2 - 5b^2|$. If $|3a^2 - 5b^2| = 1$ for some integers $a$ and $b$, then either $3a^2 - 5b^2 = 1$ or $3a^2 - 5b^2 = -1$. In the first case we have $3a^2 \\equiv 1 \\pmod{5}$, which is impossible since $3a^2 \\equiv 0 \\pmod{5}$, $3a^2 \\equiv 3 \\pmod{5}$ for $a \\equiv \\pm 1 \\pmod{5}$, and $3a^2 \\equiv 12 \\equiv 2 \\pmod{5}$ for $a \\equiv \\pm 2 \\pmod{5}$.\n\nIn the second case we have $-5b^2 \\equiv (-1) \\pmod{3}$, i.e. $5b^2 \\equiv 1 \\pmod{3}$, which is impossible since $5b^2 \\equiv 0 \\pmod{3}$ for $b \\equiv 0 \\pmod{3}$, and $5b^2 \\equiv 5 \\pmod{3}$ for $b \\equiv \\pm 1$.\n\nTherefore, $|3a^2 - 5b^2| \\neq 1$ for all integers $a$ and $b$, hence $|3a^2 - 5b^2| \\geq 2$ for all integers $a$ and $b$. So\n$$\n\\begin{aligned}\n|a\\sqrt{3} - b\\sqrt{5}| &= \\frac{|3a^2 - 5b^2|}{a\\sqrt{3} + b\\sqrt{5}} \\ge \\frac{2}{a\\sqrt{3} + b\\sqrt{5}} = \\frac{4}{2a\\sqrt{3} + 2b\\sqrt{5}} \\\\\n&\\ge [2\\sqrt{3} < 4,\\ 2\\sqrt{5} < 5] \\ge \\frac{4}{4a + 5b}\n\\end{aligned}\n$$\nas required.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25191,
    "subject": "Mathematics (Multi-modal)",
    "question": "Integers from $1$ to $80$ are written on a blackboard. Tom and Jerry play the following game. They, in turn (Tom is the first), erase these numbers. Per move it is allowed to erase exactly $1$, or exactly $5$, or exactly $8$ numbers. The player wins if he erases the last number.\nWho wins if both of the players play to win?",
    "options": [],
    "answer": "Jerry",
    "solution": "It is easy to see that if either $1$ or $5$ numbers remain on the blackboard, then the player who must move wins, but if $2$ or $4$ numbers remain, then this player loses. So we will solve the problem moving backward. We write all numbers from $1$ to $80$ and mark them with \"+\" or \"-\". If $k$ numbers remain on the blackboard before the move of a player and he can win, then we write $+k$, otherwise we write $-k$. We have the following table:\n\n| +1 | -2 | +3 | -4 | +5 | -6 | +7 | +8 | +9 | +10 | +11 | +12 | -13 |\n|----|----|----|----|----|----|----|----|----|-----|-----|-----|-----|\n| 1  | -  | 1  | -  | 5  | -  | 1  | 8  | 5  | 8   | 5   | 8   | -   |\n\n| +14 | -15 | +16 | -17 | +18 | -19 | +20 | +21 | +22 | +23 | +24 | +25 | -26 | ... |\n|-----|-----|-----|-----|-----|-----|-----|-----|-----|-----|-----|-----|-----|-----|\n| 1   | -   | 1   | -   | 5   | -   | 1   | 8   | 5   | 8   | 5   | 8   | -   | ... |\n\nThe second row of the table contains the possible winning moves (to win the player can erase so many numbers that after his move the quantity of the numbers remained on the blackboard would be marked with \"+\"). We see that the signs in the first row of the table are repeated with the period equaled $13$ (it is easy to prove by induction).\n\nSince $80$ leaves remainder $2$ when divided by $13$, it follows that $80$ and $2$ are marked with the same sign, i.e. \"-\". Thus the number $80$ is the losing number of the numbers for the beginning player. To win Jerry can erase so many numbers as it is written in the second row of the table.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25192,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find $\\left\\{ \\frac{2009!}{2011!} \\right\\}$. (Here $\\{x\\}$ means the fractional part of $x$.)",
    "options": [],
    "answer": "1/2011",
    "solution": "Answer: $\\frac{1}{2011}$.\nSince $2011$ is a prime number, we see that $2010! \\equiv -1 \\pmod{2011}$ (Wilson's theorem). Hence $2010! \\equiv 2010 \\pmod{2011}$, which gives",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25193,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $r$ be some positive real number. It is known that for some positive integer $n$ the following condition holds: all positive real numbers $a_1, \\dots, a_n$ satisfying the equality $a_1 + \\dots + a_n = r\\left(\\frac{1}{a_1} + \\dots + \\frac{1}{a_n}\\right)$, $a_i \\ne \\sqrt{r}$, $i = 1, \\dots, n$, also satisfy the equality $\\frac{1}{\\sqrt{r} - a_1} + \\dots + \\frac{1}{\\sqrt{r} - a_n} = \\frac{1}{\\sqrt{r}}$. Find $n$.\n(D. Bazylev)",
    "options": [],
    "answer": "n = 2",
    "solution": "Note that if $a_1, \\dots, a_n$ satisfy the equality\n$$\na_1 + \\dots + a_n = r \\left( \\frac{1}{a_1} + \\dots + \\frac{1}{a_n} \\right), \\quad (*)\n$$\nthen the numbers $b_1 = \\frac{r}{a_1}, \\dots, b_n = \\frac{r}{a_n}$ satisfy the equality\n$$\nb_1 + \\dots + b_n = r \\left( \\frac{1}{b_1} + \\dots + \\frac{1}{b_n} \\right).\n$$\nBy condition, we have\n$$\n\\frac{1}{\\sqrt{r} - b_1} + \\dots + \\frac{1}{\\sqrt{r} - b_n} = \\frac{1}{\\sqrt{r}},\n$$\nwhich gives $\\frac{1}{\\sqrt{r} - \\frac{r}{a_1}} + \\dots + \\frac{1}{\\sqrt{r} - \\frac{r}{a_n}} = \\frac{1}{\\sqrt{r}}$. The last equality is equivalent to the equality\n$$\n\\frac{a_1}{\\sqrt{r} - a_1} + \\dots + \\frac{a_n}{\\sqrt{r} - a_n} = -1. \\quad (1)\n$$\nBy condition,\n$$\n\\frac{1}{\\sqrt{r} - a_1} + \\dots + \\frac{1}{\\sqrt{r} - a_n} = \\frac{1}{\\sqrt{r}}. \\quad (2)\n$$\nSubtracting (1) from (2) multiplied by $\\sqrt{r}$, we obtain\n$$\n\\frac{\\sqrt{r} - a_1}{\\sqrt{r} - a_1} + \\dots + \\frac{\\sqrt{r} - a_n}{\\sqrt{r} - a_n} = 1 - (-1) = 2.\n$$\nIt follows that $n=2$.\n\nIt remains to verify that the problem condition holds for $n = 2$. Indeed\n$$\na_1 + a_2 = r \\left( \\frac{1}{a_1} + \\frac{1}{a_2} \\right) \\iff (a_1 + a_2) \\left( 1 - \\frac{r}{a_1 a_2} \\right) = 0.\n$$\nBy condition, $a_1, a_2 \\ge 0$ hence the last equality is equivalent to the equality $a_1 a_2 = r$. Then\n$$\n\\frac{1}{\\sqrt{r} - a_1} + \\frac{1}{\\sqrt{r} - a_2} = \\frac{2\\sqrt{r} - (a_1 + a_2)}{r - \\sqrt{r}(a_1 + a_2) + a_1 a_2} = \\frac{2\\sqrt{r} - (a_1 + a_2)}{2r - \\sqrt{r}(a_1 + a_2)} = \\frac{1}{\\sqrt{r}},\n$$\nas required.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25194,
    "subject": "Mathematics (Multi-modal)",
    "question": "Fifteen points are marked on a plane. Some of them are painted red, some others are painted blue, and all remained points are painted green. It is known that the number of the red points is the largest. The sum of the distances between the red points and the blue points is $5$, the sum of the distances between the red points and the green points is $31$, and the sum of the distances between the blue points and the green points is $25$.\nFind the number of points of each color.\n(I. Voronovich)",
    "options": [],
    "answer": "red 6, blue 5, green 4",
    "solution": "Let $m, n, k$ be respectively the numbers of red, green, and blue points. Let $A, B, C$ be respectively the sums of the distances between blue and green points, red and blue points, red and green points. From the inequality of triangle it follows that the following inequalities are necessary for the existence of the points on a plane:\n$$\nmA - kC \\le nB \\le mA + kC, \\quad mA - nB \\le kC \\le mA + nB, \\quad nB - kC \\le mA \\le nB + kC \\quad (*)\n$$\nIn our case\n$$\n25m - 5n \\le 31k, \\quad (1)\n$$\n$$\n31k \\le 25m + 5n. \\quad (2)\n$$\nBy condition $m+n+k=15$. Since the number of the red points is the largest, we have $m \\ge 6, k \\le 6, n \\le 6$.\nSuppose that $m \\ge 7$. Then from (1) it follows $26k + (5k + 5n + 5m) \\ge 30m$, hence $26k + 75 \\ge 30m$, which gives $k \\ge 6$. Therefore, $k=6$, so $m+n=9$, and we have\n$$\n31k \\le 25m + 5n \\Rightarrow 31 \\cdot 6 \\le 25m + 5n = 20m + 5(m + n) = 20m + 45.\n$$\n\nTherefore, $m=6, k+n=9$. Now equality (1) has the form\n$$\n150 \\le 31k + 5n = 26k + 5(k + n) = 26k + 45 \\Rightarrow 26k \\ge 105,\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25195,
    "subject": "Mathematics (Multi-modal)",
    "question": "4 blue, 10 green and several (at least one) red points are marked on a plane. All points are distinct. It is known that the sum of the distances between the red points and the blue points is $21$, the sum of the distances between the red points and the green points is $2$.\nCan the sum of the distances between the blue points and the green points be equal to \na) $20$? \nb) $18$?",
    "options": [],
    "answer": "a) no; b) yes",
    "solution": "a) Let $n$ denote the number of red points. Then the following inequality holds (see (*) in the solution of Problem 2, Category C)\n$$\n10 \\cdot 21 - 4 \\cdot 2 \\le n \\cdot 20 \\le 10 \\cdot 21 + 4 \\cdot 2 \\iff 202 \\le 20n \\le 218.\n$$\nIt is easy to see that there are no integers $n$ satisfying this condition.\n\nb) We show that it is possible to mark $4$ blue, $10$ green, and $12$ red points on a plane so that the problem conditions hold.\nFor the convenience we multiply all distances by $4 \\cdot 10 \\cdot 12 = 480$. Let $A, B, C$ denote respectively the sums of the distances between red and green points, between blue and red points, green and blue points.\nWe have to construct the example with $A = 2 \\cdot 480$, $B = 21 \\cdot 480$, $C = 18 \\cdot 480$. All points we place on the number axes.\n\nFirst, we place $4$ blue points at the point $0$, $10$ green points at the point $216$, $6$ red points at the point $202$, and the other $6$ red points at the point $218$. Then $C = 4 \\cdot 10 \\cdot 216 = 18 \\cdot 480$, $B = 4 \\cdot 6 \\cdot 202 + 4 \\cdot 6 \\cdot 218 = 21 \\cdot 480$, and $A = 10 \\cdot 6 \\cdot (216 - 202) + 10 \\cdot 6 \\cdot (218 - 216) = 2 \\cdot 480$.\nIt remains to move slightly the points so that all points become distinct but $A, B, C$ keep their values. We can replace $4$ blue points with the common coordinate $0$ by the points $-2\\epsilon$, $-\\epsilon$, $\\epsilon$, $2\\epsilon$; $10$ green points with the common coordinate $216$ by the points\n$$\n216 - 5\\epsilon, 216 - 4\\epsilon, \\dots, 216 - \\epsilon, 216 + \\epsilon, \\dots, 216 + 5\\epsilon;\n$$\n$6$ red points with the common coordinate $202$ by the points\n$$\n202 - 3\\epsilon, 202 - 2\\epsilon, 202 - \\epsilon, 202 + \\epsilon, 202 + 2\\epsilon, 202 + 3\\epsilon;\n$$\nand, finally, $6$ red points with the common coordinate $218$ by the points\n$$\n218 - 3\\epsilon, 218 - 2\\epsilon, 218 - \\epsilon, 218 + \\epsilon, 218 + 2\\epsilon, 218 + 3\\epsilon.\n$$\nWe can choose $\\epsilon$ sufficiently small to keep the order of the points on the number axes.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25196,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two cats, Bill and Tom, play the following game. They, in turn (Bill starts), eat fishes from the heap of 50 fishes. Per move it is allowed to eat exactly 1, or exactly 4, or exactly 7 fishes. The player wins if he eats the last fish of the heap.\nWho of the cats wins if both of them play to win?",
    "options": [],
    "answer": "Tom",
    "solution": "It is easy to see that if either $1$ or $4$ fishes remain in the heap, then the player who must move wins, but if $2$ fishes remain, then this player loses. So we will solve the problem moving backward. We write all numbers from $1$ to $50$ and mark them with \"+\" or \"-\". If $k$ fishes remain in the heap before the move of a player and he can win, then we write $+k$, otherwise we write $-k$. We have the following table\n\n| +1 | -2 | +3 | +4 | -5 | +6 | +7 | -8 | +9 | -10 | +11 | +12 | -13 | +14 | +15 | -16 | ... |\n|----|----|----|----|----|----|----|----|----|-----|-----|-----|-----|-----|-----|-----|-----|\n| 1  | -  | 1  | 4  | -  | 4  | 7  | -  | 1  |  -  | 1   | 4   |  -  | 4   | 7   |  -  | ... |\n\nThe second row of the table contains the possible winning moves (to win the player can eat so many fishes that after his move the number of the fishes remained in the heap would be marked with \"+\" ). We see that the signs in the first row of the table are repeated with the period equaled $8$ (it is easy to prove by induction).\n\nSince $50$ leaves remainder $2$ when divided by $8$, it follows that $50$ and $2$ are marked with the same sign, i.e. \"-\". Thus the number $50$ is the losing number of the fishes for the beginning player. To win Tom can eat so many fishes as it is written in the second row of the table.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25197,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let a six-digit number $A$ be divisible by $101$.\nProve that one can transpose two digits of $A$ so that the obtained number is also divisible by $101$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "It is possible to transpose either $a_5$ and $a_1$ or $a_4$ and $a_0$ to obtain the number which is divisible by $101$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25198,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a trapezium $ABCD$ ($AD \\parallel BC$) with $AD = 3BC$. A circle $\\Gamma_1$ with the center $B$ passes through the midpoint of $BD$, and a circle $\\Gamma_2$ with the center $C$ passes through the midpoint of $AC$.\nProve that the line through the points of intersection of $\\Gamma_1$ and $\\Gamma_2$ meets the side $AD$ at its midpoint.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $R$ be the midpoint of $BD$, $M$ be the midpoint of $AD$, $P$ be the point of intersection of $AC$ and $BD$. We show that $LD = 2LP$.\n\n![](attached_image_1.png)\n\nBy condition, $BP : PD = 1 : 3$. So $BP = PR = 0.5RD$, hence $LP$ is the median of the triangle $BLP$, and $LR$ is the median of the triangle $BLD$.\n\nUsing the formula for the median length, we obtain\n$$\n4LP^2 = 2LB^2 + 2LR^2 - BR^2, \\quad 4LR^2 = 2LB^2 + 2LD^2 - BD^2.\n$$\nThus\n$$\n\\begin{aligned}\n4LP^2 &= 2LB^2 + 2LR^2 - BR^2 = 2LB^2 + 0.5(2LB^2 + 2LD^2 - BD^2) - BR^2 = \\\\\n&= 3LB^2 + LD^2 - 0.5(2BR)^2 - BR^2 = LD^2 - 3(BL^2 - BR^2) = [BL = BR] = LD^2.\n\\end{aligned}\n$$\nSimilarly, $LA = 2LP$. Therefore, $LA = LD$, i.e. $L$ lies on the perpendicular bisector of $AD$. Similarly, $K$ lies on the perpendicular bisector of $AD$. Hence the line $LK$ meets $AD$ at its midpoint.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25199,
    "subject": "Mathematics (Multi-modal)",
    "question": "Point $M$ is the midpoint of the side $AB$ of the acute-angled triangle $ABC$, points $P$ and $Q$ are the feet of the altitudes $AP$ and $BQ$. A circle passing through $B, M, P$, touches the side $AC$. Prove that a circle passing through $A, M, Q$ touches the extension of the side $BC$.\n\n(I. Voronovich)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First solution. Let $\\Gamma$ be a circle passing through $M, P, B$. Let $\\Gamma$ touch the side $AC$ at $S$, and $T$ be the intersection of the lines $MS$ and $BC$. Since point $S$ lies on the side $AC$, it follows that $T$ lies on the extension of the side $BC$ (over $C$).\n\n![](attached_image_1.png)\n\nSince $AM = MB = MP$ and $MSPB$ is an inscribed quadrilateral, we see that $\\angle MBP = \\angle MPB = \\angle MSB = \\beta$, $\\angle MBS = \\angle ASM = \\angle TSC = \\varphi$. Let $\\angle CAB = \\alpha$. From the triangle $ASB$ we have $\\alpha + \\beta + 2\\varphi = \\pi$. Since $\\angle SMB$ is an external angle of $\\triangle ASM$, we have $\\angle SMB = \\alpha + \\varphi$. Hence $\\angle MTB = \\pi - (\\alpha + \\varphi) - \\beta = \\varphi$. Thus, $\\angle CTS = \\angle CST(= \\varphi)$, so $CS = CT$. By Menelaus' theorem ($\\triangle ACB$ and points $M, S, T$),\n$$\n\\frac{AM}{MB} \\cdot \\frac{BT}{TC} \\cdot \\frac{CS}{SA} = 1 \\iff BT = SA.\n$$\nBy the power of point theorem, $BT^2 = AS^2 = AM \\cdot AB = BM \\cdot BA$. It is equivalent to the fact that $BT$ is tangent to $\\Gamma_1$, passing through $A, M, T$. Let $Q'$ be the intersection point of $\\Gamma_1$ and the line $AC$. Since $\\Gamma_1$ touches the line $BT$, we see that $Q'$ lies on the side $AC$ (not on the extension). Further\n$$\n\\begin{align*}\n\\sim AM + \\sim Q'T &= 2\\angle TSC = 2\\varphi = 2\\angle STC = \\\\\n&= [TC \\text{ is tangent to } \\Gamma_1] = \\sim MQ'T \\iff \\\\\n\\sim MA + \\sim Q'T &= \\sim MQ'T = \\sim MQ' + \\sim Q'T \\iff \\sim AM = \\sim MQ'.\n\\end{align*}\n$$\n\nSo $AM = MQ'$ (chords subtending the equal arcs). Therefore, $BM = AM = MQ'$, which gives that $AQ'B$ is a right-angled triangle ($\\angle AQ'B = 90^\\circ$). Thus $BQ' \\perp AC$, i.e. points $Q$ and $Q'$ coincide, which completes the proof.\n\nSecond solution. Let $\\Gamma$ be a circle passing through $M, P, B$. Let $\\Gamma$ touch the side $AC$ at $S$. Since $APB$ and $AQB$ are right-angled triangles, we have $MA = MB = MQ = MP$, so $A, Q, P, B$ lie on the circle $\\Gamma_0$ with the center at $M$. Consider the inversion (with $M$ as center) with respect to $\\Gamma_0$. Since $\\Gamma$ passes through the center of the inversion, its image is a line $\\ell$, but $B$ and $P$ are fixed points, the line $\\ell$ coincide with the line $BC$. Similarly, the image of the line $AC$ ($M \\notin AC$) is a circle $\\Gamma_1$ passing through $A, Q, M$, since $A$ and $Q$ are fixed points.\nTherefore, $S$ as the tangency point of $\\Gamma$ and the side $AC$ is transformed to $S'$ which is the tangency point of the line $BC$ and $\\Gamma_1$. To complete the proof, it suffices to note that $S'$ lies on the ray $MS$, i.e. is the intersection point of the ray $MS$ and the line $BC$. Since $S$ lies on the side $AC$, we see that $S'$ lies on the extension of the side $BC$ (over $C$).",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25200,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a right-angled triangle $ABC$ with $\\angle C = 90^\\circ$. Its perimeter is equal to $30$ cm. $CH$ is the altitude of this triangle, $CK$ and $CL$ are the bisectors of the angles $ACH$ and $BCH$ respectively.\nFind the length of the hypotenuse $AB$ if $KL = 4$ cm.",
    "options": [],
    "answer": "13",
    "solution": "Let $a = BC$, $b = AC$, $c = AB$, $\\alpha = \\angle BAC$, $\\beta = \\angle ABC$, $P = a+b+c$. Then $\\angle ACH = \\beta$, $\\angle BCH = \\alpha$, thus $\\angle ACK = \\angle KCH = \\beta/2$, $\\angle HCL = \\angle LCB = \\alpha/2$. Since $\\angle HKC = \\angle KAC + \\angle ACK$ (exterior angle of triangle $ACK$), we have\n$$\n\\angle HKC = \\alpha + \\beta/2 = \\angle BCH + \\angle LCH = \\angle BCK.\n$$\nHence triangle $KCB$ is isosceles and $BK = BC = a$. Similarly, $AL = AC = b$. Therefore, $AK = AB - BK = c - a$, so $KL = AL - AK = b - (c-a) = a+b-c$. Hence\n$$\nc = \\frac{(a+b+c) - (a+b-c)}{2} = \\frac{P - KL}{2} = \\frac{30-4}{2} = 13 \\text{ (cm).}\n$$\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25201,
    "subject": "Mathematics (Multi-modal)",
    "question": "Basil considers all quadratic polynomials $y = ax^2 + bx + c$ with negative discriminants, where $a$, $b$, $c$ are positive integers not greater than $50$. For each polynomial Basil writes its minimal value.\nFind the largest and the smallest numbers among the written numbers.",
    "options": [],
    "answer": "largest = 50 - 1/200, smallest = 3/196",
    "solution": "The distance between the parabola $y = ax^2 + bx + c$ and the axis of abscissae is equal to\n$$\nd = \\frac{4ac - b^2}{4a}\n$$\nTherefore, the problem is equivalent to the following problem: find the greatest and the smallest value of the expression above if positive integers $a$, $b$, $c$ satisfy the inequalities $1 \\le a \\le 50$, $1 \\le b \\le 50$, $1 \\le c \\le 50$ and $b^2 - 4ac < 0$.\n\nWe find the greatest value. We have\n$$\nd = \\frac{4ac - b^2}{4a} = c - \\frac{b^2}{4a} \\le 50 - \\frac{b^2}{4a} \\le 50 - \\frac{1}{4a} \\le 50 - \\frac{1}{200}.\n$$\nThe first inequality follows from the inequality $c \\le 50$, the second one follows from the inequalities $a \\le 50$ and $b^2 \\ge 1$. We see that all inequalities become equalities for $c = 50$, $b = 1$ and $a = 50$. Moreover, for these $a$, $b$, $c$ the inequality $b^2 - 4ac = 1^2 - 4 \\cdot 50 \\cdot 50 < 0$ is valid, too. Therefore the required greatest value is equal to $50 - 1/200$.\n\nNow we find the smallest value. In the fraction above the numerator and the denominator are positive integers. We find the smallest possible value of the numerator. The numerator can be neither $1$ nor $2$. Indeed, in these cases we have either $b^2 = 4ac - 1$ or $b^2 = 4ac - 2$, i.e. $b^2 = 4(ac - 1) + 3$ or $b^2 = 4(ac - 1) + 2$. It follows that $b^2$ is congruent to either $2$ or $3$ modulo $4$, which is impossible.\n\nSo the smallest value of the numerator is $3$, i.e. $b^2 = 4ac - 3$. Now our next goal is to determine the greatest possible value of the denominator. From the equality $b^2 = 4ac - 3$ it follows that $b$ is odd, i.e. $b = 2k + 1$ for some nonnegative integer $k$. So $(2k+1)^2 = 4ac - 3$, i.e. $k^2 + k + 1 = ac$. It follows that $a$ and $c$ are odd. Thus the possible values of $a$ are $a = 49, 47, 45, \\dots$. We verify these values (starting with the greatest one).\n\nSet $a = 49$. We have $c = \\frac{b^2+3}{4a} \\le \\frac{2503}{4 \\cdot 49} < 13$. Now we must verify whether $4 \\cdot 49 \\cdot c - 3 = 196c - 3$ is a perfect square for some of $c = 1, 3, 5, 7, 9, 11$. First, for $c = 1, 5, 11$ the last digit of the number $196c - 3$ is either $3$ or $7$, which is impossible for a perfect square. Further, for $c = 3, 7$ and $9$ we respectively have $196c - 3 = 585, 1369$ and $1671$. We see that $1369 = 37^2$, therefore for $c = 7$, $b = 37$ and $a = 49$ the value of $d$ is equal to $\\frac{3}{4 \\cdot 49} = \\frac{3}{196}$.\n\nHence, if the denominator is equal to $3$, then the smallest value of $d$ is $3/196$. If the denominator is no smaller than $4$, then we have $d = \\frac{4ac - b^2}{4a} \\ge \\frac{4}{4 \\cdot 50} = \\frac{1}{50} > \\frac{3}{196}$. Therefore, the required smallest value is $\\frac{3}{196}$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 25202,
    "subject": "Mathematics (Multi-modal)",
    "question": "Points $A_1, B_1$ are marked on the sides $AC$ and $BC$ of the triangle $ABC$, respectively, so that $A_1B_1 \\parallel AB$. Points $A_2, B_2$ are the feet of perpendiculars from $A_1, B_1$ onto $AB$, respectively.\nProve that $AC = AB_2 + CB_1$ if and only if $BC = BA_2 + CA_1$.\n(I. Voronovich)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $A_1A_2 = B_1B_2 = 2x$, $\\angle A = \\alpha$, $\\angle B = \\beta$, $AB = c$, $AC = b$, $BC = a$. Then we have (see Fig. 1)\n$$\nAC = AB_2 + CB_1 \\iff b = c - BB_2 + a - BB_1 = c + a - 2x \\operatorname{ctg} \\beta - \\frac{2x}{\\sin \\beta} \\iff\n$$\n$$\na + c - b = 2x \\left( \\operatorname{ctg} \\beta + \\frac{1}{\\sin \\beta} \\right) = 2x \\operatorname{ctg} \\frac{\\beta}{2}.\n$$\n![](attached_image_1.png)\n\n![](attached_image_2.png)\n\nNote that (see Fig. 2) $\\frac{a+c-b}{2} = r \\operatorname{ctg} \\frac{\\beta}{2}$, where $r$ is the inradius of the triangle $ABC$. Therefore, the condition $AC = AB_2 + CB_1$ is equivalent to $A_1A_2 = B_1B_2 = 2r$.\nSimilarly, the conditions $BC = BA_2 + CA_1$ and $A_1A_2 = B_1B_2 = 2r$ are also equivalent.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25203,
    "subject": "Mathematics (Multi-modal)",
    "question": "There is a heap of 330 stones. Nick and Mike play the following game. They, in turn (Nick is the first), remove the stones from the heap. Per move it is allowed to remove exactly 1 or exactly $m$ or exactly $n$ stones. The player wins if he removes the last stone. Before the start Nick fixes the value of $n$ ($1 < n < 10$). After that Mike fixes the value of $m$ ($m \\ne n$, $1 < m < 10$), and Nick begins the game.\nCan somebody of the players fix his number to win if both of them play to win?",
    "options": [],
    "answer": "Mike",
    "solution": "We separate the numbers from 2 to 9 into the pairs (2, 7), (3, 8), (5, 6), (4, 9). In order to win Mike can use the following rule: if Nick fixes one of the numbers from any of the pairs, then Mike fixes the other number from the same pair.\n\nNow we will solve the problem moving backward. We write all numbers from 1 to 330 and mark them with \"+\" or \"-\". If $k$ stones remain in the heap before the move of a player and he can win, then we write $+k$, otherwise we write $-k$.\n\nLet $n, m = 2, 7$. We have the following table\n\n| +1 | +2 | -3 | +4 | +5 | -6 | +7 | +8 | -9 | +10 | +11 | -12 | +13 | +14 | -15 | +16 | ... |\n\nWe see that the signs in the table are repeated with the period equaled 3 (it is easy to prove by induction).\nSince $330 \\div 3$ it follows that 330 is marked with the sign \"-\". Thus the number 330 is the losing number of the stones for the beginning player.\n\nLet $n, m = 3, 8$. We have the following table\n\n| +1 | -2 | +3 | -4 | +5 | -6 | +7 | +8 | +9 | +10 | -11 |\n|  1 |  - |  1 |  - |  1 |  - |  1 |  8 |  3 |  8 |  -  |\n| +12 | -13 | +14 | -15 | +16 | -17 | +18 | +19 | +20 | +21 | -22 | ... |\n|  1  |  -  |  1  |  -  |  1  |  -  |  1  |  8  |  3  |  8  |  -  | ... |\n\nWe see that the signs in the table are repeated with the period equaled 11 (it is easy to prove by induction).\nSince $330 \\div 11$ it follows that 330 is marked with the sign \"-\". Thus the number 330 is the losing number of the stones for the beginning player.\n\nLet $n, m = 5, 6$. We have the following table\n\n| +1 | -2 | +3 | -4 | +5 | +6 | +7 | +8 | +9 | +10 | -11 |\n|  1 |  - |  1 |  - |  5 |  6 |  5 |  6 |  5 |  6  |  -  |\n| +12 | -13 | +14 | -15 | +16 | +17 | +18 | +19 | +20 | +21 | -22 | ... |\n|  1  |  -  |  1  |  -  |  5  |  6  |  5  |  6  |  5  |  6  |  -  | ... |\n\nWe see that the signs in the table are repeated with the period equaled 11 (it is easy to prove by induction).\nSince $330 \\div 11$ it follows that 330 is marked with the sign \"-\". Thus the number 330 is the losing number of the stones for the beginning player.\n\nNow, let $n, m = 4, 9$. We have the table\n\n| +1 | -2 | +3 | +4 | -5 | +6 | -7 | +8 | +9 | -10 |\n|  1 |  - |  1 |  4 |  - |  1 |  - |  1 |  9 |  -  |\n| +11 | -12 | +13 | +14 | -15 | +16 | -17 | +18 | +19 | -20 | ... |\n|  1  |  -  |  1  |  4  |  -  |  1  |  -  |  1  |  9  |  -  | ... |\n\nWe see that the signs in the table are repeated with the period equaled 10 (it is easy to prove by induction).\nSince $330 \\div 10$ it follows that 330 is marked with the sign \"-\". Thus the number 330 is the losing number of the stones for the beginning player.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25204,
    "subject": "Mathematics (Multi-modal)",
    "question": "The bisectors of the angles $BAD$ and $CDA$ of the trapezium $ABCD$ ($BC \\parallel AD$) meet at a point on the perpendicular bisector of the side $BC$.\nProve that either $AB = CD$ or $AB + CD = AD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $P$ and $M$ be marked on $AD$ and $BC$ so that $MP \\perp AD$ and $BM = MC$. By condition, $K$ belongs to $MP$. Let $L$ and $N$ be respectively the feet of the perpendiculars from $K$ to the lines $AB$ and $DC$. Two cases are possible:\n\n1) Both the points lie on the sides $AB$ and $DC$ (or on their extensions).\n\n2) One of the points lies on the side and the other point lies on the extension of the other side.\n\nSince $AK$ and $DK$ are bisectors, we have $KL = KP = KN$. Since $K$ lies on the perpendicular bisector, we have $BK = KC$. Therefore the right-angled triangles $BLK$ and $CNK$ are equal, so $\\angle LBK = \\angle NCK$, $BL = CN$.\n\nFor the first case (see Fig. 1) from the isosceles triangle $BKC$ it follows that $\\angle KBC = \\angle KCB$. So\n$$\n\\begin{aligned}\n\\angle ABC &= \\angle ABK + \\angle KBC = (180^\\circ - \\angle LBK) + \\angle KBC = \\\\\n&= (180^\\circ - \\angle NCK) + \\angle KCB = \\angle KCD + \\angle KCB = \\angle BCD.\n\\end{aligned}\n$$\nTherefore, $\\angle ABC = \\angle DCB$, and so the trapezium $ABCD$ is isosceles. (Similarly we can consider the cases when both $L$ and $N$ lie on the sides $AB$ and $DC$).\n\n![](attached_image_1.png)\nFig. 1\n\nIf we have the second case (see Fig. 2), then $AL = AP$, $PD = DN$ (since $AK$ and $DK$ are bisectors). Therefore,\n$$\nAB + DC = (AL + LB) + (DN - NC) = AP + PD + (LB - NC) = AP + PD = AD,\n$$\nas required.\n\n![](attached_image_1.png)\nFig. 2",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25205,
    "subject": "Mathematics (Multi-modal)",
    "question": "Points $M$ and $N$ are the midpoints of the sides $AC$ and $BC$ of the triangle $ABC$, respectively.\nProve that a circle passing through $C, M, N$ touches the side $AB$ if and only if $AB = \\frac{AC + BC}{\\sqrt{2}}$.\n(I. Voronovich)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $I$ be the center of the circle $\\Gamma$ passing through $C, M, N$; let $r$ be the radius of $\\Gamma$. Let $P$ and $Q$ be the intersection points of segments $AI, BI$ and $\\Gamma$; let $H$ be the foot of the perpendicular from $I$ on $AB$. Set $a = BC, b = AC, c = AB, x = AP, y = BQ, l = IH$.\n\n![](attached_image_1.png)\n\n![](attached_image_2.png)\nFig. 3\nIt is evident that $l = r$ if and only if $\\Gamma$ touches the line $AB$ ($H$ is a point of tangency. If $H$ lies on the side $AB$, then $c = AH + HB$ (see Fig. 1-3). By the power of a point theorem,\n$$\nb \\cdot \\frac{b}{2} = x(x + 2r), \\quad a \\cdot \\frac{a}{2} = y(y + 2r), \\qquad (1)\n$$\n\nIf $H$ lies on the side $AB$, then $c = AH + HB$. So\n$$\n\\begin{aligned}\nc = & \\sqrt{(x+r)^2 - l^2} + \\sqrt{(y+r)^2 - l^2} = \\sqrt{x^2 + 2xr + r^2 - l^2} + \\\\\n& \\sqrt{y^2 + 2yr + r^2 - l^2} = \\sqrt{x(x+2r) + r^2 - l^2} + \\sqrt{y(y+2r) + r^2 - l^2}\n\\end{aligned} \\quad (1)\n$$\n$$\n= \\sqrt{\\frac{a^2}{2} + r^2 - l^2} + \\sqrt{\\frac{b^2}{2} + r^2 - l^2}. \\quad (2)\n$$\nIt is evident that for $l > r$ from (2) it follows $c < \\frac{a}{\\sqrt{2}} + \\frac{b}{\\sqrt{2}}$, for $l < r$ from (2) it follows $c > \\frac{a}{\\sqrt{2}} + \\frac{b}{\\sqrt{2}}$. Therefore, $c = \\frac{a}{\\sqrt{2}} + \\frac{b}{\\sqrt{2}}$ if and only if $l = r$, i.e. the circle passing through $C, M, N$, touches the side $AB$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25206,
    "subject": "Mathematics (Multi-modal)",
    "question": "Is it possible to mark 8 distinct points (1 red, 3 blue, and 4 green points) on a plane so that the sum of the distances between the red point and the blue points is 6, the sum of the distances between the red point and the green points is 16, and the sum of the distances between the blue points and the green points is 56?",
    "options": [],
    "answer": "Yes; for example, place the blue points at -3, -1, 2; the red point at 0; and the green points at 3, 3.5, 4.5, 5.",
    "solution": "For example, the points can be marked on the straight line in the following way: the blue points have the coordinates $-3$, $-1$, $2$, the red point has the coordinate $0$, and the green points have the coordinates $3$, $3.5$, $4.5$, $5$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25207,
    "subject": "Mathematics (Multi-modal)",
    "question": "Is it possible to mark $6$ distinct points ($1$ red, $2$ blue, and $3$ green points) on a plane so that the sum of the distances between the red point and the blue points is $8$, the sum of the distances between the red point and the green points is $6$, and the sum of the distances between the blue points and the green points is $9$?\n\n(S. Mazanik, I. Voronovich)",
    "options": [],
    "answer": "No",
    "solution": "Suppose, contrary to our claim, that there exist points satisfying the problem condition. We label the red point, two blue points, and three green points $R$, $B_1$, $B_2$, and $G_1$, $G_2$, $G_3$ respectively. By triangle inequality,\n$$\nB_1G_1 + RG_1 \\ge RB_1, \\quad B_2G_1 + RG_1 \\ge RB_2, \\quad B_1G_2 + RG_2 \\ge RB_1,\n$$\n$$\nB_2G_2 + RG_2 \\ge RB_2, \\quad B_1G_3 + RG_3 \\ge RB_1, \\quad B_2G_3 + RG_3 \\ge RB_2.\n$$\nSumming these inequalities, we obtain\n$$\n(B_1G_1 + B_1G_2 + B_1G_3 + B_2G_1 + B_2G_2 + B_2G_3) + 2(RG_1 + RG_2 + RG_3) \\ge 3(RB_1 + RB_2),\n$$\ni.e. $9 + 2 \\cdot 6 \\ge 3 \\cdot 8$, which is impossible.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25208,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given the set of $N$ four-digit numbers formed from the digits $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$. Find the smallest possible value of $N$ if the set satisfies the following condition:\n\na) For any two different digits from $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$ there exists a number from the set which contains both of them;\n\nb) For any two different digits from $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$ there exist no greater than two numbers from the set which contain both of them.\n\n(E. Barabanov)",
    "options": [],
    "answer": "a) 6; b) 7",
    "solution": "a.\nSee Problem B.4.\n\nb.\nLet it be possible to choose six four-digit numbers satisfying the problem condition. By the above, any of digits $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$ must be exactly in three of these six numbers.\n\nFurther, it is not hard to prove that no two of the chosen numbers have the same three digits.\n\nLet $a_1, a_2, a_3, a_4, a_5, a_6$ be the chosen six numbers. Let $P_i$ denote the set consisting of six pairs formed by the digits of $a_i$, $i = 1, \\dots, 6$. It is easy to see that the set $\\bigcup_{i=1}^6 P_i$ has exactly $28$ elements. Moreover, either $P_i \\cap P_k = \\emptyset$ or $P_i \\cap P_k$ has exactly one element, and $P_i \\cap P_j \\cap P_k = \\emptyset$, for all distinct $i, k, j$.\n\nBy inclusion-exclusion formula ($N(X)$ is the number of elements of the set $X$)\n$$\nN\\left(\\bigcup_{i=1}^{6} P_i\\right) = \\sum_{i=1}^{6} N(P_i) - \\sum_{\\substack{i,j=1 \\\\ i<j}}^{6} N(P_i \\cap P_j). \\quad (*)\n$$\nSince $N\\left(\\bigcup_{i=1}^{6} P_i\\right) = 28$ and $N(P_i) = 6$ for any $i = 1, \\dots, 6$, equality $(*)$ can be written as\n$$\n\\sum_{\\substack{i,j=1 \\\\ i < j}}^{6} N(P_i \\cap P_j) = 8. \\qquad (**)\n$$\nLet $x$ be the number of pairs $(P_i, P_j)$, $i < j$, for which $N(P_i \\cap P_j) = 0$, then for the remaining $15-x$ pairs we have $N(P_i \\cap P_j) = 1$. Thus $(*)$ can be rewritten as $15-x=8$, or $x=7$.\n\nIt is not very hard to prove that $x \\neq 7$. Thus $x \\leq 6$, and so $N \\geq 7$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25209,
    "subject": "Mathematics (Multi-modal)",
    "question": "Fifteen red, blue and green points are marked on a plane. It is known that the sum of the distances between the red points and the blue points is $51$, the sum of the distances between the red points and the green points is $39$, the sum of the distances between the blue points and the green points is $1$.\nHow many points of each color are marked? (Determine all possibilities.)",
    "options": [],
    "answer": "Either red 8, green 3, blue 4; or red 13, green 1, blue 1.",
    "solution": "Let $n$, $k$, $m$ denote respectively the number of red, green, blue points. Then $m + n + k = 15$. From (*) in the solution of Problem 2, Category C, we have\n$$\n39m \\le 51k + n, \\quad (1)\n$$\n$$\n51k - n \\le 39m. \\quad (2)\n$$\nSubstitute for $k$ in (1) to obtain (after simple manipulations)\n$$\n18m + 10n \\le 153 \\Rightarrow 9m + 5n \\le 76. \\quad (3)\n$$\nSubstitute for $k$ in (2) to obtain $51 \\cdot 15 \\le 90m + 52n$. Using $n \\le 13$, we obtain $51 \\cdot 15 \\le 90m + 50n + 26$, hence\n$$\n74 \\le 9m + 5n. \\quad (4)\n$$\nIf $n \\le 7$, then we have $51 \\cdot 15 \\le 90m + 50n + 14$ or $76 \\le 9m + 5n$. From (3), it follows that $9m + 5n = 76$. However, it is easy to verify that the last equality cannot be valid for $n \\le 7$.\nVerifying the values $n = 8, 9, \\dots, 13$, we see that inequalities (3) and (4) can be valid only if $n = 8$ or $n = 13$. Then we have respectively $m = 4$, $k = 3$ and $m = k = 1$. Examples for these values can be constructed similarly to the example of Problem 2, Category B.\n\nTherefore, the answer is $(n, k, m) = (8, 3, 4)$ or $(13, 1, 1)$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25210,
    "subject": "Mathematics (Multi-modal)",
    "question": "There is a heap of $25201$ stones. Nick and Mary play the following game. They, in turn, remove the stones from the heap. Per move it is allowed to remove either exactly $m$ or exactly $n$ stones. The player wins if he removes the last stone. If the last stone can not be removed by any of the players, then the result of the game is a draw. Before the start Nick fixes the value of $n$ ($1 \\le n \\le 10$) and defines the player to start the game. After that Mary fixes the value of $m$ ($m \\ne n, 1 \\le m \\le 10$), then the game starts.\nCan somebody of the players fix his/her number to win if both of them play to win?\n(V. Kaskevich)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If Nick fixes the number $n$ and defines himself as the first player, then Mary wins if she sets $m = 11 - n$. Indeed, if Nick removes $n$ ($m$) stones, then Mary removes $m$ ($n$) stones. So exactly $11 = m + n$ stones are removed from the heap after each pair of moves (Nick - Mary). Since $25201 = 2291 \\cdot 11$, Mary wins.\n\nIf Nick fixes the number $n$ and defines Mary as the first player, then Mary wins if she sets $m = 1$ for $n = 2, 3, 4, 5, 6, 7, 8, 9$ and her first move is to remove 1 stone from the heap. $25200$ stones remain in the heap. It is easy to see that $2520$ is divide by $3, 4, 5, 6, 7, 8, 9, 10$, i.e. $25200$ is divided by $n+m$ for the fixed value of $n$ and $m$. Therefore, Mary wins if she uses the symmetric strategy: when Nick removes $n$ ($m$) stones Mary removes $m$ ($n$) stones. So exactly $m+n$ stones are removed from the heap after each pair of moves (Nick - Mary).\n\nIf Nick sets $n = 1$, then Mary can choose any number from $2, 3, 4, 5, 6, 7, 8, 9$ for $m$, and play in the same way as above (first move is $1$ stone, and symmetric strategy).\n\nIf Nick sets $n = 10$, then Mary wins if she sets $m = 7$ and her first move is to remove $7$ stones from the heap. $25194$ stones remain in the heap. It is easy to see that $25194 = 1482 \\cdot 17$ is divided by $17$, i.e. $25194$ is divided by $n+m$. To win Mary can use the symmetric strategy.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25211,
    "subject": "Mathematics (Multi-modal)",
    "question": "From the digits $1, 2, 3, 4, 5, 6, 7, 8$, find the smallest possible value of $N$ such that, for any two distinct digits from this set, there exists a number among the $N$ numbers (each number is a four-digit number formed from these digits) which contains both of them.",
    "options": [],
    "answer": "6",
    "solution": "Answer: $N = 6$.\n\nLet some digit, say $1$, appear exactly in $k$ numbers from $N$ given numbers. Hence, $1$ forms at most $3$ distinct pairs with the remaining $3$ digits of any of these $k$ numbers. Since the total number of all distinct pairs formed by $1$ and the other $7$ numbers ($2,3,\\ldots,8$) is equal to $7$, we see that $3k \\ge 7$. So $k \\ge 3$. Therefore, each of the digits $1, 2, \\ldots, 8$ must appear in at least $3$ numbers. Thus, the total number of all digits in $N$ numbers is greater than or equal to $8 \\cdot 3 = 24$. But $N$ numbers contains exactly $4N$ digits. Therefore, $4N \\ge 24$, so $N \\ge 6$.\n\nThe following example shows that there are $6$ four-digit numbers satisfying the ...",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25212,
    "subject": "Mathematics (Multi-modal)",
    "question": "There is a heap of $360$ stones. Nick and Mary play the following game. They, in turn (Mary is the first), remove the stones from the heap. Per move it is allowed to remove exactly $1$ or exactly $m$ or exactly $n$ stones. The player wins if he/she removes the last stone. Before the start Nick fixes the value of $n$ ($2 \\le n \\le 9$). After that Mary fixes the value of $m$ ($m \\ne n, 2 \\le m \\le 9$) and begins the game.\nCan somebody of the players fix his/her number to win if both of them play to win?\n(V. Kaskevich)",
    "options": [],
    "answer": "Nick can guarantee a win by choosing n = 2.",
    "solution": "Nick wins if he sets $n = 2$. We consider all possibilities for $m$. Let $m = 3$. If Mary removes $k$ stones then Nick removes $4 - k$. In this case exactly $4$ stones are removed from the heap after each pair of moves (Mary - Nick). Since $360 : 4$, Nick wins.\n\nNow we will solve the problem moving backward. We write all numbers from $1$ to $360$ and mark them with \"+\" or \"-\". If $k$ stones remain in the heap before the move of a player and he/she can win, then we write $+k$, otherwise we write $-k$. For $m = 4, 5, 7, 8$ we have the following table\n\n| +1 | +2 | -3 | +4 | +5 | -6 | +7 | +8 | -9 | +10 | +11 | -12 | +13 | +14 | -15 | +16 | ... |\n\nWe see that the signs in the table are repeated with the period equaled $3$ (it is easy to prove by induction).\nSince $360 : 3$ it follows that $360$ is marked with the sign \"-\". Thus the number $360$ is the losing number of the stones for the beginning player.\n\nFor $m = 6$ we have the table\n\n| +1 | +2 | -3 | +4 | +5 | +6 | -7 | +8 | +9 | -10 | +11 | +12 | +13 | -14 | ... |\n|  1 |  2 |  - |  1 |  2 |  6 |  - |  1 |  2 |  -  |  1  |  2  |  6  |  -  | ... |\n\nThe second row of the table contains the possible winning moves (to win the player can remove so many stones that after his move the number of the stones remained in the heap would be marked with \"+\"). We see that the signs in the first row of the table are repeated with the period equaled $7$ (it is easy to prove by induction).\nSince $360$ is congruent to $3$ modulo $7$, it follows that $360$ and $3$ are marked with the same sign, i.e. \"-\". Thus the number $360$ is the losing number of the stones for the beginning player. To win Nick can remove so many stones as it is written in the second row of the table.\n\nIn a similar way, for $m = 9$ we have the table\n\n| +1 | +2 | -3 | +4 | +5 | -6 | +7 | +8 | +9 | -10 |\n|  1 |  2 |  - |  1 |  2 |  - |  1 |  2 |  9 |  -  |\n\n| +11 | +12 | -13 | +14 | +15 | -16 | +17 | +18 | +19 | -20 | ... |\n|  1  |  2  |  -  |  1  |  2  |  -  |  1  |  2  |  9  |  -  | ... |\n\nThe second row of the table contains the possible winning moves (to win the player can remove so many stones that after his move the number of the stones remained in the heap would be marked with \"+\"). We see that the signs in the first row of the table are repeated with the period equaled $10$ (it is easy to prove by induction).\nSince $360$ is divided by $10$, it follows that $360$ and $10$ are marked with the same sign, i.e. \"-\". Thus the number $360$ is the losing number of the stones for the beginning player. To win Nick can remove so many stones as it is written in the second row of the table.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25213,
    "subject": "Mathematics (Multi-modal)",
    "question": "Some elements of an eight-element set are colored white while the others are colored black. Let $N$ be a number of all subsets containing the elements of both colors.\nFind the greatest possible value of $N$.",
    "options": [],
    "answer": "225",
    "solution": "Suppose $k$ elements are colored white and $8 - k$ elements are colored black. A subset contains elements of both colors if it contains at least one white and at least one black element.\n\nThe total number of subsets of the eight-element set is $2^8 = 256$.\n\nThe subsets that do not contain elements of both colors are:\n- The empty set (contains no elements).\n- Subsets containing only white elements: there are $2^k - 1$ such subsets (all non-empty subsets of the $k$ white elements).\n- Subsets containing only black elements: there are $2^{8 - k} - 1$ such subsets (all non-empty subsets of the $8 - k$ black elements).\n\nTherefore,\n$$\nN = 2^8 - 1 - (2^k - 1) - (2^{8 - k} - 1) = 256 - 1 - (2^k - 1) - (2^{8 - k} - 1)\n$$\n$$\n= 255 - (2^k - 1) - (2^{8 - k} - 1) = 255 - 2^k - 2^{8 - k} + 2\n$$\n$$\n= 257 - 2^k - 2^{8 - k}\n$$\n\nTo maximize $N$, we need to minimize $2^k + 2^{8 - k}$. Since $k$ is an integer between $1$ and $7$ (since both colors must be present), we compute $2^k + 2^{8 - k}$ for $k = 1, 2, ..., 7$:\n\n- $k = 1$: $2^1 + 2^7 = 2 + 128 = 130$\n- $k = 2$: $2^2 + 2^6 = 4 + 64 = 68$\n- $k = 3$: $2^3 + 2^5 = 8 + 32 = 40$\n- $k = 4$: $2^4 + 2^4 = 16 + 16 = 32$\n- $k = 5$: $2^5 + 2^3 = 32 + 8 = 40$\n- $k = 6$: $2^6 + 2^2 = 64 + 4 = 68$\n- $k = 7$: $2^7 + 2^1 = 128 + 2 = 130$\n\nThe minimum is $32$ when $k = 4$.\n\nTherefore, the greatest possible value of $N$ is:\n$$\nN = 257 - 32 = 225\n$$\n\nSo, the answer is $225$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25214,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that\n$$\n\\frac{a^2}{a+b} + \\frac{b^2}{b+c} \\ge \\frac{3a+2b-c}{4}\n$$\nfor all positive real $a, b, c$.\n\n(D. Pirshtuk)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "$$\n(2x - y)^2 \\ge 0 \\iff \\frac{x^2}{y} \\ge \\frac{4x - y}{4}\n$$\nfor any positive $x, y$.\n\nTherefore,\n$$\n\\frac{a^2}{a+b} + \\frac{b^2}{b+c} \\ge \\frac{4a - (a+b)}{4} + \\frac{4b - (b+c)}{4} = \\frac{3a + 2b - c}{4},\n$$\nas required.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25215,
    "subject": "Mathematics (Multi-modal)",
    "question": "Points $B$ and $C$ are marked on the half-hyperbola $y = 1/x$ which lies in the first quadrant of the Cartesian plane. The abscissa of $C$ is greater than the abscissa of $B$. Let $A$ be the intersection point of the other half-hyperbola and the line passing through the origin and $B$.\n\nProve that the angle $BAC$ is equal to one of the angles between the line $BC$ and the tangent to the hyperbola at point $B$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\ell$ be tangent to the hyperbola at point $B$, $F$ be the intersection of $x$-axis and $BC$, $E$ be the intersection point of the $x$-axis and $\\ell$,\n\n![](attached_image_1.png)\n\n$D$ be the intersection point of $\\ell$ and the line through $A$ parallel to $y$-axis (see the Fig.). Since $A$ and $B$ are symmetric with respect to the origin, we see that $A(-x_1; -1/x_1)$. Further,\n\n$$\n\\tan \\angle \\text{BEO} = \\tan(\\pi - \\angle \\text{BEF}) = -\\tan \\angle \\text{BEF} = - (1/x)' \\bigg|_{x=x_1} = \\frac{1}{x_1^2}.\n$$\n\nOn the other hand, $\\tan \\angle BAD = \\tan \\angle BOE = 1/x_1 : x_1 = 1/x_1^2$, i.e.\n$$\n\\angle BAD = \\angle BEO. \\tag{1}\n$$\n\nFurther,\n$$\n\\tan \\angle BFE = \\tan \\angle BCG = \\frac{1/x_1 - 1/x_2}{x_2 - x_1} = \\frac{1}{x_1 x_2}\n$$\nand\n$$\n\\tan \\angle CAD = \\frac{1/x_2 + 1/x_1}{x_2 + x_1} = \\frac{1}{x_1 x_2}.\n$$\nHence $\\angle CAD = \\angle CFE = \\angle BFE$.\n\nSo,\n$$\n\\angle BAC = \\angle BAD - \\angle CAD \\stackrel{(1)}{=} \\angle BEO - \\angle BFE = \\angle FBE = \\angle CBE,\n$$\n\nas required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25216,
    "subject": "Mathematics (Multi-modal)",
    "question": "For all admissible $a$, $b$, $c$ find all possible values of the expression\n$$\n\\frac{(a+b-c)^2}{(a-c)(b-c)} + \\frac{(b+c-a)^2}{(b-a)(c-a)} + \\frac{(c+a-b)^2}{(c-b)(a-b)}\n$$",
    "options": [],
    "answer": "4",
    "solution": "Set $A = a + b + c$. Then\n$$\n\\begin{aligned}\nM &= \\frac{(a+b-c)^2}{(a-c)(b-c)} + \\frac{(b+c-a)^2}{(b-a)(c-a)} + \\frac{(c+a-b)^2}{(c-b)(a-b)} = \\\\\n&= \\frac{(A-2c)^2}{(a-c)(b-c)} + \\frac{(A-2a)^2}{(b-a)(c-a)} + \\frac{(A-2b)^2}{(c-b)(a-b)} = \\\\\n&= \\frac{(A-2c)^2(b-a) + (A-2a)^2(c-b) + (A-2b)^2(a-c)}{(a-b)(b-c)(c-a)} = \\frac{L}{(a-b)(b-c)(c-a)}\n\\end{aligned}\n$$\n\nWe have\n$$\n\\begin{aligned}\nL &= (A^2 - 4Ac + 4c^2)(b-a) + (A^2 - 4Aa + 4a^2)(c-b) + \\\\\n&\\quad (A^2 - 4Ab + 4b^2)(a-c) = A^2((b-a) + (c-b) + (a-c)) - \\\\\n&\\quad -2A(c(b-a) + a(c-b) + b(a-c)) + 4(c^2(b-a) + a^2(c-b) + b^2(a-c)) = \\\\\n&= -4A(cb - ca + ac - ab + ba - bc) + 4N = 4N.\n\\end{aligned}\n$$\n\nFurther\n$$\n\\begin{aligned}\nN &= c^2(b-a) + a^2(c-b) + b^2(a-c) = c^2(b-a) + (a^2c - a^2b + b^2a - b^2c) = \\\\\n&= c^2(b-a) + (a^2c - b^2c) - (a^2b - b^2a) = c^2(b-a) + c(a+b)(a-b) - ab(a-b) = \\\\\n&= (a-b)(ca + cb - ab - c^2) = (a-b)(a(c-b) + c(b-c)) = (a-b)(b-c)(c-a).\n\\end{aligned}\n$$\nTherefore, $M = 4$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 25217,
    "subject": "Mathematics (Multi-modal)",
    "question": "Circles $\\Gamma_1$ and $\\Gamma_2$ touch each other externally at point $M_3$. They touch a circle $\\Gamma_3$ internally at points $M_1$ and $M_2$ respectively. Let $S$ be the circumcenter of the triangle $M_1M_2M_3$. Prove that the line $SM_1$ touches $\\Gamma_3$.\n(M. Karpuk)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $O_1, O_2, O_3$ be respectively the centers of the circles $\\Gamma_1, \\Gamma_2, \\Gamma_3$, and $R_1, R_2, R_3$ be their radii.\nLet $\\Gamma$ be the circumscribed circle of the triangle $O_1O_2O_3$ (see Fig. 1), and let $\\Gamma$ touch the lines $O_3O_1, O_3O_2, O_1O_2$ at points $K_1, K_2, K_3$ respectively. Then\n$$\nO_1K_1 = O_1K_3, \\quad O_2K_2 = O_2K_3 \\implies O_1K_1 + O_2K_2 = O_1O_2 = R_1 + R_2. \\quad (1)\n$$\n\nMoreover\n$$\n(R_3 - R_1) + O_1K_1 = O_3O_1 + O_1K_1 = O_3K_1 = O_3K_2 = O_3O_2 + O_2K_2 = (R_3 - R_2) + O_2K_2.\n$$\n\n---\n\nTherefore, if $O_1K_1 > R_1$ ($O_1K_1 < R_1$), then $O_2K_2 = O_3K_2 - (R_3 - R_2) = (R_3 - R_1) + O_1K_1 - (R_3 - R_2) = R_2 + (O_1K_1 - R_1) > R_2$, ($O_2K_2 < R_2$), which gives $O_1K_1 + O_2K_2 > R_1 + R_2$ ($O_1K_1 + O_2K_2 < R_1 + R_2$), contrary to (1). Therefore, $O_1K = R_1$, $O_2K_2 = R_2$. So $K_1, K_2, K_3$ coincide with $M_1, M_2, M_3$ respectively. It means that $\\Gamma$ is circumcircle of the triangle $M_1M_2M_3$ (see Fig. 2). So $SM_1 \\perp O_3O_1$ (as the radius of the inscribed circle), thus $SM_1$ is the tangent to $\\Gamma_3$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25218,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that\n$$\n\\frac{a^3(a+c)(a+b)}{(a-c)(a-b)} + \\frac{b^3(b+a)(b+c)}{(b-a)(b-c)} + \\frac{c^3(c+b)(c+a)}{(c-b)(c-a)} = abc\n$$\nfor all admissible $a$, $b$, $c$ such that $a+b+c=0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "In view of $a+b+c=0$, we rewrite the left hand side of the required inequality in the form\n$$\n\\begin{aligned}\nM &= \\frac{a^3(a+c)(a+b)}{(a-c)(a-b)} + \\frac{b^3(b+a)(b+c)}{(b-a)(b-c)} + \\frac{c^3(c+b)(c+a)}{(c-b)(c-a)} = \\\\\n&= \\frac{a^3(-b)(-c)}{(a-c)(a-b)} + \\frac{b^3(-c)(-a)}{(b-a)(b-c)} + \\frac{c^3(-a)(-b)}{(c-b)(c-a)} = \\\\\n&= abc \\left( \\frac{a^2}{(a-c)(a-b)} + \\frac{b^2}{(b-a)(b-c)} + \\frac{c^2}{(c-b)(c-a)} \\right) = abcL.\n\\end{aligned}\n$$\nWe have\n$$\n\\begin{aligned}\nL &= \\frac{a^2}{(a-c)(a-b)} + \\frac{b^2}{(b-a)(b-c)} + \\frac{c^2}{(c-b)(c-a)} = \\\\\n&= \\frac{a^2(b-c) + b^2(c-a) + c^2(a-b)}{(a-c)(b-a)(c-b)} = \\frac{N}{(a-c)(b-a)(c-b)}.\n\\end{aligned}\n$$\nFurther,\n$$\n\\begin{aligned}\nN &= a^2(b-c) + b^2(c-a) + c^2(a-b) = a^2(b-c) + b^2c - b^2a + c^2a - c^2b = \\\\\n&= a^2(b-c) + (b^2c - c^2b) - (b^2a - c^2a) = a^2(b-c) + bc(b-c) - a(b+c)(b-c) =\n\\end{aligned}\n$$\n$$\n=(b-c)(a^2+bc-ab-ac)=(b-c)(a(a-b)-c(a-b))=(b-c)(a-b)(a-c)=(a-c)(b-a)(c-b).\n$$\nSo $L = 1$, hence $M = abc$, as required.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25219,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$ denote integers. Prove that $49$ divides $10a^2 + 23ab + 12b^2$, if $7$ divides $3a^2 + 2ab - 2b^2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $3a^2 + 2ab - 2b^2$ is divisible by $7$, we see that\n$$10a^2 + 23ab + 12b^2 = (3a^2 + 2ab - 2b^2) + 7(a^2 + 3ab + 2b^2)$$\nis divisible by $7$.\n\nWe have\n$$10a^2 + 23ab + 12b^2 = 10a^2 + 15ab + 8ab + 12b^2 = 5a(2a + 3b) + 4b(2a + 3b) = (2a + 3b)(5a + 4b)$$\nwhich is divisible by $7$.\n\nSince $7$ is a prime number, we see that the product $(2a + 3b)(5a + 4b)$ is divisible by $7$ if at least one of the co-factors is divisible by $7$.\n\nNote that the sum of these co-factors\n$$(2a + 3b) + (5a + 4b) = 7(a + b)$$\nis divisible by $7$, so if one of the numbers is divisible by $7$, then the other is also divisible by $7$.\n\nThus the product of these numbers is divisible by $7 \\cdot 7 = 49$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25220,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all triples of positive integers $(x, y, z)$ satisfying the equality $3^x + 7^y = 4^z$.",
    "options": [],
    "answer": "(2,1,2)",
    "solution": "$z \\ge 3$, then $2^z - 1 \\vdash 3$, but it is possible only if $z$ is even. Let $z = 2c$ for some nonnegative integer $c$. Then $4^c - 3^a = 1$. If $a = 1$, we find $c = 1$ and obtain $x = 1, z = 2, y = 1$. If $a > 1$, then $4^c = 3^a + 1$ has remainder 1 when divided by 9. The power of 4 has this remainder only if $c$ is divisible by 3. But in this case $4^3 \\equiv 1 \\pmod 7$ and hence $4^c \\equiv 1 \\pmod 7$. It follows that $3^a = 4^c - 1 \\vdash 7$, which is impossible.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25221,
    "subject": "Mathematics (Multi-modal)",
    "question": "Bob has the collection of $n$ coins. Each of them weights an integer number of grams, and their total weight is equal to 300 grams.\nFind the smallest possible value of $n$ for which Bob always (independently of coin's weights) can partition his collection into three groups so that the total weight the coins in each of the groups is equal to 100 grams.",
    "options": [],
    "answer": "201",
    "solution": "Answer: 201.\nLet $a_1 \\le a_2 \\le \\dots \\le a_n$ be the weights of the coins in Bob's collection. If there exists an $i$ such that $a_i \\ge 101$, and $k = l = m = 100$, then obviously Bob\n\nThen $S_j - S_i \\doteq 100$, and, on the other hand, $S_j - S_i < S_{100} + a_{201} = 200$. It follows that $S_j - S_i = 100$. Therefore, $a_{i+1} + \\dots + a_j = 100$, and the corresponding coins combine into the second group. The remaining coins combine into the third group.\n\nIf one of these sums is equal to $m$, then the corresponding coins combine into the third group, and the remaining coins combine into the second group with the weight equaled $l$. Otherwise, there exists an $j$, such that $S_j < m$ and $S_{j+1} > m$. In particular, $S_{j+1} - S_j \\ge 2$, i. e. $x_{201-k-j} \\ge 2$. This means that among the numbers $x_{201-k}$, $x_{201-k-1}$, ..., $x_{201-k-j}$ there are no 1's. Moreover, this means that $m - S_j < x_{201-k-j} \\le x_{201-k} \\le t$. So\n$$\nx_1 + \\dots + x_{m-s_j} + x_{201-k-j+1} + \\dots + x_{201-k} = m,\n$$\nand the corresponding coins combine into the third group. The remaining coins combine into the second group with weight equaled $l$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 25222,
    "subject": "Mathematics (Multi-modal)",
    "question": "All cells of a $7 \\times 7$ table are painted black and white. Per move it is allowed to choose any $n \\times n$ square, $1 < n < 7$ (with the sides coinciding with the sides of the cells) and to change the color of all its cells (from black to white and vice versa).\nIs it possible to get the table with all white cells from the table with the arbitrary number of black cells?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We separate the table into five parts: the central $3 \\times 3$ square, two $2 \\times 7$ rectangles, and two $3 \\times 2$ rectangles (see Fig. 1).\n\n![](attached_image_1.png)\nFig. 1\n![](attached_image_2.png)\n![](attached_image_3.png)\n![](attached_image_4.png)\n![](attached_image_5.png)\n![](attached_image_6.png)\n\nFig. 2\n\n![](attached_image_7.png)\n\nFig. 3\n\n![](attached_image_8.png)\n\nFig. 4\nFig. 5\n\nWe can paint white all black cells of the $2 \\times 7$ rectangles. It suffices to show how we can paint white any black cell of the rectangle so that all other cells of this rectangle keep their color. The corresponding procedures using $2 \\times 2$ and $3 \\times 3$ squares are shown in Fig. 2 and Fig. 3.\n\nFurther, we can paint white all black cells of the $2 \\times 3$ rectangles. It suffices to show how we can paint white any black cell of these rectangles so that all other cells of these rectangles and all cells of the $2 \\times 7$ rectangles keep their color. The corresponding procedures are shown in Fig. 4 and Fig. 5.\n\nIt remains to paint white all black cells of the central $3 \\times 3$ square. It suffices to show how we can paint white any black cell of this square so that all other cells of the table keep their color. The corresponding procedures are shown in Fig. 6.\n![](attached_image_9.png)\nFig. 6",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25223,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\tau(n)$ be the number of the divisors of positive integer $n$ (including $1$ and $n$) and $\\sigma(n)$ be their sum. Prove that $\\sqrt{n} \\le \\frac{\\sigma(n)}{\\tau(n)} \\le \\frac{n+1}{2}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note that if $d$ takes on the values of all divisors of positive integer $n$, then $n/d$ takes on the values of all divisors of $n$, too. Let $d_1, \\dots, d_{\\tau(n)}$ denote the divisors of $n$. We have\n$$\n\\sigma(n) = \\frac{1}{2} \\left( \\left( d_1 + \\frac{n}{d_1} \\right) + \\left( d_2 + \\frac{n}{d_2} \\right) + \\dots + \\left( d_{\\tau(n)} + \\frac{n}{d_{\\tau(n)}} \\right) \\right). \\quad (*)\n$$\nWe prove that\n$$\n\\frac{\\sigma(n)}{\\tau(n)} \\le \\frac{n+1}{2}. \\quad (1)\n$$\nFirst, we show that $x + \\frac{n}{x} < y + \\frac{n}{y}$, if $0 < y < x \\le \\sqrt{n}$. Indeed, this inequality is equivalent to the inequality $x-y < \\frac{n(x-y)}{xy}$, i.e. $xy < n$. The last inequality is valid because $xy < \\sqrt{n} \\cdot \\sqrt{n} = n$. Since either $d_i \\le \\sqrt{n}$, or $n/d_i \\le \\sqrt{n}$, we obtain $d_i + \\frac{n}{d_i} \\le 1 + \\frac{n}{1} = n+1$.\nTherefore, substituting $n+1$ for $d_i+n/d_i$ in $(*)$, we get $\\sigma(n) \\le \\frac{1}{2}(n+1)\\tau(n)$. Inequality (1) is proved. From the proof it follows that the equality is achieved only if $n$ is a prime (otherwise there exist $d_i \\ne 1, n$, and then $d_i + \\frac{n}{d_i} < n+1$).\n\nNow we prove that\n$$\n\\frac{\\sigma(n)}{\\tau(n)} \\ge \\sqrt{n}. \\quad (2)\n$$\nBy Cauchy's inequality, $x + \\frac{n}{x} \\ge 2\\sqrt{n}$ for any $x > 0$. Therefore, substituting $2\\sqrt{n}$ for $d_i+n/d_i$ in $(*)$, we get $\\sigma(n) \\ge \\frac{1}{2} \\cdot 2\\sqrt{n}\\tau(n) = \\sqrt{n}\\tau(n)$. Inequality (2) is proved. From the proof it follows that the equality is achieved only if $n=1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25224,
    "subject": "Mathematics (Multi-modal)",
    "question": "Bob has the collection of $n$ coins. Each of them weights an integer number of grams, and their total weight is equal to $300$ grams. Let $k, l, m$ be positive integers such that $k + l + m = 300$.\nFind the smallest possible number of $n$ for which Bob always (independently of coin's weights and for all possible $k, l, m$) can partition his collection into three groups so that the total weight of the coins in the first, second, and third groups is equal to $k, l$, and $m$ grams, respectively.",
    "options": [],
    "answer": "201",
    "solution": "Answer: $201$.\n\nLet $a_1 \\le a_2 \\le \\dots \\le a_n$ be the weights of the coins in Bob's collection. If there exists an $i$ such that $a_i \\ge 101$, and $k = l = m = 100$, then, obviously, Bob cannot partition his collection into three equally weighted groups. It is easy to see that if $n \\le 200$, then there can exist an $a_i \\ge 101$ (e.g., $a_1 = \\dots = a_{n-1} = 1$, $a_n = 300 - (n-1) \\ge 300 - 200 = 100$). Therefore, $n \\ge 201$.\n\nWe now show that the smallest possible value of $n$ is $201$.\n\nFirst, note that if $n = 201$, then $a_1 = \\dots = a_i = 1$, where $i \\ge 102$. Indeed, if $i \\le 101$, then $2 \\le a_{i+1} \\le \\dots \\le a_n$, and in this case $a_1 + \\dots + a_n \\ge 1 \\cdot i + 2 \\cdot (201 - i) = 402 - i \\ge 402 - 101 = 301$, contrary to the problem condition.\n\nWithout loss of generality, we suppose that $k \\le l \\le m$. Thus $k \\le 100$. So the first $k$ coins combine into the first group ($a_1 + \\dots + a_k = k$).\n\nWe renumber the remaining $201-k$ weights: $1 = x_1 = x_2 = \\dots = x_t < x_{t+1} \\le \\dots \\le x_{201-k}$. The total weight of these coins is equal to\n$$\n300 - k \\le x_{201-k} + (201 - k - 1 - t) \\cdot 2 + t\n$$\nwhich implies\n$$\nt \\ge x_{201-k} + 100 - k.\n$$\nIn particular, $t \\ge x_{201-k}$ since $k \\le 100$.\n\nNow we show how to form the third group with the weight equal to $m$.\n\nConsider the sums:\n$$\nS_1 = a_{201-k}, \\quad S_2 = a_{201-k} + a_{201-k-1}, \\quad \\dots, \\\\\nS_i = a_{201-k} + a_{201-k-1} + \\dots + a_{201-k-i+1}, \\quad i = 1, 2, \\dots\n$$\nIf one of these sums is equal to $m$, then the corresponding coins combine into the third group, and the remaining coins combine into the second group with the weight equal to $l$. Otherwise, there exists a $j$, such that $S_j < m$ and $S_{j+1} > m$. In particular, $S_{j+1} - S_j \\ge 2$, i.e. $x_{201-k-j} \\ge 2$. This means that among the numbers $x_{201-k}$, $x_{201-k-1}$, ..., $x_{201-k-j}$ there are no $1$'s. Moreover, this means that $m - S_j < x_{201-k-j} \\le x_{201-k} \\le t$. So\n$$\nx_1 + \\dots + x_{m-S_j} + x_{201-k-j+1} + \\dots + x_{201-k} = m,\n$$\nand the corresponding coins combine into the third group. The remaining coins combine into the second group with weight equal to $l$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 25225,
    "subject": "Mathematics (Multi-modal)",
    "question": "Exactly one integer number is written in each cell of an $8 \\times 8$ square table. Per move it is allowed to choose any $n \\times n$ square, $1 < n < 8$, and either to increase by 1 or to decrease by 1 all numbers in the cells of the chosen square.\nIs it possible to get the table with zeros in all its cells from the arbitrary initial table?\n(V. Kaskevich)",
    "options": [],
    "answer": "Yes",
    "solution": "We separate the table into five parts: the central $4 \\times 4$ square, two $2 \\times 8$ rectangles, and two $4 \\times 2$ rectangles (see Fig. 1).\nWe can obtain the 0's in all cells of the $2 \\times 8$ rectangles. It suffices to show how we can change (increase by 1 or decrease by 1) the value in any cell of the rectangle so that all other cells of this rectangle keep their value. The corresponding procedures using $2 \\times 2$ and $3 \\times 3$ squares are shown in Fig. 2 and Fig. 3.\n\n![](attached_image_1.png)\n\n![](attached_image_2.png)\nFig. 3\n\nIn similar way we can change (increase by 1 or decrease by 1) the value in any cell of the $2 \\times 4$ rectangles. It suffices to show how we can change the value in any cell of these rectangles so that all other cells of these rectangles and all cells of the $2 \\times 8$ rectangles keep their value. The corresponding procedures are shown in Fig. 4 (see, in addition, Fig 3).\n\n![](attached_image_3.png)\n\nIt remains to change the numbers in the cells of the central $4 \\times 4$ square. It suffices to show how we can change the value of any cell of this square so that all other cells of the table keep their value. The corresponding procedures are shown in Fig. 5.\n\n![](attached_image_4.png)\nFig. 5",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25226,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that there exist infinitely many positive integers $n$ so that $n$ and the sum of its digits are perfect squares and\n\na) the decimal representation of $n$ contains at most one $0$;\n\nb) the decimal representation of $n$ does not contain $0$. (T. Lasy)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note that the sum of the digits of this number is equal to\n$$\nS(X_n) = 1 \\cdot n + 5 \\cdot (n - 1) + 6 = 6n + 1.\n$$\nFurther,\n$$\n\\begin{aligned}\nX_n &= \\frac{10^{2n} - 1}{9} + \\frac{4 \\cdot (10^n - 1)}{9} + 1 \\\\\n&= \\frac{10^{2n} - 1 + 4 \\cdot (10^n - 1) + 9}{9} \\\\\n&= \\frac{10^{2n} + 4 \\cdot 10^n + 4}{9} \\\\\n&= \\left( \\frac{10^n + 2}{3} \\right)^2\n\\end{aligned}\n$$\nis a perfect square. (Since $S(10^n + 2) = 3$, we have $10^n + 2 \\vdots 3$.) It suffices to note that if $n = 6m^2 + 2m$ ($m \\in \\mathbb{N}$), then $S(X_n) = 6n + 1 = 6(6m^2 + 2m) + 1 = (6m + 1)^2$. Therefore, the numbers $X_{6m^2+2m}$ satisfy the problem condition for all $m \\in \\mathbb{N}$, and we have infinitely many such numbers.\n\n*Remark.* Also we can consider\n$$(2 \\cdot 10^n + 1)^2$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25227,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given cyclic quadrilateral $ABCD$ with $\\frac{CD}{BD} > \\frac{AB}{AC}$.\nProve that $CD \\cdot BD > AB \\cdot AC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $ABCD$ is cyclic, $\\angle ABD = \\angle ACD$, $\\angle BAC = \\angle BDC$. For the area of the triangles $ABD$ and $ACD$ we have\n$$\nS(ABD) = AB \\cdot BD \\sin \\angle ABD = S_1 + S_4,\n$$\n$$\nS(ACD) = AC \\cdot CD \\sin \\angle ACD = S_3 + S_4.\n$$\nBy condition, $AC \\cdot CD > AB \\cdot BD$, so $S(ACD) > S(ABD)$, which gives $S_3 > S_1$.\n\nSimilarly, for the area of $BAC$ and $BDC$ we have\n$$\nS(BAC) = AB \\cdot AC \\sin \\angle BAC = S_1 + S_2, \\quad S(BDC) = BD \\cdot CD \\sin \\angle BDC = S_3 + S_2.\n$$\nSince $S_3 > S_1$, we have $S(BAC) > S(BDC)$, which gives $CD \\cdot BD > AB \\cdot AC$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25228,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a $3n \\times 3n$ table with one of two signs \"+\" or \"-\" in any of its cells. Per move one can replace the signs in all cells of some row (or of some column) by the opposite signs. At the beginning there is exactly one minus in the table (all other signs are pluses). After some moves the table with exactly 36 minuses is obtained.\nFind all possible values of $n$.",
    "options": [],
    "answer": "n = 3",
    "solution": "Answer $n = 3$.\n\nNote that if changing of signs is performed even number of times to some row (column), then it is equivalent that the operation is not applied at all. So we can assume that the operation is applied exactly once to some rows (columns), and is not applied to the remaining rows (columns). Let the operation be applied to $x$ rows and $y$ columns. Then the total number of the cells in which the signs are changed is equal to $c = 3n x + 3n y - 2 x y$. We rewrite the equality as\n$$\n|3n - 2x| \\cdot |3n - 2y| = |9n^2 - 2c| \\quad (1)\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25229,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an $n \\times n$ table with one of two signs \"+\" or \"-\" in any of its cells. Per move one can replace the signs in all cells of some row (or of some column) by the opposite signs. At the beginning there are exactly two minuses in the table (all other signs are pluses). After some moves the table with exactly 9 minuses is obtained.\nFind the smallest and the greatest values of $n$.\n(I. Voronovich)",
    "options": [],
    "answer": "n = 5, n = 11",
    "solution": "Answer: $n = 5, n = 11$.\n\nNote that if the operation of the sign changes is applied even times to some row (column), then it is equivalent that the operation is not applied at all. If the operation of the sign changes is applied odd times to some row (column), then it is equivalent that the operation is applied one time only. So we suppose that the operation is applied exactly once to some rows (columns), and is not applied to the remaining rows (columns). Let the operation of sign changes be applied to $x$ rows and $y$ columns. Then the total number of the cells in which the signs is changed is equal to $c = nx + ny - 2xy$. We rewrite the equality as\n$$\n(n - 2x) \\cdot (n - 2y) = n^2 - 2c. \\quad (1)\n$$\n\nNote that $c$ may admit the values 7, 9 or 11 depending on the number 0, 1 or 2 of initial minuses which is changed by pluses. If $n$ is even, then the number of minuses after any operation keeps its parity, but after all operations this parity is changed ($2 \\rightarrow 9$), a contradiction. So $n$ is odd. Therefore, both the co-factors $(n - 2x)$ and $(n - 2y)$ are odd and are no more than $n$ (since $0 \\le x \\le n$, $0 \\le y \\le n$). If at least one of them is equal to $\\pm n$, then the right-hand side of (1) is a multiple of $n$, so $c \\ne n$, which gives $n \\ge c \\ge 11$. If $|n-2x| < n$, $|n-2y| < n$, then $|n-2x| \\le n-2$, $|n-2y| \\le n-2$ ($n$ is odd). Then $|n^2 - 2c| \\le (n-2)^2$, in particular, $n^2 - 2c \\le n^2 - 4n + 4$, so $n \\le \\frac{c+2}{2} \\le \\frac{13}{2}$. It follows that $n \\le 5$ ($n$ is odd). In any case $n \\le 11$. Note that the case $n=1$ is impossible, since the $1 \\times 1$ board has not contain 9 minuses. It remains to consider the following cases:\n\n1) $n=3$. We have $(3-2x) \\cdot (3-2y) = 9-14=-5$ or $(3-2x) \\cdot (3-2y) = 9-18=-9$, or $(3-2x) \\cdot (3-2y) = 9-22=-13$. Here only -9 can be presented as product of two numbers no greater than 3. It is possible only if $x=3$, $y=0$ or $x=0$, $y=3$. It is evident that both these cases cannot be realized.\n\n2) $n=5$. It is easy to see that if both the minuses are in the left bottom cells of $5 \\times 5$ board and we change the signs in the first column and in the first two rows, the new board will contain exactly 9 minuses. It follows that the smallest value of $N$ is equal to 5.\n\n3) $n=11$. It is easy to see that if both the minuses are in the left bottom cells of $11 \\times 11$ board and we change the signs in the first column, the new board will contain exactly 9 minuses. It follows that the greatest value of $N$ is equal to 11.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25230,
    "subject": "Mathematics (Multi-modal)",
    "question": "$n$ distinct points ($n \\ge 3$) are marked on a plane so that no three of them lie on the same line. All points are connected with the segments. All segments are painted one of the four colors so that if in some triangle (with the vertices at the marked points) two sides have the same color, then all its sides have the same color (each of the four colors is used).\nWhat is the largest possible value of $n$?\n(S. Sobolevskii)",
    "options": [],
    "answer": "9",
    "solution": "Note that at most two segments of the same color start from any point. Otherwise there exist 4 points connected with the segments of the same color. But then either any of the other points are connected with these 4 points with the segments of the same color (it follows that all segments have the same color, which is impossible) or there exists a point connected with given 4 points with the segments of the remaining three colors. In this case this point is connected with at least two points with the segments of the same color (different from the color of the segment connected these two points), which is also impossible. So at most $2 \\times 4 = 8$ segments start from any point. Since $(n-1)$ start from any point, we have $n-1 \\le 8$, which gives $n \\le 9$.\n\nThe example for $n = 9$. To each point assign one of the pairs $(1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3)$, and to each color assign one of the numbers $1,2,3,4$. Then three segments between points $(1,1), (1,2), (1,3)$, three segments between points $(2,1), (2,2), (2,3)$, and three segments between points $(3,1), (3,2), (3,3)$ we paint color 1; three segments between points $(1,1), (2,1), (3,1)$, three segments between points $(1,2), (2,2), (3,2)$, and three segments between points $(1,3), (2,3), (3,3)$ we paint color 2; three segments between points $(1,1), (2,2), (3,3)$, three segments between points $(1,2), (2,3), (3,1)$, and three segments between points $(1,3), (2,1), (3,2)$ we paint color 3; finally, three segments between points $(1,1), (2,3), (3,2)$, three segments between points $(1,2), (2,1), (3,3)$, and three segments between points $(1,3), (2,2), (3,1)$ we paint color 4.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25231,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $M$ be the midpoint of the side $AB$ of the acute-angled non-isosceles triangle $ABC$, $H$ be the orthocenter of $ABC$, and $I$ be the incenter of $ABC$.\nProve that if $M$, $I$, and $H$ are collinear, then the length of the segment $CH$ is equal to the length of the radius of the incircle of the triangle $ABC$.\n(Folklore)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\Gamma$ be incircle of the triangle $ABC$. Let $K$ be the point of tangency of $\\Gamma$ and the side $AB$. Let the line $CL$ meet the side $AB$ at $N$. It is easy to see that $AK = BN$ (it suffices to consider the homothety with the center $C$ which transform $\\Gamma$ into excircle touching $AB$ at $N$, and use the power-point theorem). Since $M$ is the midpoint of $AB$, we have $KM = AM - AK = BM - BN = MN$. So $IM \\parallel LN$. If $H, I, M$ are collinear, then $HI \\parallel CL$, and since $CH \\perp AB$ and $LI \\perp AB$, we see that $CH \\parallel LI$. Therefore, $CHIL$ is a parallelogram. It follows that $CH = LI$. But $LI = r$, where $r$ is the radius of incircle of $ABC$. Thus, $CH = r$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25232,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P$ be the intersection point of the diagonals of inscribed quadrilateral $ABCD$. Points $K$ and $L$ are marked on the bisectors of $\\angle APD$ and $\\angle BPC$, respectively, so that $AP = PK$ and $BP = PL$. Let $M$ be the intersection point of the lines $AK$ and $BL$, and $N$ be the intersection point of the lines $KD$ and $LC$.\nProve that the lines $KL$ and $MN$ are perpendicular.\n(D. Pirshtuk)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\angle APD = \\angle BPC = \\varphi$. Then $\\angle PKA = \\angle PLB = \\frac{\\pi-\\varphi}{2}$ since the triangles $APK$ and $BPL$ are isosceles. Therefore, $MK = ML$.\n\n![](attached_image_1.png)\n\nFrom the equalities $AP = PK$, $BP = PL$ and the power of a point theorem it follows that $PK \\cdot PC = AP \\cdot PC = BP \\cdot PD = PL \\cdot PD$. Hence, $\\frac{PK}{PL} = \\frac{PD}{PC}$. Further, $\\angle KPD = \\angle LPC = \\frac{\\varphi}{2}$. Therefore, the triangles $KPD$, $LPC$ are similar, so $\\angle PKD = \\angle PLC$. Hence, $NK = NL$.\nNow we have $MK = ML$ and $NK = NL$. It follows that $M$ and $N$ lie on the perpendicular bisector of the segment $KL$, which gives the required statement.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25233,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are three chess boards in a chess club, one of them is peaceful, the other one is fair, and the third board is strange. If two chess players use the peaceful board, then the result of their game is always a draw, if they use the fair board, then a stronger player wins, but if players use the strange board, then a stronger player loses.\nIs it possible to organize the chess tournament between six chess players (none of them is equal to another in power) so that the following conditions hold: the tournament consists of five rounds, each round consists of three games and all these three chess boards are used, each participant plays exactly one game with every other participant, and the position of each player in the final table of the results of the tournament corresponds to the power of the player?\n(A participant receives 1 point for a win, 1/2 point for a draw, and 0 point for a loss.)\n(I. Bliznets)",
    "options": [],
    "answer": "No, it is impossible.",
    "solution": "We number chess players in accordance with their power: $P_1, P_2, P_3, P_4, P_5, P_6$ (if the $i$'th player is stronger than the $j$'th one, then $i < j$). Suppose, contrary to our claim, that the tournament can be organized. Let chess player $P_i$ get $a_i$ points ($i = 1, \\dots, 6$). By conditions, $a_i - a_{i+1} \\ge 0.5$ for $i = 1, \\dots, 5$. In particular\n$$\na_i - a_{i+3} = (a_i - a_{i+1}) + (a_{i+1} - a_{i+2}) + (a_{i+2} - a_{i+3}) \\ge 0.5 + 0.5 + 0.5 = 1.5. \\quad (1)\n$$\nLet $A = \\{P_1, P_2, P_3\\}$, $B = \\{P_4, P_5, P_6\\}$, $S_A = a_1 + a_2 + a_3$, $S_B = a_4 + a_5 + a_6$. It is easy to see that $S_A + S_B = 15$. Moreover $S_A - S_B \\ge 4.5$. Indeed\n$$\nS_A - S_B = (a_1 + a_2 + a_3) - (a_4 + a_5 + a_6) = (a_1 - a_4) + (a_2 - a_5) + (a_3 - a_6) \\ge 1.5 + 1.5 + 1.5 = 4.5. \\quad (2)\n$$\nLet $S_Y(X)$ denote the total number of points got by the chess players of group X in the games with the chess players of group Y (here $X, Y \\in \\{A, B\\}$).\nIt is evident that $S_A(A) = S_B(B) = 3$. It follows that $S_B(A) - S_A(B) = (S_A(A) + S_B(A)) - (S_A(B) + S_B(B)) = S_A - S_B$. The total number of the games between the players of A and B is equal to $3 \\cdot 3 = 9$, so $9 \\cdot 1 = 9$ points are distributed in these games. So for $S_A - S_B \\ge 4.5$ it is necessary to have $S_B(A) \\ge 7$. Otherwise, $S_B(A) \\le 6.5$ gives $S_A(B) = 9 - S_B(A) \\ge 2.5$, which implies $S_A - S_B \\le 6.5 - 2.5 = 4$, contrary to (2).\nFor $S_B(A) \\ge 7$ it is necessary to use the fair board at least 5 times, otherwise, since only using the fair board gives 1 point to a stronger player, $S_B(A) \\le 4 \\cdot 1 + 5 \\cdot 0.5 = 6.5$, which is impossible.\nSince the fair board is used exactly 5 times, the players from A must use the peaceful board for the remaining $4 = 9-5$ games with the players from B to obtain $S_B(A) \\ge 7 = 5+4 \\cdot 0.5$.\nTherefore the peaceful board is used at most 1 time for the games between the players of the same group. It follows that $a_6 \\ge 1 + 0.5 = 1.5$. Then\n$$\n15 = S_A + S_B = a_6 + a_5 + a_4 + a_3 + a_2 + a_1 \\ge 1.5 + 2 + 2.5 + 3 + 3.5 + 4 = 16.5,\n$$\na contradiction. Hence it is impossible to organize the tournament satisfying the problem conditions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25234,
    "subject": "Mathematics (Multi-modal)",
    "question": "Points $K$ and $L$ are marked on the side $AB$ of the triangle $ABC$ so that $\\angle ACK = \\angle KCL = \\angle LCB$. Point $M$ is marked on the side $BC$ so that $\\angle MKC = \\angle BKM$.\nFind the value of $\\angle MLC$, if $ML$ is a bisector of $\\angle KMB$.\n(S. Mazanik)",
    "options": [],
    "answer": "30°",
    "solution": "Answer: $30^\\circ$.\n\n![](attached_image_1.png)\n\nSince $L$ lies on the bisector of $\\angle KCB$, $L$ is equidistant from the lines $CP$ and $CB$. Similarly, since $L$ lies on the bisector of $\\angle KMB$, $L$ is equidistant from the rays $MK$ and $MB$. Therefore, $L$ is an equidistant point for the rays $KM$ and $KP$, so $L$ lies on the bisector of $\\angle PKM$. Thus, $\\angle PKL = \\angle BKM = \\angle MKC$. Therefore, each of these angles is equal to $60^\\circ$. Let $\\angle ACB = 3x$. Since $\\angle LKC = 120^\\circ$, we have $\\angle KAC = \\angle LKC - \\angle ACK = 120^\\circ - x$. Therefore\n$$\n\\angle ABC = 180^\\circ - \\angle BAC - \\angle BCA = 180^\\circ - (120^\\circ - x) - 3x = 60^\\circ - 2x.\n$$\nSo,\n$$\n\\begin{aligned}\n\\angle LMB &= 0.5\\angle KMB = 0.5(180^\\circ - \\angle LKM - \\angle ABC) = \\\\\n&= 0.5(180^\\circ - 60^\\circ - (60^\\circ - 2x)) = 30^\\circ + x.\n\\end{aligned}\n$$\nSince $\\angle LMB = \\angle LCM + \\angle MLC$, we obtain $\\angle MLC = 30^\\circ + x - x = 30^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25235,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $I$ be the incenter of the acute-angled non-isosceles triangle $ABC$. Let the incircle touch the side $AB$ at point $Q$. Point $T$ is marked on the side $AB$ so that $IT \\parallel CQ$. The line through $T$ touches the incircle at point $K$ (different from $Q$) and meets the lines $CA$ and $CB$ at points $L$ and $N$, respectively.\n\nProve that $T$ is the midpoint of the segment $LN$.\n\n(Ya. Konstantinovski)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $x = \\angle CQI$, $y = \\angle QCI$, $2\\alpha = \\angle BAC$, $2\\beta = \\angle ABC$, $2\\gamma = \\angle ACB$. Let $r$ be the inradius of the triangle $ABC$. It is easy to see that\n\n![](attached_image_1.png)\n\n$$\n\\angle CIQ = 360^\\circ - 90^\\circ - 2\\alpha - \\gamma =\n$$\n$$\n= 270^\\circ - (\\alpha + \\beta + \\gamma) - \\alpha + \\beta = 180^\\circ - \\alpha + \\beta.\n$$\nSo,\n$$\nx + y = 180^\\circ - \\angle CIQ = \\alpha - \\beta. \\quad (1)\n$$\nFurther,\n$$\n\\angle QTL = \\angle NTB = 180^\\circ - 2(90^\\circ - x) = 2x,\n$$\n\nhence\n$$\n\\begin{aligned}\n\\angle ILT &= 0.5\\angle ALT = 0.5(2\\alpha - 2x) = \\alpha - x, \\angle INT = 0.5\\angle CNT = \\\\\n&= 0.5(2\\beta + 2x) = \\beta + x.\n\\end{aligned}\n$$\nThen\n$$\n\\begin{aligned}\nLT &= LK - KT = r \\cot \\angle ILT - r \\tan x = r(\\cot(\\alpha - x) - \\tan x), \\\\\nNT &= NK + KT = r \\cot \\angle INT + r \\tan x = r(\\cot(\\beta + x) + \\tan x).\n\\end{aligned}\n$$\nTherefore, from\n$$\n\\cot(\\alpha - x) - \\tan x = \\cot(\\beta + x) + \\tan x \\quad (2)\n$$\nit will follow the problem condition.\nIndeed, since\n$$\n\\begin{aligned}\n\\cot(\\alpha - x) - \\tan x &= \\frac{1 + \\tan \\alpha \\tan x}{\\tan \\alpha - \\tan x} - \\tan x = \\\\\n&= \\frac{1 + \\tan \\alpha \\tan x - \\tan \\alpha \\tan x + \\tan^2 x}{\\tan \\alpha - \\tan x} = \\frac{1 + \\tan^2 x}{\\tan \\alpha - \\tan x}, \\\\\n\\cot(\\beta + x) - \\tan x &= \\frac{1 - \\tan \\beta \\tan x}{\\tan \\beta + \\tan x} + \\tan x = \\\\\n&= \\frac{1 - \\tan \\beta \\tan x + \\tan \\beta \\tan x + \\tan^2 x}{\\tan \\beta + \\tan x} = \\frac{1 + \\tan^2 x}{\\tan \\beta + \\tan x},\n\\end{aligned}\n$$\nwe see that (2) is equivalent to\n$$\n\\tan \\alpha - \\tan x = \\tan \\beta + \\tan x. \\quad (3)\n$$\nBy law of sines for $\\triangle CQT$, $\\frac{r}{\\sin y} = \\frac{CI}{\\sin x}$. Since $CI = r/\\sin \\gamma$, we have $\\sin y = \\sin \\gamma \\cdot \\sin x$, i.e. from (1) and the equality $\\alpha + \\beta + \\gamma = 90^\\circ$ it follows that $\\sin(\\alpha - \\beta - x) = \\sin x \\cdot \\cos(\\alpha + \\beta)$. Then\n$\\sin x \\cdot \\cos(\\alpha + \\beta) = \\sin(\\alpha - \\beta) \\cos x - \\cos(\\alpha - \\beta) \\sin x,$\nso $\\sin x(\\cos(\\alpha + \\beta) + \\cos(\\alpha - \\beta)) = \\sin(\\alpha - \\beta) \\cos x$, which is equivalent to\n$2 \\sin x \\cos \\alpha \\cos \\beta = (\\sin \\alpha \\cos \\beta - \\cos \\alpha \\sin \\beta) \\cos x$, $\\iff 2 \\tan x = \\tan \\alpha - \\tan \\beta$,\ni.e. is equivalent to (3), as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25236,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a $20 \\times 20$ table with one of two signs \"+\" or \"-\" in any of its cells. Per move one can replace the signs in all cells of some row (or of some column) by the opposite signs. At the beginning there are $8$ minuses in the table (all other signs are pluses). After some moves the table with exactly $50$ minuses is obtained.\nProve that exactly one initial minus is in the same cell.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "than $10$ (since $0 \\le x \\le 10$, $0 \\le y \\le 10$). It follows that exactly one number, namely $72$, may be presented as the product of two positive integers no greater than $10$ ($72 = 8 \\cdot 9$, e.g., $x = 1, y = 2$). It corresponds to the case when $7$ initial minuses are changed. This means that exactly one initial minus keeps its position.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25237,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that there is no function $f: \\mathbb{R} \\to \\mathbb{R}$ such that $f(f(x)) = 1 - x f(x)$ for all real $x$.\n(I. Voronovich)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose that there exists a function $f: \\mathbb{R} \\to \\mathbb{R}$ satisfying\n$$\nf(f(x)) = 1 - x f(x) \\quad (*)\n$$\nfor any $x \\in \\mathbb{R}$.\n\nSet $c = f(0)$. Using $(*)$ for $x = 0$, we obtain $f(c) = f(f(0)) = 1 - 0 \\cdot f(0) = 1 - 0 \\cdot c = 1$.\nSince\n$$\n1 = f(c), \\quad (1)\n$$\nwe have $f(1) = f(f(c)) = 1 - c \\cdot f(c) = 1 - c \\cdot 1 = 1 - c$. Similarly, since\n$$\n1 - c = f(1), \\quad (2)\n$$\nwe have\n$$\nf(1-c) = f(f(1)) = 1 - f(1) = 1 - (1-c) = c. \\quad (3)\n$$\nIt follows that $f(c) = f(f(1-c))_{(*)} = 1 - (1-c) \\cdot f(1-c)_{(3)} = 1 - (1-c)c = c^2 - c + 1$, i.e.\n$$\nc^2 - c + 1 = f(c). \\quad (4)\n$$\nFrom (1) and (4) it follows that $c^2 - c + 1 = 1$, so either $c = 0$ or $c = 1$.\nWe show that both the equalities $c = 0$ and $c = 1$ give a contradiction. Indeed, if $c = 0$, then from definition of $c$ it follows that $0 = f(0)$, but (1) gives $1 = f(0)$, a contradiction. If $c = 1$, then (1) and (2) give $1 = f(1)$ and $0 = f(1)$ respectively, a contradiction. Therefore there is no function satisfying $(*)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25238,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a triangle $ABC$ with $\\angle ABC = 120^\\circ$, $BC = 2AB$. Find the angle between its medians $AM$ and $BK$.\n(S. Mazanik)",
    "options": [],
    "answer": "60°",
    "solution": "Let $P$ be the point of intersection of the medians $AM$ and $BK$. Draw the line $l$ through $A$ parallel to $BC$. Let $L$ be the point of intersection of $l$ and the line $BK$. The triangles $AKL$ and $BKC$ are equal ($AK = KC$, $\\angle AKL = \\angle BKC$, $\\angle KAL = \\angle BCK$), so $AL = BC = 2AB$. Since $\\angle BAL = 180^\\circ - \\angle ABC = 180^\\circ - 120^\\circ = 60^\\circ$, we see that $ABL$ is a right-angled triangle and $\\angle ABP = \\angle ABL = 90^\\circ$.\n\nSince $BM = MC = 0.5BC = AB$, it follows that $ABM$ is an isosceles triangle and\n![](attached_image_1.png)\n$$\n\\angle BAM = \\angle BMA = 0.5(180^\\circ - \\angle ABC) = 0.5(180^\\circ - 120^\\circ) = 30^\\circ.\n$$\n$$\n\\text{Therefore, } \\angle BPA = 180^\\circ - \\angle PAB - \\angle ABP = 180^\\circ - 30^\\circ - 90^\\circ = 60^\\circ.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25239,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $BK$ be the bisector of the angle $B$ of a triangle $ABC$. Find the angles of the triangle $ABC$ if $AK = 1$, $BK = KC = 2$.",
    "options": [],
    "answer": "∠A = 90°, ∠B = 60°, ∠C = 30°",
    "solution": "We mark the point $L$ on the extension of $BA$ over $A$ so that $AL = AB$. Let $M$ be the point of intersection of the medians of $LBC$. Since $AK : KC = 1 : 2$, point $K$ of the lines $BK$ and $CL$. Then $CA$ is the point of intersection of the medians of $LBC$. Therefore, $BM$ is also the median of $LBC$. Since $BK = KC$, the medians $BM$ and $CA$ are equal. Hence the triangle $LBC$ is isosceles and $LB = LC$. Since $BK$ is the bisector of the triangle of $ABC$, we have $AB : BC = AK : KC = 1 : 2$, so $BC = 2AB = BL$. Therefore, the triangle $LBC$ is equilateral and $\\angle ABC = 60^\\circ$. Since $\\angle ABC = 60^\\circ$ and $BC = 2AB$, we see that $ABC$ is a right-angled triangle, hence $\\angle CAB = 90^\\circ$, $\\angle BCA = 30^\\circ$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25240,
    "subject": "Mathematics (Multi-modal)",
    "question": "Four points $A$, $B$, $C$, $D$ are marked on the parabola $y = x^2$ so that the quadrilateral $ABCD$ is a trapezoid ($AD \\parallel BC$, $AD > BC$). Let $m$ and $n$ be the distances between the intersection point of the diagonals of the trapezoid and the midpoints of its bases $AD$ and $BC$, respectively.\nFind the area of $ABCD$.\n(D. Bazylev, I. Voronovich)",
    "options": [],
    "answer": "S = (m+n)^2 / sqrt(m-n)",
    "solution": "Answer: $S = \\frac{(m+n)^2}{\\sqrt{m-n}}$.\n\nLet $x_R, y_R$ denote the coordinates of point $R$. Let $y = kx + a$ and $y = kx + b$ be the equations of the lines $AD$ and $BC$, respectively. Then\n$$\n\\begin{cases} x_A^2 = y_A = kx_A + a, \\\\ x_D^2 = y_D = kx_D + a, \\end{cases} \n\\begin{cases} x_B^2 = y_B = kx_B + b, \\\\ x_C^2 = y_C = kx_C + b. \\end{cases}\n$$\nSo,\n\n![](attached_image_1.png)\n\n$$\nk = x_A + x_D = x_B + x_C, \\quad (1)\n$$\n$$\na = -x_A x_D, \\quad b = -x_B x_C. \\quad (2)\n$$\nSince $K$ and $L$ are the midpoints of the sides $AD$ and $BC$ respectively, we see that\n$$\nx_K = 0.5(x_A + x_D), \\quad x_L = 0.5(x_B + x_C). \\quad (3)\n$$\nIt is well-known fact that point $M$ of intersection of the diagonals of the trapezoid $ABCD$ lies on the segment joining the midpoints of the trapezoid bases. Taking into account (3) we see that the segment $KL$ containing $M$ is perpendicular to the axis $Ox$, and $x_M = x_K = x_L$. Let $\\alpha$ be the angle between the lines $AD$, $BC$ and the positive direction of $Ox$, then $\\tan \\alpha = k$. If $LN$ is an altitude of the trapezoid $ABCD$, then $\\angle NLK = \\alpha$ ($LN \\perp AD, KL \\perp Ox$). Therefore $LN = KL \\cos \\alpha$. On the other hand, $AD = (y_D - y_A)/\\cos \\alpha$, $BC = (y_C - y_B)/\\cos \\alpha$. Therefore, the required area is equal to\n$$\n\\begin{aligned}\nS &= \\frac{1}{2}(AD + BC)LN = \\frac{1}{2}(y_D - y_A + y_C - y_B)KL \\cdot \\frac{\\cos \\alpha}{\\sin \\alpha} = [KL = m+n] = \\\\\n&= \\frac{1}{2}(y_D - y_A + y_C - y_B) \\cdot \\frac{m+n}{k} = \\frac{1}{2} \\left( \\frac{y_D - y_A}{k} + \\frac{y_C - y_B}{k} \\right) (m+n) \\stackrel{(1)}{=} \\\\\n&= \\frac{1}{2}(x_D - x_A + x_C - x_B)(m+n)\n\\end{aligned} \n\\quad (4)\n$$\nLet $y = k_1x + c$ be the equation of the line $BD$. Then\n$$\n\\left\\{ \n\\begin{array}{l}\nx_B^2 = y_B = k_1 x_B + c, \\\\\nx_D^2 = y_D = k_1 x_D + c,\n\\end{array}\n\\right. \n\\quad \\Rightarrow \\quad \nk_1 = x_B + x_D, \\quad c = -x_B x_D. \n\\quad (5)\n$$\nSo, $y_M = 0.5k_1(x_A + x_D) + c = 0.5k_1(x_B + x_C) + c$. Therefore, from (1) - (3), and (5) it follows that\n$$\n\\begin{aligned}\nm &= y_K - y_M = \\frac{1}{2}(x_A+x_D)(x_A+x_D) - x_A x_D - \\frac{1}{2}(x_B+x_D)(x_A+x_D) + x_B x_D = \\\\\n&= \\frac{1}{2}(x_A + x_D)(x_A - x_B) - x_D(x_A - x_B) = \\frac{1}{2}(x_A - x_B)(x_A - x_D).\n\\end{aligned}\n$$\n$$\n\\begin{aligned}\nn &= y_M - y_L \\stackrel{(1)}{=} \\frac{1}{2}(x_B+x_D)(x_A+x_D) - x_B x_D - \\frac{1}{2}(x_A+x_D)(x_A+x_D) + x_B x_C = \\\\\n&= \\frac{1}{2}(x_A + x_D)(x_B - x_A) - x_B(x_D - x_C) = [x_D - x_C \\stackrel{(1)}{=} x_B - x_A] = \\\\\n&= \\frac{1}{2}(x_B - x_A)(x_A + x_D - 2x_B) \\stackrel{(1)}{=} \\frac{1}{2}(x_B - x_A)(x_C - x_B).\n\\end{aligned}\n$$\nThen $m+n = \\frac{1}{2}(x_B - x_A)(x_C - x_B + x_D - x_A)$. Hence, (4) can be presented in the form\n$$\nS = \\frac{1}{2} \\cdot \\frac{2(m+n)}{x_B - x_A} (m+n) = \\frac{(m+n)^2}{x_B - x_A}.\n$$\nNote that\n$$\nm-n = \\frac{1}{2}(x_B - x_A)(x_D - x_A - x_C + x_B) = [x_D - x_C \\stackrel{(1)}{=} x_B - x_A] = (x_B - x_A)^2.\n$$\n\nTherefore, finally, we have $S = \\frac{(m+n)^2}{x_B - x_A} = \\frac{(m+n)^2}{\\sqrt{m-n}}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25241,
    "subject": "Mathematics (Multi-modal)",
    "question": "A function $f: \\mathbb{R} \\to \\mathbb{R}$ satisfies the equality $f(f(x)) = x f(x) + x - 1$ for all real numbers $x$.\n\na) Find $f(-1)$.\n\nb) Find all possible values of $f(1)$.",
    "options": [],
    "answer": "f(-1) = -1; f(1) ∈ {-1, 1}",
    "solution": "Let the function $f: \\mathbb{R} \\to \\mathbb{R}$ satisfy the equality\n$$\nf(f(x)) = x f(x) + x - 1 \\quad (*)\n$$\nfor all $x \\in \\mathbb{R}$.\n\na) Set $c = f(0)$. We have $f(c) = f(f(0)) = 0 \\cdot f(0) + 0 - 1 = -1$. So\n$$\n-1 = f(c). \\quad (1)\n$$\nTherefore, taking into account $(*)$ for $x = c$ and (1), we obtain the required value\n$$\nf(-1) = f(f(c)) = c \\cdot f(c) + c - 1 = -c + c - 1 = -1.\n$$\n\nb) Let $a = f(1)$. Taking into account $(*)$ for $x = 1$, we obtain\n$$\nf(a) = f(f(1)) = 1 \\cdot f(1) + 1 - 1 = a.\n$$\nSince\n$$\na = f(a), \\quad (2)\n$$\nit follows that\n$$\nf(a) = f(f(a)) = a \\cdot f(a) + a - 1 = a^2 + a - 1. \\quad (3)\n$$\nFrom (2) and (3) it follows that $a^2 + a - 1 = a$, i.e. either $a = -1$ or $a = 1$.\n\nBoth the values are achieved. For example,\n$$\nf(x) = -1 \\quad \\text{and} \\quad f(x) = \\begin{cases} -1, & \\text{if } x \\neq 1, \\\\ 1, & \\text{if } x = 1. \\end{cases} \\quad (3)\n$$\nIt is easy to verify that both the functions satisfy the initial equation $(*)$ for all real $x$.\n\nSome more examples:\n$$\nf(x) = \\begin{cases} \\frac{1}{x}, & \\text{if } x \\neq 0, \\\\ -1, & \\text{if } x = 0 \\end{cases} \\quad \\text{or} \\quad f(x) = \\begin{cases} -1, & \\text{if } x \\neq 0, \\\\ \\alpha, & \\text{if } x = 0, \\end{cases}\n$$\nwhere $\\alpha$ is any real number different from $0$.\nAnother function: $f(x) = \\frac{1}{x}$, if $x \\in A$, and $f(x) = -1$ for all other $x$, where $A$ is an arbitrary subset of $\\mathbb{R} \\setminus \\{0\\}$, such that $A^{-1} \\subset A$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25242,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all real $a$ such that there exists a function $f : \\mathbb{R} \\to \\mathbb{R}$ satisfying the equality $f(\\sin x) + a f(\\cos x) = \\cos 2x$ for all real $x$. (I. Voronovich)",
    "options": [],
    "answer": "all real numbers except 1",
    "solution": "Answer: $a \\in \\mathbb{R} \\setminus \\{1\\}$.\n\nFirst, if $a \\neq 1$ then it is easy to verify that the function\n$$f(x) = \\frac{2x^2 - 1}{a - 1}$$\nsatisfies the given equation.\n\nOn the other hand, for $a = 1$ we have the functional equation\n$$f(\\sin x) + f(\\cos x) = \\cos 2x,$$\nwhich after changing $x$ by $\\pi/2 - x$ becomes\n$$f(\\cos x) + f(\\sin x) = - \\cos 2x,$$\ni.e. $\\cos 2x = - \\cos 2x$ for all real $x$, a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25243,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $S(n)$ be the sum of all digits in the decimal representation of $n \\in \\mathbb{N}$. Do there exist integers $n$ such that\n$$\na)\\quad n - S(n) = 3 \\cdot 2010? \\qquad b)\\quad n - S(n) = 3 \\cdot 2011?\n$$\n(I. Gorodnin)",
    "options": [],
    "answer": "a) Yes, for example n = 6042. b) No.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25244,
    "subject": "Mathematics (Multi-modal)",
    "question": "Non-zero real numbers $a, b, c$ satisfy the equality\n$$\n\\frac{ab}{b-c} + \\frac{bc}{c-a} + \\frac{ca}{a-b} = \\frac{ab}{b+c} + \\frac{bc}{c+a} + \\frac{ca}{a+b} + 6abc.\n$$\nFind all possible values of the expression\n$$\n\\frac{1}{(a^2 - b^2)^2} + \\frac{1}{(b^2 - c^2)^2} + \\frac{1}{(c^2 - a^2)^2}\n$$",
    "options": [],
    "answer": "9",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 25245,
    "subject": "Mathematics (Multi-modal)",
    "question": "In an acute-angled triangle $ABC$ the orthocenter is $H$. $I_H$ is the incenter of $\\triangle BHC$. The bisector of $\\angle BAC$ intersects the perpendicular from $I_H$ to the side $BC$ at point $K$. Let $F$ be the foot of the perpendicular from $K$ to $AB$.\n\nProve that $2KF + BC = BH + HC$.\n\n(A. Voidelevich)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First, easy counting of angles shows that $AK \\parallel HI_H$. By condition, $KI_H \\parallel AH$, so we conclude that $AKIH_H$ is a parallelogram. Hence $AK = HI_H$. Let $T$ be the tangency point of the incircle of $\\triangle BHC$ with the side $BH$, then $I_HT \\perp HT$; thus from the above $\\triangle EKA = \\triangle THI_H$. So $EK = TH$. By the well-known formula $TH = 0.5(BH + HC - BC)$ whence the required equality follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25246,
    "subject": "Mathematics (Multi-modal)",
    "question": "Three distinct real numbers satisfy the following condition: the sum of cubes of any two of them is $\\frac{3}{4}$ smaller than the square of the remaining number.\nFind all possible values of the product of these numbers.",
    "options": [],
    "answer": "1/8",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25247,
    "subject": "Mathematics (Multi-modal)",
    "question": "Non-zero real numbers $a$, $b$, $c$ satisfy the equality\n$$\n\\frac{ab}{b-c} + \\frac{bc}{c-a} + \\frac{ca}{a-b} = \\frac{ab}{b+c} + \\frac{bc}{c+a} + \\frac{ca}{a+b} + 6abc.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25248,
    "subject": "Mathematics (Multi-modal)",
    "question": "$AB$ and $CD$ are two parallel chords of a parabola. Circle $S_1$ passing through points $A$, $B$ intersects circle $S_2$ passing through $C$, $D$ at points $E$, $F$.\n\nProve that if $E$ belongs to the parabola, then $F$ also belongs to the parabola.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First note that all parabolas are similar, thus we may consider the parabola $y = x^2$. Let $a$, $b$, $c$, $d$, $e$ be the abscissae of the points $A$, $B$, $C$, $D$, $E$ respectively. We use the following easy lemmas.\n\n**Lemma 1.** The chords $AB$ and $CD$ of the parabola are parallel if and only if $a + b = c + d$.\n\n**Lemma 2.** The points $A$, $B$, $C$, $D$ of the parabola (at least three of them let be different) are concyclic if and only if $a + b + c + d = 0$.\n\nFrom Lemma 2 it follows that in addition to $A$, $B$, $E$ the circle $S_1$ has one more common point $F_1$ with the parabola, the abscissa of $F_1$ being $f_1 = -a - b - e$. Similarly, $S_2$ has common point $F_2$ with the parabola, its abscissa $f_2 = -c - d - e$. Since $AB \\parallel CD$, we have $a + b = c + d$, so $f_1 = f_2$. This means that $F_1 = F_2 = F$ belongs to the parabola.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25249,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$ be the sum of all 10 pairwise products of the sides of a convex pentagon, $S$ be the area of the pentagon.\n\na) Prove that $S \\le \\frac{1}{5}A$.\n\nb) Does there exist a constant $c < 1/5$ such that $S \\le cA$?\n\n(I. Voronovich)",
    "options": [],
    "answer": "yes",
    "solution": "Answer: b) yes.\n\nWe use the following well-known\n\n**Lemma**. Let $a, b, c, d$ be the lengths of the sides of some quadrilateral, and $S$ be its area. Then $2S \\le ab + cd$ and $2S \\le ac + bd$.\n\nLet now $a, b, c, d, e$ be the lengths of the sides of the given pentagon, $f$ be the length of one of its diagonals (see the Fig.), $S$ be its area. Then $2S \\le ab + cf + de < ab + c(d + e) + de$, i.e. $2S \\le (ab + cd + de) + ce$. Note that all summands in parentheses are the products of the neighboring sides of the pentagon, while $ce$ is the product of non-neighboring sides.\n\nSumming all similar inequalities which we can obtain by the cyclic permutation $a \\to b \\to c \\to d \\to e \\to a$, we get\n\n$$\n10S \\le 3B + C, \\quad (1)\n$$\n\n![](attached_image_1.png)\n\nwhere $B$ is the sum of all five pairwise products of the neighboring sides, $C$ is the sum of all five pairwise products of the non-neighboring sides.\n\nFurther, $2S \\le ac + bf + de < ac + b(d+e) + de = (ac + bd + be) + de$. In the same way we obtain from the last inequality the following one\n\n$$\n10S \\le 3C + B. \\quad (2)\n$$\n\nSumming (1) and (2) gives the required inequality.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25250,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given natural number $a > 1$ and different odd prime numbers $p_1, \\ldots, p_n$, with\n$$\na^{p_1} \\equiv 1 \\pmod{p_2},\\ a^{p_2} \\equiv 1 \\pmod{p_3},\\ \\ldots,\\ a^{p_n} \\equiv 1 \\pmod{p_1}.\n$$\nProve that\n\na) $(a-1)$ is divisible by $p_i$ for some $i = 1, \\ldots, n$.\n\nb) Can $(a-1)$ be divisible by $p_i$ for exactly one $i$ of $i = 1, \\ldots, n$?",
    "options": [],
    "answer": "a) There exists an index i such that p_i divides a−1. b) Yes; it is possible for a−1 to be divisible by exactly one of the primes.",
    "solution": "a) (Solution of A.Ivanin, O.Volod'ko, A.Zhuk.) (Further we put $p_{n+1} = p_1$.)\nLet $p_k$ be the greatest number among all $p_i$ ($i = 1, 2, \\ldots, n$). The condition $a^{p_i} \\equiv 1 \\pmod{p_{i+1}}$ implies that $a$ and $p_i$ are relatively prime for all $i$. Then $a^{p_{k+1}-1} \\equiv 1 \\pmod{p_{k+1}}$ by Little Fermat Theorem. Since, by the problem condition, $a^{p_k} \\equiv 1 \\pmod{p_{k+1}}$, we also have $a^d \\equiv 1 \\pmod{p_{k+1}}$ where $d = \\text{GCD}(p_{k+1}-1, p_k)$.\nBut from $d \\mid p_k$ it follows that $d = p_k$ or $d = 1$. If $d = p_k$ then $p_{k+1} - 1 \\ge p_k$, $p_{k+1} > p_k$, a contradiction. Hence $d=1$ and $a \\equiv 1 \\pmod{p_{k+1}}$, q.e.d.\n\nb) Take any prime $p_1$. By the Dirichlet theorem, we can choose a sequence of primes $p_1, \\ldots, p_n$ such that $p_2 - 1 \\mid p_1,\\ p_3 - 1 \\mid p_2,\\ \\ldots,\\ p_n - 1 \\mid p_{n-1}$.\n\nLet $p_{k+1} - 1 = p_k b_{k+1}$, $k = 1, \\ldots, n - 1$. Further, choose for any $l = 2, \\ldots, n$ a primitive root $x_l \\pmod{p_l}$ and set $a_k = x_k^{b_k}$. In particular, $a_{k+1} \\not\\equiv 1 \\pmod{p_{k+1}}$ and $a_{k+1}^{p_k} = x_{k+1}^{p_{k+1}-1} \\equiv 1 \\pmod{p_{k+1}}$. Now by the Chinese Remainder Theorem, there exists an $a$ such that $a \\equiv 1 \\pmod{p_1}$ and $a \\equiv a_{k+1} \\pmod{p_k}$ for $k = 1, \\ldots, n-1$. This $a$ satisfies the condition.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25251,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $S(n)$ be the sum of all digits in the decimal representation of $n \\in \\mathbb{N}$. How many numbers $n$, $1 \\le n \\le 999$, satisfy the following condition $S(n) = \\text{GCD}(n, n+6)$?",
    "options": [],
    "answer": "25",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25252,
    "subject": "Mathematics (Multi-modal)",
    "question": "Is it possible to arrange the numbers $1, 2, \\ldots, 2011$ over the circle in some order so that among any $25$ successive numbers at least $8$ numbers are multiples of $5$ or $7$ (or both $5$ and $7$)?",
    "options": [],
    "answer": "not possible",
    "solution": "Answer: it is not possible.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25253,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $g(n)$ be the number of all $n$-digit natural numbers each consisting only of the digits $0, 1, 2, 3$ (but not necessarily all of them) such that the sum of no two neighboring digits equals $2$.\nDetermine whether $g(2010)$ and $g(2011)$ are divisible by $11$.\n(I. Kozlov)",
    "options": [],
    "answer": "g(2010) is not divisible by 11; g(2011) is divisible by 11.",
    "solution": "Let $g_0(n)$, $g_1(n)$, $g_2(n)$, $g_3(n)$ be the quantities of the numbers satisfying the given condition which end by the digits $0$, $1$, $2$, $3$ respectively. Then $g(n) = g_0(n) + g_1(n) + g_2(n) + g_3(n)$. By condition,\n$$\n\\begin{align*}\ng_0(n+1) &= g_0(n) + g_1(n) + g_3(n) = g(n) - g_2(n), \\\\\ng_1(n+1) &= g(n) - g_1(n), \\\\\ng_2(n+1) &= g(n) - g_0(n), \\\\\ng_3(n+1) &= g(n).\n\\end{align*}\n$$\nSumming all the equalities, we get\n$$\ng(n+1) = 3g(n) + g_3(n) = 3g(n) + g(n-1).\n$$\nNote that modulo $11$ we have $g(1) \\equiv 3$, $g(2) \\equiv 10$, $g(3) \\equiv 0$, $g(4) \\equiv 10$, $g(5) \\equiv 8$, $g(6) \\equiv 1$, $g(7) \\equiv 0$, $g(8) \\equiv 1$, $g(9) \\equiv 3$, $g(10) \\equiv 10$. Now we see that residues of the numbers $g(n)$ modulo $11$ repeat periodically with period $8$. So, $g(n) \\equiv 0 \\pmod{11}$ iff $n \\equiv 3 \\pmod{8}$ or $n \\equiv 7 \\pmod{8}$, whence $g(2010) \\not\\equiv 0 \\pmod{11}$, $g(2011) \\equiv 0 \\pmod{11}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25254,
    "subject": "Mathematics (Multi-modal)",
    "question": "Point $P$ inside an acute-angled triangle $A_1A_2A_3$ is chosen so that its projections $P_1, P_2, P_3$ onto the sides $A_1A_2$, $A_2A_3$, $A_3A_1$ respectively lie on the sides of the triangle.\nProve that for points $X_1$, $X_2$, $X_3$ on the sides $A_1A_2$, $A_2A_3$, $A_3A_1$ respectively, $$\\max \\left\\{ \\frac{X_1X_2}{P_1P_2}, \\frac{X_2X_3}{P_2P_3}, \\frac{X_3X_1}{P_3P_1} \\right\\} \\ge 1$$ if\n\na) $X_1$, $X_2$, $X_3$ are the midpoints of the corresponding sides;\nb) $X_1$, $X_2$, $X_3$ are the feet of the corresponding altitudes;\nc) $X_1$, $X_2$, $X_3$ are arbitrary points on the corresponding sides.\n\n(IMO-2010 Shortlist, Problem G3, modified)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a), b) (Solution of M. Mankevich, A. Nekrashevich, A. Tanana.) The statement from a) and b) simply follow from the general\n\n**Lemma.** If there is a point $X$ such that $X_1$, $X_2$, $X_3$ are the orthogonal projections of $X$ onto the sides $A_2A_3$, $A_3A_1$, $A_1A_2$ respectively, then the statement of the problem is valid.\n\nIndeed, $P$ belongs to one of the quadrilaterals $A_1X_2XX_3$, $A_2X_3XX_1$, $A_3X_1XX_2$; say, $P$ belongs to $A_2X_1XX_3$. Then\n$$\nP_1P_3 = A_2P \\sin \\angle A_2 \\le A_2X \\sin \\angle A_2,\n$$\ni.e. $X_1X_3 / P_1P_3 \\ge 1$. Thus the lemma is proved.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25255,
    "subject": "Mathematics (Multi-modal)",
    "question": "The lengths of the medians from the vertices of the acute angles of the right-angled triangle are equal to $19$ and $22$.\nFind the length of the hypotenuse of the triangle.",
    "options": [],
    "answer": "26",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25256,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the least positive integer $n$ for which there exists a set $\\{s_1, \\dots, s_n\\}$ consisting of $n$ distinct positive integers such that\n$$\n\\left(1 - \\frac{1}{s_1}\\right) \\left(1 - \\frac{1}{s_2}\\right) \\cdots \\left(1 - \\frac{1}{s_n}\\right) = \\frac{51}{2010}\n$$\n(IMO-2010 Shortlist, Problem N1)",
    "options": [],
    "answer": "59",
    "solution": "2. See IMO-2010 Shortlist, Problem N1.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25257,
    "subject": "Mathematics (Multi-modal)",
    "question": "The external angle bisector of the angle $A$ of an acute-angled triangle $ABC$ meets the circumcircle of $\\triangle ABC$ at point $T$. The perpendicular from the orthocenter $H$ of $\\triangle ABC$ to the line $TA$ meets the line $BC$ at point $P$. The line $TP$ meets the circumcircle of $\\triangle ABC$ at point $D$.\n\nProve that $AB^2 + DC^2 = AC^2 + BD^2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $l_a$ be the bisectrix of $\\angle A$. Then $l_a \\perp AT$ and $PH \\parallel l_a$ (as, by condition, $PH \\perp AT$). Further,\n$$ \\angle (BH, HP) = \\angle (BH, AC) + \\angle (AC, HP) = 90^\\circ + \\angle (AC, l_a) = 90^\\circ + \\angle (l_a, AB) = 90^\\circ + \\angle (HP, AB) = \\angle (HP, AB) + \\angle (AB, CH) = \\angle (HP, CH). $$\nHence, $HP$ is the bisectrix of $\\angle BHC$.\n\nLet $AH$ meet the circumcircle of $\\triangle ABC$ at $D_1 \\neq A$. We have $\\angle D_1BC = 90^\\circ - \\angle C = \\angle HBC$ and, similarly, $\\angle D_1CB = \\angle HCB$. Then the triangles $BHC$ and $BD_1C$ are equal, Hence, $D_1P$ is the bisectrix of $\\angle BD_1C$. Therefore, $D_1P$ intersects the circumcircle at the middle of the arc $BAC$, i.e. at the point $T$. So, $D_1 = D$ and $AD \\perp BC$. Thus the statement follows from the well-known\n\n*Lemma*. The diagonals of the quadrilateral $ABCD$ are perpendicular if and only if $AB^2 + CD^2 = AC^2 + BD^2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25258,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Does there exist a function $f : \\mathbb{R} \\to \\mathbb{R}$, such that for any real $x$ the following equality holds $f(\\sin x) + f(\\cos x) = 2$?\n\nb) The same question for $f(\\sin x) + f(\\cos x) = \\sin 2x$.",
    "options": [],
    "answer": "a) Yes: f(x) = 1 for all real x. b) No, such a function does not exist.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25259,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs $(m, n)$ of nonnegative integers for which\n$$\nm^2 + 2 \\cdot 3^n = m(2^{n+1} - 1).\n$$",
    "options": [],
    "answer": "((m,n) = (6,3), (9,3), (9,5), (54,5))",
    "solution": "2. See IMO-2010 Shortlist, Problem N2.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25260,
    "subject": "Mathematics (Multi-modal)",
    "question": "Point $K$ is the intersection point of medians $AL$, $BM$, and $CN$ of a triangle $ABC$.\nProve that $\\angle ABM = \\angle CAL$ if $\\angle BLA = \\angle CNA$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25261,
    "subject": "Mathematics (Multi-modal)",
    "question": "Three distinct real numbers satisfy the following condition: the square of any of them is $1$ greater than the product of the remaining two.\n\nFind all possible values of the sum of pairwise products of these numbers.",
    "options": [],
    "answer": "-1",
    "solution": "Let the three distinct real numbers be $a$, $b$, and $c$.\n\nThe condition says: the square of any of them is $1$ greater than the product of the other two.\nSo:\n\n$a^2 = bc + 1$\n$b^2 = ca + 1$\n$c^2 = ab + 1$\n\nAdd all three equations:\n$a^2 + b^2 + c^2 = ab + bc + ca + 3$\n\nRecall that $(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$, so:\n$a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ca)$\n\nLet $S = ab + bc + ca$ and $T = a + b + c$.\n\nSo:\n$(a + b + c)^2 - 2S = S + 3$\n$T^2 - 2S = S + 3$\n$T^2 - 3S = 3$\n$T^2 = 3S + 3$\n\nNow, let's try to find $S$.\n\nFrom the first equation: $a^2 = bc + 1$\nSo $a^2 - bc = 1$\nSimilarly, $b^2 - ca = 1$, $c^2 - ab = 1$\n\nNow, consider $a^2 - bc = 1$\nBut $a^2 = bc + 1$\nSo $a^2 - bc = 1$\n\nLet us try to find the possible values of $S = ab + bc + ca$.\n\nLet us try to find the numbers explicitly.\n\nLet $a$, $b$, $c$ be roots of the cubic $x^3 - p x^2 + S x - Q = 0$.\nBut perhaps it's easier to try to find the numbers directly.\n\nLet us try to subtract the equations:\n$a^2 - b^2 = (bc + 1) - (ca + 1) = bc - ca = c(b - a)$\nSo $a^2 - b^2 = c(b - a)$\nBut $a^2 - b^2 = (a - b)(a + b)$\nSo $(a - b)(a + b) = c(b - a)$\nSo $(a - b)(a + b + c) = 0$\nSo either $a = b$ (contradicts distinctness), or $a + b + c = 0$\n\nTherefore, $a + b + c = 0$\nSo $T = 0$\n\nRecall $T^2 = 3S + 3$\nSo $0 = 3S + 3$\nSo $S = -1$\n\nTherefore, the only possible value for the sum of pairwise products is $\\boxed{-1}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25262,
    "subject": "Mathematics (Multi-modal)",
    "question": "Positive real $a$, $b$, $c$ satisfy the condition\n$$\n\\frac{a}{b+c} + \\frac{b}{c+a} + \\frac{c}{a+b} = 1 + \\frac{1}{6} \\left( \\frac{a}{c} + \\frac{b}{a} + \\frac{c}{b} \\right).\n$$\nProve that\n$$\n\\frac{a^3bc}{b+c} + \\frac{b^3ca}{c+a} + \\frac{c^3ab}{a+b} \\ge \\frac{1}{6}(ab + bc + ca)^2.\n$$\n(I. Voronovich)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By the Cauchy-Bunyakovskii inequality, we have\n$$\n(a^2b + b^2c + c^2a)(a+b+c) \\ge (ab + bc + ca)^2. \\quad (*)\n$$\n\nNow consider the given equality. Multiplying both its sides by $(a+b+c)abc$, we get\n$$\nA = (a+b+c)\\left(\\frac{a^2bc}{b+c} + \\frac{b^2ca}{c+a} + \\frac{c^2ab}{a+b}\\right) = abc(a+b+c) + \\frac{(a+b+c)(a^2b + b^2c + c^2a)}{6} \\geq \\text{[see } (*)\\text{]} \\geq abc(a+b+c) + \\frac{1}{6}(ab+bc+ca)^2.\n$$\nFurther,\n$$\nA = \\sum_{cyc} \\left(\\frac{a^3bc}{b+c} + a^2bc\\right) = \\sum_{cyc} \\frac{a^3bc}{b+c} + abc(a+b+c) \\geq abc(a+b+c) = \\frac{1}{6}(ab+bc+ca)^2,\n$$\nwhence the required inequality follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25263,
    "subject": "Mathematics (Multi-modal)",
    "question": "Points $L$ and $H$ are marked on the sides $AB$ of an acute-angled triangle $ABC$ so that $CL$ is a bisector and $CH$ is an altitude. Let $P$, $Q$ be the feet of the perpendiculars from $L$ to $AC$ and $BC$ respectively.\nProve that $AP \\cdot BH = BQ \\cdot AH$. (I. Gorodnin)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We have\n$$\n\\begin{align*}\n\\frac{AP}{PH} &= \\frac{AL \\cos \\angle A}{AC \\cos \\angle A} = \\frac{AL}{AC} = [\\text{since } CL \\text{ is a bisectrix}] = \\\\\n&= \\frac{BL}{BC} = \\frac{BL \\cos \\angle B}{BC \\cos \\angle B} = \\frac{BQ}{BH}.\n\\end{align*}\n$$\nIt follows that $AP \\cdot BH = BQ \\cdot AH$, as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25264,
    "subject": "Mathematics (Multi-modal)",
    "question": "Point $M$ is the midpoint of the side $BC$ of an acute-angled triangle $ABC$, $N$ and $K$ are the feet of its altitudes $AN$ and $CK$, $H$ is the orthocenter of $ABC$. The bisector of the angle $ACB$ meets the segment $AH$ at point $T$.\nFind the length of the circumradius of the triangle $NBK$ if $CT \\parallel MH$, $TH = 10$.\n(Slovenia, 2010)",
    "options": [],
    "answer": "5",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25265,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x_1, \\ldots, x_{100}$ be nonnegative real numbers such that $x_i + x_{i+1} + x_{i+2} \\le 1$ for all $i = 1, \\ldots, 100$ (we put $x_{101} = x_1, x_{102} = x_2$).\nFind the maximal possible value of the sum $S = \\sum_{i=1}^{100} x_i x_{i+2}$.",
    "options": [],
    "answer": "25/2",
    "solution": "3. See IMO-2010 Shortlist, Problem A3.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25266,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$, $g : \\mathbb{R} \\to \\mathbb{R}$ satisfying the following equality $f(f(x+y)) = x f(y) + g(x)$ for all real $x$ and $y$. (I. Gorodnin)",
    "options": [],
    "answer": "f(x) = a, g(x) = a - a x for any real a",
    "solution": "Answer: $f(x) = a$, $g(x) = a - a x$ for all $a \\in \\mathbb{R}$.\n\nLet $a = f(0)$, $b = g(0)$. Note that due to the symmetry the given equation implies\n$$\nx f(y) + g(x) = y f(x) + g(y). \\tag{1}\n$$\nSet $y = 0$ in (1), then we have $g(x) = b - a x$. Now set $y = 1$ in (1), then $f(x) = (f(1) - a)x + a = a + c x$. Now substituting $g(x)$ and $f(x)$ in the given equation we easily obtain that $c = 0$, $b = a$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25267,
    "subject": "Mathematics (Multi-modal)",
    "question": "Points $K$ and $L$ are marked on the side $AB$ of a triangle $ABC$ so that $AK = KL = LB$. Points $M$ and $N$ are the midpoints of the sides $AC$ and $BC$, respectively. $X$ is the intersection point of the segments $AN$ and $CK$, $Y$ is the intersection point of the segments $BM$ and $CL$. Find the length of $XY$ if $AB = 36$.",
    "options": [],
    "answer": "9",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25268,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Does there exist a function $f: \\mathbb{R} \\to \\mathbb{R}$, such that for any real $x$ the following equality holds $f(\\sin x) + f(\\cos x) = 1$?\n\nb) The same question for $f(\\sin x) + f(\\cos x) = \\sin x$.\n\n(I. Gorodnin)",
    "options": [],
    "answer": "a) Yes. For example, f(x) = 1/2 for all real x works. b) No such function exists.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25269,
    "subject": "Mathematics (Multi-modal)",
    "question": "Each student of a group is friends with at least a half of the other students of this group. At the beginning, all students are partitioned into pairs according to the seats they occupy in the classroom. Per move any two students may swap their pairs.\nIs it possible that after a finite number of such moves each pair consists of friends?",
    "options": [],
    "answer": "Yes",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 25270,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a concert, 20 singers will perform. For each singer, there is a (possibly empty) set of other singers such that he wishes to perform later than all the singers from the set.\n\nCan it happen that there are exactly 2010 orders of the singers such that all their wishes are satisfied?\n\n(IMO-2010 Shortlist, Problem C1)",
    "options": [],
    "answer": "Yes",
    "solution": "3. See IMO-2010 Shortlist, Problem C1.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25271,
    "subject": "Mathematics (Multi-modal)",
    "question": "2500 chess kings have to be placed on a $100 \\times 100$ chessboard so that\n1) no king can capture any other one (i.e. no two kings are placed in two squares sharing a common vertex);\n2) each row and each column contains exactly $25$ kings;\nFind the number of such arrangements. (Two arrangements differing by rotation or symmetry are supposed to be different.)\n(IMO-2010 Shortlist, Problem C3)",
    "options": [],
    "answer": "binomial(50, 25)^100",
    "solution": "3. See IMO-2010 Shortlist, Problem C3.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 25272,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that for all real $x, y$\n$$\nf(x - f(x/y)) = x f(1 - f(1/y))\n$$\nand\na) $f(1 - f(1)) \\neq 0$;\nb*) $f(1 - f(1)) = 0$.",
    "options": [],
    "answer": "All solutions are linear: f(x) = k x for some real k. Case a) f(1 − f(1)) ≠ 0 corresponds to k ∈ ℝ \\ {0, 1}. Case b*) f(1 − f(1)) = 0 corresponds to k ∈ {0, 1}, i.e., f(x) ≡ 0 or f(x) = x.",
    "solution": "$$\nf(x - f(x/y)) = x f(1 - f(1/y)), \\quad \\forall x, y \\in \\mathbb{R}, y \\neq 0. \\quad (*)\n$$\nSet $y = 1$ in $(*)$ then\n$$\nf(x - f(x)) = x f(1 - f(1)). \\tag{1}\n$$\nIf $f(t) = 0$ for some $t$ then (1) implies $f(t - f(t)) = f(t) = 0$, hence, in view of\n(*), $t = 0$ since $f(1 - f(1)) \\ne 0$. On the other hand, (*) gives $f(0 - f(0)) = 0$\nso $0 - f(0) = 0$, i.e. $f(0) = 0$. Therefore, $0$ is only zero of $f$.\nso $0 - f(0) = 0$, i.e. $f(0) = 0$. Therefore, $0$ is only zero of $f$.\nFurther, from (1) it follows that $f$ is surjective thus there exist an $\\alpha \\neq 0$ such\nthat $f(\\alpha) = 1$. Set $y = 1/\\alpha$ in (*), then",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25273,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an $n \\times n$ square table. Exactly one beetle sits in each cell of the table. At 12.00 all beetles creep to some neighboring cell (two cells are neighboring if they have the common side).\nFind the greatest number of cells which can become empty (i.e. without beetles) if\n\na) $n = 8$;\nb) $n = 9$.\n\n(Problem Committee of BMO 2011)",
    "options": [],
    "answer": "a) 44; b) 56",
    "solution": "a) Answer: $44$.\n(Solution of A. Zhuk, O. Volod'ko.) Let's mark $20$ cells of the table as it is shown in Fig. 1. It is obvious that all beetles from the marked cells after their moves must occupy different cells. That is, at least $20$ cells will remain occupied, so at most $64 - 20 = 44$ cells can become empty. On the other hand, it is easy to see that after moves all beetles can occupy the marked cells, so the beetles can make at least $44$ empty cells.\n\n![](attached_image_1.png)\n\nb) Answer: $56$.\nAs in a), solution follows from Fig. 2 with $25$ marked cells. So the answer is $81 - 25 = 56$.\n\n![](attached_image_2.png)",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 25274,
    "subject": "Mathematics (Multi-modal)",
    "question": "All cells of a $9 \\times 10$ board are colored red, blue, and green. In any row of the board the number of red cells is no smaller than the number of blue cells and no smaller than the number of green ones. In any column of the board the number of blue cells is no smaller than the number of red cells and no smaller than the number of green ones.\nFind the greatest and the smallest possible number of green cells of the board.\n(I. Voronovich)",
    "options": [],
    "answer": "smallest = 10, greatest = 18",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25275,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given nonzero real numbers $a, b, c$, with\n$$\na + b + c = a^2 + b^2 + c^2 = a^3 + b^3 + c^3. \\quad (*)\n$$\na) Find $\\left(\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c}\\right) (a + b + c - 2)$.\nb) Do there exist pairwise different nonzero $a, b, c$ satisfying $(*)$?\n(D. Bazylev)",
    "options": [],
    "answer": "3; Yes, for example a = 1/2, b = (2 + sqrt(6))/4, c = (2 - sqrt(6))/4.",
    "solution": "a) (Solution of M.Mankevich, E.Dovgialo.) Let $\\sigma_1 = a + b + c$, $\\sigma_2 = ab + bc + ca$, $\\sigma_3 = abc$. From the given equalities we have the system\n$$\n\\sigma_1 = \\sigma_1^2 - 2\\sigma_2 \\quad (1), \\quad \\sigma_1^2 = \\sigma_1^3 - 3\\sigma_1\\sigma_2 + 3\\sigma_3. \\quad (2)\n$$\nFrom (1) we have $\\sigma_1^2 = \\sigma_1^3 - 2\\sigma_1\\sigma_2$. Then from (2) it follows that\n$$\n-2\\sigma_1\\sigma_2 = -\\sigma_1\\sigma_2 + 3\\sigma_3. \\quad (3)\n$$\nFinally, the expression we search is\n$$\n\\frac{\\sigma_2}{\\sigma_3}(\\sigma_1 - 2) = \\frac{\\sigma_1\\sigma_2 - 2\\sigma_2}{\\sigma_3} = [\\text{in view of (3)}] = \\frac{3\\sigma_3}{\\sigma_3} = 3.\n$$\n\nb) It is easy to verify, that $a = \\frac{1}{2}$, $b = \\frac{2+\\sqrt{6}}{4}$, $c = \\frac{2-\\sqrt{6}}{4}$ satisfy the given system.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25276,
    "subject": "Mathematics (Multi-modal)",
    "question": "All cells of a $7 \\times 8$ board are colored red, blue, and green. In any row of the board the number of red cells is no smaller than the number of blue cells and no smaller than the number of green ones. In any column of the board the number of blue cells is no smaller than the number of red cells and no smaller than the number of green ones. Find the greatest and the smallest possible number of green cells of the board.",
    "options": [],
    "answer": "smallest = 8, greatest = 8",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25277,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that the number $9 \\cdot 2^m$ can be presented as the sum of the squares of three positive integers for any positive integer $m$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25278,
    "subject": "Mathematics (Multi-modal)",
    "question": "Do there exist integers $x$ and $y$ satisfying the equality $2x^3 + y^3 = 2011$?",
    "options": [],
    "answer": "No",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25279,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integer $m$ so that the number $7 \\cdot 2^m$ may be presented as the sum of the squares of three positive integers.",
    "options": [],
    "answer": "All odd positive integers m",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25280,
    "subject": "Mathematics (Multi-modal)",
    "question": "Do there exist integers $x$ and $y$ satisfying the equality $2x^2 - 5y^3 = 2011$?",
    "options": [],
    "answer": "No",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25281,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given triangle $ABC$ with $\\angle ACB = 120^\\circ$. Point $L$ is marked on the side $AB$ so that $CL$ is the bisector of $\\angle ACB$. Points $N$ and $K$ are marked on the sides $AC$ and $BC$, respectively, so that $CN + CK = CL$.\nProve that the triangle $KLN$ is equilateral.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25282,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that if positive numbers $a$, $b$, $x$, $y$ satisfy the inequalities $ab \\ge xa + yb$ and $a \\ge b$, then they satisfy the inequality $x + y \\le a$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 25283,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that if positive numbers $a, b, x, y$ satisfy the inequality $ab \\ge xa + yb$, then they satisfy the inequality $\\sqrt{a+b} \\ge \\sqrt{x} + \\sqrt{y}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 25284,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that if positive numbers $a, b, x, y$ satisfy the inequality $ab \\ge xa + yb$, then they satisfy the inequality $\\sqrt{a+b} \\ge \\sqrt{x} + \\sqrt{y}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 25285,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that if positive numbers $a$, $b$, $x$, $y$ satisfy the inequalities $ab \\ge xa + yb$, then they satisfy the inequality $ab \\ge 4xy$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 25286,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that if positive numbers $a$, $b$, $c$, $k$, $l$, $m$ satisfy the inequalities $abc \\ge ka + lb + mc$, then they satisfy the inequality\n$$\na + b + c \\ge \\sqrt{3}(\\sqrt{k} + \\sqrt{l} + \\sqrt{m}).\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 25287,
    "subject": "Mathematics (Multi-modal)",
    "question": "Points $A_1$ and $B_1$ are marked on the sides $AC$ and $BC$ of a triangle $ABC$ so that $AA_1 + BB_1 = AB$. Let $I$ be incenter of the triangle $ABC$, and $O$ be circumcenter of the triangle $A_1B_1C$.\nProve that the lines $OI$ and $A_1B_1$ are perpendicular.\n(I. Voronovich)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25288,
    "subject": "Mathematics (Multi-modal)",
    "question": "Points $M$ and $K$ are marked on the side $AB$ of the triangle $ABC$ so that $AM = MK$, $CM = CB$, $\\angle AKC = \\frac{1}{2} \\angle CAK + 90^\\circ$.\nFind $AC$ if $MB = 8$.",
    "options": [],
    "answer": "8",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25289,
    "subject": "Mathematics (Multi-modal)",
    "question": "Points $M$ and $L$ are the midpoints of the sides $AB$ and $BC$ of an isosceles triangle $ABC$ ($BC = AC$). Point $N$ is marked on the side $AC$ so that $NA + AM = LN = LM$.\nFind the value of the angle $NLM$.\n(Slovenia, 2010)",
    "options": [],
    "answer": "36 degrees",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25290,
    "subject": "Mathematics (Multi-modal)",
    "question": "Several chess players took part in a chess tournament. Each participant played exactly one game with any other participant. A participant received $1$ point for a win, $0.5$ point for a draw, and $0$ point for a lose. Any two players received different numbers of points and the participant taking the last place received $k$ points.\nWhat is the smallest number of the wins of the participant taking the first place?",
    "options": [],
    "answer": "2k",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25291,
    "subject": "Mathematics (Multi-modal)",
    "question": "Several chess players took part in a chess tournament. Each participant played exactly one game with any other participant. A participant received 1 point for a win, 0.5 point for a draw, and 0 point for a lose. Any two players received different numbers of points.\nWhat is the smallest number of the wins of the participant taking the first place?\nFind the greatest possible value of the points received by the participant taking the last place.\n(Jury)",
    "options": [],
    "answer": "Let n be the number of participants.\n- Smallest number of wins of the first-place participant: ceil((n−1)/2).\n- Greatest possible points of the last-place participant: floor((n−1)/2)/2.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25292,
    "subject": "Mathematics (Multi-modal)",
    "question": "Several chess players took part in a chess tournament. Each participant played exactly one game with any other participant. A participant received $1$ point for a win, $0.5$ point for a draw, and $0$ point for a lose. Any two players received different numbers of points and the participant taking the last place received $2.5$ points.\nWhat is the smallest number of the wins of the participant taking the first place?",
    "options": [],
    "answer": "5",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25293,
    "subject": "Mathematics (Multi-modal)",
    "question": "Several chess players took part in a chess tournament. Each player played against every other player. How many games were played?",
    "options": [],
    "answer": "n(n-1)/2",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25294,
    "subject": "Mathematics (Multi-modal)",
    "question": "Several chess players took part in a chess tournament. Each participant played exactly one game with any other participant. A participant received $1$ point for a win, $0.5$ point for a draw, and $0$ point for a loss. Any two players received different numbers of points and the participant taking the last place received $2$ points.\nFind the smallest possible number of the participants of the tournament.",
    "options": [],
    "answer": "9",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25295,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all triples $(x; n; p)$ of positive integers $x$, $n$, and primes $p$, such that\n$$\nx^3 + 3x + 14 = 2 \\cdot p^n.\n$$",
    "options": [],
    "answer": "(1, 2, 3) and (3, 2, 5)",
    "solution": "Answer: $(x; n; p) = (1; 2; 3)$, $(x; n; p) = (3; 2; 5)$.\n\nIt is easy to see that $x^3 + 3x + 14 = (x+2)(x^2 - 2x + 7)$, so the initial equality can be rewritten as\n$$\n(x+2)(x^2-2x+7)=2 \\cdot p^n \\quad (1)\n$$\nIt is evident that $x^2-2x+7 > x+2$ for all $x \\in \\mathbb{N}$. So for the case $x+2 = 2 \\cdot p^k$, $x^2 - 2x + 7 = 2 \\cdot p^{n-k}$ we have $n-k \\ge k \\ge 0$. It means that for both the cases $2(x^2 - 2x + 7) : (x+2)$. Therefore, the number $2(x^2 - 2x + 7)/(x+2)$ is integer, but since $2(x^2 - 2x + 7) = 2x(x+2) - 8(x+2) - 30$, the number $30/(x+2)$ must be integer. It possible only if $(x+2)$ is a divisor of $30$. From (1) it follows that the number $x+2$ has at most two prime divisors, one of them being $2$. So for the natural number $x$ the number $x+2$ can admit only the following values $3, 5, 6, 10$, i.e. $x$ can admit only the values $1, 3, 4, 8$.\n\nFor $x=1$ we have $(x+2)(x^2-2x+7)=3 \\cdot 6=18=2 \\cdot 3^2$, so $p=3$, $n=2$.\n\nFor $x=3$ we have $(x+2)(x^2-2x+7)=5 \\cdot 10=2 \\cdot 5^2$, so $p=5$, $n=2$.\n\nFor $x=4$ we have $(x+2)(x^2-2x+7)=6 \\cdot 15=2 \\cdot 3^2 \\cdot 5$, i.e. this number cannot be $2 \\cdot p^n$ for any natural $n$ and any prime $p$.\n\nFor $x=8$ we have $(x+2)(x^2-2x+7)=10 \\cdot 55=2 \\cdot 5^2 \\cdot 11$, i.e. this number cannot be $2 \\cdot p^n$ for any natural $n$ and any prime $p$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25296,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the greatest possible value of the constant $c$ that satisfies the following condition: for any convex heptagon the sum of the lengths of all its diagonals is greater than $cP$, where $P$ is a perimeter of the heptagon.",
    "options": [],
    "answer": "2",
    "solution": "The sum of the diagonals $A_1A_3$, $A_3A_5$, $A_5A_7$, $A_7A_2$, $A_2A_4$, $A_4A_6$, $A_6A_1$ is obviously greater than the perimeter $P$ of the heptagon (see Fig. 1).\n\n![](attached_image_1.png)\nFig. 1\n\nSimilarly, the sum of diagonals $A_1A_4$, $A_4A_7$, $A_7A_3$, $A_3A_6$, $A_6A_2$, $A_2A_5$, $A_5A_1$ is greater than $P$.\n\nHence the sum of all diagonals is greater than $2P$.\n\nOn the other hand, Fig. 2 shows that the constant $c=2$ cannot be improved.\n\n![](attached_image_2.png)\nFig. 2",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25297,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a polynomial $P(x) = (x+d_1)(x+d_2) \\cdots (x+d_9)$, where $d_1, d_2, \\ldots, d_9$ are nine distinct integers.\nProve that there exists an integer $N$ such that for all integers $x \\ge N$ the number $P(x)$ is divisible by a prime number greater than 20.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25298,
    "subject": "Mathematics (Multi-modal)",
    "question": "Pedestrian, Cyclist and Motorcyclist start at 12.00 from town $A$ to town $B$ simultaneously. When each of them arrives at $B$ he whip rounds and moves to $A$, when he arrives at $A$ he again whip rounds and moves to $B$, and so on. After the start of the movement the first meeting is the meeting of Cyclist and Motorcyclist at some point $C$. By this time Pedestrian passes $1/6$ part of the distance between $A$ and $B$, and after 6 minute he meets Motorcyclist. The first meeting of Cyclist and Pedestrian is also held at $C$.\nWhen do Pedestrian, Cyclist and Motorcyclist meet each other at the same point for the first time?",
    "options": [],
    "answer": "13.30",
    "solution": "Answer: 13.30.\nLet the distance between $A$ and $B$ be equal to $S$ (km), the speeds of Pedestrian, Cyclist and Motorcyclist be equal to $a$, $b$ and $c$ (km/h), respectively. By condition (the first meeting is the meeting of Cyclist and Motorcyclist), it follows that $a < b < c$. Let Pedestrian arrive at point $D$ at that moment when Cyclist and Motorcyclist meet at point $C$. By condition, $AD = \\frac{S}{6}$. Then $\\frac{AD}{a} = t_1 = \\frac{S}{6a}$. Since $AC + (AB + BC) = 2S$ we have $t_1 = \\frac{2S}{b+c}$. So, $\\frac{S}{6a} = \\frac{2S}{b+c}$, hence\n$$\nb+c=12a. \\tag{1}\n$$\nFrom $t_1 b = \\frac{Sb}{6a}$ it follows that $AC = \\frac{Sb}{6a}$, so $CD = AC - AD = \\frac{Sb}{6a} - \\frac{S}{6} = \\frac{S(b-a)}{6a}$. Therefore,\n$$\n\\frac{S(b-a)}{6a} = \\frac{(a+c)}{10}. \\tag{2}\n$$\nBy condition, to arrive at $C$ (the point of meeting with Cyclist) Pedestrian need the time $t_2 = \\frac{AC}{a} = \\frac{Sb}{6a^2}$. On the other hand, $t_2 = \\frac{2S}{a+b}$, so $b(a+b) = 12a^2 \\Leftrightarrow b^2 + ab - 12a^2 = 0 \\Leftrightarrow (b+4a)(b-3a) = 0$. Since $b+4a > 0$ ($a$ and $b$ are positive), we have $b-3a = 0$, i.e. $b = 3a$. Now from (1) it follows $c = 9a$, and (2) gives $S = 3a$. Therefore, $AC = \\frac{Sb}{6a} = \\frac{3a \\cdot 3a}{6a} = 1.5a$. It means that Pedestrian and Cyclist meet at point $C$ in $t_2$ hours after the start, $t_2 = \\frac{1.5a}{a} = 1.5$. Moreover, for Motorcyclist we see $1.5 \\cdot c = 13.5 \\cdot a = 4 \\cdot 3a + 1.5 \\cdot a = 4S + 1.5a = AB + BA + AB + BA + AC$, which means that after 1.5 hours after the start he arrive at point $C$. Therefore, Pedestrian, Cyclist and Motorcyclist meet at point $C$ at 13.30.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25299,
    "subject": "Mathematics (Multi-modal)",
    "question": "A cubic trinomial $x^3 + px + q$ with integer coefficients $p$ and $q$ is said to be *irrational* if it has three pairwise distinct real irrational roots $\\alpha_1$, $\\alpha_2$, $\\alpha_3$.\nFind all irrational cubic trinomials for which the value of $|\\alpha_1| + |\\alpha_2| + |\\alpha_3|$ is the minimal possible.",
    "options": [],
    "answer": "x^3 - 3x - 1 and x^3 - 3x + 1",
    "solution": "First, for $R(x) = x^3 + px + q$ to have three distinct real roots it is necessary that $p < 0$ (because the derivative $R'(x) = 3x^2 + p$ cannot be nonnegative). Let now $p < 0$, then the equation $R'(x) = 0$ has two real roots $x_1 = -\\sqrt{-p/3}$, $x_2 = \\sqrt{-p/3}$. Now, the condition that $R(x)$ has three distinct real roots is equivalent to the inequality $R(x_1) \\cdot R(x_2) < 0$, which can be written as\n$$\n4p^3 + 27q^2 < 0. \\quad (*)\n$$\nLet (*) be valid; denote by $\\alpha_1 < \\alpha_2 < \\alpha_3$ the roots of $R(x)$. Since $R(x)$ is irrational, it has no zero roots. By Vieta's formula $\\alpha_1 + \\alpha_2 + \\alpha_3 = 0$. Hence we have two possibilities:\n\ni) $\\alpha_1 < \\alpha_2 < 0 < \\alpha_3$;\nii) $\\alpha_1 < 0 < \\alpha_2 < \\alpha_3$.\n\nWithout loss of generality we can assume that case i) holds (if $x^3 + px + q$ satisfies ii), then $x^3 + px - q$ satisfies i)).\n\nSo, let $\\alpha_1 < \\alpha_2 < 0 < \\alpha_3$, then $q = -\\alpha_1\\alpha_2\\alpha_3 < 0$. Further,\n$$\n|\\alpha_1| + |\\alpha_2| + |\\alpha_3| = |\\alpha_1 + \\alpha_2| + \\alpha_3 = 2\\alpha_3.\n$$\nHence we need to find $R(x)$ satisfying (*), $p, q \\in \\mathbb{Z}$, $p < 0$, $q < 0$, and for which $2\\alpha_3$ is the smallest possible.\n\nFirst, note that due to $p < 0$, $q < 0$ we have $R(1) = 1 + p + q < 0$ which implies $\\alpha_3 > 1$. Further, $R(2) = 8 + 2p + q$. If $R(2) \\le 0$, then $\\alpha_3 \\ge 2$. Hence, if $\\alpha_3 < 2$, then\n$$\n8 + 2p + q > 0. \\quad (**)\n$$\nNow, it is easy to see that all $R(x)$ satisfying (*), ($**$) with $p < 0$, $q < 0$ are the following: $x^3 - 2x - 1$ and $x^3 - 3x - 1$. The first of them is not irrational since it has $-1$ as a root. The second trinomial satisfies the condition.\n\n*Remark.* One can verify that the value of $|\\alpha_1| + |\\alpha_2| + |\\alpha_3|$ is equal to $4 \\cos 20^\\circ$ for the founded polynomials.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25300,
    "subject": "Mathematics (Multi-modal)",
    "question": "For any point $X$ inside an acute-angled triangle $ABC$ we define\n$$\nf(X) = \\frac{AX}{A_1X} \\cdot \\frac{BX}{B_1X} \\cdot \\frac{CX}{C_1X}\n$$\nwhere $A_1, B_1$, and $C_1$ are the intersection points of the lines $AX, BX$, and $CX$ with the sides $BC, AC$, and $AB$, respectively.\nLet $H, I$, and $G$ be the orthocenter, the incenter, and the centroid of the triangle $ABC$, respectively.\nProve that $f(H) \\ge f(I) \\ge f(G)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $a = BC$, $b = AC$, $c = AB$, and $\\angle A = \\alpha$, $\\angle B = \\beta$, $\\angle C = \\gamma$. Let $AH_1$, $BH_2$, $CH_3$ be altitudes, $AL_1$, $BL_2$, $CL_3$ be angle bisectors, and $AM_1$, $BM_2$, $CM_3$ be medians of $\\triangle ABC$.\nThen\n$$\n\\frac{AI}{L_1I} = \\frac{b+c}{a}, \\quad \\frac{BI}{L_2I} = \\frac{a+c}{b}, \\quad \\frac{CI}{L_3I} = \\frac{a+b}{c}.\n$$\nSince $AG : GM_1 = BG : GM_2 = CG : GM_3 = 2 : 1$, we have\n$$\nf(G) = \\frac{AG}{M_1G} \\cdot \\frac{BG}{M_2G} \\cdot \\frac{CG}{M_3G} = 2 \\cdot 2 \\cdot 2 = 8.\n$$\nHence,\n$$\n\\begin{aligned} f(I) &= \\frac{AI}{L_1I} \\cdot \\frac{BI}{L_2I} \\cdot \\frac{CI}{L_3I} = \\frac{b+c}{a} \\cdot \\frac{a+c}{b} \\cdot \\frac{a+b}{c} \\ge \\frac{2\\sqrt{bc}}{a} \\cdot \\frac{2\\sqrt{ac}}{b} \\cdot \\frac{2\\sqrt{ab}}{c} = 8 = f(G), \\end{aligned}\n$$\ni.e. $f(I) \\ge f(G)$ with the equality if and only if $a = b = c$.\n\nFurther,\n![](attached_image_1.png)\n\n$$\n\\begin{aligned} HH_1 &= BH_1 \\cdot \\tg \\angle HBH_1 = \\\\ &= (AB \\cos \\beta) \\cdot \\tg (90^\\circ - \\gamma) = \\frac{c \\cdot \\cos \\beta \\cdot \\cos \\gamma}{\\sin \\gamma} \\end{aligned}\n$$\nand\n$$\nAH = \\frac{AH_2}{\\cos \\angle HAH_2} = \\frac{AB \\cdot \\cos \\alpha}{\\cos(90^\\circ - \\gamma)} = \\frac{c \\cdot \\cos \\alpha}{\\sin \\gamma}.\n$$\nSo we get $\\frac{AH}{H_1H} = \\frac{\\cos \\alpha}{\\cos \\beta \\cos \\gamma}$. Similarly, $\\frac{BH}{H_2H} = \\frac{\\cos \\beta}{\\cos \\alpha \\cos \\gamma}$ and $\\frac{CH}{H_3H} = \\frac{\\cos \\gamma}{\\cos \\alpha \\cos \\beta}$.\n$$\n\\text{Therefore, } f(H) = \\frac{AH}{H_1H} \\cdot \\frac{BH}{H_2H} \\cdot \\frac{CH}{H_3H} = \\frac{1}{\\cos \\alpha \\cos \\beta \\cos \\gamma}.\n$$\nLet $m = \\operatorname{tg}\\frac{\\alpha}{2}$, $n = \\operatorname{tg}\\frac{\\beta}{2}$ and $k = \\operatorname{tg}\\frac{\\gamma}{2}$. By condition, $\\alpha, \\beta, \\gamma \\in (0^\\circ, 90^\\circ)$, so $m, n, k \\in (0, 1)$. Therefore,\n$$\nf(H) = \\frac{1}{\\cos \\alpha \\cos \\beta \\cos \\gamma} = \\frac{1+m^2}{1-m^2} \\cdot \\frac{1+n^2}{1-n^2} \\cdot \\frac{1+k^2}{1-k^2}\n$$\nWe have\n$$\n\\begin{aligned}\n\\frac{b+c}{a} &= \\frac{b}{a} + \\frac{c}{a} = \\frac{\\sin \\beta}{\\sin \\alpha} + \\frac{\\sin \\gamma}{\\sin \\alpha} = \\frac{2 \\sin \\frac{\\beta+\\gamma}{2} \\cos \\frac{\\beta-\\gamma}{2}}{\\sin(180^\\circ - (\\beta+\\gamma))} = \\\\\n&= \\frac{2 \\sin \\frac{\\beta+\\gamma}{2} \\cos \\frac{\\beta-\\gamma}{2}}{\\sin(\\beta+\\gamma)} = \\frac{\\cos \\frac{\\beta-\\gamma}{2}}{\\cos \\frac{\\beta+\\gamma}{2}} = \\frac{\\cos \\frac{\\beta}{2} \\cos \\frac{\\gamma}{2} + \\sin \\frac{\\beta}{2} \\sin \\frac{\\gamma}{2}}{\\cos \\frac{\\beta}{2} \\cos \\frac{\\gamma}{2} - \\sin \\frac{\\beta}{2} \\sin \\frac{\\gamma}{2}} = \\\\\n&= \\frac{1 + \\operatorname{tg} \\frac{\\beta}{2} \\operatorname{tg} \\frac{\\gamma}{2}}{1 - \\operatorname{tg} \\frac{\\beta}{2} \\operatorname{tg} \\frac{\\gamma}{2}} = \\frac{1 + nk}{1 - nk}.\n\\end{aligned}\n$$\nSimilarly, $\\frac{a+c}{b} = \\frac{1+mk}{1-mk}$ and $\\frac{a+b}{c} = \\frac{1+mn}{1-mn}$. So,\n$$\nf(I) = \\frac{a+b}{c} \\cdot \\frac{b+c}{a} \\cdot \\frac{c+a}{b} = \\frac{1+mn}{1-mn} \\cdot \\frac{1+nk}{1-nk} \\cdot \\frac{1+km}{1-km}\n$$\nThus,\n$$\nf(H) \\ge f(I) \\iff \\frac{1+m^2}{1-m^2} \\cdot \\frac{1+n^2}{1-n^2} \\cdot \\frac{1+k^2}{1-k^2} \\ge \\frac{1+mn}{1-mn} \\cdot \\frac{1+nk}{1-nk} \\cdot \\frac{1+km}{1-km}\n$$\n**Lemma.** If $x, y \\in (0,1)$, then $\\frac{1+x^2}{1-x^2} \\cdot \\frac{1+y^2}{1-y^2} \\ge \\left(\\frac{1+xy}{1-xy}\\right)^2$ with the equality if and only if $x=y$.\n**Proof.** We have\n$$\n\\begin{aligned}\n\\frac{1+x^2}{1-x^2} \\cdot \\frac{1+y^2}{1-y^2} &\\ge \\left(\\frac{1+xy}{1-xy}\\right)^2 \\\\\n&\\iff (1+x^2)(1+y^2)(1-xy)^2 \\\\\n&\\ge (1-x^2)(1-y^2)(1+xy)^2 \\\\\n&\\iff (1+x^2y^2+x^2+y^2)(1+x^2y^2-2xy)\n\\end{aligned}\n$$\n$$\n\\geq (1 + x^2 y^2 - x^2 - y^2)(1 + x^2 y^2 + 2xy).\n$$\nLet $z = 1 + x^2 y^2$, $t = x^2 + y^2$, $w = 2xy$, then\n$$\n\\begin{align*} (z+t)(z-w) &\\ge (z-t)(z+w) \\iff zt \\ge zw \\iff t \\ge w \\iff \\\\ &\\Longleftrightarrow x^2+y^2 \\ge 2xy \\iff (x-y)^2 \\ge 0, \\end{align*}\n$$\nwhich finishes the proof of the lemma.\nBy the lemma,\n$$\n\\left(\\frac{1+m^2}{1-m^2} \\cdot \\frac{1+n^2}{1-n^2}\\right) \\left(\\frac{1+n^2}{1-n^2} \\cdot \\frac{1+k^2}{1-k^2}\\right) \\left(\\frac{1+k^2}{1-k^2} \\cdot \\frac{1+m^2}{1-m^2}\\right) \\ge \\\\ \\ge \\left(\\frac{1+mn}{1-mn}\\right)^2 \\left(\\frac{1+nk}{1-nk}\\right)^2 \\left(\\frac{1+km}{1-km}\\right)^2,\n$$\ni.e.\n$$\n\\left( \\frac{1+m^2}{1-m^2} \\cdot \\frac{1+n^2}{1-n^2} \\cdot \\frac{1+k^2}{1-k^2} \\right)^2 \\ge \\left( \\frac{1+mn}{1-mn} \\cdot \\frac{1+nk}{1-nk} \\cdot \\frac{1+km}{1-km} \\right)^2,\n$$\nwith equality if and only if $m = n = k$, which gives the required statement of the problem.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25301,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs $(n; m)$ of positive integers $n$ and $m$ satisfying the equality $n^2 + n + 1 = (m^2 + m - 3)(m^2 - m + 5)$.",
    "options": [],
    "answer": "(4; 2)",
    "solution": "Answer: $(n; m) = (4; 2)$.\nBy condition,\n$$\nn^2 + n + 1 = (m^2 + n - 3)(m^2 - n + 5) = m^4 + m^2 + 8m - 15.\n$$\nConsider the obtained equation\n$$\nn^2 + n - (m^4 + m^2 + 8m - 16) = 0 \\quad (1)\n$$\nas a quadratic equation with respect to $n$. It has positive integer roots only if the determinant $D = 4m^4+4m^2+32m-63$ of this equation is a perfect square of some integer number. But\n$$\nD = 4m^4 + 4m^2 + 32m - 63 = (2m^2 + 2)^2 - 4(m - 4)^2 - 3 < (2m^2 + 2)^2\n$$\nfor any natural number $m$, and\n$$\nD = 4m^4 + 4m^2 + 32m - 63 = (2m^2 + 1)^2 + 32(m - 2) > (2m^2 + 1)^2\n$$\nfor any natural number $m > 2$. Therefore, (1) has the natural roots only if $m = 1$ or $m = 2$.\nIf $m = 1$, then $n^2 + n + 6 = 0$, so either $n = -2$ or $n \\in \\mathbb{Z}$.\nIf $m = 2$, then $n^2 + n - 20 = 0$, so either $n = -5$ or $n = 4$.\nThus, (4; 2) is a unique pair of positive integers satisfying the problem condition.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25302,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m$, $n$, $k$ be pairwise relatively prime positive integers greater than $3$.\nFind the minimal possible number of points on the plane with the following property: there are $x$ of them which are the vertices of a regular $x$-gon for $x = m$, $x = n$, $x = k$.",
    "options": [],
    "answer": "m + n + k - 4",
    "solution": "Answer: $m + n + k - 4$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25303,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all triples $(x; n; p)$ of positive integers $x$, $n$, and primes $p$, such that $2x^3 + x^2 + 10x + 5 = 2 \\cdot p^n$.",
    "options": [],
    "answer": "[[1, 2, 3], [3, 2, 7]]",
    "solution": "Answer: $(x; n; p) = (1; 2; 3)$, $(x; n; p) = (3, 2, 7)$.\nIt is easy to see that $2x^3 + x^2 + 10x + 5 = (x^2 + 5)(2x + 1)$, so the initial equality can be rewritten as\n$$\n(x^2 + 5)(2x + 1) = 2 \\cdot p^n. \\quad (1)\n$$\nSince $(2x + 1)$ is odd for all natural $x$, we have $p \\neq 2$ and $2x + 1 = p^k$, $x^2 + 5 = 2 \\cdot p^{n-k}$. It is evident that $x^2 + 5 > 2x + 1$ for all $x \\in \\mathbb{N}$ and $p \\ge 3$, then $n - k \\ge k$. Therefore, $(x^2 + 5) \\div (2x + 1)$. So the number $(x^2 + 5)/(2x + 1)$ is integer, but then the number $2(x^2 + 5)/(2x + 1)$ is also integer. Since $2(x^2 + 5) = x(2x + 1) + (10 - x)$, we see that $(10 - x)/(2x + 1)$ must be also integer, and so the number $2(10 - x)/(2x + 1)$ must be integer too. But $2(10 - x) = -(2x + 1) + 21$, so the number $21/(2x + 1)$ must be integer. It is possible only if $2x+1 = 1, 3, 7, 21$, i.e. if $x = 1, x = 3, x = 10$.\n\nFor $x = 1$ we have $(x^2 + 5)(2x + 1) = 6 \\cdot 3 = 18 = 2 \\cdot 3^2$, so $p = 3$, $n = 2$.\nFor $x = 3$ we have $(x^2 + 5)(2x + 1) = 14 \\cdot 7 = 2 \\cdot 7^2$, so $p = 7$, $n = 2$.\nFor $x = 10$ we have $(x^2 + 5)(2x + 1) = 405 \\cdot 21 = 5 \\cdot 7^2 \\cdot 3^5$, i.e. this number cannot be $2 \\cdot p^n$ for any natural $n$ and any prime $p$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25304,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all prime numbers $p$ such that the number $p^2 - p - 1$ is a cube of some positive integer.",
    "options": [],
    "answer": "2, 37",
    "solution": "We have $p^2 - p - 1 = n^3$. First, note that $p^2 > n^3 \\ge n^2 \\Rightarrow p \\ge n + 1$. If $p = n + 1$, then the equation becomes $n^3 - n^2 - n + 1 = 0 \\Rightarrow n = 1, p = 2$, which is a solution of the problem.\n\nLet now $p > n + 1$. Rewrite the equation as $p(p-1) = (n+1)(n^2-n+1)$. So, $(n+1)(n^2-n+1) \\nmid p$. Since $n+1 < p$, we have $n^2-n+1 \\nmid p$, hence\n$$\nn^2 - n + 1 = pk, \\quad k \\in \\mathbb{N}. \\qquad (1)\n$$\nThus $p-1 = k(n+1)$ or\n$$\np = kn + k + 1. \\qquad (2)\n$$\nSubstituting $p$ into (1) gives\n$$\nn^2 - (k^2 + 1)n - (k^2 + k - 1) = 0. \\qquad (3)\n$$\nThe discriminant of this equation is $D = k^4 + 6k^2 + 4k - 3$ must be a square of an integer. But if $k \\ge 4$, then it is easy to see that $(k^2 + 3)^2 < D < (k^2 + 4)^2$, a contradiction.\n\nIt remains to inspect the cases $k = 1, 2, 3$, which gives the answer.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25305,
    "subject": "Mathematics (Multi-modal)",
    "question": "Point $M$ is marked inside the convex quadrilateral $ABCD$ so that the ratio of the areas of the triangles $AMC$ and $BMD$ is equal to the ratio of the tangents of the angles $AMC$ and $BMD$, i.e. $S(AMC) : S(BMD) = \\tg \\angle AMC : \\tg \\angle BMD$.\nProve that $AM^2 + MC^2 + BD^2 = AC^2 + BM^2 + MD^2$ if $M$ does not belong to any of the diagonals of the quadrilateral.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since\n$$\nS(AMC) = 0.5 AM \\cdot MC \\sin \\angle AMC, \\quad S(BMD) = 0.5 BM \\cdot MD \\sin \\angle BMD,\n$$\nwe have\n$$\nAM \\cdot MC = 2S(AMC)/\\sin \\angle AMC, \\quad BM \\cdot MD = 2S(BMD)/\\sin \\angle BMD. \\quad (1)\n$$\n\n![](attached_image_1.png)\n\nBy the cosine law,\n$$\nAC^2 = AM^2 + MC^2 - 2AM \\cdot MC \\cos \\angle AMC,\n$$\n$$\nBD^2 = BM^2 + MD^2 - 2BM \\cdot MD \\cos \\angle BMD.\n$$\n\nFrom (1) it follows\n$$\nAC^2 = AM^2 + MC^2 - 4S(AMC) \\cos \\angle AMC / \\sin \\angle AMC = \\\\\nAM^2 + MC^2 - 4S(AMC) / \\operatorname{tg} \\angle AMC, \\qquad (2)\n$$\n$$\nBD^2 = BM^2 + MD^2 - 4S(BMD) \\cos \\angle BMD / \\sin \\angle BMD = \\\\\nBM^2 + MD^2 - 4S(BMD) / \\operatorname{tg} \\angle BMD. \\qquad (3)\n$$\nBy condition,\n$$\nS(AMC) / \\operatorname{tg} \\angle AMC = S(BMD) / \\operatorname{tg} \\angle BMD,\n$$\nso (2) and (3) gives the required equality.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25306,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose that $1000$ students are standing in a circle.\nProve that there exists an integer $k$ with $100 \\le k \\le 300$ such that in the circle there exists a contiguous group of $2k$ students, for which the first half contains the same number of girls as the second half.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25307,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two distinct points $A$ and $B$ are marked on the left half of the parabola $y = x^2$. Consider any pair of parallel lines which pass through $A$ and $B$ and intersect the right half of the parabola at points $C$ and $D$. Let $K$ be the intersection point of the diagonals $AC$ and $BD$ of the obtained trapezoid $ABCD$. Let $M, N$ be the midpoints of the bases of $ABCD$.\nProve that the difference $KM - KN$ depends only on the choice of points $A$ and $B$ but does not depend on the pair of parallel lines described above.\n(I. Voronovich)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Hint.** Show that if the given points have coordinates $A(a; a^2)$, $B(b; b^2)$, then $KM - KN = (b - a)^2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25308,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A_1A_2A_3A_4$ be a non-cyclic quadrilateral. Let $O_1$ and $r_1$ be the circumcenter and the circumradius of the triangle $A_2A_3A_4$. Define $O_2$, $O_3$, $O_4$ and $r_2$, $r_3$, $r_4$ in a similar way.\nProve that\n$$\n\\frac{1}{O_1A_1^2 - r_1^2} + \\frac{1}{O_2A_2^2 - r_2^2} + \\frac{1}{O_3A_3^2 - r_3^2} + \\frac{1}{O_4A_4^2 - r_4^2} = 0.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25309,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the greatest possible value of $n$ that satisfies the following condition: for any choice of $n$ subsets $M_1, ..., M_n$ of the set $M = \\{1, 2, ..., n\\}$ satisfying the conditions\ni) $i \\in M_i$; and\nii) $i \\in M_j \\Leftrightarrow j \\notin M_i$ for all $i \\neq j$,\nthere exist $M_k$ and $M_l$ such that $M_k \\cup M_l = M$.",
    "options": [],
    "answer": "6",
    "solution": "(Solution of A. Zhuk.) First, if $M = \\{1, 2, ..., 6\\}$, then due to the conditions i) and ii) we have $|M_1| + |M_2| + ... + |M_6| = 21$. It follows that $|M_k| \\ge 4$ for some index $k$. There is nothing to prove if $|M_k| = 6$. If $|M_k| = 5$, then there exists $l \\notin M_k$, and $M_k \\cup M_l = M$. If $|M_k| = 4$, then there exist distinct $l$, $m \\notin M_k$, and either $M_k \\cup M_l = M$ or $M_k \\cup M_m = M$.\nIt remains to show that $n = 7$ does not satisfy the problem condition. Indeed, for $M = \\{1, 2, ..., 7\\}$, the sets $M_1 = \\{1, 2, 3, 4\\}$, $M_2 = \\{2, 4, 6, 7\\}$, $M_3 = \\{2, 3, 5, 7\\}$, $M_4 = \\{3, 4, 5, 6\\}$, $M_5 = \\{1, 2, 5, 6\\}$, $M_6 = \\{1, 3, 6, 7\\}$, $M_1 = \\{1, 4, 5, 7\\}$ give the counterexample needed.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25310,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\Gamma$ be the incircle of a none-isosceles triangle $ABC$, $I$ be its center. Let $A_1$, $B_1$, $C_1$ be the tangency points of $\\Gamma$ with the sides $BC$, $AC$, $AB$, respectively. Let $A_2 = \\Gamma \\cap AA_1$, $M = C_1B_1 \\cap AI$, $P$ and $Q$ be the other (different from $A_1$, $A_2$) intersection points of $A_1M$, $A_2M$ and $\\Gamma$, respectively.\nProve that $A$, $P$, $Q$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We first show that the quadrilateral $A_1IMA_2$ is cyclic. Indeed, $C_1B_1 \\perp AI$ and $IC_1 \\perp AB$. So, $C_1M$ is an altitude in the right-angled triangle $AIC_1$, whence $AM \\cdot AI = AC_1^2$. Further, by the power of a point theorem, $AC_1^2 = A_2A \\cdot AA_1$. So,\n$$\nAM \\cdot AI = AC_1^2 = A_2A \\cdot AA_1,\n$$\nhence $A_1IMA_2$ is cyclic. Therefore, $\\angle IMA_1 = \\angle IA_2A_1$, $\\angle A_2MI + \\angle IA_1A_2 = 180^\\circ$. Since $IA_2 = IA_1$ we have $\\angle IA_2A_1 = \\angle IMA_1$. Hence,\n$$\n\\angle QMI = 180^\\circ - \\angle A_2MI = \\angle IA_1A_2 = \\angle IA_2A_1 = \\angle IMA_1. \\quad (1)\n$$\nFrom (1) it follows that the symmetrical image of the line $A_2Q$ with respect to the line $AI$ is $PA_1$. Hence $P$ is the image of $A_2$ and $Q$ is the image of $A_1$. Therefore the line $PQ$ is the image of $A_2A_1$ with respect to the line $AI$, thus the statement follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25311,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all possible values of real number $a$ such that there exist a function $f : \\mathbb{R} \\to \\mathbb{R}$, and real number $\\alpha$ satisfying the equalities $f(\\alpha) = 0$ and $f(f(x)) = x f(x) + a$ for all real $x$.",
    "options": [],
    "answer": "0",
    "solution": "Answer: $a = 0$.\nIndeed, if $a = 0$, then the function $f \\equiv 0$ satisfies the condition. Now let $a \\neq 0$. Suppose that $f(\\alpha) = 0$ for some $\\alpha$. We have $f(0) = f(f(\\alpha)) = \\alpha \\cdot f(\\alpha) + a = a$. Then $f(a) = f(f(0)) = 0 \\cdot f(0) + a = a$. Therefore, $a = f(a) = f(f(a)) = a \\cdot f(a) + a = a \\cdot a + a = a^2 + a$, i.e. $a = a^2 + a$, and then $a = 0$, a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25312,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs $(n, m)$ of integers $n$ and $m$ satisfying the equality $n^2 + m = m^2 + 2n - 9$.",
    "options": [],
    "answer": "[(-10, -11), (-10, 9), (-3, -5), (-3, 3), (2, -5), (2, 3), (9, -11), (9, 9)]",
    "solution": "Answer: $(-10, -11), (-10, 9), (-3, -5), (-3, 3), (2, -5), (2, 3), (9, -11), (9, 9)$.\n\nMultiplying the given equality by $4$, we obtain\n$$\n4n^2 + 4n = 4m^2 + 8m - 36 \\Leftrightarrow 4n^2 + 4n + 1 = 4m^2 + 8m + 4 - 39 \\Leftrightarrow\n$$\n$$\n(2n+1)^2 = (2m+2)^2 - 39 \\Leftrightarrow (2m+2)^2 - (2n+1)^2 = 39 \\Leftrightarrow (2m+2-2n-1)(2m+2+2n+1) = 39 \\Leftrightarrow (2m-2n+1)(2m+2n+3) = 39.\n$$\nSince $39 = 1 \\cdot 39 = 39 \\cdot 1 = 3 \\cdot 13 = 13 \\cdot 3 = (-1) \\cdot (-39) = (-39) \\cdot (-1) = (-3) \\cdot (-13) = (-13) \\cdot (-3)$ are all possible factorizations of $39$, it suffices to consider the following cases.\n$$\n\\begin{array}{l@{\\quad}l@{\\qquad}l}\n1) \\left\\{ \\begin{array}{l} 2m - 2n + 1 = 1, \\\\ 2m + 2n + 3 = 39 \\end{array} \\right. & \\text{which gives } \\left\\{ \\begin{array}{l} 4m + 4 = 40, \\\\ 4n + 2 = 38 \\end{array} \\right. & \\text{so } \\left\\{ \\begin{array}{l} m = 9, \\\\ n = 9. \\end{array} \\right. \\\\\n2) \\left\\{ \\begin{array}{l} 2m - 2n + 1 = 3, \\\\ 2m + 2n + 3 = 13 \\end{array} \\right. & \\Leftrightarrow & \\left\\{ \\begin{array}{l} 4m + 4 = 16, \\\\ 4n + 2 = 10 \\end{array} \\right. \\Leftrightarrow \\left\\{ \\begin{array}{l} m = 3, \\\\ n = 2. \\end{array} \\right. \\\\\n3) \\left\\{ \\begin{array}{l} 2m - 2n + 1 = 39, \\\\ 2m + 2n + 3 = 1 \\end{array} \\right. & \\Leftrightarrow & \\left\\{ \\begin{array}{l} 4m + 4 = 40, \\\\ 4n + 2 = -38 \\end{array} \\right. \\Leftrightarrow \\left\\{ \\begin{array}{l} m = 9, \\\\ n = -10. \\end{array} \\right. \\\\\n4) \\left\\{ \\begin{array}{l} 2m - 2n + 1 = 13, \\\\ 2m + 2n + 3 = 3 \\end{array} \\right. & \\Leftrightarrow & \\left\\{ \\begin{array}{l} 4m + 4 = 16, \\\\ 4n + 2 = -10 \\end{array} \\right. \\Leftrightarrow \\left\\{ \\begin{array}{l} m = 3, \\\\ n = -3. \\end{array} \\right. \\\\\n5) \\left\\{ \\begin{array}{l} 2m - 2n + 1 = -1, \\\\ 2m + 2n + 3 = -39 \\end{array} \\right. & \\Leftrightarrow & \\left\\{ \\begin{array}{l} 4m + 4 = -40, \\\\ 4n + 2 = -38 \\end{array} \\right. \\Leftrightarrow \\left\\{ \\begin{array}{l} m = -11, \\\\ n = -10. \\end{array} \\right. \\\\\n6) \\left\\{ \\begin{array}{l} 2m - 2n + 1 = -3, \\\\ 2m + 2n + 3 = -13 \\end{array} \\right. & \\Leftrightarrow & \\left\\{ \\begin{array}{l} 4m + 4 = -16, \\\\ 4n + 2 = -10 \\end{array} \\right. \\Leftrightarrow \\left\\{ \\begin{array}{l} m = -5, \\\\ n = -3. \\end{array} \\right. \\\\\n7) \\left\\{ \\begin{array}{l} 2m - 2n + 1 = -39, \\\\ 2m + 2n + 3 = -1 \\end{array} \\right. & \\Leftrightarrow & \\left\\{ \\begin{array}{l} 4m + 4 = -40, \\\\ 4n + 2 = 38 \\end{array} \\right. \\Leftrightarrow \\left\\{ \\begin{array}{l} m = -11, \\\\ n = 9. \\end{array} \\right. \\\\\n8) \\left\\{ \\begin{array}{l} 2m - 2n + 1 = -13, \\\\ 2m + 2n + 3 = -3 \\end{array} \\right. & \\Leftrightarrow & \\left\\{ \\begin{array}{l} 4m + 4 = -16, \\\\ 4n + 2 = 10 \\end{array} \\right. \\Leftrightarrow \\left\\{ \\begin{array}{l} m = -5, \\\\ n = 2. \\end{array} \\right. \\end{array}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25313,
    "subject": "Mathematics (Multi-modal)",
    "question": "Do there exist a function $f : \\mathbb{R} \\to \\mathbb{R}$, and real number $\\alpha$ such that $f(\\alpha) = -2$ and $f(f(x)) = x f(x) + 2x$ for all real $x$?",
    "options": [],
    "answer": "Such function does not exist.",
    "solution": "Answer: such function does not exist.\n\nSuppose, contrary to our claim, that there exists a function satisfying the equality $f(f(x)) = x f(x) + 2x$ for all real $x$, and $f(\\alpha) = -2$ for some $\\alpha$. We have $f(-2) = f(f(\\alpha)) = \\alpha f(\\alpha) + 2\\alpha = -2\\alpha + 2\\alpha = 0$. Then $f(0) = f(f(-2)) = -2 f(-2) - 4 = -4$. Further, $f(-4) = f(f(0)) = 0 \\cdot f(0) + 2 \\cdot 0 = 0$. Now we have $f(0) = f(f(-4)) = -4 f(-4) - 8 = -4 \\cdot 0 - 8 = -8$. But $f(0) = -4$, a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25314,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all possible values of real number $\\alpha$ such that there exist a function $f: \\mathbb{R} \\rightarrow \\mathbb{R}$, and real number $\\alpha$ satisfying the equalities $f(\\alpha) = 0$ and $f(f(x)) = x f(x) + \\alpha$ for all real $x$.",
    "options": [],
    "answer": "alpha = 0",
    "solution": "Answer: $a=0$.\n\nIndeed, if $a=0$, then the function $f \\equiv 0$ satisfies the condition.\n\nNow let $a \\ne 0$. Suppose that $f(\\alpha) = 0$ for some $\\alpha$. We have $f(0) = f(f(\\alpha)) = \\alpha \\cdot f(\\alpha) + a = a$. Then $f(a) = f(f(0)) = 0 \\cdot f(0) + a = a$. Therefore, $a = f(a) = f(f(a)) = a \\cdot f(a) + a = a \\cdot a + a = a^2 + a$, i.e. $a = a^2 + a$, and then $a=0$, a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25315,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all pairs of functions $f, g$, $f : \\mathbb{R} \\to \\mathbb{R}$, $g : \\mathbb{R} \\to \\mathbb{R}$, that satisfy the equality\n$$\ng(f(x + y)) = f(x) + (2x + y)g(y)\n$$\nfor all real $x, y$.",
    "options": [],
    "answer": "Either f(x) = x^2 + c and g(x) = x for an arbitrary real constant c, or f(x) = 0 and g(x) = 0.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25316,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given cyclic pentagon $ABCDE$ with $AB = BC$, $CD = DE$. Segments $AD$ and $BE$ intersect at $P$, segment $BD$ intersects $CA$ and $CE$ at $Q$ and $T$, respectively.\nProve that triangle $PQT$ is isosceles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $AB = BC$, $CD = DE$ we have $\\sim AB = \\sim BC$, $\\sim CD = \\sim DE$.\n\n![](attached_image_1.png)\n\nTherefore,\n$$\n\\angle BPA = \\angle DPE = \\frac{1}{2}(\\sim AB + \\sim DE) = \\frac{1}{2}(\\sim AB + \\sim CD) = \\angle BQA. \\quad (1)\n$$\n\nSo, points $A$, $B$, $Q$, $P$ belong to the same circle and we have\n$$\n\\angle QAB = \\angle QPB \\text{ (subtended by the same chord $BQ$)} \\quad (2)\n$$\nand\n$$\n\\angle QAB = \\angle CAB = \\angle CEB \\text{ (subtended by the same chord $BC$).} \\quad (3)\n$$\nThus from (2) and (3) we have $\\angle QPB = \\angle CEB$, so $PQ \\parallel CE$. Therefore,\n$$\n\\angle TQP = \\angle DTE = \\frac{1}{2}(\\sim BC + \\sim DE) = \\frac{1}{2}(\\sim AB + \\sim CD) = \\angle AQB. \\quad (4)\n$$\nSimilarly, points $P$, $T$, $D$, $E$ belong to the same circle because\n$$\n\\angle EPD = \\frac{1}{2}(\\sim AB + \\sim DE) = \\frac{1}{2}(\\sim BC + \\sim DE) = \\angle ETD.\n$$\nSo, $\\angle PTE = \\angle PDE$ (subtended by the same chord $PE$). Since $\\angle ACE = \\angle ADE$ (subtended by the same chord $AE$) we have $\\angle ACE = \\angle PTE$. Therefore, $PT \\parallel AC$ and hence $\\angle PTQ = \\angle AQB$. Now from (4) it follows that $\\angle PTQ = \\angle TQP$, thus triangle $PTQ$ is isosceles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25317,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a polynomial $P(x)$ with positive real coefficients.\nProve that $P(1)P(xy) \\ge P(x)P(y)$ for all $x \\ge 1, y \\ge 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We use weighted Chebyshev's inequality: if $a_1, \\dots, a_n > 0$, and $x_1 \\le x_2 \\le \\dots \\le x_n$, $y_1 \\le y_2 \\le \\dots \\le y_n$, then\n$$\n\\frac{(a_1x_1 + \\dots + a_nx_n)(a_1y_1 + \\dots + a_ny_n)}{a_1 + \\dots + a_n} \\le a_1x_1y_1 + \\dots + a_nx_ny_n. \\quad (*)\n$$\n\nNow let $p(x) = a_1 + a_2x + \\dots + a_nx^{n-1}$ be the given polynomial. Then we simply set in $(*)$ $x_1 = 1$, $x_2 = x$, ..., $x_i = x^{i-1}$, $y_1 = 1$, ..., $y_i = y^{i-1}$, thus finishing the proof.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 25318,
    "subject": "Mathematics (Multi-modal)",
    "question": "For each positive integer $k$, let $t(k)$ be the largest odd divisor of $k$.\nDetermine all positive integers $a$ for which there exists a positive integer $n$ such that all the differences\n$t(n+a) - t(n)$, $t(n+a+1) - t(n+1)$, ..., $t(n+2a-1) - t(n+a-1)$ are divisible by $4$.",
    "options": [],
    "answer": "a = 1 or a = 3",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25319,
    "subject": "Mathematics (Multi-modal)",
    "question": "$ABCD$ is a cyclic quadrilateral. Let the circle $\\Gamma_1$ pass through points $A$ and $B$ and touch $CD$ at point $E$; let the circle $\\Gamma_2$ pass through points $B$ and $C$ and touch $DA$ at point $F$; let the circle $\\Gamma_3$ pass through points $C$ and $D$, and touch $AB$ at point $G$; let at last the circle $\\Gamma_4$ pass through points $D$ and $A$ and touch $BC$ at point $H$.\nProve that $EG \\perp FH$.\n(A. Voidelevich)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If the quadrilateral $ABCD$ has the pair of parallel sides the result follows from the symmetry of the construction. So we suppose that $ABCD$ is different from a trapezoid and a rectangle.\nLet $X = BC \\cap AD$. Since $\\Gamma_2$ touches $AD$ at point $F$, we have $XF^2 = XB \\cdot XC$.\n![](attached_image_1.png)\nSimilarly, $XH^2 = XA \\cdot XD$. Since $ABCD$ is cyclic, we have $XB \\cdot XC = XA \\cdot XD$, so $XF = XH$, and hence $\\triangle XHF$ is an isosceles triangle. Thus $\\angle BHF = 0.5(180^\\circ - \\angle CXD) = 0.5(\\angle C + \\angle D)$. Similarly, $\\angle BGE = 0.5(\\angle A + \\angle D)$. Let $GE \\cap HF = Y$. Then $\\triangle GYH = 360^\\circ - (\\angle B + 0.5(\\angle C + \\angle D)) + 0.5(\\angle A + \\angle D)) = 360^\\circ - (\\angle B + \\angle D) - 0.5(\\angle A + \\angle C)$. Since $\\angle A + \\angle C = \\angle B + \\angle D = 180^\\circ$, we have $\\triangle GYH = 360^\\circ - 180^\\circ - 90^\\circ = 90^\\circ$, so $EG \\perp FH$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25320,
    "subject": "Mathematics (Multi-modal)",
    "question": "Ten chess players take part in a chess tournament. Each player plays exactly one game with any other player. For marking results a new system is used at this tournament: a player receives 3 points for a win, 1 point for a draw, and 0 point for a loss (in accordance with the old system a player receives 1 point for a win, 0.5 point for a draw, and 0 point for a loss). Grand master $N$ takes first place, he wins more than a half of his games and a number of his points is greater than a number of any other player points. Grand master $A$ says that in accordance with the old marking system grand master $N$ does not find himself among the first seven players (i.e. the number of his points is less than the number of points of the seventh player), while grand master $B$ says that in accordance with the old system grand master $N$ does not find himself even among the first eight players (i.e. the number of his point is less than the number of points of the eighth player).\n\na) Can grand master $A$ be right ?\nb) Can grand master $B$ be right ?",
    "options": [],
    "answer": "a) Yes. b) No.",
    "solution": "a) The following table shows the example of the tournament such that grand master $N$ takes the eighth place only. So grand master $A$ can be right.\n\n|   | N | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | Points (Σ) | Points (Σ) |\n|---|---|---|---|---|---|---|---|---|---|---|-------------|-------------|\n|   |   |   |   |   |   |   |   |   |   |   | New system  | Old system  |\n| N | ● | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 15          | 5           |\n| 1 | 1 | ● | 0.5 | 0.5 | 0.5 | 0 | 0.5 | 0.5 | 1 | 1 | 14 | 5.5         |\n| 2 | 1 | 0.5 | ● | 0.5 | 0.5 | 0.5 | 0 | 0.5 | 1 | 1 | 14 | 5.5         |\n| 3 | 1 | 0.5 | 0.5 | ● | 0 | 0.5 | 0.5 | 0.5 | 1 | 1 | 14 | 5.5         |\n| 4 | 0 | 0.5 | 0.5 | 1 | ● | 0.5 | 0.5 | 0.5 | 1 | 1 | 14 | 5.5         |\n| 5 | 0 | 1 | 0.5 | 0.5 | 0.5 | ● | 0.5 | 0.5 | 1 | 1 | 14 | 5.5         |\n| 6 | 0 | 0.5 | 1 | 0.5 | 0.5 | 0.5 | ● | 0.5 | 1 | 1 | 14 | 5.5         |\n| 7 | 1 | 0.5 | 0.5 | 0.5 | 0.5 | 0.5 | 0.5 | ● | 0.5 | 1 | 13 | 5.5         |\n| 8 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0.5 | ● | 1 | 4 | 1.5         |\n| 9 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | ● | 0 | 0           |\n\nb) By condition, grand master $N$ wins more than a half of his games, i.e. he wins at least 5 of his games, so he takes at least 5 points in accordance with the old system of marking. So if $N$ is not among the first eight players, i.e. the eighth player has more points than $N$, then at least eight players have more points than grand master $N$, i.e. any of them has at least 5.5 points. Thus the sum of the points of the first eight players is greater than or equal to $8 \\cdot 5 + 5 + 5 = 44 + 5 = 49$. But the total number of points in the tournament is $1 \\cdot \\frac{10 \\cdot 9}{2} = 45$ in accordance with the old marking system, a contradiction. Thus grand master $B$ cannot be right.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25321,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be an acute triangle. Let $\\omega$ be a circle whose center $L$ lies on the side $BC$. Suppose that $\\omega$ is tangent to $AB$ at $B_1$ and to $AC$ at $C_1$. Suppose also that the circumcenter $O$ of the triangle $ABC$ lies on the shorter arc $B_1C_1$ of $\\omega$.\nProve that the circumcircle of $ABC$ and $\\omega$ meet at two points.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25322,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the greatest positive integer $k$ that satisfies the following property: The set of positive integers can be partitioned into $k$ subsets $A_1, \\dots, A_k$ such that for all integers $n \\ge 15$ and all $i \\in \\{1, 2, \\dots, k\\}$ there exist two distinct elements of $A_i$ whose sum is $n$.",
    "options": [],
    "answer": "2",
    "solution": "3. See IMO-2011 Shortlist, Problem C4.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25323,
    "subject": "Mathematics (Multi-modal)",
    "question": "Define $M_n = \\{1, 2, \\dots, n\\}$, for any $n \\in \\mathbb{N}$. A collection of 3-element subsets of $M_n$ is said to be *fine* if for any coloring of elements of $M_n$ in two colors there is a subset of the collection all three elements of which are of the same color.\nFor any $n \\ge 5$ find the minimal possible number of the 3-element subsets of $M_n$ in the fine collection.",
    "options": [],
    "answer": "Minimal number f(n): f(5) = 10, f(6) = 10, and f(n) = 7 for all n ≥ 7.",
    "solution": "We call any 3-element subset a *triple*; a triple is said to be *monochromatic* if all its three elements are the same color. Let $f(n)$ denote the minimal possible number of the triples in a fine collection of $M_n$.\n\nFirst, we have $f(5) = 10 = \\binom{5}{3}$ — the total number of triples in $M_5$.\nIndeed, if the collection does not contain some triple $\\{a, b, c\\}$, then we color $a$, $b$, $c$ black and the remaining 2 elements we color white. This coloring shows that the collection is not fine. On the other hand, the collection contains all 10 possible triples. Since for any coloring in two colors there are at least three elements, say $x, y, z$, of the same color, then the triple $\\{x, y, z\\}$ is monochromatic, thus the collection is fine.\n\nFurther, $f(6) = 10 = \\frac{1}{2} \\binom{6}{3}$ — a half of the number of all triples in $M_6$.\nNote that we can divide all the triples in $M_6$ into 10 pairs of triples so that in any pair the union of both triples is $M_6$ itself. Now, if the collection contains less than 10 triples then there is a pair of triples $\\{a, b, c\\}$, $\\{d, e, f\\}$ ($\\{a, b, c, d, e, f\\} = M_6$) such that neither of the triples belong to the collection. If $a, b, c$ are colored black and $d, e, f$ are colored white, then there is no monochromatic triple for such coloring, hence the collection is not fine. On the other hand, the collection of 10 triples having exactly one triple from any pair is obviously fine.\n\nLet now $n \\ge 7$. We start with $M_7$, considering the following collection of triples:\n$\\{1, 2, 3\\}$, $\\{1, 4, 5\\}$, $\\{1, 6, 7\\}$, $\\{2, 4, 6\\}$, $\\{2, 5, 7\\}$, $\\{3, 4, 7\\}$, $\\{3, 5, 6\\}$\nWe can observe that any element from $M_7$ belongs to exactly three triples and any possible pair of elements is present exactly in one triple. Consider any coloring of $M_7$. There are at least 4 elements of the same color, say, black.\n\nConsider a number $x$ of white color. Let $x$ belong to the triples $\\{x, a, b\\}$, $\\{x, c, d\\}$, $\\{x, e, f\\}$. Suppose that none of the triples is monochromatic (i.e., white). Then one of them, say, $\\{x, e, f\\}$ contains black elements $e, f$; the other two of them contain (except for $x$) one black and one white element.\nLet $a$ and $c$ be black, $b$ and $d$ be white. Note that if $f$ belongs to the triple $\\{f, a, c\\}$ then we have a monochromatic triple for this coloring. If not, then $f$, besides $\\{x, e, f\\}$, can belong only to the triples $\\{f, a, d\\}$, $\\{f, b, c\\}$. In this case element $e$, besides $\\{x, e, f\\}$, belongs to the triples $\\{b, d, e\\}$ and $\\{a, c, e\\}$. The latter is monochromatic. Thus the collection is fine. Hence $f(7) \\le 7$.\n\nFurther, note that $f(n+1) \\le f(n)$, because any fine collection of $M_n$ is obviously a fine collection of $M_{n+1}$ as well. It follows that $f(n) \\le 7$ for any $n \\ge 7$.\n\nLet's show that $f(n) = 7$ for any $n \\ge 7$. Suppose, contrary to our claim, that $f(n) \\le 6$ for some $n \\ge 7$, and choose the minimal $n \\ge 7$ satisfying the inequality. Thus there is a fine collection of $M_n$ containing $k \\le 6$ triples. Then there are at most $18 = 3 \\cdot 6$ pairs of elements in the collection. Further, $n \\ge 7$, so there are at least $21 = \\frac{7 \\cdot 6}{2}$ possible pairs of elements in $M_n$. Since $21 > 18$, there are two elements $a, b \\in M_n$, such that the pair $\\{a, b\\}$ does not belong to any triple from the collection. We may suppose that $\\{a, b\\} = \\{n-1, n\\}$. Let's now identify the elements $a$ and $b$. Also we identify all the pairs of triples of the kind $\\{a, x, y\\}$, $\\{b, x, y\\}$, if any. The collection thus obtained is obviously fine, and it contains at most $k$ triples. But this is a fine collection for $M_{n-1}$, contrary to the minimality of $n$. Thus the proof is finished.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 25324,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all triples $(a; b; c)$ of real numbers for which there exists a non-zero function $f$, $f: \\mathbb{R} \\to \\mathbb{R}$, such that\n$$\naf(xy + f(z)) + bf(yz + f(x)) + cf(zx + f(y)) = 0\n$$\nfor all real $x, y, z$.",
    "options": [],
    "answer": "All real triples with a + b + c = 0",
    "solution": "First, note that if $a + b + c = 0$, then the function $f \\equiv \\lambda \\neq 0$ satisfies the condition.\n\nNow suppose that $a + b + c \\neq 0$. In particular, at least one of $a, b, c$ is not zero. Say, $a \\neq 0$. Let $(*)$ denote the given functional equation.\n\nSet $x = y = 0$ in $(*)$, then $af(f(z)) = 0$, or $f(f(z)) = 0$ for all $z \\in \\mathbb{R}$.\n\nNow set $z = 0$ in $(*)$, then $af(xy + f(0)) = 0$, or $f(xy + f(0)) = 0$. But $xy + f(0)$ takes all real values, hence $f$ is a zero constant. This contradiction does prove that $a + b + c = 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25325,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two squares with the centers $C_1$ and $B_1$ are constructed on the sides $AB$ and $AC$ outside of the acute-angled triangle $ABC$, respectively. The square $C_1B_1DE$ is constructed on the segment $C_1B_1$ so that $A$ and $D$ lie in the different half-planes with respect to $C_1B_1$.\nProve that the center of the square $C_1B_1DE$ belongs to the line $BC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $H$ be a foot of perpendicular from $A$ onto $BC$. Since $\\angle AB_1C = 90^\\circ$, we have $\\angle AHC + \\angle AB_1C = 90^\\circ + 90^\\circ = 180^\\circ$. Thus, points $A, B_1, C, H$ lie on the same circle. Since $AB_1 = B_1C$, we see that $HB_1$ is a bisector of $\\angle AHC$, so $\\angle AHB_1 = \\angle B_1HC = 90^\\circ/2 = 45^\\circ$. In the same way we obtain $\\angle AHC_1 = \\angle C_1HB = 45^\\circ$.\n\nFurther, consider the circle $\\omega$ with the diameter $C_1B_1$. Since $\\angle C_1HB_1 = \\angle C_1HA + \\angle AHB_1 = 45^\\circ + 45^\\circ = 90^\\circ$, we obtain $H \\in \\omega$. Let $T$ be the second point of intersection of $\\omega$ and $BC$. Then $\\angle C_1B_1T = \\angle C_1HT = 45^\\circ$ and $\\angle TC_1B_1 = \\angle B_1HC = 45^\\circ$. Therefore, $\\triangle C_1TB_1$ is an isosceles right-angled triangle, so point $T \\in BC$ coincides with the center of the square $C_1B_1DE$.\n\nIn the case when point $T$ coincides with $H$, i.e. $\\omega$ touches $BC$, we have $\\angle C_1B_1H = \\angle C_1HB = 45^\\circ$ and $\\angle B_1C_1H = \\angle B_1HC = 45^\\circ$. Therefore, point $H$ is the center of the square $C_1B_1DE$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25326,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f, f: \\mathbb{Q} \\to \\mathbb{Q}$, such that\n$$\nf(x + f(y + f(z))) = y + f(x + z)\n$$\nfor all $x, y, z \\in \\mathbb{Q}$.",
    "options": [],
    "answer": "f(x) = x for all rational x; and f(x) = -x + a for any fixed rational a",
    "solution": "(Solution of S. Dabryneuski, A. Tanana, A. Zhuk.) Set $x = 0$ and $z = 0$ in\n$$\nf(x + f(y + f(z))) = y + f(x + z), \\quad (*)\n$$\n\nthen $f(f(y + f(0))) = y + f(0)$, or\n$$\nf(f(x)) = x, \\quad \\forall x \\in \\mathbb{Q}, \\quad (1)\n$$\nhence $f$ is bijective. Let $f(0) = a$, then $f(a) = 0$. Further, the right hand side of (*) is symmetric with respect to $x$ and $z$, hence $f(x + f(y + f(z))) = f(z + f(y + f(x)))$. Applying the bijectivity of $f$ we conclude that\n$$\nx + f(y + f(z)) = z + f(y + f(x)). \\quad (2)\n$$\n\n$$\nf(y + z) = f(z) + f(y + f(x)) - x. \\quad (3)\n$$\n\nSet $x = a = f(0)$ in (3) then $f(y+z) = f(z) + f(y) - a$, or $f(y+z) - a = (f(y) - a) + (f(z) - a)$. We see that the function $f(x) - a$ is additive, thus $f(x) - a = \\alpha x$, or $f(x) = \\alpha x + a$. Substituting $f$ in (*) gives $f(x) = x$; $f(x) = -x + a, a \\in \\mathbb{Q}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25327,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for every positive integer $n$, the set $\\{2, \\dots, 3n+1\\}$ can be partitioned into $n$ triples in such a way that the numbers from each triple are the lengths of the sides of some obtuse triangle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25328,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a right-angled triangle $ABC$ with $\\angle C = 90^\\circ$.\n![](attached_image_1.png)\nTriangle $AMN$ equal to $ABC$ is constructed on the hypotenuse of $ABC$, $\\angle ANM = 90^\\circ$, $AN = BC$ (see the fig.). The incircle $\\Gamma_1$ of triangle $AMN$ touches the hypotenuse $AM$ at point $P$, and the incircle $\\Gamma_2$ of triangle $ABC$ touches the side $BC$ at point $Q$.\nProve that the segment $PQ$, the hypotenuse $AB$, and the segment connecting the centers of $\\Gamma_1$ and $\\Gamma_2$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note that $\\angle MAN = 90^\\circ - \\angle BAC = \\angle ABC$. So $MN = AC$,\n\n![](attached_image_2.png)\nFig. 1\n\n![](attached_image_3.png)\nFig. 2\n\n$AN = BC$ and, moreover, $AP = BQ$. Let $R$ be the midpoint of $AB$. Then $\\triangle APR = \\triangle BQR$ ($AP = BQ$, $AR = BR$ and $\\angle PAR = \\angle QBR$). So\n\n$\\angle PRA = \\angle BRQ$, hence the angles $PRA$ and $BRQ$ are vertical. Therefore, $PQ$ meets $AB$ at point $R$ (see fig. 1). On the other hand, let $T$ be an intersection point of $AB$ and the segment connecting the centers $Q_1$ and $Q_2$ of $\\Gamma_1$ and $\\Gamma_2$. Then $\\angle O_1AT = 0.5\\angle MAN = 0.5\\angle ABC = \\angle TBO_2$, which gives $AO_1 \\parallel BO_2$. Since, moreover, $AO_1 = BO_2$, we conclude that $AO_1BO_2$ is a parallelogram. Since $T$ is a point of intersection of its diagonals we have $AT = TB$. It follows that $T$ is the midpoint of $AB$, i.e. points $T$ and $R$ coincide (see fig. 2). Thus, all three segments $AB$, $PQ$, $O_1O_2$ meet at the midpoint of $AB$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25329,
    "subject": "Mathematics (Multi-modal)",
    "question": "Ten points are marked in the plane so that no three of them lie on the same straight line. All points are connected with segments. Each of these segments is painted one of the $k$ colors.\nFor what positive integer $k$ ($1 \\le k \\le 5$) is it possible to paint the segments so that for any $k$ of the given 10 points there are $k$ segments with the ends at these $k$ points, all of these segments being painted $k$ different colors?",
    "options": [],
    "answer": "5",
    "solution": "Answer: $k=5$.\n\nFirst we show that for $1 \\le k \\le 4$ the required colouring does not exist.\n\n1. There is nothing to prove for $k=1$ and $k=2$.\n\n2. Let $k=3$. Consider one (say $A$) of these 10 points. We have 3 colours and 9 segments connecting $A$ with other points. So there are two segments (say $AB$ and $AC$) having the same colour. Therefore, the edges of the graph with vertices $A$, $B$, and $C$ are painted at most 2 colours, thus the required colouring does not exist.\n\n3. Let $k=4$. Suppose that there is a point $A$ such that at least 4 segments (say, $AB$, $AC$, $AD$, and $AE$, see Fig. 1) have the same colour (call it blue). Then among the segments connecting points $B$, $C$, $D$, and $E$ there is at least one blue segment (say, $BD$). In this case, there are four blue segments $AB$, $AC$, $AD$, $BD$ and we have to use three different colours to paint two segments $BC$ and $CD$, which is impossible.\n\n![](attached_image_1.png)\nFig. 1\n![](attached_image_2.png)\nFig. 2\n\nHence if the required colouring is possible, then at most 3 segments from any point have the same colour. This yields that for any point (take one of them, say, $A$) there is a colour (say, blue) such that there are exactly three (say, $AB$, $AC$, $AD$, see Fig. 2) blue segments from $A$. Then none of the segments $BC$, $CD$, $BD$ is blue, otherwise $A$, $B$, $C$, $D$ would give a contradiction. For any three points of points $K$, $E$, $F$, $G$, and $H$ among the segments connecting these three points there is a blue segment, otherwise there is no blue segment among the segments connecting $A$ and these three points. (This is a variant of the well-known lemma: among any six persons there are either three pairwise acquainted or three pairwise not acquainted.) Thus there are three points (say, $K$, $E$, $F$) such that all segments $KE$, $KF$, $EF$ are blue. There is at least one blue segment (say, $BK$) among segments $BK$, $CK$, $DK$ (otherwise there is no blue segment among the segments connecting points $B$, $C$, $D$, and $K$). It is sufficient to consider points $B$, $K$, $E$, and $F$ to obtain a contradiction. Therefore the required colouring is impossible for $k = 4$.\n\n4. It remains to consider the case when $k = 5$. We present a complete graph with 10 vertices as a disjoint union of 5 graphs (Fig. 3) that are isomorphic to the graph on Fig. 4 and paint each of these graphs its own colour (different from others). It is easy to see that 1) each vertex is the end of the segments of all 5 colours: there are 3 segments having the first colour, 3 segments having the second colour, and 3 segments painted with remaining 3 colours; 2) there are exactly 9 segments of the same colour; 3) each monochromatic graph is isomorphic to the graph on Fig. 4. Therefore for $k = 5$ the required colouring is possible.\n\n<table>\n<thead>\n<tr><th></th><th>0</th><th>1</th><th>2</th><th>3</th><th>4</th><th>5</th><th>6</th><th>7</th><th>8</th><th>9</th></tr>\n</thead>\n<tbody>\n<tr><th>0</th><td></td><td>5</td><td>1</td><td>4</td><td>4</td><td>1</td><td>1</td><td>4</td><td>3</td><td>2</td></tr>\n<tr><th>1</th><td>5</td><td></td><td>2</td><td>3</td><td>2</td><td>3</td><td>4</td><td>1</td><td>2</td><td>3</td></tr>\n<tr><th>2</th><td>1</td><td>2</td><td></td><td>5</td><td>2</td><td>1</td><td>1</td><td>3</td><td>2</td><td>4</td></tr>\n<tr><th>3</th><td>4</td><td>3</td><td>5</td><td></td><td>4</td><td>3</td><td>2</td><td>4</td><td>1</td><td>3</td></tr>\n<tr><th>4</th><td>4</td><td>2</td><td>2</td><td>4</td><td></td><td>5</td><td>3</td><td>4</td><td>2</td><td>1</td></tr>\n<tr><th>5</th><td>1</td><td>3</td><td>1</td><td>3</td><td>5</td><td></td><td>1</td><td>2</td><td>4</td><td>3</td></tr>\n<tr><th>6</th><td>1</td><td>4</td><td>1</td><td>2</td><td>3</td><td>1</td><td></td><td>5</td><td>5</td><td>5</td></tr>\n<tr><th>7</th><td>4</td><td>1</td><td>3</td><td>4</td><td>4</td><td>2</td><td>5</td><td></td><td>5</td><td>5</td></tr>\n<tr><th>8</th><td>3</td><td>2</td><td>2</td><td>1</td><td>2</td><td>4</td><td>5</td><td>5</td><td></td><td>5</td></tr>\n<tr><th>9</th><td>2</td><td>3</td><td>4</td><td>3</td><td>1</td><td>3</td><td>5</td><td>5</td><td>5</td><td></td></tr>\n</tbody>\n</table>\nFig. 3",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25330,
    "subject": "Mathematics (Multi-modal)",
    "question": "$k$ pawns stand in some $k$ cells of an $n \\times n$ table, $n \\ge 2$, (exactly one pawn stands in each cell of these $k$ cells) so that each $2 \\times 2$ square contains exactly $2$ pawns.\nDetermine all possible values of $k$.",
    "options": [],
    "answer": "If n is even, the only possible value is k = n^2 / 2. If n is odd, all integers k with n(n − 1)/2 ≤ k ≤ n(n + 1)/2 are possible.",
    "solution": "If $n = 2k$ is even, the table can be partitioned into $k^2 = n^2/4$ of $2 \\times 2$ squares, and there are exactly two pawns in each of them. So the total number of the pawns in the table is equal to $2k^2 = n^2/2$.\n\nFig. 1\n![](attached_image_1.png)\nFig. 2\n![](attached_image_2.png)\nFig. 3\n![](attached_image_3.png)\nFig. 4\n\nNow let $n = 2k + 1$. We partition the table (see Fig. 2) into the $2 \\times 2$ squares and the figures which is shown in Fig. 1. At least one pawn must stand in each of these figures, otherwise at least $3$ pawns stand in the $2 \\times 2$ square adjoining with the given figure (see Fig. 3), a contradiction. So the total number of the pawns in the table is greater than or equal to $2k^2 + k$.\nOn the other hand, since exactly $2$ empty cells must be in any $2 \\times 2$ square, the total number of the empty cells is also greater than or equal to $2k^2 + k$. Hence the number of the pawns in the table is less than or equal $(2k + 1)^2 - (2k^2 + k) = 2k^2 + 3k + 1$. The following example (see Fig. 4) shows that any value of the number of the pawns from the stated interval is admissible. (In this example the total number of the pawns is equal to $2k^2 + k + l$, where $0 \\le l \\le 2k + 1$.)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25331,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given $0 < a < b < c$ prove that\n$$\na^{20}b^{12} + b^{20}c^{12} + c^{20}a^{12} < b^{20}a^{12} + a^{20}c^{12} + c^{20}b^{12}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Lemma. Let $0 < x < y < z$ and $\\lambda > 1$. Then $\\sum_{cyc} x^{\\lambda}y < \\sum_{cyc} x y^{\\lambda}$.\n\nProof. Consider the function $f(y) = \\sum_{cyc} (x^{\\lambda}y - x y^{\\lambda})$. It takes the value $0$ for $y = x$ and $y = z$. We have also $f''(y) = \\lambda(\\lambda - 1)y^{\\lambda}(z-x) > 0$, hence $f$ is convex. It follows that $f(y) < 0$ for $y \\in (x; z)$. The lemma is proved.\n\nNow use the lemma for $x = a^{12}$, $y = b^{12}$, $z = c^{12}$, and $\\lambda = 20/12$ thus obtaining the inequality needed.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25332,
    "subject": "Mathematics (Multi-modal)",
    "question": "Exactly one integer stands in each cell of an $m \\times n$ table ($m \\ge 4, n \\ge 4$). The number in any cell is equal to the arithmetic mean of the numbers in some two neighboring cells (i.e. in the cells having the common side with the given cell).\nFind the greatest possible number of all distinct integers in this table.",
    "options": [],
    "answer": "mn - 6",
    "solution": "**Answer: $mn-6$.**\nLet $A$ be the maximal number in the table and $a$ be the minimal number. Consider any cell $k$ of the table where $A$ stands. Since $A$ is the arithmetic mean of the number in some two neighboring cells $k_1$ and $k_2$. But all numbers in the table are no greater than $A$, the numbers in $k_1$ and $k_2$ are equal to $A$. Since $k_1$ and $k_2$ have the common sides with $k$, they have no common side, so they are not neighbors. Since $A$ stands in $k_1$, number $A$ must stand in two cells having the common sides with $k_1$. One of them is cell $k$ and the other cell is different from $k_2$ ($k_1$ and $k_2$ are not neighbors). It follows that $A$ stands in at least 4 different cells.\n\nSimilarly, $a$ stands in at least 4 different cells.\n\nThus, the maximal and minimal numbers stand in at least 8 different cells of the table. The number of all cells is $mn$, so at most $mn - 8 + 2 = mn - 6$ different numbers can stand in the table. We construct the example for $mn - 6$ different numbers.\n\n![](attached_image_1.png)\n\nWe mark two $2 \\times 2$ neighboring squares (see the fig.) and join these squares with the line passing through any cell of the table exactly one time as it is shown in the figures depending on the parity of $m$ and $n$. We stand 1 in all cells of the first square and stand $mn - 6$ in all cells of the second square. We move along the line from the cell with 1 to the cell with $mn - 6$. We put in each cell of the line the number which is 1 greater than the number in the previous cell. It is easy to see that the obtained table satisfies the condition and has exactly $mn - 6$ different numbers.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 25333,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Is it possible for some $n$ to arrange 100 pawns in the cells of an $n \\times n$ table so that at most one pawn stands in each cell and exactly 2 pawns stand in each $2 \\times 2$ square?\n\nb) Is it possible to arrange 110 pawns in the same way?",
    "options": [],
    "answer": "a) No; b) Yes",
    "solution": "a) It is impossible.\n\nFor $n = 14$ we can divide the table into 49 of $2 \\times 2$ squares. If there are exactly 2 pawns in each of them, then the total number of the pawns in the table is equal to $2 \\cdot 49 = 98$. Since we have more than 98 pawns, we have $n > 14$. On the other hand, if $n \\geq 15$, then there are at least 105 pawns in the table (see solution of Problem A.8). It follows that 100 pawns cannot be arranged in the table.\n\nb) It is possible.\n\nThe required arrangement of the 110 pawns for $n = 15$ is given in the figure.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25334,
    "subject": "Mathematics (Multi-modal)",
    "question": "Six teams take part in a football tournament. Each team plays exactly one game with any other team. A team receives 3 points for a win, 1 point for a draw, and 0 point for a loss. After the tournament is over, the teams have 10, 9, 6, 6, 4, and 2 points.\n\na) Prove that the team taking the second place (i.e. having 9 points) does not lose the game with the team winning the first place (i.e. having 10 points).\n\nb) Is it possible uniquely to determine the result of the game between the teams taking the second and the first places?",
    "options": [],
    "answer": "a) The second-place team did not lose to the first-place team. b) No; the result is not uniquely determined.",
    "solution": "The following tables contain all possible results of the tournament.\n\n| Wins | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 |\n|------|---|---|---|---|---|---|---|---|---|---|\n| Draws | 0 | 1 | 2 | 3 | 4 | 5 | 0 | 1 | 2 | 3 |\n| Losses | 5 | 4 | 3 | 2 | 1 | 0 | 4 | 3 | 2 | 1 |\n| Points | 0 | 1 | 2 | 3 | 4 | 5 | 3 | 4 | 5 | 6 |\n\n| Wins | 2 | 2 | 2 | 2 | 3 | 3 | 3 | 4 | 4 | 5 |\n|------|---|---|---|---|---|---|---|---|---|---|\n| Draws | 0 | 1 | 2 | 3 | 0 | 1 | 2 | 0 | 1 | 0 |\n| Losses | 3 | 2 | 1 | 0 | 2 | 1 | 0 | 1 | 0 | 0 |\n| Points | 6 | 7 | 8 | 9 | 9 | 10 | 11 | 12 | 13 | 15 |\n\nWe see that the first team getting 10 points and having the first place wins 3 games, loses 1 game, and ends 1 game in a draw; so this team has 4 effective games. If the second team getting 9 points and having the second place loses the game with the first team, then it wins 3 games and loses 2 games, i.e. 5 of its games are effective. Therefore the total number of effective games of the first and the second teams is equal to 9, but only one of these games is common. So there are at least 8 effective games in the tournament, which contradicts the equality $x = 7$. Hence, the second team does not lose the game with the first team.\n\nb) There are tournaments with distinct results (see the tables).\n\n\n\n|   | Nº1 | Nº2 | Nº3 | Nº4 | Nº5 | Nº6 | Σ |\n|---|-----|-----|-----|-----|-----|-----|---|\n| Nº1 | • | 0 | 3 | 3 | 3 | 1 | 10 |\n| Nº2 | 3 | • | 1 | 1 | 1 | 3 | 9 |\n| Nº3 | 0 | 1 | • | 1 | 1 | 3 | 6 |\n| Nº4 | 0 | 1 | 1 | • | 1 | 3 | 6 |\n| Nº5 | 0 | 1 | 1 | 1 | • | 1 | 4 |\n| Nº6 | 1 | 0 | 0 | 0 | 1 | • | 2 |\n\nTable 1\n\n\n\n|   | Nº1 | Nº2 | Nº3 | Nº4 | Nº5 | Nº6 | Σ |\n|---|-----|-----|-----|-----|-----|-----|---|\n| Nº1 | • | 1 | 0 | 3 | 3 | 3 | 10 |\n| Nº2 | 1 | • | 3 | 1 | 1 | 3 | 9 |\n| Nº3 | 3 | 0 | • | 1 | 1 | 1 | 6 |\n| Nº4 | 0 | 1 | 1 | • | 1 | 3 | 6 |\n| Nº5 | 0 | 1 | 1 | 1 | • | 1 | 4 |\n| Nº6 | 0 | 0 | 1 | 0 | 1 | • | 2 |\n\nTable 2",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25335,
    "subject": "Mathematics (Multi-modal)",
    "question": "A parabola $y = a x^2$, $a > 0$, and the hyperbola $y = 1/x$ meet at point $T$. The common tangent of these curves touches the hyperbola at point $Q$ and touches the parabola at point $P$.\na) Prove that some two medians of the triangle $PQT$ are perpendicular.\nb) Determine the area of the triangle $PQT$.\n(V. Karamzin)",
    "options": [],
    "answer": "27/4",
    "solution": "b) Answer: $S = 27/4$.\n![](attached_image_1.png)\nWe find the coordinates $(x_T; y_T)$ of the point of intersection of the parabola $y = a x^2$ and the hyperbola $y = 1/x$. We have $a x_T^2 = 1/x_T$, so $x_T = 1/\\sqrt{a} = 1/t$ and $y_T = 1/x_T = t$. The equation of the tangent to the parabola at point $P(x_P; y_P)$ has the form $y - y_P = 2a x_P(x - x_P)$, and the equation of the tangent to the hyperbola at point $Q(x_Q; y_Q)$ has the form $y - y_Q = (-1/x_Q^2)(x - x_Q)$. Since the tangent is a common tangent for both the graphs, we have\n$$\n2a x_P = -1/x_Q^2, \\qquad (1)\n$$\n$$\n-2a x_P^2 + y_P = 1/x_Q + y_Q. \\qquad (2)\n$$\nSince $y_P = a x_P^2$ and $y_Q = 1/x_Q$, the obtained equality has the form\n$$\n-a x_P^2 = 2/x_Q. \\qquad (2)\n$$\nFrom equations (1), (2) it follows that $x_Q = -1/(2t)$, $x_P = -2/t$, and so $y_Q = -2t$, $y_P = 4t$. Therefore,\n$$\nT(\\frac{1}{t}; t), \\quad P(-\\frac{2}{t}; 4t), \\quad Q(-\\frac{1}{2t}; -2t). \\qquad (3)\n$$\nLet $M, K, L$ be the midpoints of the sides $PQ, PT, TQ$ respectively. Then\n$$\nM(-\\frac{5}{4t}; t), \\quad K(-\\frac{1}{2t}; \\frac{5}{2}t); \\quad L(\\frac{1}{4t}; -\\frac{t}{2}).\n$$\nComparing the obtained values with (3), we note that $MT$ parallel to the axis $Ox$, since $y_M = y_T$, and $QK$ parallel to $Oy$, since $y_Q = y_K$. Therefore, medians $MT$ and $QK$ in the triangle $PQT$ are perpendicular, as required.\nLet $H$ be a centroid of the triangle $PQT$. Then\n$$\nS(PQT) = 2S(QKT) = [TH \\perp QK] = 2 \\cdot \\frac{1}{2} QK \\cdot TH = \\\\\n= [TH = \\frac{2}{3}MT] = \\frac{2}{3}QK \\cdot TM.\n$$\nSince $TM \\parallel Ox$, we have $TM = 1/t - (-5/(4t)) = 9/(4t)$, Similarly, since $QK \\parallel Oy$, we have $QK = 5t/2 - (-2t) = 9t/2$. Therefore,\n$$\nS(PQT) = \\frac{2}{3} \\cdot \\frac{9}{4t} \\cdot \\frac{9t}{2} = \\frac{27}{4}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25336,
    "subject": "Mathematics (Multi-modal)",
    "question": "The graphs of the functions $f(x) = ax^2 + bx + c$, $g(x) = \\frac{cx + b}{cx + a}$, $h(x) = cx + b$ have the unique common point of the intersection, and the graph of $h(x)$ has no another common points with the graphs of $f(x)$ and $g(x)$. All $a, b, c$ are non-zero pairwise distinct real numbers.\nFind all possible values of $a, b, c$.",
    "options": [],
    "answer": "a = 3/11, b = -8/11, c = 4/11",
    "solution": "Since the graphs of $f(x)$ and $h(x)$ has exactly one common point, the equation $ax^2 + bx + c = cx + b$, i.e. the equation\n$$\nax^2 - (c-b)x + (c-b) = 0, \\quad (1)\n$$\nhas exactly one root. So, its discriminant is equal to $0$, i.e. $(c-b)^2-4a(c-b) = 0 \\Leftrightarrow (c-b)(c-b-4a) = 0$. By condition, $c \\neq b$, i.e. $c-b \\neq 0$, so we have $c-b-4a=0$ or\n$$\nc - b = 4a. \\quad (2)\n$$\n\nFrom (1) and (2) it follows $ax^2 - 4ax + 4a = 0$, whence $x^2 - 4x + 4 = 0 \\Leftrightarrow (x-2)^2 = 0 \\Leftrightarrow x = 2$. Thus $x = 2$ is an abscissa of the common point of all three graphs.\n\nSince the graphs of $g(x)$ and $h(x)$ has also exactly one common point, the equation $\\frac{cx+b}{cx+a} = cx+b$, i.e. the equation $cx+b = (cx+b)(cx+a)$ has exactly one root. Therefore, $x = -\\frac{b}{c} = -\\frac{1-a}{c}$. But $x = 2$, so\n$$\n-\\frac{b}{c} = 2, \\quad \\text{and} \\quad 2 = \\frac{1-a}{c}. \\qquad (3)-(4)\n$$\nFrom (2), (3) and (4) we obtain $a = \\frac{3}{11}$, $b = -\\frac{8}{11}$ and $c = \\frac{4}{11}$. It is easy to see that for these values of $a$, $b$ and $c$ the graphs of all three functions have exactly one common point.",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 25337,
    "subject": "Mathematics (Multi-modal)",
    "question": "Non-zero real numbers $a$, $b$, $c$, $d$ satisfy the equalities\n$$\na + b + c + d = 0, \\quad \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{1}{d} + \\frac{1}{abcd} = 0.\n$$\nFind all possible values of the product $(ab - cd)(c + d)$.",
    "options": [],
    "answer": "-1",
    "solution": "Answer: $-1$.\nBy condition,\n$$\n\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{1}{d} = -\\frac{1}{abcd} \\Rightarrow bcd + cda + dab + abc = -1.\n$$\nSo\n$$\n\\begin{aligned}\n-1 = bcd + cda + dab + abc &= (bcd + cda) + (dab + abc) = cd(b + a) + ab(c + d) = \\\\\n&= [a + b + c + d = 0 \\Rightarrow a + b = -(c + d)] = ab(c + d) - cd(c + d) = (ab - cd)(c + d).\n\\end{aligned}\n$$\nTherefore, $(ab - cd)(c + d) = -1$ for all admissible values of $a$, $b$, $c$, $d$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25338,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a trapezoid $ABCD$ ($BC \\parallel AD$) with $CD = AO$ and $BC = OD$, where $O$ is a point of intersection of the diagonals of the trapezoid. $CA$ is a bisectrix of $\\angle BCD$.\nFind the angles of $ABCD$.",
    "options": [],
    "answer": "∠A = 54°, ∠B = 126°, ∠C = 72°, ∠D = 108°",
    "solution": "Answer: $\\angle A = 54^\\circ$, $\\angle B = 126^\\circ$, $\\angle C = 72^\\circ$, $\\angle D = 108^\\circ$.\nLet $CD = AO = a$, $BC = OD = b$ and $\\angle BCD = 2\\alpha$. By condition,\n\n![](attached_image_1.png)\n\n$BD = y$, $\\angle DCA = \\angle ACB = \\alpha$. Since $BC \\parallel AD$, we have $\\angle CAD = \\angle ACB = \\alpha$. So $\\triangle ADC$ is an isosceles triangle and $AD = CD = a$. Then $\\triangle AOD$ is also an isosceles triangle ($AO = a$ and $AD = a$). Therefore, $\\angle AOD = \\angle ADO$.\n\nFurther, since $\\angle BOC = \\angle AOD$ (vertical angles) and $\\angle OBC = \\angle ODA$ (alternate angles for $BC \\parallel AD$), we see that $\\triangle BOC$ is an isosceles triangle, so $OC = BC = b$. Thus, $\\triangle COD$ is also an isosceles triangle ($OC = b$ and $OD = b$). Therefore, $\\angle ODC = \\angle OCD = \\alpha$.\n\nThen in $\\triangle COD$ the external angle $\\angle BOC = \\angle OCD + \\angle ODC = \\alpha + \\alpha = 2\\alpha$, hence, $\\angle OBC = \\angle BOC = 2\\alpha$. It means that $\\triangle BCD$ is an isosceles triangle ($\\angle DBC = 2\\alpha$ and $\\angle BC = 2\\alpha$). Then $BD = CD = a = AD$ and so $\\triangle ADB$ is also an isosceles triangle.\n\nIn $\\triangle BCD$ the sum $\\angle BCD + \\angle CDB + \\angle DBC = 2\\alpha + \\alpha + 2\\alpha = 5\\alpha$. Since the sum of all angles of a triangle is equal to $180^\\circ$, we have $5\\alpha = 180^\\circ$, hence $\\alpha = 36^\\circ$.\n\nThen for the trapezoid $ABCD$ we have $\\angle C = 2\\alpha = 72^\\circ$, $\\angle D = 180^\\circ - \\angle C = 180^\\circ - 72^\\circ = 108^\\circ$. Since $\\triangle ADB$ is isosceles, we have $\\angle A = \\frac{1}{2}(180^\\circ - \\angle ADB) = \\frac{1}{2}(180^\\circ - 2\\alpha) = 90^\\circ - \\alpha = 90^\\circ - 36^\\circ = 54^\\circ$. Hence, $\\angle B = 180^\\circ - \\angle A = 180^\\circ - 54^\\circ = 126^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25339,
    "subject": "Mathematics (Multi-modal)",
    "question": "A quadratic trinomial $x^2 + p x + q$ with integer coefficients $p$ and $q$ is said to be *irrational* if it has irrational roots $\\alpha_1$ and $\\alpha_2$.\nFind the smallest value of the sum $|\\alpha_1| + |\\alpha_2|$ among all irrational trinomials.",
    "options": [],
    "answer": "sqrt(5)",
    "solution": "Answer: $\\sqrt{5}$.\nBy condition, the trinomial $x^2 + p x + q$ has the roots, so its discriminant $D = p^2 - 4q \\ge 0$. Since $p^2 = 4q + D$ and $p$ and $q$ are integer, we have $D \\ne 2$ and $D \\ne 3$ because the square of the integer number is congruent neither to 2 nor to 3 modulo 4.\nMoreover, the trinomial $x^2 + p x + q$ has the roots\n$$\n\\alpha_1 = 2^{-1}(-p - \\sqrt{D}) \\quad \\text{и} \\quad \\alpha_2 = 2^{-1}(-p + \\sqrt{D}). \\quad (*)\n$$\nIf the trinomial is irrational, but $\\alpha_1$ and $\\alpha_2$ are irrational numbers if and only if $D$ is different from the square of some integer number. Thus, in particular, $D \\ne 0, D \\ne 1$ and $D \\ne 4$. Therefore, $D \\ge 5$.\nSince the roots $\\alpha_1$ and $\\alpha_2$ of the irrational trinomial are different from 0, there are only two possibilities:\n\n1) $\\alpha_1 \\cdot \\alpha_2 > 0$\n\n2) $\\alpha_1 \\cdot \\alpha_2 < 0$\n\nFor case 1) we have $q = \\alpha_1 \\cdot \\alpha_2 > 0$ (the Vieta theorem). So $q \\ge 1$. From (*) it follows that $|\\alpha_1| + |\\alpha_2| = |p|$. Since $D \\ge 5$, we have $p^2 = 4q + D \\ge 4 \\cdot 1 + 5 = 9$, i.e. $|p| \\ge 3$. Thus, $|\\alpha_1| + |\\alpha_2| \\ge 3$ for case 1).\n\nFor case 2) from (*) it follows that $|\\alpha_1| + |\\alpha_2| = \\sqrt{D} \\ge \\sqrt{5}$.\nIt suffices to show that the estimate $\\sqrt{5}$ is admissible.\n\nWe find all irrational trinomials with the discriminants $D = 5$. By Vieta's theorem, $\\alpha_1 \\cdot \\alpha_2 < 0$ if and only if $q = \\alpha_1 \\cdot \\alpha_2 < 0$, i.e. case 2) holds if and only if $q \\le -1$. If $q = -1$, then $D = 5$ only if $p^2 = 1$, but if $q \\le -2$, then $D = p^2 - 4q \\ge -4q \\ge -4 \\cdot (-2) = 8$. Therefore, there exist exactly two irrational trinomials with $D = 5$ and $\\alpha_1 \\cdot \\alpha_2 < 0$: $x^2 - x - 1$ and $x^2 + x - 1$.\n\nTherefore, the smallest value of the sum of the modules of the roots of the irrational trinomial is equal to $\\sqrt{5}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25340,
    "subject": "Mathematics (Multi-modal)",
    "question": "The lengths of some three sides of a quadrilateral are equal to $2$, $7$, and $11$.\nFind the area of the quadrilateral if it has the greatest area among all quadrilaterals with the mentioned lengths of their sides.",
    "options": [],
    "answer": "30√3",
    "solution": "Answer: $30\\sqrt{3}$.\n\nIt is easy to show that if the area of the quadrilateral with three given sides is a maximum, then the quadrilateral is convex. Let $AB$, $BC$, $CD$ be given and the quadrilateral $ABCD$ have the maximal area. We have $S(ABCD) = S(ABD) + S(BCD)$ and $S(ABD) = 0.5 \\cdot AB \\cdot BD \\sin \\angle ABD$. If $\\angle ABD \\ne 90^\\circ$, then there exists a quadrilateral with given sides having the greatest possible area. Hence $AB \\perp BD$. In the same manner, we obtain $CD \\perp AC$. So the right angle $ABD$ and $ACD$ subtend the segment $AD$, so the quadrilateral $ABCD$ is inscribed in the circle with the diameter $AC$.\n\nLet $\\angle CAD = \\beta$, $\\angle BDA = \\alpha$, $AB = a$, $BC = b$, $CD = c$, $AC = x$, $BD = y$, $AD = z$. Then\n![](attached_image_1.png)\n$$\na = z \\sin \\alpha, \\quad c = z \\sin \\beta, \\quad x = z \\cos \\beta, \\quad y = z \\cos \\alpha,\n$$\n$$\n\\begin{aligned}\nb &= z \\sin \\angle CDA = z \\sin(90^\\circ - \\beta - \\alpha) = z \\cos(\\beta + \\alpha) = \\\\\n&= z(\\cos \\beta \\cos \\alpha - \\sin \\beta \\sin \\alpha).\n\\end{aligned}\n$$\n\nIt is easy to see that $bz + ac = xy$. (Note that this equality follows from the Ptolemaeus theorem.) On the other hand, using the Pythagoras theorem for triangles $ABD$ and $ACD$, we obtain $ac + zb = \\sqrt{(z^2 - c^2)(z^2 - a^2)}$. So $z^4 - (a^2 + b^2 + c^2)z^2 - 2abc z = 0$, and since $z \\neq 0$, we have\n$$\nz^{3} - (a^{2} + b^{2} + c^{2})z - 2abc = 0.\n$$\nNote that the value of $z$ is independent of the lengths of the sides $AB$, $BC$ and $CD$, so without loss of generality, we set $a = 2$, $b = 7$, $c = 11$. We have\n$$\nz^{3} - 174z - 308 = 0. \\qquad (1)\n$$\nWe find one of the roots of this equation: $z = 14$. Two other roots of (1) are negative numbers.\n\nUsing the sines law for the triangle $BCD$, we obtain $7 = b = z \\sin \\angle BDC = 14 \\sin \\angle BDC$, so $\\sin \\angle BDC = 1/2$, i.e. $\\angle BDC = 30^\\circ$. Therefore the angle between the diagonals $AC$ and $BD$ is equal to $\\angle COD = 90^\\circ - \\angle BDC = 90^\\circ - 30^\\circ = 60^\\circ$. Now we find\n$$\nAC = x = \\sqrt{z^2 - c^2} = \\sqrt{196 - 121} = \\sqrt{75} = 5\\sqrt{3},\n$$\n$$\nBD = y = \\sqrt{z^2 - a^2} = \\sqrt{196 - 4} = \\sqrt{192} = 8\\sqrt{3}.\n$$\nTherefore the required area is equal to\n$$\nS(ABCD) = 0.5 \\cdot AC \\cdot BD \\cdot \\sin \\angle COD = 0.5 \\cdot 5\\sqrt{3} \\cdot 8\\sqrt{3} \\cdot \\sqrt{3}/2 = 30\\sqrt{3}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25341,
    "subject": "Mathematics (Multi-modal)",
    "question": "Pirate Bob has 14 silver, 15 gold, and 16 platinum coins, and Pirate Bill has 16 silver, 15 gold, and 14 platinum coins. From time to time they exchange their coins using the following rule: one of the pirates gives to the other pirate two coins of the same metal and instead of them gets two coins from the other two metals. At some moment Bill has no gold coins.\nHow many platinum coins can Bill have at this moment?",
    "options": [],
    "answer": "17, 20, 23, 26, 29",
    "solution": "Let $(S, G, P)$ be the set of gold, silver and platinum coins of Bill at some moment. The initial set is $(16, 15, 14)$. Note that Bob and Bill have together $30$ gold, $30$ silver and $30$ platinum coins. So, $S \\le 30$, $G \\le 30$, and $P \\le 30$ at any moment. By condition, $S + G + P = 16 + 15 + 14 = 45$ at any time. Therefore, if $G = 0$ at some moment, then $P = 45 - S \\ge 45 - 30 = 15$ at the same moment, so $15 \\le P \\le 30$.\n\nBy condition, after any interchange of coins the set $(S, G, P)$ can be one of the following sets\n1) $(S+2, G-1, P-1)$,\n2) $(S-2, G+1, P+1)$,\n3) $(S-1, G+2, P-1)$,\n4) $(S+1, G-2, P+1)$,\n5) $(S-1, G-1, P+2)$,\n6) $(S+1, G+1, P-2)$.\nIt is easy to see that the difference $G - P$ in the old and in any new set are congruent modulo $3$ in any case. We have $G - P = 15 - 14 = 1$ for the initial set, so $G - P \\equiv 1 \\pmod 3$ at any time. For $G = 0$ we have $P \\equiv 2 \\pmod 3$. We see that the numbers $17, 20, 23, 26,$ and $29$ are congruent $2$ modulo $3$ (among the numbers from $15$ to $30$). Therefore, when Bill has $0$ gold coins he can have only one of these five values of platinum coins. On the other hand, Bill can have any of these five numbers of platinum coins.\n\nIndeed, if Bill gives two gold coins to Bill seven times successively, then $(16, 15, 14) \\rightarrow (23, 1, 21)$. The first table shows how Bill can get the smallest number of platinum coins ($17$), and the second table shows how Bill can get the greatest number of platinum coins ($29$). We also see that Bill can also get $20, 23, 26$ coins.\n\n| $S$ | 23 | 21 | 22 | 23 | 24 | 25 | 26 | 27 | 28 |\n|-----|----|----|----|----|----|----|----|----|----|\n| $G$ |  1 |  2 |  0 |  1 |  2 |  0 |  1 |  2 |  0 |\n| $P$ | 21 | 22 | **23** | 21 | 19 | **20** | 18 | 16 | **17** |\n\n| $S$ | 23 | 21 | 22 | 20 | 18 | 19 | 17 | 15 | 16 |\n|-----|----|----|----|----|----|----|----|----|----|\n| $G$ |  1 |  2 |  0 |  1 |  2 |  0 |  1 |  2 |  0 |\n| $P$ | 21 | 22 | **23** | 24 | 25 | **26** | 27 | 28 | **29** |",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25342,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the greatest possible value of the area of a quadrilateral $ABCD$ if the length of broken line $ABDC$ is equal to $L$.",
    "options": [],
    "answer": "L^2/8",
    "solution": "Answer: $S(ABCD) = L^2/8$.\nLet the area of $ABCD$ be a maximum for some $AB = x$, $BD = y$, $CD = z$, $x + y + z = L$. Since $S(ABCD) = S(ABD) + S(DBC) = \\frac{1}{2} AB \\cdot BD \\sin \\angle ABD + \\frac{1}{2} BD \\cdot DC \\sin \\angle BDC$, we see that the area of the quadrilateral with fixed values of $x$, $y$, $z$ is maximum if $\\angle ABD = \\angle CDB = 90^\\circ$. Therefore,\n$$\n\\begin{aligned}\nS(ABCD) &= \\frac{1}{2}xy + \\frac{1}{2}yz = \\frac{1}{2}y(x+z) = \\\\\n&= [x+y+z = L \\Rightarrow x+z = L-y] = \\frac{1}{2}y(L-y).\n\\end{aligned}\n$$\n\n![](attached_image_1.png)\n\nIt is easy to see that for $0 < y < L$ the value of the product $y(L-y)$ is a maximum if $y = L/2$ and it is equal to $L^2/4$. Therefore, the maximal value of $ABCD$ with given sum of its sides $AB$, $CD$ and diagonal $BD$, $AB + BD + DC = L$, is equal to $L^2/8$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25343,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the greatest possible value of the area of a quadrilateral $ABCD$ if the length of broken line $ABCD$ is equal to $L$.\n(I. Gorodnin)",
    "options": [],
    "answer": "L^2/8",
    "solution": "**Answer:** $S(ABCD) = \\frac{L^2}{8}$.\nLet the area of $ABCD$ be a maximum for some $AB = x$, $BD = y$, $CD = z$, $x + y + z = L$. Since $S(ABCD) = S(ABD) + S(DBC) = \\frac{1}{2} AB \\cdot BD \\sin \\angle ABD + \\frac{1}{2} BD \\cdot BC \\sin \\angle BDC$, we see that the area of the quadrilateral with fixed values of $x$, $y$, $z$ is maximum if $\\angle ABD = \\angle CDB = 90^\\circ$. Therefore,\n$$S(ABCD) = \\frac{1}{2} x y + \\frac{1}{2} y z = \\frac{1}{2} y(x+z) = \\frac{1}{2} y(L-y).$$\nIt is easy to see that for $0 < y < L$ the value of the product $y(L-y)$ is a maximum if $y = \\frac{L}{2}$ and it is equal to $\\frac{L^2}{4}$. Therefore, the maximal value of $ABCD$ with given sum of its sides $AB$, $CD$ and diagonal $BD$, $AB + BD + DC = L$, is equal to $\\frac{L^2}{8}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25344,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a quadrilateral $ABCD$ with $AB + CD = 6$, $BC + DA = 8$.\nFind the area of $ABCD$ if it has the greatest area among all quadrilaterals with the mentioned sums of the opposite sides.",
    "options": [],
    "answer": "12",
    "solution": "Answer: $12$.\nWe use the following obvious inequalities. If the quadrilateral $ABCD$ has the sides $AB = a$, $BC = b$, $CD = c$, $DA = d$, and the area $S$, then\n$$S \\le \\frac{ab + cd}{2} \\text{ because}$$\n$$\nS = S(ABC) + S(ACD) = \\\\\n= \\frac{1}{2}ab \\sin \\beta + \\frac{1}{2}cd \\sin \\delta \\le \\frac{ab + cd}{2}.\n$$\nSimilarly, $S \\le \\frac{ad + bc}{2}$.\nSumming these inequalities, we easily obtain $S \\le \\frac{(a+c)(b+d)}{4}$. Hence\n$$S \\le \\frac{6 \\cdot 8}{4} = 12.$$ It is easy to see that equality occurs for a rectangle.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25345,
    "subject": "Mathematics (Multi-modal)",
    "question": "$N$ boys ($N \\ge 3$), no two of them having the same height, are arranged along a circle. A boy in the given arrangement is said to be *tall* if he is taller than both of his neighbors; a boy is said to be *short* if he is shorter than both of his neighbors.\nProve that the number of tall boys is equal to the number of short boys in any arrangement of the boys along the circle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nConsider arbitrary arrangement of the boys along the circle. We put the signs \"+\" or \"-\" before any boy in accordance with the following rule: we move clockwise along the circle and put the sign \"+\" before the boy if he is taller than the previous boy and we put the sign \"-\" if he is shorter than the previous one (see the fig.). It is evident that the boy is tall if and only if the sign \"+\" stands before him and the sign \"-\" stands after him; the boy is short if and only if the sign \"-\" stands before him and the sign \"+\" stands after him. (Note that since the tallest boy among all $N$ boys is tall, there exist the signs \"+\" as well as the signs \"-\" in any arrangement.) Therefore, the number of tall boys in the arrangement is equal to the number of alternations of \"+\" and \"-\"; the number of short boys in the arrangement is equal to the number of alternations of \"-\" and \"+\". Regarding all successive signs \"+\" as one sign \"+\" and all successive signs \"-\" as one sign \"-\", we construct a new arrangement of signs. It is easy to see that the number of alternations of \"+\" and \"-\" (of \"-\" and \"+\") in the initial arrangement is equal to the number of alternations of \"+\" and \"-\" (of \"-\" and \"+\") in the new arrangement. But it is evident that the number of alternations of \"+\" and \"-\" is equal to the number of alternations of \"-\" and \"+\" in the new arrangement. Therefore, the numbers of tall and short boys in any arrangement are equal.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25346,
    "subject": "Mathematics (Multi-modal)",
    "question": "$N$ boys ($N \\ge 3$), no two of them having the same height, are arranged along a circle. A boy in the given arrangement is said to be *middle* if he is taller than one of his neighbors and shorter than the other one.\nProve that if $N$ is odd number then there exists at least one middle boy in any arrangement of the boys along the circle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider arbitrary arrangement of the boys along the circle. We put\n![](attached_image_1.png)\nthe signs \"+\" or \"-\" before any boy in accordance with the following rule: we move clockwise along the circle and put the sign \"+\" before the boy if he is taller than the previous boy and we put the sign \"-\" if he is shorter than the previous one (see the fig.). It is evident that the boy is middle if and only if both the signs before and after him are either \"+\" or \"-\". If there is no middle boy in the arrangement, then the signs \"+\" and \"-\" are alternate, so the total number of the signs (equaled the number of the boys) is even, contrary to the condition. Therefore, there is at least one middle boy in any arrangement.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25347,
    "subject": "Mathematics (Multi-modal)",
    "question": "$N$ boys ($N \\ge 3$), no two of them having the same height, are arranged along a circle. A boy in the given arrangement is said to be *tall* if he is taller than both of his neighbors.\nFind all possible numbers of tall boys in the arrangement.",
    "options": [],
    "answer": "All integers from 1 to floor(N/2)",
    "solution": "Answer: any integer number from $1$ to $[N/2]$.\n\nConsider arbitrary arrangement of the boys along the circle. We put the signs \"+\" or \"-\" before any boy in accordance with the following rule: we move clockwise along the circle and put the sign \"+\" before the boy if he is taller than the previous boy and we put the sign \"-\" if he is shorter than the previous one. It is evident that the boy is tall if and only if the sign \"+\" stands before him and the sign \"-\" stands after him. (Note that since the tallest boy among all $N$ boys is tall, there exist the signs \"+\" as well as the signs \"-\" in any arrangement.) Therefore, the number of tall boys in the arrangement is equal to the number of alternations of \"+\" and \"-\". It is evident that the number of these alternations is less than or equal to $[N/2]$.\n\nNow we show that for any $b$, $1 \\le b \\le [N/2]$, there exists an arrangement with exactly $b$ tall boys.\n\nWe number all boys in accordance with their heights: the shortest boy has the number $1$, and the tallest boy has the number $N$. We partition all boys into three groups $A$, $B$, and $C$: group $A$ contains the shortest boys, i.e. the boys with the numbers $1, 2, ..., b$; group $B$ contains the tallest boys, i.e. the boys with the numbers $N-b+1, N-b+2, ..., N$ (since $b \\le [N/2]$, i.e. $2b \\le N$, there exists such partition); finally, group $C$ consists of all remaining boys (if $N$ is even and $b = N/2$, then $C$ is empty).\n\nWe also number the places on the circle with numbers from $1$ to $N$. We place the boys from $A$ on the places with the numbers $2k-1$, $k=1, ..., b$; the boys from $B$ are placed on the places with the numbers $2k$, $k=1, ..., b$; the boys from $C$ are placed on the remaining places (with the numbers $2b+1, ..., N$) so that the boy with the number $N-b$ is placed on the place with the number $2b+1, ...$, the boy with the number $b+1$ is placed on the place with the number $N$ (see the fig.). It is easy to see that there exist exactly $b$ tall boys in this arrangement.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25348,
    "subject": "Mathematics (Multi-modal)",
    "question": "$N$ boys ($N \\ge 3$), no two of them having the same height, are arranged along a circle. A boy in the given arrangement is said to be *middle* if he is taller than one of his neighbors and shorter than the other one.\nFind all possible numbers of middle boys in the arrangement.",
    "options": [],
    "answer": "Any integer from 0 to N−2 with the same parity as N.",
    "solution": "Answer: any integer number from $0$ to $N-2$ of the same parity as $N$.\n\nConsider arbitrary arrangement of the boys along the circle. We say that a boy in the given arrangement is *tall* if he is taller than both of his neighbors, and a boy is *short* if he is shorter than both of his neighbors.\n\nThe numbers of tall and short boys are equal in any arrangement (see solution of Problem C.8). Let $b$ be the number of the tall boys in the arrangement. Then the number $s$ of the middle boys is equal to $N - 2b$ and has the same parity with $N$. Since the number $b$ of the tall boys in the arrangement can admit any value from $1 \\le b \\le [N/2]$ (see solution of Problem C.8), the number $s$ of the middle boys can admit any value from $[0; N - 2]$ and must be the same parity as $N$. The corresponding examples for any $s$ see in solution of Problem B.8.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25349,
    "subject": "Mathematics (Multi-modal)",
    "question": "$N$ boys ($N \\ge 3$), no two of them having the same height, are arranged along a circle. A boy in the given arrangement is said to be *tall* if he is taller than both of his neighbors.\nFind all possible numbers of tall boys in the arrangement.",
    "options": [],
    "answer": "all integers from 1 to floor(N/2)",
    "solution": "Answer: any integer number from $1$ to $[N/2]$.\n\nConsider arbitrary arrangement of the boys along the circle. We put the signs \"+\" or \"-\" before any boy in accordance with the following rule: we move clockwise along the circle and put the sign \"+\" before the boy if he is taller than the previous boy and we put the sign \"-\" if he is shorter than the previous one. It is evident that the boy is tall if and only if the sign \"+\" stands before him and the sign \"-\" stands after him. (Note that since the tallest boy among all $N$ boys is tall, there exist the signs \"+\" as well as the signs \"-\" in any arrangement.) Therefore, the number of tall boys in the arrangement is equal to the number of alternations of \"+\" and \"-\". It is evident that the number of these alternations is less than or equal to $[N/2]$.\n\nNow we show that for any $b$, $1 \\le b \\le [N/2]$, there exists an arrangement with exactly $b$ tall boys.\n\nWe number all boys in accordance with their heights: the shortest boy has the number $1$, and the tallest boy has the number $N$. We partition all boys into three groups $A, B$, and $C$: group $A$ contains the shortest boys, i.e. the boys with the numbers $1, 2, \\dots, b$; group $B$ contains the tallest boys, i.e. the boys with the numbers $N-b+1, N-b+2, \\dots, N$ (since $b \\le [N/2]$, i.e. $2b \\le N$, there exists such partition); finally, group $C$ consists of all remaining boys (if $N$ is even and $b = N/2$, then $C$ is empty).\n\nWe also number the places on the circle with numbers from $1$ to $N$. We place the boys from $A$ on the places with the numbers $2k-1$, $k=1, \\dots, b$; the boys from $B$ are placed on the places with the numbers $2k$, $k=1, \\dots, b$; the boys from $C$ are placed on the remaining places (with the numbers $2b+1, \\dots, N$) so that the boy with the number $N-b$ is placed on the place with the number $2b+1, \\dots$, the boy with the number $b+1$ is placed on the place with the number $N$ (see the fig.). It is easy to see that there exist exactly $b$ tall boys in this arrangement.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25350,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a right-angled triangle $ABC$ with $AC = BC$, and $\\angle C = 90^\\circ$. Let points $M$ and $N$ belong to the sides $AC$ and $BC$ respectively; $MN = BC$. For each pair of such points $M$ and $N$ a circle passing through $M$, $N$ and touching the hypotenuse $AB$ is constructed.\nFind the locus of the centers of these circles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Answer: The segment $XY$ such that $AXYB$ is a rectangle and the midpoint of $XY$ coincides with $C$.\n\n(Solution of E. Dauhiala, B. Gilevich, K. Kostevich.) Let $O$ be the point in the same half-plane as $C$ with respect to the line $MN$ such that $\\angle MON = 90^\\circ$ and $OM = ON$. Then $OM = ON = \\dfrac{MN}{\\sqrt{2}} = CH$ (the altitude of the triangle $ABC$). Since $\\angle MON = \\angle MCN = 90^\\circ$, the quadrilateral $MOCN$ is cyclic and hence $\\angle OCM = \\angle ONM = 45^\\circ$. It follows that $OC \\parallel AB$, so the distance between $O$ and $AB$ equals $OK = CH = OM = ON$. Therefore $O$ is the center of the circle passing through $M$, $N$ and touching $AB$. Hence the needed locus is contained in the segment $XY$ described in the answer.\n\n![](attached_image_1.png)\n\nConversely, one can verify that any point of this segment is a center of some circle which touches $AB$ and intersects $CA$ and $CB$ at points $M$ and $N$ such that $MN = BC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25351,
    "subject": "Mathematics (Multi-modal)",
    "question": "Points $X$ and $Y$ are marked on the sides $AB$ and $AD$ of the convex quadrilateral $ABCD$ respectively.\nFind $AX : BX$ if $CX \\parallel DA$, $DX \\parallel CB$, $BY \\parallel CD$, $CY \\parallel BA$.",
    "options": [],
    "answer": "(1+sqrt(5))/2",
    "solution": "Answer: $(\\sqrt{5} + 1)/2$.\nLet $\\lambda = AX : BX$ be the required ratio. On one hand, by Thales' theorem\n$$\n\\begin{aligned}\n\\lambda &= \\frac{AX}{BX} = \\frac{YR}{RB} = \\\\\n&= [YR = CD, \\text{ since } CDYR \\text{ is a parallelogram}] = \\\\\n&= \\frac{CD}{RB}.\n\\end{aligned} \n\\quad (1)\n$$\n![](attached_image_1.png)\nOn the other hand,\n$$\n\\begin{aligned}\n\\lambda &= \\frac{AX}{BX} = [AX = CY, \\text{ since } CYAX \\text{ is a parallelogram}] = \\frac{CY}{BX} = \\\\\n&= [BX = CP, \\text{ since } CPXB \\text{ is a parallelogram}] = \\frac{CY}{CP} = \\\\\n&= [\\text{Thales' theorem for } \\angle CYB, PQ \\parallel BC] = \\frac{BY}{BQ} = \\\\\n&= [BQ = CD, \\text{ since } BCDQ \\text{ is a parallelogram}] = \\frac{BY}{CD} = \\frac{YR + BR}{CD} = \\\\\n&= [YR = CD, \\text{ since } RCDY \\text{ is a parallelogram}] = \\frac{CD + BR}{CD} = \\\\\n&= \\frac{1 + BR/CD}{1} = [\\text{see (1)}] = 1 + \\frac{1}{\\lambda}.\n\\end{aligned}\n$$\nSo, $\\lambda = 1 + \\frac{1}{\\lambda}$ or $\\lambda^2 - \\lambda - 1 = 0$. From this quadratic equation, taking into account that $\\lambda > 0$, we obtain $\\lambda = (\\sqrt{5} + 1)/2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25352,
    "subject": "Mathematics (Multi-modal)",
    "question": "Three points $A$, $B$, $C$, are marked on the hyperbola $y = 1/x$ so that the triangle $ABC$ is equilateral.\nFind all possible values of the product of the sum of abscissae and the sum of ordinates of the vertices of $ABC$.",
    "options": [],
    "answer": "9",
    "solution": "Answer 9.\n\nLet $A(a; 1/a)$, $B(b; 1/b)$, $C(c; 1/c)$ be the marked points. Then the required product is equal to\n$$\nT = (a+b+c) \\left( \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\right).\n$$\n\n![](attached_image_1.png)\n\nSince any vertical and any horizontal line meets the hyperbola $y = 1/x$ at most at one point we see that the numbers $a, b, c$ are pairwise distinct. By condition, the triangle $ABC$ is regular so\n$$\n\\begin{aligned}\nAB = BC = CA &\\Leftrightarrow (b-a)^2 + (1/b - 1/a)^2 = \\\n&= (c-b)^2 + (1/c - 1/b)^2 = (a-c)^2 + (1/a - 1/c)^2.\n\\end{aligned} \n\\quad (1)\n$$\n\nFrom (1) we obtain\n$$\n\\begin{aligned}\nb^2 - 2ab + a^2 + 1/b^2 - 2/(ab) + 1/a^2 &= c^2 - 2bc + b^2 + 1/c^2 - 2/(bc) + 1/b^2 \\\n&\\implies (a^2 - c^2) - 2b(a-c) + \\frac{c^2 - a^2}{a^2c^2} - \\frac{2(c-a)}{abc} = 0.\n\\end{aligned}\n$$\n\nSince $a \\neq c$ we have\n$$\na+c-2b-\\frac{a+c}{a^2c^2}+\\frac{2}{abc}=0.\n$$\nIn a similar way from (1) we obtain two more equalities\n$$\nb+a-2c-\\frac{b+a}{b^2a^2}+\\frac{2}{abc}=0 \\quad \\text{and} \\quad c+b-2a-\\frac{c+b}{c^2b^2}+\\frac{2}{abc}=0.\n$$\nSumming all three equalities, we obtain $\\frac{6}{abc} - \\frac{a+c}{a^2c^2} - \\frac{b+a}{b^2a^2} - \\frac{c+b}{c^2b^2} = 0$, so $\\frac{b}{c} + \\frac{b}{a} + \\frac{c}{a} + \\frac{c}{b} + \\frac{a}{b} + \\frac{a}{c} = 6$. Thus\n$$\nT = (a+b+c) \\left( \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\right) = 1+\\frac{a}{b}+\\frac{a}{c}+\\frac{b}{a}+1+\\frac{b}{c}+\\frac{c}{a}+\\frac{c}{b}+1 = 3+6=9.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25353,
    "subject": "Mathematics (Multi-modal)",
    "question": "Some businessmen decide to establish a firm and divide all profits in equal parts. However, some day after a good pennyworth, a head of the firm transfers a part of the funds from firm's account on his individual account. This part is three times as many as the part of each of the others if they divided the rest of the funds in equal parts. After that he leaves the firm. Next head of the firm gets the right to command of all remaining funds, and he deals with the funds just as the previous one, and so on. Finally, next to the last head of the firm transfers a part of the remaining funds from firm's account on his individual account and this part is also three times as many as he leaves to the last of the co-funder of the firm. As a result of this revenue sharing the profit of the last businessman is 210 times as little as the profit of the first head of the firm.\nHow many businessmen have established this firm?",
    "options": [],
    "answer": "20",
    "solution": "Answer: 20.\nLet $n$ be the number of co-funders and $d_i$ be the value of the $i$-th director, $i = 1, \\ldots, n$. By condition,\n$$\nd_i = 3 \\cdot \\frac{d_{i+1} + d_{i+2} + \\dots + d_n}{n-i}.\n$$\nSo,\n$$\n\\begin{align*}\nd_{i-1} &= 3 \\cdot \\frac{d_i + d_{i+1} + \\dots + d_n}{n-i+1} = \\\\\n&= 3 \\cdot \\frac{3 \\cdot \\frac{d_{i+1} + d_{i+2} + \\dots + d_n}{n-i} + d_{i+1} + \\dots + d_n}{n-i+1} = \\\\\n&= 3 \\cdot \\frac{(n-i+3)(d_{i+1} + d_{i+2} + \\dots + d_n)}{(n-i)(n-i+1)}.\n\\end{align*}\n$$\nTherefore, $\\frac{d_{i-1}}{d_i} = \\frac{n-i+3}{n-i+1}$, $i = 2, \\ldots, n$.\nMultiplying these equalities, we obtain\n$$\n\\frac{d_1}{d_n} = \\frac{d_1}{d_2} \\cdot \\frac{d_2}{d_3} \\cdot \\dots \\cdot \\frac{d_{n-1}}{d_n} = \\frac{n+1}{n-1} \\cdot \\frac{n}{n-2} \\cdot \\dots \\cdot \\frac{4}{2} \\cdot \\frac{3}{1} = \\frac{(n+1)n}{2}.\n$$\nBy condition, $d_1/d_n = 210$, so $(n+1)n = 420$ which gives $n = 20$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 25354,
    "subject": "Mathematics (Multi-modal)",
    "question": "Three points $A$, $B$, $C$, are marked on the hyperbola $y = 1/x$ so that the triangle $ABC$ is equilateral.\nFind all possible values of the product of the sum of abscissae and the sum of ordinates of the vertices of $ABC$.",
    "options": [],
    "answer": "9",
    "solution": "Answer 9.\nLet $A(a; 1/a)$, $B(b; 1/b)$, $C(c; 1/c)$ be the marked points. Then the required product is equal to\n$$\nT = (a + b + c) \\left( \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\right).\n$$\n![](attached_image_1.png)\nSince any vertical and any horizontal line meets the hyperbola $y = 1/x$ at most at one point we see that the numbers $a, b, c$ are pairwise distinct. By condition, the triangle $ABC$ is regular so\n$$\n\\begin{aligned}\nAB = BC = CA &\\iff (b-a)^2 + (1/b - 1/a)^2 = \\\\\n&= (c-b)^2 + (1/c - 1/b)^2 = (a-c)^2 + (1/a - 1/c)^2.\n\\end{aligned} \n\\quad (1)\n$$\nFrom (1) we obtain\n$$\n\\begin{aligned}\nb^2 - 2ab + a^2 + 1/b^2 - 2/(ab) + 1/a^2 &= c^2 - 2bc + b^2 + 1/c^2 - 2/(bc) + 1/b^2 \\\\\n&\\implies (a^2 - c^2) - 2b(a-c) + \\frac{c^2 - a^2}{a^2c^2} - \\frac{2(c-a)}{abc} = 0.\n\\end{aligned}\n$$\nSince $a \\neq c$ we have\n$$\na+c-2b-\\frac{a+c}{a^2c^2}+\\frac{2}{abc}=0.\n$$\nIn a similar way from (1) we obtain two more equalities\n$$\nb+a-2c-\\frac{b+a}{b^2a^2}+\\frac{2}{abc}=0 \\quad \\text{and} \\quad c+b-2a-\\frac{c+b}{c^2b^2}+\\frac{2}{abc}=0.\n$$\nSumming all three equalities, we obtain $\\frac{6}{abc} - \\frac{a+c}{a^2c^2} - \\frac{b+a}{b^2a^2} - \\frac{c+b}{c^2b^2} = 0$, so $\\frac{b}{c} + \\frac{b}{a} + \\frac{c}{a} + \\frac{c}{b} + \\frac{a}{b} + \\frac{a}{c} = 6$. Thus\n$$\nT = (a+b+c) \\left( \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\right) = 1 + \\frac{a}{b} + \\frac{a}{c} + \\frac{b}{a} + 1 + \\frac{b}{c} + \\frac{c}{a} + \\frac{c}{b} + 1 = 3 + 6 = 9.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25355,
    "subject": "Mathematics (Multi-modal)",
    "question": "Four points $A$, $B$, $C$, $D$ are marked on the hyperbola $y = 1/x$ so that the quadrilateral $ABCD$ is a parallelogram ($AB \\parallel CD$) and $AB = 2 \\cdot BC$.\nFind all possible values of the area of $ABCD$.",
    "options": [],
    "answer": "16/3",
    "solution": "Answer: $\\dfrac{16}{3}$.\nLet $A(a; 1/a)$, $B(b; 1/b)$, $C(c; 1/c)$, $D(d; 1/d)$ be the marked points (see Fig. 1). Since any vertical and any horizontal line meets the hyperbola $y = 1/x$ at most at one point we see that the numbers $a$, $b$, $c$, $d$ are pairwise distinct. Since the opposite sides of the parallelogram are equal, we have\n$$\n\\begin{aligned}\nAB = CD &\\iff (b-a)^2 + (1/b - 1/a)^2 = (d-c)^2 + (1/d - 1/c)^2 \\\n&\\iff (b-a)^2 \\left(1 + \\frac{1}{(ab)^2}\\right) = (d-c)^2 \\left(1 + \\frac{1}{(cd)^2}\\right). \\quad (1)\n\\end{aligned}\n$$\nMoreover, since opposite sides of the parallelogram are equal and parallel we see that their projections on any line are equal, in particular, their projections on $Ox$-axis are equal, i.e. $|b-a| = |d-c|$. Then from (1) it follows that\n$$\n1 + \\frac{1}{a^2 b^2} = 1 + \\frac{1}{c^2 d^2} \\implies a^2 b^2 = c^2 d^2.\n$$\nSimilarly, from $BC = AD$ we obtain $b^2c^2 = a^2d^2$. Therefore, $a^2 = c^2$ and $b^2 = d^2$, and so, taking into consideration $a \\neq c$ and $b \\neq d$, we obtain $a = -c$, $d = -b$. Without loss of generality we suppose that $a < 0 < b$, and then $b < c$. Since $AB = 2BC$ we have\n$$\n(b+c)^2 \\left(1+\\frac{1}{(bc)^2}\\right) = (b-a)^2 \\left(1+\\frac{1}{(ab)^2}\\right) = AB^2 = (2BC)^2 =\n$$\n$$\n= 4(c-b)^2 \\left(1 + \\frac{1}{(bc)^2}\\right),\n$$\nso $(b+c)^2 = 4(b-c)^2$. Then $b+c = 2(c-b)$, so we have $c = 3b$.\n![](attached_image_1.png)\nFig. 1\n![](attached_image_2.png)\nFig. 2\nConsider the pentagon $B_1BCC_2O$ (see Fig. 2). Since $B_1(0, 1/b)$, $B_2(b, 0)$, $C_1(0, 1/c)$, $C_2(c, 0)$ we have\n$$\n\\begin{align*}\nS(B_1BCC_2O) &= S(B_1BB_2O) + S(BCC_2B_2) = b \\cdot \\frac{1}{b} + \\frac{1/b + 1/c}{2} \\cdot (c-b) = \\\\\n&= [c=3b] = 1 + \\frac{4}{3b} \\cdot \\frac{1}{2} \\cdot 2b = \\frac{7}{3}.\n\\end{align*}\n$$\nOn the other hand,\n$$\n\\begin{align*}\nS(B_1BCC_2O) &= S(OB_1B) + S(OBC) + S(OCC_2) = \\\\\n&= \\frac{1}{2} \\cdot \\frac{1}{b} \\cdot b + S(OBC) + \\frac{1}{2} \\cdot \\frac{1}{c} \\cdot c = 1 + S(OBC),\n\\end{align*}\n$$\nso $S(OBC) = \\frac{4}{3}$. Since $S(ABCD) = 4S(OBC)$ the required area of the parallelogram $ABCD$ is equal $\\dfrac{16}{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25356,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve the equation in prime numbers $p$ and $q$: $q^3 = p^2 - p + 1$.",
    "options": [],
    "answer": "p = 19, q = 7",
    "solution": "Answer: $p = 19, q = 7$.\n\n(Solution of E. Dauhiala, B. Gilevich, A. Zhuk, A. Semchankau.) We have\n$$\nq^3 = p^2 - p + 1 \\Leftrightarrow (q-1)(q^2+q+1) = p(p-1). \\quad (1)\n$$\nIf $(q-1) \\nmid p$, then $q \\ge p+1$, so $q^3 > p^2 - p + 1$.\nHence, $(q^2+q+1) \\nmid p$, i.e.\n$$\nq^2 + q + 1 = k p \\quad (2)\n$$\nfor some $k \\in \\mathbb{N}$. It follows that $k(q-1) = p-1$, or $p = kq - k + 1$. Substituting this expression in (2) we obtain $q^2 + (1-k^2)q + (k^2-k+1) = 0$. The discriminant of this quadratic on $q$ equation\n$$\nD = (k^2 - 1)^2 - (4k^2 - k + 1) = k^4 - 6k^2 + 4k - 3\n$$\nmust be a perfect square. But it is easy to check that\n$$\n(k^2 - 3)^2 < k^4 - 6k^2 + 4k - 3 < (k^2 - 1)^2 \\quad \\text{for any } k > 3.\n$$\nMoreover, the equation $k^4 - 6k^2 + 4k - 3 = (k^2 - 2)^2$ has no solutions.\nHence, $k \\le 3$. For $k = 1$ we have $D = -4$, and for $k = 2$ we have $D = -3$, which is impossible. For $k = 3$ we obtain $D = 6^2$, so $q = (8 \\pm 6)/2$. Thus $q = 7$ and $p = k(q-1) + 1 = 19$. It remains to note the primes $p = 19, q = 7$ do satisfy the equation.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25357,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two parallel chords $AB$ and $CD$ are constructed on the hyperbola $y = 1/x$. The lines $AC$ and $BD$ meet ordinate axis $Oy$ at points $A_1$ and $D_1$ respectively, and meet abscissae axis $Ox$ at points $C_1$ and $B_1$ respectively.\nProve that the areas of the triangles $A_1OC_1$ and $D_1OB_1$ are equal.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $A_1(0; a_1)$, $B_1(b_1; 0)$, $C_1(c_1; 0)$, $D_1(0; d_1)$, $A(a; 1/a)$, $B(b; 1/b)$, $C(c; 1/c)$, $D(d; 1/d)$ be the marked points. Since any vertical and any horizontal line meets the hyperbola $y = 1/x$ at most at one point we see that the numbers $a, b, c, d$ are pairwise distinct.\n\n![](attached_image_1.png)\n\nThe equation of the line $AB$ has the form $y - \\frac{1}{a} = k_1(x - a)$. Since point $B$ belongs to the line $AB$ its coordinates must satisfy the line equation, i.e. $\\frac{1}{b} - \\frac{1}{a} = k_1(b - a)$, so $k_1 = -\\frac{1}{ab}$.\n\n$k_2 = -\\frac{1}{cd}$. Since $AB \\parallel CD$ we have $k_1 = k_2$, so\n\nSimilarly, for the line $CD$ we have $y - c = k_2(x - c)$, and so\n\n$$\nab = cd \\implies \\frac{a}{c} = \\frac{b}{d}. \\qquad (1)\n$$\n\nThe equation of the line $AC$ has the form $y - \\frac{1}{a} = k(x - a)$. Setting $x = 0$ in this equation, we find the ordinate of $A_1$: $a_1 = \\frac{1}{a} - ka$, and setting\n\ny $= 0$ in the equation we find the abscissa of $C_1$: $c_1 = a - \\frac{1}{ka}$. Since point $C$ belongs to the line $AC$ its coordinates must satisfy the line equation, i.e. $\\frac{1}{c} - \\frac{1}{a} = k(c-a)$, so $k = -\\frac{1}{ac}$. Therefore $a_1 = \\frac{1}{a} + \\frac{1}{c}$ and $c_1 = a + c$.\n\nSimilarly, from the equation of the line $BD$ we find $d_1 = \\frac{1}{b} + \\frac{1}{d}$ and $b_1 = b + d$.\n\nTherefore,\n\n$$\nS(A_1OC_1) = \\frac{1}{2}|a_1c_1| = \\frac{1}{2} \\cdot \\frac{(a+c)^2}{|ac|} = \\frac{1}{2} \\cdot \\frac{(a/c+1)^2}{|a/c|},\n$$\n$$\nS(D_1OB_1) = \\frac{1}{2}|b_1d_1| = \\frac{1}{2} \\cdot \\frac{(b+d)^2}{|bd|} = \\frac{1}{2} \\cdot \\frac{(b/d+1)^2}{|b/d|}.\n$$\n\nNow from (1) it follows that $S(A_1OC_1) = S(D_1OB_1)$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25358,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve the equation in prime numbers $p$ and $q$: $q^3 = p^2 - p + 1$.",
    "options": [],
    "answer": "p = 19, q = 7",
    "solution": "Answer: $p = 19$, $q = 7$.\n\nWe have\n$$\nq^3 = p^2 - p + 1 \\Leftrightarrow (q-1)(q^2 + q + 1) = p(p-1). \\quad (1)\n$$\nIf $(q-1) \\nmid p$, then $q \\ge p+1$, so $q^3 > p^2 - p + 1$.\nHence, $(q^2 + q + 1) \\nmid p$, i.e.\n$$\nq^2 + q + 1 = k p \\quad (2)\n$$\nfor some $k \\in \\mathbb{N}$. It follows that $k(q-1) = p-1$, or $p = kq - k + 1$. Substituting this expression in (2) we obtain $q^2 + (1-k^2)q + (k^2-k+1) = 0$. The discriminant of this quadratic on $q$ equation\n$$\nD = (k^2 - 1)^2 - (4k^2 - k + 1) = k^4 - 6k^2 + 4k - 3\n$$\nmust be a perfect square. But it is easy to check that\n$$\n(k^2 - 3)^2 < k^4 - 6k^2 + 4k - 3 < (k^2 - 1)^2 \\quad \\text{for any } k > 3.\n$$\nMoreover, the equation $k^4 - 6k^2 + 4k - 3 = (k^2 - 2)^2$ has no solutions.\nHence, $k \\le 3$. For $k = 1$ we have $D = -4$, and for $k = 2$ we have $D = -3$, which is impossible. For $k = 3$ we obtain $D = 6^2$, so $q = (8 \\pm 6)/2$. Thus $q = 7$ and $p = k(q-1) + 1 = 19$. It remains to note the primes $p = 19$, $q = 7$ do satisfy the equation.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25359,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two parallel chords $AB$ and $CD$ are constructed on the hyperbola $y = 1/x$. The lines $AC$ and $BD$ meet ordinate axis $Oy$ at points $A_1$ and $D_1$ respectively, and meet abscissae axis $Ox$ at points $C_1$ and $B_1$ respectively.\nProve that the areas of the triangles $A_1OC_1$ and $D_1OB_1$ are equal.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $A_1(0; a_1)$, $B_1(b_1; 0)$, $C_1(c_1; 0)$, $D_1(0; d_1)$, $A(a; 1/a)$, $B(b; 1/b)$, $C(c; 1/c)$, $D(d; 1/d)$ be the marked points. Since any vertical and any horizontal line meets the hyperbola $y = 1/x$ at most at one point we see that the numbers $a$, $b$, $c$, $d$ are pairwise distinct.\n\n![](attached_image_1.png)\n\nThe equation of the line $AB$ has the form $y - \\frac{1}{a} = k_1(x - a)$. Since point $B$ belongs to the line $AB$ its coordinates must satisfy the line equation, i.e. $\\frac{1}{b} - \\frac{1}{a} = k_1(b - a)$, so $k_1 = -\\frac{1}{ab}$.\nSimilarly, for the line $CD$ we have $y - \\frac{1}{c} = k_2(x - c)$, and so\n$$\nk_2 = -\\frac{1}{cd}. \\text{ Since } AB \\parallel CD \\text{ we have } k_1 = k_2, \\text{ so}\n$$\n$$\nab = cd \\implies \\frac{a}{c} = \\frac{b}{d}. \\qquad (1)\n$$\n\nThe equation of the line $AC$ has the form $y - \\frac{1}{a} = k(x - a)$. Setting $x = 0$ in this equation, we find the ordinate of $A_1$: $a_1 = \\frac{1}{a} - ka$, and setting $y = 0$ in the equation we find the abscissa of $C_1$: $c_1 = a - \\frac{1}{ka}$. Since point $C$ belongs to the line $AC$ its coordinates must satisfy the line equation, i.e. $\\frac{1}{c} - \\frac{1}{a} = k(c-a)$, so $k = -\\frac{1}{ac}$. Therefore $a_1 = \\frac{1}{a} + \\frac{1}{c}$ and $c_1 = a + c$.\nSimilarly, from the equation of the line $BD$ we find $d_1 = \\frac{1}{b} + \\frac{1}{d}$ and $b_1 = b + d$.\nTherefore,\n$$\nS(A_1 OC_1) = \\frac{1}{2} |a_1 c_1| = \\frac{1}{2} \\cdot \\frac{(a+c)^2}{|ac|} = \\frac{1}{2} \\cdot \\frac{(a/c+1)^2}{|a/c|},\n$$\n$$\nS(D_1 OB_1) = \\frac{1}{2} |b_1 d_1| = \\frac{1}{2} \\cdot \\frac{(b+d)^2}{|bd|} = \\frac{1}{2} \\cdot \\frac{(b/d+1)^2}{|b/d|}.\n$$\nNow from (1) it follows that $S(A_1 OC_1) = S(D_1 OB_1)$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25360,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that $\\left(C_{2n}^{n+1}\\right)^2 \\le C_{n^2+n}^{n+1}$ for any $n \\in \\mathbb{N}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note that\n$$\n(t + k)^2 \\le (k + 1)(t^2 + k) \\quad (*)\n$$\nfor all integers $k$ and $t$, $0 \\le k < t$. Indeed, we have\n$$\n(*) \\Leftrightarrow t^2+2tk+k^2 \\le kt^2+t^2+k^2+k \\Leftrightarrow kt^2-2tk+k \\ge 0 \\Leftrightarrow k(t-1)^2 \\ge 0.\n$$\nIt is obvious that the last inequality holds for $0 \\le k \\le t$. Multiplying inequalities $(*)$ for $k = 0, \\dots, t$, we obtain\n$$\n(t(t+1)(t+2)\\cdots (2t))^2 \\le 1 \\cdot 2 \\cdots (t+1)t^2(t^2+1) \\cdots (t^2+t) \\Leftrightarrow\n$$\n$$\n\\left( \\frac{(2t)!}{(t-1)!} \\right)^2 \\le (t+1)! \\cdot \\frac{(t^2+t)!}{(t^2-1)!} \\Leftrightarrow \\\\\n\\left( \\frac{(2t)!}{(t-1)!(t+1)!} \\right)^2 \\le \\frac{(t^2+t)!}{(t^2-1)!(t+1)!} \\Leftrightarrow (C_{2t}^{t+1})^2 \\le C_{t^2+t}^{t+1}\n$$\nNow, set $t = n$ to obtain the required result.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25361,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A_1$, $B_1$, $C_1$ be midpoints of sides $BC$, $CA$, $AB$ of an acute-angled triangle $ABC$ respectively. Let $C_1L$ and $C_1K$ be bisectrices of $\\triangle ACC_1$ and $\\triangle BCC_1$ respectively. The line $C_1B_1$ intersects the line $LK$ at $A_2$, and the line $C_1A_1$ intersects the line $LK$ at $B_2$.\nProve that the lines $AA_2$, $BB_2$, $CC_1$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First show that the line $LK$ is parallel to the side $AB$. Indeed, since $C_1L$ and $C_1K$ are the bisectrices of the angles $AC_1C$ and $BC_1C$ respectively we have\n$$\n\\frac{AL}{LC} = \\frac{AC_1}{CC_1} = [AC_1 = BC_1] = \\frac{C_1B}{CC_1} = \\frac{BK}{KC}.\n$$\n![](attached_image_1.png)\n\nSo $LK \\parallel AB$. Since $A_1$ and $B_1$ are the midpoints of $BC$ and $AC$ respectively, we have $B_1A_1 \\parallel AB \\parallel LK$. Therefore\n$$\n\\frac{B_1A_2}{B_1C_1} = \\frac{A_1B_2}{A_1C_1}. \\qquad (1)\n$$\nSince $LA_2 \\parallel AC_1$ we have $\\frac{LA_2}{AC_1} = \\frac{B_1A_2}{B_1C_1}$, so $LA_2 = \\frac{B_1A_2}{B_1C_1} AC_1$.\nSimilarly, $KB_2 = \\frac{A_1B_2}{A_1C_1} C_1B$.\nSince $AC_1 = C_1B$, it follows from (1) that $LA_2 = KB_2$.\nLet $CC_1$ meets $LK$ at $C_2$. Since $CC_1$ is the median of $ABC$ and $LK \\parallel AB$, we have $LC_2 = C_2K$. Therefore,\n$$\n|A_2C_2| = |LC_2 - LA_2| = |KC_2 - B_2K| = |C_2B_2|.\n$$\nThus the line $C_1C_2$ passes through the midpoints $C_1$ and $C_2$ of the bases of the trapezoid $AA_2B_2B$. It is well-known fact that the line $C_1C_2$ and the extensions of the sides $AA_2$, $BB_2$ of the trapezoid $AA_2B_2B$ are concurrent.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25362,
    "subject": "Mathematics (Multi-modal)",
    "question": "On planet Automoria there live a (possible infinite) number Automorians. Any Automorian have feelings like love and respect. It is known that\n1) each Automorian loves exactly one Automorian and respects exactly one Automorian;\n2) if $A$ loves $B$, then every Automorian respecting $A$ also loves $B$;\n3) if $A$ respects $B$, then every Automorian loving $A$ also respects $B$;\n4) for every Automorian there is somebody loving him.\nIs it true that every Automorian respects the Automorian he loves?",
    "options": [],
    "answer": "yes",
    "solution": "Answer: yes.\nDenote by $f(x)$ and $g(x)$ the Automorians whom the Automorian $x$ loves and respects, respectively. The first condition of the problem implies that $f$ and $g$ are well defined functions on the set of Automorians. We have to prove that these functions are equal.\n\nTake any Automorian $x$, then he respects $g(x)$, who in turn loves $f(g(x))$. Using the second condition of the problem we see that this Automorian is also loved by $x$, thus\n$$\nf(g(x)) = f(x). \\qquad (1)\n$$\nSimilarly we get from the third condition that\n$$\ng(f(x)) = g(x). \\qquad (2)\n$$\nThe fourth condition implies that the function $f$ must be surjective.\n\nEquation (2) gives $f(g(f(x))) = f(g(x))$ and using (1) we obtain $f(f(x)) = f(x)$. Since $f$ is surjective, we must have $f(x) = x$ for every Automorian $x$. Then (1) now gives $g(x) = f(g(x)) = f(x) = x$ and hence $f(x) = g(x) = x$.\n\nThus, every Automorian loves and respects himself only.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25363,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that\n$$\nf(x + y) + y \\le f(f(f(x)))\n$$\nfor all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "All functions are f(x) = a - x for an arbitrary real constant a.",
    "solution": "Answer: $f(x) = a - x$ for arbitrary real constant $a$.\n\nFirst set $y = 0$ in the initial inequality\n$$\nf(x + y) + y \\le f(f(f(x))) \\quad (*)\n$$\nthus obtaining\n$$\nf(x) \\leq f(f(f(x))) \\quad \\forall x \\in \\mathbb{R}. \\qquad (1)\n$$\nFurther, set $y = f(f(x)) - x$ in $(*)$. Then\n$$\nf(f(x)) \\leq x \\quad \\forall x \\in \\mathbb{R}. \\qquad (2)\n$$\nReplacing $x$ by $f(x)$ in (2) we obtain $f(f(f(x))) \\leq f(x)$ which together with (1) gives $f(f(f(x))) = f(x)$.\n\nNow $(*)$ becomes\n$$\nf(x + y) + y \\leq f(x) \\quad \\forall x, y \\in \\mathbb{R}. \\qquad (3)\n$$\nSet $x = 0$ in (3), then\n$$\nf(y) \\leq a - y \\quad \\forall y \\in \\mathbb{R}, \\qquad (4)\n$$\nwhere $a = f(0)$.\nFinally, set $y = -x$ in (3) thus getting\n$$\na - x \\leq f(x). \\qquad (5)\n$$\nComparing (4) and (5) we obtain $f(x) = a - x$.\n\nIt is easy to verify that the function $f(x) = a - x$ satisfies the given inequality for any real number $a$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25364,
    "subject": "Mathematics (Multi-modal)",
    "question": "For two positive integers $a$ and $b$ the number $\\overline{a.b}$ is equal to the decimal fraction which we have if after the number $a$ we put the decimal point and then write the number $b$. For example, for $a = 20$, $b = 13$ we get $\\overline{a.b} = 20.13$, and $\\overline{b.a} = 13.2$.\nProve that there are infinite number of natural $n$, such that the equation $\\overline{a,b} \\cdot \\overline{b,a} = n$ has positive integer solutions $a$ and $b$. (D. Bazylev)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the decimal representation of $a$ consist of $k$ digits. Set $b = 5 \\cdot 10^m$, $m \\in \\mathbb{N}$. Then\n$$\n\\begin{align*}\n\\overline{a,b} \\cdot \\overline{b,a} &= n \\iff \\left(a + \\frac{b}{10^k}\\right)\\left(b + \\frac{a}{10^m}\\right) = n \\iff \\\\\n&\\iff \\frac{(2a + 1) \\cdot 5 \\cdot 10^m}{2} + \\frac{a(2a + 1)}{2 \\cdot 10^k} = n \\iff \\\\\n&\\iff (2a + 1) \\cdot 5^{m+1} \\cdot 2^{m-1} + \\frac{a(2a + 1)}{2 \\cdot 10^k} = n.\n\\end{align*}\n$$\nWe see that the last equality holds for some positive integer $n$ if and only if the number $a(2a + 1)$ is divisible by $2 \\cdot 10^k = 2^{k+1} \\cdot 5^k$. So if we find at least one such number $a$, then the problem statement will be proved, since for this $a$ and any $b = 5 \\cdot 10^m$, $m \\in \\mathbb{N}$ the number $a, b \\cdot b, a$ will be integer.\nConsider $k=4$. Then the required relation has the form $a(2a+1) \\mid 2^5 \\cdot 5^4$.\nIt is easily followed from two relations $a \\mid 2^5$ and $(2a+1) \\mid 5^4$. So, to solve the problem it suffices to find at least one $a$ with 4 digits in its decimal representation and satisfying these two relations. Such $a$ does exist, for example, $a = 5312$. Show how we can find such $a$.\nFirst we show that there exist infinitely many positive integers $a$ such that $a \\mid 2^5$ and $(2a+1) \\mid 5^4$. The first relation is equivalent to the equality $a = 32c$, where $c \\in \\mathbb{N}$, and then the second relation has the form $2 \\cdot 32c + 1 \\mid 5^4$ or $64c + 1 \\mid 625$. We have\n$$\n\\begin{align*}\n64c + 1 \\mid 625 &\\iff 64c + 1 - 625 \\mid 625 &\\iff 64c - 624 \\mid 625 &\\iff \\\\\n16(4c - 39) \\mid 625 &\\iff 4c - 39 \\mid 625 &\\iff 4c - 39 - 625 \\mid 625 &\\iff \\\\\n4c - 664 \\mid 625 &\\iff 4(c - 166) \\mid 625 &\\iff c - 166 \\mid 625 &\\iff \\\\\nc = 625l + 166, l \\in \\mathbb{Z}.\n\\end{align*}\n$$\nIn particular, for $l = 0$ we have $c = 166$, and so $a = 32c = 32 \\cdot 166 = 5312$ has 4 digits in its decimal representation, as required.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 25365,
    "subject": "Mathematics (Multi-modal)",
    "question": "For two positive integers $a$ and $b$ the number $\\overline{a.b}$ is equal to the decimal fraction which we have if after the number $a$ we put the decimal point and then write the number $b$. For example, for $a = 20, b = 13$ we get $\\overline{a.b} = 20.13$, and $\\overline{b.a} = 13.2$.\nProve that there are infinite number of natural $n$, such that the equation $\\overline{a, b} \\cdot \\overline{b, a} = n$ has no positive integer roots $a$ and $b$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Show that if $n = 9k \\pm 3$, $k \\in \\mathbb{N}$, then the given equation has no natural solutions.\nLet the decimal representations of $a$ and $b$ consist of $m$ and $l$ digits respectively. Then the initial equation is equivalent to the equation\n$$\n\\left(a + \\frac{b}{10^l}\\right) \\left(b + \\frac{a}{10^m}\\right) = 9k \\pm 3 \\quad \\text{or} \\quad ab + \\frac{ab}{10^{m+l}} + \\frac{a^2}{10^m} + \\frac{b^2}{10^l} = 9k \\pm 3,\n$$\ni. e.\n$$\n10^{m+l}ab + ab + a^2 10^l + b^2 10^m = 10^{m+l} \\cdot (9k \\pm 3). \\quad (*)\n$$\nSince $10^t - 1 = \\underbrace{9\\dots9}_{t \\text{ times}}$ for any positive integer $t$, i. e. $10^t = 9A + 1$ for some positive integer $A$, we can replace all powers of 10 in (*) by their presentations, then we obtain\n$$\n(9A_1 + 1)ab + ab + (9A_2 + 1)a^2 + (9A_3 + 1)b^2 = (9A_4 + 1) \\cdot (9k \\pm 3)\n$$\nor $(a+b)^2 = 9B \\pm 3$ for some positive integer $B$. But this equality is impossible because the left hand side of it is the square number and the left hand side is a number which is divisible by 3 but is not divisible by 9.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 25366,
    "subject": "Mathematics (Multi-modal)",
    "question": "Point $X$ is marked inside an acute-angled triangle $ABC$ such that $\\angle AXC = \\angle A + \\angle B$, $\\angle BXA = \\angle B + \\angle C$, $\\angle BXC = \\angle C + \\angle A$.\n\nProve that $\\frac{BC \\cdot AX}{AB} = \\frac{AB \\cdot CX}{AC}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Extend the cevians $AX$, $BX$, $CX$; let their intersection points with the sides $BC$, $AC$, $BA$ be $A_1$, $B_1$, $C_1$ respectively. We have\n$$\n\\angle AXB_1 = 180^\\circ - \\angle BXA = 180^\\circ - (\\angle B + \\angle C) = \\angle A = \\alpha.\n$$\nSimilarly, $\\angle BXC_1 = \\angle B = \\beta$, $\\angle CXA_1 = \\angle C = \\gamma$. Let $\\angle CAA_1 = \\varphi$. Then $\\angle ABB_1 = \\angle ABX = 180^\\circ - (\\alpha - \\varphi) - (\\alpha + \\beta) = \\varphi$ and, similarly, $\\angle BCC_1 = \\varphi$. That is, $X$ is the Brakar point of the triangle $ABC$.\n\nNow, using the law of sines for the triangles $ABC$, $AXB$, $AXC$, we have respectively\n$$\n\\frac{BC}{AC} = \\frac{\\sin \\alpha}{\\sin \\gamma}, \\quad \\frac{AX}{AB} = \\frac{\\sin \\varphi}{\\sin \\angle BXA} = \\frac{\\sin \\varphi}{\\sin \\alpha},\n$$\n$$\n\\frac{AC}{CX} = \\frac{\\sin \\angle CXA}{\\sin \\varphi} = \\frac{\\sin \\gamma}{\\sin \\varphi}.\n$$\nMultiplying all three equalities we obtain\n$$\n\\frac{BC}{AB} \\cdot \\frac{AX}{AB} \\cdot \\frac{AC}{CX} \\text{ which is equivalent to } \\frac{BC \\cdot AX}{AB} = \\frac{AB \\cdot CX}{AC}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25367,
    "subject": "Mathematics (Multi-modal)",
    "question": "Three of six segments (three sides and three medians of a triangle) are painted red, and three others are painted blue.\nCan one construct a triangle using the segments of the same color as its sides?",
    "options": [],
    "answer": "Yes",
    "solution": "Answer: yes, one can.\n\nLet $G$ be a gravicenter of the triangle $ABC$, and $A_1$, $B_1$, $C_1$ be the midpoints of the sides $BC$, $AC$, $AB$ respectively. Denote the sides and the medians of the triangle $ABC$ in the following way: $AB = c$, $BC = a$, $CA = b$, $AA_1 = d$, $BB_1 = e$, $CC_1 = f$.\n\nSuppose that one can paint three of six segments mentioned red and three others blue so that it is not possible to construct a triangle using the segments of the same color. Then the length of some red segment is not less than the sum of two other red segments; the same is true for blue segments.\n\nSo, there are two of six segments, the sum of which is not less than the sum of four others: $x + y \\ge z + u + v + w$. We show, however, that this situation is not possible. Consider the following three cases.\n\n1) Both $x$ and $y$ are some sides of $\\triangle ABC$. Without loss of generality, let $a + b \\ge c + d + e + f$. But from the triangles $BGC$ and $AGC$ we have $2f/3 + 2e/3 > a$ and $2f/3 + 2d/3 > b$. Summing all three inequalities we get $f/3 > c + d/3 + e/3$ which contradicts to the triangle inequality for medians: $d + e > f$.\n\n2) Both $x$ and $y$ are some medians of $\\triangle ABC$. Without loss of generality, let $d + e \\ge a + b + c + f$. But from the triangles $BGC_1$ and $AGC_1$ we have $f/3 + c/2 > 2e/3$ and $f/3 + c/2 > 2d/3$, or $f/2 + 3c/4 > e$ and $f/2 + 3c/4 > d$. It follows that $f + 3c/2 > d + e \\ge a + b + c + f$, or $c/2 > a + b$, a contradiction.\n\n3a) $x$ and $y$ are a side and a median of $\\triangle ABC$ having the common endpoint. Without loss of generality, let $a + e \\ge b + c + d + f$. But this inequality contradicts to the inequalities $a < b + c$ and $e < d + f$.\n\n3b) $x$ is a median of $\\triangle ABC$ with the endpoint in the middle of the side $y$. Without loss of generality, let $a + d \\ge b + c + e + f$. But this inequality contradicts to the inequalities $a < b + c$ and $d < e + f$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25368,
    "subject": "Mathematics (Multi-modal)",
    "question": "Three of six segments (three sides and three medians of a triangle) are painted red, and three others are painted blue.\nCan one construct a triangle using the segments of the same color as its sides?",
    "options": [],
    "answer": "yes",
    "solution": "Answer: yes, one can.\n\nLet $G$ be a gravicenter of the triangle $ABC$, and $A_1$, $B_1$, $C_1$ be the midpoints of the sides $BC$, $AC$, $AB$ respectively. Denote the sides and the medians of the triangle $ABC$ in the following way: $AB = c$, $BC = a$, $CA = b$, $AA_1 = d$, $BB_1 = e$, $CC_1 = f$.\n\nSuppose that one can paint three of six segments mentioned red and three others blue so that it is not possible to construct a triangle using the segments of the same color. Then the length of some red segment is not less than the sum of two other red segments; the same is true for blue segments.\n\nSo, there are two of six segments, the sum of which is not less than the sum of four others: $x + y \\ge z + u + v + w$. We show, however, that this situation is not possible. Consider the following three cases.\n\n1) Both $x$ and $y$ are some sides of $\\triangle ABC$. Without loss of generality, let $a + b \\ge c + d + e + f$. But from the triangles $BGC$ and $AGC$ we have $2f/3 + 2e/3 > a$ and $2f/3 + 2d/3 > b$. Summing all three inequalities we get $f/3 > c + d/3 + e/3$ which contradicts the triangle inequality for medians: $d + e > f$.\n\n2) Both $x$ and $y$ are some medians of $\\triangle ABC$. Without loss of generality, let $d + e \\ge a + b + c + f$. But from the triangles $BGC_1$ and $AGC_1$ we have $f/3 + c/2 > 2e/3$ and $f/3 + c/2 > 2d/3$, or $f/2 + 3c/4 > e$ and $f/2 + 3c/4 > d$. It follows that $f + 3c/2 > d + e \\ge a + b + c + f$, or $c/2 > a + b$, a contradiction.\n\n3a) $x$ and $y$ are a side and a median of $\\triangle ABC$ having the common endpoint. Without loss of generality, let $a+e \\ge b+c+d+f$. But this inequality contradicts the inequalities $a < b+c$ and $e < d+f$.\n\n3b) $x$ is a median of $\\triangle ABC$ with the endpoint in the middle of the side $y$. Without loss of generality, let $a + d \\ge b + c + e + f$. But this inequality contradicts the inequalities $a < b+c$ and $d < e+f$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25369,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two parallelograms $ABCD$ and $NMKD$ are placed in the plane as it is shown in the figure; $T$ is the intersection point of the segments $BC$ and $MN$. Prove that the points $D$, $T$, and the intersection point of $AM$ and $BK$ are collinear if the points $A$, $T$, and $K$ are collinear.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Construct the segment $AK$. By condition, $T$ lies on $AK$. Let $R$ be the intersection point of lines $DT$ and $KM$ (see the figures). Let $AN = x$, $ND = y$, $RM = z$, $AT : TK = 1 : k$. By Thales' theorem, $k = TK : AT = y : x$ (since $NT \\parallel DK$). Since $NMKD$ is a parallelogram, we have $MK = ND = y$, and then\n$$\nz : y = RM : MK = [TM \\parallel DK] = RT : TD = [TC \\parallel RK] = KC : CD =\n$$\n$$\n= [TC \\parallel AD] = KT : TA = k. \\quad (1)\n$$\nSo, $y = kx$, $z = ky = k^2x$. Let $P_1$ be the point of intersection of $AM$ and $RD$ (Fig. 1).\n\n![](attached_image_2.png)\n\n![](attached_image_3.png)\n\nConstruct a line passing through $A$ parallel to $AM$, and let $Q_1$ be the point of intersection of this line and the line $MK$. By Thales' theorem,\n$$\n\\begin{gathered}\nRP_1 : P_1D = RM : MQ_1 = [AMQ_1D \\text{ is a parallelogram,} \\\\\nMQ_1 = AD = x + y] = z : (x + y) = k^2x : (x + kx) = \\frac{k^2}{1+k}.\n\\end{gathered}\n$$\nTherefore,\n$$\nRP_1 = \\frac{k^2}{1+k} P_1D = \\frac{k^2}{1+k} (RD - RP_1) \\Rightarrow RP_1 = \\frac{k^2}{1+k+k^2} RD. \\quad (2)\n$$\n\nLet $P_2$ be the point of intersection of the segments $BK$ and $RD$ (Fig. 2). Construct a line passing through $T$ parallel to $BK$, and let $Q_2$ be the point of intersection of this line with the line $MK$. By Thales' theorem,\n$$\nRP_2 : P_2T = RK : KQ_2 = [BKQ_2T, BTNA, \\text{NMKD is a parallelogram,}]\n$$\n$$\nKQ_2 = BT = AN = x, MK = ND = y] = (z + y) : x = (k^2 + k).\n$$\nTherefore,\n$$\nRP_2 = (k^2 + k)P_2T = k(k + 1)(RT - RP_2) \\implies\n$$\n$$\nRP_2 = \\frac{k(k+1)}{k^2+k+1}RT = [RT = kTD = k(RD - RT) \\implies\n$$\n$$\nRT = \\frac{k}{k+1}RD] = \\frac{k^2}{k^2+k+1}RD.\n$$\nComparing this result with (2), we see that $P_1$ coincides with $P_2$, and so they coincide with $P$, which gives the required statement.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25370,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\mathbb{Z}$ and $\\mathbb{Q}$ be the sets of integers and rationals respectively.\n\na) Does there exist a partition of $\\mathbb{Z}$ into three non-empty subsets $A, B, C$ such that the sets $A+B$, $B+C$, $C+A$ are disjoint?\n\nb) Does there exist a partition of $\\mathbb{Q}$ into three non-empty subsets $A, B, C$ such that the sets $A+B$, $B+C$, $C+A$ are disjoint?\n\nHere $X + Y$ denotes the set $\\{x + y \\mid x \\in X, y \\in Y\\}$, for $X, Y \\subseteq \\mathbb{Z}$ and $X, Y \\subseteq \\mathbb{Q}$.",
    "options": [],
    "answer": "a) Yes. For example, take the three residue classes modulo three.\nb) No.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25371,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given positive real numbers $a$, $b$, $c$. Find the greatest real number $x$ such that there exist positive real numbers $p$, $q$, $r$ ($p + q + r = 1$) with $x \\le \\min\\{ap/q, bq/r, cr/p\\}$.",
    "options": [],
    "answer": "sqrt[3]{abc}",
    "solution": "Answer: $\\sqrt[3]{abc}$.\nWe can consider $x > 0$. Multiplying three inequalities $x \\le ap/q$, $x \\le bq/r$, $x \\le cr/p$, we obtain $x \\le \\sqrt[3]{abc}$.\nIt remains to show that the number $\\sqrt[3]{abc}$ satisfies the problem condition. It suffices to verify that the system of the equations (with unknown $p, q, r$)\n$$\n\\begin{cases} \\sqrt[3]{abc} = a \\frac{p}{q}, \\\\ \\sqrt[3]{abc} = b \\frac{q}{r}, \\\\ p + q + r = 1 \\end{cases}\n$$\nhas a positive solution $(p; q; r)$. From the first two equations we present $p$ and $r$ as the functions of $q$ and then replace $p$ and $q$ by these presentations in the third equation. So we find the positive solution\n$$\np = \\frac{yq}{a}, \\quad r = \\frac{bq}{y}, \\quad q = \\frac{1}{\\left(\\frac{y}{a} + \\frac{b}{y} + 1\\right)}, \\quad \\text{where} \\quad y = \\sqrt[3]{abc},\n$$\nas required.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25372,
    "subject": "Mathematics (Multi-modal)",
    "question": "A triangle $ABC$ is inscribed in the parabola $y = x^2$. Let $a$, $b$, $c$ be the abscissae of the midpoints of its sides.\nFind the radius of the circumcircle of $\\triangle ABC$.",
    "options": [],
    "answer": "R = 0.5*sqrt((1 + 4a^2)(1 + 4b^2)(1 + 4c^2))",
    "solution": "Answers: $R = 0.5\\sqrt{(1 + 4a^2)(1 + 4b^2)(1 + 4c^2)}$.\n\nLet $A(l; l^2)$, $B(m; m^2)$, $C(n; n^2)$ be coordinates of the vertices of the triangle $ABC$. Then\n$$\nAB = \\sqrt{(m-l)^2 + (m^2-l^2)^2} = |m-l|\\sqrt{1+(m+l)^2} = |m-l|\\sqrt{1+4c^2},\n$$\nsince $c = 0.5(m + l)$.\nSimilarly, $BC = |m - n|\\sqrt{1 + 4a^2}$ and $AC = |n - l|\\sqrt{1 + 4b^2}$.\n\nFurther, it is easy to calculate that $S(ABC) = 0.5|(m - l)(l - n)(n - m)|$.\n\nNow we find the circumradius using the formula\n$$\nR = \\frac{AB \\cdot BC \\cdot CA}{4S(ABC)} = \\frac{1}{2} \\sqrt{(1 + 4a^2)(1 + 4b^2)(1 + 4c^2)}.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25373,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given real positive numbers $x_1, \\dots, x_n$ ($n \\ge 3$) such that $x_1 \\cdot \\dots \\cdot x_n = 1$, prove that\n$$\n\\frac{x_1^8}{(x_1^4 + x_2^4)x_2} + \\frac{x_2^8}{(x_2^4 + x_3^4)x_3} + \\dots + \\frac{x_n^8}{(x_n^4 + x_1^4)x_1} \\ge \\frac{n}{2}.\n$$\n(I. Voronovich)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We use the following\n\n**Lemma.** For any positive $a$ and $b$ the following inequality is valid\n$$\n(a^3 + b^3)^2 \\geq 2ab(a^4 + b^4). \\qquad (*)\n$$\nIndeed,\n$$\n(*) \\Leftrightarrow a^6 - 2a^5b + 2a^3b^3 - 2ab^5 + b^6 \\geq 0 \\Leftrightarrow (a-b)^2(a^4 - a^2b^2 + b^4) \\geq 0\n$$\nwhich is true.\n\nNow, using the lemma, we have\n$$\n\\sum_{i=1}^{n} \\frac{x_i^8}{(x_i^4 + x_{i+1}^4)x_{i+1}} = \\sum_{i=1}^{n} \\frac{x_i^9}{(x_i^4 + x_{i+1}^4)x_i x_{i+1}} \\geq \\sum_{i=1}^{n} \\frac{2x_i^9}{(x_i^3 + x_{i+1}^3)^2} =\n$$\n$$\n= 2 \\sum_{i=1}^{n} \\frac{(x_i^3)^3}{(x_i^3 + x_{i+1}^3)^2} \\ge [\\text{the H\\\"older inequality}] \\ge 2 \\frac{\\left(\\sum_{i=1}^{n} x_i^3\\right)^3}{\\left(2 \\sum_{i=1}^{n} x_i^3\\right)^2} = \\frac{1}{2} \\sum_{i=1}^{n} x_i^3 \\ge \\frac{1}{2} n \\left(\\sqrt[n]{x_1 \\cdots x_n}\\right)^3 = \\frac{n}{2},\n$$\nas was to be proved.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25374,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all polynomials $P(x)$ such that the equality\n$$\n(x - 1)P(x + 1) - (x + 1)P(x - 1) = 4P(x)\n$$\nholds for all real $x$.",
    "options": [],
    "answer": "P(x) = a x(x-1)(x+1) for any real a",
    "solution": "Answer: $P(x) = a x(x-1)(x+1)$, where $a \\in \\mathbb{R}$.\n\nSet $x = 1$ and $x = -1$ in the initial identity\n$$\n(x-1)P(x+1) - (x+1)P(x-1) = 4P(x). \\quad (1)\n$$\nThus we obtain $-2P(0) = 4P(1)$ and $-2P(0) = 4P(-1)$ respectively. Setting $x = 0$ in (1), we obtain $-P(1) - P(-1) = 4P(0)$, so, taking into account two previous equalities, we have $P(-1) = P(0) = P(1) = 0$.\n\nTherefore $0$, $1$, and $-1$ are zeroes of the polynomial $P(x)$, hence $P(x) = x(x-1)(x+1)Q(x)$, where $Q(x)$ is some polynomial. Setting this representation of $P(x)$ in (1), we obtain\n$$\n(x-1)(x+1)x(x+2)Q(x+1) - (x+1)(x-1)(x-2)xQ(x-1) = \\\\\n= 4x(x-1)(x+1)Q(x),\n$$\nwhence\n$$\n(x+2)Q(x+1) - (x-2)Q(x-1) = 4Q(x) \\quad \\text{for all } x \\in \\mathbb{R}. \\quad (2)\n$$\nSet $x = 2$ in (2), then we obtain $Q(3) = Q(2)$. By induction on $k$, show that $Q(k) = Q(2)$ for all positive integers $k \\ge 2$. Indeed, let $Q(k) = Q(2)$ for all $k = 2, 3, \\dots, m$, where $m \\ge 3$. Then setting $x = m$ in (2), we obtain $(m+2)Q(m+1) - (m-2)Q(m-1) = 4Q(m)$, so, taking into account our assumption, we obtain $(m+2)Q(m+1) = (m+2)Q(2)$, i.e. $Q(m+1) = Q(2)$. It follows that $Q(k) = Q(2)$ for all positive integers $k \\ge 2$.\n\nSince the polynomial $Q(x) - Q(2)$ has zero value at infinitely many points, we conclude that this polynomial is equal to zero identically, i.e. $Q(x) = Q(2) = a$ for some real number $a$.\n\nTherefore, the polynomial satisfying (1) needs to have the form $P(x) = a x(x-1)(x+1)$. It is easy to verify that any polynomial of such form (for any real $a$) satisfies (1).",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 25375,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $M$ and $N$ be the midpoints of the sides $AB$ and $BC$ of the triangle $ABC$ respectively. Points $K$ and $L$ are the tangency points of the inscribed circle of the triangle $ABC$ with the sides $AB$ and $AC$ respectively. Let $T$ be the intersection point of the lines $MN$ and $KL$. Prove that $T$ belongs to the bisector of the angle $ACB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $P$ be the tangency point of the inscribed circle of $\\triangle ABC$ with the side $BC$. Let $x = BK = BP$, $y = AK = AL$, $z = CP = CL$.\nIf $BC = AC$, then the statement of the problem holds because the points $M$, $K$, $T$ are coincide and $CT$ is a bisector of the angle $ACB$.\n\n![](attached_image_1.png)\nFig. 1\n![](attached_image_2.png)\nFig. 2\n\nLet $AC > BC$ (see Fig. 1). Since $MN$ is the midline we have $MN \\parallel AC$ and so $\\angle KTM = \\angle KLA$. But $\\angle AKL = \\angle KLA$ (whence $AK = AL$),\ntherefore $\\angle MKT = \\angle AKL = \\angle KTM$, and\n$$\nMT = MK = 0.5 \\cdot AB - BK = 0.5(x + y) - x = 0.5(y - x).\n$$\nThus, $TN = MN - MT = 0.5(y + z) - 0.5(y - x) = 0.5(z + x) = CN$. Therefore $\\triangle TNC$ is an isosceles triangle $\\angle NTC = \\angle NCT$. Since $MN \\parallel AC$ we have $\\angle NTC = \\angle TCA$. From these two equalities we obtain $\\angle NCT = \\angle TCA$, i.e. $CT$ is the bisector of $\\angle BCA$.\n\nSimilar consideration can be applied for the case $AC < BC$ (see Fig. 2).",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25376,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all triples $(x, y, z)$ of nonnegative integers $x$, $y$, $z$ such that $7^x = 3^z - 2^y$.",
    "options": [],
    "answer": "[[0, 1, 1], [1, 1, 2], [0, 3, 2], [2, 5, 4]]",
    "solution": "Answer: $\\{(x, y, z)\\} = \\{(0, 1, 1), (1, 1, 2), (0, 3, 2), (2, 5, 4)\\}$.\n\nWe rewrite the equation in the form $7^x + 2^y = 3^z$. Note that for $y = 0$ the left-hand side of the equation is an even integer while the right-hand side is an odd integer. So, $y > 0$.\n\nConsider three cases: $y = 1$, $y = 2$ and $y \\ge 3$.\n\n1)\n$y = 1$. In this case we have\n$$\n7^x + 2 = 3^z. \\tag{1}\n$$\nIf $x = 0$, then $z = 1$; if $x = 1$, then $z = 2$. Thus, it remains to consider the case $x > 2$, $z > 2$. In this case $(7^x + 2) \\div 3^3$. Consider the residues of $7^n$ modulo $27$:\n$$\n7 \\rightarrow -5 \\rightarrow -8 \\rightarrow -2 \\rightarrow 13 \\rightarrow 10 \\rightarrow -11 \\rightarrow 4 \\rightarrow 1.\n$$\nIt follows that $x \\equiv 4 \\pmod{9}$. Note that $7^6 + 7^3 + 1 = 117993 = 3 \\cdot 37 \\cdot 1063$, so $(7^9 - 1) \\div (7^6 + 7^3 + 1) \\div 37$. Thus, $7^9 \\equiv 1 \\pmod{37}$ and then\n$$\n7^x \\equiv 7^4 \\equiv 49^2 \\equiv 12^2 \\equiv 144 \\equiv 33 \\pmod{37}.\n$$\nHence $7^x + 2 \\equiv 35 \\pmod{37}$.\n\nOn the other hand, considering the residues of $3^n$ modulo $37$, we have\n$$\n3 \\rightarrow 9 \\rightarrow 27 \\rightarrow 7 \\rightarrow 21 \\rightarrow 26 \\rightarrow 4 \\rightarrow 12 \\rightarrow 36 \\rightarrow 34 \\rightarrow 28 \\rightarrow 10 \\rightarrow 30 \\rightarrow \\\\\n\\rightarrow 16 \\rightarrow 11 \\rightarrow 33 \\rightarrow 25 \\rightarrow 1.\n$$\nWe see that $3^z \\not\\equiv 35 \\pmod{37}$ for all $z$. Therefore, there are no solutions of (1) for $x > 2$, $z > 2$.\n\n2) $y = 2$. We have $3^z = 7^x + 4 \\equiv 2 \\pmod{3}$, a contradiction.\n\n3) $y \\ge 3$. Then $7^x \\equiv 3^z \\pmod{8}$, whence $x \\div 2$ and $z \\div 2$. Let $x = 2x_1$, $z = 2z_1$. Now the initial equation can be rewritten in the form\n$$\n(3^{z_1} - 7^{x_1})(3^{z_1} + 7^{x_1}) = 2^y.\n$$\nSo, it follows that\n$$\n3^{z_1} - 7^{x_1} = 2^a, \\qquad (2)\n$$\n$$\n3^{z_1} + 7^{x_1} = 2^b, \\qquad (3)\n$$\nand, since $(3^{z_1} - 7^{x_1}) \\div 2$, we have $b > a \\ge 1$. Summing (2) and (3), we obtain $2^a + 2^b = 2 \\cdot 3^{z_1}$. Hence $a = 1$. Then (2) can be rewritten as $3^{z_1} - 7^{x_1} = 2$, or $3^{z_1} = 7^{x_1} + 2$, which coincides with (1). Therefore, we have $\\{(x_1, z_1)\\} = \\{(0, 1), (1, 2)\\}$, and $b$ is equal to $2$ and $4$ respectively.\n\nFinally, in this case we have $\\{(x, y, z)\\} = \\{(0, 3, 2), (2, 5, 4)\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25377,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all triples $(x, y, z)$ of nonnegative integers $x$, $y$, $z$ such that $7^x = 3^z - 2^y$.",
    "options": [],
    "answer": "[[0, 1, 1], [1, 1, 2], [0, 3, 2], [2, 5, 4]]",
    "solution": "$\\{(x, y, z)\\} = \\{(0, 1, 1), (1, 1, 2), (0, 3, 2), (2, 5, 4)\\}$.\nWe rewrite the equation in the form $7^x + 2^y = 3^z$. Note that for $y = 0$ the left-hand side of the equation is an even integer while the right-hand side is an odd integer. So, $y > 0$.\nConsider three cases: $y = 1$, $y = 2$ and $y \\ge 3$.\n1) $y = 1$. In this case we have\n$$\n7^x + 2 = 3^z. \\quad (1)\n$$\nIf $x = 0$, then $z = 1$; if $x = 1$, then $z = 2$. Thus, it remains to consider the case $x > 2$, $z > 2$. In this case $(7^x + 2) \\ge 3^3$. Consider the residues of $7^n$ modulo 27:\n$$\n7 \\rightarrow -5 \\rightarrow -8 \\rightarrow -2 \\rightarrow 13 \\rightarrow 10 \\rightarrow -11 \\rightarrow 4 \\rightarrow 1.\n$$\nIt follows that $x \\equiv 4 \\pmod 9$. Note that $7^6 + 7^3 + 1 = 117993 = 3 \\cdot 37 \\cdot 1063$, so $(7^9 - 1) \\equiv (7^6 + 7^3 + 1) \\equiv 37$. Thus, $7^9 \\equiv 1 \\pmod{37}$ and then\n$$\n7^x \\equiv 7^4 \\equiv 49^2 \\equiv 12^2 \\equiv 144 \\equiv 33 \\pmod{37}.\n$$\nHence $7^x + 2 \\equiv 35 \\pmod{37}$.\nOn the other hand, considering the residues of $3^n$ modulo 37, we have\n$$\n3 \\rightarrow 9 \\rightarrow 27 \\rightarrow 7 \\rightarrow 21 \\rightarrow 26 \\rightarrow 4 \\rightarrow 12 \\rightarrow 36 \\rightarrow 34 \\rightarrow 28 \\rightarrow 10 \\rightarrow 30 \\rightarrow 16 \\rightarrow 11 \\rightarrow 33 \\rightarrow 25 \\rightarrow 1.\n$$\nWe see that $3^z \\not\\equiv 35 \\pmod{35}$ for all $z$. Therefore, there are no solutions of (1) for $x > 2$, $z > 2$.\n2) $y = 2$. We have $3^z = 7^x + 4 \\equiv 2 \\pmod 3$, a contradiction.\n3) $y \\ge 3$. Then $7^x \\equiv 3^z \\pmod 8$, whence $x \\equiv 2 \\pmod 2$ and $z \\equiv 2 \\pmod 2$. Let $x = 2x_1$, $z = 2z_1$. Now the initial equation can be rewritten in the form\n$$\n(3^{z_1} - 7^{x_1})(3^{z_1} + 7^{x_1}) = 2^y.\n$$\nSo, it follows that\n$$\n3^{z_1} - 7^{x_1} = 2^a, \\quad (2)\n$$\n$$\n3^{z_1} + 7^{x_1} = 2^b, \\quad (3)\n$$\nand, since $(3^{z_1} - 7^{x_1}) \\equiv 2$, we have $b > a \\ge 1$. Summing (2) and (3), we obtain $2^a + 2^b = 2 \\cdot 3^{z_1}$. Hence $a = 1$. Then (2) can be rewritten as $3^{z_1} - 7^{x_1} = 2$, or $3^{z_1} = 7^{x_1} + 2$, which coincides with (1). Therefore, we have $\\{(x_1, z_1)\\} = \\{(0, 1), (1, 2)\\}$, and $b$ is equal to 2 and 4 respectively.\nFinally, in this case we have $\\{(x, y, z)\\} = \\{(0, 3, 2), (2, 5, 4)\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25378,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest positive integer $n$ such that the number $2013n$ can be presented as the difference of two cubes of positive integer numbers.",
    "options": [],
    "answer": "39",
    "solution": "Answer: $n = 39$.\n(Solution of A. Semchankau, A. Zhuk.) Let\n$$\n2013n = a^3 - b^3. \\qquad (1)\n$$\nThen\n$$\n(1) \\Leftrightarrow 61 \\cdot 11 \\cdot 3n = 2013n = (a-b)^3 + 3ab(a-b) \\Rightarrow (a-b) \\vdash 3,\n$$\ni.e. $(a-b) = 3k, k \\in \\mathbb{N}$. So $61 \\cdot 11 \\cdot 3n = 3^3 k^3 + 3^2 abk$, whence $n \\vdash 3$, i.e. $n = 3m, m \\in \\mathbb{N}$. Then (1) can be rewritten as\n$$\n3k^3 + abk = 61 \\cdot 11m. \\qquad (2)\n$$\nNote that if $k = 11, b = 10, a = 43$, then $m = 13$, i.e. $n = 39$. Show that $n = 39$ is the smallest possible value of $n$, i.e. $m = 13$ is the smallest possible value of $m$. Indeed, from (2) it follows that $k(3k^2 + 3kb + b^2) \\vdash 11$, and it is easy to show that $k \\vdash 11$. But for $k \\ge 22$ we have $m > 13$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25379,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all integers $n$ and $x_1, \\ldots, x_n$ ($1 \\le x_i \\le 50$) such that\n$$ \\sum_{i=1}^{n} x_i(100 - x_i) = 1515. $$",
    "options": [],
    "answer": "n = 2; {x1, x2} = {4, 13}",
    "solution": "Answer: $n = 2$, $\\{x_1, x_2\\} = \\{4, 13\\}$.\n\nIt is clear that there are no solutions for $n = 1$. Hence consider the case $n \\ge 2$.\nLet $s = \\sum_{i} x_i$, $S = \\sum_{i} x_i^2$. Then $100s - S = 1515$ and hence $100s > 1515 > 100s - s^2$. Taking into account that $s$ is a nonnegative integer, we obtain from these inequalities that $s \\ge 16$ and $|50-s| \\ge 32$, i.e. $s \\ge 82$ or $16 \\le s \\le 18$.\n\nWe shall now show that $s \\ge 82$ cannot hold. Suppose it does hold, then among the numbers $x_i$ we can find two elements $50 > x_k \\ge x_l > 1$ — if this were not true, then either there is some $x_i = 50$, implying\n$$\n\\sum_{i=1}^{n} x_i (100 - x_i) \\ge 50 \\cdot (100 - 50) > 1515,\n$$\nor in view of $s \\ge 82$, we would have $x_i = 1$ for at least $82 - 49 = 33$ indices $i$, implying\n$$\n\\sum_{i=1}^{n} x_i (100 - x_i) \\ge 33 \\cdot 1 \\cdot (100 - 50) = 33 \\cdot 50 > 1515.\n$$\nHence, we have $50 > x_k \\ge x_l > 1$ for some $k \\ne l$ and we shall prove that\n$x_k(100-x_k)+x_l(100-x_l) > (x_k+1)(100-(x_k+1))+(x_l-1)(100-(x_l-1))$.\n\nExpanding, gathering similar terms and simplifying we get that this inequality is equivalent to the inequality $2x_k-2x_l > -2$, which is true since $x_k \\ge x_l$. So, replacing $x_k$ and $x_l$ by $x_k+1$ and $x_l-1$, we decrease the sum $\\sum_i x_i(100-x_i)$ and after a finite number of such steps either one of the elements $x_i$ becomes equal to 50 or we have $x_i = 1$ for at least 33 indices $i$, yielding a contradiction in both cases as it is shown above.\n\nTherefore, we have $16 \\le s \\le 18$. Since $S = 100s - 1515$ is odd, we have $s$ is also odd because $s$ and $S$ have the same parity. So $s = 17$ and $S = 185$.\n\nIf $n = 2$ we easily get the solution $\\{x_1, x_2\\} = \\{4, 13\\}$.\n\nSuppose now that $n \\ge 3$. Then for any $k$ we have $x_k^2 \\le 185$ and\n$$\n185 = x_k^2 + \\sum_{i \\ne k} x_i^2 < x_k^2 + (17 - x_k)^2\n$$\n(since the sum of squares is less than the square of the sum), implying $x_k < 14$ and $(x_k - 4)(x_k - 13) > 0$. Therefore, all elements $x_i$ are not greater than 3. But then we have $S = \\sum x_i^2 \\le n \\cdot 9 \\le 17 \\cdot 9 < 185$, a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25380,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all polynomials $P(x)$ such that the equality\n$$\n(x - 1)P(x + 1) - (x + 1)P(x - 1) = 4P(x)\n$$\nholds for all real $x$.",
    "options": [],
    "answer": "P(x) = a x(x - 1)(x + 1), where a ∈ ℝ",
    "solution": "Answer: $P(x) = a x(x - 1)(x + 1)$, where $a \\in \\mathbb{R}$.\nSet $x = 1$ and $x = -1$ in the initial identity\n$$\n(x - 1)P(x + 1) - (x + 1)P(x - 1) = 4P(x). \\quad (1)\n$$\nThus we obtain $-2P(0) = 4P(1)$ and $-2P(0) = 4P(-1)$ respectively. Setting $x = 0$ in (1), we obtain $-P(1) - P(-1) = 4P(0)$, so, taking into account two previous equalities, we have $P(-1) = P(0) = P(1) = 0$.\n\nTherefore $0$, $1$, and $-1$ are zeroes of the polynomial $P(x)$, hence $P(x) = x(x - 1)(x + 1)Q(x)$, where $Q(x)$ is some polynomial. Setting this representation of $P(x)$ in (1), we obtain\n$$\n(x - 1)(x + 1)x(x + 2)Q(x + 1) - (x + 1)(x - 1)(x - 2)xQ(x - 1) = \\\\\n= 4x(x - 1)(x + 1)Q(x),\n$$\nwhence\n$$\n(x + 2)Q(x + 1) - (x - 2)Q(x - 1) = 4Q(x) \\quad \\text{for all } x \\in \\mathbb{R}. \\quad (2)\n$$\nSet $x = 2$ in (2), then we obtain $Q(3) = Q(2)$. By induction on $k$, show that $Q(k) = Q(2)$ for all positive integers $k \\ge 2$. Indeed, let $Q(k) = Q(2)$ for all $k = 2, 3, \\dots, m$, where $m \\ge 3$. Then setting $x = m$ in (2), we obtain $(m + 2)Q(m + 1) - (m - 2)Q(m - 1) = 4Q(m)$, so, taking into account our assumption, we obtain $(m+2)Q(m+1) = (m+2)Q(2)$, i.e. $Q(m+1) = Q(2)$. It follows that $Q(k) = Q(2)$ for all positive integers $k \\ge 2$.\n\nSince the polynomial $Q(x) - Q(2)$ has zero value at infinitely many points, we conclude that this polynomial is equal to zero identically, i.e. $Q(x) = Q(2) = a$ for some real number $a$.\n\nTherefore, the polynomial satisfying (1) needs to have the form $P(x) = a x(x-1)(x+1)$. It is easy to verify that any polynomial of such form (for any real $a$) satisfies (1).",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 25381,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs $(m, n)$ of positive integers $m$ and $n$ such that for all polynomial $P(x)$ with real coefficients and $\\deg P(x) = m$ there exists a polynomial $Q(x)$ with real coefficients and $\\deg Q(x) = n$ such that $Q(P(x))$ is divisible by $Q(x)$.\n(A. Mirotin, S. Mazanik, I. Voronovich)",
    "options": [],
    "answer": "All pairs with m odd and any positive n; or with m even and n even.",
    "solution": "Answer: all $(m, n)$ with odd $m$ and arbitrary $n$; or $(m, n)$ with even $m$ and even $n$.\n(Solution of A. Zhuk.)\n\n1. Let $m$ be odd. Show that any positive $n$ is appropriate.\nIndeed, if $P(x) \\equiv x$, then $Q(P(x)) \\not\\equiv Q(x)$ for any $Q \\in \\mathbb{R}[x]$.\nIf $P(x) \\not\\equiv x$, then $P(x) - x$ is a polynomial of odd degree, hence it has a real root, say, $a$. Then for any $n \\in \\mathbb{N}$ consider $Q(x) = (x-a)^n$. We have $Q(P(x)) = (P(x)-a)^n$. Note that $P(x)-a = (P(x)-x+(x-a)) \\not\\equiv (x-a)$. Hence $(P(x)-a)^n \\not\\equiv (x-a)^n = Q(x)$, as we need.\n\n2. Now, let $m$ be even. First, show that $n$ cannot be odd. Set, for example, $P(x) = x^m + x + 1$. Then $P(x) - x > 0$ for any $x \\in \\mathbb{R}$. If $n$ is odd, then $Q(x)$ has real roots. Let $c$ be the largest real root of $Q$. That is\n$$\nQ(x) = b(x-c) \\prod_{i=1}^{k} (x-a_i) \\cdot \\prod_{j=1}^{l} q_j(x),\n$$\nwhere all $q_j(x)$ are monic quadratic polynomials with negative discriminants, $c \\ge a_i$ for all $i = 1, 2, ..., k$, $b \\ne 0$.\nThen\n$$\nQ(P(c)) = b(P(c)-c) \\prod_{i=1}^{k} (P(c)-a_i) \\cdot \\prod_{j=1}^{l} q_j(P(c)).\n$$\nNote that $P(c)-c > 0$, $P(c)-a_i > c-a_i \\ge 0$, $(\\forall i = 1, ..., k)$, $q_j(P(c)) > 0$ $(\\forall j = 1, ..., l)$, that is $Q(P(c)) \\ne 0$. Hence, $c$ is not a root of $Q(P(x))$, so $Q(P(x)) \\not\\equiv Q(x)$.\n\nLet now both $m$ and $n$ be even, $n = 2k$.\nIf $P(x) - x$ has a real root $a$, then, as above, we set $Q(x) = (x-a)^n$, and we are done.\nLet $P(x) - x$ has no real roots. Let $z$ and $\\bar{z}$ be any complex-conjugate roots of $P(x)$, that is $P(x) - x$ is divisible by $p(x) = (x-z)(x-\\bar{z}) \\in \\mathbb{R}[x]$. Set $Q(x) = (p(x))^k$. Then $Q(P(x)) = (p(P(x)))^k$. It suffices to prove that $p(P(x)) : p(x)$. Note that $p(P(x)) = p(P(x)) - p(x) + p(x)$ and\n$$\n(p(P(x)) - p(x)) : (P(x) - x) : p(x).\n$$\nTherefore,\n$$\nQ(P(x)) = (p(P(x)))^k : (p(x))^k = Q(x).\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25382,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest positive integer $n$ such that the number $2013n$ can be presented as the difference of two cubes of positive integer numbers.",
    "options": [],
    "answer": "39",
    "solution": "Answer: $n=39$. (Solution of A. Semchankau, A. Zhuk.) Let\n$$\n2013n = a^3 - b^3. \\qquad (1)\n$$\nThen\n$$\n(1) \\Leftrightarrow 61 \\cdot 11 \\cdot 3n = 2013n = (a-b)^3 + 3ab(a-b) \\Rightarrow (a-b) \\ge 3,\n$$\ni.e. $(a-b) = 3k, k \\in \\mathbb{N}$. So $61 \\cdot 11 \\cdot 3n = 3^3 k^3 + 3^2 abk$, whence $n \\ge 3$, i.e. $n = 3m, m \\in \\mathbb{N}$. Then (1) can be rewritten as\n$$\n3k^3 + abk = 61 \\cdot 11m. \\qquad (2)\n$$\nNote that if $k=11, b=10, a=43$, then $m=13$, i.e. $n=39$. Show that $n=39$ is the smallest possible value of $n$, i.e. $m=13$ is the smallest possible value of $m$. Indeed, from (2) it follows that $k(3k^2+3kb+b^2) \\ge 11$, and it is easy to show that $k \\ge 11$. But for $k \\ge 22$ we have $m > 13$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25383,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs $(f, h)$ of functions $f$ and $h$, $f : \\mathbb{R} \\to \\mathbb{R}$, $h : \\mathbb{R} \\to \\mathbb{R}$, such that the equality $f(x^2 + y h(x)) = x h(x) + f(xy)$ holds for all real $x$ and $y$.",
    "options": [],
    "answer": "Either (i) f(x) = c for all x, and h(x) = 0 for x ≠ 0 with h(0) = a arbitrary; or (ii) f(x) = x + b for all x and h(x) = x for all x, where a, b, c are arbitrary real constants.",
    "solution": "Answer: either $f(x) = c$, $h(x) = \\begin{cases} 0, & x \\neq 0, \\\\ a, & x = 0, \\end{cases}$ where $a$ and $c$ are arbitrary constants or $f(x) = x + b$, $h(x) = x$, where $b$ is an arbitrary constant.\n\nLet $h(0) = a$. Set $x = 0$ in the initial equation\n$$\nf(x^2 + y h(x)) = x h(x) + f(xy) \\quad \\text{for all } x, y \\in \\mathbb{R}. \\quad (*)\n$$\nWe obtain $f(a y) = f(0) = c$ for all $y$.\nIf $a \\neq 0$, then $a y$ admits all real values, so the function $f(x) = c$ is identically constant. So the equation has the form $c = x h(x) + c$ or $x h(x) = 0$, therefore $h(x) = 0$ for $x \\neq 0$, and $h(0) = a$ admits any value. As it is easy to verify the obtained pair of the functions $(f(x), h(x))$ satisfies $(*)$.\n\nLet $a = 0$, i.e. $h(0) = 0$. Note that if $h(x_0) \\neq x_0$ for some $x_0$ ($x_0 \\neq 0$), then there exists $y_0$ such that $x_0^2 + y_0 h(x_0) = x_0 y_0$ (indeed, it suffices to set $y_0 = x_0^2/(x_0 - h(x_0))$). Setting $x = x_0$, $y = y_0$, in the initial equation, we obtain $x_0 h(x_0) = 0$, i.e. $h(x_0) = 0$. Now setting $x = x_0$ in the initial equation, we have $f(x_0^2) = f(x_0 y)$ for all $y$. Since $x_0 \\neq 0$, we see that $x_0 y$ admits any real values, so $f(x) = c$ is identically constant. But this case is already considered above.\n\nIt remains to suppose that $h(x) = x$ for all $x$. Then $(*)$ can be rewritten in the form\n$$\nf(x^2 + y x) = x^2 + f(x y) \\quad \\text{for all } x, y \\in \\mathbb{R}. \\quad (**)\n$$\nLet $f(0) = b$. Setting $y = 0$ in $(**)$, we obtain $f(x^2) = x^2 + b$ for all $x$, i.e. $f(x) = x + b$ for all nonnegative $x$. If we set $y = -x$, then we obtain $f(-x^2) = -x^2 + b$, i.e. $f(x) = x + b$ for all nonpositive $x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25384,
    "subject": "Mathematics (Multi-modal)",
    "question": "Ann and Bob play the following game. They, in turn (Ann starts), replace one of the asterisks in the expression $\\overline{*****}$ by one of the digits from $0$ to $9$ (each digit can be used exactly one time). Ann wins if the obtained $5$-digit number is divisible by $11$, otherwise Bob wins.\nWho of the players wins if both of them play to win?",
    "options": [],
    "answer": "Bob wins",
    "solution": "Answer: Bob wins.\n\nIt is well-known that a natural number $n$ is divisible by $11$ if and only if $(S_o - S_e) \\div 11$, where $S_o, S_e$ are the sums of the digits on the odd and even positions respectively in the decimal representation of $n$.\n\nLet Ann use the numbers $a$, $c$, and $y$ for her first, second, and third moves respectively; let Bob use the numbers $b$ and $x$ for his first and second moves respectively.\n\nThe following cases are possible after two Ann's moves and one Bob's move:\n\n1. $a$ and $c$ occupy two odd positions, $b$ occupies the even position of the asterisks;\n2. $a$ and $c$ occupy two even positions, $b$ occupies the odd position of the asterisks (different from the first position);\n3. $a$ and $c$ occupy one even and one odd positions, $b$ occupies the even position of the asterisks.\n\nLet $X = \\{x_1, x_2, \\dots, x_7\\}$ be the set of unused digits before the second Bob's move.\n\nIn the first case Bob replaces the asterisk on the even position by $x$. Then $S_o = a + c + y$, $S_e = b + x$. Let $r = b - a - c$. Now the congruence $S_o \\equiv S_e \\pmod{11}$ is equivalent to the congruence $y \\equiv x + r \\pmod{11}$.\n\nIf $r \\equiv 0 \\pmod{11}$, then $B$ can use any digit from the set $X$ as the number $x$ because the congruence $y \\equiv x \\pmod{11}$ is impossible if $x$ and $y$ are the distinct digits.\n\nLet $r \\not\\equiv 0 \\pmod{11}$. Suppose that for any digit $x_i \\in X$ there exists a digit $y_i \\ne x_i$, $y_i \\in X$ such that $y_i \\equiv x_i + r \\pmod{11}$, $i = 1, 2, \\dots, 7$. Summing these seven congruences we obtain\n$$\ny_1 + y_2 + \\dots + y_7 \\equiv x_1 + x_2 + \\dots + x_7 + 7r \\pmod{11}.\n$$\nSince it is evident that $y_1 + y_2 + \\dots + y_7 = x_1 + x_2 + \\dots + x_7$, we see that the obtained congruence is equivalent to the congruence $7r \\equiv 0 \\pmod{11}$, which is impossible since $r \\not\\equiv 0 \\pmod{11}$. Therefore, there exists a number $x \\in X$ such that $y \\not\\equiv x + r \\pmod{11}$ for any digit $y \\in X$ different from $x$. So Bob can use this number $x$ for his second move to win.\n\nConsider two other cases. In these cases Bob replaces the asterisk on the odd position (different from the first one) by $x$. Then we have\n$$\nS_o - S_e = (b + x + y) - (a + c) \\quad (1)\n$$\nfor the second case, and\n$$\nS_o - S_e = (c + x + y) - (a + b) \\quad (2)\n$$\nor\n$$\nS_o - S_e = (a + x + y) - (c + b) \\quad (3)\n$$\nfor the third one.\n\nSet $r = a + c - b$ in (1), $r = a + b - c$ in (2), and $r = b + c - a$ in (3).\n\nTherefore, the congruence $S_o \\equiv S_e \\pmod{11}$ is equivalent to the congruence $x+y \\equiv r \\pmod{11}$ for all cases. Suppose that for any digit $x_i \\in X$ there exists a digit $y_i \\in X$ such that $x_i + y_i \\equiv r \\pmod{11}$, $y_i \\neq x_i$, $i = 1, 2, \\dots, 7$. It follows that all digits from $X$ are partitioned into the pairs of distinct digits with the sum which is congruent to $r$ modulo $11$, but it is impossible since the number of digits in $X$ is odd. Therefore, there exists a number $x \\in X$ such that $y+x \\not\\equiv r \\pmod{11}$ for any digit $y \\in X$ different from $x$. So Bob can use this number $x$ for his second move to win.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25385,
    "subject": "Mathematics (Multi-modal)",
    "question": "$N$ segments are arranged inside a unit circle $\\Gamma$. The sum of the lengths of all these segments is equal to $2\\sqrt{N}$.\nProve that there exists a concentric with $\\Gamma$ circumference intersecting at least two of these segments.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider $360^\\circ$ rotation of $\\Gamma$ (together with all segments) about its center. Under this rotation each of the segments covers some ring with the center at the center of $\\Gamma$. If we show that the sum of the areas of all these rings is not less than $\\pi$, then the statement will be proved.\n\nWe estimate the area $S_{AB}$ of the ring covered by the segment $AB$ under the rotation. Let the altitude from the center $O$ of $\\Gamma$ onto the line containing $AB$ meet the line at point $C$. Consider two cases depending on the position of $C$.\n\nLet $C$ belong to $AB$. Then the area $S_{AB}$ is equal to\n$$\nS_{AB} = \\pi \\max\\{OA^2 - OC^2, OB^2 - OC^2\\} = \\pi \\max\\{AC^2, BC^2\\} \\ge \\pi AB^2/4.\n$$\n\nNow let $C$ do not belong to $AB$. Then\n$$\n\\begin{align*} S = \\pi |OB^2 - OA^2| &= \\pi |CB^2 - CA^2| = \\pi (AB^2 + 2AB \\cdot AC) \\ge \\\\ &\\ge \\pi AB^2 > \\pi AB^2/4. \\end{align*}\n$$\n\nLet $d_1, \\dots, d_N$ be the lengths of the segments. Then the sum of the areas of all rings is not less than $\\frac{\\pi}{4}(d_1^2 + \\dots + d_N^2) \\ge \\frac{\\pi}{4N}(d_1 + \\dots + d_N)^2 \\ge \\pi$, as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25386,
    "subject": "Mathematics (Multi-modal)",
    "question": "The odd number of the asterisks are written on the blackboard: $\\underbrace{**\\dots*}_{2n+1}$.\nAnn and Bob play the following game. They, in turn (Ann starts), replace one of the asterisks in the expression $\\underbrace{**\\dots*}_{2n+1}$ by any of the digits from $0$ to $9$ (the first left asterisk cannot be replaced by $0$). Ann wins if the obtained number is divisible by $11$, otherwise Bob wins.\nWho of the players wins if both of them play to win?",
    "options": [],
    "answer": "Bob",
    "solution": "Answer: Bob wins.\nIt is well-known that a natural number $n$ is divisible by $11$ if and only if $(S_o - S_e) \\div 11$, where $S_o$, $S_e$ are the sums of the digits on the odd and even positions respectively in the decimal representation of $n$.\nLet $2n + 1$ ($n \\in \\mathbb{N}$) asterisks be written on the blackboard: $\\underbrace{**\\dots*}_{2n+1}$.\nShow Bob's winning strategy. If Ann replaces some asterisk (different from the first one) by $c$, then, in answer, Bob replaces asterisk of the opposite parity by the same digit $c$ (note that Bob always chooses the asterisk different from the first one). Thus, if at the end Ann replaces the first asterisk by some digit $d$, then $S_o - S_e = d$. Since $d \\neq 0$ we see that the obtained number is not divisible by $11$.\nIf Ann replaces the first asterisk by some digit $c \\neq 0$ and it is not her last move, then Bob replaces some even asterisk by $c-1$, and further he keeps the strategy described above. As the result in the end Ann must replace some odd asterisk by some digit $d$. Therefore, in this case $S_o - S_e = d + c - (c - 1) = d + 1$. Since $d$ is a digit we have $0 < d + 1 < 11$, so the obtained number is not divisible by $11$, and Bob wins.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25387,
    "subject": "Mathematics (Multi-modal)",
    "question": "Several small circles are arranged inside a unit circle $\\Gamma$. The sum of the perimeters of all these small circles is not less than $\\pi$ and none of them includes the center of $\\Gamma$.\nProve that there exists a concentric with $\\Gamma$ circumference intersecting at least two of these small circles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $r_1, r_2, \\dots, r_k$ be radii of small circles. By condition,\n$$\n2\\pi(r_1 + r_2 + \\dots + r_k) \\ge \\pi,\n$$\ni. e.\n$$\nr_1 + r_2 + \\dots + r_k \\ge 1/2. \\quad (*)\n$$\nConsider $360^\\circ$ rotation of $\\Gamma$ (together with all small circles) about its center. Under this rotation each of small circles covers some ring with the center at the center of $\\Gamma$. The width $d_i$ of the ring covered by the small circle with the radius $r_i$ is equal to $2r_i$. If all covered rings have no common points, then\n$$\nd_1 + d_2 + \\dots + d_k = 2(r_1 + r_2 + \\dots + r_k) < 1,\n$$\ncontrary to $(*)$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25388,
    "subject": "Mathematics (Multi-modal)",
    "question": "A code lock has 10 keys labeled by digits from 0 to 9. The lock is opened by entering a code of 4 keys that are not necessarily different. It is opened as soon as the correct keys are pressed consequently, no matter what keys have been pressed earlier. For example, if the code happens to be $2002$, the sequence $45677892002$ will open the lock, but the sequence $20202$ will not.\nWhat is the minimal length of a sequence that will guarantee opening the lock regardless of what code is?",
    "options": [],
    "answer": "10003",
    "solution": "Answer: $10003$.\nWe need to find the shortest sequence of decimal digits that includes every 4-digit sequence as a (consecutive) subsequence. Since there are $10^4$ possible codes, and an $n$-digit sequence has at most $n-3$ distinct 4-digit subsequences, the answer must be at least $10^4 + 3 = 10003$.\n\nWe will construct a sequence of length $10^4+3$ that fulfills the condition. We define a directed graph $G = (V, E)$ whose vertex set $V$ consists of all 3-digit sequences. The edge set $E$ contains all ordered pairs of the form $(abc, bcd)$. (This implies that also pairs $(aaa, aaa)$ are edges.) The in-degree of a vertex is defined as the number of edges directed toward the vertex, obviously, the in-degree of every vertex in $G$ is $10$. Symmetrically, the out-degree of every vertex is also $10$. Note also that $G$ is (strongly) connected.\n\nBy a well-known argument, a connected directed graph has an Euler circuit if and only if the in-degree of any vertex equals its out-degree. We can now construct the required sequence as follows. Start from any vertex, say $000$, and begin the sequence by the digits of that vertex. Follow the Euler circuit, and at each edge $(abc, bcd)$ add $d$ to the sequence. It is easily proved by induction that after the edge $(abc, bcd)$ has been followed, the four last digits of the sequence are $ABCD$. Since the Euler circuit passes through every edge, the resulting sequence includes every 4-digit sequence. There are $10^3$ vertices, each of which is the starting point of $10$ edges, so all in all there are $10^4$ edges. Thus the length of the sequence is $3 + 10^4$ as required.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 25389,
    "subject": "Mathematics (Multi-modal)",
    "question": "$N$ segments are arranged inside a unit circle $\\Gamma$. The sum of the lengths of all these segments is equal to $2\\sqrt{N}$.\nProve that there exists a concentric with $\\Gamma$ circumference intersecting at least two of these segments.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider $360^\\circ$ rotation of $\\Gamma$ (together with all segments) about its center. Under this rotation each of the segments covers some ring with the center at the center of $\\Gamma$. If we show that the sum of the areas of all these rings is not less than $\\pi$, then the statement will be proved.\n\nWe estimate the area $S_{AB}$ of the ring covered by the segment $AB$ under the rotation. Let the altitude from the center $O$ of $\\Gamma$ onto the line containing $AB$ meet the line at point $C$. Consider two cases depending on the position of $C$.\n\nLet $C$ belong to $AB$. Then the area $S_{AB}$ is equal to\n$$\nS_{AB} = \\pi \\max\\{OA^2 - OC^2, OB^2 - OC^2\\} = \\pi \\max\\{AC^2, BC^2\\} \\geq \\pi AB^2/4.\n$$\n\nNow let $C$ do not belong to $AB$. Then\n$$\n\\begin{align*}\nS &= \\pi|OB^2 - OA^2| = \\pi|CB^2 - CA^2| = \\pi(AB^2 + 2AB \\cdot AC) \\geq \\\\\n&\\geq \\pi AB^2 > \\pi AB^2/4.\n\\end{align*}\n$$\n\nLet $d_1, \\dots, d_N$ be the lengths of the segments. Then the sum of the areas of all rings is not less than $\\frac{\\pi}{4}(d_1^2 + \\dots + d_N^2) \\ge \\frac{\\pi}{4N}(d_1 + \\dots + d_N)^2 \\ge \\pi$, as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25390,
    "subject": "Mathematics (Multi-modal)",
    "question": "Each of twenty two sets contains five elements. An intersection of any two of these sets contains exactly two elements.\nProve that the intersection of all these sets contains exactly two elements.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume there are no two such elements. Let $A_1, \\ldots, A_{22}$ be the initial sets, $|A_i| = 5$, $i = 1, \\ldots, 22$.\nPut $S = \\bigcup_{i=1}^{22} A_i$ and $a(x, y) = |\\{i \\mid \\{x, y\\} \\subset A_i\\}|$ for $x, y \\in S$.\nAssume there are five sets that contain $x$ and $y$, without loss of generality, $A_1, A_2, A_3, A_4, A_5$. Then a set $A_j$ with $\\{x, y\\} \\not\\subset A_j$ must contain an element from $A_i \\setminus \\{x, y\\}$ for $i = 1, 2, 3, 4, 5$. Hence, $\\{1, 2\\} \\cap A_j = \\emptyset$ and a contradiction $|A_1 \\cap A_j| = 2$ follows.\nConsequently, we can assume that $a(x, y) \\le 4$ for all $x, y \\in S$.\nThere are exactly 10 unordered pairs contained in $A_1$, and every $A_i$, $i = 2, 3, \\ldots, 22$, contains exactly one of them. Hence, one of the pairs is contained in $A_1$ and three other sets, i.e. there are $x, y \\in S$ with $a(x, y) = 4$.\nWithout loss of generality, we assume that $x = 1, y = 2$ and that the four sets containing 1 and 2 are:\n$$\nA_1 = \\{1, 2, 3, a, b\\}, \\quad A_2 = \\{1, 2, 4, c, d\\}, \\quad A_3 = \\{1, 2, 5, e, f\\}, \\quad A_4 = \\{1, 2, 6, g, h\\}.\n$$\nFurthermore, we can assume that $A_5 = \\{1, 3, 4, 5, 6\\}$. To meet the condition $|A_i \\cap A_j| = 2$ for $i = 1, 2, 3, 4$ every $A_j$ with $j > 4$ must contain either 1 or 2. Assume that $1 \\in A_j$ for $j = 5, 6, \\ldots, 14$. Each of the nine sets $A_j$, $j = 6, \\ldots, 14$, must contain $3, 4, 5$, or 6 because of $|A_5 \\cap A_j| = 2$. Hence, there is an $x \\in \\{3, 4, 5, 6\\}$ with $a(1, x) \\ge 5$, a contradiction. Consequently, at most nine of the sets $A_j$, $j = 5, \\ldots, 22$, contain 1. Analogously, at most nine of the sets $A_j$, $j = 5, \\ldots, 22$, contain 2. It follows that exactly nine of the sets $A_j$, $j = 5, \\ldots, 22$, contain 1, without loss of generality, the sets $A_5, \\ldots, A_{13}$.\nTaking into account that $a(x, y) \\le 4$ for all $x, y \\in S$ and $|A_i \\cap A_j| = 2$ for $1 \\le i < j \\le 13$, without loss of generality, we can assume\n$$\nA_6 = \\{1, 3, c, e, g\\}, \\quad A_7 = \\{1, 3, d, f, h\\}, \\quad A_8 = \\{1, 4, a, e, h\\},\n$$\n$$\nA_9 = \\{1, 4, b, f, g\\}, \\quad A_{10} = \\{1, 5, a, d, g\\}, \\quad A_{11} = \\{1, 5, b, c, h\\},\n$$\nand there is no proper choice left for $A_{12}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25391,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs $(p; q)$ of prime numbers $p$ and $q$ satisfying the equality\n$$\np^{5} + p^{3} + 2 = q^{2} - q.\n$$",
    "options": [],
    "answer": "[[2, 7], [3, 17]]",
    "solution": "Answer: $(p; q) = (2; 7)$ and $(p; q) = (3; 17)$.\nIf $p = 2$ or $p = 3$, then the obtained quadratic equations with respect to $q$ have two prime solutions: $q = 7$ and $q = 17$.\n\nNow let $p > 3$. We rewrite the initial equation in the form $p^3(p^2 + 1) = (q + 1)(q - 2)$. Note that the greatest common divisor of the numbers $q + 1$ and $q - 2$ is equal to either $1$ or $3$. Since $(q + 1)(q - 2) \\div p^3$ and $p \\neq 3$ we see that exactly one of the numbers $q + 1$ and $q - 2$ can be divisible by $p$, and so exactly one of them is divisible by $p^3$. If $(q + 1) \\div p^3$, then $q + 1 \\geq p^3$,\n\nand if $(q - 2) \\nmid p^3$, then $q - 2 \\ge p^3$. In any case $q \\ge p^3 - 1$. Then we obtain $p^5 + p^3 = (q + 1)(q - 2) \\ge p^3(p^3 - 3)$, so $p^2 + 1 \\ge p^3 - 3$, i.e. $0 \\ge p^3 - p^2 - 4 = (p - 2)(p^2 + p + 2)$, which is impossible for $p > 2$. Therefore, there are no solutions different from the solutions mentioned above.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25392,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs $(n; p)$ of natural numbers $n$ and prime numbers $p$ satisfying the equality $p^8 - p^4 = n^5 - n$.",
    "options": [],
    "answer": "(n, p) = (3, 2)",
    "solution": "Answer: $(n; p) = (3; 2)$.\n\nIt is clear that $p \\ne n$. If $p = 2$, then $n \\ge 3$ and we have $2^8 - 2^4 = 240 = 3^5 - 3$, i.e. $p = 2$, $n = 3$ is a solution. On the other hand, if $n > 3$, then $n^5 - n = n(n^4 - 1) > 3(3^4 - 1) = 240$, i.e. for $p = 2$ there are no $n$ different from $3$ satisfying the initial equality.\n\nNow let $p > 2$. Then $p$ is an odd prime number and $n \\ge 3$. We rewrite the initial equality in the form\n$$\nn(n-1)(n+1)(n^2+1) = p^4(p^4-1). \\quad (*)\n$$\nNote that exactly one of four co-factors in the left-hand side of the equation can be divisible by $p$. Indeed, $n$ is coprime with any of numbers $n-1, n+1, n^2+1$. From the equalities $n+1 = (n-1)+2$, $n^2+1 = (n-1)^2+2(n-1)+2$, $n^2+1 = (n+1)^2 - 2(n+1) + 2$ it follows that the greatest common divisor of any two of three numbers $n-1, n+1, n^2+1$ is equal to $1$ or $2$. Therefore, any two of them have not $p$ as a common divisor.\n\nThus, exactly one of four co-factors in the left-hand side of $(*)$ is divisible by $p$, and so, it is divisible by $p^4$. Then this co-factor is not less than $p^4$. In any case $n^2 + 1 \\ge p^4$ or $n^2 \\ge p^4 - 1$. So $p^4(p^4 - 1) = n(n^2 - 1)(n^2 + 1) \\ge n(p^4 - 2)p^4$, whence $p^4 - 1 \\ge n(p^4 - 2) > 2(p^4 - 1)$, which is impossible. Therefore, the pair $(n; p) = (3; 2)$ is a unique solution of the given equation.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25393,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs $(n; p)$ of natural numbers $n$ and prime numbers $p$ satisfying the equality $p(p - 1) = 2(n^3 + 1)$.",
    "options": [],
    "answer": "(20, 127)",
    "solution": "Answer: $(n; p) = (20; 127)$.\nIt is easy to see that the given equality\n$$\np(p-1) = 2(n^3 + 1) \\qquad (1)\n$$\ndoes not hold for $p=2$ and positive integer $n$. So $p \\ge 3$ is an odd prime number. Then $(n+1)(n^2 - n + 1)$ is divisible by $p$.\n\n1. If $(n+1) \\nmid p$, then $n+1 = kp$ for some positive integer $k$. In particular, $n+1 \\ge p$. From (1) we have $p(p-1) = 2(n+1)(n^2-n+1) \\ge 2p(n^2-n+1)$, and so $p-1 \\ge 2n^2-2n+2$. Then $n \\ge p-1 \\ge 2n^2-2n+2$ or $2n^2-3n+2 \\le 0$, which is impossible.\n\n2. Therefore $n^2 - n + 1 \\nmid p$, i. e.\n$$\nn^2 - n + 1 = kp \\qquad (2)\n$$\nfor some positive integer $k$. Substituting (2) in (1), we obtain $p-1 = 2k(n+1)$ or\n$$\np = 2kn + 2k + 1. \\qquad (3)\n$$\nSubstituting (3) in (2), we obtain $n^2 - n + 1 = 2k^2n + 2k^2 + k$ or\n$$\nn^2 - (2k^2 + 1)n - (2k^2 + k - 1) = 0. \\qquad (4)\n$$\nThe discriminant $D$ of this quadratic equation with respect to $n$ is equal to $D = (2k^2 + 1)^2 + 4(2k^2 + k - 1)$ and $D$ is evidently an odd number, $D > (2k^2 + 1)^2$; on the other hand $D < (2k^2 + 5)^2$ (indeed,\n$$\n(2k^2+1)^2+4(2k^2+k-1) < (2k^2+5)^2 \\iff 4(2k^2+k-1) < (4k^2+6) \\cdot 4 \\iff k-7 < 2k^2.\n$$\nTherefore, $D = (2k^2+1)^2 + 4(2k^2+k-1) = (2k^2+3)^2$ ($D$ must be perfect square, otherwise (4) has no integer roots). Thus,\n$$\n4(2k^2+k-1) = (2k^2+3)-(2k^2+1) = ((4k^2+4) \\cdot 2 \\iff 2k^2+k-1 = 2k^2+2\n$$\n$$\n\\iff k = 3.\n$$\nThen (4) has the form $n^2 - 19n - 20 = 0$, so $n = 20$. From (3) we obtain $p = 2 \\cdot 3 \\cdot 20 + 2 \\cdot 3 + 1 = 127$, and $p$ is the prime number indeed.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25394,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs $(n; p)$ of natural numbers $n$ and prime numbers $p$ satisfying the equality $p^8 - p^4 = n^5 - n$.",
    "options": [],
    "answer": "n = 3, p = 2",
    "solution": "Answer: $(n; p) = (3; 2)$.\nIt is clear that $p \\ne n$. If $p=2$, then $n \\ge 3$ and we have $2^8 - 2^4 = 240 = 3^5 - 3$, i.e. $p=2$, $n=3$ is a solution. On the other hand, if $n > 3$, then $n^5 - n = n(n^4 - 1) > 3(3^4 - 1) = 240$, i.e. for $p=2$ there are no $n$ different from $3$ satisfying the initial equality.\n\nNow let $p > 2$. Then $p$ is an odd prime number and $n \\ge 3$. We rewrite the initial equality in the form\n$$\nn(n-1)(n+1)(n^2+1) = p^4(p^4-1). \\quad (*)\n$$\nNote that exactly one of four co-factors in the left-hand side of the equation can be divisible by $p$. Indeed, $n$ is coprime with any of numbers $n-1$, $n+1$, $n^2+1$. From the equalities $n+1 = (n-1)+2$, $n^2+1 = (n-1)^2+2(n-1)+2$, $n^2+1 = (n+1)^2 - 2(n+1) + 2$ it follows that the greatest common divisor of any two of three numbers $n-1$, $n+1$, $n^2+1$ is equal to $1$ or $2$. Therefore, any two of them have not $p$ as a common divisor.\n\nThus, exactly one of four co-factors in the left-hand side of $(*)$ is divisible by $p$, and so, it is divisible by $p^4$. Then this co-factor is not less than $p^4$. In any case $n^2 + 1 \\ge p^4$ or $n^2 \\ge p^4 - 1$. So $p^4(p^4 - 1) = n(n^2 - 1)(n^2 + 1) \\ge n(p^4 - 2)p^4$, whence $p^4 - 1 \\ge n(p^4 - 2) > 2(p^4 - 1)$, which is impossible. Therefore, the pair $(n; p) = (3; 2)$ is a unique solution of the given equation.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25395,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest real number $x$ such that the inequality $x + c \\le (x + a)(x + b)$ holds for any triangle, where $a \\le b \\le c$ are the sides of the triangle.",
    "options": [],
    "answer": "1",
    "solution": "$x = 1$.\n\nFirst, we prove that if $a$, $b$, $c$ are the sides of a triangle, then the inequality\n$$\nx + c \\le (x + a)(x + b) \\quad (*)\n$$\nholds for $x = 1$. Indeed, we can rewrite $(*)$ as $x + c \\le x^2 + (a + b)x + ab$. It is easy to see that this inequality holds for $x = 1$ since $x^2 = x = 1$ and, by the triangle inequality, $(a + b)x = a + b > c$.\n\nNow we show that for any $x < 1$ there exists a triangle such that $(*)$ does not hold. Indeed, if $x < 0$, then it suffices to consider the triangle with the sides $a = 1 - x$, $b = 1 - x$, and $c = 2 - x$. If $0 \\le x < 1$, then it suffices to consider the regular triangle with the side $\\frac{1 - x}{2}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25396,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $I$ be the center of inscribed circle of the non-isosceles triangle $ABC$. The ray $AI$ meets circumscribed circle of the triangle $ABC$ at point $D$. The circle passing through $C$, $D$, and $I$ meets again the ray $BI$ at point $K$.\nProve that $BK = CK$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $O$ be the circumcenter of the triangle $\\triangle ABC$. Let $\\angle BAC = \\alpha$, $\\angle ABC = \\beta$, $\\angle ACB = \\gamma$. We construct the line passing through $D$ and $O$. Let $L$ be the point of intersection of this line and the ray $BI$. Since $AI$ is a bisector of the angle $BAC$, we have $BD = DC$, and so the line $DO$ is a perpendicular bisector of the segment $BC$. Therefore, $BL = CL$, i.e. the triangle $BLC$ is isosceles and $\\angle LBC = \\angle LCB = \\beta/2$ ($BI$ is the bisector of the angle $ABC$). Hence,\n\n![](attached_image_1.png)\n\n$$\n\\angle ILC = \\angle BLC = 180^{\\circ} - (\\angle LBC + \\angle LCB) = 180^{\\circ} - \\beta.\n$$\n\nOn the other hand,\n$$\n\\begin{align*}\n\\angle CDI &= \\angle CDA = 180^{\\circ} - (\\angle DAC + \\angle ACD) = \\\\\n&= [\\angle DCB = \\angle DAB = \\angle DAC = 0.5\\alpha, \\angle ACB = \\gamma, \\angle ACD = \\\\\n&= \\angle ACB + \\angle DCB = \\gamma + 0.5\\alpha] = 180^{\\circ} - (0.5\\alpha + \\gamma + 0.5\\alpha) = 180^{\\circ} - \\alpha - \\gamma = \\beta.\n\\end{align*}\n$$\n\nTherefore, $\\angle ILC + \\angle CDI = 180^{\\circ}$. It follows that the points $I, D, C$, and $L$ lie on the same circumference (passing through $I, D, C$). Thus, $K$ and $L$ coincide. Hence, $CK = CL = BL = BK$, as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25397,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a decimal fraction $d$ such that there are only digits 0, 1, and 2 in its decimal representation. It is known that if each 0 in the decimal representation of $d$ are replaced by 1, then the obtained decimal fraction is periodic; if each 1 in the decimal representation of $d$ are replaced by 2, then the obtained decimal fraction is periodic too.\nCan one claim that the decimal fraction $d$ is periodic ?\n(M. Karpuk)",
    "options": [],
    "answer": "Yes, the decimal fraction is periodic.",
    "solution": "Answer: yes, one can.\nSuppose that if we replace each 0 in the decimal fraction $d$ by 1 we obtain the decimal fraction $a$, and if we replace each 1 in the decimal fraction $d$ by 2 we obtain the decimal fraction $b$. Since both the fractions $a$ and $b$ are periodic, we see that both these numbers are rational numbers.\nIf we replace each 2 in the decimal fraction $a$ by 0 we obtain some decimal fraction $\\alpha$ which is evident periodic and so $\\alpha$ is a rational number. Similarly, if we replace each 0 in the decimal fraction $b$ by 1 we obtain some decimal fraction $\\beta$ which is also periodic, and so $\\beta$ is a rational number too.\nIt is easy to see that $d = \\beta - \\alpha$. Therefore, $d$ is a rational number and so the decimal fraction $d$ is periodic.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25398,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given triangle $ABC$ with $AC = (AB + BC)/2$. Let $BL$ be the bisector of the angle $ABC$; let $K$ and $M$ be the midpoints of $AB$ and $BC$ respectively.\nFind the value of the angle $KLM$ if $\\angle ABC = \\beta$.",
    "options": [],
    "answer": "90° - β/2",
    "solution": "Answer: $90^\\circ - \\beta/2$.\nLet point $N$ be marked on the side $AC$ such that $AN = 0.5 AB$. Then, by condition, $NC = AC - AN = 0.5(AB + BC) - 0.5 AB = 0.5 BC$. Therefore, $AN : NC = AB : BC$. Since $BL$ is a bisector of the angle $ABC$ we have $AL : LC = AB : BC$, so $L$ and $N$ coincide. Therefore the triangles $KAL$ and $MCL$ are isosceles and\n$$\n\\angle AKL = \\angle ALK = 0.5(180^\\circ - \\angle BAC), \\quad \\angle CML = \\angle CLM = 0.5(180^\\circ - \\angle BCA).\n$$\nThus,\n$$\n\\begin{aligned}\n\\angle KLM &= 180^\\circ - \\angle ALK - \\angle CLM \\\\\n&= 180^\\circ - 0.5(180^\\circ - \\angle BAC) - 0.5(180^\\circ - \\angle BCA) \\\\\n&= 0.5(\\angle BAC + \\angle BCA) \\\\\n&= 0.5(180^\\circ - \\angle ABC) \\\\\n&= 0.5(180^\\circ - \\beta) \\\\\n&= 90^\\circ - \\beta/2.\n\\end{aligned}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25399,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest real number $x$ such that the inequality $x + c \\le (x + a)(x + b)$ holds for any triangle, where $a \\le b \\le c$ are the sides of the triangle.",
    "options": [],
    "answer": "1",
    "solution": "**Answer:** $x = 1$.\n\nFirst, we prove that if $a$, $b$, $c$ are the sides of a triangle, then the inequality\n$$\nx + c \\le (x + a)(x + b) \\quad (*)\n$$\nholds for $x = 1$. Indeed, we can rewrite (*) as $x + c \\le x^2 + (a + b)x + ab$. It is easy to see that this inequality holds for $x = 1$ since $x^2 = x = 1$ and, by the triangle inequality, $(a + b)x = a + b > c$.\n\nNow we show that for any $x < 1$ there exists a triangle such that (*) does not hold. Indeed, if $x < 0$, then it suffices to consider the triangle with the sides $a = 1 - x$, $b = 1 - x$, and $c = 2 - x$. If $0 \\le x < 1$, then it suffices to consider the regular triangle with the side $\\frac{1 - x}{2}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25400,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $I$ be the center of inscribed circle of the non-isosceles triangle $ABC$. The ray $AI$ meets circumscribed circle of the triangle $ABC$ at point $D$. The circle passing through $C$, $D$, and $I$ meets again the ray $BI$ at point $K$.\nProve that $BK = CK$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $O$ be the circumcenter of the triangle $\\triangle ABC$. Let $\\angle BAC = \\alpha$, $\\angle ABC = \\beta$, $\\angle ACB = \\gamma$. We construct the line passing through $D$ and $O$. Let $L$ be the point of intersection of this line and the ray $BI$. Since $AI$ is a bisector of the angle $BAC$, we have $BD = DC$, and so the line $DO$ is a perpendicular bisector of the segment $BC$. Therefore, $BL = CL$, i.e. the triangle $BLC$ is isosceles and $\\angle LBC = \\angle LCB = \\beta/2$ ($BI$ is the bisector of the angle $ABC$). Hence,\n\n![](attached_image_1.png)\n\n$$\n\\angle ILC = \\angle BLC = 180^\\circ - (\\angle LBC + \\angle LCB) = 180^\\circ - \\beta.\n$$\n\nOn the other hand,\n$$\n\\begin{align*}\n\\angle CDI &= \\angle CDA = 180^\\circ - (\\angle DAC + \\angle ACD) = \\\\\n&= [\\angle DCB = \\angle DAB = \\angle DAC = 0.5\\alpha, \\angle ACB = \\gamma, \\angle ACD = \\\\\n&= \\angle ACB + \\angle DCB = \\gamma + 0.5\\alpha] = 180^\\circ - (0.5\\alpha + \\gamma + 0.5\\alpha) = 180^\\circ - \\alpha - \\gamma = \\beta.\n\\end{align*}\n$$\nTherefore, $\\angle ILC + \\angle CDI = 180^\\circ$. It follows that the points $I$, $D$, $C$, and $L$ lie on the same circumference (passing through $I$, $D$, $C$). Thus, $K$ and $L$ coincide. Hence, $CK = CL = BL = BK$, as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25401,
    "subject": "Mathematics (Multi-modal)",
    "question": "Three of six segments (four sides and two diagonals of a parallelogram) are painted red, and three others are painted green.\n\nProve that one can construct a triangle using the segments of the same color as its sides.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $BC = AD = a$, $AB = CD = b$, $AC = d_1$, $BD = d_2$. Let for the definiteness $\\angle ADC \\ge 90^\\circ$. Then $d_1 > a$, $d_1 > b$. Without loss of generality we suppose that $a \\ge b$. Then $a+a \\ge a+b > d_1 \\ge d_2$. Exactly two cases are possible.\n\n1. The diagonals have the different colors. If two adjacent sides $a$ and $b$ are painted the same color, then together with the diagonal of the same color they form the triangle congruent with (by three sides) either the triangle $ADB$ or the triangle $ADC$. Now let two opposite sides $a$ and $b$ have the same color with one of the diagonals. Then one can construct the triangle with these sides and the diagonal as its sides since $a+a > d_1$ and $a+a > d_2$.\n\n![](attached_image_1.png)\n\n2. Both the diagonals are painted the same color, say green. Then one of the side is also green, and the side equaled green side and two remaining sides are painted red. If the sides $a$, $a$, $b$ are painted the same color, then one can construct the triangle with $a$, $a$, $b$ as its sides. It follows from the inequalities $a+a \\ge a+b > b$. Suppose that $d_1, d_2, a$ are painted green and $b, b, a$ are painted red. One cannot form a triangle from the first triple only if $d_1 \\ge a+d_2$ or $d_2 \\ge a+d_1$ (since the inequality $d_1+d_2 > a$ certainly holds); one cannot form a triangle from the second triple only if $a \\ge 2b$. Let $d_1 \\ge a+d_2$. Then\n\nd_1 \\ge 2b + d_2, i.e. AC \\ge AB + BD + DC > AD + DC, a contradiction. Similarly, if $d_2 \\ge a + d_1$, then $d_2 \\ge 2b + d_1$, i. e. $BD \\ge BA + CD + AC > BA + AD$, a contradiction.\n\nTherefore, there always exist three segments of the same color such that one can construct the triangle with these segments as its sides.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25402,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Three of six segments (four sides and two diagonals of an isosceles trapezoid) are painted red, and three others are painted green.\nProve that one can construct a triangle using the segments of the same color as its sides.\n\nb) Is the previous statement true if these six segments are four sides and two diagonals of an arbitrary trapezoid?",
    "options": [],
    "answer": "No",
    "solution": "b) The statement may not hold.\n\na) See the solution of Problem D.4.\n\nb) Consider, for example, a trapezoid $ABCD$ such that its diagonal $AC$ is perpendicular to the bases $AD$ and $BC$ (see the Fig.). Let $CA = 1$. Let $AB, AC, CD$ be painted green and $BC, BD, AD$ be painted red. For these red segments the following inequality holds $BD > B'D = B'A + AD = BC + AD$\n\n(see the Fig.), so one cannot construct the triangle with these red segments as its sides.\nWe select the lengths of the bases $AD = a$, $BC = b$ so that one cannot construct the triangle with these green segments as its sides. It suffices to choose $a$ and $b$ such that the following inequality",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25403,
    "subject": "Mathematics (Multi-modal)",
    "question": "$2n$ girls and $2n$ boys take part in a dancing party. It is known that for any two girls the number of the boys who have a dance with exactly one of these two girls is equal to $n$.\nProve that for any two boys the number of the girls who have a dance with exactly one of these two boys is equal to $n$ too.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Number girls and boys by numbers from $1$ to $2n$. For each $i$-th girl we correspond a vector $S_i$ with $j$-th entry equal to $1$ if the $i$-th girl has a dance with the $j$-th boy, and equal to $-1$ otherwise. Then the condition is equivalent to $S_i \\cdot S_j^T = 0$ for every $i \\neq j$, because $S_i$ and $S_j$ differ at exactly $n$ positions.\n\nDefine the $2n \\times 2n$ matrix $S$ with rows $S_i$, $i = 1...2n$. Then $S \\cdot S^T = 2nI$, where $I$ is the identity $2n \\times 2n$ matrix. This implies, that $\\frac{1}{2n}S^T = S^{-1}$. Therefore $S^T \\cdot S = 2nI$, which means that every two columns differ at exactly $n$ positions, which is equivalent to the required statement.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25404,
    "subject": "Mathematics (Multi-modal)",
    "question": "Exactly one integer number from $1$ to $25$ is written in each cell of the $5 \\times 5$ square table (see the figure). Per move it is allowed:\n\n| 1 | 2 | 3 | 4 | 5 |\n|---|---|---|---|---|\n| 6 | 7 | 8 | 9 | 10 |\n| 11 | 12 | 13 | 14 | 15 |\n| 16 | 17 | 18 | 19 | 20 |\n| 21 | 22 | 23 | 24 | 25 |\n\n1) to choose any two cells with a common side then to increase (to decrease) the number in one of the cells by $2$ and to decrease (to increase respectively) the number in the other cell by $1$;\n2) to choose any two cells with a common vertex then to increase the number in one of the cells by $1$ and to decrease the number in the other cell by $1$;\n\na) Is it possible to obtain the table with the same numbers in all its cells after the finite number of these moves?\nb) Is it possible to obtain the table with $2$ in all its cells after the finite number of these moves?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Answer: **a)** it is possible; **b)** it is impossible.\n\na) We show that using the allowed moves we can decrease any number in the table by $3$ so that all other numbers in the table keep their values. We will not consider the whole table but only the number $x$ which will be decreased by $3$ and three more numbers $a$, $b$, and $c$ which occupy (together with $x$) the cells of some $2 \\times 2$ square).\n![](attached_image_1.png)\nSo we can consecutively decrease all numbers in the table so that to obtain their residues modulo $3$, i.e. to obtain the table:\n![](attached_image_2.png)\nConsider the $2 \\times 2$ corner squares:\n![](attached_image_3.png)\nAll these squares have the same form and per one move (+1 - 1) they can be transformed into the squares with $1$ in all their cells. So we can obtain the following table:\n![](attached_image_4.png)\nFurther, we transform the third row of the table:\n![](attached_image_5.png)\nIn a similar way we can transform the third column of the table. As the result we obtain the table with $1$ in all its cells.\n\nb) Consider the chess coloring of the table. For the definiteness we suppose that the corner cells are white. So, there are $13$ white and $12$ black cells in the table. All white cells are occupied with odd numbers, and all black cells are occupied with even numbers. Therefore, the sum $S_w$ of the numbers in the white cells is equal to $\\frac{1+25}{2} \\cdot 13 = 13 \\cdot 13$, and the sum $S_b$ of the numbers in the black cells is equal to $\\frac{2+24}{2} \\cdot 12 = 13 \\cdot 12$. So, $S_w - S_b = 13$.\nIt is easy to see that any allowed move does not change the residue modulo $3$ of the difference between the sums of the numbers in white and black cells. If all cells in the table are occupied with $2$, then this difference is equal to $2$. But $2 \\not\\equiv 13 \\pmod{3}$, so we cannot obtain the table with $2$ in all its cells.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 25405,
    "subject": "Mathematics (Multi-modal)",
    "question": "Exactly one integer number from $1$ to $49$ is written in each cell of the $7 \\times 7$ square table (see the figure). Per move it is allowed to choose any cell then simultaneously to increase (to decrease) the number in this cell by $1$ and to decrease (to increase respectively) the numbers in some two adjacent cells by $1$ (two cells are adjacent if they have a common side).\n\n<table><thead><tr><th>1</th><th>2</th><th>3</th><th>4</th><th>5</th><th>6</th><th>7</th></tr></thead><tbody><tr><td>8</td><td>9</td><td>10</td><td>11</td><td>12</td><td>13</td><td>14</td></tr><tr><td>15</td><td>16</td><td>17</td><td>18</td><td>19</td><td>20</td><td>21</td></tr><tr><td>22</td><td>23</td><td>24</td><td>25</td><td>26</td><td>27</td><td>28</td></tr><tr><td>29</td><td>30</td><td>31</td><td>32</td><td>33</td><td>34</td><td>35</td></tr><tr><td>36</td><td>37</td><td>39</td><td>39</td><td>40</td><td>41</td><td>42</td></tr><tr><td>43</td><td>44</td><td>45</td><td>46</td><td>47</td><td>48</td><td>49</td></tr></tbody></table>\n\na) Is it possible to obtain the table with the same numbers in all its cells after the finite number of these moves?\n\nb) Is it possible to obtain the table with $2013$ in all its cells after the finite number of these moves?",
    "options": [],
    "answer": "a) Yes. b) No.",
    "solution": "**a)** We show that using the allowed moves we can decrease any number in the table by $3$ so that all other numbers in the table keep their values. We will not consider the whole table but only the number $x$ which will be decreased by $3$ and three more numbers $a$, $b$, and $c$ which occupy (together with $x$) the cells of some $2 \\times 2$ square).\n\n![](attached_image_1.png)\n\nSo we can consecutively decrease all numbers in the table so that to obtain their residues modulo $3$, i.e. to obtain the table:\n\n![](attached_image_2.png)\n\nConsider the $6 \\times 6$ down-left corner square. It consists of the same four $3 \\times 3$ squares:\n\n![](attached_image_3.png)\n\nIt is easy to see that after two moves ($-1+1+1$) each of these $3 \\times 3$ squares is transformed into the square with $1$'s in all its cells. So we obtain the table:\n\n<table>\n<tr>\n<td>1</td>\n<td>2</td>\n<td>0</td>\n<td>1</td>\n<td>2</td>\n<td>0</td>\n<td>1</td>\n</tr>\n<tr>\n<td>1</td>\n<td>1</td>\n<td>1</td>\n<td>1</td>\n<td>1</td>\n<td>1</td>\n<td>2</td>\n</tr>\n<tr>\n<td>1</td>\n<td>1</td>\n<td>1</td>\n<td>1</td>\n<td>1</td>\n<td>1</td>\n<td>0</td>\n</tr>\n<tr>\n<td>1</td>\n<td>1</td>\n<td>1</td>\n<td>1</td>\n<td>1</td>\n<td>1</td>\n<td>1</td>\n</tr>\n<tr>\n<td>1</td>\n<td>1</td>\n<td>1</td>\n<td>1</td>\n<td>1</td>\n<td>1</td>\n<td>2</td>\n</tr>\n<tr>\n<td>1</td>\n<td>1</td>\n<td>1</td>\n<td>1</td>\n<td>1</td>\n<td>1</td>\n<td>0</td>\n</tr>\n<tr>\n<td>1</td>\n<td>1</td>\n<td>1</td>\n<td>1</td>\n<td>1</td>\n<td>1</td>\n<td>1</td>\n</tr>\n</table>\n\nFurther, we transform the first row of the table:\n\n$$\n\\begin{tabular}{|c|c|c|c|c|c|c|}\n\\hline\n1 & 2 & 0 & 1 & 2 & 0 & 1 \\\\\n\\hline\n\\multicolumn{7}{c}{$+1-1-1$} \\\\\n\\hline\n1 & 1 & 1 & 0 & 2 & 0 & 1 \\\\\n\\hline\n\\multicolumn{7}{c}{$-1+1+1$} \\\\\n\\hline\n1 & 1 & 1 & 1 & 1 & 1 & 1 \\\\\n\\hline\n\\end{tabular}\n$$\n\nIn a similar way we can transform the last column of the table. As the result we obtain the table with $1$ in all its cells.\n\n**b)** Consider the chess coloring of the table. For the definiteness we suppose that the corner cells are white. So, there are $25$ white and $24$ black cells in the table. All white cells are occupied with odd numbers, and all black cells are occupied with even numbers. Therefore, the sum $S_w$ of the numbers in the white cells is equal to $\\frac{1+49}{2} \\cdot 25 = 25 \\cdot 25$, and the sum $S_b$ of the numbers in the black cells is equal to $\\frac{2+48}{2} \\cdot 24 = 25 \\cdot 24$. So, $S_w - S_b = 25$.\n\nIt is easy to see that any allowed move does not change the residue modulo $3$ of the difference between the sums of the numbers in the white and black cells. If all cells in the table are occupied with the number $2013$, then this difference is equal to $2013$. But $2013 \\ne 25 \\pmod{3}$, so we cannot obtain the table with $2013$ in all its cells.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25406,
    "subject": "Mathematics (Multi-modal)",
    "question": "$2n$ girls and $2n$ boys take part in a dancing party. It is known that Bob has a dance with every girl, and Ann has a dance with every boy. Moreover, for any two girls the number of the boys who have a dance with exactly one of these two girls is equal to $n$.\n\nProve that\n\na) any girl, except for Ann, has a dance with exactly $n$ boys;\n\nb) any boy, except for Bob, has a dance with exactly $n$ girls.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We use the solution of Problem A.8.\n\na) Let Ann get number $1$. Each vector $S_i$, $i \\neq 1$ differs from $S_1$ at exactly $n$ positions and all entries of $S_1$ are equal to $1$. Therefore, $S_i$ have exactly $n$ entries equal to $1$, which proves the statement.\n\nb) The same proof as in a). We consider the columns instead of the rows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25407,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $I$ be the incenter of a triangle $ABC$. The circle passing through $I$ and centered at $A$ meets the circumference of the triangle $ABC$ at points $M$ and $N$.\nProve that the line $MN$ touches the incircle of the triangle $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\angle BAC = 2\\alpha$, $\\angle ABC = 2\\beta$, $\\angle BCA = 2\\gamma$. Since $AM = AN$, we obtain $\\angle AMN = \\angle ANM$. We have $\\angle AMN = \\angle ACN$ and $\\angle MNC = \\angle MAC$ (as inscribed angles). Further,\n$$\n\\begin{aligned}\n\\angle AKN &= \\angle MAK + \\angle AMK = \\\\\n&= \\angle MAC + \\angle AMN = \\\\\n&= \\angle MNC + \\angle ANM = \\angle ANC.\n\\end{aligned}\n$$\nIt follows that the triangles $AKN$ and $ANC$ are similar. Similarly, the triangles $ALM$ and $AMB$ are similar. Then $\\frac{AN}{AC} = \\frac{AK}{AN}$, so $AI^2 = AN^2 = AK \\cdot AC$.\nIn the same manner we see that $AI^2 = AL \\cdot AB$, so $AK \\cdot AC = AL \\cdot AB$, which shows that $L$, $K$, $C$, $B$ are concyclic. Then $\\angle AKL = \\angle ABC = 2\\beta$. Moreover, since $AI : AK = AC : AI$, it follows that the triangles $AIK$ and $ACI$ are similar. Then $\\angle AIK = \\angle ICA = 0.5\\angle BCA = \\gamma$. Since $\\angle IKC = \\angle IAK + \\angle AIK = \\alpha + \\gamma$, we have\n$$\n\\begin{aligned}\n\\angle IKL &= 180^\\circ - \\angle AKL - \\angle IKC = 180^\\circ - 2\\beta - \\alpha - \\gamma = \\\\\n&= 180^\\circ - 2(\\alpha + \\beta + \\gamma) + \\alpha + \\gamma = [\\alpha + \\beta + \\gamma = 90^\\circ] = \\alpha + \\gamma = \\angle IKC.\n\\end{aligned}\n$$\nSimilarly, $\\angle KLI = \\angle ILB$. Therefore, all bisectors of the quadrilateral $LBCK$ intersect at the same point $I$. Hence $LBCK$ is a circumscribed quadrilateral. Thus $MN$ touches the inscribed circle of the triangle $ABC$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25408,
    "subject": "Mathematics (Multi-modal)",
    "question": "The graph of the parabola $y = x^2$ is drawn on the Cartesian plane $Oxy$. A triangle $ABC$ is inscribed in the parabola so that its side $AB$ is parallel to the $Ox$ axis and point $C$ lies between the line $AB$ and the $Ox$ axis. It is known that the length of the side $AB$ is 1 less than the length of the altitude $CH$ of the triangle $ABC$.\nFind the value of the angle $ACB$.",
    "options": [],
    "answer": "45 degrees",
    "solution": "Answer: $\\angle ACB = 45^\\circ$.\nLet $A(a; a^2)$ and $C(c; c^2)$ (see the Fig.). Since $AB \\parallel Ox$, we see that $A$ and $B$ are symmetric with respect to the axis $Oy$. Then their ordinates are equal and abscissae differ by sign.\nHence $B(-a; a^2)$. Since $H$ belongs to $AB$ and $CH \\parallel Oy$, we have $H(c; a^2)$.\nThen $CH = a^2 - c^2$ and (without loss of generality we assume $a > 0$) $AB = 2a$. By condition,\n$$\na^2 - c^2 = 2a + 1. \\quad (*)\n$$\nMoreover, $AH = a - c$ and $BH = c + a$. By Pythagorean theorem, from the right-angled triangle $HCA$ and $HCB$ we obtain\n$$\nCA^2 = (a^2 - c^2)^2 + (a - c)^2 \\quad \\text{and} \\quad CB^2 = (a^2 - c^2)^2 + (c + a)^2.\n$$\nBy the law of cosines, from $\\triangle ABC$ we obtain\n$$\nAB^2 = CA^2 + CB^2 - 2CA \\cdot CB \\cdot \\cos \\angle ACB.\n$$\n![](attached_image_1.png)\n\nReplacing the lengths of $AB$, $CA$ and $CB$ by their obtained expressions and taking into account $(*)$, we obtain\n$$\n\\begin{align*} \n4a^2 &= (a^2 - c^2)^2 + (a-c)^2 + (a^2 - c^2)^2 + (c+a)^2 - \\\n&\\quad 2\\sqrt{(a^2 - c^2)^2 + (a-c)^2}\\sqrt{(a^2 - c^2)^2 + (c+a)^2} \\cos \\angle ACB = 2(a^2 - c^2)^2 \\\n&\\quad + 2(a^2 + c^2) - 2(a-c)\\sqrt{(a+c)^2 + 1}(a+c)\\sqrt{(a-c)^2 + 1} \\cos \\angle ACB = \\\n&\\quad \\phantom{+ 2(a^2 + c^2)} 2(a^2 - c^2)^2 + 2(a^2 + c^2) - \\\n&\\quad -2(a^2 - c^2)\\sqrt{(a^2 - c^2)^2 + (a+c)^2 + (a-c)^2 + 1} \\cos \\angle ACB \\\n&\\stackrel{(*)}{=} 2(2a+1)^2 + \\\n&\\quad +2(a^2 + a^2 - 2a - 1) - 2(2a+1)\\sqrt{(2a+1)^2 + 2a^2 + 2c^2 + 1} \\cos \\angle ACB = \\\n&\\stackrel{\\text{not, then from (3)}}{=} 12a^2 + 4a - 2(2a+1)\\sqrt{4a^2 + 4a + 1 + 2a^2 + 2(a^2-2a-1) + 1} \\cos \\angle ACB = \\\n&\\stackrel{\\text{not, then from (3)}}{=} 12a^2 + 4a - 2(2a+1)\\sqrt{8a^2} \\cos \\angle ACB = 12a^2 + 4a - 4\\sqrt{2a}(2a+1) \\cos \\angle ACB. \n\\end{align*}\n$$\nThus,\n$$\n4\\sqrt{2}a(2a+1) \\cos \\angle ACB = 8a^2 + 4a,\n$$\nwhence $\\cos \\angle ACB = 1/\\sqrt{2}$, and therefore, $\\angle ACB = 45^\\circ$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 25409,
    "subject": "Mathematics (Multi-modal)",
    "question": "The graph of the hyperbola $y = \\frac{1}{x}$ is drawn on the Cartesian plane $Oxy$. Three snails start simultaneously from the origin $O$ and move along the abscissa axis $Ox$ (each snail has its constant speed). Let $A(t)$, $B(t)$ and $C(t)$ be points on the graph of the hyperbola such that their abscissae are equal to the abscissae of the first, the second and the third snails at the moment $t$, respectively.\nProve that the area of the triangle $ABC$ is independent of time.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $a(t)$, $b(t)$, $c(t)$ be the abscissae of the first, the second and the third snails at the moment $t$, respectively, and $u_a, u_b, u_c$ be their speeds respectively.\nFirst, suppose that all snails move in the same direction. Without loss of generality we assume that the snails move along the positive part of the axis $Ox$ and $u_a \\le u_b \\le u_c$. Then (see Fig. 1)\n$$\nS(ABC) = S(ACC_1A_1) - S(ABB_1A_1) - S(BCC_1B_1) \\quad (1)\n$$\n![](attached_image_1.png)\nFig. 1\n![](attached_image_2.png)\nFig. 2\nSince the quadrilaterals $ACC_1A_1$, $ABB_1A_1$, $BCC_1B_1$ are trapezia and we know the coordinates of their vertices $A_1(a; 0)$, $B_1(b; 0)$, $C_1(c; 0)$ and $A\\left(a; \\frac{1}{a}\\right)$, $B\\left(b; \\frac{1}{b}\\right)$, $C\\left(c; \\frac{1}{c}\\right)$, we can find their areas:\n$$\n\\begin{align*} \nS(ACC_1A_1) &= \\frac{1}{2} A_1 C_1 \\cdot (AA_1 + CC_1) = \\frac{1}{2} (c-a) \\left(\\frac{1}{a} + \\frac{1}{c}\\right) = \\frac{1}{2} \\frac{c^2 - a^2}{ac} = \\\\ \n&= \\frac{1}{2} \\left(\\frac{c}{a} - \\frac{a}{c}\\right) = \\frac{1}{2} \\left(\\frac{u_c t}{u_a t} - \\frac{u_a t}{u_c t}\\right) = \\frac{1}{2} \\left(\\frac{u_c}{u_a} - \\frac{u_a}{u_c}\\right). \n\\end{align*}\n$$\nSince the speeds of the snails are constant, we see that the area $S(ACC_1A_1)$ is independent of time. Similarly, we conclude that the areas of the trapezia $ABB_1A_1$ and $BCC_1B_1$ are independent of time, too.\n\n![](attached_image_1.png)\n\n![](attached_image_2.png)\n\nNow we consider the case when two snails (say, the second and the third snails) move in the same direction and the first snail moves in the opposite direction. Without loss of generality we assume that the first snail moves along the negative part of the axis $Ox$ (see Fig. 2) and $u_b \\le u_c$. Then\n$$\nS(ABC) = S(ABB_1) + S(BCC_1B_1) - S(ACC_1).\n$$\nWe have\n$$\n\\begin{align*} \nS(ABB_1) &= \\frac{1}{2} B_1 A \\cdot BB_1 = \\frac{1}{2} (b-a) \\left(\\frac{1}{b} - \\frac{1}{a}\\right) = \\\\ \n&= \\frac{1}{2} \\left(2 - \\frac{b}{a} - \\frac{a}{b}\\right) = \\frac{1}{2} \\left(2 - \\frac{u_b t}{-u_a t} - \\frac{-u_a t}{u_b t}\\right) = 1 + \\frac{1}{2} \\left(\\frac{u_a}{u_b} + \\frac{u_b}{u_a}\\right). \n\\end{align*}\n$$\nSince the speeds of the snails are constant we see that the areas of the triangles $ABB_1$ and $ACC_1$ are independent of time.\nIt remains to show that the area of the trapezium $BCC_1B_1$ is also independent of time. Indeed,\n$$\n\\begin{align*} \nS(BCC_1B_1) &= \\frac{1}{2} C_1 B_1 \\cdot (CC_1 + BB_1) = \\frac{1}{2} (c-b) \\left(\\frac{1}{c} - \\frac{1}{a} + \\frac{1}{b} - \\frac{1}{a}\\right) = \\\\ \n&= \\frac{1}{2} \\left(\\frac{c}{b} - \\frac{b}{c} - 2 \\frac{c}{a} + 2 \\frac{b}{a}\\right) = \\frac{1}{2} \\left(\\frac{u_c}{u_b} - \\frac{u_b}{u_c} + 2 \\frac{u_c}{u_a} - 2 \\frac{u_b}{u_a}\\right). \n\\end{align*}\n$$\nTherefore, the area of the triangle ABC is independent of time.",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 25410,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $O$ be the circumcenter of an acute-angled triangle $ABC$. Let $AH$ be the altitude of this triangle, $M$, $N$, $P$, $Q$ be the midpoints of the segments $AB$, $AC$, $BH$, $CH$, respectively.\n\nLet $\\omega_1$ and $\\omega_2$ be the circumcircles of the triangles $AMN$ and $POQ$.\nProve that one of the intersection points of $\\omega_1$ and $\\omega_2$ belongs to the altitude $AH$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $M$, $N$ and $R$ be the midpoints of the sides $AB$, $AC$, $BC$, respectively. Let $X$ denote the foot of the perpendicular from $O$ on $AH$. We claim that $X$ is the point of intersection of $\\omega_1$ and $\\omega_2$.\n\n![](attached_image_1.png)\n\nSince $O$ is the center of the circumcircle of the triangle $ABC$, we have $OR \\perp BC$, $ON \\perp AC$ and $OM \\perp AB$. So $O$, $M$, $A$, $N$ and $X$ lie on the same circle $\\omega_1$, and $OA$ is the diameter of $\\omega_1$.\n\nShow that $PR = HQ$. Indeed,\n$$\nPR = BR - BP = \\frac{1}{2}(BC - BH) = \\frac{1}{2}HC = HQ.\n$$\nFrom $PR = HQ$ and $OX \\parallel BC$ it follows that $POXQ$ is an isosceles trapezium. Therefore $X$ belongs to $\\omega_2$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25411,
    "subject": "Mathematics (Multi-modal)",
    "question": "The graph of the parabola $y = x^2$ is drawn on the Cartesian plane $Oxy$. The vertices of a triangle $ABC$ belong to the parabola. The median $BM$ of the triangle is parallel to the ordinate axis and is equal to $2$.\nFind the area of the triangle $ABC$.",
    "options": [],
    "answer": "2√2",
    "solution": "Answer: $2\\sqrt{2}$.\n\nLet $A(a; a^2)$, $B(b; b^2)$, $C(c; c^2)$. Without loss of generality we assume that $a < c$. Since $M$ is the midpoint of $AC$, we have $M\\left(\\frac{a+c}{2}; \\frac{a^2+c^2}{2}\\right)$. Since $BM \\parallel Oy$, we see that the abscissae of $M$ and $B$ are equal, i.e. $\\frac{a+c}{2} = b$, so $BM = \\frac{a^2+c^2}{2} - b^2$. Therefore\n$$\n\\begin{aligned}\n2 &= \\frac{a^2 + c^2}{2} - b^2 = \\frac{a^2 + c^2}{2} - \\frac{(a+c)^2}{4} \\\\\n&= \\frac{(a-c)^2}{4}.\n\\end{aligned}\n$$\nHence, $(a-c)^2 = 8$, i.e. $c-a = 2\\sqrt{2}$.\n\n$$\nS(ABC) = 2S(ABM) = 2 \\cdot 0.5AK \\cdot BM,\n$$\n\n![](attached_image_1.png)\n\nwhere $AK$ is the altitude of the triangle $ABM$. Since $BM \\parallel Oy$, we have $AK \\parallel Ox$, so the length of the segment $AK$ is equal to the difference of the abscissae of $A$ and $C$, i.e. $c-a = 2\\sqrt{2}$.\n\nTherefore,\n$$\nS(ABC) = 2 \\cdot 0.5 \\cdot 2\\sqrt{2} \\cdot 2 = 2\\sqrt{2}.\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 25412,
    "subject": "Mathematics (Multi-modal)",
    "question": "All vertices of triangles $ABC$ and $A_1B_1C_1$ lie on the hyperbola $y = 1/x$. It is known that $AB \\parallel A_1B_1$ and $BC \\parallel B_1C_1$. Prove that $AC_1 \\parallel A_1C$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the coordinates of the given points be $A(a; 1/a)$, $B(b; 1/b)$, $C(c; 1/c)$, $A_1(a_1; 1/a_1)$, $B_1(b_1; 1/b_1)$, $C_1(c_1; 1/c_1)$. It is easy to calculate the slope of the line $AB$: $k = -1/(ab)$. Similarly, the slopes of $A_1B_1$, $BC$, $B_1C_1$ are $-1/(a_1b_1)$, $-1/(bc)$, $-1/(b_1c_1)$, respectively. Now, the conditions $AB \\parallel A_1B_1$ and $BC \\parallel B_1C_1$ are equivalent to the equalities $-1/(ab) = -1/(a_1b_1)$ and $-1/(bc) = -1/(b_1c_1)$. These equalities imply the equality $-1/(ac_1) = -1/(a_1c)$ which is equivalent to the condition $AC_1 \\parallel A_1C$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25413,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\Gamma_B$ and $\\Gamma_C$ be excircles of an acute-angled triangle $ABC$ opposite to its vertices $B$ and $C$, respectively. Let $C_1$ and $L$ be the tangent points of $\\Gamma_C$ and the side $AB$ and the line $BC$ respectively. Let $B_1$ and $M$ be the tangent points of $\\Gamma_B$ and the side $AC$ and the line $BC$, respectively. Let $X$ be the point of intersection of the lines $LC_1$ and $MB_1$.\nProve that $AX$ is equal to the inradius of the triangle $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\angle BAC = \\alpha$, $\\angle CBA = \\beta$, $\\angle ACB = \\gamma$. Let $AH$ be the altitude of the triangle $ABC$ and $X'$ be the point of intersection of $AH$ and the line $MB_1$. Then $\\angle X'AB_1 = 90^\\circ - \\gamma$. Since $\\gamma = \\angle ACB = \\angle CB_1M + \\angle B_1MC$ and $CB_1 = CM$ we have $\\angle CB_1M = \\angle B_1MC = \\gamma/2$.\n\n![](attached_image_1.png)\n\nCalculate the value of the angle $\\angle B_1X'A$:\n$$\n\\begin{align*} \\angle B_1X'A &= 180^\\circ - \\angle X'AB_1 - \\angle X'B_1A = [\\angle X'B_1A = \\angle CB_1M] = \\\\ &= 180^\\circ - (90^\\circ - \\gamma) - \\frac{1}{2}\\gamma = 90^\\circ + \\frac{1}{2}\\gamma. \\end{align*}\n$$\nBy the law of sines for $\\triangle B_1X'A$ we obtain\n$$\n\\frac{AX'}{\\sin \\frac{\\gamma}{2}} = \\frac{AB_1}{\\sin (90^\\circ + \\gamma/2)} = \\frac{AB_1}{\\cos \\frac{\\gamma}{2}},\n$$\ni.e. $AX' = AB_1 \\tan \\frac{\\gamma}{2}$. Let $I$ be the center of the incircle $\\Gamma$ of the triangle $ABC$ and $\\Gamma$ touch the side $BC$ at $A_1$. Let $IA_1 = r$. Since $A_1IC$ is a right-angled triangle we have $IA_1 = CA_1 \\tan \\frac{\\gamma}{2}$. It is well-known fact that $CA_1 = (AC + CB - AB)/2 = AB_1$. Therefore, $AX' = IA_1 = r$.\nSimilarly, if $X''$ is the point of intersection of $AH$ and the line $LC_1$, we show that $AX'' = r$ which gives the required statement.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25414,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given $m, n \\in \\mathbb{N}$ such that $m > n^{n-1}$ and the numbers $m+1, m+2, \\dots, m+n$ are composite.\nProve that there exist distinct primes $p_1, p_2, \\dots, p_n$ such that $m+k$ is divisible by $p_k$ for any $k = 1, 2, \\dots, n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We call the number $m + k$ from the problem condition convenient if it has at least $n$ distinct prime divisors, otherwise we call the number $m + k$ inconvenient. It is easy to see that we may consider inconvenient numbers only. Take one of them, $m + k = q_1^{a_1} q_2^{a_2} \\dots q_l^{a_l}$, where all $q_i$ are primes, $l \\le n - 1$. Since\n$$\nq_1^{a_1} q_2^{a_2} \\dots q_l^{a_l} = m + k > n^{n-1} \\ge n^l,\n$$\nthere exists an $i$ with $q_i^{a_i} > n$, so we can choose $p_k = q_i$. In a similar way we can choose prime divisors for other inconvenient numbers. It remains to note that for different inconvenient numbers the chosen prime divisors are different.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25415,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $AA_1$, $BB_1$ be the altitudes of an acute non-isosceles triangle $ABC$. The circumcircle of triangle $ABC$ meets that of triangle $A_1B_1C$ at point $N$ (different from $C$). Let $M$ be the midpoint of $AB$ and $K$ be the intersection point of $CN$ and $AB$.\n\nProve that the line of centers of the circumcircles of triangles $ABC$ and $KMC$ is parallel to the line $AB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $O$ and $L$ be the circumcenters of the triangles $ABC$ and $KMC$, respectively, $H$ the orthocenter of the triangle $ABC$. Let $S$ and $P$ be the midpoints of the segments $KM$ and $CH$, respectively. Let $R$ denote the circumradius of the triangle $ABC$. Since $A$, $B$, $A_1$, $B_1$ lie on the circle with $AB$ as its diameter, we have $\\angle CB_1A_1 = 180^\\circ - \\angle AB_1A_1 = \\angle ABC$, so $\\triangle A_1B_1C \\sim \\triangle ABC$ and the ratio of similitude is equal to $k = CB_1/CB = \\cos \\angle ACB$. Since $CH$ is the diameter of circumcircle of the triangle $CB_1A_1$ ($\\angle CB_1H = 90^\\circ$), we have $CH = 2R \\cos \\angle ACB$. Since $\\triangle AOB$ is isosceles and $\\angle AOB = 2\\angle ACB$, we have\n$$\nOM = AO \\cos \\angle AOM = R \\cos \\angle ACB = CH/2 = CP = PH.\n$$\n![](attached_image_1.png)\n\nTaking into account that $PH \\perp AB$ and $OM \\perp AB$, we obtain $PH \\parallel OM$, so $PHMO$ is a parallelogram. Since $PO$ is the line of centers of circumcircles of the triangles $CB_1A_1$ and $ABC$, we have $PO \\perp CN$. Then $MH \\perp CN$. Since the angle $CNH$ subtends the diameter of the circumcircle of the triangle $CB_1A_1$, we have $NH \\perp CN$, so $N$, $H$, $M$ lie on the same line. Moreover, $MN$ is the altitude of $\\triangle KMC$, so $H$ is the orthocenter of $\\triangle KMC$ ($CH \\perp AB$). Therefore $LS = CH/2 = PH = OM$, hence the lines $LO$ and $KB$ are parallel, as required.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25416,
    "subject": "Mathematics (Multi-modal)",
    "question": "Three cyclists start from town $A$ simultaneously. They move along the closed route consisting of three straight-line segments $AB$, $BC$ and $CA$. The speeds of the first cyclist on these segments are $12$, $10$ and $15$ kilometers per hour, respectively. The speeds of the second cyclist are $15$, $15$ and $10$ (km/h) and the speeds of the third cyclist are $10$, $20$ and $12$ (km/h).\nFind the value of the angle $ABC$ if all three cyclists finish at town $A$ at the same time.",
    "options": [],
    "answer": "90°",
    "solution": "Answer: $90^\\circ$.\nLet $AB = a$, $BC = b$, $CA = c$ (km). We find the time of each cyclist to cover the route (this time is independent of the moving direction). By condition,\n$$\n\\frac{a}{12} + \\frac{b}{10} + \\frac{c}{15} = \\frac{a}{15} + \\frac{b}{15} + \\frac{c}{10} = \\frac{a}{10} + \\frac{b}{20} + \\frac{c}{12},\n$$\nthus\n$$\n5a + 6b + 4c = 4a + 4b + 6c = 6a + 3b + 5c. \\quad (*)\n$$\nConsequently, $2c = a + 2b$ and $c = 2a - b$. So $a + 2b = 2(2a - b)$ which implies $3a = 4b$. Similarly, from $(*)$ it follows that $3c = 5b$. Then $a = \\frac{4b}{3}$ and $c = \\frac{5b}{3}$. Setting $b = 3x$ we obtain $a = 4x$ and $c = 5x$. Now it is easy to see that $a^2 + b^2 = c^2$. Hence, the triangle $ABC$ is a right-angled triangle with $c$ as hypotenuse. Therefore, $\\angle ABC = 90^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25417,
    "subject": "Mathematics (Multi-modal)",
    "question": "Circles $\\Gamma_1$ and $\\Gamma_2$ meet at points $X$ and $Y$. A circle $S_1$ touches $\\Gamma_1$ internally at $A$ and $\\Gamma_2$ externally at $B$. A circle $S_2$ touches $\\Gamma_2$ internally at $C$ and $\\Gamma_1$ externally at $D$.\n\nProve that the points $A$, $B$, $C$, $D$ are either collinear or concyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $O_1$ be the center of $\\Gamma_1$ and $R_1$ be its radius. Let $O_2$ be the center of $\\Gamma_2$ and $R_2$ be its radius. Homothety with center at $A$ and with a positive coefficient transforms $\\Gamma_1$ to $\\omega_1$. Homothety with center at $B$ and with a negative coefficient transforms $\\omega_1$ to $\\Gamma_2$. Let $Z$ be the center of the homothety with the negative coefficient which transforms $\\Gamma_1$ to $\\Gamma_2$. By the theorem of three homotheties, $Z$ lies on the line $AB$. Similarly, $Z$ lies on the line $CD$. It is evident that $Z$ lies on the segment $O_1O_2$ and $O_1Z : ZO_2 = R_1 : R_2$.\n\n![](attached_image_1.png)\n\n![](attached_image_2.png)\n\nShow that $ZB \\cdot ZA = ZC \\cdot ZD$. Let $E$ be the intersection point (different from $D$) of the line $CD$ and $\\Gamma_1$. The power of point $Z$ with respect to $\\Gamma_1$ is equal to $R_1^2 - O_1Z^2$, on the other hand, it is equal to $ZC \\cdot ZE$. From homothety's properties we have $EZ = \\frac{R_1}{R_2} ZD$. Then $ZC \\cdot ZE = ZC \\cdot \\frac{R_1}{R_2} ZD = R_1^2 - O_1Z^2$, i.e. $ZC \\cdot ZD = \\frac{R_2}{R_1}(R_1^2 - O_1Z^2)$. Since $O_1Z = \\frac{R_1}{R_1+R_2} O_1O_2$, we have\n$$\nZC \\cdot ZD = \\frac{R_2}{R_1} \\left( R_1^2 - R_1^2 \\frac{O_1 O_2^2}{(R_1 + R_2)^2} \\right) = R_1 R_2 \\left( 1 - \\frac{O_1 O_2^2}{(R_1 + R_2)^2} \\right).\n$$\nSimilarly, we obtain that $ZB \\cdot ZA = R_1 R_2 \\left( 1 - \\frac{O_1 O_2^2}{(R_1 + R_2)^2} \\right)$. Therefore, if $A$, $B$, $C$, and $D$ do not lie on the same line, then they lie on the same circle.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25418,
    "subject": "Mathematics (Multi-modal)",
    "question": "We have $2^m$ sheets of paper, with the number $1$ written on each of them. We perform the following operation. In every step we choose two distinct sheets: if the numbers on the two sheets are $a$ and $b$, then we erase these numbers and write the number $a+b$ on both sheets.\n\nProve that after $m2^{m-1}$ steps, the sum of the numbers on all the sheets is at least $4^m$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "1. See IMO-2014 Shortlist, Problem C2.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25419,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a triangle $ABC$. Let $S$ be the circle passing through $C$ centered at $A$. Let $X$ be a variable point on $S$, and let $K$ be the midpoint of the segment $CX$.\nFind the locus of the midpoints of $BK$, when $X$ moves along $S$.",
    "options": [],
    "answer": "A circle centered at N, the midpoint of BM (where M is the midpoint of AC), with radius equal to one quarter of AC (equivalently, R/4 where R is the radius of S).",
    "solution": "Let $R$ be the radius of the given circle, $M$ the midpoint of $AC$, $N$ the midpoint of $BM$, $T$ the midpoint of $BK$. Let $a, b, c, m, n, t, k$, $x$ be the (complex) coordinates of the points $A, B, C, M, N, T, X$, respectively. Then $m = (a+c)/2$, $n = (b+m)/2$, $k = (c+x)/2$, $t = (k+b)/2$. So we can easily find $x = 4t - 2b - c$, $n = (2b + c + a)/2$. The complex equation of the given circle is $|x - a| = R$, i.e. $|4t - 2b - c - a| = R$, or $|t - (2b + c + a)/4| = R/4$, or $|t - n| = R/4$. That is the locus of points $T$ is the circle of radius $R/4$ with the center at $N$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25420,
    "subject": "Mathematics (Multi-modal)",
    "question": "Pairwise distinct prime numbers $p$, $q$, $r$ satisfy the equality\n$$\nrp^3 + p^2 + p = 2rq^2 + q^2 + q.\n$$",
    "options": [],
    "answer": "2014",
    "solution": "Answer: 2014.\nBy condition, $rp^3 = -(p^2 + p) + 2rq^2 + (q^2 + q)$, and $p^2 + p$, $q^2 + q$, $2rq^2$ are even for all natural $p$ and $q$, then $rp^3$ is also even. Therefore, since $r$ and $p$ are prime, we see that either $p = 2$ or $r = 2$.\n\nIf $p = 2$, then the initial equality has the form $8r + 4 + 2 = 2rq^2 + q^2 + q$. But this is impossible because the right-hand side of this expression is greater than its left-hand side. Indeed, since $p = 2$ and $q \\neq p$ is prime, we have $q > 2$, so $2rq^2 + q^2 + q > 8r + 4 + 2$.\n\nThus $r = 2$ and the initial equality has the form $2p^3 + p^2 + p = 4q^2 + q^2 + q$, or\n$$\np(2p^2 + p + 1) = q(5q + 1). \\quad (1)\n$$\nSince $p$ and $q$ are distinct prime numbers, we see that $p$ and $q$ are coprime. Then from (1) it follows that $2p^2 + p + 1$ is divisible by $q$, and $5q + 1$ is divisible by $p$. Therefore $2p^2 + p + 1 = mq$ and $5q + 1 = mp$ for some $m \\in \\mathbb{N}$. We put $q = (mp - 1)/5$ into the former equality, then\n$$\n10p^2 + (5 - m^2)p + (m + 5) = 0. \\quad (2)\n$$\nConsider this equality as a quadratic equation with respect to $p$. Since the coefficients of (2) are integer and its root $p$ is integer, we obtain that the discriminant of (2) is necessarily a perfect square, i.e.\n$$\nD = (m^2 - 5)^2 - 4 \\cdot 10 \\cdot (m + 5) = m^4 - 10m^2 - 40m - 175 = n^2 \\quad (3)\n$$\nfor some nonnegative integer $n$. Since $D < m^4 - 175$, we see that for $m = 1, 2, 3$ the discriminant is negative. Moreover, for $m = 4$ we have $D = 256 - 160 - 160 - 175 < 0$. Therefore $m \\ge 5$.\n\nFrom (3) it follows that $(m^2 - 5)^2 > n^2$. Show that $n^2 > (m^2 - 11)^2$. If not, then from (3) it follows that\n$$\n(m^2 - 5)^2 - 40(m + 5) \\le (m^2 - 11)^2 \\Leftrightarrow 3m^2 - 10m - 74 \\le 0. \\quad (4)\n$$\nIt is easy to see that the latter inequality holds only for $m \\le 6$. But $m \\ge 5$, so to prove that $n^2 > (m^2 - 11)^2$ it remains to show that for $m = 5$ and $m = 6$ equality (2) does not hold for prime $p$.\n\nIf $m = 5$, then (2) has the form $10p^2 - 20p + 10 = 0$ and has exactly one root $p = 1$; but $1$ is not prime number. If $m = 6$, then the discriminant of (2) is equal to $D = 1296 - 360 - 240 - 175 = 521$; but $521$ is not perfect square.\n\nTherefore $(m^2 - 11)^2 < n^2 < (m^2 - 5)^2$, i.e., taking into account that $m \\ge 7$ and $n \\ge 0$, we have $m^2 - 11 < n < m^2 - 5$. Note that from (3) it follows that $m$ and $n$ have different parity, so $n$ can admit only two values $n = m^2 - 9$ or $n = m^2 - 7$.\n\nIf $n = m^2 - 9$, then from (3) we obtain $m^4 - 10m^2 - 40m - 175 = (m^2 - 9)^2$ or $m^2 - 5m - 32 = 0$. However, it is easy to see that the discriminant of this equation is equal to $153$, so the equation has no integer solutions.\n\nIf $n = m^2 - 7$, then from (3) we get $m^4 - 10m^2 - 40m - 175 = (m^2 - 7)^2$ or $m^2 - 10m - 56 = 0$. We see that $m = -4$ and $m = 14$ are the roots of this equation. Since $m \\in \\mathbb{N}$, we have $m = 14$.\n\nTherefore $m = 14$, then $n = m^2 - 7 = 189$. From (2) we obtain $p = (m^2 - 5 \\pm n)/20 = (191 \\pm 189)/20$. Since $p$ is integer, we have $p = 19$ which is prime. Hence $q = (mp - 1)/5 = (14 \\cdot 19 - 1)/5 = 265/5 = 53$.\n\nThus the required value of the product is $pqr = 19 \\cdot 53 \\cdot 2 = 2014$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25421,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for all even positive integers $n$ the following inequality holds\n$$\na) \\{n\\sqrt{6}\\} > \\frac{1}{n}; \\quad b) \\{n\\sqrt{6}\\} > \\frac{1}{n - 1/(5n)}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $m = [n\\sqrt{6}]$. Then $n\\sqrt{6} > m$, or $6n^2 - m^2 > 0$. Note that $6n^2 - m^2 \\ne 1$ or $4 \\pmod{3}$, $6n^2 - m^2 \\ne 2 \\pmod{4}$ (recall that $n$ is even), $6n^2 - m^2 \\neq 3 \\pmod{9}$, so $6n^2 - m^2 \\geq 5$. Now we have\n$$\n\\{n\\sqrt{6}\\} = n\\sqrt{6} - m = \\frac{6n^2 - m^2}{n\\sqrt{6} + m} \\geq \\frac{5}{n\\sqrt{6} + m} > \\frac{5}{2n\\sqrt{6}} > \\frac{1}{n},\n$$\nthus a) is proved. In particular, $m < n\\sqrt{6} - \\frac{1}{n}$, so\n$$\n\\{n\\sqrt{6}\\} \\geq \\frac{5}{n\\sqrt{6} + m} > \\frac{5}{2n\\sqrt{6} - 1/6} > \\frac{1}{n - \\frac{1}{5n}},\n$$\nthus b) is proved.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 25422,
    "subject": "Mathematics (Multi-modal)",
    "question": "$$\nf(x) = \\begin{cases} x + \\frac{1}{2}, & \\text{if } x < \\frac{1}{2}, \\\\ x^2, & \\text{if } x \\ge \\frac{1}{2}. \\end{cases}\n$$\nLet $a$ and $b$ be two real numbers such that $0 < a < b < 1$. We define the sequences $a_n$ and $b_n$ by $a_0 = a, b_0 = b$, and\n$$\na_n = f(a_{n-1}),\\ b_n = f(b_{n-1}) \\text{ for } n > 0.\n$$\nShow that there exists a positive integer $n$ such that\n$$\n(a_n - a_{n-1})(b_n - b_{n-1}) < 0.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "2. See IMO-2014 Shortlist, Problem A2.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25423,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all sequences $(a_n)$ of positive integers satisfying the equality $a_n = a_{n-1} + a_{n+1}$\na) for all $n \\ge 2$;\nb) for all $n \\ge 3$.",
    "options": [],
    "answer": "a) No such sequences exist.\nb) The unique sequence is a1 = 1, a2 = 1, and ai = 2 for all i at least 3.",
    "solution": "a) there are no such sequences;\n\nb) there exists a unique sequence: $a_1 = a_2 = 1$, $a_i = 2$ for all $i \\ge 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25424,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given positive real numbers $a$, $b$, $c$, with $ab + bc + ca \\ge a + b + c$, prove that\n\n$$(a+b+c)(ab+bc+ca) + 3abc \\ge 4(ab+bc+ca).$$\n\n(I. Gorodnin)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By the Cauchy-Buniakowski inequality,\n$$\n(\\sqrt{a}^2 + \\sqrt{b}^2 + \\sqrt{c}^2)((\\sqrt{a}(b+c))^2 + (\\sqrt{b}(c+a))^2 + (\\sqrt{c}(a+b))^2) \\geq (a(b+c) + b(c+a) + c(a+b))^2 = 4(ab + bc + ca)^2.\n$$\nHence,\n$$\n(a+b+c)(a(b+c)^2+b(c+a)^2+c(a+b)^2) \\geq 4(ab+bc+ca)^2 \\geq 4(ab+bc+ca)(a+b+c) \\Leftrightarrow (a(b+c)^2+b(c+a)^2+c(a+b)^2) \\geq 4(ab+bc+ca).\n$$\nIt remains to note that\n$$\na(b+c)^2 + b(c+a)^2 + c(a+b)^2 = (a+b+c)(ab+bc+ca) + 3abc,\n$$\nwhich gives the required inequality.\n\n\nAlternative solution:\n\nLet $\\sigma_1 = a+b+c$, $\\sigma_2 = ab+bc+ca$, $\\sigma_3 = abc$. By the problem condition, $\\sigma_2 \\ge \\sigma_1$. Since $\\sigma_1^2 \\ge 3\\sigma_2$ (well-known inequality), we have $\\sigma_1^2 \\ge 3\\sigma_2 \\ge 3\\sigma_1$, i.e. $\\sigma_1 \\ge 3$.\nNow we use Schur's inequality $\\sigma_1^3 + 9\\sigma_3 \\ge 4\\sigma_1\\sigma_2$. We have $\\sigma_1^2\\sigma_2 \\ge \\sigma_1^3$ and $3\\sigma_1\\sigma_3 \\ge 9\\sigma_3$, hence $\\sigma_1^2\\sigma_2 + 3\\sigma_1\\sigma_3 \\ge 4\\sigma_1\\sigma_2$ and then $\\sigma_1\\sigma_2 + 3\\sigma_3 \\ge 4\\sigma_2$, as required.\n\n\nAlternative solution:\n\nSince $ab+bc+ca \\ge a+b+c$, it suffices to prove the inequality $(a+b+c)^2(ab+bc+ca) + 3abc(a+b+c) \\ge 4(ab+bc+ca)^2$. The last inequality can be easily transformed to\n$$\na^3b + b^3a + a^3c + c^3a + b^3c + c^3b \\ge 2(a^2b^2 + b^2c^2 + c^2a^2),\n$$\nwhich is true by Muirhead's theorem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25425,
    "subject": "Mathematics (Multi-modal)",
    "question": "Points $K$, $L$, $M$ lie on the sides $BC$, $CA$, $AB$ of a triangle $ABC$, respectively, so that $AK$, $BL$, $CM$ meet at a common point. Let $r_1$, $r_2$, $r_3$, $r$ be the inradii of the triangles $ALM$, $BMK$, $CKL$, $ABC$, respectively.\nProve that $r_1 + r_2 + r_3 \\ge r$ or $r_2 + r_3 \\ge r$ or $r_1 \\ge r$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "2. See IMO-2014 Shortlist, Problem G2.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25426,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a cyclic quadrilateral, $P = AB \\cap CD$, $Q = AD \\cap BC$.\nProve that the distance between the orthocenters of the triangles $APD$ and $AQB$ is equal to that of the triangles $CQD$ and $BPC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We refer to the following well-known fact: in any cyclic quadrilateral all perpendiculars drawn through the midpoints of the sides to the opposite sides meet at the same point $U$. It follows that the altitudes $DD_1$ and $CC_1$ of the triangle $ADB$ and $BCP$, respectively, are symmetric with respect to $U$. The same is true for the altitudes $AA_1$ and $BB_1$. Therefore the intersection points $DD_1 \\cap AA_1$ and $CC_1 \\cap BB_1$ are symmetric with respect to $U$. That is, the orthocenters $H_{ADP}$ and $H_{BCP}$ are symmetric. Similarly, the orthocenters $H_{AQB}$ and $H_{DQC}$ are symmetric with respect to $U$. So the segments $H_{ADP}H_{AQB}$ and $H_{CQD}H_{BCP}$ are symmetric, so they have the same length.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25427,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{N} \\to \\mathbb{N}$, such that $mf(m) + n$ is divisible by $m^2 + f(n)$ for all $m, n \\in \\mathbb{N}$.",
    "options": [],
    "answer": "f(n) = n for all n in the positive integers",
    "solution": "### 2. See IMO-2013 Shortlist, Problem N1.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25428,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all pairs $(x, y)$ of positive integers such that\n$$\n\\sqrt[3]{7x^2 - 13xy + 7y^2} = |x - y| + 1.\n$$",
    "options": [],
    "answer": "{(x, y) = (1, 1)} ∪ { (x, y) = (n^3 + 2n^2 − n − 1, n^3 + n^2 − 2n − 1), (n^3 + n^2 − 2n − 1, n^3 + 2n^2 − n − 1) for integers n ≥ 2 }",
    "solution": "2. See IMO-2014 Shortlist, Problem N2.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25429,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x$, $y$, $z$ be pairwise distinct real numbers such that $x^2 - 1/y = y^2 - 1/z = z^2 - 1/x$. Given $z^2 - 1/x = a$, prove that\n$$\n(x+y+z)xyz = -a^2. \\qquad \\text{(I. Voronovich)}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First, note that $a \\neq 0$, $x \\neq -y$, etc. (this easily follows from the given relation). Now, the equality $x^2 - 1/y = y^2 - 1/z$ can be written in two other ways:\n$$\n\\begin{aligned}\nx^2 - y^2 + \\frac{1}{x} - \\frac{1}{y} &= \\frac{1}{x} - \\frac{1}{z} \\quad \\Leftrightarrow \\quad (x-y)(x+y-\\frac{1}{xy}) = \\frac{z-x}{xz} \\\\\n(x-y)(x^2+xy-\\frac{1}{y}) &= \\frac{z-x}{z} \\quad \\Leftrightarrow \\quad (x-y)(a+xy) = \\frac{z-x}{z}\n\\end{aligned}\n\\quad (1)\n$$\nand\n$$\n\\begin{aligned}\nx^2 - y^2 - \\frac{1}{x} - \\frac{1}{y} &= \\frac{1}{x} - \\frac{1}{z} \\quad \\Leftrightarrow \\quad (x+y)(x-y-\\frac{1}{xy}) = -\\frac{z+x}{xz} \\\\\n(x+y)(x^2 - xy - \\frac{1}{y}) &= -\\frac{z+x}{z}.\n\\end{aligned}\n\\quad (2)\n$$\nNow, multiplying (1) by two similar equalities and reducing by $(x-y)(y-z)(z-x) \\neq 0$, we obtain $(a+xy)(a+yz)(a+zx) = \\frac{1}{xyz}$\n$$\na^3 + a^2(xy + yz + zx) + a(x + y + z)xyz + (xyz)^2 = \\frac{1}{xyz}. \\quad (3)\n$$\nProceeding similarly with (2), we obtain\n$$\na^3 - a^2(xy + yz + zx) + a(x + y + z)xyz - (xyz)^2 = -\\frac{1}{xyz}. \\quad (4)\n$$\nSumming (3) and (4) and reducing by $2a \\neq 0$, we obtain the required equality.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25430,
    "subject": "Mathematics (Multi-modal)",
    "question": "Bob cuts an apple into either $20$ or $14$ pieces. Then he cuts one of these pieces into either $20$ or $14$ pieces. He repeats this procedure several times.\nCan Bob obtain $1! + 2! + 3! + \\dots + 1013! + 2014!$ small bits of the apple?",
    "options": [],
    "answer": "yes",
    "solution": "Answer: yes, he can.\nIf Bob cuts a piece of the apple into $20$ pieces, then the total number of the pieces increases by $19$. If Bob cuts a piece of the apple into $14$ pieces, then the total number of the pieces increases by $13$. So, if Bob makes $x$ cuts into $20$ pieces and $y$ cuts into $14$ pieces, then the apple ($1$ piece) is cut into exactly $1 + 19x + 13y$ pieces.\n\nIt remains to show that there exist nonnegative integer numbers $x$ and $y$ such that $1 + 19x + 13y = 1! + 2! + 3! + \\dots + 1013! + 2014!$. This equality is equivalent to the equality $19x + 13y = 2! + 3! + 4! + \\dots + 1013! + 2014!$.\n\nWe divide the summands in the right-hand side of the last equality into some groups:\n$$\n\\begin{align*}\n2! + 3! + 4! + \\dots + 1013! + 2014! &= (2! + 3! + 4! + 5!) + (6! + 8!) + \\\\\n&\\quad (7! + 9! + 10!) + (11! + 12!) + (13! + 14! + \\dots + 1013! + 2014!).\n\\end{align*}\n$$\nWe show that the sum of the numbers in each group is divisible either by $13$ or by $19$. Indeed, $2! + 3! + 4! + 5! = 152 = 8 \\cdot 19$, $6! + 8! = 6! (1 + 7 \\cdot 8) = 6! \\cdot 57 = 6! \\cdot 3! \\cdot 19$, $7! + 9! + 10! = 7! (1 + 8! \\cdot 9 + 8! \\cdot 9! \\cdot 10) = 7! \\cdot 793 = 7! \\cdot 61! \\cdot 13$, $11! + 12! = 11! (1 + 12) = 11! \\cdot 13$, and the last sum $13! + 14! + \\dots + 1013! + 2014!$ is divisible by $13$ since each summand of this sum is divisible by $13$.\n\nTherefore, the right-hand side of the equality has the form $19a + 13b$. Thus Bob can cut the apple into $1! + 2! + 3! + \\dots + 1013! + 2014!$ pieces (it is sufficient to make $a$ cuts into $20$ pieces and $b$ cuts into $14$ pieces).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25431,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prime numbers $p$, $q$, $r$ ($p + q < 111$) satisfy the equality\n$$\n\\frac{p + q}{r} = p - q + r.\n$$\nFind the largest possible value of the product $pqr$.",
    "options": [],
    "answer": "2014",
    "solution": "We have\n$$\n\\frac{p + q}{r} = p - q + r \\Leftrightarrow q(r + 1) - p(r - 1) = r^2.\n$$\nIf $r > 2$, then $r$ is odd (since $r$ is prime). Thus the left-hand side of the latter equation is an even number while its right-hand side is an odd number, a contradiction. Therefore, $r = 2$. Then from the initial equality it follows that $p = 3q - 4$. So, if $q$ increases, then $p$ also increases. Thus, the product $pqr = (3q - 4) \\cdot q \\cdot 2$ has maximal value as $q$ has maximal value.\n\nSince $p = 3q - 4$ and, by condition, $p + q < 111$, we have $3q - 4 + q = 4q - 4 < 111$, whence $q < 29$. Then $q$, as a prime, can admit only the following values $23$, $19$, $17$, $...$, $2$.\n\nIf $q = 23$, then $p = 3q - 4 = 3 \\cdot 23 - 4 = 65$ is a composite number.\nIf $q = 19$, then $p = 3q - 4 = 3 \\cdot 19 - 4 = 53$ is a prime number.\n\nThus, the largest possible value of the product $pqr = 53 \\cdot 19 \\cdot 2 = 2014$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25432,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs $(a; b)$ of integers $a$ and $b$ satisfying the equality $(b^2 + 7(a - b))^2 = a^3 b$.",
    "options": [],
    "answer": "All pairs with a equal to b for any integer, plus the pairs (-18, -2), (12, 3), and (0, 7).",
    "solution": "Answer: $a = b = t$, $t \\in \\mathbb{Z}$, $(a; b) = (-18; -2)$, $(a; b) = (0; 7)$, $(a; b) = (12; 3)$.\n\nWe rearrange the initial equation $(b^2 + 7(a-b))^2 = a^3 b$:\n$$\n\\begin{align*}\n(b^2 + 7(a-b))^2 &= a^3 b \\Leftrightarrow b^4 + 14b^2(a-b) + 49(a-b)^2 = a^3 b \\Leftrightarrow \\\\\n&\\qquad a^3 b - b^4 - 14b^2(a-b) - 49(a-b)^2 = 0 \\Leftrightarrow \\\\\n&\\qquad b(a^3 - b^3) - 14b^2(a-b) - 49(a-b)^2 = 0 \\Leftrightarrow \\\\\n&\\qquad (a-b)(ba^2 + ab^2 + b^3 - 14b^2 - 49(a-b)) = 0. \\tag{1}\n\\end{align*}\n$$\nIf $a = b$, then it is evident that (1) holds. Therefore we have the infinite set of the integer solutions: $a = b = t$, where $t \\in \\mathbb{Z}$.\n\nLet $a \\neq b$, then from (1) it follows that\n$$\nba^2 + ab^2 + b^3 - 14b^2 - 49(a-b) = 0. \\quad (2)\n$$\nIf $b=0$, then from (2) it follows that $a=0$, i.e. $a=b=0$, a contradiction. Thus $b \\neq 0$. Consider (2) as a quadratic equation with respect to $a$.\n$$\nba^2 + (b^2 - 49)a + b^3 - 14b^2 + 49b = 0. \\quad (3)\n$$\nThe discriminant $D(b)$ of this equation is\n$$\n\\begin{align*}\nD(b) = (b^2 - 49)^2 - 4b(b^3 - 14b^2 + 49b) &= ((b-7)(b+7))^2 - 4b^2(b-7)^2 = \\\\\n&= (b-7)^2((b+7)^2 - 4b^2) = -(b-7)^3(3b+7).\n\\end{align*}\n$$\nEquation (3) has real solutions iff $D(b) \\ge 0$, i.e. for $b \\in [-7/3; 7]$. The segment $[-7/3; 7]$ contains exactly ten integers: $-2, -1, 0, 1, 2, 3, 4, 5, 6$ and $7$. Since $b \\neq 0$ and $a_{1,2} = (49 - b^2 \\pm \\sqrt{D})/(2b)$, we have exactly four integers ($b = -2, 3, 6, 7$) such that $D(b)$ is a perfect square (we need integer solutions of (3)).\n\n1) If $b = -2$, then $D = 27^2$ and either $a_1 = -9/2$ or $a_2 = -18$, so, in this case $(a; b) = (-18; -2)$ is a required solution.\n\n2) If $b = 3$, then $D = 2^{10}$ and either $a_1 = 4/3$ or $a_2 = 12$, so $(a; b) = (12; 3)$ is a required solution.\n\n3) If $b = 6$, then $D = 5^2$ and either $a_1 = 13/12$ or $a_2 = 85/12$, so there are no solutions in this case.\n\n4) If $b = 7$, then $D = 0$ and $a_{1,2} = 0$, so $(a; b) = (0; 7)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25433,
    "subject": "Mathematics (Multi-modal)",
    "question": "Do there exist functions $f$ and $g$, $f: \\mathbb{R} \\to \\mathbb{R}$, $g: \\mathbb{R} \\to \\mathbb{R}$, such that $f(x + f(y)) = y^2 + g(x)$ for all real $x$ and $y$?",
    "options": [],
    "answer": "No, there are no such functions.",
    "solution": "Answer: there are no such functions.\n\nSuppose that there exist functions $f, g$ satisfying the equality\n$$\nf(x + f(y)) = y^2 + g(x). \\quad (*)\n$$\nFirst, suppose that $f(y) = f(z)$ for some $y, z \\in \\mathbb{R}$. Then $z^2 + g(x) = f(x + f(z)) = f(x + f(y)) = y^2 + g(x)$, whence\n$$\nz^2 = y^2. \\quad (1)\n$$\nNow, $f(x+f(y)) = y^2 + g(x) = (-y)^2 + g(x) = f(x+f(-y))$. Using (1), we get $(x+f(y))^2 = (x+f(-y))^2$ for all $x, y$. That is $x+f(y) = -x-f(-y)$, or $x+f(y) = x+f(-y)$. Of these two equalities, the former is impossible since the equality $2x = -f(-y) - f(y)$ implies that $2x$ is a constant which is not true. Hence\n$$\nf(-y) = f(y). \\qquad (2)\n$$\nSet $f(0) = a$. Putting $y = 0$ in (*), we have\n$$\ng(x) = f(x + a). \\qquad (3)\n$$\nNow,\n$$\n\\begin{aligned}\nf(x + f(y)) &= y^2 + f(x + a) = y^2 + f(-x - a) = \\\\\n&= y^2 + f((-x - 2a) + a) = f(-x - 2a + f(y)).\n\\end{aligned}\n$$\nHence from (1) it follows that\n$$\n(x + f(y))^2 = (-x - 2a + f(y))^2 \\Leftrightarrow (2f(y) - 2a)(2x + 2a) = 0\n$$\nfor all $x, y \\in \\mathbb{R}$, which implies $f(y) = a$, and (3) gives $g(x) = a$. So, (*) becomes $a = y^2 + a$ for all $y \\in \\mathbb{R}$, a contradiction. Therefore there are no such functions $f$ and $g$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25434,
    "subject": "Mathematics (Multi-modal)",
    "question": "$n$ points are marked on a plane. Each pair of these points is connected with a segment. Each segment is painted one of four different colors.\nFind the largest possible value of $n$ such that one can paint the segments so that for any four points there are four segments (connecting these four points) of four different colors.",
    "options": [],
    "answer": "9",
    "solution": "Answer: the largest possible value of $n$ is $9$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25435,
    "subject": "Mathematics (Multi-modal)",
    "question": "$N$ cells are marked on an $n \\times n$ table so that at least one marked cell is among any four cells of the table which form the figure $\\Box\\Box$ (the figure may be rotated). Find the smallest possible value of $N$.",
    "options": [],
    "answer": "(n - 3)([n/3] + 1) + 1",
    "solution": "Answer: $(n - 3)([n/3] + 1) + 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25436,
    "subject": "Mathematics (Multi-modal)",
    "question": "Construct a tetramino by attaching two $2 \\times 1$ dominoes along their sides such that the midpoint of the longer side of one domino is a corner of other domino. This construction yields two kinds of the tetraminoes with opposite orientations. Let us call them S- and Z-tetraminoes, respectively.\n![](attached_image_1.png)\nS-tetraminoes Z-tetraminoes\n\nAssume that a lattice polygon $P$ can be tiled with S-tetraminoes.\nProve that no matter how we tile $P$ using only S- and Z-tetraminoes, we always use even number of Z-tetraminoes.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider the following arrangement of numbers in the cells of the lattice (see Fig.1).\n\nIt is evident that the sum of the numbers in any S-tetramino is always zero, so the sum of the numbers in all cells of the polygon $P$ from the problem condition equals zero.\n![](attached_image_2.png)\nFig. 2\n![](attached_image_3.png)\nFig. 3\n\nNow we can see that the sum of the numbers in any vertical Z-tetramino (see Fig. 2) is always zero, while the sum of the numbers in any horizontal Z-tetramino (see Fig. 3) is $2$ or $-2$. Therefore in any tiling of $P$ with tetraminoes the number of horizontal Z-tetraminoes should be even (the numbers of tetraminoes with sum $2$ equals that of with the sum $-2$). In the same way we can conclude that the number of vertical Z-tetraminoes is also even, which finishes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25437,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a fixed circle $\\Gamma$ with three fixed points $A$, $B$, and $C$ on it. Also let us fix a real number $\\lambda \\in (0,1)$. For a variable point $P \\notin \\{A, B, C\\}$ on $\\Gamma$, let $M$ be the point on the segment $CP$ such that $CM = \\lambda \\cdot CP$. Let $Q$ be the second point of intersection of the circumcircles of the triangles $AMP$ and $BMC$.\n\nProve that as $P$ varies, the point $Q$ lies on a fixed circle.\n\n(IMO-2014 Shortlist, Problem G4)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "3. See IMO-2014 Shortlist, Problem G4.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25438,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all functions $f: \\mathbb{Z} \\to \\mathbb{Z}$ satisfying\n$$\nf(f(m) + n) + f(m) = f(n) + f(3m) + 2014\n$$\nfor all integers $m$ and $n$.",
    "options": [],
    "answer": "f(n) = 2n + 1007 for all integers n",
    "solution": "(Solution by I. Voronovich.) Rewrite the given equation as\n$$\nf(f(m) + n) = f(n) - f(m) + f(3m) + c, \\quad (1)\n$$\nwhere $c = 2014$; all we need in this solution is that $c$ is even and coprime with 3. Let $f(0) = a$.\n\n1) Setting $m = 0$ we get $f(n+a) = f(n)+c$, then by standard induction to both sides $f(n+ka) = f(n) + kc$, $k \\in \\mathbb{Z}$; in particular, $f(ka) = a + kc$ and $f(a) = a + c$.\n\n2) Set $m = n$ in (1), then\n$$\nf(n) + n = f(3n) + c = [\\text{in view of 1}] = f(3n + a).\n$$\n\n3) Setting $m = a$ in (1) we get\n$$\nf(n + f(a)) = f(n) - f(a) + f(3a) + c,\n$$\nor, in view of 1) $f(n+c) + c = f(n+c+a) = f(n) - c - a + a + 3c + c$, or $f(n+c) = f(n) + 2c$. Again, by easy induction $f(n+kc) = f(n) + 2kc$.\n\n4) Searching on injectivity: suppose that $f(s) = f(t)$. Then setting $m = s$ and $m = t$ in (1) we obtain $f(3s) = f(3t)$, and thus $f(3^l s) = f(3^l t)$ for all $l \\in \\mathbb{N}$. Since $(3;c)=1$, there exists an $l$ such that $3^l - 1 = nc$ for some $n \\in \\mathbb{N}$. Then in view of 3) we have\n$$\nf(3^l s) = f(3^l t) \\iff f(s + (3^l - 1)s) = f(t + (3^l - 1)t),\n$$\nor $f(s + nsc) = f(t + ntc)$, or $f(s) + 2nsc = f(t) + 2ntc$, $2nsc = 2ntc$, so $s = t$. Thus $f$ is injective. Since in view of 2) $f(f(n) + n) = f(3n + a)$, we obtain $f(n) + n = 3n + a$, $f(n) = 2n + a$. Easy verification shows that the function is a solution of (1) if and only if $a = c/2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25439,
    "subject": "Mathematics (Multi-modal)",
    "question": "The angles at the vertices $A$ and $C$ in the convex quadrilateral $ABCD$ are not acute. Points $K, L, M$ and $N$ are marked on the sides $AB, BC, CD$ and $DA$ respectively.\nProve that the perimeter of $KLMN$ is not less than the double length of the diagonal $AC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Lemma.** Let $CC_1$ be the median of a triangle $ABC$. Then the inequality $CC_1 \\le 0.5 AB$ is equivalent to the inequality $\\angle ACB \\ge 90^\\circ$.\n\nConsider the parallelogram $ADBC$ (see Fig. 1). By the cosine law, from the $\\triangle ABC$ and $\\triangle CBD$, it follows that\n$$\n\\cos \\angle ACB = \\frac{BC^2 + AC^2 - 2AB^2}{2BC \\cdot AC}\n$$\nand\n$$\n\\cos \\angle CBD = \\frac{BC^2 + AC^2 - 2CD^2}{2BC \\cdot AC}.\n$$\nSince the sum of these angles is equal to $180^\\circ$, we see that $\\angle ACB \\ge 90^\\circ$ iff $\\cos \\angle CBD \\ge \\cos \\angle ACB$. But this inequality is equivalent to the inequality $CD \\le AB$ or, since $CD = 2CC_1$, to the inequality $CC_1 \\le 0.5AB$. The proof is complete.\n\n![](attached_image_1.png)\nFig. 1\n\n![](attached_image_2.png)\nFig. 2\n\nLet $P, Q, T$ be the midpoints of the segments $KN, KM, LM$, respectively (see Fig. 2). We have $AC \\le AP + PQ + QT + TC$. By condition, $\\angle NAK \\ge 90^\\circ$ and $\\angle MCL \\ge 90^\\circ$, then from the lemma it follows that $AP \\le 0.5KN$ and $CT \\le 0.5LM$. Moreover, $PQ = 0.5NM$ and $QT = 0.5KL$ as the midlines of the respective triangles. Therefore $AC \\le 0.5KN + 0.5NM + 0.5KL + 0.5LM$, i.e. the perimeter of $KLMN$ is greater than or equal to $2AC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25440,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider all polynomials $P(x)$ with real coefficients that have the following property: for all real $x$ and $y$ one has\n$$\n|y^2 - P(x)| \\le 2x \\iff |x^2 - P(y)| \\le 2|y|.\n$$\n\nDetermine all possible values of $P(0)$.",
    "options": [],
    "answer": "(-∞, 0)",
    "solution": "3. See IMO-2014 Shortlist, Problem A5.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 25441,
    "subject": "Mathematics (Multi-modal)",
    "question": "Points $X$, $Y$ and $Z$ are marked on the sides $AD$, $AB$ and $BC$ of the rectangular $ABCD$, respectively.\nGiven $AX = CZ$ prove that $XY + YZ \\ge AC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $ZM$ be parallel to $AC$ (see the figure).\n\nIt is evident that $MZCA$ is a parallelogram, so $MZ = AC$ and $MA = CZ$. By condition, $AX = CZ$, so $AX = MA$. Therefore, the altitude $YA$ of the triangle $MYX$ is a median, hence this triangle is an isosceles triangle and $MY = YX$.\n\nSince the length of the segment $MZ$ is less than or equal to the length of the broken line $MYZ$, we obtain $XY + YZ = MY + YZ \\ge MZ = AC$, as required.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25442,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n > 1$ be a given integer. Prove that infinitely many terms of the sequence $(a_k)_{k \\ge 1}$, defined by\n$$\na_k = \\left\\lfloor \\frac{n^k}{k} \\right\\rfloor,\n$$\nare odd. (For a real number $x$, $\\lfloor x \\rfloor$ denotes the largest integer not exceeding $x$.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "3. See IMO-2014 Shortlist, Problem N4.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25443,
    "subject": "Mathematics (Multi-modal)",
    "question": "Points $X$, $Y$ and $Z$ are marked on the sides $AD$, $AB$ and $BC$ of a parallelogram $ABCD$, respectively. It is known that $AX = CZ$.\n\na) Prove that at least one of the inequalities holds\n$$\nXY + YZ \\geq AC \\text{ or } XY + YZ \\geq BD.\n$$\n\nb) Is it true that $XY + YZ \\geq \\frac{AC + BD}{2}$?",
    "options": [],
    "answer": "No",
    "solution": "b) It is not true.\n\na) Without loss of generality we assume that $\\alpha = \\angle BAD \\le 90^\\circ$. In this case we have\n$$\n\\begin{aligned}\nBD &= \\sqrt{AB^2 + AD^2 - 2AB \\cdot AD \\cos \\alpha} \\le \\\\\n&\\le \\sqrt{AB^2 + AD^2 + 2AB \\cdot AD \\cos \\alpha} = [AD = BC] = \\\\\n&= \\sqrt{AB^2 + BC^2 - 2AB \\cdot BC \\cos(\\pi - \\alpha)} = AC.\n\\end{aligned}\n$$\nSo, it is sufficient to show that $XY + YZ \\ge BD$. Let $YM \\parallel AD$, $DM \\parallel XY$ (see the Fig.). It is evident that $XYMD$ is a parallelogram, so $YM = XD$ and $XY = MD$. Since $CZ = AX$, we have $XD = AD - AX = BC - CZ = ZB$. Since $ZB \\parallel XD \\parallel YM$, we see that $YBZM$ is a parallelogram.\n\n![](attached_image_1.png)\n\nIn the parallelogram $YBZM$ we have $\\angle BYM = \\angle BAD = \\alpha \\le 90^\\circ$, therefore,\n$$\n\\begin{aligned}\nYZ &= \\sqrt{YB^2 + BZ^2 - 2YB \\cdot BZ \\cos(\\pi - \\alpha)} \\ge [BZ = YM] \\ge \\\\\n&\\ge \\sqrt{YB^2 + YM^2 - 2YB \\cdot YM \\cos \\alpha} = BM.\n\\end{aligned}\n$$\nThus $XY + YZ \\ge XY + BM = MD + BM \\ge BD$, as required.\n\nb) We show that there exist a parallelogram $ABCD$ and a point $Y$ such that the inequality\n$$\nXY + YZ \\ge \\frac{AC + BD}{2} \\quad (*)\n$$\nis not valid. Let a parallelogram $ABCD$ is different from a rectangle. Then its diagonals are not equal. Without loss of generality we assume that $AC > BD$. Let $YZ \\parallel AC$ (see the Fig.). Let $AX/AD = CZ/BC = \\lambda$. By Thales' theorem, we have $AY/AB = CZ/BC = \\lambda$. Hence $AY/AB = AX/AD = \\lambda$. It follows that the triangles $AYX$ and $ABD$ are similar and $XY/BD = \\lambda$, i.e. $XY = \\lambda BD$.\n\nMoreover, by construction of $Y$, the triangles $YBZ$ and $ABC$ are similar and $YZ/AC = BZ/BC = (BC - CZ)/BC = 1 - \\lambda$, hence $YZ = (1 - \\lambda)AC$.\n\nTherefore, $XY + YZ = \\lambda BD + (1 - \\lambda)AC$ and (*) has the form $\\lambda BD + (1 - \\lambda)AC \\ge \\frac{1}{2}BD + \\frac{1}{2}AC$ or $(\\lambda - \\frac{1}{2})BD \\ge (\\lambda - \\frac{1}{2})AC$. But this inequality is not valid if $\\lambda > \\frac{1}{2}$.\n\n![](attached_image_2.png)",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 25444,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ points be given inside a rectangle $R$ such that no two of them lie on a line parallel to one of the sides of $R$. The rectangle $R$ is to be dissected into smaller rectangles with sides parallel to the sides of $R$ in such a way that none of these rectangles contains any of the given points on its interior.\nProve that we have to dissect $R$ into at least $n + 1$ smaller rectangles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25445,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given triangle $ABC$ with $AB = c$, $BC = a$, $CA = b$. The pairs of points $C_1$ and $C_2$, $A_1$ and $A_2$, $B_1$ and $B_2$ are marked on the sides $AB$, $BC$, $CA$, respectively so that the following equalities are valid:\n\n![](attached_image_1.png)\n\n$$\n\\begin{aligned}\n\\frac{CA_1}{a} &= \\frac{CB_2}{b} = \\frac{a+b}{a+b+c}, \\\\\n\\frac{AB_1}{b} &= \\frac{AC_2}{c} = \\frac{b+c}{a+b+c}, \\\\\n\\frac{BC_1}{c} &= \\frac{BA_2}{a} = \\frac{a+c}{a+b+c}.\n\\end{aligned}\n$$\n\nProve that the points of intersection of the lines $A_1C_2$, $C_1B_2$ and $B_1A_2$ belong to the circumcircle of the triangle $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $K, L, N$ be points of intersection of the bisectors $AK, BL, CN$ and the circumcircle of a triangle $ABC$, respectively (see the Fig.). Let $I$ be the incenter of $\\triangle ABC$ and let the segment $NL$ meet $AC$ at $B'_2$. Let $M$ be the intersection point of $CN$ and $AB$. By the trefoil theorem $AL = IL$, i.e. $\\triangle AIL$ is isosceles. Since $\\angle ALN = \\angle ACN = \\angle NCB = \\angle BLN$ (as inscribed angles subtending equal arcs), the ray $LN$ is the bisector of the isosceles triangle $AIL$, hence $LN$ is the perpendicular bisector of the side $AI$. Therefore, $\\triangle AB'_2I$ is also isosceles and $\\angle IAB'_2 = \\angle B'_2I A$. Since $AK$ is the bisector of the angle $BAC$, we have $\\angle IAB'_2 = \\angle BAI$, so $AB \\parallel B'_2I$.\nThen we have $\\frac{CB'_2}{B'_2A} = [\\text{the Thales theorem}] = \\frac{CI}{MI} = \\frac{AC}{AM} =$\n$= \\left[ \\frac{AM}{AB} = \\frac{AC}{AC + BC} \\right] = \\frac{AC + BC}{AB} = \\frac{a+b}{c}$, so $\\frac{CB'_2}{AC} = \\frac{a+b}{a+b+c}$.\nTherefore $B'_2$ coincide with $B_2$. Thus, $B_2$ and $C_1$ lie on the segment $NL$. Similarly, we see that $B_1$ and $A_2$ lie on $KL$, $A_1$ and $C_2$ lie on $KN$, which gives the required statement.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25446,
    "subject": "Mathematics (Multi-modal)",
    "question": "Point $L$ is marked on the side $AB$ of a triangle $ABC$. The incircle of the triangle $ABC$ meets the segment $CL$ at points $P$ and $Q$.\nIs it possible that the equalities $CP = PQ = QL$ hold if $CL$ is\n\na) the median?\nb) the bisector?\nc) the altitude?\nd) the segment joining vertex $C$ with the point $L$ of tangency of the excircle of the triangle $ABC$ with $AB$?",
    "options": [],
    "answer": "a) Yes; for triangles with AC = 5x, AB = 10x, BC = 13x. b) No; impossible. c) No; impossible. d) Yes; for triangles with AC = 3x, AB = 4x, BC = 5x.",
    "solution": "a) yes, it is possible; this is valid for the triangles with $AC = 5x$, $AB = 10x$, $BC = 13x$, where $x$ is any positive real number;\n\nb) it is impossible;\n\nc) it is impossible;\n\nd) yes, it is possible; this is valid for the triangles with $AC = 3x$, $AB = 4x$, $BC = 5x$, where $x$ is any positive real number.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 25447,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all integers $a$ and $b$ satisfying the equality\n$$\n3^a - 5^b = 2.\n$$",
    "options": [],
    "answer": "(a, b) = (1, 0) and (a, b) = (3, 2)",
    "solution": "(Solution of D. Babrou.) If $a \\le 3$ or $b \\le 2$, then we see that only the pairs $(a; b) = (1; 0)$, $(a; b) = (3; 2)$ satisfy the equation $3^a - 5^b = 2$.\n\nLet now $a \\ge 4$ and $b \\ge 3$. We rewrite the equation in the form $3^3(3^{a-3} - 1) = 5^2(5^{b-2} - 1)$. Setting $x = a - 3$, $y = b - 2$ ($x, y > 0$) we obtain $3^3(3^x - 1) = 5^2(5^y - 1)$.\n\nIt is easy to verify that the least possible $n$ with $3^n \\equiv 1 \\pmod{25}$ is $n = 20$, hence we have $x \\equiv 20$.\n\nSimilarly we conclude that $y \\equiv 18 \\pmod{15}$ because the least $n$ with $5^n \\equiv 1 \\pmod{27}$ is $n = 18$. So, $5^y - 1 \\equiv 19 \\pmod{19}$.\n\nFurther, the smallest positive $x$ such that $3^x \\equiv 1 \\pmod{19}$ is 18, so $x \\equiv 180 \\pmod{100}$. Since $x \\equiv 10 \\pmod{10}$, we have $3^x - 1 \\equiv 11 \\pmod{11}$, then $5^y - 1 \\equiv 11 \\pmod{11}$.\n\nThe smallest positive $y$ such that $5^y - 1 \\equiv 11 \\pmod{5}$, hence\n$y \\equiv 90 \\equiv 5 \\cdot 18$. Since $x \\equiv 12 \\pmod{12}$, we have $3^x - 1 \\equiv 13 \\pmod{13}$. Then $5^y - 1 \\equiv 13 \\pmod{13}$.\n\nThe order of 5 modulo 13 is 4, hence $y \\equiv 4 \\pmod{13}$, thus $y \\equiv 180 \\pmod{12}$. Then $y \\equiv 12 \\pmod{12}$.\n\nSince $5^{12} - 1 = (5^6 - 1)(5^6 + 1)$, $5^6 + 1 \\equiv 601 \\pmod{601} = 5^4 - 5^2 + 1$, where 601 is a prime number, we have $5^y - 1 \\equiv 601 \\pmod{601}$ hence also $3^y - 1 \\equiv 601 \\pmod{601}$. Note that $\\varphi(601) = 600 = 5^2 \\cdot 24$.\n\nLet $\\alpha$ be the order 3 modulo 601. If $\\alpha \\nmid 25$, then $\\alpha$ is a divisor of 120 (since $600 \\nmid \\alpha$). So if $3^{120} - 1 \\nmid 601$, then $\\alpha \\equiv 25 \\pmod{601}$, and $x \\equiv 25 \\pmod{601}$.\n\nWe have $3^{120} - 1 \\equiv 729^{20} - 1 \\equiv 128^{20} - 1 \\equiv 2^{140} - 1 \\pmod{601}$.\n\nIf $2^{20} \\not\\equiv 1$, then $2^{140} \\not\\equiv 1 \\pmod{601}$ since 7 is not a divisor of 600. Next $2^{20} = 1024^2 = 423^2 \\not\\equiv 1 \\pmod{601}$, indeed.\n\nTherefore $x \\equiv 25 \\pmod{601}$, so $x \\equiv 100 \\pmod{601}$.\n\nFurther, since $\\varphi(125) = 100$, it follows that $3^x - 1 \\equiv 125 \\pmod{125}$, but if $y > 0$, then $5^2(5^y - 1) \\nmid 125$. Thus $x = y = 0$ contrary to $x, y > 0$. Therefore the only pairs mentioned above are the solutions of given equation.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25448,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the largest possible number of three-element sets that can be formed so that any two of these sets have exactly one common element, but there is not an element that belongs to all these sets.",
    "options": [],
    "answer": "7",
    "solution": "Answer: 7 sets.\nLet $N$ be the required number of 3-element sets satisfying the problem condition. Suppose that $N \\ge 8$. Let $M = \\{x, y, z\\}$ be one of these sets. Since the number of the remaining sets is greater than or equal to 7 and each such set has exactly one common element with $M$, we see that there exists an element of $M$ (say, $x$) that belongs to at least $[7/3] = 3$ sets. Let the sets $M_1, M_2, M_3$ contain $x$. By condition, there is not an element that belongs to all sets, so there exists a set (say, $M_0$) such that there is an element from $M$ (say, $y$), which belongs to $M_0$. It follows that $x \\notin M_0$. Since any two of the sets $M_1, M_2, M_3$ and $M$ have not common elements except for $x$ (and $x \\notin M_0$), so all four elements of the intersection of $M_0$ with $M_1, M_2, M_3$, $M$ are distinct. Therefore $M_0$ consists of at least four elements, a contradiction. Thus, $N \\le 7$.\n\nIt remains to show the example for $N = 7$. The required sets (see the Fig.) are $\\{A, K, B\\}$, $\\{A, P, M\\}$, $\\{A, N, C\\}$, $\\{B, M, C\\}$, $\\{B, P, N\\}$, $\\{C, P, K\\}$, $\\{K, M, N\\}$.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25449,
    "subject": "Mathematics (Multi-modal)",
    "question": "Some family pairs are friends with each other. Every St. Valentine's Day each husband of these pairs presents some roses to each wife of these pairs (including his own wife). Any wife will be offended by her husband if the number of roses that she obtains from her husband is less than or equal to the number of roses that he presents to all other wives together. This year, it turned out, that for any wife there is a partition of all husbands into two groups such that the total numbers of roses that are presented to this wife by the men from each group are equal (the groups may be different for different women).\nProve that at least one wife will be offended.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider $n$ ($n \\ge 2$) pairs. Let $M$ denote the set $\\{1, 2, ..., n\\}$. Let $a_{ij}$ be the number of roses presented by the husband from the $i$-th pair to the wife from the $j$-th pair. By condition, the wife from the $j$-th pair will be offended if\n$$\na_{jj} \\le \\sum_{i \\in M \\setminus \\{j\\}} a_{ji}. \\quad (1)\n$$\nOn the other hand, for any $j \\in M$ there exists a partition of $M$ such that $M = M_j^1 \\cup M_j^2$ and\n$$\n\\sum_{i \\in M_j^1} a_{ij} = \\sum_{i \\in M_j^2} a_{ij} \\quad \\forall j \\in M. \\quad (2)\n$$\nSince each $j \\in M$ belongs either to $M_j^1$ or to $M_j^2$, we have $\\sum_{i \\in M_j^1} a_{ij} = \\sum_{i \\in M_j^2} a_{ij} \\ge a_{jj}$. Now from (2) it follows that\n$$\n\\sum_{i \\in M} a_{ij} = \\sum_{i \\in M_j^1} a_{ij} + \\sum_{i \\in M_j^2} a_{ij} \\ge 2a_{jj} \\quad \\forall j \\in M. \\quad (3)\n$$\nIf we suppose that there are no offended wives then from (1) it follows that\n$$\n2a_{jj} > \\sum_{i \\in M \\setminus \\{j\\}} a_{ji} + a_{jj} = \\sum_{i \\in M} a_{ji} \\quad \\forall j \\in M. \\quad (4)\n$$\nFrom (3) and (4) it follows that\n$$\n2 \\sum_{j \\in M} a_{jj} \\le \\sum_{j \\in M} \\sum_{i \\in M} a_{ij} = \\sum_{i \\in M} \\sum_{j \\in M} a_{ij} < 2 \\sum_{i \\in M} a_{ii}.\n$$\nThe obtained contradiction yields that there exists at least one offended wife.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 25450,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $n$ towns in a country. Some of them are connected with domestic flights. Each flight connects exactly two towns. If there is a flight from town $A$ to town $B$, then there is a flight from $B$ to $A$. It is known that for any two towns there exists a unique route such that one can reach the second town from the first one using at most two flights. There is no town that is connected with all other towns of the country with the direct flights.\nProve that $n-1$ is a square of some integer number.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We use the graph theory language and consider the towns as the vertices of the graph $\\Gamma$. We connect two vertices by the edge iff there exists a flight from one of them to the other town. The problem conditions are equivalent to the following conditions for the graph $\\Gamma$:\n\ni) any two different vertices are connected with a unique path consisting of at most two edges;\nii) there are cyclic paths of length neither 3 nor 4 (see Fig. 1);\niii) there is no vertex connected with all others.\n\n![](attached_image_1.png)\nFig. 1\n\nFix any vertex $v$ of $\\Gamma$ and partition all the vertices of $\\Gamma$ into three sets: $\\Gamma_0$, $\\Gamma_1$ and $\\Gamma_2$ (the vertices of $\\Gamma_i$ are called the vertices of the $i$-th level, $i = 0, 1, 2$). $\\Gamma_0$ consists of exactly one vertex $v$; $\\Gamma_1$ consists of all vertices connected with $v$ by the edge; $\\Gamma_2$ consists of all remaining vertices of $\\Gamma$. Let $\\Gamma_1$ consist of vertices $v_1, v_2, ..., v_k$ (see Fig. 2). From ii) it follows that no two vertices $v_p, v_q \\in \\Gamma_1$ are connected by the edge (otherwise there is a cyclic path of length 3: $vv_pv_qv$). From iii) it follows that $\\Gamma_2 \\ne \\emptyset$.\n\n![](attached_image_2.png)\nFig. 2\n\nShow that each vertex of the 1st level is connected with at least one vertex of the 2nd level, and the distinct vertices of the 1st level cannot be connected with the same vertex of the 2nd level. Indeed, if some vertex $v_p \\in \\Gamma_1$ is connected with none of the vertices of the 2nd level, then ($v_p$ is connected with none of the vertices of the 1st level) the path from $v_p$ to any of the vertices of the 2nd level passes through $v$ (see Fig. 2), i.e. this path consists of at least two edges, contrary to i). If two vertices $v_p, v_q \\in \\Gamma_1$ are connected with the same vertex $w \\in \\Gamma_2$, then we have the cyclic path $vv_pwv_qv$ of length 4 (see Fig. 2), contrary to ii).\n\nNow it follows that $\\Gamma_2 = \\{G_1, G_2, ..., G_k\\}$, $G(i) \\cap G(j) = \\emptyset$ for all $i \\ne j$, and $G_p$ consists of those vertices of the 2nd level which are connected with $v_p$, $p = 1, 2, ..., k$, (see Fig. 2). Prove that all $G_1, G_2, ..., G_k$ contain the same number of the vertices. Consider any two sets $G_p$ and $G_q$ ($p \\ne q$). Suppose that there is $w \\in G_p$ which is connected with none of the vertices of $G_q$. Then there is no path from $w$ to $v_q$ with at most two edges. Indeed, we can reach $v_q$ either from $v$ or from any vertex of $G_q$. But the shortest path from $w$ to $v$ consists of two edges, and the path from $w$ to each vertex of $G_q$ consists of at least two edges. The obtained contradiction shows that each vertex of $G_p$ is connected by the edge with at least one of the vertices of $G_q$. Moreover, each vertex $w \\in G_p$ is connected by the edge with exactly one vertex of $G_q$ (otherwise if $w$ is connected by the edges with $u_1, u_2 \\in G_q$, then we have the cyclic path $wu_1v_qu_2w$ of length 4). Since we choose $G_p$ and $G_q$ arbitrarily, it follows that all sets $G(i)$, ($i = 1, ..., k$) have the same number of the vertices. Let $m$ denote the number of the vertices in each set of the sets $G_1, G_2, ..., G_k$.\n\nConsider any $G_p$ and some of its vertex $w$. Since no two vertices of $G_p$ are connected by the edge (otherwise there exists a cyclic path of length 3), $w$ is connected by the edge with exactly one vertex of each of the $k-1$ of the sets $G_m$, $m \\neq p$; $w$ is connected by the edge with $v_p$ and there are no edges from $w$ different from the mentioned edges, we see that the degree of $w$ is equal to $(k-1)+1=k$. Thus, the degree of each vertex of the 2nd level is equal to $k$, i.e. these degrees are equal to the number of the vertices of the 1st level.\n\nNote that vertex $v$ is chosen arbitrarily. Hence, applying the same arguments for $v_1$, we obtain that the 1st level for $v_1$ consists of $m=1$ vertices (the 1st level for $v_1$ consists of $v$ and all vertices of $G_1$ (see Fig. 2)). Therefore, the degree of each vertex of the 2nd level for $v_1$ is equal to $m+1$.\n\nIn particular, the degree of each vertex of, for example, $G(2)$ is equal to $m+1$. On the other hand, by the above, the degree of this vertex is equal to $k$. Therefore, we obtain $m=k-1$. Thus, $\\Gamma$ consists of exactly $1+k+k(k-1)=k^2+1$ vertices. The required statement is proved.\n\n![](attached_image_3.png)\n![](attached_image_4.png)\n\nFig. 3\n\n*Remark.* There exist graphs described in the problem conditions and solution, e.g., a 5-cyclic graph 5 or the Petersen graph (see Fig. 3).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25451,
    "subject": "Mathematics (Multi-modal)",
    "question": "Thirty rays with the origin at the same point are constructed on a plane. Consider all angles between any two of these rays. Let $N$ be the number of acute angles among these angles.\nFind the smallest possible value of $N$.",
    "options": [],
    "answer": "135",
    "solution": "Answer: 135.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25452,
    "subject": "Mathematics (Multi-modal)",
    "question": "A sequence $(a_n)$, $n \\in \\mathbb{N}$ is defined as\n$$\na_1 = 1,\\ a_2 = 2,\\ a_3 = 3 \\text{ and } a_n = \\frac{a_{n-1} a_{n-2} + 7}{a_{n-3}}, \\text{ for } n \\ge 4.\n$$\nProve that all terms of this sequence are integers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "It is evident that all terms of the sequence are positive. Find $a_4$. Since $a_1 = 1$, $a_2 = 2$, $a_3 = 3$, we have\n$$\na_4 = \\frac{a_3 a_2 + 7}{a_1} = 13. \\quad (1)\n$$\nBy condition,\n$$\n\\begin{aligned}\na_{n+1} &= \\frac{a_n a_{n-1} + 7}{a_{n-2}} = [7 = a_n a_{n-3} - a_{n-1} a_{n-2}] = \\\\\n&= \\frac{a_n a_{n-1} + a_n a_{n-3} - a_{n-1} a_{n-2}}{a_{n-2}} = a_n \\frac{a_{n-1} + a_{n-3}}{a_{n-2}} - a_{n-1}, \\quad n \\ge 4.\n\\end{aligned} \n\\quad (2)\n$$\nIf we prove that all numbers $\\frac{a_{n-1} + a_{n-3}}{a_{n-2}}$, $n \\ge 4$, are integers, then we obtain\nthat all numbers $a_n$, $n \\ge 5$, are integers, since the numbers $a_1, a_2, a_3, a_4$ are integers. Let $b_n = (a_{n-1} + a_{n-3})/a_{n-2}$, $n \\ge 4$. In particular\n$$\nb_4 = \\frac{a_3 + a_1}{a_2} = \\frac{3+1}{2} = 2 \\quad \\text{and} \\quad b_5 = \\frac{a_4 + a_2}{a_3} = \\frac{13+2}{3} = 5. \\quad (3)\n$$\nShow that the sequence $(b_n)$, $n \\ge 4$, is periodic with 2 as its period, i.e. $b_n = b_{n-2}$ for all $n \\ge 6$. From definition of the sequences $(b_n)$ and $(a_n)$ it follows that\n$$\n\\begin{aligned}\nb_n &= \\frac{a_{n-1} + a_{n-3}}{a_{n-2}} = \\frac{\\frac{a_{n-2} a_{n-3} + 7}{a_{n-4}} + a_{n-3}}{a_{n-2}} = \\\\\n&= \\frac{a_{n-2} a_{n-3} + 7 + a_{n-3} a_{n-4}}{a_{n-4} a_{n-2}} = [7 = a_{n-2} a_{n-5} - a_{n-3} a_{n-4}] = \\\\\n&= \\frac{a_{n-2} a_{n-3} + (a_{n-2} a_{n-5} - a_{n-3} a_{n-4}) + a_{n-3} a_{n-4}}{a_{n-4} a_{n-2}} = \\\\\n&= \\frac{a_{n-2} a_{n-3} + a_{n-2} a_{n-5}}{a_{n-4} a_{n-2}} = \\frac{a_{n-3} + a_{n-5}}{a_{n-4}} = b_{n-2}, \\quad n \\ge 6.\n\\end{aligned}\n$$\nTherefore, taking into account (3), we have $b_{2k}=b_4=2$ and $b_{2k+1}=b_5=5$ for all $k=2,3,\\dots$. Since (see (2)) $a_{n+1} = a_n b_n - a_{n-1}$ for all $n \\ge 4$, and all numbers $b_n$, $n \\ge 4$, and numbers $a_1, a_2, a_3, a_4$ are integers, it follows that all terms of the sequence $(a_n)$ are integers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25453,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that\n$$\n\\frac{1}{x+y+1} - \\frac{1}{(x+1)(y+1)} < \\frac{1}{11}\n$$\nfor all positive $x$ and $y$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By the Cauchy inequality,\n$$\n(x+1)(y+1) \\le \\frac{((x+1) + (y+1))^2}{4} = \\frac{(x+y+2)^2}{4}.\n$$\nSo the required inequality\n$$\n\\frac{1}{x+y+1} - \\frac{1}{(x+1)(y+1)} < \\frac{1}{11}, \\quad \\text{if } x > 0 \\text{ and } y > 0, \\quad (1)\n$$\nis a consequence of the inequality\n$$\n\\frac{1}{x+y+1} - \\frac{4}{(x+y+2)^2} < \\frac{1}{11}, \\quad \\text{if } x > 0 \\text{ and } y > 0. \\quad (2)\n$$\nPutting $x+y+1 = t$ ($t > 1$, since $x$ and $y$ are positive) into (2), we obtain the inequality\n$$\n\\frac{1}{t} - \\frac{4}{(t+1)^2} < \\frac{1}{11}, \\quad \\text{if } t > 1 \\quad (3)\n$$\nwhich is equivalent to (2). This inequality can be represented as\n$$\nt^3 - 9t^2 + 23t - 11 > 0, \\quad \\text{if } t > 1. \\quad (4)\n$$\nInequality (1) will be proved once we prove (4). We rearrange (4) as\n$$\nt^3 - 9t^2 + 23t - 11 = (t-3)^3 - 4t + 16 = (t-3)^3 - 4(t-3) + 4 = (t-1)(t-3)(t-5) + 4.\n$$\nTherefore, (4) is equivalent to the inequality\n$$\n(t-1)(t-3)(t-5) + 4 > 0, \\quad \\text{if } t > 1. \\quad (5)\n$$\nWe see that for $t > 1$ the polynomial $(t-1)(t-3)(t-5) \\le 0$ only if $t \\in [3; 5]$. But in this case $0 < t - 1 \\le 4$ and $(t-3)(t-5) \\ge -1$, therefore, $(t-1)(t-3)(t-5) \\ge -4$ for $t \\in [3; 5]$. Thus (5) is proved, which implies that (1) holds.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25454,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x = \\sqrt{ab}$ and $y = \\sqrt{\\frac{a^2 + b^2}{2}}$.\nCompare the arithmetic mean of positive numbers $a$ and $b$ with the arithmetic mean of $x$ and $y$.",
    "options": [],
    "answer": "(a + b)/2 ≥ (x + y)/2, with equality if and only if a = b",
    "solution": "Answer: $\\frac{a+b}{2} \\ge \\frac{x+y}{2}$.\nWe prove that\n$$\n\\frac{a+b}{2} \\ge \\frac{x+y}{2}.\n$$\n(*)\nIndeed,\n$$\n\\frac{a+b}{2} \\ge \\frac{x+y}{2} \\Leftrightarrow a+b \\ge \\sqrt{ab} + \\sqrt{\\frac{a^2+b^2}{2}} \\Leftrightarrow\n$$\n$$\n\\Leftrightarrow (a+b)^2 \\ge ab + \\frac{a^2+b^2}{2} + 2\\sqrt{ab \\cdot \\frac{a^2+b^2}{2}} \\Leftrightarrow\n$$\n$$\n\\Leftrightarrow (a+b)^2 \\ge 4\\sqrt{ab \\cdot \\frac{a^2+b^2}{2}} \\Leftrightarrow (a+b)^4 \\ge 8ab(a^2+b^2) \\Leftrightarrow\n$$\n$$\n\\Leftrightarrow \\left(\\frac{(a+b)^2}{ab}\\right)^2 \\ge \\frac{8(a^2+b^2)}{ab} \\Leftrightarrow \\left(\\frac{a}{b} + \\frac{b}{a} + 2\\right)^2 \\ge 8 \\cdot \\left(\\frac{a}{b} + \\frac{b}{a}\\right).\n$$\nLet $t = \\frac{a}{b} + \\frac{b}{a}$; then\n$$\n\\left(\\frac{a}{b} + \\frac{b}{a} + 2\\right)^2 \\ge 8 \\cdot \\left(\\frac{a}{b} + \\frac{b}{a}\\right) \\Leftrightarrow (t+2)^2 \\ge 8t \\Leftrightarrow (t-2)^2 \\ge 0.\n$$\nNote that the equality occurs in (*) if $a = b$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25455,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $H$ be an intersection point of the altitudes $AA_1$, $BB_1$, $CC_1$ of an acute-angled triangle $ABC$. Let $M$ and $N$ be the midpoints of the segments $BC$ and $AH$, respectively.\nProve that $MN$ is the perpendicular bisector of the segment $B_1C_1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note that the segments $B_1M$ and $C_1M$ are medians in the right-angled triangles $B_1BC$ ($\\angle B_1 = 90^\\circ$) and $C_1CB$ ($\\angle C_1 = 90^\\circ$) respectively. So, $B_1M = 0.5CB$ and $C_1M = 0.5CB$, therefore $B_1M = C_1M$, i.e. $M$ is equidistant from the ends of the segment $B_1C_1$.\n\nSimilarly, we have $B_1N = 0.5AH = C_1N$, so $N$ is equidistant from the ends of the segment $B_1C_1$.\n\nTherefore $MN$ is the perpendicular bisector of the segment $B_1C_1$, as required.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25456,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\Omega$ be the circumcircle of a triangle $ABC$. A circle passing through the vertex $A$ and touching $BC$ at point $X$ meets $\\Omega$ at point $Y$ (different from $A$). Let point $Z$ (different from $Y$) be the intersection point of the ray $YX$ and $\\Omega$.\n$$\n\\angle CAX = \\angle ZAB.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\omega$ be the circle passing through $A$ and touching the side $BC$ at $X$. Since the angle $YAX$ subtends the arc $YX$, we have $\\angle YAX = \\frac{1}{2} \\angle YX$.\n\nFurther, $\\angle YXC = \\frac{1}{2} \\angle YX$ (as the angle between the tangent $BC$ and the chord $XY$). So, $\\angle YAX = \\angle YXC$. Let $W$ be the intersection point of the ray $AX$ and $\\Omega$. Then $\\angle YAX = \\angle YAW$, so $\\angle YAW = \\angle YXC$. We have $\\angle BXZ = \\angle YXC$ as the vertical angles, so $\\angle BXZ = \\angle YAW$. Further, $\\angle YAW = \\angle YZW$ as inscribed angles of $\\Omega$ subtending the same arc. Therefore, $\\angle BXZ = \\angle YZW$, so $BC \\parallel WZ$. Hence $CW = ZB$, and then $\\angle CAW = \\angle ZAB$ as inscribed angles of $\\Omega$ subtending the equal arcs.\n\n![](attached_image_1.png)",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 25457,
    "subject": "Mathematics (Multi-modal)",
    "question": "$n$ $2 \\times 2$ squares are drawn on the Cartesian plane. The sides of these squares are parallel to the coordinate axes. It is known that the center of any square is not an inner point of any other square. Let $\\Pi$ be a rectangle such that it contains all these $n$ squares and its sides are parallel to the coordinate axes.\nProve that the perimeter of $\\Pi$ is no smaller than $4(\\sqrt{n} + 1)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "7. See solution of Problem A7a.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25458,
    "subject": "Mathematics (Multi-modal)",
    "question": "An $n \\times n$ ($n \\ge 4$) square is divided into $n^2$ unit cells. Find all possible values of $n$ such that this square can be covered with some layers of 4-cell figures of the following shape [ ] (i.e. each cell of the square must be covered with the same number of these figures).\n(The sides of each figure must coincide with the sides of the cells; the figures may be rotated but none of them can go beyond the bounds of the square.)\n(Jury)",
    "options": [],
    "answer": "n = 4k",
    "solution": "Answer: $n = 4k$, $k \\in \\mathbb{N}$.\nIf $n = 4k$, $k \\in \\mathbb{N}$, then this square can be covered with one layer (and then with any number of layers) of the figures [ ] [ ] [ ] [ ].\n\nLet $n = 2m + 1$, $m \\ge 2$. In this case we use chess coloring of the square. Suppose that the square is covered with $k$ layers of given figures. We write 1 and $-1$ in all black and in all white cells of the square respectively. Let $B(n)$ and $W(n)$ be the sums of the numbers in black and white cells of the square, respectively. Without loss of generality we can assume that $B(n) > W(n)$. It is evident that $B(n) + W(n) = 1$. Now we sum up the numbers in all cells of the square so that we count each number in the cell as many times as the number of the figures covering this cell. Let $S$ be this sum. Since $k$ is the number of the figures covering each cell, we have $S = k(B(n) + W(n)) = k$. On the other hand, any figure [ ] [ ] [ ] [ ] covers the same number of black and white cells, so the sum of the numbers in the cells covered with this figure is equal to 0. Therefore, $S = 0$, a contradiction.\nLet $n = 4m + 2$ ($m \\in \\mathbb{N}$). Using four colors we paint all cells of the square as it is shown in the figure (the number in the cell corresponds to the number of the color). Note that there is no cell with 4th color in the $2 \\times 2$ square in the right lower corner of the square. We place 1 in all cells painted second and third colors and place $-1$ in all remaining cells. As above we sum up the numbers in all cells of the square, then $S = 2k$, where $k$ is a number of layers. On the other hand, each figure covers exactly one cell of each color, so the sum of the numbers in the cells covered with this figure is equal to 0. Therefore, $S = 0$, a contradiction.\n\n![](attached_image_1.png)\n\n**Remark.** Similar arguments can be applied for the following coloring: see Fig. 1 for $n = 4m + 2$, and Fig. 2 for $n = 2m + 1$.\n![](attached_image_2.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25459,
    "subject": "Mathematics (Multi-modal)",
    "question": "$N$ numbers are marked in the set $\\{1, 2, ..., 2000\\}$ so that any pair of the numbers $(1, 2), (2, 4), ..., (1000, 2000)$ contains at least one marked number.\nFind the least possible value of $N$.",
    "options": [],
    "answer": "666",
    "solution": "For any odd number $a$, $a < 2000$, define the set\n$$\nS_a = \\{a \\cdot 2^i \\mid i \\ge 0,\\ a \\cdot 2^i \\le 2000\\}.\n$$\nNote that $\\{1, 2, ..., 2000\\} = \\bigcup_a S_a$, where $a$ varies through all odd numbers less than $2000$. Both numbers of the pair $(k, 2k)$ belong to one and the same $S_a$. So we need to find the least amount $m_a$ of the numbers that one should mark in any $S_a$ in order that any pair $(k, 2k)$ from $S_a$ contains at least one marked number. It is obvious that if $x_1 < x_2 < ... < x_n$ are all numbers from $S_a$, then this least amount equals $[n/2]$ (since at least one of any two neighboring numbers should be marked).\n\nFurther,\n$$\n|S_1| = 11 \\text{ since } 2^{10} < 2000 < 2^{11}, \\text{ thus } m_1 = 5;\n$$\n$$\n|S_3| = 10 \\text{ since } 3 \\cdot 2^9 \\le 2000 < 3 \\cdot 2^{10}, \\text{ thus } m_3 = 5;\n$$\n$$\n|S_5| = |S_7| = 9 \\text{ since } 7 \\cdot 2^8 \\le 2000 < 5 \\cdot 2^9, \\text{ thus } m_5 = m_7 = 4;\n$$\n$$\n|S_a| = 8 \\text{ for } a = 9, 11, 13, 15 \\text{ since } 15 \\cdot 2^7 \\le 2000 < 9 \\cdot 2^8, \\text{ thus } m_a = 4 \\text{ for these } a;\n$$\n$$\n|S_a| = 7 \\text{ for odd } a, 17 \\le a \\le 31 \\text{ (in total, 8 values) since } 31 \\cdot 2^6 \\le 2000 < 17 \\cdot 2^7, \\text{ thus } m_a = 3 \\text{ for these } a;\n$$\n$$\n|S_a| = 6 \\text{ for odd } a, 33 \\le a \\le 61 \\text{ (in total, 15 values) since } 61 \\cdot 2^5 \\le 2000 < 33 \\cdot 2^6, \\text{ thus } m_a = 3 \\text{ for these } a;\n$$\n$$\n|S_a| = 5 \\text{ for odd } a, 63 \\le a \\le 125 \\text{ (in total, 32 values) since } 125 \\cdot 2^4 \\le 2000 < 63 \\cdot 2^5, \\text{ thus } m_a = 2 \\text{ for these } a;\n$$\n$$\n|S_a| = 4 \\text{ for odd } a, 127 \\le a \\le 249 \\text{ (in total, 62 values) since } 249 \\cdot 2^3 \\le 2000 < 127 \\cdot 2^4, \\text{ thus } m_a = 2 \\text{ for these } a;\n$$\n$$\n|S_a| = 3 \\text{ for odd } a, 251 \\le a \\le 499 \\text{ (in total, 125 values) since } 499 \\cdot 2^2 \\le 2000 < 251 \\cdot 2^3, \\text{ thus } m_a = 1 \\text{ for these } a;\n$$\n$$\n|S_a| = 2 \\text{ for odd } a, 501 \\le a \\le 999 \\text{ (in total, 250 values) since } 999 \\cdot 2^1 \\le 2000 < 501 \\cdot 2^2, \\text{ thus } m_a = 1 \\text{ for these } a.\n$$\nSo we need to mark at least $1 \\cdot (250 + 125) + 2 \\cdot (62 + 32) + 3 \\cdot (15 + 8) + 4 \\cdot (4 + 2) + 5 \\cdot (1 + 1) = 375 + 188 + 69 + 24 + 10 = 666$ numbers. On the other hand, one can easily mark exactly $666$ numbers to satisfy the problem condition.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25460,
    "subject": "Mathematics (Multi-modal)",
    "question": "Do there exist numbers $a, b \\in \\mathbb{R}$ and a surjective function $f: \\mathbb{R} \\to \\mathbb{R}$ such that\n$$\nf(f(x)) = b x f(x) + a\n$$\nfor all real $x$?",
    "options": [],
    "answer": "No; there are no real a, b and surjective f satisfying the equation.",
    "solution": "(Solution by A. Asanau, Y. Dubovik, Y. Laurenau, V. Vityaz.) Suppose that such $a$, $b$, $f$ exist. Obviously $b \\neq 0$. Further, since $f$ is surjective, there exists a $\\lambda$ such that $f(\\lambda) = 0$. Set $x = \\lambda$ in\n$$\nf(f(x)) = b x f(x) + a, \\tag{1}\n$$\nthen $f(0) = f(f(\\lambda)) = b \\lambda f(\\lambda) + a = b \\lambda \\cdot 0 + a = a$. Further,\n$$\nf(a) = f(f(0)) = b \\cdot 0 \\cdot f(0) + a = a.\n$$\nNow $a = f(a) = f(f(a)) = b a f(a) + a = b a^2 + a$ whence $a = 0$. Thus (1) becomes\n$$\nf(f(x)) = b x f(x). \\tag{2}\n$$\nFurther, there exists a $\\sigma \\neq 0$ such that $f(\\sigma) = -1/b$. Set $x = \\sigma$ in (2), then\n$$\nf(-1/b) = f(f(\\sigma)) = b \\sigma f(\\sigma) = b \\sigma(-1/b) = -\\sigma. \\quad (3)\n$$\nHence\n$$\nf(-\\sigma) = f(f(-1/b)) = b(-1/b)f(-1/b) = \\sigma.\n$$\nTherefore,\n$$\n\\frac{1}{b} = f(\\sigma) = f(f(-\\sigma)) = b(-\\sigma)f(-\\sigma) = -b \\sigma^2\n$$\nand so $\\sigma = \\pm 1/b$.\nIf $\\sigma = -1/b$, then $\\sigma = -1/b = f(\\sigma) = f(-1/b)$, contrary to (3).\nIf $\\sigma = 1/b$, then in view of (3) we have\n$$\nf(-\\sigma) = -\\sigma \\implies -\\sigma = f(-\\sigma) = f(f(-\\sigma)) = b(-\\sigma)f(-\\sigma) = b \\sigma^2 \\implies \\\\\n\\qquad \\sigma = -1/b,\n$$\na contradiction.\n\nTherefore, there are no $a$, $b$, $f$ satisfying the problem condition.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25461,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given two hyperbolae $H_1$ and $H_2$ with the equations $y = 1/x$ and $y = -1/x$, respectively. A straight line meets $H_1$ at points $A$ and $B$, and meets $H_2$ at points $C$ and $D$. The lines tangent to $H_1$ at points $A$ and $B$ intersect at point $M$, and the lines tangent to $H_2$ at points $C$ and $D$ intersect at point $N$.\nProve that $M$ and $N$ are symmetric with respect to the origin of coordinates.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Without loss of generality we may assume that the positions of all hyperbolae, lines, and points look like in the figure (otherwise we can rotate the plane by the angle which is a multiple of $90^{\\circ}$, and rename the points).\n\n![](attached_image_1.png)\n\nLet $A(a; 1/a)$, $B(b; 1/b)$, $C(c; -1/c)$, $D(d, -1/d)$. Note that all numbers $a$, $b$, $c$, $d$ are pairwise distinct and $c < 0$, $d > a > b > 0$. Since the derivative of the function $y(x) = 1/x$ is equal to $y'(x) = -1/x^2$, the equations of the tangents to $H_1$ at $A$ and $B$ have the forms\n$$\ny = -\\frac{1}{a^2}(x-a) + \\frac{1}{a} \\quad \\text{and} \\quad y = -\\frac{1}{b^2}(x-b) + \\frac{1}{b}.\n$$\nLet $M(x_M; y_M)$ be the point of intersection of these tangents, then\n$$\n-\\frac{1}{a^2}(x-a) + \\frac{1}{a} = -\\frac{1}{b^2}(x-b) + \\frac{1}{b}, \\quad \\text{so } x_M = \\frac{2ab}{a+b}. \\quad \\text{Therefore}\n$$\n$$\ny_M = -\\frac{1}{a^2}(x_M - a) + \\frac{1}{a} = \\frac{2}{a+b}.\n$$\n\nSimilarly, since the derivative of the function $y(x) = -1/x$ is equal to $y'(x) = 1/x^2$, the equations of the tangents to $H_2$ at $C$ and $D$ have the forms\n$$\ny = \\frac{1}{c^2}(x-c) - \\frac{1}{c} \\quad \\text{and} \\quad y = \\frac{1}{d^2}(x-d) - \\frac{1}{d}.\n$$\nLet $N(x_N; y_N)$ be the point of intersection of these tangents, then\n$$\n\\frac{1}{c^2}(x-c) - \\frac{1}{c} = \\frac{1}{d^2}(x-d) - \\frac{1}{d}, \\quad \\text{so } x_N = \\frac{2cd}{c+d}. \\quad \\text{Therefore}\n$$\n$$\ny_N = \\frac{1}{c^2}(x_N - c) - \\frac{1}{c} = -\\frac{2}{c+d}.\n$$\nSince $A$, $B$, $C$, $D$ belong to the same line, we have $a+b = c+d$ and $ab = -cd$ (see the solution of Problem C.1). Hence $x_M = -x_N$ and $y_M = -y_N$, i.e., $M$ and $N$ are symmetric with respect to the origin.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 25462,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given two hyperbolae $H_1$ and $H_2$ with the equations $y = 1/x$ and $y = -1/x$, respectively. A straight line meets $H_1$ at points $A$ and $B$, and meets $H_2$ at points $C$ and $D$. Let $O$ be the origin of coordinates.\nProve that the areas of the triangles $OAC$ and $OBD$ are equal. (S. Mazanik)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Without loss of generality we can assume that the positions of all hyperbolae, lines, and points look like in the figure (otherwise we can rotate the plane by the angle which is a multiple of $90^\\circ$, and rename the points).\n![](attached_image_1.png)\nLet $A(a; 1/a)$, $B(b; 1/b)$, $C(c; -1/c)$, $D(d; -1/d)$. Note that all numbers $a$, $b$, $c$, $d$ are pairwise distinct and $c < 0$, $d > a > b > 0$. We write the equations of the line $l_{AB}$ and $l_{CD}$ passing through the pairs of points $A, B$ and $C, D$:\n$$\nl_{AB}: \\quad \\frac{x-a}{b-a} = \\frac{y-1/a}{1/b-1/a} \\Leftrightarrow y = -\\frac{1}{ab}x + \\frac{1}{a} + \\frac{1}{b},\n$$\n\n$$\nl_{CD}: \\frac{x-c}{d-c} = \\frac{y+1/c}{-1/d+1/c} \\Leftrightarrow y = \\frac{1}{cd}x - \\frac{1}{c} - \\frac{1}{d}\n$$\nSince these lines coincide, $l = l_{AB} = l_{CD}$, we have\n$$\nab = -cd, \\quad \\frac{1}{a} + \\frac{1}{b} = -\\frac{1}{c} - \\frac{1}{d} \\Rightarrow a+b=c+d. \\qquad (1)\n$$\nThe abscissa of the midpoint $M$ of the segment $AB$ is equal to $x_M = (a+b)/2$, and from (1) it follows that it is equal to the abscissa $(c+d)/2$ of the midpoint of the segment $CD$. Since the points $A$, $B$, $C$, $D$ lie on the same line, we see that $M$ is the midpoint of the segments $AB$ and $CD$. So $AC = CM + MA = MD + MB = BD$. Since the lengths of the altitudes of the triangles $OAC$ and $OBD$ from the vertex $O$ are equal, we see that the area of these triangles are equal.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25463,
    "subject": "Mathematics (Multi-modal)",
    "question": "Numbers $x_1$, $x_2$, $x_3$ ($x_1 < x_2 < x_3$) are the roots of the equation\n$$\nx^3 - 3x^2 + (a+2)x - a = 0,\n$$\nwhere $a$ is a real number.\nFind all possible values of the expression $4x_1 - x_1^2 + x_3^2$.",
    "options": [],
    "answer": "4",
    "solution": "Answer: 4.\nNote that 1 is the root of the equation\n$$\nx^3 - 3x^2 + (a+2)x - a = x^3 - x^2 - 2x^2 + 2x + ax - a = (x-1)(x^2 - 2x + a), \\quad (1)\n$$\nTherefore, since\n$$\nx^3 - 3x^2 + (a+2)x - a = x^3 - x^2 - 2x^2 + 2x + ax - a = (x-1)(x^2 - 2x + a),\n$$\nwe obtain the equivalent equation $(x-1)(x^2 - 2x + a) = 0$. So, two other roots of equation (1) are the roots of the equation\n$$\nx^2 - 2x + a = 0, \\quad (2)\n$$\ntheir sum is equal to 2, so one of them is less than 1 and another is greater than 1. Hence $x_2 = 1$, and $x_1$, $x_3$ are the roots of (2). Therefore,\n$$\n4x_1 - x_1^2 + x_3^2 = (x_3 + x_1)(x_3 - x_1) + 4x_1 = \\\\ = 2(x_3 - x_1) + 4x_1 = 2(x_3 + x_1) = 2 \\cdot 2 = 4.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 25464,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ such that\n$$\nn = q(q^2 - q - 1) = r(2r + 1)\n$$\nfor some primes $q$ and $r$.",
    "options": [],
    "answer": "2015",
    "solution": "(Solution by V. Vityaz.) First, $q \\neq r$ otherwise from\n$$\nq(q^2 - q - 1) = r(2r + 3) \\quad (1)\n$$\nwe would have $r^2 - r - 1 = 2r + 3$ which gives $r = 4$, which is not prime.\n\nHence (1) implies $q^2 - q - 1 = kr$, so $q^2 - q - 1 = kr$, $2r + 3 = kq$. Eliminating $r$ we obtain\n$$\n2q^2 - (2 + k^2)q + 3k - 2 = 0. \\quad (2)\n$$\nThe discriminant of this equation is equal to $D = k^4 + 4k^2 - 24k + 20$ and $D$ must be a perfect square. Note that $(k^2)^2 < D < (k^2 + 2)^2$ for $k > 5$. Hence\n$$\nD = (k^2 + 1)^2 \\Longleftrightarrow 2k^2 - 24k + 19 = 0.\n$$\nBut the last equation has no integer roots.\n\nIf $k = 1$, then (2) becomes $2k^2 - 3q + 1 = 0$, $q = 1$ which is not prime. Note that $k$ is odd since $kr = q^2 - q - 1$ is odd. If $k = 3$, then (2) has no integer solutions. If $k = 5$, then (2) becomes $2q^2 - 27q + 13 = 0$, then $q = 13$, $r = 31$, and $n = r(2r + 3) = 2015$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25465,
    "subject": "Mathematics (Multi-modal)",
    "question": "A circle intersects a parabola at four distinct points. Let $M$ and $N$ be the midpoints of the arcs of the circle which are outside the parabola.\nProve that the line $MN$ is perpendicular to the axis of the parabola.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We may assume that the parabola is defined by the equation $y = x^2$, while the circle is defined by the equation $(x-a)^2 + (y-b)^2 = R^2$. Let $A(a, b)$ be the center of the circle, $X_i(x_i, x_i^2)$, $i = 1, 2, 3, 4$, be common points of the circle and the parabola. Then $x_1, x_2, x_3, x_4$ are four roots of the equation\n$$\n(x - a)^2 + (x^2 - b)^2 = R^2 \\implies \n$$\n$$\nx^4 - (2b - 1)x^2 - 2a x + (a^2 + b^2 - R^2) = 0.\n$$\n![](attached_image_1.png)\nIt follows that $x_1 + x_2 + x_3 + x_4 = 0$ (Vieta's formula). Denote by $M(k, l)$, $N(m, n)$ the coordinates of the midpoints of the arcs $X_1X_2$, $X_3X_4$, respectively. Then, in particular, $\\overrightarrow{X_1X_2} \\perp \\overrightarrow{AM}$, which gives\n$$\n(x_1 - x_2)(k - a) + (x_1^2 - x_2^2)(l - b) = 0 \\Rightarrow k - a = -(x_1 + x_2)(l - b).\n$$\nSince $M$ belongs to the circle, we have $(k-a)^2 + (l-b)^2 = R^2$ hence\n$$\n((x_1+x_2)^2+1)(l-b)^2 = R^2,\n$$\nso $(l-b)^2 = R^2/((x_1+x_2)^2+1)$. Similarly,\n$$\n(n-b)^2 = R^2/((x_3+x_4)^2+1).\n$$\nSince $x_1 + x_2 = -(x_3+x_4)$, we get\n$$\n(l - b)^2 = (n - b)^2. \\tag{1}\n$$\nIt is not difficult to see that the slope of the line $MA$ is positive (because that of the line $X_1X_2$ is negative) hence $b > l$; similarly, $b > n$. Therefore from (1) it follows that $b - l = b - n$, or $l = n$, which yields $MN \\perp Oy$ as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25466,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve the equation in nonnegative integers $a$, $b$, $c$:\n$$\n3^a + 2^b + 2015 = 3c!\n$$",
    "options": [],
    "answer": "(4, 6, 6)",
    "solution": "(Solution by A. Sheremet.) Since $3c! > 2015$, we have $c \\ge 6$.\n\n1) For $a = 0$ from the given equation\n$$\n3^a + 2^b + 2015 = 3c! \\quad (1)\n$$\nwe have $2^b = 3(c! - 672)$, which is impossible.\n\n2) Let $a = 1$. If $c = 6$, then $2^b = 142$ -- there are no solutions. If $c \\ge 7$, then (1) implies $2^b + 2018 \\ge 7$, or $2^b + 2 \\equiv 0 \\pmod 7$, which is impossible.\n\n3) Let now $a \\ge 2$. Then $3^a \\ge 9$, hence from (1) it follows that $2^b \\equiv 1 \\pmod 9$. Therefore $b \\ge 6$.\nIf $b = 0$, then we have $3^a + 2016 = 3c!$, which is impossible for $c = 6$; for $c \\ge 7$ the obtained equality implies $3^a \\ge 7$, which is false.\nTherefore $b \\ge 6$ and from (1) it follows that $3^a \\equiv 1 \\pmod{16}$ which implies $a \\ge 4$. Let $a = 4t$, $b = 6q$. Then (1) becomes\n$$\n81^t + 64^q + 2015 = 3c! \\quad (2)\n$$\nThe case $c = 6$ leads to $t = q = 1$, i.e., $a = 4$, $b = 6$. The case $c \\ge 7$ leads to the congruence $4^t + 1^q + 6 \\equiv 0 \\pmod 7$, which is impossible. Thus the only solution is $(a, b, c) = (4, 6, 6)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 25467,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given two hyperbolae $H_1$ and $H_2$ with the equations $y = 1/x$ and $y = -1/x$, respectively. Let $M$ be an arbitrary point on $H_1$. Let $M_1$ and $M_2$ be the tangency points of the lines $MM_1$ and $MM_2$ with the hyperbola $H_2$.\nProve that the line $M_1M_2$ is tangent line of $H_1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Without loss of generality we may assume that the point $M(a; 1/a)$ lies on the upper-right half-hyperbola $H_1$, $y = 1/x$. $M_1(b_1; -1/b_1)$ and $M_2(b_2; -1/b_2)$ be the points of tangency with $H_2$, mentioned in the problem condition (see the Fig.).\n\n![](attached_image_1.png)\n\nSince the derivative of the function $y(x) = -1/x$ is equal to $y'(x) = \\frac{1}{x^2}$, the equation of the line tangent to $H_2$ at $(b; -1/b)$ has the form\n$$\ny = \\frac{1}{b^2}(x - b) - \\frac{1}{b}.\n$$\nSince $M(a; 1/a)$ belongs to this tangent, we have $\\frac{1}{a} = \\frac{1}{b^2}(a-b) - \\frac{1}{b}$. Therefore, the abscissae $b_1$ and $b_2$ of $M_1$ and $M_2$ satisfy the equation\n$$\nb^2 + 2ab - a^2 = 0. \\quad (1)\n$$\nThe equation of the straight line passing through $M_1$ can be written as $y = k(x - b_1) - \\frac{1}{b_1}$, and if this line passes through $M_2$, then\n\n$\\frac{1}{b_2} = k(b_2 - b_1) - \\frac{1}{b_1}$, so $k = \\frac{1}{b_1 b_2}$ since $b_1 \\neq b_2$. Therefore, the equation of the line $M_1 M_2$ has the form\n$$\ny = \\frac{1}{b_1 b_2} (x - b_1) - \\frac{1}{b_1} \\iff y = \\frac{1}{b_1 b_2} x - \\frac{1}{b_2} - \\frac{1}{b_1} \\iff y = \\frac{x - (b_1 + b_2)}{b_1 b_2}.\n$$\nSince $b_1$ and $b_2$ are the roots of (1), by Vieta's theorem, we have $b_1 b_2 = -a^2$, $b_1 + b_2 = -2a$. Then the equation of the line $M_1 M_2$ has the form\n$$\ny = \\frac{x + 2a}{-a^2}.\n$$\nNote that $y = -1/a$, for $x = -a$, i.e., the point $N(-a; -1/a)$ lying on the hyperbola $H_1$ belongs to the line $M_1 M_2$. It remains to note that the slope of the line tangent to $H_1$ at $N(-a; -1/a)$ is equal to $y'(-a) = -\\frac{1}{(-a)^2}$, i.e., coincides with the slope of the line $M_1 M_2$. Therefore, the line $M_1 M_2$ touches the hyperbola $H_1$ (at the point $N$).",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 25468,
    "subject": "Mathematics (Multi-modal)",
    "question": "Does there exist a function $f$, $f: \\mathbb{R} \\to \\mathbb{R}$, such that\n$$\n\\begin{cases}\n\\{f(x)\\} \\sin^2 x + \\{x\\} \\cos f(x) \\cos x = f(x), \\\\\n f(f(x)) = f(x),\n\\end{cases}\n$$\nfor all real $x$.\n(Here $\\{y\\}$ stands for the fractional part of $y$.)",
    "options": [],
    "answer": "No",
    "solution": "Assume that there exists a function $f(x)$ satisfying the problem condition:\n$$\n\\begin{cases}\n\\{f(x)\\} \\sin^2 x + \\{x\\} \\cos f(x) \\cos x = f(x), \\\\\n f(f(x)) = f(x),\n\\end{cases}\n$$\nfor all real $x$.\nReplacing $x$ by $f(x)$ in the first equality, we obtain\n$$\n\\{f(f(x))\\} \\sin^2 f(x) + \\{f(x)\\} \\cos^2 f(x) = f(f(x)).\n$$\nSince $f(f(x)) = f(x)$, from the obtained equality it follows that $\\{f(x)\\} = f(x)$. So $f: \\mathbb{R} \\rightarrow [0; 1]$.\nReplacing $x$ by $\\pi$, we have $-\\{\\pi\\} \\cos f(\\pi) = f(\\pi)$. Since $f(\\pi) \\in [0, 1]$ and $\\{\\pi\\} \\neq 0$, we see that the left-hand side of the last equality is negative, whereas the right-hand side is nonnegative, a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25469,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) After division of a positive integer $n$ by two positive integers one has two remainders different from zero.\nIs it possible for $n$ to be the sum of these two remainders?\n\nb) After division of a positive integer $n$ by $29$, $39$, and $59$ one has three nonzero remainders such that their sum is equal to $n$.\nFind all possible values of $n$.",
    "options": [],
    "answer": "a) No. b) 112.",
    "solution": "a) Let $n = a q_1 + r_1 = b q_2 + r_2$, $a, b \\in \\mathbb{N}$, $a < b$, $r_1 \\neq 0$, $r_2 \\neq 0$. Suppose that $n = r_1 + r_2$. Then $n = a q_1 + r_1 = b q_2 + r_2 = r_1 + r_2$, ($q_1 \\ge 0$, $q_2 \\ge 0$) and $r_1 < a$, $r_2 < b$, so $b q_2 = r_1 < a \\le b$. The inequality $b q_2 < b$ holds only if $q_2 = 0$. But then $r_1 = b q_2 = b \\cdot 0 = 0$, a contradiction.\n\nb) Let, by condition,\n$$\nn = 29 q_1 + r_1 = 39 q_2 + r_2 = 59 q_3 + r_3 = r_1 + r_2 + r_3,\n$$\n$r_1 < 29$, $r_2 < 39$, $r_3 < 59$. From these inequalities it follows that $59 q_3 = r_1 + r_2 \\le 28 + 38 = 66$, so $q_3 = 1$. Then\n$$\nr_1 + r_2 = 59. \\quad (1)\n$$\nFurther, $39 q_2 = r_1 + r_3 \\le 28 + 58 = 86$, so $q_2 \\le 2$.\nConsider two cases:\n1) $q_2 = 1$. Then\n$$\nr_1 + r_3 = 39. \\quad (2)\n$$\n$98 = 59 + 39 = r_1 + r_2 + r_1 + r_3 = n + r_1 = 29 q_1 + 2 r_1,$\ni.e., $29 q_1 + 2 r_1 = 98$, so $q_1$ is even and $29 q_1 < 98$, hence $q_1 = 2$. Then $r_1 = \\frac{1}{2}(98 - 2 \\cdot 29) = 20$. But from (1) it follows that $r_2 = 59 - 20 = 39$, which is impossible since $r_2 < 39$.\n\n2) $q_2 = 2$. Then\n$$\nr_1 + r_3 = n - r_2 = 39 q_2 = 78. \\qquad (3)\n$$\nThen from (1) and (3) we obtain $29 q_1 + 2 r_1 = 137$, so $q_1$ is odd and $29 q_1 < 137$, hence either $q_1 = 1$ or $q_1 = 3$. If $q_1 = 1$, then $2 r_1 = 108$, i.e., $r_1 = 54$, a contradiction. If $q_1 = 3$, then $r_1 = 25$, $r_2 = 34$, $r_3 = 53$, and $n = 25 + 34 + 53 = 112$, which satisfies the problem condition.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25470,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the sequence of digits $2, 0, 2, 9, 3, \\dots$ any digit is equal to the last digit in the decimal representation of the sum of four previous digits.\n\nDo the four numbers $2, 0, 1, 5$ in that order occur in the sequence? (Folklore)",
    "options": [],
    "answer": "Yes",
    "solution": "In the given sequence consider all possible quadruple of successive digits. There are finite number of possible quadruples (no more than $10^4$). So if we proceed the sequence sufficiently long, some of the quadruple will occur more than once:\n$$\n2029 \\dots d\\ c\\ b\\ a \\dots d_1\\ c_1\\ b_1\\ a_1 \\dots\n$$\n(here $d = d_1$, $c = c_1$, $b = b_1$, $a = a_1$). Note that any quadruple uniquely defines the preceding digits. That is, if $e$ and $e_1$ precede to $d\\ c\\ b\\ a$ and $d_1\\ c_1\\ b_1\\ a_1$, respectively, then $e = e_1$:\n$$\n2029 \\dots e\\ d\\ c\\ b\\ a \\dots e_1\\ d_1\\ c_1\\ b_1\\ a_1 \\dots\n$$\nHence the quadruple $2029$ should occur in the given sequence somewhere between $d\\ c\\ b\\ a$ and $d_1\\ c_1\\ b_1\\ a_1$:\n$$\n2029 \\dots d\\ c\\ b\\ a \\dots 2029 \\dots d_1\\ c_1\\ b_1\\ a_1 \\dots\n$$\nNow proceeding to the left we can easily see which digits precede the second quadruple $2029$:\n$$\n2029 \\dots d\\ c\\ b\\ a \\dots 2015\\ 84\\ 85\\ 52\\ 029\\ d_1\\ c_1\\ b_1\\ a_1 \\dots\n$$\nWe see that the quadruple $2015$ occurs in the sequence.\n\n*Remark.* Considering the given sequence modulo $2$ we see that it is periodic with $5$ as a period. Further, considering the sequence modulo $5$, starting from $2024$ ($\\equiv 2029 \\pmod{5}$) after $78$ moves (it takes about ten minutes) we can get $1012$ which are $3 \\cdot 2, 3 \\cdot 0, 3 \\cdot 2, 3 \\cdot 4$ modulo $5$. So after $4 \\cdot 78$ moves we get $3^4 \\cdot 2, 3^4 \\cdot 0, 3^2, 3^4 \\cdot 4$ which is the starting quadruple $2024$. Thus we conclude that the sequence modulo $5$ has period $4 \\cdot 78 = 312$. So the initial sequence is periodic with period $5 \\cdot 312 = 1560$. So to get the quadruple $2015$ in the sequence we need to perform $1560 - 9 = 1551$ moves.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25471,
    "subject": "Mathematics (Multi-modal)",
    "question": "After division of a positive integer $n$ by $29$, $41$, and $59$ one has three nonzero remainders such that their sum is equal to $n$.\nFind all possible values of $n$.",
    "options": [],
    "answer": "79 and 114",
    "solution": "Let, by condition,\n$$\nn = 29q_1 + r_1 = 41q_2 + r_2 = 59q_3 + r_3 = r_1 + r_2 + r_3,\n$$\n$r_1 < 29$, $r_2 < 41$, $r_3 < 59$. From these inequalities it follows that $59q_3 = r_1 + r_2 \\le 28 + 40 = 68$, so $q_3 = 1$. Then\n$$\nr_1 + r_2 = 59. \\quad (1)\n$$\n\nFurther, $41q_2 = r_1 + r_3 \\le 28 + 58 = 86$, so $q_2 \\le 2$. Consider two cases:\n$$\n1) \\quad q_2 = 1. \\quad \\text{Then} \\qquad r_1 + r_3 = 41. \\qquad (2)\n$$\nFrom (1) and (2) it follows that\n$$\n100 = 59 + 41 = r_1 + r_2 + r_1 + r_3 = n + r_1 = 29q_1 + 2r_1,\n$$\ni.e., $29q_1 + 2r_1 = 100$. Hence $q_1$ is even and $29q_1 < 100$, so $q_1 = 2$. Then $r_1 = \\frac{1}{2}(100 - 2 \\cdot 29) = 21$, $r_2 = 38$, $r_3 = 20$, and $n = 21 + 38 + 20 = 79$, which satisfies the problem condition.\n\n2) $q_2 = 2$. Then\n$$\nr_1 + r_3 = n - r_2 = 41q_2 = 82. \\qquad (3)\n$$\nFrom (1) and (3) it follows that $29q_1 + 2r_1 = 141$, so $q_1$ is odd and $29q_1 < 141$. Hence either $q_1 = 1$ or $q_1 = 3$. If $q_1 = 1$, then $2r_1 = 122$, i.e., $r_1 = 61$, a contradiction. If $q_1 = 3$, then $r_1 = 27$, $r_2 = 32$, $r_3 = 55$, and $n = 27 + 32 + 55 = 114$, which satisfies the problem condition.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25472,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs $(n; m)$ of positive integers $n$ and $m$ satisfying the equality\n$$\nn! + 505 = m^2.\n$$",
    "options": [],
    "answer": "(4, 23), (5, 25), (6, 35)",
    "solution": "Answer: $(4; 23)$, $(5; 25)$, $(6; 35)$.\nIt is easy to see that the numbers $1! + 505 = 506$, $2! + 505 = 507$, $3! + 505 = 511$ are not perfect squares. Further, $4! + 505 = 529 = 23^2$, $5! + 505 = 625 = 25^2$, $6! + 505 = 1225 = 35^2$. So we have three pairs $(4; 23)$, $(5; 25)$, $(6; 35)$ satisfying the problem conditions.\n\nShow that there are no other solutions of the initial equation. Indeed, the numbers $7! + 505 = 5545$ and $9! + 505 = 363385$ are not perfect squares since they are divisible by $5$ but they are not divisible by $25$. The number $8! + 505 = 40825 = 25 \\cdot 1633$ is not a perfect square since $1633$ is not a perfect square. At last, if $n \\ge 10$, then the decimal representation of $n!$ ends by at least two zeros, so the decimal representation of the number $n! + 505 = m^2$ ends with the digits $05$. Therefore, this number is divisible by $5$ but is not divisible by $25$, so this number is not a perfect square.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25473,
    "subject": "Mathematics (Multi-modal)",
    "question": "The medians $AM$ and $BN$ of a triangle $ABC$ are the diameters of the circles $\\omega_1$ and $\\omega_2$.\nIf $\\omega_1$ touches the altitude $CH$ prove that $\\omega_2$ also touches $CH$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The statement easily follows from the following fact: if points $A_1$, $B_1$ are chosen on the sides $AC$, $BC$ of the triangle $ABC$, respectively, then the radical axis of the circles with the diameters $AA_1$ and $BB_1$ contains the orthocenter of the triangle $ABC$.\n\nIn case when $AA_1$ and $BB_1$ are medians the centerline of $\\omega_1$ and $\\omega_2$ is obviously parallel to the side $AB$, then the radical axis is perpendicular to $AB$, thus the statement follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25474,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a cyclic quadrilateral $ABCD$ with $AB = AD$. Points $M$ and $N$ are marked on the sides $CD$ and $BC$, respectively, so that $DM + BN = MN$.\n\nProve that the circumcenter of the triangle $AMN$ belongs to the segment $AC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "On the prolongation of the segment $CD$ over $D$ we mark the point $K$ such that $DK = BN$. Then the triangles $KDA$ and $NBA$ are equal since $\\angle KDA = \\angle NBA$, $KD = BN$, $DA = AB$. Hence $KA = NA$, $KM = DK + DM = BN + DM = NM$, so that the triangles $KMA$ and $NMA$ are equal. It easily follows that $\\angle MAN = 0.5\\angle DAB$. Let $O$, $C$ be the intersection points of $AC$ and the circumcircle of the triangle $MCN$.\n\nSince $\\angle MCO = \\angle DCA = \\angle BCA = \\angle NCO$, we have $MO = NO$. Besides,\n$$\n\\angle MON = 180^\\{\\circ\\} - \\angle MCN = 180^\\{\\circ\\} - \\angle DCB = \\angle DAB = 2\\angle MAN.\n$$\nThis equality along with $MO = NO$ means that $O$ is the circumcenter of the triangle $MAN$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25475,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let the incircle of the triangle $ABC$ touch the side $AB$ at point $Q$; the incircles of the triangles $QAC$ and $QBC$ touch $AQ$, $AC$ and $BQ$, $BC$ at points $P$, $T$ and $D$, $F$, respectively.\nProve that $PDFT$ is a cyclic quadrilateral.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(Solution by A. Gaponenko, D. Voynov.) First, note that incircles of the triangles $QAC$ and $QBC$ touch $CQ$ at the same point $X$ (well-known fact). Hence $CF = CX = CT$. Also $AP = AT$, $BF = BD$. Now we have\n$$\n\\angle TFD = 180^\\circ - \\angle TFC - \\angle BFD =\n$$\n$$\n= 180^\\circ - (90^\\circ - \\frac{1}{2} \\angle C) - (90^\\circ - \\frac{1}{2} \\angle B) =\n$$\n$$\n= \\frac{1}{2}(\\angle C + \\angle B) = 90^\\circ - \\frac{1}{2} \\angle A = \\angle APT.\n$$\nIt follows that\n$$\n\\angle TFD + \\angle TPD = \\angle TFD + (180^\\circ - \\angle APT) = 180^\\circ\n$$\nwhich finishes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25476,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $AM$ be the median of the triangle $ABC$, $B_1$ be the foot of the perpendicular $BB_1$ from $B$ onto the bisector of the angle $BMA$, $C_1$ be the foot of the perpendicular $CC_1$ from $C$ onto the bisector of the angle $AMC$. The ray $MA$ intersects the segment $B_1C_1$ at a point $A_1$.\nFind the value of the ratio $B_1A_1/A_1C_1$.",
    "options": [],
    "answer": "1",
    "solution": "Answer: 1.\nLet $\\angle BMA = 2x$, then $\\angle AMC = 180^\\circ - 2x$. Since $MB_1$ is the bisector of the angle $BMA$, we have $\\angle BMB_1 = \\angle B_1MA = x$.\nSimilarly, $\\angle AMC_1 = \\angle C_1MC = 90^\\circ - x$. The triangle $BMB_1$ is a right-angled triangle, so $\\angle B_1BM = 90^\\circ - \\angle B_1MB = 90^\\circ - x$. The right-angled triangles $BMB_1$ and $MCC_1$ are equal ($BM = MC$, $\\angle B_1BM = \\angle C_1MC = 90^\\circ - x$). So $BB_1 = MC_1$. Since $\\angle B_1BM = \\angle C_1MC$, we have $BB_1 \\parallel MC_1$. Therefore $BMC_1B_1$ is a parallelogram. Since $B_1C_1 \\parallel BM$, we have $\\angle BMB_1 = \\angle A_1B_1M = \\angle B_1MA_1$.\nSo, the triangle $B_1MA_1$ is an isosceles triangle: $MA_1 = A_1B_1$. Similarly, the triangle $MC_1A_1$ is an isosceles triangle: $MA_1 = C_1A_1$. Therefore, we have $B_1A_1/A_1C_1 = MA_1/MA_1 = 1$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25477,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sides $AB$ and $CD$ of the trapezoid $ABCD$ ($BC \\parallel AD$) are the diameters of the circles $\\Gamma_{AB}$ and $\\Gamma_{CD}$. $\\Gamma_{AB}$ meets the segments $AC$ and $BD$ at points $M$ and $N$, respectively. $\\Gamma_{CD}$ meets the segments $AC$ and $BD$ at points $K$ and $L$, respectively.\nProve that $NK \\parallel ML$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Construct the circumference $\\Gamma_{BC}$ with the diameter $BC$.** $\\Gamma_{BC}$ passes through $M$ and $L$. Indeed, $\\angle AMB = 90^\\circ$ ($AB$ is the diameter of $\\Gamma_{AB}$) then $\\angle BMC = 180^\\circ - \\angle AMB = 90^\\circ$, so $M$ belongs to $\\Gamma_{BC}$. Similarly\n$$\n\\angle DLC = 90^\\circ \\Rightarrow \\angle BLC = 180^\\circ - \\angle DLC = 90^\\circ,\n$$\nso $L$ also belongs to $\\Gamma_{BC}$. Now construct the circumference $\\Gamma_{AD}$ with the diameter $AD$. We conclude similarly that $N$ and $K$ belong to $\\Gamma_{AD}$.\n\nNote that\n$$\n\\angle NKM = \\angle NKA = 0.5 \\sim AN = \\angle NDA = \\angle BDA\n$$\nand\n$$\n\\angle KML = \\angle CML = 0.5 \\sim CL = \\angle CBL = \\angle CBD.\n$$\nSince $BC \\parallel AD$, we have $\\angle BDA = \\angle CBD$, and so $\\angle NKM = \\angle KML$, which gives $NK \\parallel ML$, as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25478,
    "subject": "Mathematics (Multi-modal)",
    "question": "The incircle of the triangle $ABC$ touches the sides $AC$ and $BC$ at points $P$ and $Q$, respectively, $N$ and $M$ are the midpoints of $AC$ and $BC$, respectively. Let $X = AM \\cap BP$, $Y = BN \\cap AQ$.\nGiven $C, X, Y$ are collinear prove that $CX$ is the angle bisector of the angle $ACB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(Solution by V. Vityaz, A. Gaponenko.) Let without loss of generality $CX \\ge CY$. Then $AP \\le AN, BM \\le BQ$ (see the Fig.) By the Menelaus theorem, we have: for the triangle $AQM$ and the line $XC$\n\n![](attached_image_1.png)\n\n$$\n\\frac{AY}{YQ} \\cdot \\frac{QC}{CM} \\cdot \\frac{MX}{XA} = 1, \\quad (1)\n$$\n\nfor the triangle $AQC$ and the line $NB$\n$$\n\\frac{AY}{YQ} \\cdot \\frac{QB}{BC} \\cdot \\frac{CN}{NA} = 1, \\quad (2)\n$$\n\nfor the triangle $AMC$ and the line $PB$\n$$\n\\frac{AX}{XM} \\cdot \\frac{MB}{BC} \\cdot \\frac{CP}{PA} = 1. \\qquad (3)\n$$\nLet $AC = b$, $AB = c$, $BC = a$, $p$ be the semiperimeter of the triangle $ABC$. Then (2) and (3) give $AY/YQ = a/(p-b)$ and $MX/XA = 0.5(p-c)/(p-a)$, respectively. Substituting these equalities in (1) we get\n$$\n\\frac{a}{p-b} \\cdot \\frac{2(p-c)}{a} \\cdot \\frac{p-c}{2(p-a)} = 1 \\iff (p-c)^2 = (p-a)(p-b) \\iff\n$$\n$$\na^2 + b^2 = c(a+b) \\iff a(p-c) = b(p-b) \\iff \\frac{YA}{YQ} = \\frac{CA}{CQ},\n$$\nthe last proportion is equivalent to the fact that $CY$ bisects the angle $ACQ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25479,
    "subject": "Mathematics (Multi-modal)",
    "question": "Alice has $n^2$ sweets. These sweets are distributed among $n$ boxes ($n > 2$). Alice chooses some two of these boxes and if the total number of the sweets in these two boxes is even, then she redistributes the sweets so that the numbers of the sweets in these two boxes will be equal. Otherwise she chooses another pair of the boxes and tries to realize the same procedure.\nDetermine all values of $n$ for which Alice can equalize the numbers of the sweets in all boxes using this procedure regardless of the initial distribution of the sweets.",
    "options": [],
    "answer": "n = 2^m with m > 1",
    "solution": "Answer: $n = 2^m$, where $m \\in \\mathbb{N}$, $m > 1$.\n\nThe goal of Alice is to obtain exactly $n$ sweets in any box. The number $a-n$ is said to be the distance if $a$ is a number of the sweets in the box. Note that the sum of the distances is equal to zero at any moment. Now we see that Alice's goal is to make all distances to be zeros. If there are two boxes with the distances $d_1$ and $d_2$ of the same parity, then Alice can decrease the absolute values of the distances for these boxes if and only if $d_1 \\neq d_2$. The necessity of this inequality is obvious. To prove the sufficiency we suppose that $|d_1| \\ge |d_2|$:\n\n1) if $|d_1| > |d_2|$, then new distances of these boxes satisfy the inequality $|(d_1 + d_2)/2| \\le (|d_1| + |d_2|)/2 < (|d_1| + |d_1|)/2 = |d_1|$;\n\n2) if $d_1 = -d_2$, then new distances of these boxes are equal to zero.\n\nTherefore, Alice cannot continue decreasing the absolute values of the distances if and only if at some moment either all distances are equal to zeros (and the goal is attained) or $k$ distances are even and equal to some $\\alpha > 0$, and all other distances are odd and equal to some $\\beta < 0$.\n\nLet $n = 2^m$ ($m > 1$). The sum of all distances is equal to zero at any moment. So, if $k$ distances are even and equal to $\\alpha$, and all other $n-k$ distances are odd and equal to $\\beta$, then $\\alpha k + \\beta(n-k) = 0$ which yields $k(\\alpha - \\beta) = 2^m\\beta$. But the last equality is impossible since $\\alpha - \\beta$ is odd and $k < 2^m$. Therefore, if $n = 2^m$ ($m > 1$), then Alice can equalize the numbers of the sweets in all boxes.\n\nNow let $n \\neq 2^m$. Then $n = (2p + 1) \\cdot 2^m$ for some positive integer $p$ and nonnegative integer $m$. Suppose that there are $k = 2^m$ boxes with the distance $\\alpha = 2p$ and all other $n-k = 2p2^m$ boxes with the distance $\\beta = -1$. Note that this distribution of the sweets is possible since the equality $\\alpha k + \\beta(n-k) = 0$ holds. Since the numbers $\\alpha$ and $\\beta$ have the different parities, Alice cannot equalize the numbers of the sweets in all boxes.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25480,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of polynomials $p(x), q(x) \\in \\mathbb{R}[x]$ satisfying the equality\n$$\np(x^2) = p(x)q(1-x) + p(1-x)q(x)\n$$\nfor all real $x$.",
    "options": [],
    "answer": "All solutions are given by constant p(x) = a. If a = 0, q(x) is arbitrary in R[x]. If a ≠ 0, then q must satisfy q(x) + q(1 − x) = 1 for all real x, equivalently q(x) = (x − 1/2) r((x − 1/2)^2) + 1/2 for some polynomial r(x) in R[x].",
    "solution": "(Solution by A. Asanau, D. Voynov.) Replace $x$ by $1-x$ in\n$$\np(x^2) = p(x)q(1-x) + p(1-x)q(x), \\quad (1)\n$$\nthen we have $p((x-1)^2) = p(1-x)q(x) + p(x)q(1-x) = p(x^2)$. It follows that the polynomial $p(x^2)$ is periodic, i.e., $p(x)$ is a constant polynomial, $p(x) = a$ for some $a \\in \\mathbb{R}$.\nIf $a=0$, then (1) is valid for any arbitrary polynomial $q(x)$.\nIf $a \\neq 0$, then (1) becomes $q(1-x) + q(x) = 1$ for all $x$. Replacing $x$ by $x+0.5$ we can rewrite the last equality as\n$$\nq(0.5 - x) - 0.5 + q(0.5 + x) - 0.5 = 0. \\quad (2)\n$$\nLet $h(x) = q(0.5 + x) - 0.5$. Then (2) shows that $h(-x) = -h(x)$, so we may write $h(x) = xr(x^2)$ for some polynomial $r(x)$.\nHence $q(0.5 + x) = xr(x^2) + 0.5$. Replace $x$ by $x - 0.5$ here, we obtain $q(x) = (x - 0.5)r((x - 0.5)^2) + 0.5$ which obviously satisfies the condition.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 25481,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that $(a+b+c)^5 \\ge 81(a^2+b^2+c^2)abc$ for any positive real numbers $a$, $b$, $c$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Write $\\sigma_1 = a + b + c$, $\\sigma_2 = ab + bc + ca$, $\\sigma_3 = abc$. Then the inequality we have to prove becomes\n$$\n\\sigma_1^5 \\ge 81(\\sigma_1^2 - 2\\sigma_2)\\sigma_3 \\iff \\sigma_1^5 + 162\\sigma_2\\sigma_3 \\ge 81\\sigma_1^2\\sigma_3. \\quad (1)\n$$\n\nNow using the AM-GM inequality for three summands $\\sigma_1^5$, $81\\sigma_2\\sigma_3$, $81\\sigma_2\\sigma_3$ we get\n$$\n\\sigma_1^5 + 81\\sigma_2\\sigma_3 + 81\\sigma_2\\sigma_3 \\ge 3\\sqrt[3]{81^2\\sigma_1^5\\sigma_2^2\\sigma_3^2}.\n$$\nSo for (1) to be proved it suffices to verify that\n$$\n3\\sqrt[3]{81^2\\sigma_1^5\\sigma_2^2\\sigma_3^2} \\ge 81\\sigma_1^2\\sigma_3 \\iff 81^2\\sigma_1^5\\sigma_2^2\\sigma_3^2 \\ge 27^3\\sigma_1^6\\sigma_3^3 \\iff \\sigma_2^2 \\ge 3\\sigma_1\\sigma_3.\n$$\nThe last inequality is well known.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25482,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f, g: [0; 1] \\to [0; 1]$, satisfying the following conditions:\n$$\n\\begin{cases}\n\\{f(x)\\} \\sin^2 x + \\{x\\} \\cos f(x) \\cos x = f(x), \\\\\n f(f(x)) = f(x),\n\\end{cases}\n$$\nfor all real $x$.\n(Here $\\{y\\}$ stands for the fractional part of $y$.)",
    "options": [],
    "answer": "f(x) = x for all x in [0,1]",
    "solution": "Let $f(x)$ satisfies the problem condition:\n$$\n\\left\\{ \n\\begin{array}{l} \n\\{f(x)\\} \\sin^2 x + \\{x\\} \\cos f(x) \\cos x = f(x), \\\\\n f(f(x)) = f(x), \n\\end{array} \n\\right. \n\\qquad (1) \n\\qquad (2)\n$$\nfor all real $x$.\nReplacing $x$ by $f(x)$ in (1) we obtain:\n$$\n\\{f(f(x))\\} \\sin^2 f(x) + \\{f(x)\\} \\cos^2 f(x) = f(f(x)).\n$$\nFrom (2) it follows that $\\{f(x)\\} = f(x)$. So $f: [0, 1] \\to [0, 1)$.\nFrom (1) and the equality $\\{f(x)\\} = f(x)$ it follows that:\n$$\n\\{x\\} \\cos f(x) \\cos x = f(x)(1 - \\sin^2 x) = f(x) \\cos^2 x. \\quad (3)\n$$\nWe have $\\cos x > 0$ and $\\{x\\} = x$ for $x \\in [0, 1]$. Therefore from (3) we have $x \\cos f(x) = f(x) \\cos x$ for $x \\in [0, 1]$. Dividing the last equality by $\\cos x \\cos f(x)$ we obtain:\n$$\n\\frac{x}{\\cos x} = \\frac{f(x)}{\\cos f(x)}. \\qquad (4)\n$$\nConsider the function $g(x) = x/\\cos(x)$, $g: [0, 1] \\to \\mathbb{R}$. It is increasing since $x$ increases and $\\cos x$ decreases. Since in view of (4) $g(x) = g(f(x))$ for $x \\in [0, 1]$, we obtain $f(x) = x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25483,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given positive real numbers $a, b, c, d$, with\n$$\n\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{1}{d} = 1,\n$$\nprove that\n$$\n\\frac{a+b}{a^2 - ab + b^2} + \\frac{b+c}{b^2 - bc + c^2} + \\frac{c+d}{c^2 - cd + d^2} + \\frac{d+a}{d^2 - da + a^2} \\le 2.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $x = 1/a$, $y = 1/b$, $z = 1/c$, $u = 1/d$. Then $x + y + z + u = 1$, $xyzu \\ne 0$, and the required inequality has the form\n$$\n\\frac{xy(x+y)}{x^2-xy+y^2} + \\frac{yz(y+z)}{y^2-yz+z^2} + \\frac{zu(z+u)}{z^2-zu+u^2} + \\frac{ux(u+x)}{u^2-ux+x^2} \\le 2, \\quad (1)\n$$\nsince\n$$\n\\frac{a+b}{a^2-ab+b^2} = \\frac{ab(b^{-1}+a^{-1})}{a^2b^2(b^{-2}-a^{-1}b^{-1}+a^{-2})} = \\frac{xy(y+x)}{y^2-xy+x^2}.\n$$\n\n(similar arguments apply to the remaining summands). Since for $x \\ne 0$ and $y \\ne 0$ we have $x^2-xy+y^2 = x^2-xy+y^2/4+3y^2/4 = (x-y/2)^2+3y^2/4 > 0$ and $0 \\le (x-y)^2 = x^2-2xy+y^2 = (x^2-xy+y^2)-xy$. So $x^2-xy+y^2 \\ge xy$ and $\\frac{xy}{ux} \\le 1$. Similarly, $y^2-yz+z^2 \\le 1$, $\\frac{zu}{z^2-zu+u^2} \\le 1$,\n$\\frac{u^2-ux+x^2}{ux} \\le 1$. Therefore\n$$\n\\frac{xy(x+y)}{x^2-xy+y^2} + \\frac{yz(y+z)}{y^2-yz+z^2} + \\frac{zu(z+u)}{z^2-zu+u^2} + \\frac{ux(u+x)}{u^2-ux+x^2} \\le\n$$\n$$\n\\le (x+y) + (y+z) + (z+u) + (u+x) = 2(x+y+z+u),\n$$\nand since $x+y+z+u=1$, we obtain the required inequality (1).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25484,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that\n$$\n|(a-b)(b-c)(c-d)(d-a)| \\le \\frac{abcd}{4},\n$$\nif real numbers $a, b, c, d$ belong to the segment $[1; 2]$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "It is evident that\n$$\n|(a-b)(b-c)(c-d)(d-a)| \\le \\frac{abcd}{4} \\iff\n$$\n$$\n\\frac{(a-b)^2}{ab} \\cdot \\frac{(b-c)^2}{bc} \\cdot \\frac{(c-d)^2}{cd} \\cdot \\frac{(d-a)^2}{da} \\le \\frac{1}{16}. \\qquad (*)\n$$\n\nNote that\n$$\n\\frac{(a-b)^2}{ab} \\le \\frac{1}{2}. \\qquad (1)\n$$\nIndeed, we have\n$$\n\\frac{(a-b)^2}{ab} \\le \\frac{1}{2} \\Leftrightarrow 2(a-b)^2 \\le ab \\Leftrightarrow 2\\left(\\frac{a}{b}\\right)^2 - 5\\left(\\frac{a}{b}\\right) + 2 \\le 0 \\Leftrightarrow \\\\\n2\\left(\\frac{a}{b} - 2\\right)\\left(\\frac{a}{b} - \\frac{1}{2}\\right) \\le 0.\n$$\nThe last inequality holds since $a, b \\in [1, 2]$.\nSimilarly we have\n$$\n\\frac{(b-c)^2}{bc} \\le \\frac{1}{2}, \\qquad (2)\n$$\n$$\n\\frac{(c-d)^2}{cd} \\le \\frac{1}{2}, \\qquad (3)\n$$\n$$\n\\frac{(d-a)^2}{da} \\le \\frac{1}{2}. \\qquad (4)\n$$\nMultiplying (1), (2), (3), (4), we obtain the required inequality (*).\n\nNote that the equality occurs when\n$$\n\\{(a, b, c, d) = (2, 1, 2, 1), (1, 2, 1, 2)\\}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25485,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given integers $a$, $b$, $c$, with\n$$\n\\frac{a^2}{a^2 + b^2} + \\frac{c^2}{a^2 + c^2} = \\frac{2c}{b+c},\n$$\nprove that the product $bc$ is a perfect square.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $b = 0$ or $c = 0$, then $bc$ is a perfect square. If $a = 0$, then from\n$$\n\\frac{a^2}{a^2 + b^2} + \\frac{c^2}{a^2 + c^2} = \\frac{2c}{b+c} \\quad (1)\n$$\nit follows that $b = c$, so $bc$ is a perfect square. Therefore we can assume that all $a$, $b$, $c$ are different from $0$. Then we can rewrite (1) as\n$$\n\\frac{1}{1 + \\frac{b^2}{a^2}} + \\frac{1}{1 + \\frac{a^2}{c^2}} = \\frac{2}{1 + \\frac{b}{a} \\cdot \\frac{c}{c}}\n$$\nSet $x = \\frac{b}{a}$ and $y = \\frac{a}{c}$, we have\n$$\n\\frac{1}{1 + x^2} + \\frac{1}{1 + y^2} = \\frac{2}{1 + xy} \\quad \\Leftrightarrow \\quad \\frac{2 + x^2 + y^2}{(1 + x^2)(1 + y^2)} = \\frac{2}{1 + xy} \\Leftrightarrow\n$$\n$$\n(2 + x^2 + y^2)(1 + xy) = 2(1 + x^2)(1 + y^2) \\Leftrightarrow\n$$\n$$\n2 + x^2 + y^2 + 2xy + x^3y + xy^3 = 2 + 2x^2 + 2y^2 + 2x^2y^2 \\Leftrightarrow\n$$\n$$\nx^3y - 2x^2y^2 + xy^3 - x^2 + 2xy - y^2 = 0 \\Leftrightarrow xy(x-y)^2 - (x-y)^2 = 0 \\Leftrightarrow (x-y)^2(xy-1) = 0.\n$$\nThen, if $x - y = 0$, that is $x = y$, then $\\frac{b}{a} = \\frac{a}{c}$, and $bc = a^2$ is a perfect square. If $xy = 1$, that is $\\frac{b}{a} \\cdot \\frac{a}{c} = 1 \\Leftrightarrow \\frac{b}{c} = 1$, then $b = c$, so $bc$ is a perfect square.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25486,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all real $x \\ge -1$ such that the inequality\n$$\n\\frac{a_1 + x}{2} \\cdot \\frac{a_2 + x}{2} \\cdot \\dots \\cdot \\frac{a_n + x}{2} \\le \\frac{a_1 a_2 \\dots a_n + x}{2}\n$$\nholds for all $a_1, \\dots, a_n \\ge 1$, where $n \\ge 2$.",
    "options": [],
    "answer": "x ∈ [-1, 1]",
    "solution": "If $x \\ge -1$ satisfies the problem condition, then setting $a_1 = a_2 = \\dots = a_n = 1$ in the given inequality:\n$$\n\\frac{a_1 + x}{2} \\cdot \\frac{a_2 + x}{2} \\cdot \\dots \\cdot \\frac{a_n + x}{2} \\le \\frac{a_1 a_2 \\dots a_n + x}{2} \\quad (1)\n$$\nwe obtain the inequality $\\left(\\frac{1 + x}{2}\\right)^n \\le \\frac{1 + x}{2}$, which yields $x \\le 1$.\nWe prove, by induction on $n \\ge 2$, that for $x \\in [-1, 1]$ inequality (1) holds for all $n \\in \\mathbb{N}$ and all $a_1, \\dots, a_n \\ge 1$.\nFor $n=2$ inequality (1) has the form:\n$$\n\\frac{a_1 + x}{2} \\cdot \\frac{a_2 + x}{2} \\le \\frac{a_1 a_2 + x}{2}, \\quad (2)\n$$\nwhich is equivalent to:\n$$\nx^2 + (a_1 + a_2 - 2)x - a_1 a_2 \\le 0.\n$$\nThe last inequality holds for all $x \\in [-1, 1]$ and all $a_1, a_2 \\ge 1$. Indeed,\n$$\n\\begin{aligned}\n& x^2 + (a_1 + a_2 - 2)x - a_1 a_2 \\le 1 + (a_1 + a_2 - 2) \\cdot 1 - a_1 a_2 \\\\\n& = a_1 + a_2 - a_1 a_2 - 1 = -(a_1 - 1)(a_2 - 1) \\le 0.\n\\end{aligned}\n$$\nThe base of induction is proved.\nIf we suppose that inequality (1) holds for some $n \\ge 2$, then from (1) and (2) it follows that:\n$$\n\\begin{aligned}\n& \\frac{a_1 + x}{2} \\cdot \\frac{a_2 + x}{2} \\cdot \\dots \\cdot \\frac{a_n + x}{2} \\cdot \\frac{a_{n+1} + x}{2} \\le \\frac{a_1 a_2 \\dots a_n + x}{2} \\cdot \\frac{a_{n+1} + x}{2} \\le \\\\\n& \\le \\frac{a_1 a_2 \\dots a_n a_{n+1} + x}{2},\n\\end{aligned} \\quad (3)\n$$\nas required.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 25487,
    "subject": "Mathematics (Multi-modal)",
    "question": "Real numbers are written in the cells of the $7 \\times 7$ table so that the product of the numbers in any $3 \\times 3$ square is equal to the product of the numbers in any $4 \\times 4$ square.\nIs it possible for the product of all numbers in the table to be 2015?\n(V. Kaskevich)",
    "options": [],
    "answer": "Yes",
    "solution": "Answer: yes, it is possible.\nLet $a$ and $b$ be real numbers such that $ab = 1$. Consider the following table\n\n| $a$ | $b$ | $a$ | $b$ | $a$ | $b$ | $a$ |\n|-----|-----|-----|-----|-----|-----|-----|\n| $b$ | $a$ | $b$ | $a$ | $b$ | $a$ | $b$ |\n| $1$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$ |\n| $a$ | $b$ | $a$ | $b$ | $a$ | $b$ | $a$ |\n| $b$ | $a$ | $b$ | $a$ | $b$ | $a$ | $b$ |\n| $1$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$ |\n| $a$ | $b$ | $a$ | $b$ | $a$ | $b$ | $a$ |\n\nIt is easy to see that the products of the numbers in all $3 \\times 3$ squares and $4 \\times 4$ squares are equal to $1$. Moreover, the product of all numbers in the table is equal to $a$, so this product can admit any value different from $0$, in particular this product can be equal to $2015$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25488,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $M$ be the subset of all numbers from $\\{1, 2, \\dots, 2015\\}$ which are not perfect squares.\n\na) Prove that $\\{\\sqrt{n}\\} > 0.011$ for any $n \\in M$.\n\nb) Prove that there exists a number $n \\in M$ such that $\\{\\sqrt{n}\\} < 0.0115$.\n\n(Here $\\{y\\}$ stands for the fractional part of $y$.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**a.)** To prove the required statement it suffices to find the number $n \\in M$ such that $\\{\\sqrt{n}\\} = \\min_{k \\in M}\\{\\sqrt{k}\\}$. Any number $n \\in M$ can be uniquely presented as\n$$\nn = k^2 + r, \\qquad (1)\n$$\nwhere $1 \\le r \\le 2k$, and $k = [\\sqrt{n}]$ ($\\cdot$ is the whole part of a number). Then\n$$\n\\{\\sqrt{n}\\} = \\sqrt{k^2 + r} - k = \\frac{r}{\\sqrt{k^2 + r} + k}.\n$$\nSince the function $y = \\sqrt{x}$ is increasing on any segment $[l^2, (l+1)^2]$, $l \\in \\mathbb{N}$, and its values belong to $[l, l+1]$, we see that $\\{\\sqrt{x}\\}$ has its minimal value for positive integers $x \\in (l^2, (l+1)^2)$ when $x = l^2 + 1$. So the minimal value of $\\{\\sqrt{n}\\}$ one can look for among $n$ such that $r = 1$ in representation (1), i.e., in other words, one must find the minimum of the function\n$$\n\\{\\sqrt{n}\\} = \\frac{1}{\\sqrt{k^2 + 1} + k}, \\qquad (2)\n$$\nwhere $n \\in M$, $k = [\\sqrt{n}]$. From (2) it follows that $\\{\\sqrt{n}\\}$, $n \\in M$, takes the minimal value when $k$ takes the maximal possible value. Since $44^2 = 1936 < 2015 < 2025 = 45^2$, $k$ takes its maximal value when $n = 44 \\in M$. Hence for any $n \\in M$ we have\n$$\n\\{\\sqrt{n}\\} \\ge \\{\\sqrt{44^2 + 1}\\} = \\frac{1}{\\sqrt{44^2 + 1} + 44} > \\frac{1}{89} > 0.011.\n$$\n\n**b.)** We have\n$$\n\\{\\sqrt{44^2 + 1}\\} = \\frac{1}{\\sqrt{44^2 + 1} + 44} < \\frac{1}{88} < 0.0115.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25489,
    "subject": "Mathematics (Multi-modal)",
    "question": "A sequence is called *N-nice* if it consists of a finite number of the consecutive positive integers and the sum of all its terms is equal to $N$.\nFind the total number of all $4030^k$-nice sequences, where $k$ is a positive integer. (*N*-nice sequence may consist of a single term.)",
    "options": [],
    "answer": "(k+1)^3",
    "solution": "Let $P(N)$ be the number of $N$-nice sequences, $D(N)$ be the number of odd divisors of $N$. Then $P(N) = D(N)$ (see the solution of Problem C.7). Since $D(4030^k) = D(2^k \\cdot 5^k \\cdot 13^k \\cdot 31^k) = (k+1)^3$, we see that the number of the required $4030^k$-nice sequences is equal to $(k+1)^3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25490,
    "subject": "Mathematics (Multi-modal)",
    "question": "The circle $\\omega$ passes through the vertices $B$ and $C$ of a triangle $ABC$, and meets its sides $AB$ and $AC$ at points $C_1$ and $B_1$ respectively. Let $X$ be the midpoint of the arc $\\sim B_1BC$, and $Y$ be the midpoint of the arc $\\sim BCC_1$ of $\\omega$.\nFind the angle between the line $XY$ and the bisector of the angle $CAB$.",
    "options": [],
    "answer": "90 degrees",
    "solution": "Let $\\angle ABC = \\beta$ and $\\angle BCA = \\gamma$. Then $\\angle CAB = 180^\\circ - \\beta - \\gamma$. By $\\xi$ denote the degree measure of the arc $C_1B$ that does not contain the vertex $C$, by $\\eta$ denote the degree measure of the arc $\\tilde{C}B_1$ that does not contain the vertex $B$. Then $\\gamma = (\\xi + \\eta)/2$ and $\\beta = (\\eta + \\zeta)/2$. Let $Z$ be the point of intersection of the lines $XY$ and $BC$. Let the figure shows the positions of $X$, $Y$, and $Z$. The angle $BZX$ is the external angle for the triangle $ZXC$, so $\\angle BZX = \\angle ZCX + \\angle CXZ$. We have\n$$\n\\angle ZCX = \\frac{1}{2} \\left( \\xi + \\eta - \\frac{1}{2} (360^\\circ - \\zeta) \\right)\n$$\nand\n$$\n\\angle CXZ = \\frac{1}{2} \\left( \\frac{1}{2} (360^\\circ - \\xi) - (\\eta + \\zeta) \\right).\n$$\n\nTherefore,\n$$\n\\angle BZX = \\angle ZCX + \\angle CXZ = (\\xi - \\zeta)/4 = ((\\xi + \\eta) - (\\zeta + \\eta))/4 = (\\gamma - \\beta)/2.\n$$\nLet the bisector of the angle $CAB$ meet the side $BC$ at $L$. The angle $ALC$ is the external angle for the triangle $LBA$, so\n$$\n\\angle ALC = \\angle ABC + \\angle LAB = \\beta + (90^\\circ - (\\gamma + \\beta)/2) = 90^\\circ - (\\gamma - \\beta)/2.\n$$\n$$\n\\text{Since } \\angle BZX + \\angle ALC = 90^\\circ, \\text{ we have } AL \\perp XY.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25491,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $H$ be the orthocenter of an acute triangle $ABC$, points $A_1$, $B_1$, $C_1$ be the feet of the altitudes $AA_1$, $BB_1$, $CC_1$, respectively. Let points $D$, $K$, $L$, $M$, $N$ be the midpoints of the segments $AC$, $BC_1$, $BA_1$, $HA_1$, $HC_1$, respectively, and $P$, $Q$, $R$ be the intersection points of the lines $AL$ and $CK$, $AN$ and $CK$, $AL$ and $CM$, respectively.\nProve that the line $DQ$ touches the circumcircle of the triangle $PQR$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note that all right-angled triangles $ABA_1$, $AHC_1$, $CBC_1$, $CHA_1$ are pairwise similar. Therefore, since $L$, $N$, $K$, $M$ are the midpoints of the corresponding sides, we see that $\\triangle ALB \\sim \\triangle ANH \\sim \\triangle CKB \\sim \\triangle CMH$. It follows that $\\angle LAB = \\angle NAH = \\angle KCB = \\angle MCH = \\varphi$. Let $\\angle CAB = \\alpha$. Then\n$$\n\\begin{aligned}\n\\angle RCA + \\angle RAC &= (\\angle RCC_1 + \\angle C_1CA) + (\\angle CAB - \\angle BAL) = \\\\\n&= (\\varphi + 90^\\circ - \\alpha) + (\\alpha - \\varphi) = 90^\\circ,\n\\end{aligned}\n$$\nso $\\angle ARC = 90^\\circ$. Similarly we have $\\angle AQC = 90^\\circ$.\n\nIt follows that $A$, $R$, $Q$, $C$ lie on the same circle, and $AC$ is the diameter of this circle. So $DC = DQ$ and $\\angle PQR = \\angle CAP$. Therefore, $\\triangle DQC$ is an isosceles triangle. So\n$$\n\\begin{aligned}\n\\angle RQD &= 180^\\circ - \\angle PQR - \\angle DQC = \\\\\n&= 180^\\circ - \\angle CAP - \\angle ACP = \\angle APC = \\angle RPQ,\n\\end{aligned}\n$$\nwhich yields that the line $DQ$ touches the circumcircle of the triangle $PQR$ (at point $Q$).",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25492,
    "subject": "Mathematics (Multi-modal)",
    "question": "A straight line $l$ and a point $A$ are on a plane. $A$ does not belong to the $l$. For each point $M$ on the line $l$ a point $N$ is marked on the plane so that the triangle $AMN$ is equilateral (the vertices of the triangle $AMN$ are mentioned clockwise).\nFind the locus of the vertices $N$ of the equilateral triangles $AMN$ when the vertex $M$ moves along the line $l$.\n(S. Mazanik)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $A$ lies under the line $\\ell$, then the required locus is the line $BC$, where $B \\in \\ell$, $C \\in m$, $m \\parallel \\ell$, $m \\nparallel A$, and the triangle $ABC$ is an equilateral triangle.\n\nWithout loss of generality suppose that $A$ lies under the line $\\ell$. Show that the required locus is the line $BC$, where $B \\in \\ell$, $C \\in m$, $m \\parallel \\ell$, $m \\nparallel A$, and the triangle $ABC$ is an equilateral triangle.\n\nFirst show that all points $N$ lie on $BC$. If $M$ coincides with $B$, then it is evident that $N$ coincides with $C$. Let $M$ be an arbitrary point of $\\ell$ different from $B$. In the right half-plane with respect to the line $AM$ we construct the ray with the origin at $A$ such that the angle between this ray and the line $AM$ is equal to $60^\\circ$. Let $K$ be the point of intersection of this ray and the line $BC$ (see the Fig.). Let $D$ be the point on the line $\\ell$ (at the left of $B$) such that the triangle $ABD$ is equilateral. Note that\n![](attached_image_1.png)\n$$\n\\begin{aligned}\n\\angle MAD &= \\angle MAK - \\angle DAK = 60^\\circ - \\angle DAK = \\\\\n&= \\angle DAB - \\angle DAK = \\angle KAB.\n\\end{aligned}\n\\quad (1)\n$$\n\nMoreover,\n$$\n\\begin{aligned}\n\\angle MDA &= 180^\\circ - \\angle BDA = 180^\\circ - 60^\\circ = 120^\\circ = \\\\\n&= 180^\\circ - \\angle ABC = \\angle KBA.\n\\end{aligned}\n\\qquad (2)\n$$\nBy construction, $AD = AB$, so from (1) and (2) it follows that the triangles $MAD$ and $KAB$ are equal. Therefore, $AM = AK$, and since, by construction, $\\angle MAK = 60^\\circ$, we see that the triangle $MAK$ is equilateral, i.e., $K$ coincides with $N$, which yields $N \\in BC$.\n\nIt remains to show that for any point $N$, $N \\in BC$ there exists a point $M \\in l$ such that the triangle $NAM$ is an equilateral triangle.\n\nIndeed, let $N \\in BC$. In the left half-plane with respect to the line $AN$ we construct the ray with the origin at $A$ such that the angle between this ray and the line $AN$ is equal to $60^\\circ$. Let $M$ be the point of intersection of this ray and the line $l$ (see the Fig.). Let $D$ be the point on the line $l$ (at the left of $B$) such that the triangle $ABD$ is equilateral. Note that\n$$\n\\begin{aligned}\n\\angle MAD &= \\angle MAN - \\angle DAN = 60^\\circ - \\angle DAN = \\\\\n&= \\angle DAB - \\angle DAN = \\angle NAB\n\\end{aligned}\n\\qquad (3)\n$$\n\nMoreover,\n$$\n\\begin{aligned}\n\\angle MDA &= 180^\\circ - \\angle BDA = 180^\\circ - 60^\\circ = 120^\\circ = \\\\\n&= 180^\\circ - \\angle ABC = \\angle NBA.\n\\end{aligned}\n\\qquad (4)\n$$\nBy construction, $AD = AB$, so from (3) and (4) it follows that the triangles $MAD$ and $NAB$ are equal. Therefore, $AM = AN$, and since, by construction, $\\angle MAN = 60^\\circ$, we see that the triangle $MAN$ is equilateral, as required.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 25493,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $I$ be the incenter of a triangle $ABC$. Let $A_1$, $B_1$, $C_1$ be the tangency points of the incircle of the triangle $ABC$ with the sides $BC$, $CA$, $AB$, respectively. Let the circumcircle of the triangle $BC_1B$ intersect the line $BC$ at points $B$ and $K$, and the circumcircle of the triangle $CB_1C_1$ intersect the line $BC$ at points $C$ and $L$.\nProve that the lines $LC_1$, $KB_1$, $IA_1$ are concurrent.\n(A. Voidelevich)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $X$ be the intersection point of the lines $LC_1$ and $KB_1$. To prove the problem statement it suffices to show that points $X$, $I$ and $A_1$ lie on the same line. Let $\\angle CAB = \\alpha$, $\\angle ABC = \\beta$, $\\angle BCA = \\gamma$. Since $LC_1B_1C$ is an inscribed quadrilateral, we have\n$$\n\\angle C_1LC = 180^\\circ - \\angle C_1B_1C = \\angle C_1B_1A = \\angle B_1C_1A = 90^\\circ - \\frac{\\alpha}{2} = (\\beta + \\gamma)/2.\n$$\nSimilarly, $\\angle B_1KB = (\\beta + \\gamma)/2$. Therefore the triangle $LXK$ is isosceles ($XL = XK$) and\n$$\n\\angle C_1XB_1 = \\angle LXK = 180^\\circ - (\\beta + \\gamma) = \\alpha.\n$$\nSince $IC_1 \\perp AB$ and $IB_1 \\perp AC$, we have $\\angle C_1IB_1 = 180^\\circ - \\alpha$.\n\n![](attached_image_1.png)\n\nTherefore $I$, $C_1$, $A$, $X$, $B_1$ lie on the same circle $\\omega$. The bisector of the angle $C_1AB_1$ meets $\\omega$ at points $A$ and $I$. So\n$$\n\\angle C_1XI = \\angle C_1AI = \\angle IAB_1 = \\angle IXB_1,\n$$\nwhich yields that $XI$ is the bisector of the angle $C_1XB_1$. Since the triangle $LXK$ is an isosceles triangle, we see that $XI$ is the altitude of this triangle. On the other hand, $IA_1 \\perp BC$, so points $X$, $I$, $A$ lie on the same line, as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25494,
    "subject": "Mathematics (Multi-modal)",
    "question": "Is it possible to partition a set $M$ of positive integers into some disjoint subsets so that in each subset one of the numbers is equal to the sum of all other numbers of this subset if\n$$\n\\text{a) } M = \\{1, 2, \\dots, 20\\} \\text{?}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Yes, it is possible; the desired partition exists. For example, see the following table:\n\n| 20 | 19 | 18 | 17 | 16 | 15 |\n|----|----|----|----|----|----|\n| 14 | 12 | 10 | 13 | 11 | 9  |\n| 6  | 7  | 8  | 4  | 5  | 1, 2, 3 |\n\nIn each subset, one of the numbers is equal to the sum of the others.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25495,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ ($n < 24$) such that the set $M = \\{n, n + 1, \\dots, 24\\}$ can be partitioned into some disjoint subsets so that in each subset one of the numbers is equal to the sum of all other numbers of this subset.\n(V. Kaskevich)",
    "options": [],
    "answer": "n = 1 and n = 4",
    "solution": "A number $a$ from a subset of the desired partition is called major if $a$ is equal to the sum of all other numbers of this subset. All numbers in each subset must be distinct, since one of them is major we see that there exist at least three numbers in each subset. Let $k$ be the number of the subsets of the desired partition. Since we have $25-n$ numbers in the initial set $M = \\{n, n+1, \\dots, 24\\}$, we have $k \\le \\frac{25-n}{3}$. On the other hand, if $a$ is the major number of some subset, then the sum of the numbers of this subset is equal to $2a$. So the sum $S(n)$ of all numbers of the set $M$ must be even than or equal to\n$$\n2 \\cdot (24 + 23 + 22 + \\dots + (25 - k)) = 2 \\cdot \\frac{(24 + 25 - k) \\cdot k}{2} = (49 - k) \\cdot k \\le \\\\\n\\le \\left(49 - \\frac{25-n}{3}\\right) \\cdot \\frac{25-n}{3} = \\frac{(122+n)(25-n)}{9}.\n$$\n(The last inequality holds because $k \\le 25-n/3 \\le 49/2$, and the function $(49-x)x$ increases for $x \\le 49/2$.) Therefore, the following condition is necessary to exist the desired partition:\n$$\nS(n) \\le \\frac{(122+n)(25-n)}{9} \\iff \\frac{(n+24)(25-n)}{2} \\le \\frac{(122+n)(25-n)}{9} \\iff 9(n+24) \\le 2(122+n) \\iff n \\le 4.\n$$\nFor $n=2$ and $n=3$ we see that both the sums\n$$\nS(2) = \\frac{(2+24)(25-2)}{2} = 13 \\cdot 23, \\quad S(3) = \\frac{(3+24)(25-3)}{2} = 27 \\cdot 11\n$$\nare odd, so the desired partitions do not exist.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25496,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ ($n < 60$) such that the set $M = \\{n, n+1, \\dots, 60\\}$ can be partitioned into disjoint subsets so that in each subset one of the numbers is equal to the sum of all other numbers of this subset.\n(V. Kaskevich)",
    "options": [],
    "answer": "1, 4, 5",
    "solution": "A number $a$ from a subset of the desired partition is called major if $a$ is equal to the sum of all other numbers of this subset. All numbers in each subset must be distinct, since one of them is major we see that there exist at least three numbers in each subset. Let $k$ be the number of the subsets of the desired partition. Since we have $61 - n$ numbers in the initial set $M = \\{n, n+1, \\dots, 60\\}$, we have $k \\le \\frac{61-n}{3}$. On the other hand, if $a$ is the major number of some subset, then the sum of the numbers of this subset is equal to $2a$. So the sum $S(n)$ of all numbers of the set $M$ must be even\n\n($S(n) = \\frac{(n+60)(61-n)}{2}$, and the sum of all numbers of all subsets is less than or equal to\n$$\n2 \\cdot (60 + 59 + 58 + \\dots + (61-k)) = 2 \\cdot \\frac{(60+61-k) \\cdot k}{2} = (121-k) \\cdot k \\le\n$$\n$$\n\\le \\left(121 - \\frac{61-n}{3}\\right) \\cdot \\frac{61-n}{3} = \\frac{(302+n)(61-n)}{9}.\n$$\n(The last inequality holds because $k \\le (61-n)/3 \\le 121/2$ and the function $(121-x)x$ increases for $x \\le 121/2$.) Therefore, the following condition is necessary to exist the desired partition:\n$$\nS(n) \\le \\frac{(302+n)(61-n)}{9} \\Leftrightarrow \\frac{(n+60)(61-n)}{2} \\le \\frac{(302+n)(61-n)}{9} \\Leftrightarrow\n$$\n$$\n9(n+60) \\le 2(302+n) \\Leftrightarrow 7n \\le 64,\n$$\ni.e., $n \\le 9^{1/7}$, and since $n$ is an integer number, we have $n \\le 9$.\nThe sum $S(n) = \\frac{(n+60)(61-n)}{2}$ is even, so either $n = 4m$ or $n = 4m+1$, i.e., $n \\in \\{1, 4, 5, 8, 9\\}$.\nIf $n=9$, then $S(9) = \\frac{(9+60)(61-9)}{2} = 69 \\cdot 26 = 1794$. The number $k$ of the subsets is less than or equal to $k \\le \\frac{61-9}{3} = 17\\frac{1}{3}$, i.e., $k \\le 17$. But for these $k$ the sum of all numbers in the subsets is less than or equal to $(121-k) \\cdot k = 104 \\cdot 17 = 1768 < S(9)$, a contradiction. Similarly, if $n=8$, then $S(8) = 1804$, $k = [53:3] = 17$, and $1768 < S(8)$, a contradiction.\n\n<table><tr><td>60</td><td>59</td><td>58</td><td>57</td><td>56</td><td>55</td><td>54</td><td>53</td><td>52</td></tr><tr><td>43</td><td>41</td><td>39</td><td>37</td><td>35</td><td>33</td><td>31</td><td>29</td><td>27</td></tr><tr><td>17</td><td>18</td><td>19</td><td>20</td><td>21</td><td>22</td><td>23</td><td>24</td><td>25</td></tr></table>\n<table><tr><td>51</td><td>50</td><td>49</td><td>48</td><td>47</td><td>46</td><td>45</td><td>44</td><td>26</td></tr><tr><td>42</td><td>40</td><td>38</td><td>36</td><td>34</td><td>32</td><td>30</td><td>28</td><td>8,6</td></tr><tr><td>9</td><td>10</td><td>11</td><td>12</td><td>13</td><td>14</td><td>15</td><td>16</td><td>7,5</td></tr></table>\n\n<table><tr><td>60</td><td>59</td><td>58</td><td>57</td><td>56</td><td>55</td><td>54</td><td>53</td><td>52</td></tr><tr><td>43</td><td>41</td><td>39</td><td>37</td><td>35</td><td>33</td><td>31</td><td>29</td><td>27</td></tr><tr><td>17</td><td>18</td><td>19</td><td>20</td><td>21</td><td>22</td><td>23</td><td>24</td><td>25</td></tr><tr><td>51</td><td>50</td><td>49</td><td>48</td><td>47</td><td>46</td><td>45</td><td>44</td><td>28</td></tr><tr><td>42</td><td>40</td><td>38</td><td>36</td><td>34</td><td>32</td><td>30</td><td>26</td><td>16</td></tr><tr><td>9</td><td>10</td><td>11</td><td>12</td><td>13</td><td>14</td><td>15</td><td>8,6,4</td><td>7,5</td></tr></table>\n$n=4$\n\nIf we add the subset $\\{1,2,3\\}$ to the given partition for $n=4$ we obtain the desired partition for $n=1$.\n\nThe desired partitions exist for $n=1,4,5$. The following tables give the examples of the desired partition for $n=5$ and $n=4$:\n<table><tr><td>60</td><td>59</td><td>58</td><td>57</td><td>56</td><td>55</td><td>54</td><td>53</td><td>52</td></tr><tr><td>43</td><td>41</td><td>39</td><td>37</td><td>35</td><td>33</td><td>31</td><td>29</td><td>27</td></tr><tr><td>17</td><td>18</td><td>19</td><td>20</td><td>21</td><td>22</td><td>23</td><td>24</td><td>25</td></tr></table>\n<table><tr><td>51</td><td>50</td><td>49</td><td>48</td><td>47</td><td>46</td><td>45</td><td>44</td><td>26</td></tr><tr><td>42</td><td>40</td><td>38</td><td>36</td><td>34</td><td>32</td><td>30</td><td>28</td><td>8,6</td></tr><tr><td>9</td><td>10</td><td>11</td><td>12</td><td>13</td><td>14</td><td>15</td><td>16</td><td>7,5</td></tr></table>\n\n$n=5$\n\nIf we add the subset $\\{1,2,3\\}$ to the given partition for $n=4$ we obtain the desired partition for $n=1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25497,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$, $d$, $x$, $y$ denote the lengths of the sides $AB$, $BC$, $CD$, $DA$ and the diagonals $AC$, $BD$ of a cyclic quadrilateral $ABCD$, respectively.\n\n$$\n\\left(\\frac{1}{a} + \\frac{1}{c}\\right)^2 + \\left(\\frac{1}{b} + \\frac{1}{d}\\right)^2 \\ge 8 \\left(\\frac{1}{x^2} + \\frac{1}{y^2}\\right).\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(Solution by Y. Dubovik, U. Kazlouski.) By the Cosine Law for the triangles $DAB$ and $DCB$,\n$$\ny^2 = a^2 + d^2 - 2ad \\cos A, \\quad (1)\n$$\n$$\ny^2 = b^2 + c^2 + 2bc \\cos A. \\quad (2)\n$$\nMultiplying (1) and (2) by $bc$ and $ad$, respectively, and summing the obtained equalities, we get\n$$\n(bc + ad)y^2 = (a^2 + d^2)bc + (b^2 + c^2)ad\n$$\n![](attached_image_1.png)\n$$\n\\geq 2adbc + 2bcad = 4abcd.\n$$\n\n$$\n\\frac{8}{y^2} \\le \\frac{2(bc + ad)}{abcd} = \\frac{2}{ad} + \\frac{2}{bc}. \\quad (3)\n$$\nIn the same way we can obtain\n$$\n\\frac{8}{x^2} \\le \\frac{2}{ab} + \\frac{2}{cd}. \\quad (4)\n$$\nSo,\n$$\n\\begin{aligned}\n8 \\left( \\frac{1}{x^2} + \\frac{1}{y^2} \\right) &\\le \\frac{2}{ad} + \\frac{2}{bc} + \\frac{2}{ab} + \\frac{2}{cd} = 2 \\left( \\frac{1}{a} + \\frac{1}{c} \\right) \\left( \\frac{1}{b} + \\frac{1}{d} \\right) \n\\\\ &\\le \\left( \\frac{1}{a} + \\frac{1}{c} \\right)^2 + \\left( \\frac{1}{b} + \\frac{1}{d} \\right)^2,\n\\end{aligned}\n$$\nas required.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25498,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be an acute triangle with the orthocenter $H$. Let $D$ be the point such that $HABD$ is a parallelogram ($AB \\parallel HD$, $AH \\parallel BD$). Let $E$ be the point on the line $DH$ such that $AC$ bisects $HE$. The line $AC$ meets the circumcircle of the triangle $DCE$ at $C$ and $F$.\nProve that $EF = AH$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "1. See IMO-2015 Shortlist, Problem G1.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25499,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that\n$$\n\\left(a^2 + \\frac{c^2}{b^2}\\right) \\left(b^2 + \\frac{a^2}{c^2}\\right) \\left(c^2 + \\frac{b^2}{a^2}\\right) \\ge \\frac{abc}{\\sqrt{3}} \\left(a + \\frac{1}{a}\\right) \\left(b + \\frac{1}{b}\\right) \\left(c + \\frac{1}{c}\\right)\n$$\nfor all positive real $a, b, c$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(Solution by S. Shaban, A. Vasileuski) By the Cauchy-Bunyakovsky-Schwarz inequality,\n$$\n\\sqrt{a^2 + \\frac{c^2}{b^2}} \\sqrt{b^2 + \\frac{a^2}{c^2}} \\geq \\left( ab + \\frac{a}{b} \\right) = a \\left( b + \\frac{1}{b} \\right),\n$$\n$$\n\\sqrt{b^2 + \\frac{a^2}{c^2}} \\sqrt{c^2 + \\frac{b^2}{a^2}} \\geq \\left( bc + \\frac{b}{c} \\right) = b \\left( c + \\frac{1}{c} \\right),\n$$\n$$\n\\sqrt{c^2 + \\frac{b^2}{a^2}} \\sqrt{a^2 + \\frac{c^2}{b^2}} \\geq \\left( ca + \\frac{c}{a} \\right) = c \\left( a + \\frac{1}{a} \\right).\n$$\nMultiplying these inequalities, we obtain the required inequality.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25500,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$, $g: \\mathbb{R} \\to \\mathbb{R}$ such that\n$$\nf(x - 2f(y)) = xf(y) - yf(x) + g(x)\n$$\nfor all real $x, y$.",
    "options": [],
    "answer": "All solutions are given by f(x) = a(x + 2a) and g(x) = a(1 − 2a)(x + 2a) for real a.",
    "solution": "Answer: $f(x) = a(x + 2a)$, $g(x) = a(1 - 2a)(x + 2a)$, $a \\in \\mathbb{R}$.\n\n(Solution by A. Goloubitskaya, L. Manzhulina.) First, we claim that $f$ takes the value $0$. If $f(0) = 0$, then there is nothing to prove. If $f(0) = b \\neq 0$, then setting $x = 0$ in the given equality\n$$\nf(x - 2f(y)) = xf(y) - yf(x) + g(x), \\quad (1)\n$$\nwe obtain $f(-2f(y)) = -by + g(0)$. The right-hand side attains all real values since $b \\neq 0$. So $f$ attains all real values including $0$.\nLet $c$ be such that $f(c) = 0$. Set $y = c$ in (1): $f(x) = -cf(x) + g(x)$, whence $g(x) = (c+1)f(x)$. Therefore, (1) becomes\n$$\nf(x - 2f(y)) = xf(y) + (c + 1 - y)f(x). \\quad (2)\n$$\nLet $a = f(c + 1)$. Substituting $c + 1$ for $y$ in (2), we obtain $f(x - 2a) = ax$, whence $f(x) = ax + 2a^2$. Since $f(c) = 0$, we have $ac + 2a^2 = 0$ and\n$$\ng(x) = (c+1)(ax + 2a^2) = (-2a^2 + a)x - 4a^3 + 2a^2.\n$$\nEasy verification shows that these functions $f$ and $g$ satisfy the given equality for all real $a$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25501,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that there are at most a finite number of primes $p$ such that the equation $a^3 + b^3 = 2016pab$ has a solution in positive integers $a$ and $b$ which are not divisible by $p$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $d$ denote the greatest common divisor of $a$ and $b$, i.e., $a = d a_1$, $b = d b_1$, where $\\gcd(a_1, b_1) = 1$. Then the given equality can be presented in the form\n$$\nd(a_1^3 + b_1^3) = 2016p a_1 b_1.\n$$\nIt follows that $d \\mid a_1^3 + b_1^3$, and, since $a_1$ and $b_1$ are coprime, $d \\mid b_1^3$ and $d \\mid a_1^3$. Similarly, $d \\mid a_1^3$, so $d \\mid a_1 b_1$ because $a_1$ and $b_1$ are coprime. Let $d = d_1 a_1 b_1$, then the equation can be presented in the form\n$$\nd_1(a_1^3 + b_1^3) = 2016p. \\qquad (1)\n$$\nThere are exactly two possible cases:\n\n1) $a_1 + b_1 \\neq p$.\n\n2) $a_1 + b_1 = p$.\n\n1) If $a_1 + b_1 \\neq p$, then\n$$\n\\begin{aligned}\np \\le a_1 + b_1 &\\implies p^3 \\le (a_1 + b_1)^3 \\le 4(a_1^3 + b_1^3) \\le 4 \\cdot 2016p \\\\\n&\\implies p^2 \\le 4 \\cdot 2016 \\\\\n&\\implies p \\le 2 \\cdot \\sqrt{2016} \\le 2 \\cdot 45 = 90.\n\\end{aligned}\n$$\nTherefore there are only a finite number of $p$ satisfying the problem condition.\n\n2) By condition, $d_1 \\nmid p$ (otherwise $a$ and $b$ are divisible by $p$), then from (1) it follows that $a_1^3 + b_1^3 \\neq p$. Since\n$$\na_1^3 + b_1^3 = (a_1 + b_1)(a_1^2 - a_1 b_1 + b_1^2) = (a_1 + b_1)((a_1 + b_1)^2 - 3 a_1 b_1) \\quad (2)\n$$\nand $a_1 + b_1 \\nmid p$, we have $(a_1 + b_1)^2 - 3 a_1 b_1 \\neq p$. Hence\n$$\np \\le (a_1 + b_1)^2 - 3 a_1 b_1. \\qquad (3)\n$$\nSince $a_1 + b_1 \\nmid p$, from (1) and (2) it follows that $a_1 + b_1$ is a divisor of $2016$, i.e., $a_1 + b_1 \\le 2016$. Then from (3) we obtain\n$$\np \\le (a_1 + b_1)^2 - 3 a_1 b_1 < (a_1 + b_1)^2 \\le 2016^2.\n$$\nThus, in this case there are also only a finite number of $p$ satisfying the problem condition.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25502,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all positive integers $M$ for which the sequence $a_0, a_1, a_2, \\ldots$, defined by $a_0 = 2M + \\frac{1}{2}$ and $a_{k+1} = a_k[a_k]$ for $k = 0, 1, 2, \\ldots$, contains at least one integer term.",
    "options": [],
    "answer": "All positive integers M",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25503,
    "subject": "Mathematics (Multi-modal)",
    "question": "Students of two groups decided to organize a chess tournament where each student from the first group plays exactly one game with each student from the other group. But one student from the first group and one student from the other one, due to some reasons, failed to participate in the tournament, so the total number of the games in the tournament has been 20% smaller than that of the games planned.\nFind all possible numbers of the students participated in the tournament.",
    "options": [],
    "answer": "17, 20, 29",
    "solution": "Answer: 17, 20, 29.\nLet $n$ and $m$ be the numbers of students of the first, and respectively the second group that were initially supposed to participate in the tournament. Then $S = (n-1)+(m-1)$ students have taken part in the tournament. By condition,\n$$\n(n-1)(m-1) = 0.8 \\text{ nm} \\iff (m-5)(n-5) = 20.\n$$\nTherefore, the pair $(n-5, m-5)$ is one of three pairs: (1, 20), (2, 10), (4, 5). Thus, $(n-5) + (m-5)$ is equal to $1+20=21$, or $2+10=12$, or $4+5=9$. Thus $m+n$ is equal to 31, or 22, or 19.\nIt is easy to see that all these pairs $(m, n)$ satisfy the problem conditions.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 25504,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given real numbers $a, b, c, d$ such that $\\sin a + b > \\sin c + d$, $a + \\sin b > c + \\sin d$, prove that $a + b > c + d$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By condition, we have\n$$\na + \\sin b > c + \\sin d, \\qquad (1)\n$$\n$$\n\\sin a + b > \\sin c + d. \\qquad (2)\n$$\nSuppose, contrary to our claim, that\n$$\nc+d \\geq a+b. \\tag{3}\n$$\nSumming (1) and (3), we get\n$$\na + \\sin b + c + d > c + \\sin d + a + b \\implies d - \\sin d > b - \\sin b. \\quad (4)\n$$\nSumming (2) and (3), we get\n$$\n\\sin a + b + c + d > \\sin c + d + a + b \\implies c - \\sin c > a - \\sin a. \\quad (5)\n$$\nLet $f(x) = x - \\sin x$. Since $f'(x) = 1 - \\cos x \\ge 0$ and there are no intervals such that $f'(x) = 0$, we see that the function $f$ is strictly increasing. By the same argument, the function $g(x) = x + \\sin x$ is also strictly increasing.\nTherefore, from (4) and (5) we have\n$$\nd > b \\Rightarrow d + \\sin d > b + \\sin b, \\qquad (6)\n$$\nand\n$$\nc > a \\implies c + \\sin c > a + \\sin a. \\tag{7}\n$$\nSumming (1) and (6), we get\n$$\na + \\sin b + d + \\sin d > c + \\sin d + b + \\sin b \\Rightarrow a + d > b + c,\n$$\nand summing (2) and (7), we get\n$$\n\\sin a + b + c + \\sin c > \\sin c + d + a + \\sin a \\implies b + c > a + d.\n$$\nThis contradiction proves the required statement.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25505,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Determine all functions $f: \\mathbb{Z} \\to \\mathbb{Z}$ such that\n$$\nf(x - f(y)) = f(f(x)) - f(y) - 1\n$$\nfor all integers $x$ and $y$.\n\nb) The same question if\n$$\nf(x - f(y)) = f(f(x)) - f(y) - 2\n$$\nfor all integers $x$ and $y$.",
    "options": [],
    "answer": "a) All solutions: f(x) = -1 for all integers x; and f(x) = x + 1 for all integers x.\nb) All solutions: f(x) = -2 for all integers x; f(x) = x + 2 for all integers x; and, for any odd integer d, the function defined by f(x) = x + 2 for even x and f(x) = x + d for odd x.",
    "solution": "**a)** $f(x) = -1$ or $f(x) = x + 1$.\n\n(Alternative solution by I. Voronovich)\nSetting $y = f(x)$ in the given equation\n$$\nf(x - f(y)) = f(f(x)) - f(y) - 1, \\qquad (1)\n$$\nwe obtain $f(x - f(f(x))) = -1$, i.e., there exists an integer $\\lambda$ such that $f(\\lambda) = -1$. Set $y = \\lambda$ in (1), then\n$$\nf(f(x)) = f(x + 1). \\qquad (2)\n$$\nSubstituting $f(x) + f(y)$ for $x$ in (1), we obtain\n$$\nf(f(x)) = f(f(x) + f(y)) - f(y) - 1.\n$$\nDue to symmetry\n$$\nf(f(y)) = f(f(y) + f(x)) - f(x) - 1.\n$$\nTherefore $f(f(x)) - f(x) = f(f(y)) - f(y)$, so $f(f(x)) - f(x) = c$ or, in view of (2),\n$$\nf(x+1) - f(x) = c, \\quad c \\in \\mathbb{Z} \\qquad (3)\n$$\n\nIf $c = 0$, then $f(x+1) = f(x)$ for all integer $x$, so $f(x)$ is a constant function, thus from (1) it follows that $c = -1$, i.e., $f(x) = -1$ for all integer $x$.\n\nIf $c > 0$ or $c < 0$, then we conclude from (3) that $f(x)$ is increasing or decreasing, respectively. Anyway, $f$ is injective, so (2) gives $f(x) = x + 1$.\n\nBoth the functions obviously satisfy the initial equation (1).\n\n\n**b)** $f(x) = -2$ or $f(x) = x + 2$.\n\n(Solution by A. Asanau.) First, in the same way as in a) we can easily establish for any $f: \\mathbb{Z} \\to \\mathbb{Z}$ satisfying the given equation\n$$\nf(x - f(y)) = f(f(x)) - f(y) - 2 \\quad (1)\n$$\nthe following equalities:\n$$\nf(x - f(f(x))) = -2, \\quad (2)\n$$\n$$\nf(f(x)) = f(x + 2), \\quad (3)\n$$\n$$\nf(x + 2) - f(x) = c, \\quad (4)\n$$\nwhere $c$ is a constant. In particular, (2) shows that $-2 \\in E(f)$, the range of $f$. Now we have two cases.\n\nI. $c = 0$. Let $f(0) = a$, $f(1) = b$. Then from (4) we have $f(x+2) = f(x)$, so $f(x) = a$ for all even $x$ and $f(x) = b$ for all odd $x$. Note that one of $a$, $b$ equals $-2$ in view of (2). We claim that $a = b = -2$. Suppose that $a \\neq b$. Then\n$$\nf(x) = f(y) \\iff x \\equiv y \\pmod{2}.\n$$\nTherefore (3) implies $f(x) \\equiv x + 2 \\equiv x \\pmod{2}$. In particular, $f(0) \\equiv 0 \\pmod{2}$, whence $f(0) = a = -2$.\nNow we set $x = 0$, $y = 1$ in (1); then we get $f(-b) = f(-2) - b - 2 = -b - 4$. But $f(-b)$ equals either $b$ or $-2$. In both cases we have $b = -2$, contrary to $a \\neq b$.\nSo $a = b = -2$, and $f(x) = -2$ for all $x$. It is easy to see that this function satisfies the initial equation.\n\nII. $c \\neq 0$. Let again $f(0) = a$, $f(1) = b$. Then from (4) we have\n$$\nf(2n) = cn + a, \\quad f(2n + 1) = cn + b. \\quad (5)\n$$\nIn particular, (5) implies that, on the one hand, $f(x) \\to \\infty$ as $x \\to \\infty$, and, on the other hand, $\\lim_{x \\to \\infty} \\frac{f(x)}{x} = \\frac{c}{2}$. Then from (3) we have\n$$\n\\frac{f(f(x))}{f(x)} = \\frac{f(x+2)}{f(x)} \\to 1 \\text{ as } x \\to \\infty,\n$$\nbut $\\frac{f(f(x))}{f(x)} \\to \\frac{c}{2}$. Hence, $c/2 = 1$, i.e. $c = 2$. Then (5) becomes\n$$\n\\begin{aligned} f(x) &= x + a & \\text{for even } x, \\\\ f(x) &= x + d & \\text{for odd } x \\quad (\\text{where } d = b - 1). \\end{aligned} \\quad (6)\n$$\n1) Now, if $f$ is an injective function, then (3) immediately implies $f(x) = x + 2$, and this function satisfies (1).\n\n2) Suppose that $f$ is not injective. But from (6) it easily follows that $f$ is injective on the set of even numbers and also injective on the set of odd numbers. Thus we see that some $f(2n)$ and $f(2m+1)$ are equal. From (6) it follows that $2n - 2m - 1 = d - a$, i.e., $a$ and $d$ are of different parity. Then (6) implies that all values of $f$ are of the same parity. Since $-2 \\in E(f)$, all the values of $f(x)$ are even. Then from (3) we have $f(x+2) = f(f(x)) = f(x) + a$. Thus (4) gives $a = c = 2$. In particular, $d$ is odd. So, (6) becomes\n$$\n\\begin{aligned} f(x) &= x + 2 & \\text{for even } x, \\\\ f(x) &= x + d & \\text{for odd } x. \\end{aligned}\n$$\nOne can verify that for any odd $d$ this function is a solution of (1).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25506,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $n \\geq 1$ cities on a horizontal line. Each city is guarded by a pair of stationary elephants, one just to the left and one just to the right of the city, and facing away from it. The $2n$ elephants are of different sizes. If an elephant walks forward, it will knock aside any elephant that it approaches from behind, and in a face-to-face meeting, the smaller elephant will be knocked aside. A moving elephant will keep walking in the same direction until it is knocked aside.\n\nShow that there is a unique city with the property that if any one of the other cities orders its elephants to walk, then that city will not be invaded by an elephant.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25507,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of positive integers $(a; b)$, $a \\le b$, satisfying the equality $a^3 + b^3 = 1911ab$.",
    "options": [],
    "answer": "(756, 1008), (600, 960)",
    "solution": "Answer: $(756; 1008)$, $(600; 960)$.\n\nLet $d$ denote the greatest common divisor of $a$ and $b$, i.e., $a = da_1$, $b = db_1$, where $\\gcd(a_1, b_1) = 1$. Then the given equality can be presented in the form\n$$\nd(a_1^3 + b_1^3) = 1911a_1b_1.\n$$\nIt follows that $d a_1^3 : b_1$, and, since $a_1$ and $b_1$ are coprime, $d : b_1$. Similarly, $d : a_1$. Hence, $d : a_1b_1$ because $a_1$ and $b_1$ are coprime. Let $d = d_1 a_1 b_1$, then the equation can be presented in the form\n$$\nd_1(a_1^3 + b_1^3) = 1911, \\quad a_1 \\le b_1. \\qquad (1)\n$$\nNote that $a_1^3 + b_1^3 \\ge (a_1 + b_1)^3/4$, whence $4 \\cdot 1911 \\ge d_1 \\cdot (a_1 + b_1)^3$; therefore, $8000 > (a_1 + b_1)^3$, or $a_1 + b_1 < 20$.\n\nFurther, $a_1^3 + b_1^3 = (a_1 + b_1)^3 - 3 a_1 b_1 (a_1 + b_1)$, and (1) can be rewritten as\n$$\nd_1(a_1 + b_1)((a_1 + b_1)^2 - 3 a_1 b_1) = 1911. \\qquad (2)\n$$\nWe see that $a_1 + b_1$ is a divisor (less than 20) of $1911 = 3 \\cdot 7^2 \\cdot 13$. Hence exactly three cases are possible.\n\n1. $a_1 + b_1 = 3$, $a_1 = 1$, $b_1 = 2$. Then $d_1 \\cdot 3 \\cdot (9 - 6) = 1911$, but $9$ is not a divisor of $1911$.\n\n2. $a_1 + b_1 = 7$. If $a_1 = 1$, $b_1 = 6$, then $d_1 \\cdot 7 \\cdot (49 - 18) = 1911$, but $49 - 18 = 31$ is not a divisor of $1911$.\nIf $a_1 = 2$, $b_1 = 5$, we have $d_1 \\cdot 7 \\cdot (49 - 30) = 1911$, but $49 - 30 = 19$ is not a divisor of $1911$.\nFinally, if $a_1 = 3$, $b_1 = 4$, then $d_1 \\cdot 7 \\cdot (49 - 36) = 1911$, which gives $d_1 = 21$, whence $d = 21 \\cdot 3 \\cdot 4 = 252$, i.e., the pair $a = 756$, $b = 1008$ is a solution.\n\n3. $a_1 + b_1 = 13$. If $a_1 = 1$, $b_1 = 12$, then $d_1 \\cdot 13 \\cdot (169 - 36) = 1911$, but $169 - 36 = 133$ does not divide $1911$.\nIf $a_1 = 2$, $b_1 = 11$, then $d_1 \\cdot 13 \\cdot (169 - 66) = 13 d_1 \\cdot 103$ does not divide $1911$.\nIf $a_1 = 3$, $b_1 = 10$, then $d_1 \\cdot 13 \\cdot (169 - 90) = 13 d_1 \\cdot 79$ does not divide $1911$.\nIf $a_1 = 5$, $b_1 = 8$, then $d_1 \\cdot 13 \\cdot (169 - 120) = d_1 \\cdot 13 \\cdot 49 = 1911$, which gives $d_1 = 3$, whence $d = 3 \\cdot 5 \\cdot 8 = 120$, i.e., the pair $a = 600$, $b = 960$ is a solution.\nFinally, if $a_1 = 6$, $b_1 = 7$, then $d_1 \\cdot 13 \\cdot (169 - 126) = 13 d_1 \\cdot 43$ does not divide $1911$.\n\nThus, the equation has two solutions mentioned above.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25508,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given real numbers $x, y, z$, with\n$$\nx + y + z = xy + yz + zx = -1.\n$$\nProve that $(xy - z^2)(yz - x^2)(zx - y^2) = xyz - 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $xyz = a$. Then\n$$\n\\begin{aligned}\n(xy - z^2)(yz - x^2)(zx - y^2) &= \\\\\n&= \\left(\\frac{a}{z} - z^2\\right) \\left(\\frac{a}{x} - x^2\\right) \\left(\\frac{a}{y} - y^2\\right) = \\frac{1}{a}(a - z^3)(a - x^3)(a - y^3) = \\\\\n&= \\frac{1}{a}(a^3 - (x^3 + y^3 + z^3)a^2 + (x^3y^3 + y^3z^3 + z^3x^3)a - x^3y^3z^3) = \\\\\n&= [x^3y^3z^3 = a^3] = (x^3y^3 + y^3z^3 + z^3x^3) - (x^3 + y^3 + z^3)a = A - Ba, \\quad (1)\n\\end{aligned}\n$$\nwhere\n$$\nA = x^3y^3 + y^3z^3 + z^3x^3 \\quad \\text{and} \\quad B = x^3 + y^3 + z^3.\n$$\nLet $xy^2 + x^2y + yz^2 + y^2z + zx^2 + xz^2 = C$. Find the values of $A$, $B$, $C$. First,\n$$\nx^2 + y^2 + z^2 = (x + y + z)^2 - 2(xy + yz + zx) = 1 - 2 \\cdot (-1) = 3.\n$$\nSecondly,\n$$\nB = x^3 + y^3 + z^3 = (x+y+z)(x^2+y^2+z^2) - C = (-1) \\cdot 3 - C = -3 - C. \\quad (2)\n$$\nFurther,\n$$\nC = (x + y + z)(xy + yz + zx) - 3xyz = -1 \\cdot (-1) - 3a = 1 - 3a.\n$$\nFrom (2) it follows that\n$$\nB = -3 - (1 - 3a) = 3a - 4.\n$$\nThen\n$$\nx^2y^2 + y^2z^2 + z^2x^2 = (xy + yz + zx)^2 - 2xyz(x+y+z) = (-1)^2 - 2 \\cdot (-1) = 2a + 1.\n$$\nSo,\n$$\nA = (xy+yz+zx)(x^2y^2+y^2z^2+z^2x^2)-Cxyz = (-1)(2a+1)-(1-3a)a = 3a^2-3a-1.\n$$\nSubstituting $A$ and $B$ in (1), we obtain\n$$\nA - Ba = 3a^2 - 3a - 1 - (3a - 4)a = a - 1 = xyz - 1,\n$$\nas required.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25509,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sequence $a_1, a_2, \\dots$ of positive real numbers satisfies\n$$\na_{k+1} \\ge \\frac{ka_k}{a_k^2 + k - 1}\n$$\nfor every positive integer $k$.\nProve that $a_1 + a_2 + \\dots + a_n \\ge n$ for every $n \\ge 2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "2. See IMO-2015 Shortlist, Problem A1.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25510,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $K$ and $L$ be the centers of excircles of a non-isosceles triangle $ABC$ lying opposite vertices $B$ and $C$, respectively. Let $B_1$ and $C_1$ be the midpoints of the sides $AC$ and $AB$, respectively. Let $M$ and $N$ be symmetric to $B$ and $C$ about $B_1$ and $C_1$, respectively.\nProve that the lines $MK$ and $LN$ meet the line $BC$ at the same point.\n\n(D. Voinov)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $P$ and $Q$ be the projections of $L$ and $N$ onto the line $BC$, respectively. Let $a = BC$, $b = CA$, $c = AB$, let $p$ and $S$ be the half-perimeter and the area of the triangle $ABC$, respectively. Let $X$, $Y$ be the intersection points of the lines $LN$, $MK$ and $BC$, respectively. Without loss of generality we assume that $X$, $Y$ lie on the segment $BC$ (all other positions of these points can be considered in similar way).\n\n![](attached_image_1.png)\n\nIt is well known that the radius of the excircle with the center lying opposite to the vertex $C$ is equal to $S/(p-c)$, so $LP = S/(p-c)$. By condition, $AC_1 = BC_1$ and since $N$ is symmetric to $C$ about $C_1$, we have $NC_1 = C_1C$, so $ANBC$ is a parallelogram. If $AH$ is the altitude of $ABC$, then $NQ = AH = 2S/a$. Since $CP = p$, $CB = a$, we have $BP = p-a$. From the equality of the triangles $CHA$ and $BQN$ it follows that $BQ = CH = b \\cos \\angle C$. Since $LP \\parallel NQ$, we see that $\\triangle XLP \\sim \\triangle XNQ$. Then $\\frac{XP}{XQ} = \\frac{LP}{NQ}$. Let $BX = x$, then\n\n$$\n\\frac{x+p-a}{x+b \\cos \\angle C} = \\frac{S/(p-c)}{2S/a} \\Rightarrow x = \\frac{2(p-c)(p-a) - ab \\cos \\angle C}{c-b}\n$$\n\nSimilarly, if $y = CY$, then $y = \\frac{2(p-b)(p-a) - ac \\cos \\angle B}{b-c}$. To prove the required statement, it suffices to show that $x+y=a$. Indeed,\n\n$$\n\\begin{align*}\nx+y &= \\frac{2(p-c)(p-a) - ab \\cos \\angle C}{c-b} + \\frac{2(p-b)(p-a) - ac \\cos \\angle B}{b-c} = \\\\\n&= \\frac{2(p-c)(p-a) - 2(p-b)(p-a) + ac \\cos \\angle B - ab \\cos \\angle C}{c-b} = \\\\\n&= \\frac{2(b-c)(p-a) + ac \\cdot \\frac{a^2+c^2-b^2}{2ac} - ab \\cdot \\frac{a^2+b^2-c^2}{2ab}}{c-b} = \\\\\n&= \\frac{(b-c)(b+c-a) + c^2-b^2}{c-b} = \\frac{a(c-b)}{c-b} = a.\n\\end{align*}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25511,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all real numbers $a$ such that there exists a function $f: \\mathbb{R} \\to \\mathbb{R}$ satisfying the following conditions:\n1) $f(f(x)) = x f(x) - a x$ for all real $x$;\n2) $f$ is not constant;\n3) $f$ takes the value $a$.",
    "options": [],
    "answer": "a = 0 or a = -1",
    "solution": "Answer: $a = 0$ or $a = -1$.\n\nBy condition, there is $c$ such that $f(c) = a$. Substituting $c$ for $x$ in the equality\n$$\nf(f(x)) = x f(x) - a x, \\quad (1)\n$$\nwe obtain $f(a) = f(f(c)) = c f(c) - a c = c a - a c = 0$.\n\nSubstituting $a$ for $x$ in (1), we obtain\n$$\nf(0) = f(f(a)) = a f(a) - a^2 = -a^2 \\quad (2)\n$$\n\nSubstituting $0$ for $x$ in (1), we obtain $f(-a^2) = f(f(0)) = 0 \\cdot f(0) - a \\cdot 0 = 0$.\n\nSubstituting $-a^2$ for $x$ in (1), we obtain\n$$\nf(0) = f(f(-a^2)) = -a^2 f(-a^2) - a \\cdot (-a^2) = a^3. \\quad (3)\n$$\n\nFrom (2) and (3) it follows that $-a^2 = a^3$, hence either $a = 0$ or $a = -1$.\n\nIf $a = -1$, then it is easy to see that the function\n$$\nf(x) = \\begin{cases} -1 & \\text{for } x \\ne -1, \\\\ 0 & \\text{for } x = -1 \\end{cases}\n$$\nsatisfies the problem condition.\n\nIf $a = 0$, then it is easy to see that the function\n$$\nf(x) = \\begin{cases} 0 & \\text{for } x \\ne 1, \\\\ 1 & \\text{for } x = 1 \\end{cases}\n$$\nsatisfies the problem condition.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25512,
    "subject": "Mathematics (Multi-modal)",
    "question": "We say that a diagonal of a convex pentagon is *good* if it divides the pentagon into a triangle and a circumscribed quadrilateral.\nFind the greatest number of good diagonals in a convex pentagon. (I. Gorodnin)",
    "options": [],
    "answer": "2",
    "solution": "Answer: 2.\n\nShow that any two intersecting diagonals of the pentagon cannot be good at the same time. Suppose, contrary to our claim, that there are two good intersecting diagonals. Without loss of generality, we assume that $AD$ and $BE$ are good diagonals of the pentagon $ABCDE$ (see Fig. 1). Then $BCDE$ and $ACDE$ are circumscribed quadrilaterals, therefore the sums of their opposite sides are equal. Hence,\n$$\nAC + DE = AE + CD \\quad \\text{and} \\quad BE + CD = BC + DE,\n$$\nwhich gives\n$$\nAC + BE = AE + BC. \\quad (1)\n$$\nLet $M$ be the intersection point of $AC$ and $BE$. Then, by the triangle inequality, we have\n$$\nAC + BE = (AM + MC) + (BM + ME) = \\\\\n= (AM + ME) + (BM + MC) > AE + BC,\n$$\ncontrary to (1). Similarly, the crossing diagonals $AC$ and $BD$ of the pentagon $ABCDE$ cannot be good at the same time. Since there are no more than two noncrossing diagonals in the pentagon, the number of good diagonals in the pentagon is less than or equal to 2.\n\n![](attached_image_1.png)\nРис. 1\n![](attached_image_2.png)\nРис. 2\n\nShow that there exist a pentagon with two good diagonals. We mark five points $A, B, C, D, E$ in the plain such that $AB = AC = AD = AE$ and $\\angle BAC = \\angle CAD = \\angle DAE = \\alpha < 60^\\circ$ (see Fig. 2). Then it is evident that the triangles $BAC, CAD, DAE$ are equal. Therefore, $BC = CD = DE$ and $\\angle ABC = \\angle ACB = \\angle ACD = \\angle ADC = \\angle ADE = \\angle AED = (180^\\circ - \\alpha)/2 = \\beta < 90^\\circ$. The constructed pentagon is convex since, by construction, $\\angle A = 3\\alpha < 180^\\circ$, $\\angle B = \\angle E = \\beta < 90^\\circ$, $\\angle C = \\angle D = 2\\beta < 180^\\circ$. Two diagonals $AC$ and $AD$ are good since the quadrilaterals $ACDE$ and $ABCD$ are circumscribed quadrilaterals because $AC + DE = AE + CD$ and $AB + CD = AD + BC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25513,
    "subject": "Mathematics (Multi-modal)",
    "question": "Points $B_1$ and $C_1$ are marked respectively on the sides $AB$ and $AC$ of an acute isosceles triangle $ABC$ ($AB = AC$) such that $BB_1 = AC_1$. The points $B$, $C$, and $S$ lie in the same half-plane with respect to the line $B_1C_1$ so that\n$$\n\\angle SB_1C_1 = \\angle SC_1B_1 = \\angle BAC.\n$$\nProve that $B$, $C$, $S$ are collinear if and only if the triangle $ABC$ is equilateral.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Necessity. Let $B$, $S$, $C$ be collinear. Let $X$ be the midpoint of the segment $B_1C_1$. Then the bisector of $B_1C_1$ passes through $X$ and intersects the side $BC$ at $S$ since $S$ lies on $BC$ and from the condition it follows that $SB_1 = SC_1$ (see Fig. 1).\nLet $O$ be the circumcenter of the triangle $ABC$. Since the triangle $ABC$ is isosceles, $AO$ is the bisector of the angle $BAC$. Since $OB = OA$, it follows that $\\angle OBB_1 = \\angle OAB_1 = \\angle OAC_1$. By condition, $BB_1 = AC_1$, hence the triangles $OBB_1$ and $OAC_1$ are equal. Then $\\angle OBB_1B = \\angle OAC_1A$, so\n$$\n\\angle OBB_1A + \\angle OC_1A = 180^\\circ - \\angle OBB_1B + \\angle OC_1A = 180^\\circ,\n$$\nwhich means that the quadrilateral $OBB_1AC_1$ is cyclic. Since $\\angle B_1AO = \\angle C_1AO$, we see that $OB_1 = OC_1$ as the chords of the equal arcs. Hence, $OX$ is the median of the isosceles triangle $B_1OC_1$ and so $OX$ is the bisector of the angle $B_1OC_1$, i.e., $\\angle XOC_1 = 0.5\\angle B_1OC_1 = 0.5(180^\\circ - \\angle B_1AC_1)$. (Note that $OX$ is also the altitude of $B_1OC_1$ and since $SX \\perp B_1C_1$, we see that $O$ lies on $SX$.)\nTherefore,\n$$\n\\angle XOC_1 = 90^\\circ - 0.5\\angle B_1AC_1 = 90^\\circ - 0.5\\angle BAC = 90^\\circ - \\angle OAC =\n$$\n$$\n= [OA \\perp BC] = \\angle ACB = \\angle C_1CS,\n$$\nwhence the quadrilateral $OC_1CS$ is cyclic. Therefore $\\angle OSC_1 = \\angle OCC_1$. Then we have\n$$\n\\frac{1}{2}\\angle BAC = \\angle OAC = [OA=OC \\Rightarrow \\angle OAC = \\angle OCA] = \\angle OCA = \\angle OCC_1 = \\angle OSC_1 = \\angle XSC_1 = 90^\\circ - \\angle SC_1X = 90^\\circ - \\angle SC_1B_1 = 90^\\circ - \\angle BAC.\n$$\nHence $\\frac{3}{2}\\angle BAC = 90^\\circ$, so $\\angle BAC = 60^\\circ$, which means that the triangle $ABC$ is equilateral as required.\n\n![](attached_image_1.png)\n\nSufficiency. It is easy to see that there exists a unique point $S$ satisfying the problem condition. Mark the point $A_1$ on the side $BC$ such that $A_1C = AC_1$ (see Fig. 2). Show that $A_1$ and $S$ coincide if the triangle $ABC$ is equilateral. Indeed, we have\n$$\nBB_1 = AC_1 = CA_1, \\quad A_1B = B_1A = C_1C,\n$$\n$$\n\\angle A_1BB_1 = \\angle B_1AC_1 = \\angle C_1CA_1 = \\angle BAC = 60^\\circ.\n$$\nIt follows that $\\triangle A_1BB_1 \\cong \\triangle B_1AC_1 \\cong \\triangle C_1CA_1$. Hence, $A_1B_1 = B_1C_1 = C_1A_1$, so the triangle $A_1B_1C_1$ is equilateral and $\\angle A_1B_1C_1 = \\angle B_1C_1A_1 = 60^\\circ = \\angle BAC$, which means that $A_1$ and $S$ coincide. Therefore, if the triangle $ABC$ is equilateral, then the points $B$, $S$, $C$ are collinear.\n\n![](attached_image_2.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25514,
    "subject": "Mathematics (Multi-modal)",
    "question": "A point $A_1$ is marked inside an acute non-isosceles triangle $ABC$ such that $\\angle A_1AB = \\angle A_1BC$ and $\\angle A_1AC = \\angle A_1CB$.\n\nPoints $B_1$ and $C_1$ are defined in the same way. Let $G$ be the gravity center of the triangle $ABC$.\n\nProve that the points $A_1, B_1, C_1, G$ are concyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "There is nothing to prove if some two of four points $A_1$, $B_1$, $C_1$, $G$ coincide. So, we can assume that all these points are pairwise distinct.\n\nLet $\\Gamma$ and $\\Gamma(A_1)$ denote the circumcircles of the triangles $ABC$ and $AA_1B$, respectively. Let $M_A$ be the intersection point of the lines $AA_1$ and $BC$. Since $\\angle A_1AB = \\angle A_1BC$, it follows that $\\Gamma(A_1)$ touches $BC$ at $B$. By the Power of a Point Theorem, we have $M_AB^2 = M_AA_1 \\cdot M_AA$, Similarly, $M_AC^2 = M_AA_1 \\cdot M_AA$, hence $M_AB^2 = M_AC^2$, so $M_AB = M_AC$, i.e., $M_A$ is the midpoint of the side $BC$. Thus, $A_1$ lies on the median of the triangle $ABC$ from the vertex $A$. By the same arguments, $B_1$ and $C_1$ lie on the medians of the triangle $ABC$ from the vertices $B$ and $C$, respectively.\n\n![](attached_image_1.png)\n\nLet $H$ be the orthocenter of the triangle $ABC$. It is well known that the points which are symmetric to $H$ with respect to the sides of the triangle $ABC$ lie on the circumcircle of $ABC$. If $H_A$ is symmetric to $H$ with respect to the side $BC$, then $H_A$ belongs to $\\Gamma$ and we have\n$$\n\\begin{aligned}\n\\angle BHC &= \\angle BH_A C = 180^\\circ - \\angle BAC = 180^\\circ - (\\angle A_1 AB + \\angle A_1 AC) = \\\\\n&= 180^\\circ - (\\angle A_1 BC + \\angle A_1 CB) = \\angle BA_1 C,\n\\end{aligned}\n$$\nhence the quadrilateral $BA_1HC$ is cyclic and $A_1$ belongs to the circle passing through the points $B$, $H$, $C$.\n\nFurther,\n$$\n\\begin{aligned}\n\\angle HA_1M_A &= \\angle HA_1B + \\angle BA_1M_A = \\angle BCH + (\\angle A_1BA + \\angle A_1AB) = \\\\\n&= (90^\\circ - \\angle ABC) + (\\angle A_1BA + \\angle A_1BC) = 90^\\circ - \\angle ABC + \\angle ABC = 90^\\circ.\n\\end{aligned}\n$$\nTherefore, $A_1$ is the projection of $H$ on the median $AM_A$. Similarly, $B_1$ and $C_1$ are the projections of $H$ on the median of the triangle $ABC$ from the vertices $B$ and $C$, respectively.\n\nSince $G$ is the intersection point of the medians of $ABC$, we have $\\angle HA_1G = \\angle HB_1G = \\angle HC_1G = 90^\\circ$. Thus $A_1, B_1, C_1$ lie on the circle with the diameter $GH$, so all points $A_1, B_1, C_1, G$ lie on the same circle.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25515,
    "subject": "Mathematics (Multi-modal)",
    "question": "Points $K$ and $N$ are marked on the sides $AB$ and $AC$ of the triangle $ABC$, respectively, such that $KB = KN$. The bisector of the angle $ACB$ meets the circumcircle of $ABC$ at points $C$ and $R$. The perpendicular from $R$ on $AB$ intersects the segment $BN$ at $D$.\nProve that the points $A$, $K$, $D$, $N$ are concyclic.\n\n(S. Mazanik)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $R$ is the midpoint of the arc $\\angle AB$, the diameter of the circumcircle passing through $R$ is perpendicular to the chord $AB$. So, $RD$ is a perpendicular bisector of the segment $AB$. Therefore, $AD = BD$ and $\\angle ABD = \\angle BAD$.\n\nBy condition, $KB = KN$, then $\\angle KBN = \\angle KNB$. Hence, $\\angle KAD = \\angle BAD = \\angle ABD = \\angle KBN = \\angle KNB = \\angle KND$.\n\nThe points $A$, $K$, $D$, and $N$ are concyclic since the angles $KAD$ and $KND$ subtend the same segment $KD$ and lie in the same half-plane with respect to the line $KD$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25516,
    "subject": "Mathematics (Multi-modal)",
    "question": "Any diagonal connecting two opposite vertices of a convex hexagon divides this hexagon into two quadrilaterals. Six quadrilaterals can be obtained in this way.\nFind the greatest number of these quadrilaterals that can occur circumscribed quadrilaterals.\n(S. Mazanik, I. Voronovich)",
    "options": [],
    "answer": "3",
    "solution": "Answer: 3.\n\nWe show that at most three circumscribed quadrilaterals can be obtained. Suppose, contrary to our claim, that there are at least four circumscribed quadrilaterals. Then some two of them are obtained when one diagonal (say, $AD$) is constructed.\n\n![](attached_image_1.png)\n\nFig. 1\n\n![](attached_image_2.png)\n\nFig. 2\n\nConsider one of the circumscribed quadrilaterals remained. Without loss of generality we assume that this is the quadrilateral $BCDE$. For this quadrilateral we have $BC + DE = CD + BE$. For the quadrilateral $ABCD$ we have $AB + CD = BC + AD$ since, by our assumption, this quadrilateral is circumscribed. Summing these two equalities we obtain $AB + DE = AD + BE$. The last equality is impossible because $AD + BE > AB + DE$ for the convex quadrilateral $ABDE$. Indeed, if $M$ is the intersection point of $AD$ and $BE$, then, by the triangle inequality,\n\n$$\nAD + BE = (AM + MD) + (BM + ME) = (AM + BM) + (DM + EM) > \\\\\n> [AM + BM > AB, DM + EM > DE] > AB + DE.\n$$\n\nThus the number of the circumscribed quadrilaterals is less than or equal to 3.\n\nThere are many ways to construct the convex hexagon with three circumscribed quadrilaterals. One of such hexagons is shown in Fig. 2, where $A_1B_1A_2B_3$, $A_2B_2A_3B_1$, $A_3B_3A_1B_2$ are the circumscribed quadrilaterals.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25517,
    "subject": "Mathematics (Multi-modal)",
    "question": "A number $\\alpha$ is a root of the equation $x^3 - 12x + 8 = 0$.\nProve that the number $2 - \\frac{4}{\\alpha}$ is also the root of this equation.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Set $x = 2 - \\frac{4}{\\alpha}$ in the left hand side of the given equation:\n$$\n\\begin{aligned}\n& \\left(2 - \\frac{4}{\\alpha}\\right)^3 - 12\\left(2 - \\frac{4}{\\alpha}\\right) + 8 = 8\\left(\\left(1 - \\frac{2}{\\alpha}\\right)^3 - 3\\left(1 - \\frac{2}{\\alpha}\\right) + 1\\right) = \\\\\n& = 8\\left(1 - \\frac{6}{\\alpha} + \\frac{12}{\\alpha^2} - \\frac{8}{\\alpha^3} - 3 + \\frac{6}{\\alpha} + 1\\right) = 8\\left(-1 + \\frac{12}{\\alpha^2} - \\frac{8}{\\alpha^3}\\right) = \\\\\n& = -\\frac{8}{\\alpha^3}\\left(\\alpha^3 - 12\\alpha + 8\\right) = 0\n\\end{aligned}\n$$\nsince by the problem condition $\\alpha^3 - 12\\alpha + 8 = 0$. Therefore, the number $2 - \\frac{4}{\\alpha}$ is a root of the given equation.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 25518,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$, $B$, $C$ denote the intersection points of the diagonals $A_1A_4$ and $A_2A_5$, $A_1A_6$ and $A_2A_7$, $A_1A_9$ and $A_2A_{10}$ of the regular decagon $A_1A_2...A_9A_{10}$, respectively.\nFind the angles of the triangle $ABC$.\n(Folklore)",
    "options": [],
    "answer": "∠ABC = 90°, ∠BCA = 36°, ∠CAB = 54°",
    "solution": "(Solution by A. Goloubitskaya.) Let $\\Gamma$ be the circumcircle of the given regular hexagon. It is evident that $B$ is the center of $\\Gamma$. Since all sides of the regular hexagon are equal, we have $2\\alpha = \\text{arc } A_1A_2 = \\text{arc } A_2A_3 = \\dots = \\text{arc } A_9A_{10} = \\text{arc } A_{10}A_1 = 36^\\circ$. Since $A_1A_2 = A_9A_{10}$, $\\angle A_{10}A_2A_1 = \\angle A_1A_9A_{10} = \\alpha$ and $\\angle A_2A_1A_9 = \\angle A_2A_{10}A_9 = 7\\alpha$, it follows that the triangles $A_1A_2C$ and $A_{10}A_9C$ are equal, so $CA_1 = CA_{10}$. Hence the line $BC$ is the bisector of the segment $A_1A_{10}$. Then the line $BC$ contains the altitude, the median, and the bisectrix of the isosceles triangle $A_{10}BA_1$ ($BA_{10} = BA_1$). Hence, $\\angle A_1BC = 0.5 \\text{ arc } A_{10}A_1 = \\alpha = 18^\\circ$. Similarly, $\\angle A_2BA = 0.5 \\text{ arc } A_2BA_4 = 0.5 \\text{ arc } A_2A_4 = 2\\alpha = 36^\\circ$. Since $\\angle A_2BA_1 = 2\\alpha = 72^\\circ$, we have\n![](attached_image_1.png)\n$$\n\\angle ABC = \\angle ABA_2 + \\angle A_2BA_1 + \\angle A_1BC = 36^\\circ + 36^\\circ + 18^\\circ = 90^\\circ.\n$$\n\nSince $\\angle AA_2C = \\angle A_5A_2A_{10} = 5\\alpha = 90^\\circ$, we have $\\angle ABC + \\angle AA_2C = 180^\\circ$, so the points $A$, $B$, $C$, $A_2$ are concyclic. Then $\\angle BCA = \\angle BA_2A = 2\\alpha = 36^\\circ$ and $\\angle CAB = 180^\\circ - \\angle ABC - \\angle BCA = 54^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25519,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a graph with $n$ ($n \\ge 4$) vertices. It is known that for any two of the vertices there is a vertex connected with none of these two vertices.\nFind the greatest possible number of the edges in the graph.\n(E. Barabanov)",
    "options": [],
    "answer": "((n-2)(n-1))/2 - floor(n/2)",
    "solution": "(Solution by Y. Dubovik, L. Manzhulina, I. Pchalintsau, B. Serankou.)\nFirst, for the sake of convenience, we reformulate the problem as follows. Any\ntwo of $n$ vertices of the graph ($n \\ge 4$) are connected with an edge either of\nred or of blue color. It is known that for any two vertices there exists a vertex\nconnected to both the vertices with blue edges. Find the maximum number of\nred edges (or, equivalently, the minimum number of blue edges) in the graph.\nWe find the minimum number $k$ of blue edges then the maximum number of\nred edges is $\\frac{n(n-1)}{2} - k$. In the\nfigures blue edges are shown, all\nthe absent edges are red. These\nfigures show that the minimum\nnumber of blue edges cannot be\ngreater than $k = n - 1 + \\lfloor \\frac{n}{2} \\rfloor$.\nShow that this number is indeed\nthe smallest possible. We call the\nnumber of blue edges containing\na vertex a degree of this vertex. If the degrees of all the graph vertices are at least 3, then the total number of blue edges is at least $k_1 = \\frac{3n}{2} > k$. So we may assume that there is a vertex $A$ of degree 2. Let $AB$ and $AC$ be the corresponding blue edges. Applying the problem condition to $A$ and $B$, we conclude that $B$ and $C$ are certainly connected with blue edge. Let $M$ be the set of remaining (other than $A$, $B$, $C$) vertices. Then any vertex from $M$ must be connected with blue edge either to $B$ or to $C$. For any vertex from $M$ we mark this edge. From the problem condition it follows that in addition to marked edges at least one blue edge should outgo from any vertex of $M$.\nThe number of these additional edges is not less than $\\lfloor \\frac{n-2}{2} \\rfloor = \\lfloor \\frac{n}{2} \\rfloor - 1$.\nTherefore there are at least\n$3 + n - 3 + \\lfloor \\frac{n}{2} \\rfloor - 1 = n - 1 + \\lfloor \\frac{n}{2} \\rfloor$\nblue edges as was claimed.\n\nn is odd\n![](attached_image_1.png)\nn is even\n![](attached_image_2.png)\n\nSo, the maximum number of the red edges is\n$$\n\\frac{n(n-1)}{2} - (n-1) - \\lfloor \\frac{n}{2} \\rfloor = \\frac{(n-2)(n-1)}{2} - \\lfloor \\frac{n}{2} \\rfloor .\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 25520,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with $\\angle C = 90^\\circ$, and let $H$ be the foot of the altitude from $C$. A point $D$ is chosen inside the triangle $CBH$ so that $CH$ bisects $AD$. Let $P$ be the intersection point of the lines $BD$ and $CH$. Let $\\omega$ be the semicircle with diameter $BD$ that meets the segment $CB$ at an interior point. A line through $P$ is tangent to $\\omega$ at $Q$.\n\nProve that the lines $CQ$ and $AD$ meet on $\\omega$.\n\n(IMO-2015 Shortlist, Problem G3)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25521,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a finite set $A$ of positive integers, we call a partition of $A$ into two disjoint nonempty subsets $A_1$ and $A_2$ good if the least common multiple of the elements in $A_1$ is equal to the greatest common divisor of the elements in $A_2$.\n\nDetermine the minimum value of $n$ such that there exists a set of $n$ positive integers with exactly $2015$ good partitions.\n\n(IMO-2015 Shortlist, Problem C3)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25522,
    "subject": "Mathematics (Multi-modal)",
    "question": "We say that a diagonal of a convex hexagon is *good* if it divides the hexagon into two circumscribed quadrilaterals.\nFind the greatest number of good diagonals in a convex hexagon.",
    "options": [],
    "answer": "1",
    "solution": "Answer: 1.\n\nShow that among any two intersecting main diagonals of the hexagon at most one diagonal can be good. Suppose, contrary to our claim, that there are two good intersecting main diagonals. Without loss of generality, we assume that $AD$ and $BE$ are good diagonals of the hexagon $ABCDEF$ (see Fig. 1). Then $ABCD$ and $BCDE$ are circumscribed quadrilaterals, therefore the sums of their opposite sides are equal. Hence, $AB + CD = BC + AD$, $BC + DE = BE + CD$. Summing both the equalities we obtain\n$$\nAB + DE = AD + BE. \\qquad (1)\n$$\nLet $M$ be the intersection point of $AD$ and $BE$. Then by the triangle inequality we have\n$$\nAD + BE = (AM + MD) + (BM + ME) = (AM + BM) + (DM + EM) > \\\\ > AB + DE,\n$$\ncontrary to (1).\n\n![](attached_image_1.png)\nFig. 1\n![](attached_image_2.png)\nFig. 2\n\nIf the diagonal of the hexagon divides it into two quadrilaterals, then this diagonal connects two opposite vertices of the hexagon (\"main\" diagonal). But any two such diagonals intersect each other, hence the hexagon has at most one good diagonal.\n\nThere are many ways to construct the hexagon with one good diagonal. For example, we draw the circle and circumscribe the trapezoid $ABCD$, $AD \\parallel BC$, $AD > BC$ (see Fig. 2). Let $E$ and $F$ be symmetric to $C$ and $B$, respectively, with respect to the line $AD$. It is evident that $AD$ is a good diagonal of the hexagon $ABCDEF$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25523,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve the equation $2^a - 5^b = 3$ in positive integers $a, b$.",
    "options": [],
    "answer": "(a, b) = (3, 1) or (7, 3)",
    "solution": "Answer: $(a; b) = (3; 1)$ or $(7; 3)$.\n\n(Solution by Y. Dubovik.) Note that if $a \\le 7$ or $b \\le 3$, then it is easy to verify that the only solutions are $(a; b) = (3; 1)$ and $(a; b) = (7; 3)$.\n\nNow it remains to prove that there are no solutions with $a > 7, b > 3$. In this case we can write $a = 7 + \\alpha, b = 3 + \\beta$, where $\\alpha, \\beta \\in \\mathbb{N}$. Then the given equation $2^{7+\\alpha} - 5^{3+\\beta} = 3$ transforms to\n$$\n2^7(2^\\alpha - 1) = 5^3(5^\\beta - 1). \\qquad (1)\n$$\nSuppose that (1) has a solution $(\\alpha; \\beta)$ in positive integers. Set $A = 2^7(2^\\alpha - 1) = 5^3(5^\\beta - 1)$.\n\n1. From (1) it follows that $A \\supseteq 2^7$, so $5^\\beta \\equiv 1 \\pmod{2^7}$. One can easily deduce that $\\beta \\supseteq 32$.\n\n2. Also, $A \\supseteq 5^3$, so $2^\\alpha \\equiv 1 \\pmod{125}$ and we deduce that $\\alpha \\supseteq 100$. In particular, $A \\supseteq (2^{100} - 1)$, and so $A \\supseteq (2^5 - 1) = 31$. It follows that $5^\\beta - 1 \\supseteq 31$, thus $\\beta \\supseteq 3$.\n\n3. Thus we have $\\beta \\supseteq 96$, whence $A \\supseteq (5^{96} - 1)$, so $A \\supseteq 97$, which implies $2^\\alpha \\equiv 1 \\pmod{97}$. Therefore, $\\alpha \\supseteq 48$. In particular, $A \\supseteq (2^{48} - 1)$, then $A \\supseteq (2^{16} - 1)$ and $A \\supseteq (2^8 + 1) = 257$.\n\n4. Finally, $5^\\beta - 1 \\supseteq 257$, which gives $\\beta \\supseteq 256$, whence $A \\supseteq 2^8$, contrary to (1).",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 25524,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m$ and $n$ be positive integers such that $m > n$. Define $x_k = \\frac{m + k}{n + k}$ for $k = 1, 2, \\ldots, n + 1$.\nProve that if all the numbers $x_1, x_2, \\ldots, x_{n+1}$ are integers,\nthen $x_1 x_2 \\ldots x_{n+1} - 1$ is divisible by an odd prime.\n(IMO-2015 Shortlist, Problem N3)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25525,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\mathcal{A}$ be a nonempty set of positive integers. We say that a positive integer $n$ is *special* if there exists a unique subset $\\mathcal{B}$ of the set $\\mathcal{A}$ such that\n(i) the number of the elements in $\\mathcal{B}$ is odd;\n(ii) the sum of all elements of $\\mathcal{B}$ is equal to $n$.\n\nProve that there exist infinitely many positive integers that are not special.\n\n(IMO-2015 Shortlist, Problem C6)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "3. See IMO-2015 Shortlist, Problem C6.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25526,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $D$, $E$, $F$ denote the tangent points of the incircle of $ABC$ with the sides $BC$, $AC$, $AB$, respectively. Let $M$ be the midpoint of the segment $EF$. Let $L$ be the intersection point of the circle passing through $D$, $M$, $F$ and the segment $AB$, $K$ be the intersection point of the circle passing through $D$, $M$, $E$ and the segment $AC$.\nProve that the circle passing through $A$, $K$, $L$ touches the line $BC$.\n(V. Voinov)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let *I* be the incenter of the triangle *ABC*. Let $\\Gamma$ denote the circle passing through $A$, $K$, $L$. We have\n$$\n\\begin{align*}\n\\angle AKD = \\angle EKD &= 180^\\circ - \\angle DME = \\angle DMF = 180^\\circ - \\angle FLD = \\\\\n&= 180^\\circ - \\angle ALD.\n\\end{align*}\n$$\nTherefore, $D$ lies on $\\Gamma$. So, to prove that $\\Gamma$ touches $BC$ it suffices to show that $\\angle DAL = \\angle BDL$.\n\n![](attached_image_1.png)\n\nSince $AE = AF$ and $EM = MF$, it follows that $AM \\perp EF$ and $AM$ is the bisector of the angle $EAF$, i.e., the points $A$, $M$, $I$ lie on the same line.\nWe have\n$$\n\\begin{align*}\n\\angle BDL &= 90^\\circ - \\angle IDL = 90^\\circ - (\\angle IDM + \\angle MDL) = \\\\\n&= (90^\\circ - \\angle MDL) - \\angle IDM = [\\angle MDL = 180^\\circ - \\angle MFL = \\angle MFA] = \\\\\n&= (90^\\circ - \\angle MFA) - \\angle IDM = \\angle FAM - \\angle IDM = \\\\\n&= \\angle DAL + \\angle DAM - \\angle IDM. \\tag{1}\n\\end{align*}\n$$\n\nSince the triangle $AFI$ is the right-angled triangle and $FM \\perp AI$, we have $IF^2 = IM \\cdot IA$. Since $IF = ID$, we obtain $ID^2 = IM \\cdot IA$. By the Power of a Point Theorem, the circle passing through the points $A$, $M$, $D$ touches the line $ID$ at $D$, then $\\angle IDM = \\angle DAM$. From (1) it follows that $\\angle DAL = \\angle BDL$ as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25527,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two points $A$ and $B$ are marked on the right branch of the hyperbola $y = 1/x$ ($x > 0$). The straight line $\\ell$ passing through the origin $O$ is perpendicular to the line $AB$ and meets $AB$ and the given branch of the hyperbola at points $D$ and $C$, respectively. The circle $S$ passes through the points $A, B, C$ and meets $\\ell$ at $F$.\nFind all possible values of the ratio $OD:CF$.",
    "options": [],
    "answer": "1/2",
    "solution": "Let $A(a; 1/a)$, $B(b; 1/b)$, $C(c; 1/c)$. It is easy to see that $y = -\\frac{x}{ab} + \\frac{1}{a} + \\frac{1}{b}$ is the equation of the line $AB$ and $y = abx$ is the equation of the line $OD$ because $OD$ is perpendicular to $AB$. Then we easily calculate the coordinates of $C$: $C = (\\frac{1}{\\sqrt{ab}}, \\sqrt{ab})$.\n\nLet $(x-t)^2 + (y-l)^2 = r^2$ be the equation of the circle $S$. Since $A, B, C$ lie on the hyperbola, we have\n$$\n(x-t)^2 + \\left(\\frac{1}{x}-l\\right)^2 = r^2 \\implies x^4 - 2tx^3 + (t^2+l^2-r^2)x^2 - 2lx + 1 = 0. \\quad (1)\n$$\nEquation (1) has even number of real roots. Since $S$ passes through $A, B, C$, equation (1) has four real roots (not necessarily different). Let $\\alpha$ be the fourth root of (1), then\n$$\n1 = \\alpha abc = \\alpha \\frac{ab}{\\sqrt{ab}} = \\alpha \\sqrt{ab} \\implies \\alpha = \\frac{1}{\\sqrt{ab}}.\n$$\nBy Vieta's theorem,\n$$\n2t = a + b + 2\\frac{1}{\\sqrt{ab}}, \\quad 2l = \\frac{1}{a} + \\frac{1}{b} + 2\\sqrt{ab}, \\quad t^2 + l^2 - r^2 = ab + \\frac{1}{ab} + 2\\frac{a+b}{\\sqrt{ab}}.\n$$\nHence\n$$\nr^2 = \\left(\\frac{a+b}{2} + \\frac{1}{\\sqrt{ab}}\\right)^2 + \\left(\\frac{a+b}{2ab} + \\sqrt{ab}\\right)^2 - ab - \\frac{1}{ab} - 2\\frac{a+b}{\\sqrt{ab}} =\n$$\n$$\n= \\left(\\frac{a+b}{2}\\right)^2 \\left(1+\\frac{1}{a^2b^2}\\right).\n$$\n\nLet $a+b=n$, $\\sqrt{ab}=m$, then\n$$\n\\left(x - \\frac{n}{2} - \\frac{1}{m}\\right)^2 + \\left(y - \\frac{n}{2m^2} - m\\right)^2 = \\left(\\frac{n}{2}\\right)^2 \\left(1 + \\frac{1}{m^4}\\right).\n$$\nFind the coordinates of the intersection points of $S$ and the line $OD$. Since the equation of $OD$ is $y = abx = m^2x$, we have\n$$\n\\left(x - \\frac{n}{2} - \\frac{1}{m}\\right)^2 + \\left(m^2x - \\frac{n}{2m^2} - m\\right)^2 = r^2. \\quad (2)\n$$\nThis equation has two roots, one of them is the abscissa of $C$, i.e., $\\frac{1}{m}$ and the other is the abscissa of $F$ (it is easy to verify that $x = \\frac{1}{m}$ satisfies (2)). Let $F = (f; g)$. Then, by Vieta's theorem,\n$$\n\\frac{1}{m} + f = \\frac{2\\left(\\frac{n}{2} + \\frac{1}{m} + \\frac{n}{2} + m^3\\right)}{1 + m^4} = \\frac{2n}{m^4 + 1} + \\frac{2}{m} \\implies f = \\frac{2n}{m^4 + 1} + \\frac{1}{m}.\n$$\nThen $g = abf = m^2f = m + \\frac{2nm^2}{m^4 + 1}$. Therefore\n$$\nCF^2 = \\left(\\frac{1}{m} - \\frac{1}{m} - \\frac{2n}{m^4 + 1}\\right)^2 + \\left(m - m - \\frac{2nm^2}{m^4 + 1}\\right)^2 = \\\\\n= \\left(\\frac{2n}{m^4 + 1}\\right)^2 (m^4 + 1) = \\frac{4n^2}{m^4 + 1}.\n$$\nLet $D = (d; e)$. Since $D$ is the intersection point of $OD$ and $AB$, we have\n$$\nabd = -\\frac{d}{ab} + \\frac{1}{a} + \\frac{1}{b} \\implies \\frac{a+b}{ab} = d \\left(ab + \\frac{1}{ab}\\right) \\implies\n$$\n$$\nd = \\frac{a+b}{1+a^2b^2} = \\frac{n}{m^4+1},\n$$\n$$\ne = abd = m^2d = \\frac{m^2n}{m^4 + 1}.\n$$\nTherefore\n$$\nOD^2 = \\frac{n^2}{(m^4 + 1)^2} + \\frac{m^4 n^2}{(m^4 + 1)^2} = \\frac{n^2}{m^4 + 1}.\n$$\nThen\n$$\n\\frac{OD^2}{CF^2} = \\frac{n^2}{m^4 + 1} \\cdot \\frac{m^4 + 1}{4n^2} = \\frac{1}{4} \\implies \\frac{OD}{CF} = \\frac{1}{2}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25528,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve the equation $p^3 - q^3 = pq^3 - 1$ in primes $p, q$.",
    "options": [],
    "answer": "p = 19, q = 7",
    "solution": "Answer: $q = 7$, $p = 19$.\n\nNote that $q \\le p$, otherwise LHS $< 0$, RHS $> 0$. We have\n$$\n\\begin{aligned}\np^3 - q^3 &= pq^3 - 1 \\Leftrightarrow p^3 + 1 = pq^3 + q^3 \\\\\n&\\Leftrightarrow (p+1)(p^2-p+1) = q^3(p+1) \\\\\n&\\Leftrightarrow p(p-1) = q^3 - 1 \\\\\n&\\Leftrightarrow p(p-1) = (q-1)(q^2+q+1).\n\\end{aligned}\n$$\nSince $p(p-1) \\nmid p$, we have $(q-1)(q^2+q+1) \\nmid p$. But $q-1 \\nmid p$ since $q-1 < p$, so $q^2+q+1 \\nmid p$, i.e., $q^2+q+1 = kp$, $k \\in \\mathbb{N}$. Therefore\n$$\np(p-1) = (q-1)kp \\Leftrightarrow p-1 = k(q-1) \\Leftrightarrow p = k(q-1)+1.\n$$\nThen\n$$\nq^2 + q + 1 = kp \\Leftrightarrow q^2 + q + 1 = k^2(q - 1) + k. \\quad (1)\n$$\nIf $k > 3$ we have\n$$\n\\begin{aligned}\nq^2 + q + 1 = k^2(q - 1) + k &\\implies (q - 1)(q + 2) + 3 = k^2(q - 1) + k \\\\\n&\\implies (q - 1)(q + 2) - k^2(q - 1) = k - 3.\n\\end{aligned}\n$$\nHence,\n$$\n\\begin{gathered}\n(k-3) \\mid (q-1) \\implies k \\ge q-1+3=q+2 \\implies \\\\\n(q-1)(q+2) = k^2(q-1) + k - 3 \\ge (q+2)^2(q-1) + q - 1 > (q-1)(q+2), \\\\\n\\text{which is impossible. So, we have no solutions when } k > 3. \\text{ It remains to} \\\\\n\\text{consider three cases: } k=1,2,3.\n\\end{gathered}\n$$\nIf $k=1$ from (1) we have $q^2 + q + 1 = q$. Hence $q^2 + 1 = 0$, which is impossible.\n\nIf $k=2$ we have $q^2 + q + 1 = 4(q - 1) + 2$, which gives $q^2 - 3q + 3 = 0$. It is easy to see that this equation has no real solution.\n\nIf $k=3$ from (1) we have\n$$\nq^2 + q + 1 = 9(q - 1) + 3 \\Leftrightarrow q^2 - 8q + 7 = 0.\n$$\nIt is easy to see that either $q=1$ or $q=7$. But 1 is not prime, so $q=7$, then $p=3(q-1)+1=19$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25529,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given the triangle $ABC$ with $AB = 2AC$. If points $M$ and $N$ belong to the sides $BC$ and $AB$, respectively, and the perimeter of the trapezoid $CMNA$ is the sum of the lengths of the sides $AB$ and $AC$, construct $M$ using compasses and ruler.\n\n(S. Mazanik)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "$M$ is the intersection point of the line $\\ell \\parallel AC$ passing through the intersection point of the bisector of the angle $ACB$ and the side $AB$.\n\nLet $M$ be the point we search for and $NM \\parallel AC$ (see the Fig.).\n\nLet $P(CMNA)$ denote the perimeter of the trapezoid $CMNA$. Then $P(CMNA) = AN + NM + MC + CA$ and, by condition, $P(CMNA) = AB + AC = AN + NB + AC$, whence\n$$\nNM + MC = NB. \\quad (1)\n$$\nSince $MN \\parallel AC$, the triangles $NBM$ and $ABC$ are similar, so $NB : NM = AB : AC = 2$, i.e., $NB = 2NM$. From (1) it follows that $MC = MN$, hence, the triangle $NMC$ is isosceles. Therefore, $\\angle MNC = \\angle MCN$. Since $NM \\parallel AC$, we have $\\angle MNC = \\angle NCA$. Thus, $\\angle MCN = \\angle NCA$, i.e., $NC$ is the bisector of the angle $ACB$ of the given triangle $ABC$.\n\n![](attached_image_1.png)\n\nThe construction of the required point $M$: draw the bisector of the angle $ACB$ (the standard rule-compass construction) and let $N$ be the intersection point of this bisector and the side $AB$. Draw the line $\\ell$ passing through $N$ parallel to $AC$ (the standard rule-compass construction). The required point $M$ is the intersection point of the line $\\ell$ and the side $BC$. Indeed, it is easy to see that the perimeter of the trapezoid $ANMC$ thus obtained is equal to the sum of the lengths of the sides $AB$ and $AC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25530,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ and $b$ be positive integers such that $a!b!$ is a multiple of $a! + b!$.\nProve that $3a \\ge 2b + 2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "3. See IMO-2015 Shortlist, Problem N2.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25531,
    "subject": "Mathematics (Multi-modal)",
    "question": "An infinite sequence $(a_n)$, $n \\in \\mathbb{N}$, of positive numbers is called *lacunar* if there exists a number $q > 1$ such that $a_{n+1}/a_n \\ge q$ for all $n \\in \\mathbb{N}$. Also, the sequence is called *solitary* if there exists a number $r > 1$ such that the interval $(x, rx)$ contains at most one term of this sequence for any positive $x$.\n\na) Is it true that any lacunar sequence is solitary?\n\nb) Is it true that any solitary sequence is lacunar?",
    "options": [],
    "answer": "a) Yes; b) No",
    "solution": "a) Let the sequence $(a_n)$, $n \\in \\mathbb{N}$, be lacunar. Then there exists a number $q > 1$ such that\n$$\na_{n+1} \\ge q a_n \\quad \\forall n \\in \\mathbb{N}. \\quad (1)\n$$\nIn particular, any lacunar sequence is increasing. From (1) it follows that any interval $(x, qx)$ contains at most one term of this sequence. Indeed, if we assume that $a_n$ and $a_{n+1}$ belong to this interval, then $\\frac{a_{n+1}}{a_n} < \\frac{qx}{x} = q$, contrary to (1). Therefore, any lacunar sequence is solitary.\n\nb) Consider the lacunar sequence $a_n = 2^n$, $n \\in \\mathbb{N}$. This sequence is solitary as was shown in item a): every interval $(x, 2x)$, where $x > 0$, contains at most one term of this sequence. We construct a new sequence $(x_n)$, $n \\in \\mathbb{N}$, as $x_{2n-1} = a_{2n}$ and $x_{2n} = a_{2n-1}$ for any $n \\in \\mathbb{N}$. This sequence $(x_n)$ is solitary because the set of its terms and the set of the terms of the sequence $(a_n)$ coincide. But the sequence $(x_n)$ is not lacunar since this sequence is not increasing.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25532,
    "subject": "Mathematics (Multi-modal)",
    "question": "An infinite sequence $(a_n)$, $n \\in \\mathbb{N}$, of positive numbers is called *lacunar* if there exists a number $q > 1$ such that $a_{n+1}/a_n \\ge q$ for all $n \\in \\mathbb{N}$. Also, the sequence is called *rare* if there exists a positive integer $k$ such that the interval $(x, 2x)$ contains at most $k$ terms of this sequence for any positive $x$.\n\na) Is it true that any lacunar sequence is rare?\n\nb) Is it true that any increasing rare sequence is lacunar?",
    "options": [],
    "answer": "a) Yes. b) No.",
    "solution": "a) Let the sequence $(a_n)$, $n \\in \\mathbb{N}$, be lacunar. Then there exists a number $q > 1$ such that\n$$\na_{n+1} \\ge q a_n \\quad \\forall n \\in \\mathbb{N}. \\qquad (1)\n$$\nIn particular, any lacunar sequence is increasing. From (1) it follows that any half-interval $(x, qx]$ contains at most one term of this sequence. Indeed, if we assume that $a_n$ and $a_{n+1}$ belong to this half-interval, then $\\frac{a_{n+1}}{a_n} < \\frac{qx}{x} = q$, contrary to (1).\n\nConsider any positive integer $k$ such that $q^k > 2$ (e.g., $k = [\\log_2 2] + 1$, where $[y]$ stands for the integer part of $y$). It is evident that\n$$\n(x, 2x) \\subset \\bigcup_{i=1}^{k} (q^{i-1}x, q^i x]. \\qquad (2)\n$$\nSince at most one term of $(a_n)$ belongs to the half-interval $(q^{i-1}x, q^i x]$, from inclusion (2) it follows that at most $k$ terms of $(a_n)$ belong to the interval $(x, 2x)$.\n\nb) We define the sequence $(a_n)$, $n \\in \\mathbb{N}$, as\n$$\na_{2n-1} = 2^n, \\quad a_{2n} = 2^n \\left(1 + \\frac{1}{n+1}\\right), \\quad n \\in \\mathbb{N}. \\qquad (3)\n$$\nThis sequence is increasing since from (3) it follows that $a_{2n-1} < a_{2n} < a_{2n+1}$ for all $n \\in \\mathbb{N}$. Moreover, this sequence is rare since at most one number of the form $2^m$, $m \\in \\mathbb{N}$, belongs to any interval $(x, 2x)$, and so at most three terms of this sequence belong to this interval.\n\nOn the other hand, we have\n$$\n\\frac{a_{2n}}{a_{2n-1}} = 1 + \\frac{1}{n+1}, \\quad n \\in \\mathbb{N},\n$$\nbut the numbers $1 + \\frac{1}{n+1}$ may be arbitrary close to $1$ (for $n$ large enough), hence, the sequence $(a_n)$, $n \\in \\mathbb{N}$, is not lacunar.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25533,
    "subject": "Mathematics (Multi-modal)",
    "question": "An array of $N$ ($N \\ge 3$) integer numbers is written on a blackboard. Per move one change the array using the following rule: let $s$ be the arithmetic mean of all numbers of the current array, then each number which is less than or equal to $s-1$ is increased by 1 and each number which is greater than or equal to $s+1$ is decreased by 1; all other numbers are not changed.\nProve that after finite number of the moves the numbers on the blackboard cannot be changed.\n(E. Zhibrik)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $S_i$ denote the arithmetic mean of all $n$ numbers written on the blackboard after $i$ moves, $M_i$ and $m_i$ be the greatest and the smallest of the written numbers.\nShow that if $M_i - m_i \\ge 2$ then\n$$\nM_{i+1} - m_{i+1} < M_i - m_i. \\quad (1)\n$$\nIndeed, since $M_i - m_i \\ge 2$, we see that the segment $[m_i, M_i]$ along with its ends, contains at least one positive integer number. Hence, at least one of the inequalities $|m(i) - S(i)| \\ge 1$ or $|M(i) - S(i)| \\ge 1$ holds ($S_i \\in (m_i, M_i)$). If exactly one of these inequalities holds, then $M_{i+1} - m_{i+1} = M_i - m_i - 1$; if both the inequalities hold, then $M_{i+1} - m_{i+1} = M_i - m_i - 2$. Hence, inequality (1) holds.\nSince the difference $M_i - m_i$ decreases when $i$ increases, after some moves this difference will be equal either to 0 or to 1. If $M_i - m_i = 0$, then all numbers on the blackboard are equal, hence they coincide with their arithmetic mean, therefore they will not be changed ever after the $i$-th move.\n\nIf $M_i - m_i = 1$, then there are only two distinct numbers on the blackboard: $m_i$ and $M_i - m_i + 1$. So their arithmetic mean $S(i)$ is a number between $m_i$ and $m_i + 1$, hence, by condition, the numbers on the blackboard will not be changed after the $i$-th move.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25534,
    "subject": "Mathematics (Multi-modal)",
    "question": "The vertices of the regular $2n+1$-gon are marked on a circle. Two players play the following game. They, in turn, delete exactly one of the vertices. The player wins if after his move all triangles with the vertices in the remained points are obtuse.\nWho of the player wins if both of them play to win?",
    "options": [],
    "answer": "Second player",
    "solution": "4. See the 38th All-Russian Mathematical Olympiad, Final Round, Problem 11.7.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25535,
    "subject": "Mathematics (Multi-modal)",
    "question": "Vertices of a regular $n$-gon are marked on a circumference. Integer numbers are placed at the vertices such that the difference of any two adjacent numbers equals $\\pm 1$ or $0$. At a moment, simultaneously, the half-sums of each pair of the adjacent numbers are placed at the midpoints of the arcs subtending the corresponding sides of the $n$-gon.\nFind all possible values of $n$ for which after some such moves there exist a pair of opposite points, among $2n$ points thus obtained, with the same numbers at them.\n(M. Karpuk)",
    "options": [],
    "answer": "all even n",
    "solution": "**Answer:** all even $n$.\nThe following example shows that $n = 2k + 1$ does not satisfy the problem condition. We number clockwise the vertices of the $n$-gon as $A_0, A_1, \\dots, A_k, A_{-k}, \\dots, A_{-1}$ and place the number $|i|$ at the vertex $A_i$. Since $n$ is odd, the pairs of antipodal points have the form «vertex-midpoint of an arc». But the integer numbers are placed at the vertices while the fractional numbers are placed at the midpoints of all arcs (except for the arc $A_k A_{-k}$ at the midpoint of which the number $k \\neq 0$ is placed).\nShow that for $n = 2k$ there exist antipodal points with the same numbers. We number clockwise the vertices of the $n$-gon as $A_1, A_2, \\dots, A_{2k}$, and let $a_1, a_2, \\dots, a_{2k}$ be the numbers at these vertices, respectively. For convenience we assume that $A_{2k+1} = A_1$ and $a_{2k+1} = a_1$. Consider the numbers\n$$\na_{k+1} - a_1, a_{k+2} - a_2, \\dots, a_{k+i} - a_i, \\dots, a_{k+(k+1)} - a_{k+1}. \\quad (1)\n$$\nThe absolute value of the difference between any two adjacent terms of sequence (1) is less than or equal to 2. Indeed,\n$$\n|(a_{k+i} - a_i) - (a_{k+i+1} - a_{i+1})| \\le |a_{k+i+1} - a_{k+i}| + |a_{i+1} - a_i| \\le 1+1=2, \\quad (2)\n$$\nsince, by condition, $|a_{k+i+1} - a_{k+i}| \\le 1$, $|a_{i+1} - a_i| \\le 1$ for all $i$ and $k$. If there is zero between the terms of (1), then there are antipodal points with the same numbers.\nSuppose that there is no zero in sequence (1). Since the first and the last terms of (1) have different signs (their sum is equal to 0), there are two adjacent terms $a_{k+i} - a_i$ and $a_{k+i+1} - a_{i+1}$ with different signs. Since all terms of (1) are integer, 0 does not belongs to (1), and (2) holds, we see that exactly one of these two adjacent numbers is equal to 1 while the other is equal to -1. Therefore, their sum $a_{k+i} - a_i + a_{k+i+1} - a_{i+1} = 1 + (-1) = 0$, i.e., $a_i + a_{i+1} = a_{k+i} + a_{k+i+1}$. Hence the numbers placed at the antipodal points (midpoints of the arcs $A_{k+i}A_{k+i+1}$ and $A_iA_{i+1}$) are equal. Thus all even $n$ satisfy the problem condition.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25536,
    "subject": "Mathematics (Multi-modal)",
    "question": "There is a graph with 30 vertices. If any 26 of its vertices with their outgoing edges are deleted, then the remained graph is a connected graph with 4 vertices.\nWhat is the smallest number of the edges in the initial graph with 30 vertices?\n(???)",
    "options": [],
    "answer": "405",
    "solution": "Answer: 405.\n\n(Solution by B. Serankou, A. Yuran.) Let $A$ be the vertex of the graph such that the number of outgoing edges from $A$ is less than 27. Hence there exist three vertices $B$, $C$, $D$ which are not connected with $A$. If we delete all vertices except for $A$, $B$, $C$, $D$ then the obtained graph is not connected. Therefore, each vertex of the graph has at least 27 outgoing edges, so the number of the edges in the graph is greater than or equal to $(27 \\cdot 30)/2 = 405$.\n\nConsider the graph with 30 vertices: $A_1, A_2, \\dots, A_{30}$. We connect each vertex $A_i$ with all others except for the vertices $A_{i-1}, A_{i+1}$, $i = 1, 2, \\dots, 30$ (here we assume that $A_0 = A_{30}$, $A_{31} = A_1$). This graph has exactly 405 edges and satisfies the problem condition. Indeed, suppose contrary to our claim that there exists a non-connected graph with the vertices $A_a, A_b, A_c, A_d$, $0 < a < b < c < d < 31$. It is evident that the vertices $A_a$ and $A_c$ are adjacent since $|c-a| > 1$ and $c < 30$. Similarly, $A_b$ and $A_d$ are adjacent since $|d-b| > 1$ and $b > 1$. If $A_a$ and $A_d$ are not adjacent, then there is only one possibility: $a=1$ and $d=30$ since $d > a+1$. If $A_a$ and $A_b$ are not adjacent, then $b=2$ since $a < b < 30$. If $A_b$ and $A_c$ are not adjacent, then $c=3$ since $b < c < 31$. Then $A_c = A_3$ and $A_d = A_{30}$ are adjacent; therefore the graph with the vertices $A_a, A_b, A_c, A_d$ is connected, a contradiction. Thus, 405 is the smallest number of the edges of the graph satisfying the problem condition.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 25537,
    "subject": "Mathematics (Multi-modal)",
    "question": "Three positive integers are written on a blackboard. Per move one replaces the set of these numbers by the new set in accordance with the following rule: each number of the set is replaced by the quotient of the sum of the squares of two other numbers and this number.\nWhat is the maximum value of the sum of the initial numbers written on the blackboard if after 5 moves the sum of the numbers on the blackboard is equal to 2016?\n(V. Karamzin)",
    "options": [],
    "answer": "63",
    "solution": "Let $a_0, b_0, c_0$ be the initial numbers on the blackboard, and $a_i, b_i, c_i$ be the numbers written on the blackboard after the $i$-th move; let $S(a_0, b_0, c_0)$ and $S(a_i, b_i, c_i)$ be the sums of these numbers, respectively. Then after the $i+1$-th move we obtain the following numbers\n$$\na_{i+1} = \\frac{b_i^2 + c_i^2}{a_i}, \\quad b_{i+1} = \\frac{c_i^2 + a_i^2}{b_i}, \\quad c_{i+1} = \\frac{a_i^2 + b_i^2}{c_i}.\n$$\nWe have the lower estimate for $S(a_{i+1}, b_{i+1}, c_{i+1})$:\n$$\n\\begin{align*}\nS(a_{i+1}, b_{i+1}, c_{i+1}) &= \\frac{b_i^2 + c_i^2}{a_i} + \\frac{c_i^2 + a_i^2}{b_i} + \\frac{a_i^2 + b_i^2}{c_i} \\ge \\frac{2b_i c_i}{a_i} + \\frac{2c_i a_i}{b_i} + \\frac{2a_i b_i}{c_i} = \\\\\n&= \\frac{b_i c_i}{a_i} + \\frac{b_i c_i}{a_i} + \\frac{c_i a_i}{b_i} + \\frac{c_i a_i}{b_i} + \\frac{a_i b_i}{c_i} + \\frac{a_i b_i}{c_i} = \\\\\n&= \\left( \\frac{b_i c_i}{a_i} + \\frac{a_i b_i}{c_i} \\right) + \\left( \\frac{b_i c_i}{a_i} + \\frac{c_i a_i}{b_i} \\right) + \\left( \\frac{a_i b_i}{c_i} + \\frac{c_i a_i}{b_i} \\right) = \\\\\n&= b_i \\left( \\frac{c_i}{a_i} + \\frac{a_i}{c_i} \\right) + c_i \\left( \\frac{b_i}{a_i} + \\frac{a_i}{b_i} \\right) + a_i \\left( \\frac{b_i}{c_i} + \\frac{c_i}{b_i} \\right) \\ge \\left[ \\frac{\\alpha}{\\beta} + \\frac{\\beta}{\\alpha} \\ge 2 \\right] \\ge \\\\\n&\\ge 2(a_i + b_i + c_i) = 2S(a_i, b_i, c_i).\n\\end{align*}\n$$\nTherefore, $S(a_i, b_i, c_i) \\le \\frac{1}{2} S(a_{i+1}, b_{i+1}, c_{i+1})$ for all $i = 0, 1, 2, 3, 4$. Hence,\n$$\n\\begin{align*}\nS(a_0, b_0, c_0) &\\le \\frac{1}{2} S(a_1, b_1, c_1) \\le \\frac{1}{2^2} S(a_2, b_2, c_2) \\le \\dots \\le \\frac{1}{2^5} S(a_5, b_5, c_5) = \\\\\n&= \\frac{2016}{32} = 63.\n\\end{align*}\n$$\nThus, the sum of the initial numbers on the blackboard is less than or equal to 63.\nThe following example shows that the sum of the initial numbers on the blackboard can be equal to 63. Let $a_0 = b_0 = c_0 = 21$, then\n$$\n\\begin{align*}\n(a_0, b_0, c_0) = (21, 21, 21) &\\longrightarrow (42, 42, 42) = (2 \\cdot 21, 2 \\cdot 21, 2 \\cdot 21) &\\longrightarrow \\\\\n&\\longrightarrow (2^2 \\cdot 21, 2^2 \\cdot 21, 2^2 \\cdot 21) &\\longrightarrow \\dots &\\longrightarrow (2^5 \\cdot 21, 2^5 \\cdot 21, 2^5 \\cdot 21) = (a_5, b_5, c_5),\n\\end{align*}\n$$\nand the sum\n$$\nS(a_5, b_5, c_5) = S(2^5 \\cdot 21, 2^5 \\cdot 21, 2^5 \\cdot 21) = 2^5 S(21, 21, 21) = 32 \\cdot 63 = 2016.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25538,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a prime number $p$ such that the number $2p$ is equal to the sum of the squares of some four consecutive positive integers. Prove that $p-7$ is divisible by 36.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $n-1$, $n$, $n+1$, and $n+2$ denote four consecutive positive integers with the sum of the squares equaled $2p$. Then\n$$\n2p = (n-1)^2 + n^2 + (n+1)^2 + (n+2)^2 = 4n^2 + 4n + 6,\n$$\nwhence $p = 2n^2 + 2n + 3 = 2n(n+1) + 3$. Now if one of the numbers $n$ and $n+1$ is divisible by $3$, then $p$ is divisible by $3$ hence $p$ is not a prime. So, $n$ and $n+1$ are not divisible by $3$, whence $n = 3k+1$, where $k \\in \\mathbb{N}$. Then $p = 2(3k+1)(3k+2)+3 = 18k(k+1)+7$, so $p-7 = 18k(k+1)$ is divisible by $18$. Since at least one of the numbers $k$ and $k+1$ is even, we obtain that $p-7$ is divisible by $36$, as required.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25539,
    "subject": "Mathematics (Multi-modal)",
    "question": "Do there exist three two-digit prime numbers such that all six digits in their decimal representation are distinct and if one chooses any two of the numbers and swaps their unit digits, then two numbers obtained will be prime, too?",
    "options": [],
    "answer": "No, such numbers do not exist.",
    "solution": "**Answer:** there are no such numbers.\nSuppose, contrary to our claim, that there are such numbers: $\\overline{ab}$, $\\overline{xy}$ and $\\overline{pq}$. The problem condition is equivalent to the following statement: any two-digit number with tens digit $a$, $x$ or $p$, and with unit digit $b$, $y$ or $q$ is a prime number. So $b, y, q \\in \\{1, 3, 7, 9\\}$, hence, at least one of the digits 3 and 9 belongs to $\\{b, y, q\\}$. Therefore, none of the digits $a, x, p$ is divisible by 3, so $a, x, p \\in M = \\{1, 2, 4, 5, 7, 8\\}$. Thus, we can consider two-digit numbers with tens digits from the set $M$ only. Among numbers 11, 21, 41, 51, 71, 81 there are exactly two prime numbers with distinct tens and unit digits: 41 and 71. Therefore, none of the digits $b, y, q$ is equal to 1. It follows that $\\{b, y, q\\} = \\{3, 7, 9\\}$, which is impossible. Indeed, exactly two numbers 17 and 47 are prime among two-digit numbers 17, 27, 47, 57, 77, 87, while there should be at least three prime numbers among these numbers.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25540,
    "subject": "Mathematics (Multi-modal)",
    "question": "A trapezoid $ABCD$ ($AB \\parallel CD$) is inscribed into the parabola $y = x^2$. The line containing the midline $MN$ of this trapezoid meets the parabola at points $K$ and $L$.\nProve that the lengths of the segments $KM$ and $NL$ are equal.\n(I. Gorodnin)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $A(a, a^2)$, $B(b, b^2)$, $C(c, c^2)$, $D(d, d^2)$, $K(k, k^2)$, $L(l, l^2)$, $M(m, m^2)$, $N(n, n^2)$. Without loss of generality we can assume that the points are arranged as shown in the figure (all other cases are similar).\n\n![](attached_image_1.png)\n\nIt is easy to write the equations of the straight lines $AB$, $CD$, $KL$, $AD$, $BC$\n$$\nAB: y = (a+b)x - ab, \\quad CD: y = (c+d)x - cd, \\quad KL: y = (k+l)x - kl,\n$$\n$$\nAD: y = (a+d)x - ad, \\quad BC: y = (b+c)x - bc.\n$$\nSince the lines $AB$, $CD$, $KL$ are parallel, their slopes are equal, so\n$$\na+b=c+d=k+l=\\lambda, \\quad (1)\n$$\nwhere $\\lambda$ is some real number.\n\nThe point $M$ is the intersection point of the lines $AD$ and $KL$, so its abscissae $m$ satisfies the equation\n$$\n(k+l)m - kl = (a+d)m - ad \\Leftrightarrow \\\\\n\\Leftrightarrow (k+l-a-d)m = kl-ad.\n$$\nTaking into account (1), we obtain\n$$\n(c-a)m = kl-ad. \\quad (2)\n$$\n\nSimilarly, the abscissae $n$ of the point $N$ ($N$ is the intersection point of the lines $BC$ and $KL$) satisfies the equation\n$$\n(k+l)n - kl = (b+c)n - bc \\Leftrightarrow (k+l-b-c)n = kl-bc\n$$\nTaking into account (1), we obtain\n$$\n(a-c)n = kl - bc. \\quad (3)\n$$\nSubtracting (3) from (2), we get\n$$\n(c-a)(m+n) = bc-ad = [b \\stackrel{(1)}{=} \\lambda-a, d \\stackrel{(1)}{=} \\lambda-c] = (\\lambda-a)c-a(\\lambda-c) = \\lambda(c-a).\n$$\nSince $c - a \\neq 0$, we have $m + n = \\lambda \\stackrel{(1)}{=} k + l$, i.e., $m + n = k + l$. This equality can be written as $\\frac{m+n}{2} = \\frac{k+l}{2}$. It follows that the midpoints of the segments $MN$ and $KL$ coincide. Since $MN$ and $KL$ lie on the same straight line, we obtain $KM = NL$ as required.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25541,
    "subject": "Mathematics (Multi-modal)",
    "question": "A line $l$ meets the right branch of the hyperbola $y = 1/x$ ($x > 0$) at points $A$ and $B$. Lines $l_1$ and $l_2$ are parallel to $l$ and meet the left branch of this hyperbola ($x < 0$) at points $E$, $F$ and $C$, $D$, respectively. The line $l_1$ intersects the segments $AD$ and $BC$ at points $G$ and $H$, respectively.\nProve that $GE = HF$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $A(a, 1/a)$, $B(b, 1/b)$, $C(c, 1/c)$, $D(d, 1/d)$, $E(e, 1/e)$, $F(f, 1/f)$, and positions of all points on the hyperbola are shown in the Figure. It is easy to write the equations of the straight lines $AB$, $DC$, and $EF$:\n$$\nAB(l) : y = -\\frac{1}{ab}x + \\frac{1}{a} + \\frac{1}{b}, \\quad EF(l_1) : y = -\\frac{1}{cf}x + \\frac{1}{c} + \\frac{1}{f},\n$$\n$$\nDC(l_2) : y = -\\frac{1}{cd}x + \\frac{1}{c} + \\frac{1}{d}.\n$$\nSince all these lines are parallel, their slopes coincide. Therefore,\n$$\nab = cd = ef. \\qquad (1)\n$$\n![](attached_image_1.png)\nLet $M$, $N$, $K$ be the midpoints of the segments $AB$, $EF$, $DC$, respectively. Then $M((a+b)/2; (1/a + 1/b)/2)$, $N((e+f)/2; (1/e + 1/f)/2)$, $K((c+d)/2; (1/c+1/d)/2)$. Let $O$ be the origin. It is evident that the equation of the straight lines $MO$, $ON$ and $OK$ respectively have the forms\n$$\ny = \\frac{1}{ab}x, \\quad y = \\frac{1}{ef}x, \\quad y = \\frac{1}{cd}x.\n$$\nFrom (1) it follows that all these lines coincide, so all points $O$, $M$, $N$, $K$ belong to the same straight line $MK$ with the equation $y = \\beta x$, where $\\beta = 1/(ab) = 1/(cd) = 1/(ef)$.\nIt is easy to see that in any trapezoid the segment connecting the midpoints of its bases passes through the midpoint of any segment which is parallel to the trapezoid bases. Consider the trapezoid $ABCD$ and the segment $GH$, $GH \\parallel BC \\parallel AD$, such that $E$ and $F$ lie on this segment. Since the line $MK$ passes through the midpoints of $AB$ and $DC$, the line $MK$ passes through the midpoint of the segment $GH$, too. Since $MK$ also passes through the midpoint of the segment $EF$, therefore the midpoints of $GH$ and $EF$ coincide since these segments lie on the same straight line. Thus, $GE=HF$ as required.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25542,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a polynomial $p(x) = a_{2n+1}x^{2n+1} + a_{2n}x^{2n} + \\dots + a_1x + a_0$ of odd degree with positive coefficients $a_0, a_1, \\dots, a_{2n}, a_{2n+1}$.\nProve that there exists a permutation (may be, trivial) of coefficients of $p(x)$ such that the polynomial obtained has exactly one real root.\n(A. Voidelevich)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "$c_0, c_1, \\dots, c_{2n+1}$ of the numbers $a_0, a_1, \\dots, a_{2n+1}$ such that $c_{2n+1} = b_{2n+1}$, $c_{2n-1} = b_{2n-1}$, $\\ldots$, $c_1 = b_{n+1}$ and $c_2 = b_n$, $c_{2n-2} = b_{n-1}$, $\\ldots$, $c_0 = b_0$. For the set of $c_0, c_1, \\dots, c_{2n+1}$ we have\n$$\nc_0 \\le c_1 \\ge c_2 \\le c_3 \\ge \\dots \\ge c_{2n} \\le c_{2n+1}.\n$$\nConsider the polynomial $q(x) = c_{2n+1}x^{2n+1} + c_{2n}x^{2n} + c_{2n-1}x^{2n-1} + \\dots + c_1x + c_0$. Since the degree of $q(x)$ is odd and its leading coefficient is positive, we have $q(-x) < 0$ and $q(x) > 0$ for $x$ large enough. Therefore, $q(x)$ has at least one real root. If we show that $q(x)$ strictly increases, then $q(x)$ has at most one real root as required. It suffices to prove that the derivative of this polynomial is positive for all $x$. We have\n$$\nq'(x) = (2n + 1)c_{2n+1}x^{2n} + 2nc_{2n}x^{2n-1} + \\dots + 2c_2x + c_1.\n$$\nSince\n$$\n\\begin{aligned} & kc_{2k+1}x^{2k} + 2kc_{2k}x^{2k-1} + kc_{2k-1}x^{2k-2} \\ge \\\\ & \\ge kc_{2k}(|x|^{2k} - 2|x|^{2k-1} + |x|^{2k-2}) = kc_{2k}|x|^{2k-2}(|x|-1)^2 \\ge 0 \\end{aligned}\n$$\nfor all $k \\in \\{1, 2, \\dots, n\\}$ and real $x$, we obtain\n$$\n\\begin{aligned} q'(x) &= (n+1)c_{2n+1}x^{2n} + (nc_{2n+1}x^{2n} + 2nc_{2n}x^{2n-1} + nc_{2n-1}x^{2n-2}) + \\\\ &\\quad ((n-1)c_{2n-1}x^{2n-2} + 2(n-1)c_{2n-2}x^{2n-3} + (n-1)c_{2n-3}x^{2n-4}) + \\dots + \\\\ &\\quad +(c_3x^2 + 2c_2x + c_1) \\ge \\\\ &\\ge (n+1)c_{2n+1}x^{2n} > 0 \\end{aligned}\n$$\nfor $x \\ne 0$. Since $q'(0) = c_1 > 0$, we obtain $q'(x) > 0$ for all real $x$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 25543,
    "subject": "Mathematics (Multi-modal)",
    "question": "We call two quadratic trinomials $P(x) = x^2 + a x + b$ and $Q(x) = x^2 + c x + d$ *friendly* if each of them has distinct real roots and $x_1 + x_3$, $x_2 + x_4$ are the roots of the quadratic trinomial $x^2 + (a + c)x + b + d$, where $x_1 < x_2$ are the roots of $P(x)$, $x_3 < x_4$ are the roots of $Q(x)$.\nLet $M$ be the set of pairwise *friendly* trinomials consisting of at least three trinomials.\nProve that $0$ is a root of every trinomial from the set $M$.\n(E. Barabanov)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the trinomials $P(x) = x^2 + a x + b$ and $Q(x) = x^2 + c x + d$ with the roots $x_1 < x_2$ and $x_3 < x_4$, respectively, be friendly. By condition, $x_1 + x_3$ and $x_2 + x_4$ are the roots of the trinomial $x^2 + (a + c)x + b + d$. By Vieta's theorem, we have\n$$\nb + d = (x_1 + x_3)(x_2 + x_4) = x_1 x_2 + x_1 x_4 + x_3 x_2 + x_3 x_4. \\quad (1)\n$$\nSince, by Vieta's theorem applied to the trinomials $P(x)$ and $Q(x)$ we have $x_1 x_2 = b$ and $x_3 x_4 = d$, from (1) it follows\n$$\nx_1 x_4 = -x_2 x_3. \\quad (2)\n$$\nIf none of $x_1, x_2, x_3, x_4$ is zero, then\n$$\n\\frac{x_1}{x_2} = -\\frac{x_3}{x_4}.\n$$\nIn particular, the ratios of the roots of any trinomials from $M$ are of opposite signs, so $M$ cannot have more than two trinomials, a contradiction.\nTherefore, among the trinomials from $M$ there exists a trinomial with zero root. By condition, the other root of this trinomial differs from zero, so from (2) it follows that this trinomial cannot be friendly to the trinomial with two nonzero roots. Thus, zero is a root for all trinomials from $M$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25544,
    "subject": "Mathematics (Multi-modal)",
    "question": "6. Given a polynomial $p(x) = a_{2n}x^{2n} + a_{2n-1}x^{2n-1} + \\dots + a_1x + a_0$ of even degree with positive coefficients $a_0, a_1, \\dots, a_{2n}$.\na) Prove that there exists a permutation of these coefficients such that the polynomial obtained has no real roots.\nb) Does the statement of a) remain true if some coefficients of $p(x)$ are non-positive?",
    "options": [],
    "answer": "no",
    "solution": "**6. Answer : b) no, it does not.**\na) Let $b_0 \\le b_1 \\le \\dots \\le b_{2n}$ denote the coefficients $a_0, a_1, \\dots, a_{2n}$ of $p(x)$, which are arranged in non-decreasing order. Consider the permutation $c_0, c_1, \\dots, c_{2n}$ of the numbers $a_0, a_1, \\dots, a_{2n}$ such that $c_{2n} = b_{2n}, c_{2n-2} = b_{2n-1}, \\dots, c_0 = b_n$ and $c_{2n-1} = b_{n-1}, c_{2n-3} = b_{n-2}, \\dots, c_1 = b_0$. For the set of $c_0, c_1, \\dots, c_{2n}$ we have\n$$\nc_0 \\ge c_1 \\le c_2 \\ge c_3 \\le \\dots \\le c_{2n-2} \\ge c_{2n-1} \\le c_{2n}.\n$$\nShow that the polynomial $q(x) = c_{2n}x^{2n} + c_{2n-1}x^{2n-1} + \\dots + c_1x + c_0$ has no real roots. The coefficients of $q(x)$ are positive, so\n$$\nq(x) > 0 \\quad (1)\n$$\nfor $x \\ge 0$. Show that inequality (1) holds for negative $x$, too. Consider two cases: **1)** $-1 \\le x < 0$ and **2)** $x < -1$.\nIn case **1**) we have $|x|^{2k+1} \\le x^{2k}$ for $k \\in \\{0, 1, \\dots, n-1\\}$, hence, $c_{2k+1}x^{2k+1} + c_{2k}x^{2k} \\ge (c_{2k} - c_{2k+1})|x|^{2k} \\ge 0$. Therefore,\n$$\nq(x) = c_{2n}x^{2n} + (c_{2n-1}x^{2n-1} + c_{2n-2}x^{2n-2}) + \\dots + (c_1x + c_0) \\ge c_{2n}x^{2n} > 0.\n$$\nIn case **2**) we have $|x|^{2k} > |x|^{2k-1}$, hence,\n$$\nc_{2k}x^{2k} + c_{2k-1}x^{2k-1} > (c_{2k} - c_{2k-1})x^{2k} \\ge 0\n$$\nfor $k \\in \\{1, 2, \\dots, n\\}$. Therefore,\n$$\nq(x) = (c_{2n}x^{2n} + c_{2n-1}x^{2n-1}) + \\dots + (c_2x^2 + c_1x) + c_0 > c_0 > 0.\n$$\nThus, $q(x) > 0$ for all real $x$, so $q(x)$ has no real roots.\n\nb) Consider the polynomial $p(x) = x^{2n} + x^{2n-1} + \\dots + x - 2n$. The sum of the coefficients of $p(x)$ is equal to 0, i.e., $p(1) = 0$. Therefore, for any permutation $c_0, c_1, \\dots, c_{2n}$ of the numbers $-2n, 1, \\dots, 1$ we have $q(1) = 0$, where $q(x) = c_{2n}x^{2n} + c_{2n-1}x^{2n-1} + \\dots + c_1x + c_0$. It follows that 1 is a root of the polynomial $q(x)$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 25545,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let polynomials $P(x)$ and $Q(x)$ have the same degree. Let $P_Q(x)$ denote the polynomial such that its coefficients of even degree variables coincide with the corresponding coefficients of $P(x)$ and its coefficients of odd degree variables coincide with the corresponding coefficients of $Q(x)$. (For example, if $P(x) = x^3 + 2x^2 + 4x + 1$ and $Q(x) = 3x^3 + x^2 + 2$, then $P_Q(x) = 3x^3 + 2x^2 + 1$, and $Q_P(x) = x^3 + x^2 + 4x + 2$.)\n\na) Prove that there exist $P(x)$ and $Q(x)$ such that they have no real roots but both $P_Q(x)$ and $Q_P(x)$ have at least one real root.\n\nb) Find the smallest degree of $P(x)$ and $Q(x)$ satisfying a).",
    "options": [],
    "answer": "4",
    "solution": "a)\nShow, for example, that the polynomials\n$$\nP(x) = 4x^4 + 4x^3 + 1 \\quad \\text{and} \\quad Q(x) = x^4 + 4x + 4\n$$\nsatisfy the condition.\nBy Cauchy's inequality, for any real $x$ the following inequalities hold:\n$$\nx^4 + x^4 + x^4 + 1 \\ge 4\\sqrt[4]{x^4 \\cdot x^4 \\cdot x^4 \\cdot 1} = 4|x|^3, \\quad (1)\n$$\n$$\nx^4 + 1 + 1 + 1 \\ge 4\\sqrt[4]{x^4 \\cdot 1 \\cdot 1 \\cdot 1} = 4|x|. \\quad (2)\n$$\nFrom (1) it follows that $P(x) = 4x^4 + 4x^3 + 1 \\ge x^4 + 4(|x|^3 + x^3) \\ge 0$, the last inequality turns to the equality only for $x = 0$, but $P(0) = 1$, so the polynomial $P(x)$ has no real roots.\nSimilarly, from (2) it follows that $Q(x) = x^4 + 4x + 4 \\ge 4(|x|+x) + 1 > 0$, whence the polynomial $Q(x)$ has no real roots.\nBy definition,\n$$\nP_Q(x) = 4x^4 + 4x + 1 \\quad \\text{и} \\quad Q_P(x) = x^4 + 4x^3 + 4\n$$\nIt is easy to verify that $P_Q(-0.5) = -0.75 < 0$ and $P_Q(0) = 1 > 0$. Also, $Q_P(-2) = -12 < 0$ and $Q_P(0) = 4 > 0$. Therefore each of the polynomials $P_Q(x)$ and $Q_P(x)$ has at least one real root.\n\nb)\nIt is evident that the polynomials $P(x)$ and $Q(x)$ satisfying the condition of a) must have even degree (since any odd degree polynomial has at least one real root).\nShow that the degree of these polynomials is greater than 2. Suppose, contrary to our claim, that there are the trinomials $P(x) = a_1x^2 + b_1x + c_1$ and $Q(x) = a_2x^2 + b_2x + c_2$ having no real roots. Then at least one of the trinomials $P_Q(x) = a_1x^2 + b_1x + c_1$ and $Q_P(x) = a_2x^2 + b_2x + c_2$ has no real roots. Indeed, without loss of generality we can assume that $|b_2| \\ge |b_1|$. Since the discriminant of the trinomial $Q(x)$ is negative and $|b_2| \\ge |b_1|$, we see that $0 > b_2^2 - 4a_2c_2 \\ge b_1^2 - 4a_2c_2$. However, the last expression is a discriminant of the trinomial $Q_P(x)$, so this trinomial has no real roots.\nThe example of the polynomials from item a) shows the minimal degree of the polynomials is equal to 4.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 25546,
    "subject": "Mathematics (Multi-modal)",
    "question": "Points $A_1, B_1, C_1$ are the midpoints of the sides $BC$, $AC$, $AB$ of the triangle $ABC$ ($AB \\ge AC \\ge BC$), respectively. Points $C_2$ and $C_3$ are marked on $AB$ so that the perimeter of the triangle $BA_1C_2$ is equal to the perimeter of the quadrilateral $ACA_1C_2$ and the perimeter of the triangle $AB_1C_3$ is equal to the perimeter of quadrilateral $BCB_1C_3$. Point $B_2$ is marked on $AC$ so that the perimeter of the triangle $AB_2C_1$ is equal to the perimeter of the quadrilateral $BCB_2C_1$.\nProve that the segments $A_1C_2$, $B_1C_3$, $C_1B_2$ are concurrent.\n(V. Karamzin)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By condition, the perimeter $P(AB_2C_1)$ of the triangle $AB_2C_1$ is equal to the perimeter $P(BCB_2C_1)$ of the quadrilateral $BCB_2C_1$, i.e., (see the Fig.)\n\n![](attached_image_1.png)\n\n$$P(AB_2C_1) = AB_2 + B_2C_1 + C_1A = BC + CB_2 + B_2C_1 + C_1B = P(BCB_2C_1),$$\nso, taking into account $C_1A = C_1B$, we have\n$$\nAB_2 = BC + CB_2 = BC + (AC - AB_2).\n$$\nTherefore, $AB_2 = (AC + BC)/2$. Since $B_1C_1$ is the midline of the triangle $ABC$, we have $B_1C_1 = BC/2$ whence\n$$\nB_1B_2 = AB_2 - AB_1 = (AC + BC)/2 - AC/2 = BC/2 = B_1C_1.\n$$\nTherefore, the triangle $C_1B_1B_2$ is isosceles and $\\angle B_1B_2C_1 = \\angle B_1C_1B_2$. Further, $A_1C_1$ is the midline of the triangle $ABC$, then $A_1C_1 \\parallel AC$ and $\\angle B_1B_2C_1 = \\angle B_2C_1A_1$. Thus $\\angle B_1C_1B_2 = \\angle B_2C_1A_1$, and so $C_1B_2$ is the bisector of the angle $C_1$ of the triangle $A_1B_1C_1$. In the same way, we obtain that $B_1C_3$ is the bisector of the angle $B_1$ of the triangle $A_1B_1C_1$, and $A_1C_2$ is the bisector of the angle $A_1$ of the triangle $A_1B_1C_1$.\nTherefore, the segments $A_1C_2$, $B_1C_3$, $C_1B_2$ are the bisectors of the angles of the triangle $A_1B_1C_1$ and, hence, they are concurrent.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25547,
    "subject": "Mathematics (Multi-modal)",
    "question": "The excircle $\\omega_A$ of a triangle $ABC$ touches the side $BC$ at $P$. Let $I_1$ be the center of the excircle of the triangle $ABP$ touching the side $BP$, and $I_2$ be the center of the excircle of the triangle $APC$ touching the side $PC$.\nProve that the circumcircle of the triangle $I_1I_2P$ touches the circle $\\omega_A$.\n(A. Voidelevich)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We show that the circumcircle of the triangle $I_1I_2P$ touches the line $BC$ at the point $P$, then, in particular, it follows that this circle touches the circle $\\omega_A$. We use the following well-known lemma.\n\n**Lemma.** Let excircle touch the side $BC$ of the triangle $ABC$ at $A_1$ and touch the prolongations of the sides $AB$ and $AC$ at $C_1$ and $B_1$, respectively. Then\n$$\nBA_1 = BC_1 = \\frac{AC + CB - BA}{2} \\quad \\text{and} \\quad CA_1 = CB_1 = \\frac{CB + BA - AC}{2}.\n$$\nLet $\\omega_1$ and $\\omega_2$ denote excircles of the triangles $ABP$ and $APC$, respectively. Let $I_1$ and $I_2$ be the centers of $\\omega_1$ and $\\omega_2$, respectively. Let $F_1$\n\nand $F_2$ denote the tangent points of $\\omega_1, \\omega_2$ and the line $AP$, respectively. We show that $F_1$ and $F_2$ coincide. Indeed, from the lemma it follows that\n$$\n\\begin{aligned} 2PF_1 = PB + BA - AP &= \\frac{AC + CB - BA}{2} + BA - AP = \\\\ &= \\frac{AC + CB + BA}{2} - AP, \\end{aligned}\n$$\n$$\n\\begin{aligned} 2PF_2 = PC + CA - AP &= \\frac{CB + BA - AC}{2} + AC - AP = \\\\ &= \\frac{AC + CB + BA}{2} - AP. \\end{aligned}\n$$\nHence, $PF_1 = PF_2$, i.e., $F_1 \\equiv F_2$. So we may use $F$ to denote $F_1, F_2$. We have $I_1F \\perp AP$ and $I_2F \\perp AP$, so $I_1, F$, and $I_2$ lie on the same line.\n\nLet $\\angle BPF = 2x$, then $\\angle CPF = 180^\\circ - 2x$. Since the circle $\\omega_1$ touches the lines $PA$ and $PB$, it follows that the line $PI_1$ is the bisector of the angle $\\angle BPF$. Similarly, $PI_2$ is the bisector of the angle $\\angle CPF$. Therefore, $\\angle I_1I_2P = 90^\\circ$ and we have\n$$\n\\angle I_1I_2P = 90^\\circ - \\angle FPI_2 = 90^\\circ - (90^\\circ - x) = x = \\angle BPI_1.\n$$\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25548,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $E$ be the intersection point of the segments $AB$ and $CD$. Points $X$ and $Y$ are marked in the plane such that the quadrilaterals $AECX$ and $BEDY$ are parallelograms.\nProve that the lines $AD$, $BC$, and $XY$ are either pairwise parallel or concurrent.\n(M. Karpuk, A. Voidelevich)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the segment $XY$ meet the segments $AB$ and $CD$ at $F$ and $G$, respectively. Since $AX \\parallel DC$ and $DC \\parallel YB$, the triangles $AXF$ and $BYF$ are similar. Therefore we have $AF/FB = AX/YB$. In the same way we conclude that $DG/GC = DY/XC$.\n\n![](attached_image_1.png)\n\nSuppose that the lines $AD$ and $BC$ are parallel. Then the triangles $BEC$ and $AED$ are similar, so that $EB/AE = EC/DE$. Hence,\n$$\n\\frac{AF}{FB} = \\frac{AX}{YB} = \\frac{EC}{DE} = \\frac{EB}{AE} = \\frac{DY}{XC} = \\frac{DG}{GC}.\n$$\nBy Thales' theorem, $FG \\parallel AD$. Thus, we prove that the lines $AD$, $BC$, and $XY$ are pairwise parallel.\n\nFurther, let $AD$ and $CB$ intersect; let $Z$ be their intersection point (without loss of generality, let $Z$ be the intersection point of the rays $AD$ and $CB$). Let the line $ZY$ meet the segment $AB$ at $F'$. Show that $AF'/F'B = AF/FB$, which yields that the points $F'$ and $F$ coincide, i.e., the lines $AD$, $BC$, $XY$ are concurrent, as required. Let $H$ be the intersection point of the lines $DY$ and $CB$, and $I$ be the intersection point of the lines $XC$ and $BY$. Then since $DY \\parallel AB$, we have $AF'/F'B = DY/YH$. Since $DY \\parallel XC$ and $DG \\parallel YB$, the quadrilateral $BECI$ is a parallelogram, and the triangles $BHY$ and $BCI$ are similar. Therefore\n$$\n\\frac{AF'}{F'B} = \\frac{DY}{YH} = \\frac{CI}{YH} = [\\Delta BHY \\sim \\Delta BCI] = \\frac{BI}{YB} = \\frac{AX}{YB} = \\frac{AF}{FB},\n$$\nas required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25549,
    "subject": "Mathematics (Multi-modal)",
    "question": "A point $P$ (distinct from $B$ and $C$) are marked on the side $BC$ of a triangle $ABC$. Let $I_1$ and $I_2$ be the centers of the inscribed circles of the triangles $ABP$ and $APC$, respectively. The circumcircle of the triangle $I_1PI_2$ intersects the segment $BC$ at $P$ and $Q$.\nProve that $AB + QC = AC + QB$.\n(A. Voidelevich)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Without loss of generality we can assume that $Q$ belongs to the segment $PC$. First, note that if the incircle of the triangle $ABC$ touches the sides $AB$, $BC$, $CA$ at the points $C_1$, $A_1$, $B_1$, respectively, then\n$$\n\\begin{aligned}\nAB_1 &= AC_1 = \\frac{AB + AC - BC}{2}, \\\\\nBA_1 &= BC_1 = \\frac{BA + BC - AC}{2}, \\\\\nCA_1 &= CB_1 = \\frac{CA + CB - AB}{2}.\n\\end{aligned}\n$$\nLet $\\omega_1$ and $\\omega_2$ denote the incircles of the triangles $ABP$ and $APC$, respectively. Let internal tangent $l$ (different from $AP$) to the circles $\\omega_1$ and $\\omega_2$ meet the side $BC$ at $Q'$. We show that $I_1$, $P$, $Q'$, and $I_2$ are concyclic, which gives $Q' \\equiv Q$. Let $R$ be the intersection point of the lines $l$ and $AP$. Since $I_1$ and $I_2$ lie on the internal and external bisectors of the angle $\\angle PQ'R$, respectively, we have $\\angle I_1Q'I_2 = 90^\\circ$.\n\n![](attached_image_1.png)\n\nSimilarly, we show that $\\angle I_1PI_2 = 90^\\circ$. Therefore, the quadrilateral $I_1PQ'I_2$ is cyclic, so $Q' \\equiv Q$. Using the formulae mentioned above, we have $DP = (PB + PA - AB)/2$ and $PE = (PC + PA - AC)/2$, from lemma of Problem A.7 it follows that $DP = QE = PE - PQ$. Thus,\n$$\nPQ = PE - DP = (AB + PC - AC - PB)/2.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25550,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a $2n \\times 2n$ table all the cells being white. Per move it is allowed to change the color of any three consecutive cells in any row or in any column (white cell is replaced by black one and vice versa).\nFind all possible values of $n \\ge 2$ for which one can obtain the chess coloring of the given table.\n(E. Barabanov)",
    "options": [],
    "answer": "n is a multiple of 3",
    "solution": "**Answer:** $n \\equiv 0 \\pmod{3}$.\nIt is easy to see that we have $2n^2$ black and $2n^2$ white cells in any chess coloring of the given $2n \\times 2n$ table, in particular, we have even number of white cells and even number of black cells.\nSince each of $2n^2$ black cells must be recolored odd number of times and each of $2n^2$ white cells must be recolored even number of times (may be, zero), the total number of the recolorings is even. Per move odd number (namely, three) cells are recolored, so the number of moves is even.\n\nNow we show that if $n \\not\\equiv 0 \\pmod{3}$, then the number of moves to obtain the final coloring must be odd, so one cannot obtain the final coloring for such $n$.\nLet $A$, $B$, $C$, $D$ denote the vertices of the $(2n+1) \\times (2n+1)$ square (see Fig. 1).\n![](attached_image_1.png)\nFig. 1\n![](attached_image_2.png)\nFig. 2\n\nConsider any line parallel to the line $BD$ (see Fig. 1) and passing through the centers of some cells. We call the set of all cells with the centers on such line a *diagonal* (see Fig. 1, where 4 from 11 possible diagonals of the $6 \\times 6$ square are indicated). It is evident that there are $4n-1$ distinct diagonals starting from the cells of the half-perimeter $ABC$, and each diagonal consists of either black cells (we call such black) or white cells (we call such diagonal white).\n\nWithout loss of generality we assume that the cell with the vertex $A$ is black. We number the cells of the half-perimeter $ABC$ as shown in Fig. 2. We consider the diagonals starting from the cells with the numbers which are divisible by 3 (let $D_3$ be the set of these diagonals), and write the symbol \"*\" in the cells of these diagonals (see Fig. 3 and Fig. 4 for $n=2$ and $n=3$, respectively).\n\n![](attached_image_3.png)\nFig. 3\nFig. 4\n\nFind the parity of the number of the black cells with \"*\". It is evident that the diagonal from $D_3$ starting from the cell with even (odd) number is white (black). It is also evident that the diagonal starting from the cell with odd number consists of odd number of the cells. Hence, the parity of the number of the black cells with \"*\" is equal to the parity of the number of all odd numbers (divisible by 3) between 1 and $4n-1$.\n\nIf $n \\equiv 0 \\pmod{3}$, i.e., $n = 3k$, $k \\in \\mathbb{N}$, then $4n-1 = 12k-1$ and between 1 and $4n-1$ odd numbers which is divisible by 3 have the forms $3 \\cdot 1, 3 \\cdot 3, \\dots, 3 \\cdot (4k-3), 3 \\cdot (4k-1)$. There are $2k$ such numbers.\nIf $n \\equiv 1 \\pmod{3}$, i.e., $n = 3k+1$, $k \\in \\mathbb{N} \\cup \\{0\\}$, then $4n-1 = 12k+3$ and between 1 and $4n-1$ odd numbers which is divisible by 3 have the forms $3 \\cdot 1, 3 \\cdot 3, \\dots, 3 \\cdot (4k-1), 3 \\cdot (4k+1)$. There are $2k+1$ such numbers.\nIf $n \\equiv 2 \\pmod{3}$, i.e., $n = 3k+2$, $k \\in \\mathbb{N} \\cup \\{0\\}$, then $4n-1 = 12k+7$ and between 1 and $4n-1$ odd numbers which is divisible by 3 have the forms $3 \\cdot 1, 3 \\cdot 3, \\dots, 3 \\cdot (4k-1), 3 \\cdot (4k+1)$. There are $2k+1$ such numbers.\n\nTherefore, the symbol \"*\" is written in the odd number of black cells if $n \\not\\equiv 0 \\pmod{3}$, and the symbol \"*\" is written in the even number of black cells if $n \\equiv 0 \\pmod{3}$. Suppose one can obtain the finishing coloring after some moves. It is easy to see that per move exactly one cell with \"*\" is recolored. Hence we can divide all moves into two sets: the set $W$ consists of the moves when some white cell (with \"*\") is recolored and the set $B$ consists of the moves when some black cell (with \"*\") is recolored. There are even number of the moves from $W$ because any recolored white cell is recolored even number of times; there are odd number of moves from $B$ if $n \\not\\equiv 0 \\pmod{3}$, since in this case we have odd number of black cells with \"*\" and each of them must be recolored odd number of times. Therefore, one must make odd number of the moves, contrary to our previous statement about the parity of the moves for $n \\not\\equiv 0 \\pmod{3}$. Thus, for $n \\not\\equiv 0 \\pmod{3}$ one cannot obtain the required chess coloring of the table.\n\nIf $n \\equiv 0 \\pmod{3}$, i.e., $2n$ is divisible by 3, one can obtain the required coloring. It suffices to divide the given $2n \\times 2n$ table into the $3 \\times 3$ squares and in each such square make the moves depending on the arrangement of the black cells in this square so as indicated in Fig. 5 or Fig. 6, where the moves are shown as the segments connecting the centers of the recolored cells.\n![](attached_image_4.png)\nFig. 5\n![](attached_image_5.png)\nFig. 6",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 25551,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a $(2n + 1) \\times (2n + 1)$ table, all its unit cells being white. Per move it is allowed to change the color of any three consecutive cells in any row or in any column (white cell is replaced by black one and vice versa).\n\na) Prove that for any positive integer $n$ one can obtain the chess coloring of the given table with white corner cells (the example of such coloring is shown in the figure for $n = 3$).\n\nb) Find the minimum number of the moves one needs to obtain the chess coloring of the given table with white corner cells if $n = 3$. (E. Barabanov)\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "14",
    "solution": "a) We say that a cell is black if this cell becomes black in the final coloring, and white if this cell becomes white in the final coloring.\n\nWe divide the $(2n + 1) \\times (2n + 1)$ table into concentric zones with unit width, and the central cell (see Fig. 1). The central cell and all corner cells are white. The cells of any such zone can be recolored in chess order so that the corner cells remain white and only cells of such zone take part in the recoloring. Indeed, it is sufficient to make all possible moves (each move is made exactly one time) with the ends at the white cells (see Fig. 2 and Fig. 3, where it is shown for the zones of the length 16 and 24; each move is shown as the segment connecting the centers of the recolored white cells).\n\n![](attached_image_2.png)\nFig. 1\n![](attached_image_3.png)\nFig. 2\n![](attached_image_4.png)\nFig. 3\n\nIn this case each white cell is recolored two times and remains white, and each black cell is recolored exactly one time and turns black. The central cell is not recolored and remains white. Therefore, using this type of the moves one can obtain the chess coloring of the given table with white corner cells.\n\nb) If we use the algorithm from item a) to obtain the $7 \\times 7$ table with chess coloring, then we need $24$ moves. Show that it is sufficient $14$ moves to recolor the square.\n\nIt is easy to see that there are $24$ black and $25$ white cells in the $7 \\times 7$ table. Since each of $24$ black cells must be recolored odd number of times and each of $15$ white cells must be recolored even number of times (may be, zero), the total number of the recolorings is even. Per move odd number (namely, three) cells are recolored, so the number of moves is even.\n\nSince each of $24$ black cells must be recolored at least one time and per move at most two black cells are recolored, the total number of moves is greater than or equal to $24 : 2 = 12$.\n\nShow that $12$ moves are not sufficient. Suppose, contrary to our claim, that we obtain the finishing coloring using $12$ moves. It follows that exactly two black cells and exactly one white cell are recolored at the same move. But each white cell is recolored even number of times, so at most $12 : 2 = 6$ white cells can be recolored using these $12$ moves.\n\nNow, we mark seven black cells as it is shown in Fig. 4 and consider their adjacent white cells, that is, those sharing a common side with the marked black cells.\n\n![](attached_image_5.png)\nFig. 4\n![](attached_image_6.png)\nFig. 5\n\nWe see that none of these white cells is adjacent to two of the marked black cells (in Fig. 4 for each marked black cell we number its adjacent white cells with the same number). When any of these seven black cells is recolored at least one adjacent white cell must be recolored, hence (the marked black cells have no common adjacent white cells) at least seven different white cells must be recolored, a contradiction. Thus, $12$ moves are not sufficient to obtain the required chess coloring square. But the total number of the moves is even, so we need at least $14$ moves.\n\nThe example (see Fig. 5, where each move is shown as the segment connecting the centers of the recolored cells) shows how to obtain the required coloring with $14$ moves.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 25552,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a $(2n + 1) \\times (2n + 1)$ table, all its cells being white.\n![](attached_image_1.png)\nPer move it is allowed to change the color of any three consecutive cells in any row or in any column (white cell is replaced by black one and vice versa).\nFind all possible values of $n$ such that one can obtain the chess coloring of the given table with black corner cells (the example of such coloring is shown in the figure for $n = 2$).\n(E. Barabanov)",
    "options": [],
    "answer": "n ≡ 1 (mod 3)",
    "solution": "**Answer:** $n \\equiv 1 \\pmod{3}$.\n\nWe say that a cell is black if this cell becomes black in the final coloring, and white if this cell becomes white in the final coloring. It is easy to see that we have $2n^2 + 2n + 1$ black and $2n^2 + 2n$ white cells in the chess coloring of the given table with black corner cells. In particular, we have even number of white cells and odd number of black cells.\n\nSince each of $2n^2 + 2n + 1$ black cells must be recolored odd number of times and each of $2n^2 + 2n$ white cells must be recolored even number of times (may be, zero), the total number of the recolorings is odd. Per move odd number (namely, three) cells are recolored, so the number of moves is odd.\n\nNow we show that if $n \\not\\equiv 1 \\pmod{3}$, then the number of moves to obtain the final coloring must be even, so one cannot obtain the final coloring for such $n$.\n\n![](attached_image_2.png)\nFig. 1\n![](attached_image_3.png)\nFig. 2\n\nLet $A$, $B$, $C$, $D$ denote the vertices of the $(2n + 1) \\times (2n + 1)$ square (see Fig. 1). Consider any line parallel to the line $BD$ (see Fig. 1) and passing through the centers of some cells. We call the set of all cells with the centers on such line a *diagonal* (see Fig. 1, where 4 from 13 possible diagonals of the $7 \\times 7$ square are indicated). It is evident that there are $4n + 1$ distinct diagonals starting from the cells of the half-perimeter $ABC$, and each diagonal consists of either black cells (we call such diagonal black) or white cells (we call such diagonal white).\n\nWe number the cells of the half-perimeter $ABC$ as shown in Fig. 2. Consider the diagonals starting from the cells with the numbers which are divisible by 3 (let $D_3$ be the set of these diagonals), and write the symbol \"*\" in the cells of these diagonals (see Fig. 3 and Fig. 4 for $n = 2$ and $n = 3$, respectively). Find the parity of the number of the black cells with \"*\". It is evident that the diagonal from $D_3$ starting from the cell with even (odd) number is white (black). It is also evident that the diagonal starting from the cell with odd number consists of odd number of the cells. Hence, the parity of the number of the black cells with \"*\" is equal to the parity of the number of all odd numbers (divisible by 3) between 1 and $4n + 1$.\n\n![](attached_image_4.png)\nFig. 3\nFig. 4\n\nIf $n \\equiv 0 \\pmod{3}$, i.e., $n = 3k$, $k \\in \\mathbb{N}$, then $4n + 1 = 12k + 1$ and between 1 and $4n + 1$ odd numbers which are divisible by 3 have the forms $3 \\cdot 1, 3 \\cdot 3, \\dots, 3 \\cdot (4k - 3), 3 \\cdot (4k - 1)$. There are $2k$ such numbers.\n\nIf $n \\equiv 1 \\pmod{3}$, i.e., $n = 3k + 1$, $k \\in \\mathbb{N} \\cup \\{0\\}$, then $4n + 1 = 12k + 5$ and between 1 and $4n + 1$ odd numbers which are divisible by 3 have the forms $3 \\cdot 1, 3 \\cdot 3, \\dots, 3 \\cdot (4k - 1), 3 \\cdot (4k + 1)$. There are $2k + 1$ such numbers.\n\nIf $n \\equiv 2 \\pmod{3}$, i.e., $n = 3k + 2$, $k \\in \\mathbb{N} \\cup \\{0\\}$, then $4n + 1 = 12k + 9$ and between 1 and $4n + 1$ odd numbers which are divisible by 3 have the forms $3 \\cdot 1, 3 \\cdot 3, \\dots, 3 \\cdot (4k + 1), 3 \\cdot (4k + 3)$. There are $2k + 2$ such numbers.\n\nTherefore, the symbol \"*\" is written in the even number of black cells if $n \\not\\equiv 1 \\pmod{3}$, and the symbol \"*\" is written in the odd number of black cells if $n \\equiv 1 \\pmod{3}$. Suppose one can obtain the final coloring after some moves. It is easy to see that per move exactly one cell with \"*\" is recolored. Hence we can divide all moves into two sets: the set $W$ consists of the moves when some white cell (with \"*\") is recolored and the set $B$ consists of the moves when some black cell (with \"*\") is recolored. There are even number of the moves from $W$ because any recolored white cell is recolored even number of times; there are also even number of moves from $B$ if $n \\neq 1 \\pmod{3}$, since in this case we have even number of black cells with \"*\" and each of them must be recolored odd number of times. Therefore, one must make even number of moves, contrary to our previous statement about the parity of the moves for $n \\neq 1 \\pmod{3}$. Thus, for $n \\neq 1 \\pmod{3}$ one cannot obtain the required chess coloring of the table.\n\nIf $n \\equiv 1 \\pmod{3}$, i.e., $2n+1$ is divisible by 3, one can obtain the required coloring. It suffices to divide the $(2n+1) \\times (2n+1)$ table into the $3 \\times 3$ squares and in each such square make the moves depending on the arrangement of the black cells in this square as it is shown in Fig. 5 or Fig. 6, where the moves are shown as the segments connecting the centers of the recolored cells.\n\n![](attached_image_5.png)\nFig. 5\n![](attached_image_6.png)\nFig. 6",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 25553,
    "subject": "Mathematics (Multi-modal)",
    "question": "A positive integer is called *nice* if it is equal to the sum of the fourth powers of certain five distinct its divisors. (A divisor may be equal to $1$ or to the number itself.)\n\na) Prove that any nice number is divisible by $5$.\n\nb) Are there infinitely many nice numbers?",
    "options": [],
    "answer": "a) Every nice number is divisible by five. b) Yes, there are infinitely many nice numbers.",
    "solution": "Answer: b) there are an infinite number of nice numbers.\n\na) Let $N$ be a nice number, i.e. $N = d_1^4 + d_2^4 + d_3^4 + d_4^4 + d_5^4$, where $d_i$, $i = 1, 2, 3, 4, 5$, are the distinct divisors of $N$. If some divisor of $N$ is divisible by $5$, then $N$ is divisible by $5$.\nSo we suppose that $d_1, d_2, d_3, d_4, d_5$ are not divisible by $5$. Then their fourth powers are congruent to $1$ modulo $5$. Indeed, if $M = 5k \\pm 1$ ($k \\in \\mathbb{N}$), then $M^2 = 25k^2 \\pm 10k + 1 = 5m + 1$ ($m \\in \\mathbb{N}$) and $M^4 = (5m+1)^2 = 25m^2 + 10m + 1$, i.e. $M^4$ is congruent to $1$ modulo $5$; if $M = 5k \\pm 2$ ($k \\in \\mathbb{N}$), then $M^2 = 25k^2 \\pm 20k + 4 = 5m + 4$ ($m \\in \\mathbb{N}$) and $M^4 = (5m+4)^2 = 25m^2 + 40m + 16$, i.e. $M^4$ is congruent to $1$ modulo $5$. Therefore, the fourth power of any divisor $d_i$ is congruent to $1$ modulo $5$, i.e. $d_i^4 = 5k_i + 1$, and it follows that $N$ is divisible by $5$.\n\nb) Let there exist a nice number $N$, i.e. $N = d_1^4 + d_2^4 + d_3^4 + d_4^4 + d_5^4$, where $d_i$ are some positive integers and $p > 1$. If $d_1, d_2, d_3, d_4, d_5$ are distinct divisors of $N$, consider the number $N(p) = Np^4$, where $p$ is any positive integer greater than $1$. It is obvious that $d_1p, d_2p, d_3p, d_4p, d_5p$ are distinct divisors of $N(p)$, and\n$$\n\\begin{aligned}\nN(p) &= Np^4 = [N = d_1^4 + d_2^4 + d_3^4 + d_4^4 + d_5^4] = (d_1^4 + d_2^4 + d_3^4 + d_4^4 + d_5^4)p^4 = \\\\\n&= (d_1p)^4 + (d_2p)^4 + (d_3p)^4 + (d_4p)^4 + (d_5p)^4.\n\\end{aligned}\n$$\nSo there are infinitely many nice numbers $N(p)$ if there exists at least one nice number. It remains to note that the number $N = 1^4 + 2^4 + 3^4 + 6^4 + 34^4$ is nice. Indeed, it is evident that $N$ is divisible by $2$. Since the numbers $1^4, 2^4$, and $34^4$ are congruent $1$ modulo $3$, the number $N$ is divisible by $3$. Since $N$ is even and divisible by $3$, we see that $N$ is divisible by $6$. Besides,\n$$\nN = 1^4 + 2^4 + 3^4(1^4 + 2^4) + 34^4 = (1^4 + 2^4)(1^4 + 3^4) + 34^4 = 17 \\cdot (1^4 + 3^4) + 34^4,\n$$\nwhence we see that $N$ is divisible by $34$. Thus, $N$ is nice.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25554,
    "subject": "Mathematics (Multi-modal)",
    "question": "A positive integer is called *nice* if it is equal to the sum of the squares of its three distinct divisors. (A divisor may be equal to $1$ or to the number itself.)\n\na) Prove that any nice number is divisible by $3$.\n\nb) Are there infinitely many nice numbers?",
    "options": [],
    "answer": "All nice numbers are divisible by 3; Yes, there are infinitely many nice numbers.",
    "solution": "a) Let $N$ be a nice number, i.e. $N = d_1^2 + d_2^2 + d_3^2$, where $d_1, d_2, d_3$ are distinct divisors of $N$. If some divisor of $N$ is divisible by $3$, then $N$ is divisible by $3$.\n\nSo we suppose that $d_1, d_2, d_3$ are not divisible by $3$. Then their squares are congruent to $1$ modulo $3$, i.e. $d_1^2 = 3k_1 + 1$, $k_1 \\in \\mathbb{N}$, $i = 1, 2, 3$. Therefore\n$$\nN = d_1^2 + d_2^2 + d_3^2 = 3(k_1 + k_2 + k_3) + 3,\n$$\nand so $N$ is divisible by $3$.\n\nb) There exists a nice number $N'$. For example, if $N = 30$ and its divisors are $d_1 = 1$, $d_2 = 2$, $d_3 = 5$, then\n$$\nd_1^2 + d_2^2 + d_3^2 = 1^2 + 2^2 + 5^2 = 1 + 4 + 25 = 30 = N,\n$$\ni.e. $N$ is nice.\n\nConsider $N(p) = Np^2$, where $p$ is some positive integer and $p > 1$. If $d_1, d_2, d_3$ are distinct divisors of $N$, then it is obvious that $d_1p, d_2p, d_3p$ are distinct divisors of $N(p)$, and\n$$\nN(p) = Np^2 = [N = d_1^2 + d_2^2 + d_3^2] = (d_1^2 + d_2^2 + d_3^2)p^2 = (d_1p)^2 + (d_2p)^2 + (d_3p)^2.\n$$\nFrom this equality it follows that $N(p)$ is nice too. Since any positive integer can be used as $p$, there are infinitely many nice numbers $N(p)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25555,
    "subject": "Mathematics (Multi-modal)",
    "question": "Some positive integers are written on cards, at least two different numbers on each card. The same number may be written on several cards. Two cards are called *adjacent* if the maximum number on one of them is equal to the minimum number on the other.\nProve that if there are no adjacent cards then all written numbers can be divided into two sets so that any card contains at least one number from every set.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We construct two required groups $G_1$ and $G_2$ as follows: we choose the smallest number on each card and put it in the group $G_1$. Therefore, every card contains at least one number from $G_1$. All other numbers we put in the group $G_2$. Since there are no adjacent cards, the largest number in any card does not coincide with the smallest number on other cards, so this largest number belongs to $G_2$. Therefore, every card contains at least one number from $G_2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25556,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p = \\overline{abc}$ be the decimal representation of a three-digit prime number $p$.\n\nProve that the quadratic equation $ax^2 + bx + c = 0$ has no real roots.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $abc$ is the decimal representation of a three-digit prime number $p$, we have $a \\neq 0$, $c \\neq 0$. Suppose, contrary to our claim, that there exists a rational root of the equation $ax^2+bx+c=0$. Then the discriminant of this equation is a perfect square, i.e. $b^2-4ac=n^2$, where $n$ is a positive integer number, $n < b$. Multiplying the equality $p=100a+10b+c$ by $4a$ we have\n$$\n4np = 400a^2 + 40ub + 4ac = 400a^2 + 40ub + b^2 - n^2.\n$$\nThus $4np = (20a+b)^2 - n^2 = (20a+b+n)(20a+b-n)$. The numbers $20a+b+n$ and $20a+b-n$ have the same parity and since their product is divisible by 4 both of $10a+(b+n)/2$ are positive integers. Since their product is divisible by the prime number $10a+(b+n)/2 \\le 10a+b \\le 99 < p$. This contradiction proves the required statement.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25557,
    "subject": "Mathematics (Multi-modal)",
    "question": "Point $D$ is marked on the side $AB$ of triangle $ABC$. The bisectors of the angles $ABC$ and $ADC$ meet at point $U$, and the bisectors of the angles $BAC$ and $BDC$ meet at point $V$. Let $S$ be the midpoint of the segment $UV$.\n\nProve that the lines $SD$ and $AB$ are perpendicular if and only if the inscribed circles of the triangles $ADC$ and $BDC$ are tangent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note that the triangle $UDV$ is a right-angled triangle regardless of the position of the point $D$ since\n$$\n\\angle UDV = \\angle UDC' + \\angle VDC' = \\frac{1}{2} \\angle ADC + \\frac{1}{2} \\angle BDC = \\frac{1}{2} \\cdot 180^\\circ = 90^\\circ.\n$$\nSo $S$ is the center of the circumcircle of the triangle $UDV$, and the condition $SD \\perp AB$ is equivalent to the tangency of this circle and the line $AB$, and, by turn, is equivalent to the equality $\\angle ADU = \\angle DVU$. Let the lines $UV$ and $CD$ meet at point $T$. Since $\\angle ADU = \\angle CDU$, the condition $\\angle ADU = \\angle DVU$ is equivalent to the equality $\\angle UDT' = \\angle UVD$, i.e. is equivalent to the similarity $\\triangle UTD \\sim \\triangle UDV$; this similarity is equivalent to the equality $\\angle UTD = 90^\\circ$. Thus, $SD \\perp AB$ is equivalent to $UV \\perp CD$.\n\nNote that $U$ and $V$ are the centers of the circles $\\omega_A$ and $\\omega_B$ of the triangles $BDC$ and $ADC$, respectively. Let the circles $\\omega_A$ and $\\omega_B$ touch the side $CD$ at points $T_A$ and $T_B$, respectively. It is easy to see that the condition $UV \\perp CD$ is equivalent to $T_A = T_B = T$ that is equivalent to $DT_A = DT_B$.\nLet the circle $\\omega_A$ touch the lines $AB$ and $BC$ at points $K$ and $L$, respectively. Using the equality of the tangents to the circle $\\omega_A$, we get",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25558,
    "subject": "Mathematics (Multi-modal)",
    "question": "In an isosceles triangle $ABC$ with $AB = BC$, points $K$ and $M$ are the midpoints of the sides $AB$ and $AC$, respectively. The circumscribed circle of the triangle $CKB$ meets the line $BM$ at point $N$ different from $M$. The line passing through $N$ parallel to the side $AC$ meets the circumscribed circle of the triangle $ABC$ at points $A_1$ and $C_1$.\nProve that the triangle $A_1BC_1$ is equilateral.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $O$ be the center of the circumscribed circle of the triangle $ABC$. Since $KBCN$ is an inscribed quadrilateral and $NB$ is the bisector of the angle $KBC$, we have $KN = NC$ (see Fig. 1). Moreover, since $N$ lies on the perpendicular bisector of the segment $AC$, we have $AN = NC$, therefore, $N$ is the circumcenter of the triangle $AKC$.\n\n![](attached_image_1.png)\nFig. 1\n![](attached_image_2.png)\nFig. 2\n\nConstruct the perpendiculars from $N$ and $O$ to the line $AB$ (see Fig. 2). Since the circumcenter is the intersection point of the perpendicular bisectors, we see that $K$ and $T$ are the feet of perpendiculars from $O$ and $N$, respectively. So, $KT = \\frac{1}{2}KA = \\frac{1}{2}KB$. It follows that $BK:KT = 2:1$. Now by the Thales theorem, it follows that $BO:ON = BK:KT = 2:1$. Thus, the circumcenter $O$ of the isosceles triangle $A_1BC_1$ coincides with the point of intersection of the medians, so this triangle is equilateral, as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25559,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all possible values of a real number $a$ such that there exist pairwise distinct nonzero real numbers $x, y, z$ satisfying the equalities\n$$\nx - \\frac{y}{z} - \\frac{z}{y} = y - \\frac{z}{x} - \\frac{x}{z} = z - \\frac{x}{y} - \\frac{y}{x} = a.\n$$",
    "options": [],
    "answer": "1",
    "solution": "Answer: $a = 1$.\n\nLet $A = x - \\frac{y}{z} - \\frac{z}{y}$, $B = y - \\frac{z}{x} - \\frac{x}{z}$, and $C = z - \\frac{x}{y} - \\frac{y}{x}$. By condition, it follows that\n$$\n0 = A - B = (x - y) + \\frac{x - y}{z} - \\frac{z(x - y)}{xy} = (x - y) \\left( 1 + \\frac{1}{z} - \\frac{z}{xy} \\right) = \\frac{(x - y)(xyz + xy - z^2)}{xyz}.\n$$\nSince $x \\neq y$, we obtain $z^2 - xy - xyz = 0$. Similarly, considering the difference $B - C$, we obtain $x^2 - yz - xyz = 0$. Therefore,\n$$\n0 = (z^2 - xy - xyz) - (x^2 - yz - xyz) = (z - x)(x + y + z).\n$$\nSince $x \\neq z$, we have $x + y + z = 0$. Dividing this equality by $x$, $y$, and $z$ consequently, we have\n$$\n1 + \\frac{y}{x} + \\frac{z}{x} = \\frac{x}{y} + 1 + \\frac{z}{y} = \\frac{x}{z} + \\frac{y}{z} + 1 = 0.\n$$\nTherefore, $\\frac{y}{z} + \\frac{z}{y} + \\frac{x}{z} + \\frac{x}{y} + \\frac{y}{x} = -3$. Since $3a = A + B + C$, we obtain\n$$\na = \\frac{1}{3} \\cdot \\left( x + y + z - \\frac{y}{z} - \\frac{z}{y} - \\frac{x}{z} - \\frac{x}{y} - \\frac{y}{z} \\right) = \\frac{1}{3} \\cdot (0 + 3) = 1.\n$$\nIt is easy to verify that $x = 4$, $y = -2 - \\frac{2}{\\sqrt{5}}$, and $z = -2 + \\frac{2}{\\sqrt{5}}$ satisfy the problem condition for $a = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25560,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that\n$$\n\\frac{xyzt}{(x+y)(z+t)} \\le \\frac{(x+z)^2(y+t)^2}{4(x+y+z+t)^2}\n$$\nfor all positive numbers $x$, $y$, $z$, $t$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The required inequality\n$$\n\\frac{xyzt}{(x+y)(z+t)} \\le \\frac{(x+z)^2(y+t)^2}{4(x+y+z+t)^2}\n$$\nis equivalent to the inequality\n$$\n\\frac{(x+y)(z+t)}{xyzt} \\ge \\frac{4(x+y+z+t)^2}{(x+z)^2(y+t)^2}\n$$\n(1)\nNote that\n$$\n\\frac{2(x+y+z+t)}{(x+z)(y+t)} = \\frac{2}{x+z} + \\frac{2}{y+t}\n$$\nBy A.M. - G.M. inequality,\n$$\n\\frac{2}{x+z} \\le \\frac{1}{\\sqrt{xz}}, \\quad \\frac{2}{y+t} \\le \\frac{1}{\\sqrt{yt}}\n$$\nhence\n$$\n\\frac{4(x+y+z+t)^2}{(x+z)^2(y+t)^2} \\le \\left( \\frac{1}{\\sqrt{xz}} + \\frac{1}{\\sqrt{yt}} \\right)^2\n$$\nFurther,\n$$\n\\frac{(x+y)(z+t)}{xyzt} = \\frac{xz+xt+yz+yt}{xyzt}\n$$\n$$\n= \\frac{xz}{xyzt} + \\frac{xt}{xyzt} + \\frac{yz}{xyzt} + \\frac{yt}{xyzt}\n$$\n$$\n= \\frac{1}{yt} + \\frac{1}{yz} + \\frac{1}{xt} + \\frac{1}{xz}\n$$\nBut we can also write\n$$\n\\frac{(x+y)(z+t)}{xyzt} = \\frac{xz+xt+yz+yt}{xyzt} = \\frac{xz+yt+xt+yz}{xyzt}\n$$\n$$\n= \\frac{xz+yt}{xyzt} + \\frac{xt+yz}{xyzt}\n$$\n$$\n= \\frac{xz}{xyzt} + \\frac{yt}{xyzt} + \\frac{xt}{xyzt} + \\frac{yz}{xyzt}\n$$\n$$\n= \\frac{1}{yt} + \\frac{1}{xz} + \\frac{1}{xt} + \\frac{1}{yz}\n$$\nBut the key step is to use the AM-GM inequality:\n$$\n\\frac{(x+y)(z+t)}{xyzt} \\ge \\left( \\frac{1}{\\sqrt{xz}} + \\frac{1}{\\sqrt{yt}} \\right)^2\n$$\nFrom the previous step,\n$$\n\\frac{4(x+y+z+t)^2}{(x+z)^2(y+t)^2} \\le \\left( \\frac{1}{\\sqrt{xz}} + \\frac{1}{\\sqrt{yt}} \\right)^2\n$$\nTherefore,\n$$\n\\frac{(x+y)(z+t)}{xyzt} \\ge \\left( \\frac{1}{\\sqrt{xz}} + \\frac{1}{\\sqrt{yt}} \\right)^2 \\ge \\frac{4(x+y+z+t)^2}{(x+z)^2(y+t)^2}\n$$\nwhich proves the required inequality.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25561,
    "subject": "Mathematics (Multi-modal)",
    "question": "Every one of six pupils attends exactly two of four hobby groups. There are no pupils attending the same two hobby groups. Each hobby group is open every day. During some consecutive days, every one of these six pupils has attended one of her/his hobby group. It has been observed that each of these days each hobby group has been attended by either one or two of the pupils. Moreover, for any chosen two days, there has been a pupil who has attended different hobby groups during these two days.\nFind the largest number of the days for which the situation described above is possible.",
    "options": [],
    "answer": "24",
    "solution": "Answer: 24 days.\nNote that from four hobby groups one can form exactly $4 \\cdot 3/2 = 6$ different pairs. Since we have exactly six pupils, for each pair of the hobby group there exists exactly one pupil attending just these two hobby groups. By condition, each hobby group is attended by either one or two pupils and there are exactly four hobby group, so in each of these days there are exactly two hobby groups attended by two pupils and there are exactly two hobby groups attended by one pupil.\n\nConsider arbitrary day. Let this day hobby group $A$ and $B$ be attended by two pupils. Since, as was shown above, there is a pupil attending just these two hobby groups, without loss of generality we can assume that this pupil has attended hobby group $A$. Each of two remained hobby groups $C$ and $D$ has been attended by one pupil. Again, there exists a pupil attending just these hobby groups $C$ and $D$. Suppose that this pupil has attended hobby group $C$. Show that this information is sufficient to uniquely indicate for each hobby group who of the pupils has attended this hobby group.\n\nSince in the considered day hobby group $C$ has been attended by exactly one pupil (this pupil attends hobby groups $C$ and $D$), the pupil attending $A$ and $C$ has attended hobby group $A$, and the pupil attending $B$ and $C$ has attended hobby group $B$. Further, since hobby group $A$ has been attended by two pupils (one of them attends hobby groups $A$ and $B$, and the other one attends hobby group $A$ and $C$), the pupil attending $A$ and $D$ has attended hobby group $D$. Finally, the pupil attending hobby groups $B$ and $D$ has attended hobby group $B$, since hobby group $D$ has been attended by exactly one pupil attending hobby groups $A$ and $D$.\n\nWe have exactly $4 \\cdot 3/2 = 6$ ways to choose the pair $(A, B)$ that have been attended by two pupils, and for each such pair we have two ways to fix the hobby group has been attended by the pupil attending just these two hobby groups. For hobby groups $C$ and $D$ we have two ways to choose the hobby group that has been attended by the pupil attending these hobby groups $C$ and $D$. Thus, we obtain $6 \\times 2 \\times 2 = 24$ different days in total.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 25562,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $M$ be the midpoint of the hypotenuse $AB$ of the right triangle $ABC$. Point $P$ is chosen on the cathetus $CB$ so that $CP : PB = 1 : 2$. The straight line passing through $B$ meets the segments $AC, AP,$ and $PM$ at points $X, Y, \\text{ and } Z$, respectively.\nProve that the bisector of the angle $PZY$ passes through point $C$ if and only if the bisector of the angle $PYX$ also passes through $C$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $D$ be symmetric to $A$ with respect to the vertex $C$ (see the Fig.). The segment $BC$ is the altitude and the median in the triangle $ABD$ so the triangle $ABD$ is isosceles. Likewise, the triangle $ADP$ is isosceles. Let $P'$ be the intersection point of the medians $DM$ and $BC$ of the triangle $ABD$. Since the point of intersection divides the medians in the ratio $2:1$, points $P$ and $P'$ coincide. Since $C$ belongs to the bisector of the angle $\\angle DPA$, we see that $C$ is equidistant to the lines $DP$ and $PA$.\nIf $C$ belongs to the bisection of the angle $\\angle PZY$, then $C$ is equidistant to the lines $PZ$ and $ZY$, therefore, $C$ is equidistant to the lines $YP$ and $YX$, i.e. $C$ belongs to the bisection of the angle $\\angle PYX$.\nIn the same way, one can prove the converse proposition.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25563,
    "subject": "Mathematics (Multi-modal)",
    "question": "On the Cartesian plane $Oxy$ the parabola $y = x^2 - 2bx + a + 1$ meets the $x$-axis at points $A$ and $B$, and meets the $y$-axis at point $C$ (distinct from the origin). It appears that the point with the coordinates $(a, b)$ is a center of the circumscribed circle of the triangle $ABC$.\nFind all possible values of the numbers $a$ and $b$.\n(V. Karamzin)",
    "options": [],
    "answer": "a = b = 2",
    "solution": "Answer: $a = b = 2$.\nLet $A(\\alpha, 0)$, $B(\\beta, 0)$, $C(0, \\gamma)$, and $M(a, b)$. Since $M$ is the center of the circumscribed circle of the triangle $ABC$, $M$ lies on the perpendicular bisector of the segment $AB$. Therefore, the projection of $M$ on the $x$-axis is the midpoint of the segment $AB$, i.e. $a = (\\alpha + \\beta)/2$.\nSince the points $A$ and $B$ lie on the parabola, their abscissae satisfy the equation $x^2 - 2bx + a + 1 = 0$, then, by Vieta's theorem, $a = (\\alpha + \\beta)/2 = 2b/2 = b$.\nSubstituting $x = 0$ into the parabola equation, we find the ordinate $\\gamma$ of $C$: $\\gamma = a + 1$.\nLet $R$ be the radius of the circle $\\Gamma$. Since $C$ lies on $\\Gamma$, we obtain\n$$\nR^2 = |MC|^2 = (a-0)^2 + (b-\\gamma)^2 = [a=b, \\gamma=a+1] = a^2+1.\n$$\nSince $A$ and $B$ lie on $\\Gamma$, we get\n$$\na^2 + 1 = R^2 = |MA|^2 = |MB|^2 \\Rightarrow (a - \\alpha)^2 + (b - 0)^2 =\n$$\n$$\n= (a - \\beta)^2 + (b - 0)^2 = (a - \\alpha)^2 + a^2 = (a - \\beta)^2 + a^2 \\Rightarrow (a - \\alpha)^2 = (a - \\beta)^2 = 1.\n$$\nWithout loss of generality we can assume that $\\beta > \\alpha$, then $\\alpha = a - 1$, $\\beta = a + 1$. The numbers $\\alpha = a - 1$ and $\\beta = a + 1$ are the roots of the equation $x^2 - 2bx + a + 1 = 0$, so, by Vieta's theorem\n$$\na^2 - 1 = (a - 1)(a + 1) = \\alpha\\beta = a + 1 \\Rightarrow a^2 - a - 2 = 0.\n$$\nWe see that $a = -1$ and $a = 2$ are the roots of this equation.\n\nFor $a = -1$ we have $\\gamma = 0$, but then $C(0, 0)$, i.e. $C$ coincides with the origin, contrary to the problem condition.\nFor $a = 2$ we have $M(2, 2)$, the parabola equation has the form $y = x^2 - 4x + 3$, and the circle equation has the form $(x - 2)^2 + (y - 2)^2 = 5$. It is easy to verify that all problem conditions hold in this case.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25564,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Given the ten digits from $0$ to $9$, prove that three numbers $A$, $B$, and $C$ can be formed by combining these digits, provided that each digit is used exactly once and $A+B=C$. Notice that $0$ may not be the first digit of any of the numbers.\n\nb) Find all possible values of the sum of the digits of $C$.",
    "options": [],
    "answer": "9 or 18",
    "solution": "a)\nFor example, $765 + 324 = 1089$.\n\nb)\nSince any integer $Y$ is congruent modulo $9$ to the sum of its digits, we have\n$$\nA + B + C \\equiv S(A) + S(B) + S(C) = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 0 = 45 \\equiv 0 \\pmod{9}.\n$$\nBy condition, we have $A + B = C$, so $2C \\equiv 0 \\pmod{9}$, whence it follows that $C \\equiv 0 \\pmod{9}$, and, therefore, $S(C) \\equiv 0 \\pmod{9}$, i.e. the sum of the digits of $C$ is divisible by $9$.\n\nNote that $S(C) = S(A+B) \\leq S(A) + S(B)$. Indeed, if the sum of the digits of $A$ and $B$ does not exceed $9$ in all number positions, then when we add $A$ and $B$ there is no 'carry' from any number position, so $S(A+B) = S(A) + S(B)$. Otherwise, since the greatest possible carry is $1$, if there is a carry from some number position, then the sum $S(A+B)$ decreases by $9$.\n\nThus, $S(C) \\leq S(A) + S(B)$ and $18$ are smaller than $22.5$ and are divisible by $9$. The following examples show that $S(C)$ can be $9$ (the digits are $1+0+8+9$); $765 + 324 = 1089$ (the sum of the digits is equal to $18$).\n\nTherefore, the possible values for the sum of the digits of $C$ are $9$ or $18$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25565,
    "subject": "Mathematics (Multi-modal)",
    "question": "Point $M$ is marked inside a convex quadrilateral $ABCD$. It appears that $AM = BM$, $CM = DM$, and $\\angle AMB = \\angle CMD = 60^\\circ$. Let $K$, $L$, and $N$ be the midpoints of the segments $BC$, $AM$, and $DM$, respectively. Find the value of the angle $LKN$.\n\n(S. Mazanik)",
    "options": [],
    "answer": "60°",
    "solution": "Answer: $60^\\circ$.\n\nLet $E$ and $F$ be the midpoints of the segments $BM$ and $CM$, respectively. Since $\\angle AMB = \\angle CMD = 60^\\circ$, we have\n$$\n\\begin{aligned}\n\\angle BMC &= 360^\\circ - \\angle AMB - \\angle CMD - \\angle LMN = 360^\\circ - 60^\\circ - 60^\\circ - \\angle LMN = \\\\\n&= 240^\\circ - \\angle LMN.\n\\end{aligned}\n$$\nSince $KF$ is the midline in the triangle $CBM$, we have $KF \\parallel BM$, therefore, $\\angle KFC = \\angle BMC$. Then\n$$\n\\begin{aligned}\n\\angle KFM &= 180^\\circ - \\angle KFC = 180^\\circ - \\angle BMC = \\\\\n&= 180^\\circ - (240^\\circ - \\angle LMN) = \\angle LMN - 60^\\circ.\n\\end{aligned}\n$$\nBy condition, $CM = DM$ and $\\angle CMD = 60^\\circ$, so the triangle $CMD$ is equilateral. Since $FN$ is the midline in the triangle $CMD$, we see that the triangle $FMN$ is equilateral too and $FM = NM = FN$, $\\angle MFN = 60^\\circ$. Therefore,\n$$\n\\angle KFN = \\angle KFM + \\angle MFN = [\\angle MFN = 60^\\circ] = \\angle LMN - 60^\\circ + 60^\\circ = \\angle LMN.\n$$\nIn the same way, one can show that $\\angle KEL = \\angle LMN$ and $EM = ML = EL$. Thus, $\\triangle KEL = \\triangle NFK = \\triangle NML$ (by two sides and the angle between them), so $KL = KN = LN$. Therefore, the triangle $KLN$ is equilateral and $\\angle LKN = 60^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25566,
    "subject": "Mathematics (Multi-modal)",
    "question": "Points $K$ and $M$ are the midpoints of the sides $AB$ and $AC$ of triangle $ABC$, respectively. The equilateral triangles $AMN$ and $BKL$ are constructed on the sides $AM$ and $BK$ to the exterior of the triangle $ABC$. Point $F$ is the midpoint of the segment $LN$.\nFind the value of the angle $KFM$.",
    "options": [],
    "answer": "90°",
    "solution": "Answer: $90^\\circ$.\n\nLet points $A$, $B$, and $C$ lie in the same half-plane with respect to the line $LN$. Let $E$ and $D$ be the midpoints of the segments $AL$ and $AN$, respectively (see the Fig.). By condition, the triangle $BKL$ is equilateral and $K$ is the midpoint of the side $AB$, so $BK = KA = KL$. Therefore, the triangle $BLA$ is a right-angled triangle (the median $LK$ is half as long as the side $AB$) and $\\angle BLA = 90^\\circ$. By condition, $\\angle KBL = 60^\\circ$, so $\\angle LAB = 30^\\circ$. By condition, the triangle $ANM$ is equilateral, then $\\angle NAM = 60^\\circ$. Therefore,\n$$\n\\begin{aligned}\n\\angle LAN &= 360^\\circ - \\angle LAK - \\angle NAM - \\angle KAM = \\\\\n&= 360^\\circ - 30^\\circ - 60^\\circ - \\angle KAM = 270^\\circ - \\angle KAM.\n\\end{aligned} \\quad (1)\n$$\nSince $FD$ and $FE$ are the midlines of the triangle $LAN$, we have $FD \\parallel LA$ and $FE \\parallel NA$, so $EFDA$ is a parallelogram. Then\n$$\n\\angle FDA = \\angle FEA = 180^\\circ - \\angle LAN \\stackrel{(1)}{=} \\angle KAM - 90^\\circ. \\quad (2)\n$$\nSince $D$ is the midpoint of the side $AN$ of the equilateral triangle $ANM$, we have $\\angle MDA = 90^\\circ$. Now from (2) it follows that:\n$$\n\\angle FDM = \\angle FDA + \\angle MDA = \\angle KAM - 90^\\circ + 90^\\circ = \\angle KAM. \\quad (3)\n$$\n\nIn the similar way, we easily find that\n$$\n\\angle FEK = \\angle KAM. \\quad (4)\n$$\n\nSince $EFDA$ is a parallelogram, we have\n$$\nFD = EA = [\\angle KEA = 90^\\circ, \\angle EAK = 30^\\circ] = KA \\frac{\\sqrt{3}}{2}.\n$$\nAlso $DM = [\\angle MDA = 90^\\circ, \\angle MAD = 60^\\circ] = MA \\frac{\\sqrt{3}}{2}$. Therefore, $FD : DM = KA : MA$, so, taking into account (3), we see that the triangles $FDM$ and $KAM$ are similar, hence $\\angle MFD = \\angle MKA$, $\\angle FMD = \\angle KMA$. Thus,\n$$\n\\angle KMF = \\angle KMA + \\angle AMF = \\angle FMD + \\angle AMF = \\angle AMD = 30^\\circ.\n$$\nIn the same way, we get $\\triangle KEF \\sim \\triangle FAM$, since\n$$\nKE : EF = KE : AD = \\frac{1}{2}KA : \\frac{1}{2}MA = KA : MA, \\quad \\angle KEF = \\angle KAM.\n$$\nSo $\\angle EKF = \\angle AKM$, and then\n$$\n\\angle FKM = \\angle FKA + \\angle AKM = \\angle FKA + \\angle EKF = \\angle EKA = 60^\\circ.\n$$\n$$\n\\text{Thus, } \\angle KFM = 180^\\circ - \\angle KMF - \\angle FKM = 180^\\circ - 30^\\circ - 60^\\circ = 90^\\circ.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25567,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a convex $2n$-gon $H$ with pairwise parallel opposite sides.\n\na) Prove that there exists a pair of the opposite sides of $H$ which possesses the following property: there exists a straight line that is perpendicular to these sides and intersects each of them.\n\nb) Are there any values of $n$ such that for a convex $2n$-gon there exist two pairs of its opposite sides for each of which the property described in a) holds?",
    "options": [],
    "answer": "a) Such a pair always exists. b) No values of n.",
    "solution": "b) there are no such $n$.\n\na) A diagonal of the convex $2n$-gon is said to be main if there are $n-1$ vertices of this polygon between the vertices connected by this diagonal. Suppose that there exist a convex $2n$-gon $A_0A_1...A_{2n-1}$ such that its opposite sides are parallel ($A_iA_{i+1} \\parallel A_{i+n}A_{i+n+1}$, $i = 0, ..., n-1$, and $A_{2n} = A_n$) and there is no pair of the opposite sides possessing the property described by the problem condition (there exists a straight line that is perpendicular to these sides and intersects each of them). Consider some two opposite sides $A_iA_{i+1}$ and $A_{i+n}A_{i+n+1}$ of such $2n$-gon. Let $O_i$ be the intersection point of the lines $A_iA_{i+n}$ and $A_{i+1}A_{i+n+1}$. Consider the triangles $O_iA_iA_{i+1}$ and $O_iA_{i+n}A_{i+n+1}$ (see Fig. 1).\n\n![](attached_image_1.png)\nFig. 1\n![](attached_image_2.png)\nFig. 2\n\nSince $A_iA_{i+1} \\parallel A_{i+n}A_{i+n+1}$ we have $\\angle O_iA_iA_{i+1} = \\angle O_iA_{i+n}A_{i+n+1}$ and $\\angle O_iA_{i+n}A_{i+n+1}$ (let $\\alpha = \\angle O_iA_{i+n}A_{i+1}$ and $\\beta = \\angle O_iA_{i+n+1}A_{i+n}$). If at least one of these angles is not obtuse, then there exists a straight line $l$ that is perpendicular to the sides $A_iA_{i+1}$, $A_{i+n}A_{i+n+1}$ and intersects each of them. For example, the line passing through $O_i$ perpendicular to the side $A_iA_{i+1}$ can be considered as $l$. Therefore either $\\alpha$ or $\\beta$ is greater than $90^\\circ$.\n\nAny diagonal starting from the vertex of the angle of the $2n$-gon partitions this angle into two angles (we call that these two angles are *adjacent* to this diagonal). Construct all main diagonals of the $2n$-gon and number (in clockwise direction) all adjacent angles with the numbers from 1 to $4n$ (see Fig. 2). By the above either angle 1 or angle $4n$ is obtuse. Let, without loss of generality, angle 1 is obtuse. Then angle 2 is acute (any inner angle of a convex polygon is less than $180^\\circ$). From what has already been proved, it follows that angle 3 is obtuse, and so on. We see that all angles with odd numbers are obtuse, while all angles with even numbers are acute. So one of the adjacent angles is obtuse but the other is acute for any main diagonal. Consider the greatest of the main diagonals (one of them if there are more than one). Let $A_iA_{i+n}$ be such diagonal (see Fig. 1) and let, without loss of generality, $\\angle O_iA_iA_{i+1} = \\angle O_iA_{i+n}A_{i+n+1} > 90^\\circ$. Since in which is impossible since the diagonal $A_iA_{i+n}$ is the greatest one. This contradiction proves that any convex $2n$-gon with pairwise parallel opposite sides has a pair of the opposite sides such that there exists a straight line that is perpendicular to these sides and intersects each of them.\n\nb) Consider a regular $2n$-gon $S$ with the vertices $A_0, A_1, ..., A_{2n-1}$ ($A_{2n} = A_0$). We choose the vector $\\vec{v}$ satisfying the relations $\\angle(\\vec{v}, A_iA_{i+1}) \\neq 90^\\circ$ for $i \\in \\{0, ..., 2n-1\\}$,\n\n![](attached_image_3.png)\nFig. 3\n\nand $|\\angle(\\vec{v}, A_{2n-1}A_i)| < 180^\\circ/n$, (here $\\angle(\\vec{v}, \\vec{v})$ is the angle between the vectors $\\vec{v}$ and $\\vec{v}$). For any $\\lambda > 0$ we denote by $S_\\lambda$ the convex polygon with the vertices $A_0^\\lambda, A_1^\\lambda, ..., A_{2n-1}^\\lambda$, where $A_i^\\lambda = A_i + \\lambda\\vec{v}$ for $i$ from 0 to $n-1$ (from the inequality $|\\angle(\\vec{v}, A_{2n-1}A_i)| < 180^\\circ/n$ it follows that the polygon $S_\\lambda$ is convex). In the polygon $S_\\lambda$ for $\\lambda$ large enough there exists a unique pair $(A_{2n-1}^\\lambda, A_{n-1}^\\lambda)$ of the opposite sides such that there exists a straight line that is perpendicular to these sides and intersects each of them.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25568,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a convex hexagon $H$ with obtuse inner angles and parallel opposite sides.\n\na) Prove that there exists a pair of the opposite sides of $H$ which possesses the following property: there exists a straight line that is perpendicular to these sides and intersects each of them.\n\nb) Is it true that there exist two pairs of the opposite sides of $H$, each of which possesses the same property, as described in item a)?",
    "options": [],
    "answer": "a) Yes. b) No.",
    "solution": "b) it is not true.\n\na) Let $ABCDDEF$ be a hexagon with all obtuse inner angles and parallel opposite sides ($AB \\parallel DE$, $BC \\parallel EF$, $CD \\parallel FA$, see Fig. 1). Consider the greatest side of this hexagon (one of such sides if there are more than one). Let it be the side $AB$. Since the hexagon $ABCDDEF$ is convex, it completely lies in one of the half-planes with respect to the line $AB$; denote this half-plane by $\\Pi$.\n\n![](attached_image_1.png)\n\nFig. 1\n\nSuppose that none of the lines intersecting and perpendicular to the side $AB$ intersect the side $DE$. Construct the perpendiculars $h_A$ and $h_B$ in the half-plane $\\Pi$ (see Fig. 2). By our assumption, there are no points of the side $DE$ in the half-strip $P_0$. In particular, either point $D$ lies to the right of the ray $h_A$ or point $E$ lies to the left of the ray $h_B$. Without loss of generality, suppose that $D$ lies to the right of $h_A$ (see Fig. 2). Then the vertex $C$ lies neither to the right of the ray $h_A$ nor to the left of the ray $h_B$. Indeed, otherwise we obtain $BC > BA$ (see Fig. 3) and $CD > BA$ (see Fig. 4), respectively. But any of these inequalities contradicts to the choice of the side $AB$ as the greatest side of the hexagon. So the vertex $C$ lies in the half-strip $P_0$, but in this case the angle $ABC$ is not obtuse, contrary to the problem condition. Therefore, there exists a straight line such that it is perpendicular to the sides $AB$ and $DE$ and intersects both of them.\n\nb) Construct the convex hexagon $ABCDDEF$ satisfying the problem condition (all inner angles are obtuse and opposite sides are parallel) such that there exists exactly one pair of the opposite sides intersected by the perpendicular straight line.\n\n![](attached_image_2.png)\n\nFig. 5\nFig. 6\n\nConsider the isosceles triangle $KLM$ with the base $LM$ and the acute angle $LKM$ (see Fig. 5). Let $KH$ be the altitude of the triangle $KLM$. Let points $P$ and $Q$ be the inner points of the segments $MH$ and $LH$, respectively, such that they are symmetric with respect to $H$. Let $E$ be the foot of perpendicular from $P$ into $KL$, and let $D$ be the some point $A$ on the segment $LQ$ and some point $B$ on the segment $MP$. Construct the line $l(A)$ passing through $A$ parallel to $KM$, and construct the line $l(B)$ passing through $B$ parallel to $KL$. Let $F$ be the intersection point of $l_A$ and $KL$, and let $G$ be the intersection point of $l_B$ and $KM$ (see Fig. 5). Then (see Fig. 6) it is easy to see that the hexagon $ABCDDEF$ is required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25569,
    "subject": "Mathematics (Multi-modal)",
    "question": "The central area of a town has a form of the $(2n+1) \\times m$ rectangle, which is formed by $1 \\times 1$ tiles. To illuminate the area, one-lamp lampposts are used. The lampposts are placed at the corners of some tiles, including a corner at the lamppost position, and only those.\nFind the smallest number of the lampposts required to illuminate the whole area, even if one of the lamps should burn out.\n(E. Barabanov, M. Karpuk, A. Voidelevich)",
    "options": [],
    "answer": "2(n + 1) ⌊(m + 1)/2⌋",
    "solution": "Answer: $2(n + 1) \\lfloor \\frac{m + 1}{2} \\rfloor$, where $\\lfloor x \\rfloor$ is the greatest integer not exceeding $x$.\n\nWe paint some tiles of the town square black (if $m = 2k + 1$ is odd, then see Fig. 1, if $m = 2k$ is even, then see Fig. 2).\n\nIt is easy to see that any lamp can illuminate at most one painted tile. By condition, any tile must be illuminated by at least two lamps. It follows that the minimum number of the lampposts is greater than or equal to $2k$, where $l$ is the number of painted tiles. If the length of the square is odd, i.e. $m = 2k + 1$, where $k \\ge 0$, then $l = (n+1)(k+1) = (n+1) \\lfloor \\frac{m+1}{2} \\rfloor$; if $m = 2k$, then $l = (n+1)k = (n+1) \\lfloor \\frac{m+1}{2} \\rfloor$.\n\nOn the other hand, if we place the lampposts as it is shown in the figures (the lampposts are indicated as uncolored circles), then any of the square tiles will be illuminated by two lamps. So, $2l = 2(n+1) \\left[ \\frac{m+1}{2} \\right]$ lampposts are sufficient to illuminate the town square so that the problem condition will hold.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25570,
    "subject": "Mathematics (Multi-modal)",
    "question": "The central area of a town has a form of the $n \\times n$ square, which is formed by $1 \\times 1$ tiles. To illuminate the area, one-lamp lampposts are used. The lampposts are placed at the corners of some tiles, including the boundary of the area. The lamp on a lamppost illuminates all tiles with a corner at the lamppost position, and only those.\n\nFind the smallest number of the lampposts required to illuminate the whole area, even if one of the lamps should burn out.",
    "options": [],
    "answer": "If n is even, n(n+1)/2 (equivalently 2k^2 + k for n = 2k). If n is odd, (n+1)^2/2 (equivalently 2(k+1)^2 for n = 2k + 1).",
    "solution": "$$\n\\begin{cases}\n2k^2 + k, & \\text{if } n = 2k; \\\\\n2(k+1)^2, & \\text{if } n = 2k+1.\n\\end{cases}\n$$\n\nIf the length $n$ of the town square is odd, i.e. $n = 2k + 1$, then from the solution of Problem D8 it follows that the minimal value of the required lampposts is equal to\n$$\n2(k+1) \\left[ \\frac{n+1}{2} \\right] = 2(k+1)^2.\n$$\n\nNow let $n = 2k$ is even. We paint some tiles of the town square black (see Fig. 1).\n\n![](attached_image_1.png)\n\nFig. 1\nFig. 2\n\nConsider the tiles which form the painted \"angle\" (see Fig. 2). It is easy to see that to illuminate each of these tiles by at least two lamps we need at least three lampposts in total. From our coloring it follows that any lamp cannot illuminate two colored tiles if they are not adjacent (we call two tiles adjacent if they have either the common side or the common corner). Thus, the minimum value of the required lamppost is equal to $3r + 2l$, where $r$ is the number of the colored angles and $l$ is the number of the solitary colored tiles. Since $r = k$ and $l = k^2 - k$, we need at least $3r + 2l = 2k^2 + k$ lampposts.\n\nOn the other hand, if we place the lampposts as it is shown in Fig. 1 (the lampposts are indicated as uncolored circles), then any tile of the town square will be illuminated by two lamps. So, if $n = 2k$, then $2k^2 + k$ lampposts are sufficient to illuminate the town square so that the problem condition holds.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25571,
    "subject": "Mathematics (Multi-modal)",
    "question": "A rectangle $R$ with odd integer side lengths is divided into small rectangles with integer side lengths.\nProve that there is at least one rectangle among the small rectangles whose distances from the four sides of $R$ are either all odd or all even.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**1.** See IMO-2017 Shortlist, Problem C1.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25572,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all $a \\in \\mathbb{R}_{\\ge 0}$ such that $f(a) = 0$ for any function $f: \\mathbb{R}_{\\ge 0} \\to \\mathbb{R}_{\\ge 0}$ satisfying the equality\n$$\nf(f(x) + f(y)) = y f(1 + y f(x))\n$$\nfor all nonnegative real numbers $x$ and $y$.",
    "options": [],
    "answer": "{0} ∪ [1, ∞)",
    "solution": "Answer : $a \\in \\{0\\} \\cup [1, \\infty)$.\nLet function $f: \\mathbb{R}_{\\ge 0} \\to \\mathbb{R}_{\\ge 0}$ satisfy the equality\n$$\nf(f(x) + f(y)) = y f(1 + y f(x)) \\tag{1}\n$$\nfor all nonnegative real numbers $x$ and $y$. Set $f(0) = c$ and $f(1) = b$.\nFor $y = 0$ from equality (1) we get\n$$\nf(f(x) + c) = 0. \\tag{2}\n$$\nfor all $x \\in [0, \\infty)$.\nFor $x = f(x) + c$ from equality (1) we get\n$$\nf(f(y)) = b y. \\tag{3}\n$$\nfor all $y \\in [0, \\infty)$.\nSuppose $b \\neq 0$. From equality (3) it follows that function $f$ is injective. Hence from equality (2) it follows that $f(x) + c = f(0)$, a contradiction.\n\nTherefore $b = 0$ and equality (3) is equivalent to\n$$\nf(f(y)) = 0 \\tag{4}\n$$\n\nFor $y = 0$ from equality (4) we get $f(c) = 0$ hence $f(0) = f(f(c)) = 0$. So $c = f(0) = 0$.\nFor $y = 1$ from equality (1) we get\n$$\nf(1 + f(x)) = 0 \\qquad (5)\n$$\nfor all $x \\in [0, \\infty)$.\nSuppose that $f(a) = d \\neq 0$ for some $a > 1$. Substituting $x = a$, $y = \\frac{a-1}{d}$ to equality (1) we get\n$$\nf(f(\\frac{a-1}{d}) + d) = a - 1.\n$$\nwhich leads to a contradiction with equality (5) for $x = f(\\frac{a-1}{d}) + d$.\nHence for any $a \\in \\{0\\} \\cup [1, \\infty)$ the equality $f(a) = 0$ is proved. The next example shows that there exists a function satisfying equality (1) such that $f(x) \\neq 0$ for any $x \\in (0, 1)$:\n$$\nf(x) = \\begin{cases} 0, & x = 0; \\\\ \\frac{1}{x}, & x \\in (0, 1); \\\\ 0, & x \\geq a. \\end{cases}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25573,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all real numbers $a$ for which there exists a function $f$ defined on the set of all real numbers which takes as its values all real numbers exactly once and satisfies the equality\n$$\nf(f(x)) = x^2 f(x) + a x^2\n$$\nfor all real $x$.",
    "options": [],
    "answer": "a = 0",
    "solution": "Answer: $a = 0$.\n\nSubstituting $x$ such that $f(x) = -a$, we get $f(-a) = 0$.\n\nSubstituting $x = -a$, we get $f(0) = a^3$.\n\nFinally, substituting $x = 0$, we get $f(a^3) = 0$.\n\nSince $f$ takes all real values exactly once, $a^3 = -a$ which is equivalent to $a(a^2 + 1) = 0$, i.e. $a = 0$.\n\nClearly, for $a = 0$ the function $f(x) = x|x|$ satisfies the conditions of the problem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25574,
    "subject": "Mathematics (Multi-modal)",
    "question": "A village $M$ is on the road between the villages $A$ and $B$. The distance between $A$ and $M$ is twice as long as the distance between $B$ and $M$. Ann, Bob and Tom live in $A$, $B$, and $M$ respectively. One time Tom invites Ann and Bob to a game of chess. Ann and Bob walk along the road with constant and equal speeds and they start from their villages at the same time. Tom has a motorbike. Tom has two possibilities: he rides towards Ann and brings her to $M$ and after that he rides towards Bob and brings him to $M$, or he rides towards Bob and brings him to $M$ and after that he rides towards Ann and brings her to $M$. The total time in the first case differs from the total time in the second case by 2.4 minutes. The speed of the motorbike is nine times as fast as the speed of the pedestrians.\nHow long it takes Ann to achieve $M$ on foot?",
    "options": [],
    "answer": "2 hours",
    "solution": "**1.** Answer: 2 hours.\nLet the distance between $B$ and $M$ be $S$ (km), then the distance between $A$ and $M$ is $2S$. Denote by $v$ (km/h) the speed of Ann and Bob, then the speed of Tom on a motorbike is $9v$.\n\nConsider the case in which Tom first drives Ann. Since the distance between $A$ and $M$ is $2S$, and the rate of convergence of Ann and Tom is $v + 9v = 10v$, Ann and Tom will meet through $\\frac{2S}{10v} = \\frac{S}{5v}$ (h). After the same time Tom and Ann on a motorbike will drive from the meeting point to $M$. During this time equaled $2 \\cdot \\frac{S}{5v} = \\frac{2S}{5v}$. Bob will pass the distance $v \\cdot \\frac{2S}{5v} = \\frac{2S}{5}$. Therefore, when Tom with Ann will be in $M$, the distance between him and Bob will equal to $S - \\frac{2S}{5} = \\frac{3S}{5}$. Therefore, after leaving $M$ Tom will meet and pick up Bob in $\\frac{3S}{5} : 10v = \\frac{3S}{50v}$. The same time Tom (with Bob) will drive back to $M$. As a result, all three friends will be in $M$ in $\\frac{2S}{5v} + 2 \\cdot \\frac{3S}{50v} = \\frac{13S}{25v}$ (h) after the start of the motion.\n\nSimilarly, calculate the time in the case in which Tom first drives up Bob. Time of Tom on the way to a meeting with Bob and back to $M$ equals $\\frac{S}{5v}$. During this time, Ann will pass the distance $\\frac{S}{5}$. Therefore, when Tom (with Bob arrive) to $M$, Ann will be at a distance of $2S - \\frac{S}{5} = \\frac{9S}{5}$. Then Tom will need $2 \\cdot \\frac{9S}{5} : 10v = \\frac{18S}{50v}$ for the road from $M$ to the meeting with Ann and back. As a result, in this case, all three friends will be in $M$ in $\\frac{S}{5v} + \\frac{18S}{50v} = \\frac{14S}{25v}$ (h) after the start of the motion.\n\nBy condition, the total time for the whole path in the first case differs from the time in the second case by 2.4 minutes, i.e. by $\\frac{1}{25}$ hours. Therefore:\n$$\n\\frac{14S}{25v} - \\frac{13S}{25v} = \\frac{1}{25},\n$$\nwhence $\\frac{S}{v} = 1$. This means that Ann covers the distance $S$ by foot for 1 hour, so it takes her 2 hours for the distance from $A$ to $M$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 25575,
    "subject": "Mathematics (Multi-modal)",
    "question": "A point $X$ lies inside an isosceles right triangle $ABC$ with $\\angle B = 90^\\circ$.\nProve the inequality $AX + BX + \\sqrt{2}CX \\ge \\sqrt{5}AB$ and find all points $X$ for which the equality holds.",
    "options": [],
    "answer": "Inequality: AX + BX + sqrt(2)·CX ≥ sqrt(5)·AB. Equality holds for exactly one interior point X. In a coordinate model with A at (0,1), B at (0,0), and C at (1,0), this point is X = (2/5, 1/5). Geometrically, it is the image of C under a ninety-degree rotation about H, where H is the foot of the perpendicular from C to BD and D is the image of A under a ninety-degree rotation about C.",
    "solution": "First solution. Consider the rotation around point $C$ by $90^\\circ$ such that the images of point $A$ and point $B$ lie in the same half-plane with respect to the line $CB$ (see the Fig.). Denote the images of points $A$ and $X$ by $D$ and $Y$ respectively. By construction, the triangles $ACD$ and $CXY$ are isosceles right triangles with right angles at vertex $C$, hence $AD = \\sqrt{2}AC = 2AB$ and $XY = \\sqrt{2}CX$. From the right triangle $BAD$ one can easily find $BD = \\sqrt{5}AB$. Finally, $DY = AX$, since $DY$ is the image of $AX$ under rotation.\n\n![](attached_image_1.png)\n\nThe required inequality is equivalent to $AX+XY+YD \\ge BD$ which is true, as the length of the polyline $BXYD$ is greater or equal to the length of the segment $BD$. The equality holds if and only if points $X$ and $Y$ lie on the segment $AD$. It is easy to see that there is only one such point. Indeed, let $H$ be the foot of the perpendicular from point $C$ to the line $BD$. Point $H$ is the midpoint of the hypotenuse of an isosceles right triangle $CXY$, therefore the triangle $CHX$ is an isosceles right triangle with a right angle at vertex $H$. Thus, there is exactly one such point $X$, namely, the image of point $C$ (lying inside the triangle $ABC$) under the rotation by $90^\\circ$ with respect to the point $H$.\n\n![](attached_image_1.png)\nConsider points $A(0,1)$, $B(0,0)$, $C(1,0)$ and $X(x,y)$ on the Cartesian plane. Then the required inequality is equivalent to:\n$$\n\\sqrt{x^2 + (1-y)^2} + \\sqrt{x^2 + y^2} + \\sqrt{2\\sqrt{(1-x)^2 + y^2}} \\ge \\sqrt{5}. \\quad (1)\n$$\nNote that this equality is equivalent to\n$$\n\\sqrt{(1-y)^2 + x^2} + \\sqrt{x^2 + y^2} + \\sqrt{(1-x+y)^2 + (1-x-y)^2} \\ge \\sqrt{5}, \\quad (2)\n$$\nsince $2(1-x)^2 + 2y^2 = (1-x+y)^2 + (1-x-y)^2$.\nNow the equality (2) follows from $|\\vec{a}|+|\\vec{b}|+|\\vec{c}| \\ge |\\vec{a}+\\vec{b}+\\vec{c}|$ for vectors $\\vec{a}(1-y,x)$, $\\vec{b}(x,y)$, $\\vec{c}(1-x+y,1-x-y)$, and $\\vec{a}+\\vec{b}+\\vec{c} = (2,1)$.\nFor any point $X$ the equality in (2) means that vectors $\\vec{a}$, $\\vec{b}$ and $\\vec{c}$ are collinear. Hence $(1-y)y = x^2$ and $(1-x-y)x = (1-x+y)y$, which leads to $y-y^2 = x^2$ and $y+y^2 = x-x^2$. Adding this equations we get $2y = x$ and after the substitution find $x = 2/5$, $y = 1/5$. It is easy to check that point $X(2/5, 1/5)$ satisfy the conditions of the problem.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25576,
    "subject": "Mathematics (Multi-modal)",
    "question": "1. The extension of the median *AM* of the triangle *ABC* intersects its circumcircle at *D*. The circumcircle of the triangle *CMD* intersects the line *AC* at *C* and *E*. The circumcircle of the triangle *AME* intersects the line *AB* at *A* and *F*.\nProve that *CF* is an altitude of the triangle *ABC*.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**1.** It is enough to prove the equalities $FM = BM = MC$ from which it will follow\n![](attached_image_1.png)\nthat $M$ is the midpoint of the hypotenuse of the right triangle $CFB$ and in particular $\\angle CFB = 90^\\circ$.\n\n![](attached_image_1.png)\n\nSince the quadrilateral $ABDC$ is cyclic, it follows that $\\angle ABC = \\angle ADC$.\nSince points $M$, $D$, $E$ and $C$ lie on the same circle, $\\angle MDC = \\angle MEC$.\nFrom the cyclic quadrilateral $AFME$ we obtain the equalities $\\angle BFM = \\angle MEA = \\angle FBM$.\nTherefore the triangle $BFM$ is isosceles with $FM = BM$ and $FM = BM = MC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25577,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A = 2^7(7^{14} + 1) + 2^6 \\cdot 7^{11} \\cdot 10^2 + 2^6 \\cdot 7^7 \\cdot 10^4 + 2^4 \\cdot 7^3 \\cdot 10^6$.\nProve that the number $A$ ends with 14 zeroes.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Answer: 14 zeroes.\n\n*First solution.* Derive the maximum power of 2 as a common factor:\n$$\nA = 2^7 \\cdot (7^{14} + 1 + 2 \\cdot 5^2 \\cdot 7^{11} + 2^3 \\cdot 5^4 \\cdot 7^7 + 2^3 \\cdot 5^6 \\cdot 7^3).\n$$\nNote that $50 = 7^2 + 1$ and represent 50 in this way in all terms:\n$$\n2^{-7}A = 7^{14} + (7^2 + 1) \\cdot 7^{11} + 2 \\cdot (7^2 + 1)^2 \\cdot 7^7 + (7^2 + 1)^3 \\cdot 7^3 + 1.\n$$\nExpand the right-hand side in powers of 7:\n$$\n2^{-7}A = 7^{14} + 7^{13} + 7^{11} + 2 \\cdot 7^{11} + 4 \\cdot 7^9 + 2 \\cdot 7^7 + 7^9 + 3 \\cdot 7^7 + 3 \\cdot 7^5 + 7^3 + 1.\n$$\nOnce again, use equality $7^2 + 1 = 50$, for which derive even degrees of 7 in all terms:\n$$\n2^{-7}A = 7^{14} + 7 \\cdot 7^{12} + 7 \\cdot 7^{10} + 14 \\cdot 7^{10} + 28 \\cdot 7^8 + 14 \\cdot 7^6 + 7 \\cdot 7^8 + 21 \\cdot 7^6 + 21 \\cdot 7^4 + 7 \\cdot 7^2 + 1\n$$\nand expand the right-hand side in even powers of 7:\n$$\n2^{-7}A = 7^{14} + 7 \\cdot 7^{12} + 21 \\cdot 7^{10} + 35 \\cdot 7^8 + 35 \\cdot 7^6 + 21 \\cdot 7^4 + 7 \\cdot 7^2 + 1.\n$$\nNow it is clear that the right-hand side is the Newton binomial formula for $(7^2 + 1)^7$, hence the number $A = 2^7 \\cdot 50^7 = 10^{14}$ ends with 14 zeroes.\n\n\n*Second solution* For any $x_1$ and $x_2$ set $S_k = x_1^k + x_2^k$, $x_1 + x_2 = p$ and $x_1 x_2 = q$ (note that $S_0 = 2$, $S_1 = p$). Using the obvious recurrence formula $S_{k+2} = pS_{k+1} - qS_k$, we successively get $S_2 = p^2 - 2q$, $S_3 = p^3 - 3pq$, $S_4 = p^4 - 4p^2q + 2q^2$, $S_5 = p^5 - 5p^3q + 5pq^3$, $S_6 = p^6 - 6p^4 + 9p^2q^2 - 2q^3$ and, finally,\n$$\nS_7 = p^7 - 7p^5q + 14p^3q^2 - 7pq^3 \\quad (1)\n$$\nNow, take $x_1 = 100$, $x_2 = -2$ then $p = 2 \\cdot 7^2$, $q = -2 \\cdot 10^2$ and (1) takes the form $10^{14} - 2^7 = 2^7 \\cdot 7^{14} + 2^6 \\cdot 7^{11} \\cdot 10^2 + 2^6 \\cdot 7^7 \\cdot 10^4 + 2^4 \\cdot 7^3 \\cdot 10^6$ which is equivalent to $A = 10^{14}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25578,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the minimum prime number $p$ such that $\\{\\sqrt{p}\\} < \\frac{1}{501}$ (where $\\{\\cdot\\}$ stands for the fractional part).",
    "options": [],
    "answer": "65537",
    "solution": "Answer: $\\min p = F_4 = 2^{16} + 1$.\nIt is known that Fermat's number $F_4$ is prime. It is easy to verify that $\\{\\sqrt{F_4}\\} < \\frac{1}{501}$. Indeed, the latter inequality is equivalent to $\\sqrt{2^{16} + 1} - 256 < \\frac{1}{501}$, i.e. $\\frac{1}{\\sqrt{256^2 + 1} + 256} < \\frac{1}{501}$ which follows from $\\frac{1}{2 \\cdot 256} < \\frac{1}{501}$. So, $F_4$ satisfies the condition of the problem.\n\nNow let $p < F_4$; we prove that the inequality\n$$\n\\{\\sqrt{p}\\} < \\frac{1}{501} \\quad (*)\n$$\nis impossible. Indeed, let $n^2 < p < (n+1)^2$, then $p = n^2 + a$ where $1 \\le a \\le 2n$. Then $[\\sqrt{p}] = n$, hence\n$$\n(*) \\Leftrightarrow \\sqrt{n^2 + a} - n < \\frac{1}{501} \\Leftrightarrow n^2 + a < n^2 + \\frac{2n}{501} + \\frac{1}{501^2} \\Leftrightarrow n > \\frac{501}{2}a - \\frac{1}{1002}.\n$$\nIf $a \\ge 2$ then $n \\ge 500 \\Rightarrow p > F_4$ — a contradiction. So, $a = 1$ and $n \\ge 251$, i.e. $p \\ge 251^2+1$. Note that $251^2+1, 251^2+3, 251^2+5$ are even, so are not prime. Further, $252^2+1$ is composite, since it is divisible by 5. Finally, show that $254^2+1$ is divisible by 149. We have the following chain of congruences modulo 149:\n$$\n254^2 + 1 \\equiv (-44)^2 + 1 \\equiv \\frac{132^2 + 9}{9} \\equiv \\frac{17^2 + 9}{9} \\equiv \\frac{298}{9} \\equiv 0 \\pmod{149}.\n$$\nThus $254^2 + 1$ is composite, so $p = 256^2 + 1$ is minimal possible.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25579,
    "subject": "Mathematics (Multi-modal)",
    "question": "For every integer $n \\ge 2$ prove the inequality\n$$\n\\frac{1}{2!} + \\frac{2}{3!} + \\dots + \\frac{2^{n-2}}{n!} \\le \\frac{3}{2},\n$$\nwhere $k! = 1 \\cdot 2 \\cdot \\dots \\cdot k$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Transform the expression on the left side\n$$\n\\begin{aligned}\n\\frac{1}{2!} + \\frac{2}{3!} + \\dots + \\frac{2^{n-2}}{n!} &= \\frac{1}{2!} \\left( 1 + \\frac{2}{3} + \\frac{2^2}{3 \\cdot 4} + \\dots + \\frac{2^{n-2}}{3 \\cdot 4 \\dots n} \\right) \n\\\\\n&\\le \\frac{1}{2} \\left( 1 + \\frac{2}{3} + \\frac{2^2}{3^2} + \\dots + \\frac{2^{n-2}}{3^{n-2}} \\right) \n\\\\\n&= \\frac{1}{2} \\left( 1 + \\frac{2}{3} + \\left( \\frac{2}{3} \\right)^2 + \\dots + \\left( \\frac{2}{3} \\right)^{n-2} \\right) \n\\\\\n&\\le \\frac{1}{2} \\left( 1 + \\frac{2}{3} + \\left( \\frac{2}{3} \\right)^2 + \\dots + \\left( \\frac{2}{3} \\right)^{n-2} + \\dots \\right) \n\\\\\n&= \\frac{1}{2} \\cdot \\frac{1}{1 - \\frac{2}{3}} = \\frac{3}{2}.\n\\end{aligned}\n$$\n\n\nSecond solution:\n\nWe will prove by induction a stronger inequality\n$$\n\\frac{1}{2!} + \\frac{2}{3!} + \\dots + \\frac{2^{n-2}}{n!} \\le \\frac{3}{2} - \\frac{1}{n}, \\quad (*)\n$$\nfrom which, obviously, follows the required. It is easy to check that at $n = 2, 3, 4, 5$ the inequality $(*)$ holds — it is the base of induction.\n\nInduction step: assume that the inequality $(*)$ is true for some $n = k$. To prove $(*)$ for $n = k+1$ it is enough to show that the right side of this inequality increases faster than the left side, i.e. it is enough to prove the inequality\n$$\n\\frac{2^{k-1}}{(k+1)!} \\le \\frac{1}{k} - \\frac{1}{k+1} = \\frac{1}{k(k+1)}.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 25580,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given the isosceles triangle $ABC$ ($CA = CB$). The bisector of the angle $\\angle B$ intersects the side $AC$ at point $L$ and the circumcircle of the triangle $ABC$ at point $D$. It is known that $\\angle C > 60^\\circ$.\nProve that $DC + DL \\le BC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We prove the statement for $\\angle C \\ge 36^\\circ$. Let $\\angle ABC = \\angle BAC = 2x$. We have $DC + DL \\le BC \\Leftrightarrow \\frac{BC}{DC} \\ge 1 + \\frac{DL}{DC}$. Now, by the law of sines,\n$$\n\\frac{BC}{DC} = \\frac{\\sin \\angle BDC}{\\sin \\angle DBC} = \\frac{\\sin 2x}{\\sin x} = 2 \\cos x, \\\\\n\\frac{DL}{DC} = \\frac{\\sin 3x}{\\sin x} = \\frac{1}{3 - 4 \\sin^2 x} = \\frac{1}{4 \\cos^2 x - 1},\n$$\nso we need to prove that $2 \\cos x \\ge 1 + \\frac{1}{4 \\cos^2 x - 1} = \\frac{4 \\cos^2 x}{4 \\cos^2 x - 1}$. Note that $x \\le 36^\\circ$ since $\\angle C \\ge 36^\\circ$, so $\\cos x > \\frac{1}{2}$ and $4 \\cos^2 x - 1 > 0$. Therefore the inequality is equivalent to\n$$\n4 \\cos^2 x - 1 \\ge 2 \\cos x \\Leftrightarrow 4 \\cos^2 x - 2 \\cos x - 1 \\ge 0 \\Leftrightarrow \\cos x \\ge \\frac{-1 + \\sqrt{5}}{4},\n$$\nwhich is true for $x \\le 36^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25581,
    "subject": "Mathematics (Multi-modal)",
    "question": "The altitudes $AA_1$, $BB_1$ and $CC_1$ are drawn in the acute triangle $ABC$. The bisector of the angle $AA_1C$ intersects the segments $CC_1$ and $CA$ at $E$ and $D$ respectively. The bisector of the angle $AA_1B$ intersects the segments $BB_1$ and $BA$ at $F$ and $G$ respectively. The circumcircles of the triangles $FA_1D$ and $EA_1G$ intersect at $A_1$ and $X$.\nProve that $\\angle BXC = 90^\\circ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $\\angle BB_1C = \\angle CC_1B = 90^\\circ$, it is sufficient to prove that the points $B$, $C_1$, $X$ and $B_1$ lie on a circle. Since $A_1D$ and $A_1G$ are the bisectors of the angles $\\angle AA_1C$ and $\\angle AA_1B$ respectively, $\\angle GA_1D = 90^\\circ$. Since $\\angle FB_1D = \\angle FA_1D = 90^\\circ$, the point $B_1$ lies on the circumcircle of the triangle $FA_1D$. Similarly, the point $C_1$ lies on the circumcircle of the triangle $EA_1G$.\n\n![](attached_image_1.png)\n\nTherefore $\\angle C_1XA_1 = \\angle BGA_1 = 135^\\circ - \\angle ABC$ and $\\angle B_1XA_1 = \\angle CDA_1 = 135^\\circ - \\angle ACB$. Hence $\\angle C_1XB_1 = 270^\\circ - \\angle ABC - \\angle ACB = 90^\\circ + \\angle BAC$. Together with $\\angle B_1BA = 90^\\circ - \\angle BAC$, this leads to $\\angle B_1BC_1 + \\angle C_1XB_1 = 180^\\circ$, which means that the quadrilateral $BC_1XB_1$ is cyclic.\n\n![](attached_image_1.png)\n\nTherefore $\\angle C_1XA_1 = \\angle BGA_1 = 135^\\circ - \\angle ABC$ and $\\angle B_1XA_1 = \\angle CDA_1 = 135^\\circ - \\angle ACB$. Hence $\\angle C_1XB_1 = 270^\\circ - \\angle ABC - \\angle ACB = 90^\\circ + \\angle BAC$. Together with $\\angle B_1BA = 90^\\circ - \\angle BAC$, this leads to $\\angle B_1BC_1 + \\angle C_1XB_1 = 180^\\circ$, which means that the quadrilateral $BC_1XB_1$ is cyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25582,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p \\ge 2$ be a prime number. Alice and Bob play the following game: they, in turn, select an index $i$ in the set $\\{0, 1, 2, \\dots, p-1\\}$ that was not selected before by either of the two players and then chooses a digit $a_i$. Alice starts. The game ends after all the indices have been selected. The goal of Alice is to make the number\n$$\nM = a_0 + 10 \\cdot a_1 + 10^2 \\cdot a_2 + \\dots + 10^{p-1} a_{p-1}\n$$\ndivisible by $p$, and the goal of Bob is to prevent this.\nProve that Alice has the winning strategy.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "2. See IMO-2017 Shortlist, Problem N2.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25583,
    "subject": "Mathematics (Multi-modal)",
    "question": "Points $C_1$ and $B_1$ are marked on the sides $AB$ and $AC$ of the triangle $ABC$ respectively. Segments $BB_1$ and $CC_1$ intersect at point $X$, and segments $B_1C_1$ and $AX$ intersect at point $A_1$. The circumcircles of the triangles $BXC_1$ and $CXB_1$ intersect the side $BC$ at points $D$ and $E$ respectively. Lines $B_1D$ and $C_1E$ intersect at point $F$.\nProve that the lines $A_1F$, $B_1E$ and $C_1D$ are either parallel or concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First we prove the equality\n$$\n\\frac{B_1 E}{C_1 D} = \\frac{B_1 A_1}{C_1 A_1}.\n$$\nSince the quadrilateral $XB_1CE$ is cyclic, $\\angle XB_1E = \\angle XCE$. Therefore, the triangle $BB_1E$ is similar to the triangle $BCX$. Hence,\n$$\n\\frac{B_1 E}{X C} = \\frac{B B_1}{B C}. \\qquad (1)\n$$\nSimilarly one can prove the equality\n$$\n\\frac{C_1 D}{X B} = \\frac{C C_1}{B C}. \\qquad (2)\n$$\nHence, dividing equation (1) by equation (2), we get\n$$\n\\frac{B_1 E}{C_1 D} = \\frac{X C}{X B} \\cdot \\frac{B B_1}{C C_1}.\n$$\nApplying Menelaus' theorem to the triangle $XB_1C$ and the line $AB$ we obtain\n$$\n\\frac{X C_1}{C C_1} \\cdot \\frac{A C}{A B_1} \\cdot \\frac{B B_1}{X B} = 1. \\qquad (3)\n$$\nFurther, the same arguments applied to the triangle $CC_1B_1$ and the line $AX$ imply the equality\n$$\n\\frac{C_1 A_1}{B_1 A_1} \\cdot \\frac{A B_1}{A C} \\cdot \\frac{X C}{X C_1} = 1. \\qquad (4)\n$$\nThus, multiplying equalities (3) and (4), we obtain\n$$\n\\frac{B_1 A_1}{C_1 A_1} = \\frac{X C}{X B} \\cdot \\frac{B B_1}{C C_1} = \\frac{B_1 E}{C_1 D}.\n$$\n\n![](attached_image_1.png)\n\n$$\n\\frac{B_1F}{FD} = \\frac{B_1E}{C_1D} = \\frac{B_1A_1}{C_1A_1},\n$$\nso $A_1F \\parallel C_1D$.\nSuppose that the lines $B_1E$ and $C_1D$ intersect at point $Y$. To prove the statement of the problem it is sufficient to show that the lines $YA_1$, $B_1D$ and $C_1E$ intersect at the same point. Let us prove the equality $YD = YE$. Indeed, since the quadrilaterals $XB_1CE$ and $XC_1BD$ are cyclic,\n$$\n\\angle DEY = \\angle CEB_1 = \\angle CXB_1 = \\angle C_1XB = \\angle C_1DB = \\angle EDY.\n$$\nTherefore, the triangle $YDE$ is isosceles. Since\n$$\n\\frac{YE}{B_1E} \\cdot \\frac{B_1A_1}{C_1A_1} \\cdot \\frac{C_1D}{YD} = \\frac{B_1A_1}{C_1A_1} \\cdot \\frac{C_1D}{B_1E} = 1,\n$$\nso Ceva's theorem implies that the lines $YA_1$, $B_1D$ and $C_1E$ intersect at the same point.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25584,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given the parallelogram $ABCD$. The circle $S_1$ passes through the vertex $C$ and touches the sides $BA$ and $AD$ at points $P_1$ and $Q_1$ respectively. The circle $S_2$ passes through the vertex $B$ and touches the sides $DC$ and $AD$ at points $P_2$ and $Q_2$ respectively. Let $d_1, d_2$ be the distances from $C$ and $B$ to the lines $P_1Q_1$ and $P_2Q_2$ respectively.\nFind all possible values of the ratio $d_1 : d_2$.",
    "options": [],
    "answer": "1",
    "solution": "**Answer : 1.**\nWe prove that $d_1 = d_2$. Let $\\angle A = \\angle C = \\alpha$, $\\angle B = \\angle D = -\\alpha$. Choose points $F$ and $T$ on $P_1Q_1$ so that $\\angle P_1FC = \\angle Q_1TC = \\angle D$. Then $CF = CT$. Further, $\\triangle P_1BC \\sim \\triangle Q_1TC$, thus we have $\\frac{BC}{TC} = \\frac{PC}{QC}$. Similarly, $\\triangle Q_1DC \\sim \\triangle P_1FC$, so $\\frac{DC}{FC} = \\frac{QC}{PC}$. Multiplying these equalities we have $BC \\cdot DC = TC^2$. Note that $d_1$ is the length of the altitude in $\\triangle FCT$, so $d_1 = TC \\sin \\angle FTC = TC \\sin \\alpha$, or\n\n$$d_1 = \\sqrt{BC \\cdot DC} \\sin \\alpha.$$ In the same way we obtain that $d_2 = \\sqrt{BC \\cdot DC} \\sin \\alpha = d_1$, as required.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 25585,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine, whether there exists a function $f$ defined on the set of all positive real numbers and taking positive values such that\n$$\nf(x + y) \\geq y f(x) + f(f(x))\n$$\nfor all positive $x$ and $y$?",
    "options": [],
    "answer": "No such function exists.",
    "solution": "Answer: such function doesn't exist.\nSuppose that such a function exists. Putting $x = 1$ in the inequality\n$$\nf(x+y) \\geq y f(x) + f(f(x)) \\quad (1)\n$$\nwe obtain $f(1+y) \\geq a y + b$ where $a = f(1) > 0$, $b = f(f(1)) > 0$. So\n$$\nf(z) \\geq a z + b - a \\quad (2)\n$$\nfor all $z > 1$. From (2) it follows that $f(x) > 1$ for all sufficiently large $x$ (namely when $x \\geq \\frac{1-b+a}{a}$). Putting $x = y = \\frac{z}{2}$ in (1) we get\n$$\nf(z) \\geq \\frac{z}{2} \\cdot f\\left(\\frac{z}{2}\\right) + f\\left(f\\left(\\frac{z}{2}\\right)\\right) > \\frac{z}{2} \\cdot f\\left(\\frac{z}{2}\\right) \\geq \\frac{z}{2} \\left(a \\cdot \\frac{z}{2} + b - a\\right) = \\frac{a}{4} \\cdot z^2 + \\frac{b-a}{2} \\cdot z.\n$$\nTherefore\n$$\nf(z) > \\frac{a}{5} \\cdot z^2 > z \\quad (3)\n$$\nfor all $z$ large enough. Further, we have\n$$\nf(x+y) \\geq y f(x) + f(f(x)) > f(f(x)) > f(x)\n$$\nfor all $x$ large enough. Hence $f(x)$ is an increasing function for $x$ large enough. Therefore, from $f(x+y) > f(f(x))$ it follows that $x+y > f(x)$ for all $y > 0$ and all $x$ large enough. Together with (3) this leads $x+y > \\frac{a}{5} \\cdot x^2$ for all $x$ large enough, which is obviously wrong. The contradiction shows that there is no such function.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25586,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of positive integers $(m, n)$ such that\n$$\n9^m - 7^m = 2^n.\n$$",
    "options": [],
    "answer": "(m, n) = (1, 1) and (2, 5)",
    "solution": "**Answer:** $(m, n) = (1, 1)$, $(m, n) = (2, 5)$.\nIt is easy to see that the pair $(m, n) = (1, 1)$ satisfies the condition. For $m > 1$ consider the equation\n$$\na^m - b^m = (a - b)(a^{m-1} + a^{m-2}b + a^{m-3}b^2 + \\dots + a^{m-1}b^m).\n$$\n\nFor $a = 9$, $b = 7$, the first factor is $2$ and the second is greater than $1$. Since all $m$ summands in the second factor are odd numbers, $m$ be even. Let $m = 2k$, then $9^{2k} - 7^{2k} = (9^k - 7^k)(9^k + 7^k) = 2^n$. Therefore, $9^k - 7^k = 2^x$, $9^k + 7^k = 2^y$ where $x < y$ and $x + y = n$. Subtracting the first equation from the second, we get $2 \\cdot 7^k = 2^x(2^{y-x} - 1)$, whence $x = 1$ and $7^k = 2^{y-1} - 1$. So $7^k + 1$ is divided by $2^{y-1}$. However, if $y \\ge 5$ the remainder of $7^k + 1$ modulo $16$ (as it is easy to check) is equal to $2$ or $8$, i.e. $7^k + 1$ is not divisible by $16$, and therefore there is not such positive integers $k$ for which the pair $(m, n) = (2k, n)$ satisfy the condition $n = y + 1 \\ge 5 + 1 = 6$.\nSo $n \\le 5$ and $y = n - x = n - 1 \\le 4$, whence $7^k = 2^{y-1} - 1 \\le 7$, and thus $k \\le 1$. It is easy to see that $m = 2k = 2 \\cdot 1 = 2$ and $n = 5$ satisfy the condition. Therefore there are only two pairs $(m, n) = (1, 1)$, $(m, n) = (2, 5)$ satisfying the problem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25587,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $q \\in \\mathbb{R}$. There are 10 distinct real numbers on the blackboard. Alex writes the following three lines of numbers:\n\n1. In the first line Alex writes down every number of the form $a-b$, where $a, b$ are two (not necessarily distinct) numbers from the board;\n\n2. In the second line Alex writes down every number of the form $qab$, where $a, b$ are two (not necessarily distinct) numbers from the first line;\n\n3. In the third line Alex writes down every number of the form $a^2 + b^2 - c^2 - d^2$, where $a, b, c, d$ are four (not necessarily distinct) numbers from the first line.\n\nDetermine all values of $q$ such that, regardless of the numbers on the board, every number in the second line is also a number in the third line.\n\n(IMO-2017 Shortlist, Problem A2)",
    "options": [],
    "answer": "q = 2 or q = -2",
    "solution": "2. See IMO-2017 Shortlist, Problem A2.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25588,
    "subject": "Mathematics (Multi-modal)",
    "question": "We call a coloring of an $m \\times n$ table ($m, n \\ge 5$) in three colors a *good coloring* if the following two conditions are satisfied:\n\n1) Each cell has the same number of neighboring cells of two other colors;\n2) Each corner cell has no neighboring cells of its color.\n\nFind all pairs $(m, n)$ ($m, n \\ge 5$) for which there exists a good coloring of $m \\times n$ table.",
    "options": [],
    "answer": "All pairs where both dimensions are multiples of six.",
    "solution": "Answer: all $(m, n)$ such that $6$ divides both of them.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 25589,
    "subject": "Mathematics (Multi-modal)",
    "question": "For all pairs of $(m, n)$ positive integers that have the same number $k$ of divisors we define the operation $\\circ$. Write all their divisors in an ascending order: $1 = m_1 < \\dots < m_k = m$, $1 = n_1 < \\dots < n_k = n$ and set\n$$\nm \\circ n = m_1 \\cdot n_1 + \\dots + m_k \\cdot n_k.\n$$\nFind all pairs of numbers $(m, n)$, $m \\ge n$, such that $m \\circ n = 497$.",
    "options": [],
    "answer": "(20, 18)",
    "solution": "From the definition of $\\circ$ it is clear that $m \\circ n \\ge 1 + n^2$, so $n \\le \\sqrt{496} < 23$. Note that the minimal number that has at least 7 divisors is 24, hence it is sufficient to consider only $k$ from 1 to 6.\n\nExactly two divisors have only prime numbers, so for $k=2$ the numbers $m$ and $n$ are prime. From $m \\circ n = 497$ it follows that $mn = 496$. But $496 = 2^4 \\cdot 31$ doesn't equal to the product of two prime numbers.\n\nExactly three divisors have only squares of prime numbers. Let $m = p^2$, $n = q^2$, then $m \\cdot n = 1 + pq + (pq)^2 = 497$ and $pq(pq + 1) = 496 = 2^4 \\cdot 31$, hence $p = 31$, $q = 2$ which doesn't suit.\n\nConsider the case $k=4$. Since $1 < m_2 < m_3 < m$ are all divisors of $m$, then $m = m_2m_3$, similarly, $n_2n_3 = n$. Factor the number $m \\circ n$:\n$$\nm \\circ n = 1 + m_2n_2 + m_3n_3 + (m_2m_3)(n_2n_3) = (1 + m_2n_2)(1 + m_3n_3).\n$$\nSince $497 = 7 \\cdot 71$, then $m_2n_2 = 6$ and $m_3n_3 = 70$. Numbers $m_2$ and $n_2$ are prime, therefore one of them is equal to 2, the other is equal to 3, and each of the numbers $m_3$ and $n_3$ is either prime or a square of $m_2$ and respectively $n_2$. It is easy to check that for $70 = 2 \\cdot 5 \\cdot 7$ none of the options fits.\n\nExactly five divisors have only the fourth powers of prime numbers. Up to 23 there is exactly one such number $-n = 2^4$. Note that\n$$\n2^4 \\circ 2^4 = (4^5 - 1)/3 = 341 < 497, \\quad \\text{a} \\quad 3^4 \\circ 2^4 = (6^5 - 1)/5 = 1555 > 497.\n$$\nFor larger values of $m$ there certainly is no solutions.\n\nExactly six divisors have only numbers $p^5$ and $pq^2$, where $p$ and $q$ are prime. Up to 23 there are only three such numbers: 12, 18 and 20, consider them one by one.\n\nLet $n = 12$. Among the divisors of 12 there are exactly two odd: $n_1 = 1$ and $n_3 = 3$. The number 497 is odd too, hence $m_2 = 2$ and $m_3 = 4$. Therefore, $m$ has the form $2^5$ or $4p$, where $p$ is prime and greater than 3. $m = 2^5$ is impossible since $12 \\cdot 32 = 529$. In the case $m = 4p$ we get the equation $17+64p = 497$, which has no integer solutions.\n\nLet $n = 18$. From $18 \\circ m > 18 \\cdot m + 9 \\cdot 5 + 6 \\cdot 4 + 3 \\cdot 3 + 2 \\cdot 2 + 1 \\cdot 1$, follows the inequality $18 \\le m < 414/18 = 23$. Considering the two options $m = 18$ and $m = 20$, we obtain the only solution (20, 18).\n\nSuppose $n = 20$. Form $20 \\circ m > 20m + 10 + 5 + 4 + 2 + 1$ follows the inequality $20 \\le m < 475/20 = 23.75$. But $20 \\circ 20 > 500$, hence there is no solutions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25590,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given the hyperbola $y = \\frac{1}{x}$ and four circles $S_1, S_2, S_3, S_4$. The circle $S_1$ intersects the hyperbola at points $Z_1, Z_2, Z_3, Z_4$; $S_2$ intersects the hyperbola at points $Z_3, Z_4, Z_5, Z_6$; $S_3$ intersects the hyperbola at points $Z_5, Z_6, Z_7, Z_8$; $S_4$ intersects the hyperbola at points $Z_7, Z_8, Z_1, Z_2$. The radii of $S_1, S_2, S_3$ are equal to $R_1, R_2, R_3$.\nFind the radius of $S_4$.",
    "options": [],
    "answer": "sqrt(R1^2 + R3^2 - R2^2)",
    "solution": "Answer: $\\sqrt{R_1^2 + R_3^2 - R_2^2}$.\n\nFirst we prove the following\n\n**Lemma.** Let $x_1, x_2, x_3, x_4$ be abscissae of the intersection points of a circle with the hyperbola $y = \\frac{1}{x}$. Then\n$$\nR^2 = \\frac{1}{4}(x_1^2 + x_2^2 + x_3^2 + x_4^2 + \\frac{1}{x_1^2} + \\frac{1}{x_2^2} + \\frac{1}{x_3^2} + \\frac{1}{x_4^2}). \\quad (1)\n$$\n**Proof.** Let $(x-a)^2 + (x-b)^2 = R^2$ be the equation of $S$. Then $x_i$ are the roots of the equation $x^4 - 2a x^3 + (a^2 + b^2 - R^2)x^2 - 2b x + 1 = 0$. From Viète's formulas it follows that $\\sum x_i = 2a$, $\\prod x_i = 1$, $\\sum x_i x_j = a^2 + b^2 - R^2 = \\sum \\frac{1}{x_i x_j}$, $\\sum x_i x_j x_k = \\sum \\frac{1}{x_k} = 2b$. From this system of equalities we can easily express $R^2$ and obtain (1).\n\n---\n\nNow let $Z_1, Z_2, Z_3, Z_4$ and $Z_5, Z_6, Z_7, Z_8$ be the abscissae of the intersection points of the hyperbola with $S_1$ and $S_3$ respectively. Then $R_1^2 + R_3^2 = \\frac{1}{4}(\\sum_{i=1}^8 Z_i^2 + \\sum_{i=1}^8 \\frac{1}{Z_i^2}) = R_2^2 + R_4^2$ whence follows the answer.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25591,
    "subject": "Mathematics (Multi-modal)",
    "question": "The bisector of the angle $CAB$ of triangle $ABC$ intersects the side $CB$ at $L$. The point $D$ is the foot of the perpendicular from $C$ to $AL$ and the point $E$ is the foot of the perpendicular from $L$ to $AB$. The lines $CB$ and $DE$ meet at $F$.\nProve that $AF$ is an altitude of the triangle $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $F'$ be the foot of an altitude from $A$ in the triangle $ABC$.\n\n![](attached_image_1.png)\n\nIt is enough to prove that the points $D$, $F'$ and $E$ are colinear, since it will lead to $F = F'$ which means that $AF$ is an altitude of the triangle $ABC$. Since $\\angle CDA = \\angle CF'A = 90^\\circ$, the quadrilateral $CDF'A$ is cyclic and therefore $\\angle CF'D = \\angle CAD$. Since $\\angle AF'L = \\angle AEL = 90^\\circ$, the quadrilateral $AEF'L$ is cyclic and hence $\\angle BEF' = \\angle LAE$. Since $\\angle CAL = \\angle LAB$, then $\\angle CF'D = \\angle BF'E$ which means that $D$, $F'$ and $E$ are colinear.\nLet the lines $EL$ and $CD$ meet at $K$. Since the angles $ADK$ and $AEK$ are right, the points $A$, $K$, $D$ and $E$ lie on the circle with diameter $AK$. Hence $\\angle DKE = \\angle DAE$. Since $AD$ is the bisector, $\\angle DAE = \\angle DAC$, whence $\\angle DKE = \\angle DAC$. The latter is equivalent to $\\angle DKL = \\angle CAL$, whence $A$, $L$, $C$ and $K$ lie on a circle.\nConsider the Simson line of $A$ and the triangle $LCK$. The foot of the perpendicular from $A$ to $CK$ is $D$, and to $LK$ is $E$, therefore this line is $DE$. Since $DE$ meet $CL$ at $F$, $F$ is the foot of the perpendicular from $A$ to $CL$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25592,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a convex quadrilateral $ABCD$. The point $A_1$ is on the boundary of $ABCD$ such that the segment $AA_1$ divides $ABCD$ into two parts with equal areas. In the same way we define points $B_1$, $C_1$ and $D_1$. It is known that the lengths of all segments $AA_1$, $BB_1$, $CC_1$, and $DD_1$ do not exceed $1$.\nProve that the area $S(ABCD) < \\frac{2}{3}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $O = AC \\cap BD$; and, without loss of generality, $BO \\geq DO$, $CO \\geq AO$. If $AB \\parallel DC$ then $1 \\leq \\frac{BO}{DO} = \\frac{AO}{CO} \\leq 1$, whence $BO = OD$ and $ABCD$ is a parallelogram. Then by the problem condition $AC \\leq 1$, $BD \\leq 1$, hence $S(ABCD) \\leq \\frac{1}{2} AC \\cdot BD \\leq \\frac{1}{2} < \\frac{2}{3}$.\n\nLet now $AB \\nparallel DC$, and let $M$ be the intersection point of the rays $BA$ and $CD$; denote $BC = b$, $\\angle ABC = \\beta$, $\\angle DCB = \\gamma$. Then $S = S(ABCD) < S(BMC)$, while\n$$\nS(BMC) = \\frac{1}{2} MC \\cdot MB \\sin(\\beta + \\gamma) = \\frac{1}{2} \\cdot \\frac{b \\sin \\beta}{\\sin(\\beta + \\gamma)} \\cdot \\frac{b \\sin \\gamma}{\\sin(\\beta + \\gamma)} \\cdot \\sin(\\beta + \\gamma) = \\\\\n= \\frac{1}{2} \\cdot \\frac{b^2}{\\operatorname{ctg} \\beta + \\operatorname{ctg} \\gamma},\n$$\nhence\n$$\nS \\cdot (\\operatorname{ctg} \\beta + \\operatorname{ctg} \\gamma) < \\frac{b^2}{2}. \\qquad (1)\n$$\nIf $A_1 \\in [AB]$ and $D_1 \\in [CD]$ are the points such that $S(BA_1C) = \\frac{1}{2}S = S(BD_1C)$ then by the problem condition $CA_1 \\leq 1$, $BD_1 \\leq 1$. Thus\n$$\nCA_1^2 = b^2 + \\frac{S^2}{b^2 \\sin^2 \\beta} - 2S \\cdot \\frac{\\cos \\beta}{\\sin \\beta} = \\left(b - \\frac{S}{b} \\operatorname{ctg} \\beta\\right)^2 + \\frac{S^2}{b^2} \\leq 1,\n$$\nand, similarly, $\\left(b - \\frac{S}{b} \\operatorname{ctg} \\gamma\\right)^2 + \\frac{S^2}{b^2} \\leq 1$. Then, in view of (1),\n$$\n2 \\geq \\frac{2S^2}{b^2} + \\frac{1}{2}\\left(2b - \\frac{S}{b}\\left(\\operatorname{ctg} \\beta + \\operatorname{ctg} \\gamma\\right)\\right)^2 > \\frac{2S^2}{b^2} + \\frac{1}{2}\\left(2b - \\frac{b}{2}\\right)^2 = \\\\\n= \\frac{2S^2}{b^2} + \\frac{9b^2}{8} \\geq \\frac{1}{2}\\sqrt{\\frac{2S^2}{b} \\cdot \\frac{9b^2}{8}} = 3S.\n$$\nIt follows that $S < \\frac{2}{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25593,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a cyclic quadrilateral with the circumcircle $\\omega$. The points $B_1$ and $D_1$ are symmetric to $A$ with respect to the midpoints of $BC$ and $CD$. The circumcircle of the triangle $CB_1D_1$ intersects $\\omega$ at $C$ and $G$.\nProve that $AG$ is the diameter of $\\omega$.\n(A. Voidelevich)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $B_1$ is symmetric to $A$ with respect to the midpoint of $BC$, $ABB_1C$ is a parallelogram, so $BB_1 \\parallel AC$ and $BB_1 = AC$. Likewise $ACD_1D$ is a parallelogram, $DD_1 \\parallel AC$ and $DD_1 = AC$. Therefore the triangle $CD_1B_1$ is obtained from the triangle $ADB$ by the transfer along the line $AC$ by the length of the segment $AC$. Let $O$ be the center of $\\omega$ and $O_1$ be the circumcenter of the triangle $CD_1B_1$. Then $OO_1 \\parallel AC$. Since the line connecting the centers of two intersecting circles is perpendicular to their common chord, $OO_1 \\perp CG$ which means $AC \\perp CG$. Therefore $\\angle ACG = 90^\\circ$, i. e. $AG$ is a diameter of $\\omega$.\n\n![](attached_image_1.png)\n\nSecond solution:\n\nLet $K$, $L$ and $M$ be the midpoints of the sides $BC$, $CD$ and $CA$ respectively. Since the triangle $KLM$ is obtained from the triangle $BDA$ by a homothety with the center $C$ and the coefficient $1/2$, its circumcircle $\\omega_1$ is obtained from $\\omega$ in the same way: $H_C^{0.5}(\\omega) = \\omega_1$. The center $C$ of the homothety lies on $\\omega$, so it lies on $\\omega_1$ as well. Since the triangle $B_1D_1C$ is obtained from the triangle $KLM$ by a homothety with the center $A$ and the coefficient $2$, its circumcircle $\\omega_2$ is obtained from $\\omega_1$ in the same way: $H_A^2(\\omega_1) = \\omega_2$.\nConsider the intersection points of $\\omega$ and $\\omega_2$, namely, how they are obtained under the composition $H_A^2 \\circ H_C^{0.5}$, translating $\\omega$ into $\\omega_2$. One of them, the point $C$ is obtained from the point $A$, indeed: $H_A^2(H_C^{0.5}(A)) = H_A^2(M) = C$. The second point is obtained from the point $E$, lying on $\\omega$ diametrically opposite to $C$ since $O = H_C^{0.5}(E) \\in \\omega_1$ and $H_A^2(O) \\in \\omega_2$. Whence $H_A^2(O) = G$, so $AG$ is a diameter.\n\nthe inverse transformation: from $\\{C, G\\} = \\omega \\cap H_A^2(\\omega_1)$, it follows that\n$$\n\\{H_A^{0.5}(C), H_A^{0.5}(G)\\} = H_A^{0.5}(\\omega) \\cap \\omega_1 = H_A^{0.5}(\\omega) \\cap H_C^{0.5}(\\omega).\n$$\nNow, the intersection points of $H_A^{0.5}(\\omega)$ and $H_C^{0.5}(\\omega)$ are obviously the middle $M$ of the segment $AC$ and the center $O$ of the circle $\\omega$. It remains to make the transformation $H_C^2(\\omega)$ and get the desired points $A$ and $E$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25594,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given $n \\ge 2$. We call a group of people *n-compact* if for every person of group one can find *n* people (different from that person) which are acquainted with each other.\nFind the maximum possible $N$ such that every *n*-compact group of $N$ people contains a subgroup of $n + 1$ people acquainted with each other.",
    "options": [],
    "answer": "2n",
    "solution": "3. See III Silk Road Math. Competition 2004, Problem 4.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 25595,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a fixed integer $n \\ge 2$ consider the sequence\n$$\na_k = \\text{lcm}(k, k+1, \\dots, k+(n-1)).\n$$\nFind all integers $n \\ge 2$ for which the sequence $a_k$ increases starting from some number.",
    "options": [],
    "answer": "n = 2",
    "solution": "Answer: $n = 2$.\n\nNote that if $n = 2$, the sequence has the form $a_k = k(k+1)$ since consecutive numbers are always coprime. It is clear that $k(k+1) < (k+1)(k+2)$, so the sequence is increasing.\n\n*First solution.* Let us show that if $n \\geq 3$ the sequence is not increasing from any number. Choose $k = np$ where $p$ is an arbitrary prime number greater than $n$. All numbers $np+1, np+2, \\dots, np+n-1$ are not divisible by $p$ and there is at least one even among them. The number $n(p+1)$ is also not divisible by $p$ and $p+1$ is even. Hence\n$$\na_k = p \\cdot \\text{lcm}(n, np+1, np+2, \\dots, np+n-1), \\\\\na_{k+1} \\le \\frac{p+1}{2} \\cdot \\text{lcm}(n, np+1, np+2, \\dots, np+n-1).\n$$\nTherefore $a_k > a_{k+1}$ for $k$ large enough and the sequence $a_k$ is not increasing from any number.\n\n\n*Second solution.* We will show how else one can prove that for $n \\geq 3$ the sequence is not increasing from any moment. Suppose that $a_k$ is increasing from number $k_0$ then for all $k \\geq k_0$ holds $a_{k+1} \\geq a_k$. Consider $k = m! - n$ where $m > \\max(n! + n, k_0)$.\n\nDenote $\\text{lcm}(k+1, \\dots, k+n-1)$ by $N$. Then the inequality $a_{k+1} \\ge a_k$ is equivalent to $\\text{lcm}(N, m!) > \\text{lcm}(m! - n, N)$. Using the well known equality $\\text{lcm}(a, b) \\cdot \\text{gcd}(a, b) = a \\cdot b$ we obtain the equivalence:\n$$\n\\text{lcm}(N, m!) > \\text{lcm}(m! - n, N) \\iff \\frac{N \\cdot m!}{\\text{gcd}(N, m!)} > \\frac{(m! - n) \\cdot N}{\\text{gcd}(m! - n, N)},\n$$\nor, after equivalent transformations,\n$$\nn \\cdot \\text{gcd}(N, m!) > m! \\cdot (\\text{gcd}(N, m!) - \\text{gcd}(m! - n, N)). \\quad (1)\n$$\nIt is clear that $\\text{gcd}(m!, m! - l) = \\text{gcd}(m!, l) = l$ for all $l$ from $1$ to $n-1$, hence $(n-1)! \\ge \\text{gcd}(N, m!) \\ge \\text{lcm}(1, 2, \\dots, n-1)$. Moreover, since $m!$ is divisible by $l$, the number $\\text{gcd}(m! - n, m! - n + l) = \\text{gcd}(n, n-l)$ is a divisor of $n$. So the number $\\text{gcd}(m! - n, N)$ is a divisor of $n$ too. Indeed $\\nu_p(N) = \\max(\\nu_p(m! - n + 1), \\dots, \\nu_p(m! - n + (n-1)))$ for any prime divisor $p$ of $N$.\nFrom obtained inequalities and (1) it follows that\n$$\nn \\cdot (n-1)! > m! \\cdot (\\text{lcm}(1, 2, \\dots, n-1) - n). \\quad (2)\n$$\nSince $\\text{lcm}(1, 2, \\dots, n-1) \\ge (n-2)(n-1)$, for $n \\ge 4$ the right hand side of (2) is not less than $m!$ whence $n! > m!$, a contradiction.\nIf $n=3$ then $\\text{gcd}(N, m!) = \\text{gcd}(m! - 2, m!) = 2$. But\n$$\n\\text{gcd}(m! - n, N) = \\text{gcd}(m! - 3, (m! - 2)(m! - 1)) = 1,\n$$\nso the inequality (1) will become $6 > m!$ which is wrong.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25596,
    "subject": "Mathematics (Multi-modal)",
    "question": "A checkered polygon $A$ is drawn on the checkered plane. We call a cell of $A$ *internal* if all 8 of its adjacent cells belong to $A$. All other (non-internal) cells of $A$ we call *boundary*. It is known that 1) each boundary cell has exactly two common sides with other boundary cells; and 2) the union of all boundary cells can be divided into isosceles trapezoids of area 2 with vertices at the grid nodes (and acute angles of the trapezoids are equal to $45^\\circ$).\nProve that the area of the polygon $A$ is congruent to $1$ modulo $4$.\n( A. Yuran )",
    "options": [],
    "answer": "Detailed solution",
    "solution": "It is clear that the set of boundary cells of the polygon $A$ can be divided into trapezoids so that you can go from any of them to another, going only through the sides of the trapezoids. We call a polygon satisfying this condition *good*. We will prove by induction on $k \\ge 2$ the following statement: if the area of a good polygon is $S \\le 4k + 1$, then $S \\equiv 1 \\pmod{4}$.\n\nThe induction base is $k=2$. It is easy to see that there are at least $4$ trapezoids, and if there are exactly $4$ of them the polygon is a square $3 \\times 3$. If there are at least $5$ trapezoids, then the area $S \\ge 2 \\cdot 5 > 2 \\cdot 4 + 1$.\n\nThe step of the induction. Suppose that the statement is true for $k = n-1 \\ge 2$. Let us prove it for $k = n$, i.e. for any good polygon with the area $S \\in [4(n-1) + 2; 4n+1]$.\n\nConsider any side $AB$ of a good polygon, both adjacent angles to which are equal to $90^\\circ$. Such side always exists, for example the lowest horizontal side of a polygon.\nLet $AB = 3$ and the polygon is not a square $3 \\times 3$. Then two trapezoids adjacent to $AB$ are restored uniquely and it is possible to cut off the rectangle $3 \\times 4$ from the polygon and get a good one again. It is clear that with such a cut, the remainder of dividing $S$ by $4$ will not change.\n\n![](attached_image_1.png)\n\nNow let's assume that the good polygon has no sides of length $3$ (If the side of the polygon is $3$, then clearly both of its adjacent angles are $90^\\circ$). Then to the neighboring sides there are at least $2$ adjacent trapezoids (otherwise their length is $3$).\nLet $AB = l$. If the rectangle $CABD$ (see the fig.) of the size $l \\times 4$ has no border cells except those that have common points with the polyline $CABD$, then this polygon can be cut off, and we get a good polygon again by moving the trapezoids adjacent to $AB$ by $4$ cells up (similar to the case $AB = 3$).\n\n![](attached_image_2.png)\n\nSuppose that the rectangle $CABD$ contains other boundary cells.\nColor some cells of the plane in $2$ colors, as in the figure. Then all boundary cells of the form $\\square$ have color $1$, of the form $\\square$ have color $2$ (since from each cell of the first type the cell of the second is located through one). Consider all the boundary cells inside the rectangle $CABD$ and choose the lower one that has no common points with the polyline. Then some adjacent cell to it on the left or right should be the boundary cell, otherwise the border will not split into trapezoids (if this cell is colored, it contains part of the trapezoid with horizontal bases, otherwise the cell is not lower, and if not, the trapezoid containing this cell cannot have vertical bases for the same reasons).\nOf the two adjacent boundary cells at least one is crossed by the diagonal of the trapezoid. Without loss of generality, we assume that it has the form $\\square$ (it is dashed in the figure). Then it has color $1$ and is located two cells higher than the color $2$ cell of the $\\square$ type adjacent to $AB$, otherwise it cannot be located inside $CABD$ which is true for any of its inner color $1$ cells. Then it is clear that the part of this cell is contained in the trapezoid, located to the right of it.\n\n![](attached_image_3.png)\n\nNow we will make cuts, as shown in the figure, thus cutting the polygon into $2$ parts. The area of the polygon is $(4a+1) + (4b+1) + 3 \\equiv 1 \\pmod{4}$, as required.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 25597,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A_1H_1$, $A_2H_2$, $A_3H_3$ be altitudes and $A_1L_1$, $A_2L_2$, $A_3L_3$ be bisectors of an acute angled triangle $A_1A_2A_3$.\nProve the inequality $S(L_1L_2L_3) \\ge S(H_1H_2H_3)$ where $S$ stands for the area of a triangle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "One can prove the following\n**Lemma.** If $\\alpha$, $\\beta$, $\\gamma$ are angles of a triangle then\n$$\n\\cos \\alpha + \\cos \\beta + \\cos \\gamma + \\cos 2\\alpha + \\cos 2\\beta + \\cos 2\\gamma \\ge 0.\n$$\nNow, if $a$, $b$, $c$ are the side lengths of the triangle, $\\alpha$, $\\beta$, $\\gamma$ are corresponding angles and $S$ is the area of this triangle, then we have\n$$\nS(H_1H_2H_3) = S \\cdot (1 - \\cos^2 \\alpha - \\cos^2 \\beta - \\cos^2 \\gamma) = \\frac{1}{2} S \\cdot (-1 - \\cos 2\\alpha - \\cos 2\\beta - \\cos 2\\gamma);\n$$\nand\n$$\nS(L_1L_2L_3) = S \\cdot \\frac{2abc}{(a+b)(b+c)(c+a)} = S \\cdot \\frac{2 \\sin \\alpha \\sin \\beta \\sin \\gamma}{\\prod_{cycl} (\\sin \\alpha + \\sin \\beta)} = S \\cdot \\frac{\\sin \\alpha \\sin \\beta \\sin \\gamma}{4 \\prod_{cycl} (\\cos \\frac{\\gamma}{2} \\cos \\frac{\\alpha-\\beta}{2})}.\n$$\nFurther,\n$$\n\\begin{aligned}\n\\cos \\alpha + \\cos \\beta + (\\cos \\gamma - 1) &= 2 \\cos \\frac{\\alpha + \\beta}{2} \\cos \\frac{\\alpha - \\beta}{2} - 2 \\sin^2 \\frac{\\gamma}{2} = \\\\\n&= 2 \\sin \\frac{\\gamma}{2} (\\cos \\frac{\\alpha - \\beta}{2} - \\cos \\frac{\\alpha + \\beta}{2}) = 4 \\sin \\frac{\\alpha}{2} \\sin \\frac{\\beta}{2} \\sin \\frac{\\gamma}{2}.\n\\end{aligned}\n$$\nSo\n$$\n\\begin{aligned}\nS(L_1L_2L_3) &= \\frac{1}{2} S \\cdot \\frac{\\cos \\alpha + \\cos \\beta + (\\cos \\gamma - 1)}{\\cos \\frac{\\alpha-\\beta}{2} \\cos \\frac{\\beta-\\gamma}{2} \\cos \\frac{\\gamma-\\alpha}{2}} \\ge \\frac{1}{2} S \\cdot (\\cos \\alpha + \\cos \\beta + (\\cos \\gamma - 1)) \\ge \\\\\n&\\ge \\frac{1}{2} S \\cdot (-1 - \\cos 2\\alpha - \\cos 2\\beta - \\cos 2\\gamma) = S(H_1H_2H_3),\n\\end{aligned}\n$$\nas required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25598,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sum of several positive numbers from $(0, 1]$ is equal to $S$. It is known that one with the guarantee can divide all given numbers into two groups such that the sum of numbers in the first group does not exceed $1$ and the sum of numbers in the second group does not exceed $5$.\nFind the maximum possible value of $S$.",
    "options": [],
    "answer": "5.5",
    "solution": "Answer: $5.5$.\n\nWe first show that $S \\le 5.5$ (obviously, $S \\le 6$). Suppose that $S > 5.5$. Then we may write $S = 5.5 + 19x$, where $0 < x \\le \\frac{1}{38}$. It can occur that we have $11$ numbers, $10$ of which are equal to $0.5 + 2x$ and the last one is equal to $0.5 - x$. It is clear that it is impossible to divide such numbers into two groups as required in the problem conditions.\n\nNow prove that any $S \\le 5.5$ satisfies the problem conditions. There is nothing to prove if $S \\le 5$. So we may assume that $5 < S \\le 5.5$. Then among the given numbers there are several numbers with the sum $a < 5$ and some number $t$ such that $a + t \\ge 5$. Denote by $b$ the sum of all remaining numbers, $a + t + b = S \\le 5.5$.\n\nWe have $b = S - (a + t) \\le 0.5$.\n\nIf $a \\ge 4.5$ then $t + b \\le 1$, so we can include in the first group all numbers with sum $a$, and to the second group include all remaining numbers with sum $t + b$.\n\nFinally, if $a < 4.5$ then $S - t = a + b < 4.5 + 0.5 = 5$, hence we can include $t$ in the first group and all other numbers in the second group.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 25599,
    "subject": "Mathematics (Multi-modal)",
    "question": "Three $n \\times n$ squares form the figure $\\Phi$ on the checkered plane as\n\n![](attached_image_1.png)\n\nshown on the picture. (Neighboring squares are touching along the segment of length $1$).\nFind all $n > 1$ for which the figure $\\Phi$ can be covered with tiles $1 \\times 3$ and $3 \\times 1$ without overlapping.",
    "options": [],
    "answer": "n = 3k or n = 3k + 1 for integers k ≥ 1",
    "solution": "Answer: $n = 3k$, $n = 3k + 1$.\nIt is clear that any rectangle with one of the sides divided by $3$ can be covered with tiles. Therefore, if $n = 3k$, each of the three squares and the entire figure can be tiled.\n\nWe prove that $\\Phi$ can be covered for any $n$ of the form $3k+1$. Put two horizontal tiles so that each of them closes two cells in the central and one in the side square. Then each of the side squares can be divided into rectangles $(3k+1) \\times 3k$ and $3k \\times 1$ each of which can be tiled. The unoccupied part of the central square can be divided into rectangles of sizes $3k \\times 2$, $(3k+1) \\times (3k-3)$ and $2 \\times 3k$, each of which can be tiled.\n\nSuppose that for some $n = 3k+2$ the figure $\\Phi$ was covered with tiles. Since each square contains $n^2 = 3(3k^2 + 4k + 1) + 1$ cells, and two adjacent squares can have only one tile in common (intersecting both of them), the tiles entirely contained in the upper square cover it all, except for the right lower cell only. Paint the cells of the upper square in three colors as shown in the figure. It is easy to see that each tile covers one square of each color so the number of cells is the same for each color. But the number of cells of color $2$ is one more than the number of cells of color $1$, a contradiction.\n\n|1|2|3|1|2|3|1|2|\n|2|3|1|2|3|1|2|3|\n|3|1|2|3|1|2|3|1|\n|1|2|3|1|2|3|1|2|\n|2|3|1|2|3|1|2|3|\n|3|1|2|3|1|2|3|1|\n|1|2|3|1|2|3|1|2|\n|2|3|1|2|3|1|2| |",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25600,
    "subject": "Mathematics (Multi-modal)",
    "question": "Some cells of a checkered plane are marked so that the figure $A$ formed by marked cells satisfies the following two conditions:\n1) any cell of the figure $A$ has exactly two adjacent cells of $A$; and\n2) the figure $A$ can be divided into isosceles trapezoids of area $2$ with vertices at the grid nodes (and acute angles of the trapezoids are equal to $45^\\circ$).\n\nProve that the number of marked cells is divisible by $8$.\n\n( A. Yuran )",
    "options": [],
    "answer": "Detailed solution",
    "solution": "It is clear that it is enough to prove the statement of the problem for *connected* figures $A$, since any disconnected figure is divided into several connected parts, each of which (as we prove) has a number of cells divided by $8$. Let's color some cells of the plane into four colors as shown in the figure.\n\nWithout loss of generality, assume that some of the trapezoids is placed as shown in the figure (otherwise, rotate and transfer the coloring). Let's start to bypass the cells of the figure $A$ starting with this trapezoid so that we move on from the cell of color $1$ to the cell of color $2$ (we call such trapezoid a *trapezoid* $1 \\to 2$). It is easy to see that with this bypass types of trapezoids alternate in the order $1 \\to 2, 2 \\to 3, 3 \\to 4, 4 \\to 1$. Since at some point we will finish the bypass in the cell of color $1$, from which we started, the number of trapezoids passed is divided by $4$, which means that the total number of cells of the figure $A$ is divided by $8$.\n\n![](attached_image_1.png)",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25601,
    "subject": "Mathematics (Multi-modal)",
    "question": "The circle $S_1$ intersects the hyperbola $y = \\frac{1}{x}$ at four points $A$, $B$, $C$ and $D$, and the other circle $S_2$ intersects the same hyperbola at four points $A$, $B$, $F$ and $G$. It is known that the radii of circles $S_1$ and $S_2$ are equal.\nProve that the points $C$, $D$, $F$ and $G$ are the vertices of the parallelogram.\n( I. Voronovich )",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the abscissae of the points $A$, $B$, $C$, $D$, $F$ and $G$ be $a$, $b$, $c$, $d$, $f$ and $g$ respectively. If $O_1(\\alpha, \\beta)$ is the center of the circle $S_1$, the coordinates of the points $A$, $B$, $C$ and $D$ satisfy the system of equations\n$$\n\\begin{cases} (x - \\alpha)^2 + (y - \\beta)^2 = R^2, \\\\ y = \\frac{1}{x}. \\end{cases}\n$$\nEliminating $y$ we obtain the equation\n$$\nx^4 - 2\\alpha x^3 + (\\alpha^2 + \\beta^2 - R^2)x^2 - 2\\beta x + 1 = 0,\n$$\nwith solutions $a$, $b$, $c$ and $d$. Whence $a + b + c + d = 2\\alpha$ and $abcd = 1$. Similarly, if $O_2(\\lambda, \\mu)$ is the center of the circle $S_2$, then $a + b + f + g = 2\\lambda$ and $abfg = 1$. Since the radii of the circles $S_1$ and $S_2$ are equal, the quadrilateral $AO_1BO_2$ is a rhombus, hence the midpoints of $AB$ and $O_1O_2$ coincide, therefore $\\frac{a + b}{2} = \\frac{\\alpha + \\lambda}{2}$ which gives $a + b = \\alpha + \\lambda$. So\n$$\n2\\alpha + 2\\lambda = (a + b + c + d) + (a + b + f + g) = 2\\alpha + 2\\lambda + c + d + f + g,\n$$\nwhence $f + g = -(c + d)$. Moreover, $cd = fg = (ab)^{-1}$. Rewrite it as $f + g = (-c) + (-d)$ and $f \\cdot g = (-c) \\cdot (-d)$. Thus the pairs $(f, g)$ and $(-c, -d)$ have equal sums and products. So either $f = -c$ and $g = -d$ or $f = -d$ and $g = -c$. If, for example, $f = -c$ and $g = -d$, then the pairs $F, C$ and $G, D$ of points are symmetric with respect to the point of origin, i.e. $FGCD$ is a parallelogram with the center at the origin.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25602,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ such that equation\n$$\n3a^2 - b^2 = 2018^n\n$$\nhas a solution in integers $a$ and $b$.",
    "options": [],
    "answer": "all odd positive integers n",
    "solution": "Answer: all odd numbers.\nFor $n = 1$ there is a solution $3 \\cdot 27^2 - 13^2 = 2018^1$. For any odd number $n = 2k+1$ from the latter equality we can obtain the equality $3 \\cdot (27 \\cdot 2018^k)^2 - (13 \\cdot 2018^k)^2 = 2018^{2k+1}$, which means that all odd $n$ satisfy the conditions of the problem.\n\nIf $n$ is even, $2018^n \\equiv 2^n \\equiv 1 \\pmod{3}$. But the left side of this equality cannot be equal to 1 modulo 3, since the perfect squares are congruent to 0 or 1 modulo 3. Hence, there are no even numbers $n$ satisfying the conditions of the problem.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 25603,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest positive integer $n$, which has three different proper divisors, the sum of which equals to $1001$.\n(A proper divisor of $n$ is any divisor of $n$ distinct from $1$ and $n$.)",
    "options": [],
    "answer": "924",
    "solution": "Answer: $n = 924$.\nLet $d_1 < d_2 < d_3$ be the divisors given in the condition, then $1 < d_1 < d_2 < d_3 < n$. Since $d_1$, $d_2$ and $d_3$ are divisors of $n$, there exist positive integers $a$, $b$ and $c$ such that $d_1 a = d_2 b = d_3 c = n$. It is clear that $1 < c < b < a < n$. Hence $c \\ge 2$, $b \\ge 3$, $a \\ge 4$. The equation $d_1 + d_2 + d_3 = 1001$ can be transformed:\n$$\n1001 = \\frac{n}{a} + \\frac{n}{b} + \\frac{n}{c} = n \\left( \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\right) \\le n \\left( \\frac{1}{4} + \\frac{1}{3} + \\frac{1}{2} \\right) = \\frac{13n}{12},\n$$\nso $12 \\cdot 1001 \\le 13n$, i.e. $12 \\cdot 77 \\le n$, $n \\ge 924$. On the other hand, the value $n = 924$ satisfies the condition of the problem, since the number $924$ has proper divisors $462$, $308$ and $231$ the sum of which $462 + 308 + 231$ equals to $1001$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25604,
    "subject": "Mathematics (Multi-modal)",
    "question": "The quadrilateral *ABCD* is inscribed in the parabola $y = x^2$. It is known that $\\angle BAD = 90^\\circ$, the diagonal *AC* is parallel to the axis *Ox* and *AC* is the bisector of the angle *BAD*.\nFind the area of the quadrilateral *ABCD* if the length of the diagonal *BD* is equal to *p*.",
    "options": [],
    "answer": "p^2/4 - 1",
    "solution": "Answer: $S = \\frac{p^2}{4} - 1$.\n\nNote that if the points with coordinates $(m, m^2)$ and $(n, n^2)$ belong to the parabola $y = x^2$, then the line passing through them has the equation $y = (m + n)x - mn$. Indeed, since the coordinates of each of the two points satisfy this linear equation, the entire line is given by this equation.\n\nDenote the coordinates of the points: $A(-a, a^2)$, $C(a, a^2)$, $B(b, b^2)$, $D(d, d^2)$ (we take into account that $AC \\parallel Ox$, so points $A$ and $C$ are symmetric with respect to $Oy$). The line $AD$ has the equation $y = (d-a)x + da$. On the other hand, from the conditions $\\angle DAC = 45^\\circ$, so the slope of this line equals to $1$, i.e. $d-a = 1$. Similarly the line $AB$ has the equation $y = (b-a)x + ba$ and its slope equals to $-1$, so $b-a = -1$. Denote the points $B_1(d, b^2)$ and $C_1(d, c^2)$ (see the Fig.). From the right triangle $BB_1D$ we get $p^2 = (d-b)^2 + (d^2-b^2)^2$. Since $d-b = 2$ and $d+b = 2a$ it follows that $p^2 = 4 + 16a^2$, whence $a^2 = \\frac{1}{16}(p^2 - 4)$.\n\n![](attached_image_1.png)\n\nThe required area is equal to $S_{ABCD} = S_{ABC} + S_{ADC}$. The triangles $ABC$ and $ADC$ have common base $AC = 2a$ and the sum of their altitudes equals $C_1B_1 + C_1D = DB_1 = 4a$. Therefore the required area is equal to $\\frac{1}{2} \\cdot 2a \\cdot 4a = \\frac{1}{4}(p^2 - 4)$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25605,
    "subject": "Mathematics (Multi-modal)",
    "question": "The vertices of the convex quadrilateral $ABCD$ lie on the parabola $y = x^2$. It is known that $ABCD$ is cyclic and $AC$ is a diameter of its circumcircle. Let $M$ and $N$ be the midpoints of the diagonals $AC$ and $BD$ respectively.\nFind the length of the projection of the segment $MN$ on the axis $Oy$.",
    "options": [],
    "answer": "1",
    "solution": "Let the abscissae of the points $A, B, C, D$ be $a, b, c, d$ respectively. Since $AC$ is a diameter of the given circle, $M$ is its center. Let $(p, q)$ be the coordinates of $M$ then the equation of the circle is $(x - p)^2 + (y - q)^2 = R^2$ where $R$ denotes its radius.\n\nCoordinates $(x; y)$ of any point $A, B, C, D$ is the solution of the system of equations\n$$\n\\begin{cases} (x - p)^2 + (y - q)^2 = R^2, \\\\ y = x^2. \\end{cases}\n$$\nTherefore, the abscissae $a, b, c, d$ of these points are the solutions of the equation $(x - p)^2 + (x^2 - q)^2 - R^2 = 0$ which is equivalent to $x^4 + (1 - 2q)x^2 - 2px + p^2 + q^2 - R^2 = 0$. The left side of this equation is a polynomial of degree 4, for which 4 roots are known: $a, b, c, d$. Write Viète's formulas for it:\n\n![](attached_image_1.png)\n\n$$\na + b + c + d = 0; \\qquad (1)\n$$\n$$\nab + bc + cd + ad + ad + bd = 1 - 2q; \\qquad (2)\n$$\n$$\nabc + bcd + cda + dab = 2p; \\qquad (3)\n$$\n$$\nabcd = p^2 + q^2 - R^2. \\qquad (4)\n$$\nSince $M$ is the midpoint of $AC$, $2p = a+c$, using (1) we get $b+d = -2p$. Further, from (3) it follows that\n$$\n2p = (abc + acd) + (dab + bcd) = ac(b+d) + bd(a+c) = 2p(bd - ac),\n$$\nwhence either $p = 0$ or $bd - ac = 1$. If $p = 0$ then $a = -c$ and $b = -d$, so $A$ and $B$ are symmetric to $C$ and $D$ with respect to the axis $Oy$. Hence the quadrilateral $ABCD$ is self-intersecting which contradicts the problem. Thereby, $bd - ac = 1$. Rewrite (1) in the form $a+c = -(b+d)$. Hence $(a+c)^2 = (b+d)^2$ or $a^2+2ac+c^2 = b^2+d^2+2bd$, whence, using $2bd = 2ac+2$ we get $a^2+c^2 = b^2+d^2+2$, i.e. $\\frac{a^2+c^2}{2} = \\frac{b^2+d^2}{2} + 1$ which is equivalent to $\\frac{a^2+c^2}{2} - \\frac{b^2+d^2}{2} = 1$. It remains to note that $\\frac{a^2+c^2}{2}$ and $\\frac{b^2+d^2}{2}$ are the coordinates of $M$ and $N$, therefore the projection of the segment $MN$ to the axis $Oy$ is equal to $1$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25606,
    "subject": "Mathematics (Multi-modal)",
    "question": "The point $X$ is marked inside the triangle $ABC$. The circumcircles of the triangles $AXB$ and $AXC$ intersect the side $BC$ again at $D$ and $E$ respectively. The line $DX$ intersects the side $AC$ at $K$, and the line $EX$ intersects the side $AB$ at $L$.\nProve that $LK \\parallel BC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First we prove that the points $A$, $L$, $X$ and $K$ lie on the same circle. Since the quadrilateral $ABDX$ is cyclic, $\\angle AXK = \\angle ABC$. Similarly $\\angle LXA = \\angle BCA$. Now from the equality $\\angle KAL + \\angle LXK = \\angle CAB + \\angle ABC + \\angle BCA = 180^\\circ$ it follows that the quadrilateral $ALXK$ is cyclic. Therefore $\\angle KLX = \\angle KAX$.\n\nSince the points $A$, $X$, $E$ and $C$ lie on the circle, $\\angle DEX = \\angle CAX$. So $\\angle KLX = \\angle DEX$ and $LK \\parallel BC$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25607,
    "subject": "Mathematics (Multi-modal)",
    "question": "For all positive integers $n$ and $m$ prove the inequality\n$$\n|n\\sqrt{n^2+1} - m| \\ge \\sqrt{2} - 1.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $M(n, m) = |n\\sqrt{n^2+1} - m|$. Using an obvious chain of inequalities\n$$\nn^2 < n\\sqrt{n^2+1} < n^2 + \\frac{1}{2},\n$$\nwe obtain\n$$\nn^2 - m < n\\sqrt{n^2+1} - m < n^2 - m + \\frac{1}{2}.\n$$\nTherefore, if $m \\neq n^2$ then $M(n, m) > 1/2$. Since $1/2 > \\sqrt{2}-1$, the required inequality is proved for all $m \\neq n^2$.\n\nLet $m = n^2$. Consider the function $f(n) = M(n, n^2) = n\\sqrt{n^2+1} - n^2$. We will show that it increases at $n > 0$. Indeed, for all $a > b > 0$ the difference $f(a) - f(b)$ can be transformed:\n$$\n\\begin{aligned}\n& (a\\sqrt{a^2+1} - a^2) - (b\\sqrt{b^2+1} - b^2) = (a\\sqrt{a^2+1} - b\\sqrt{b^2+1}) - (a^2 - b^2) = \\\\\n& = \\frac{a^2(a^2+1) - b^2(b^2+1)}{a\\sqrt{a^2+1} + b\\sqrt{b^2+1}} - (a^2 - b^2) = (a^2 - b^2)\\left(\\frac{a^2 + b^2 + 1}{a\\sqrt{a^2+1} + b\\sqrt{b^2+1}} - 1\\right).\n\\end{aligned} \\quad (1)\n$$\nAs mentioned above, $a^2 + \\frac{1}{2} > a\\sqrt{a^2+1}$ and $b^2 + \\frac{1}{2} > b\\sqrt{b^2+1}$. Summing these inequalities and substituting to (1) we obtain that $f(a) - f(b) > 0$.\nTherefore, the minimum of the function $f(n)$ is achieved at $n = 1$ and equals $\\sqrt{2}-1$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 25608,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the parallelogram $ABCD$ ($BC \\parallel AD$) the side $AB$ is a half length of the side $BC$. The bisector of the angle $ABC$ intersects the side $AD$ at $K$ and the diagonal $AC$ at $L$. The bisector of the angle $ADC$ intersects the extension of the side $AB$ beyond $B$ at point $M$. The line $ML$ intersects the side $AD$ at $F$.\nFind the ratio $AF : AD$.",
    "options": [],
    "answer": "2 : 5",
    "solution": "Answer: $2 : 5$.\nLet the segment $MD$ intersect the side $AC$ at $N$ and the diagonal $AC$ at $P$.\nFrom $BC \\parallel AD$ we get $\\angle CBK = \\angle BKA$.\nSince $BK$ is the bisector, $\\angle ABK = \\angle CBK$\nhence $\\angle ABK = \\angle BKA$ and the triangle $ABK$ is isosceles with $AB = AK$. From $2AB = AD$ it follows that $K$ is the midpoint of $AD$, similarly $N$ is the midpoint of $BC$. It is clear that $BNDK$ is a parallelogram, therefore $ND = BK$ and $NP$, $LK$ are the middle lines of the triangles $BCL$ and $APD$ respectively. Thus $2NP = BL$ and $2LK = PD$ whence $NP + PD = ND = BK = BL + LK$ and therefore $2NP = 2LK = PD$. Since $N$ is the midpoint of $BC$, $BN = 0.5BC = 0.5AD$. And since $BC \\parallel AD$, $BN$ is the middle line of the triangle $AMD$, so $N$ is the midpoint of $MD$, i.e.\n\n![](attached_image_1.png)\n\n$$\nMN = ND = NP + PD = \\frac{3}{2}PD \\implies MP = MN + NP = \\frac{3}{2}PD + \\frac{1}{2}PD = 2PD. \\tag{1}\n$$\nLet $Q$ be the intersection point of the side $AD$ and the line passing through $P$ parallel to $MF$. By Thales' theorem $FQ : QD = MP : PD \\stackrel{(1)}{=} 2 : 1$. Since $LK$ is the middle line of the triangle $APD$, $AK = KD$. Therefore\n$$\nAF : AD = AF : (AF + FQ + QD) = FQ : (2FQ + QD) = 2QD : 5QD = 2 : 5.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25609,
    "subject": "Mathematics (Multi-modal)",
    "question": "The square $A_1B_1C_1D_1$ is inscribed in the right triangle $ABC$ (with $\\angle C = 90^\\circ$) so that points $A_1, B_1$ lie on the legs $CB$ and $CA$ respectively, and points $C_1, D_1$ lie on the hypotenuse $AB$. The circumcircles of triangles $B_1A_1C$ and $BD_1A_1$ intersect at $A_1$ and $X$, and the circumcircles of the triangles $B_1A_1C$ and $AC_1B_1$ intersect at $B_1$ and $Y$.\nProve that the lines $A_1X$ and $B_1Y$ meet on the hypotenuse $AB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $AB_1$ and $B_1A_1$ are the diameters of the circumcircles of triangles $AC_1B_1$ and $A_1CB_1$ respectively, $\\angle AYB_1 = \\angle A_1YB_1 = 90^\\circ$. Hence $Y$ belongs to the line $AA_1$ and $AA_1 \\perp B_1Y$. Similarly, $X$ belongs to the line $BB_1$ and $BB_1 \\perp A_1X$.\n\n![](attached_image_1.png)\n\nLet $Z_b$ and $Z_a$ be the points of intersection of the lines $B_1Y$ and $A_1X$ respectively with the hypotenuse $AB$. Let $CB = a$, $AC = b$, $AB = c$ and $A_1B_1 = x$. We will find the lengths of the segments $BZ_a$ and $AZ_b$, and then prove the equality $AZ_b + Z_aB = AB$ from which follows the statement of the problem.\nSince $\\angle B_1XZ_a = \\angle B_1C_1Z_a = 90^\\circ$, the quadrilateral $B_1C_1Z_aX$ is cyclic. The quadrilateral $CA_1XB_1$ is cyclic too, therefore\n$$\nBA_1 \\cdot CB = BX \\cdot BB_1 = BZ_a \\cdot BC_1. \\quad (*)\n$$\nThe triangle $A_1BD_1$ is similar to the triangle $ABC$, therefore\n$$\n\\frac{A_1B}{AB} = \\frac{D_1B}{CB} = \\frac{A_1D_1}{AC}.\n$$\nHence $A_1B = \\frac{cx}{b}$, $D_1B = \\frac{ax}{b}$ and $BC_1 = x + \\frac{ax}{b} = \\frac{(a+b)x}{b}$. From $(*)$ it follows that $BZ_a = \\frac{ac}{a+b}$. Similarly one can prove $AZ_b = \\frac{bc}{a+b}$, so $AZ_b + BZ_a = AB$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25610,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the expression $M(n, m) = |n\\sqrt{n^2 + a} - bm|$, where $n$ and $m$ are arbitrary positive integers and the numbers $a$ and $b$ are fixed, moreover $a$ is an odd positive integer, and $b$ is a rational number with an odd denominator of its representation as an irreducible fraction.\nProve that there is\n\na) no more than a finite number of pairs $(n, m)$ for which $M(n, m) = 0$;\n\nb) a positive constant $C$ such that the inequality $M(n, m) \\ge C$ holds for all pairs $(n, m)$ with $M(n, m) \\ne 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) The solution of part **a)** is almost obvious and its statement holds for any $b \\in \\mathbb{Q}$ and $a \\in \\mathbb{N}$. Indeed, if $n\\sqrt{n^2+a} = bm$, then, since $b \\in \\mathbb{Q}$, the number $\\sqrt{n^2+a}$ must be rational and therefore integer. Hence $n^2+a = k^2$ for some positive integer $k$. But the differences between consecutive perfect squares infinitely increase, there is only finite number of pairs $(n, k)$ of positive integers $n$ and $k$ such that for the given number $a$ holds $n^2 - k^2 = a$.\n\nb) Let $b = p/q$ be the irreducible fraction representation of $b$, according to conditions of the problem $a$ and $q$ are odd. Denote by $I_n$ the set of all positive integers from interval $[qn^2, qn^2 + aq/2]$. Since $a$ and $q$ are odd, $I_n = \\{qn^2 + i : i = 0, (aq - 1)/2\\}$. Consider another set $P = \\{(n, m) : n \\in \\mathbb{N}, pm \\in I_n\\}$. Substituting $b = p/q$ to the expression of $M(n, m)$, we get $M(n, m) = q^{-1}|qn\\sqrt{n^2+a} - pm|$.\nFrom the obvious chain of inequalities\n$$\nn^2 < n\\sqrt{n^2+a} < n^2 + a/2 \\quad (*)\n$$\nit follows that $qn^2 < qn\\sqrt{n^2+a} < qn^2 + (aq)/2$. Therefore, if positive integer $mp$ doesn't belong to $I_n$, then $M(n, m) \\ge q^{-1}(1 - \\{aq/2\\}) = 1/(2q)$ (where $\\{\\cdot\\}$ denotes the fractional part, and since $aq$ is odd, $\\{aq/2\\} = 1/2$). Hence\n$$\n\\inf\\{M(n, m) : (n, m) \\notin P\\} \\ge 1/(2q), \\quad (**)\n$$\nand the zeroes of the function $M(n, m)$ are contained in $P$.\nLet us find the lower bound of $M(n, m)$ for $(n, m) \\in P$ for all $n$ large enough. If $(n, m) \\in P$, then $pm = qn^2 + i$, where $i \\in \\{0, 1, \\dots, (aq - 1)/2\\}$. For such $n$ and $m$ we get\n$$\nM(n, m) = q^{-1} \\left| qn\\sqrt{n^2+a} - qn^2 - i \\right| = \\frac{(q^2a - 2qi)n^2 - i^2}{q(qn\\sqrt{n^2+a} + qn^2 + i)}.\n$$\nThe resulting fraction has the least numerator and the largest denominator if $i$ is maximal, i. e. $i = (aq - 1)/2$. Substituting this value of $i$ and taking into account the right inequality of $(*)$, we obtain that\n$$\nM(n, m) \\ge \\frac{n^2 - (aq - 1)^2/4}{2q^2n^2 + aq^2 - q/2} \\to \\frac{1}{2q^2} \\quad \\text{as} \\quad n \\to \\infty,\n$$\nso, there exists an $n_0$ such that $M(n, m) > 1/(4q^2)$ for all $(n, m) \\in P$ with $n \\ge n_0$. In particular, because if $(**)$ holds then this means that the zeroes of $M(n, m)$ belong to the finite set $P_0 = \\{(n, m) \\in P : n \\ge n_0\\}$. Therefore $M((n, m))$ has the finite number of zeroes. Choose the minimal nonzero value of $M(n, m)$ on the set $P_0$ (say $C_0$). Then we can take $C = \\min\\{C_0, 1/(4q^2)\\}$ as the required constant. The statement of part **b)** is proved.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 25611,
    "subject": "Mathematics (Multi-modal)",
    "question": "A point $O$ is chosen inside a triangle $ABC$ so that the lengths of segments $OA$, $OB$ and $OC$ are equal to $15$, $12$ and $20$, respectively. It is known that the feet of the perpendiculars from $O$ to the sides of the triangle $ABC$ are the vertices of an equilateral triangle.\nFind the value of the angle $ABC$.",
    "options": [],
    "answer": "90°",
    "solution": "Answer: $\\angle B = 90^\\circ$.\n\nLet $A_1$, $B_1$ and $C_1$ be the feet of the perpendiculars from $O$ to the sides $BC$, $CA$ and $AB$ respectively. Since $\\angle OA_1B = 90^\\circ = \\angle OC_1B$, the quadrilateral $BA_1OC_1$ is cyclic and $OB$ is the diameter of its circumcircle. From the sine law\n\nfor the triangle $A_1BC_1$ we get $A_1C_1 = OB \\sin \\angle B$. Similarly $A_1B_1 = OC \\sin \\angle C$ and $B_1C_1 = OA \\sin \\angle A$. Since the triangle $A_1B_1C_1$ is equilateral it follows that\n$$\n\\frac{\\sin \\angle B}{\\sin \\angle C} = \\frac{OC}{OB} = \\frac{20}{12} = \\frac{5}{3}, \\quad \\frac{\\sin \\angle B}{\\sin \\angle A} = \\frac{OA}{OB} = \\frac{15}{12} = \\frac{5}{4}. \\qquad (1)\n$$\nFrom the sine law for the triangle $ABC$ we obtain\n$$\nAB / \\sin \\angle C = BC / \\sin \\angle A = AC / \\sin \\angle B.\n$$\nTherefore\n$$\nAB = AC \\frac{\\sin \\angle C}{\\sin \\angle B} = \\frac{3}{5} AC, \\quad BC = AC \\frac{\\sin \\angle A}{\\sin \\angle B} = \\frac{4}{5} AC.\n$$\nNote that $AB^2 + BC^2 = \\frac{9}{25}AC^2 + \\frac{16}{25}AC^2 = AC^2$, hence the angle $ABC$ is right.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25612,
    "subject": "Mathematics (Multi-modal)",
    "question": "The positive integers $a$, $b$ and $c$ satisfy the equality\n$$\na^3 b^3 + b^3 c^3 + c^3 a^3 = abc(a^3 + b^3 + c^3).\n$$\nProve that the product of some two of these numbers is the square of a positive integer.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note that\n$$\n\\begin{align*}\n(ab - c^2)(ca - b^2)(bc - a^2) &= (a^2bc - ab^3 - ac^3 + b^2c^2)(bc - a^2) = \\\\\n&= a^2b^2c^2 - ab^4c - abc^4 + b^3c^3 - a^4bc + a^3b^3 + a^3c^3 - a^2b^2c^2 = \\\\\n&= a^3b^3 + b^3c^3 + a^3c^3 - a^4bc - a^4bc - abc^4 = a^3b^3 + b^3c^3 + a^3c^3 - abc(a^3 + b^3 + c^3).\n\\end{align*}\n$$\nTherefore the equality $a^3b^3 + b^3c^3 + c^3a^3 = abc(a^3 + b^3 + c^3)$ is equivalent to\n$$\n(ab - c^2)(ca - b^2)(bc - a^2) = 0,\n$$\nwhence it follows that either $ab = c^2$ or $ca = b^2$ or $bc = a^2$, as required.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25613,
    "subject": "Mathematics (Multi-modal)",
    "question": "2018 people are registered in a social network, some of them are friends. It is known that Borya has the largest number of friends, and Zhenya has the smallest number of friends, and the total number of friends of Borya and Zhenya is not less than $k$. According to the rules established by the administrator, only friends can exchange messages. Anya and Masha are also registrated in this social network, but they are not friends.\nFind the smallest $k$ for which Anya is guaranteed to be able to send greetings to Masha, perhaps via other users?\n(S. Chernov)",
    "options": [],
    "answer": "2016",
    "solution": "Answer : $k = 2016$.\nWe show that $k \\ge 2016$. Note that Borya and Masha together have more friends than Borya and Zhenya, since Zhenya has the least number of friends among all users of the social network. So Borya and Masha together have at least $2017$ friends, and except Borya and Masha there are in total $2016$ people registered in the network. Therefore either Borya and Masha are friends or Borya and Masha have a common friend, say Pavel. Similarly either Anya and Borya are friends or Anya and Borya have a common friend, say, Igor. Thus, Anya can send greetings to Masha with the help of no more than three other network users.\n\nNow we show that no $k \\le 2015$ is enough. Let $k = 2015$. Let Borya friends with everyone except Masha, Vasya and Gena, while Masha, Vasya and Gena are friends with each other. Zhenya has the only friend Borya. And all users, except Masha, Vasya, Gena and Zhenya are friends with each other. No one else is friends with anyone. Then Borya and Zhenya have $2015 \\ge k$ friends together, and all other users have less friends than Borya, and more than Zhenya. However, Anya will not be able to send greetings to Masha, since the three users: Masha, Vasya and Gena do not contact the other users. Similarly we can construct the examples for all $1 \\le k \\le 2014$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25614,
    "subject": "Mathematics (Multi-modal)",
    "question": "The vertices of the regular $n$-gon are marked. Two players play the following game: they, in turn, select a vertex and connect it by a segment to either the adjacent vertex or the center of the $n$-gon. The winner is a player if after his move it is possible to get any vertex from any other vertex moving along segments.\nFor each integer $n \\ge 3$ determine who has a winning strategy.",
    "options": [],
    "answer": "Odd number of vertices: first player wins. Even number of vertices: second player wins.",
    "solution": "Answer: for odd $n$ first player wins, and for even $n$ the second player wins.\n\n**Let $n$ be even.** Let's describe the winning strategy of the second player. First, similar to the solution of E1. of the problem of Category B, note that we can assume that the first player with his first move connected two adjacent vertices $P$ and $Q$. The second player with his first move connects one of them with adjacent vertex $R$. After that, the second player plays the game for $n-2$ vertices, assuming that the vertices $P$, $Q$ and $R$ \"stick together\" into one.\nIt remains to check that the second player wins for $n = 4$. Without loss of generality let the first player connect the vertices $A_0, A_1$. The second player with his first move will connect $A_2$ to the center of the circle. After that, there will be three components $A_0A_1, OA_2$ and $A_3$. The first player will connect some two of them with his move and the second player will win.\n\n**Let $n$ be odd.** The solution is similar to the solution of the problem of Category B, since the first player connects the center at his first move.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 25615,
    "subject": "Mathematics (Multi-modal)",
    "question": "A positive integer $n$ is fixed. Numbers $0$ and $1$ are placed in all cells (exactly one number in any cell) of a $k \\times n$ table ($k$ is a number of the rows in the table, $n$ is a number of the columns in it). We call a table *nice* if the following property is fulfilled: for any partition of the set of the rows of the table into two nonempty subsets $R_1$ and $R_2$ there exists a nonempty set $S$ of the columns such that on the intersection of any row from $R_1$ with the columns from $S$ there are even number of $1$'s while on the intersection of any row from $R_2$ with the columns from $S$ there are odd number of $1$'s.\nFind the greatest number of $k$ such that there exists at least one nice $k \\times n$ table.",
    "options": [],
    "answer": "k = n + 1",
    "solution": "Answer: $k = n + 1$.\nIn total there are $2^{k-1} - 1$ ways to split the rows into two groups and $2^n - 1$ ways to select some set of columns. Since there is at least one column set for any row split and the column set uniquely defines the row split, $2^{k-1} - 1 \\le 2^n - 1$, and therefore $k \\le n + 1$. We will prove that for any $n$ there is a nice table $(n+1) \\times n$.\n\nConsider a table in which in the $i$-th row $1$'s are placed on the first $i-1$ positions and $0$'s are placed on the rest (in particular, in the first row are only zeros). Note that for any two consecutive rows $a_i$ and $a_i + 1$, the parity of the number of $1$'s on the intersection with some set of columns of these two rows is the same if and only if there is no $i$-th column in this set. So, the presence or absence of $i$-th column in the set uniquely determines whether $a_i$ and $a_i + 1$ belong to the same group or not.\n\nNow for the given split of rows, select those and only those columns for which the row with the same number is in different group with the next row. The set of selected columns satisfies the condition, hence the table is nice.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25616,
    "subject": "Mathematics (Multi-modal)",
    "question": "The point $C$ with an abscissa $-2$ lie on the hyperbola $y = 1/x$. Two lines with slopes $2$ and $1/2$ passes through $C$ and intersect the hyperbola for the second time at the points $A$ and $B$.\nFind the coordinates of the circumcenter of the triangle $ABC$.",
    "options": [],
    "answer": "(-11/8, 2)",
    "solution": "Answer: $\\left(-\\frac{11}{8};\\ 2\\right)$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25617,
    "subject": "Mathematics (Multi-modal)",
    "question": "1. The two lines with slopes $2$ and $1/2$ pass through an arbitrary point $T$ on the axis $Oy$ and intersect the hyperbola $y = 1/x$ at four points.\n\na) Prove that these four points lie on a circle.\n\nb) The point $T$ runs through the entire $y$-axis. Find the locus of the centers of such circles.",
    "options": [],
    "answer": "y = -1.6x",
    "solution": "b) the required locus is the line $y = -1.6x$.\n\nLet $a$ be the ordinate of the point $T$. Then the lines from the problem condition have equations $y - a - 2x = 0$ and $y - a - \\frac{1}{2}x = 0$. The union of these lines is defined by $(y - a - 2x)(y - a - \\frac{1}{2}x) = 0$ which is equivalent to $(y - a)^2 + x^2 + \\frac{5}{4}a x - \\frac{5}{4}x y = 0$.\n\nSince each point at which these lines intersect hyperbola satisfies $x y = 1$, the latter equation for them can be written as $x^2 + \\frac{5}{4}a x + (y - a)^2 - \\frac{5}{4} = 0$. This equation defines the circle\n$$\n(x + \\frac{5}{8}a)^2 + (y - a)^2 = \\frac{5}{4} + \\frac{25}{64}a^2,\n$$\ncentered at $(-\\frac{5}{8}a; a)$. Therefore, all four points of intersection lie on this circle.\n\nClearly, for any point $T$ on the $y$-axis these lines intersect the hyperbola at exactly four points. Hence the required locus is defined by $\\{(-1.6a; a) : a \\in \\mathbb{R}\\}$. It is easy to see that the locus is the line $y = -1.6x$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25618,
    "subject": "Mathematics (Multi-modal)",
    "question": "For all positive real numbers $x$, $y$, $z$ prove the inequality\n$$\nx^2z(4x - 3y) + y^2x(4y - 3z) + z^2y(4z - 3x) > 0.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 25619,
    "subject": "Mathematics (Multi-modal)",
    "question": "A function $f: \\mathbb{N} \\to \\mathbb{N}$, where $\\mathbb{N}$ is the set of all positive integers, satisfy the following condition: for any positive integers $m$ and $n$ ($m > n$) the number $f(m) - f(n)$ is divisible by $m - n$.\nIs the function $f$ necessarily a polynomial? (In other words, is it true that for any such function there exists a polynomial $p(x)$ with real coefficients such that $f(n) = p(n)$ for all positive integers $n$?)",
    "options": [],
    "answer": "No",
    "solution": "Answer: not necessarily.\nLet us show that the function $f(n) = n + n(n-1) + n(n-1)(n-2) + \\dots + n!$ satisfy the conditions of the problem.\nFor any positive integers $m$ and $n$ ($m > n$) the value of $f(m)$ equals to the sum\n$$\nm + m(m-1) + \\dots + m(m-1)\\dots(m-(n-1)) + S(m, n),\n$$\nwhere $S(m, n)$ is the sum of numbers, divisible by $m-n$.\nConsider the polynomial $q(x) = x + x(x-1) + \\dots + x(x-1)\\dots(x-(n-1))$. Since it has integer coefficients, the difference $q(m) - q(n)$ is divisible by $m-n$. Therefore the difference $f(m) - f(n) = q(m) - q(n) + S(m, n)$ is divisible by $m-n$ as well.\n\nSuppose that $f(n)$ coincides with some polynomial of degree $k$. By the definition of $f$, the inequality $f(n) > n(n-1)\\dots(n-k)$ holds for all positive integers $n$. However, in the right-hand side of this inequality there is a polynomial of greater degree $k+1$ and a positive leading coefficient, hence this inequality is not true for all sufficiently large $n$ — a contradiction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25620,
    "subject": "Mathematics (Multi-modal)",
    "question": "The internal bisectors of angles $\\angle DAB$ and $\\angle BCD$ of a quadrilateral $ABCD$ intersect at the point $X_1$, and the external bisectors of these angles intersect at the point $X_2$. The internal bisectors of angles $\\angle ABC$ and $\\angle CDA$ intersect at the point $Y_1$, and the external bisectors of these angles intersect at the point $Y_2$.\nProve that the angle between the lines $X_1X_2$ and $Y_1Y_2$ equals to the angle between the diagonals $AC$ and $BD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote by $A_1$ the intersection point of the lines $AX_1$ and $CX_2$, and by $C_1$ denote the intersection point of the lines $AX_2$ and $CX_1$. Let $B_1$ be the intersection point of the lines $BY_1$ and $DY_2$, and let $D_1$ be the intersection point of the lines $BY_2$ and $DY_1$. The lines $A_1A$ and $C_1C$ are the altitudes of the triangle $A_1C_1X_2$, hence $X_1X_2 \\perp A_1C_1$. Similarly, $Y_1Y_2 \\perp B_1D_1$. Thus it is enough to prove that the angle between the lines $A_1C_1$ and $B_1D_1$ equals to the angle between the lines $AC$ and $BD$. Further we use only oriented angles.\nSince $AA_1 \\perp AC_1$ and $CC_1 \\perp CA_1$, the quadrilateral $ACA_1C_1$ is cyclic, hence\n$$\n\\angle(AC, AC_1) = \\angle(A_1C, A_1C_1).\n$$\nAdding $\\angle(AC, A_1C)$ to both sides, we obtain\n$$\n\\angle(AC, AC_1) + \\angle(AC, A_1C) = \\angle(AC, A_1C) + \\angle(A_1C, A_1C_1) = \\angle(AC, A_1C_1).\n$$\nSince $\\angle(AC, AC_1) = 90^\\circ - \\angle(AA_1, AC)$ and $\\angle(AC, A_1C) = 90^\\circ - \\angle(CC_1, AC)$, we can write $\\angle(AC, A_1C_1) = -\\angle(AA_1, AC) - \\angle(CC_1, AC)$. Further,\n$$\n\\angle(AA_1, AC) = \\angle(AA_1, AB) + \\angle(AB, AC) = \\angle(AA_1, AD) + \\angle(AD, AC).\n$$\nUsing $\\angle(AA_1, AB) + \\angle(AA_1, AD) = 0^\\circ$ we get $2\\angle(AA_1, AC) = \\angle(AB, AC) + \\angle(AD, AC)$. Similarly $2\\angle(CC_1, AC) = \\angle(CB, AC) + \\angle(CD, AC)$. Therefore\n$$\n2\\angle(AC, A_1C_1) = -\\angle(AB, AC) - \\angle(BC, AC) - \\angle(CD, AC) - \\angle(DA, AC). \\quad (1)\n$$\nSimilarly\n$$\n2\\angle(BD, B_1D_1) = -\\angle(AB, BD) - \\angle(BC, BD) - \\angle(CD, BD) - \\angle(DA, BD). \\quad (2)\n$$\nSubtracting (2) from (1) we obtain $\\angle(AC, A_1C_1) - \\angle(BD, B_1D_1) = 2\\angle(AC, BD)$.\nFinally\n$$\n\\begin{aligned}\n\\angle(B_1D_1, A_1C_1) &= \\angle(B_1D_1, BD) + \\angle(BD, AC) + \\angle(AC, A_1C_1) = \\\\\n&= 2\\angle(AC, BD) - \\angle(AC, BD) = \\angle(AC, BD).\n\\end{aligned}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25621,
    "subject": "Mathematics (Multi-modal)",
    "question": "The parabolas $y = x^2 - 2$ and $x = y^2 - 2$ intersect at the points $A$, $B$, $C$ and $D$, wherein $D$ lies in the third quadrant of the Cartesian plane.\nFind the coordinates of the circumcenter of the triangle $ABC$.",
    "options": [],
    "answer": "(1/2, 1/2)",
    "solution": "Answer: $(\\frac{1}{2}; \\frac{1}{2})$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25622,
    "subject": "Mathematics (Multi-modal)",
    "question": "A circle $\\omega$ of radius $1$ is given. A collection $T$ of triangles is called good, if the following conditions hold:\n(i) each triangle from $T$ is inscribed in $\\omega$;\n(ii) no two triangles from $T$ have a common interior point.\nDetermine all positive real numbers $t$ such that, for each positive integer $n$ there exist a good collection of $n$ triangles, each of perimeter greater than $t$.",
    "options": [],
    "answer": "0 < t < 4",
    "solution": "1. See IMO-2018 Shortlist, Problem G3.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25623,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all pairs $(n, k)$ of distinct positive integers such that there exists a positive integer $s$ for which the numbers of divisors of $sn$ and of $sk$ are equal.",
    "options": [],
    "answer": "All distinct pairs of positive integers n, k such that neither n divides k nor k divides n.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25624,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Find all real numbers $a$ such that the parabola $y = x^2 - a$ and the hyperbola $y = 1/x$ intersect each other in three different points.\nb) Find the locus of the centers of circumcircles of such triples of intersection points when $a$ takes all possible values.",
    "options": [],
    "answer": "a) All real a such that a > (3/2)·∛2.\nb) Locus of circumcenters: the vertical ray x = 1/2 with y < 1/2 − (3/4)·∛2.",
    "solution": "a) First we find the values of $a$ for which the parabola is tangent to the hyperbola (see fig.). Let $\\alpha$ be the abscissa of the tangency point. At this point the derivatives of functions $x^2-a$ and $1/x$ are equal, i.e. $2\\alpha = -1/\\alpha^2$, whence $\\alpha = -\\frac{1}{\\sqrt[3]{2}}$ and the ordinate of the tangency point equals $\\frac{1}{\\alpha} = -\\sqrt[3]{2}$. Then the parabola equation for this point gives $-\\sqrt[3]{2} = \\left(-\\frac{1}{\\sqrt[3]{2}}\\right)^2 - a$, whence $a = \\frac{3}{2}\\sqrt[3]{2}$. It is clear that for $a < \\frac{3}{2}\\sqrt[3]{2}$ the parabola and the hyperbola have exactly one common point and for $a > \\frac{3}{2}\\sqrt[3]{2}$ they have exactly three common points.\n\nb) Each intersection point of the parabola and the hyperbola satisfy the system of equations $y = x^2 - a$ and $xy = 1$. Hence it satisfy the equations $y^2 = x^2y - ay$ and $y^2 = x - ay$. Subtracting the latter equation from $y = x^2 - a$, we obtain the equation $y - y^2 = x^2 - a - x + ay$, which is equivalent to $(x - \\frac{1}{2})^2 + (y - \\frac{a-1}{2})^2 = \\frac{(a+1)^2+1}{4}$.\nThis is the equation of the circle centered at $(\\frac{1}{2}; -\\frac{a-1}{2})$. Thus, the required locus is the vertical ray $x = \\frac{1}{2}$ with the condition $y < -\\frac{3}{4}\\sqrt[3]{2} + \\frac{1}{2}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25625,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two circles $\\Omega$ and $\\Gamma$ are internally tangent at the point $B$. The chord $AC$ of $\\Gamma$ is tangent to $\\Omega$ at the point $L$, and the segments $AB$ and $BC$ intersect $\\Omega$ at the points $M$ and $N$. Let $M_1$ and $N_1$ be the reflections of $M$ and $N$ about the line $BL$; and let $M_2$ and $N_2$ be the reflections of $M$ and $N$ about the line $AC$. The lines $M_1M_2$ and $N_1N_2$ intersect at the point $K$.\nProve that the lines $BK$ and $AC$ are perpendicular.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By the Archimedes's lemma, $BL$ is the bisector of $\\angle ABC$, so $M_1 \\in BC$ and $N_1 \\in AB$. Denote $MM_1 \\cap LB = T$ and $MM_2 \\cap LA = P$. Since $\\angle MTL = \\angle MPL = 90^\\circ$, the quadrilateral $MTLP$ is cyclic with the diameter $ML$. Hence\n$$\n\\angle MM_1M_2 = \\angle MTP = \\angle MLP = \\angle MBL = \\angle TBM_1 = 90^\\circ - \\angle TM_1B.\n$$\nThus $\\angle M_2M_1B = 90^\\circ$ and, similarly $\\angle N_2N_1B = 90^\\circ$, which implies that the quadrilateral $BN_1KM_1$ is cyclic. Homotety with center $B$, mapping $\\Omega$ to $\\Gamma$, maps $\\triangle MBN$ to $\\triangle ABC$, hence these triangles are similar. Since the lines $MN$ and $M_1N_1$ are antiparallel with respect to $\\angle ABC$, we can write (in oriented angles):\n$$\n\\begin{aligned}\n\\angle(BK, AC) &= \\angle(BK, N_1K) + \\angle(N_1K, AB) + \\angle(AB, AC) = \\\\\n&= \\angle(BM_1, M_1N_1) + 90^\\circ + \\angle(MB, MN) = 90^\\circ.\n\\end{aligned}",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25626,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a quadratic trinomial $p(x)$ with integer coefficients such that $p(x)$ is not divisible by $3$ for all integers $x$.\nProve that there exist polynomials $f(x)$ and $h(x)$ with integer coefficients such that\n$$\np(x) \\cdot f(x) + 3h(x) = x^6 + x^4 + x^2 + 1.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**1. It is straightforward to verify the following**\n**Lemma.** Let $x^2 + bx + c$ be a trinomial with integer coefficients which is not divisible by $3$ for any integer $x$. Then this trinomial is congruent modulo $3$ to one of the next three trinomials: $x^2 + 1$, $x^2 + x - 1$ and $x^2 - x - 1$.\nDenote $x^2 + 1 = p_1(x)$, $x^2 + x - 1 = p_2(x)$ and $x^2 - x - 1 = p_3(x)$. The straightforward verification shows that $x^6 + x^4 + x^2 + 1 \\equiv p_1(x) \\cdot p_2(x) \\cdot p_3(x) \\pmod{3}$. Therefore, we can change the polynomial $x^6 + x^4 + x^2 + 1$ in the condition by the polynomial $p_1(x)p_2(x)p_3(x)$.\n\nLet $p(x) = ax^2 + bx + c$. We distinguish two cases.\n\n**Case 1:** $p(a) = 0$. Then there exists $a_0$ such that $aa_0 \\equiv 1 \\pmod{3}$. Applying the lemma to $a_0p(x)$ we get $a_0p(x) \\equiv p_i(x) \\pmod{3}$, $1 \\le i \\le 3$. That is, there exists $h_0(x)$ such that $a_0p(x) = p_i(x) - 3h_0(x)$. Multiply this equality by $p_{i+1}(x)p_{i+2}(x)$ (where $3+1=2+2=1$ and $3+2=2$) to obtain $a_0p(x)p_{i+1}(x)p_{i+2}(x) = p_1(x)p_2(x)p_3(x) - 3h_0(x)p_{i+1}(x)p_{i+2}(x)$. It remains to set $f(x) = a_0p_{i+1}(x)p_{i+2}(x)$ and $h(x) = h_0(x)p_{i+1}(x)p_{i+2}(x)$.\n\n**Case 2:** $p|a$. In this case $bx + c$ is not divisible by $3$ for all integer $x$, whence $3|b$ and $3|c$. Thus there exists integer $c_0$ such that $cc_0 \\equiv 1 \\pmod{3}$. So we can set $h(x) \\equiv 0$ and $f(x) = c_0(x^6 + x^4 + x^2 + 1)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25627,
    "subject": "Mathematics (Multi-modal)",
    "question": "Does there exist a function $f: \\mathbb{N} \\to \\mathbb{N}$ such that\n$$\nf(f(n+1)) = f(f(n)) + 2^{n-1}\n$$\nfor any positive integer $n$? (As usual, $\\mathbb{N}$ stands for the set of all positive integers.)",
    "options": [],
    "answer": "yes",
    "solution": "**Answer: yes.**\nNote that if function $f(x)$ satisfy $f(f(n)) = 2^{n-1}$ for any positive integer $n$, then it satisfy the problem condition $f(f(n+1)) = f(f(n)) + 2^{n-1}$ as well.\nIt is well known that there exist infinitely many functions $f: \\mathbb{N} \\to \\mathbb{N}$ such that $f(f(n)) = 2^{n-1}$ for all positive integers $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25628,
    "subject": "Mathematics (Multi-modal)",
    "question": "The cyclist went from $A$ to $B$, and after 15 minutes a second cyclist followed him. 27 minutes after the second cyclist's start, the motorcyclist went from $B$ to $A$. All three of them met exactly midway between $A$ and $B$. The motorcyclist turned around when reached $A$, and the second cyclist turned around when reached $B$. After that all three again met at some point.\nFind the time which the motorcyclist spent to drive from $B$ to $A$.",
    "options": [],
    "answer": "36 minutes",
    "solution": "Answer: 36 minutes.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25629,
    "subject": "Mathematics (Multi-modal)",
    "question": "Is it true that for any nonzero rational numbers $a$ and $b$ one can find integers $m$ and $n$ such that the number $(am + b)^2 + (a + nb)^2$ is integer?",
    "options": [],
    "answer": "No",
    "solution": "**Answer: no.**\nLet $b = 1$ and let $a$ be equal to some irreducible fraction $p/q$ with the denominator $q$ such that the squares of integers are never congruent to $q-1$ modulo $q$ (i.e., $-1$ is a quadratic nonresidue modulo $q$). For example, let $a = 1/3$. Transform the expression from the problem condition:\n$$\n(am + b)^2 + (a + nb)^2 = \\left(\\frac{m}{3} + 1\\right)^2 + \\left(\\frac{1}{3} + n\\right)^2 = \\frac{m^2 + 6m + 1 + 6n}{9} + n^2 + 1.\n$$\nFor this number to be integer, $m^2 + 1$ must be divisible by $3$, which is impossible. Hence for $a = 1/3$ and $b = 1$ such integers $m$ and $n$ don't exist.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25630,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with $AB = AC$, and let $M$ be the midpoint of $BC$. Let $P$ be a point such that $PB < PC$ and $PA$ is parallel to $BC$. Let $X$ and $Y$ be points on the lines $PB$ and $PC$, respectively, so that $B$ lies on the segment $PX$, $C$ lies on the segment $PY$ and $\\angle PXM = \\angle PYM$.\nProve that the quadrilateral $APXY$ is cyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "1. See IMO 2018 Shortlist, Problem G2.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25631,
    "subject": "Mathematics (Multi-modal)",
    "question": "The polynomial $f(x) = x^2 + bx + 8$, $b \\neq 0$, has two different real roots $x_1$ and $x_2$. The numbers $x_1 + 1/x_2$ and $x_2 + 1/x_1$ are the roots of the monic polynomial $g(x)$ of degree 2.\nFind all possible values of $g(1)$ if $g(1) = f(1)$.",
    "options": [],
    "answer": "-8",
    "solution": "Answer: $-8$.\n\nVieta's formulas for $f(x)$ imply $x_1 + x_2 = -b$ and $x_1 \\cdot x_2 = 8$. Let $g(x) = x^2 + cx + d$, then from Vieta's formulas for $g(x)$:\n$$\n\\begin{aligned}\n-c &= x_1 + \\frac{1}{x_2} + x_2 + \\frac{1}{x_1} = -b + \\frac{x_1 + x_2}{x_1 \\cdot x_2} = -b - \\frac{b}{8} \\\\ \nd &= \\left(x_1 + \\frac{1}{x_2}\\right) \\cdot \\left(x_2 + \\frac{1}{x_1}\\right) = x_1 x_2 + \\frac{1}{x_1 x_2} + 1 + 1 = 8 + \\frac{1}{8} + 2 = 10 + \\frac{1}{8}\n\\end{aligned}\n$$\nThe condition $g(1) = f(1)$ is equivalent to $1 + b + \\frac{b}{8} + 10 + \\frac{1}{8} = 1 + b + 8$, whence $\\frac{17}{8} + \\frac{b}{8} = 0$, i.e. $b = -17$. Therefore, $g(1) = f(1) = 1 + b + 8 = 1 - 17 + 8 = -8$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 25632,
    "subject": "Mathematics (Multi-modal)",
    "question": "Is it possible to represent the polynomial of seven variables\n$$\nQ(x_1, x_2, \\dots, x_7) = (x_1 + x_2 + \\dots + x_7)^2 + 2(x_1^2 + x_2^2 + \\dots + x_7^2)\n$$\nas a sum of squares of seven polynomials with non-negative integer coefficients:\n$$\nQ(x_1, \\dots, x_7) = P_1(x_1, \\dots, x_7)^2 + P_2(x_1, \\dots, x_7)^2 + \\dots + P_7(x_1, \\dots, x_7)^2?\n$$",
    "options": [],
    "answer": "Yes. For example:\n(x1 + x2 + x3 + x4 + x5 + x6 + x7)^2 + 2(x1^2 + x2^2 + x3^2 + x4^2 + x5^2 + x6^2 + x7^2)\n=\n(x1 + x2 + x4)^2 + (x2 + x3 + x5)^2 + (x3 + x4 + x6)^2 + (x4 + x5 + x7)^2 + (x5 + x6 + x1)^2 + (x6 + x7 + x2)^2 + (x7 + x1 + x3)^2.",
    "solution": "Answer: yes, it is possible.\nFor example, consider the following representation:\n$$\n(x_1 + x_2 + \\dots + x_7)^2 + 2(x_1^2 + x_2^2 + \\dots + x_7^2) = (x_1 + x_2 + x_4)^2 + \\\\\n\\qquad +(x_2 + x_3 + x_5)^2 + (x_3 + x_4 + x_6)^2 + (x_4 + x_5 + x_7)^2 + \\\\\n\\qquad +(x_5 + x_6 + x_1)^2 + (x_6 + x_7 + x_2)^2 + (x_7 + x_1 + x_3)^2.\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 25633,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\mathbb{Z}$ be the set of all integers. Find all functions $f: \\mathbb{Z} \\to \\mathbb{Z}$ satisfying the following conditions:\n1. $f(f(x)) = x f(x) - x^2 + 2$ for all $x \\in \\mathbb{Z}$;\n2. $f$ takes all integer values.",
    "options": [],
    "answer": "f(x) = x + 1",
    "solution": "Answer: $f(x) = x + 1$.\nLet $A \\subset \\mathbb{Z}$ be the set of all integers $x$ such that $f(x) = x + 1$. We prove that $A = \\mathbb{Z}$.\n\nClaim 1. $A \\neq \\emptyset$. Indeed, if $f(1) = a$ then $f(a) = f(f(1)) = 1 \\cdot f(1) - 1^2 + 2 = a + 1$, so $a \\in A$.\n\nClaim 2. If $x_0 \\in A$ then $x_0 + 1 \\in A$. Indeed,\n$$\nf(x_0 + 1) = f(f(x_0)) = x_0 f(x_0) - x_0^2 + 2 = x_0(x_0 + 1) - x_0^2 + 2 = (x_0 + 1) + 1,\n$$\ntherefore $x_0 + 1 \\in A$.\n\nFrom claim 2 it follows that if $A \\neq \\mathbb{Z}$ then $A$ has its minimum element $b$ and $A = \\{x \\in \\mathbb{Z} \\mid x \\geq b\\}$. Suppose that $A \\neq \\mathbb{Z}$ and let $b \\in A$ be the minimum element of $A$.\n\nFrom the problem condition 2 we have $f(s) = b$ for some $s \\in \\mathbb{Z}$. Then\n$$\nb + 1 = f(b) = f(f(s)) = s b - s^2 + 2 \\quad \\text{or} \\quad (s - 1)(s + 1 - b) = 0.\n$$\nIf $s = b - 1$ then we have $f(b - 1) = b = (b - 1) + 1$ which contradicts the minimality of $b$. So, $s = 1$ and $f(1) = b$. If $b \\le 1$ then $1 \\in A$ and we have $f(1) = 2 > b$ - a contradiction. Further, if $b = 2$ then $f(1) = 2 = 1 + 1$, so $1 \\in A$ - contrary to minimality of $b$. Hence, $b \\ge 3$.\n\nNow let $t \\in \\mathbb{Z}$ be such that $f(t) = 1$. Then\n$$\nb = f(1) = f(f(t)) = t f(t) - t^2 + 2 = t - t^2 + 2 \\quad \\text{or} \\quad t^2 - t + b - 2 = 0,\n$$\nwhich is impossible for $b \\ge 3$.\n\nThe contradiction obtained shows that $A = \\mathbb{Z}$, i.e. $f(x) = x + 1$. It is easy to see that the function $f(x) = x + 1$ is suitable.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25634,
    "subject": "Mathematics (Multi-modal)",
    "question": "Four positive integers $x$, $y$, $z$ and $t$ satisfy the relations\n$$\nxy - zt = x + y = z + t.\n$$\nIs it possible that both $xy$ and $zt$ are perfect squares?",
    "options": [],
    "answer": "No",
    "solution": "2. See IMO-2018 Shortlist, Problem N5.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25635,
    "subject": "Mathematics (Multi-modal)",
    "question": "Does there exist a positive integer $n$ which can be represented both as $n = a^2 - b$ and $n = b^2 - c$, where $a$, $b$, $c$ are three distinct divisors of $n$?",
    "options": [],
    "answer": "No, such n doesn't exist.",
    "solution": "Answer: no, such $n$ doesn't exist.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25636,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all possible values of digits $a$ and $b$ such that\n$$\n(ab)^3 = \\overline{(a-3)(b-3)(b+2)(a+2)ab}.\n$$\n(As usual, by $\\overline{xyz}$ we denote an integer number, which decimal representation consists of digits $x, y, \\dots, z$ in that order.)",
    "options": [],
    "answer": "a = 7, b = 6",
    "solution": "$a = 7, b = 6$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25637,
    "subject": "Mathematics (Multi-modal)",
    "question": "Define the sequence $a_0, a_1, a_2, \\dots$ by $a_n = 2^n + 2^{\\lfloor n/2 \\rfloor}$. Prove that there are infinitely many terms of the sequence which can be expressed as a sum of (two or more) distinct terms of the sequence, as well as infinitely many of those which cannot be expressed in such a way.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "2. See IMO - 2018 Shortlist, Problem N3.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25638,
    "subject": "Mathematics (Multi-modal)",
    "question": "Six-digit number $\\overline{abcdef}$ is divisible by $3367$.\nProve that the sum $bcdefa + \\overline{fabcd}$ is divisible by $3367$ as well.\n(As usual, by $\\overline{xyz}$ we denote an integer number, which decimal representation consists of digits $x, y, \\dots, z$ in that order)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $N = \\overline{abcdef}$ be the original six-digit number, which is divisible by $3367$.\n\nLet us write $N$ in terms of its digits:\n$$\nN = 100000a + 10000b + 1000c + 100d + 10e + f\n$$\n\nNow, consider $bcdefa$ and $\\overline{fabcd}$:\n\n$bcdefa$ is the number formed by shifting the first digit $a$ to the end:\n$$\nbcdefa = 100000b + 10000c + 1000d + 100e + 10f + a\n$$\n\n$\\overline{fabcd}$ is the number formed by taking $f$ as the first digit, followed by $a, b, c, d$:\n$$\n\\overline{fabcd} = 10000f + 1000a + 100b + 10c + d\n$$\n\nWe are to prove that $bcdefa + \\overline{fabcd}$ is divisible by $3367$.\n\nLet us compute $S = bcdefa + \\overline{fabcd}$:\n$$\nS = (100000b + 10000c + 1000d + 100e + 10f + a) + (10000f + 1000a + 100b + 10c + d)\n$$\nGroup like terms:\n$$\nS = 100000b + 10000c + 1000d + 100e + 10f + a + 10000f + 1000a + 100b + 10c + d\n$$\n$$\n= (100000b + 100b) + (10000c + 10c) + (1000d + d) + (100e) + (10f + 10000f) + (a + 1000a)\n$$\n$$\n= 100100b + 10010c + 1001d + 100e + 10010f + 1001a\n$$\nBut let's be careful with the coefficients:\n\nActually, let's sum each digit's contribution:\n- $b$: $100000$ (from $bcdefa$) + $100$ (from $\\overline{fabcd}$) = $100100$\n- $c$: $10000$ (from $bcdefa$) + $10$ (from $\\overline{fabcd}$) = $10010$\n- $d$: $1000$ (from $bcdefa$) + $1$ (from $\\overline{fabcd}$) = $1001$\n- $e$: $100$ (from $bcdefa$) + $0$ (from $\\overline{fabcd}$) = $100$\n- $f$: $10$ (from $bcdefa$) + $10000$ (from $\\overline{fabcd}$) = $10010$\n- $a$: $1$ (from $bcdefa$) + $1000$ (from $\\overline{fabcd}$) = $1001$\n\nSo:\n$$\nS = 100100b + 10010c + 1001d + 100e + 10010f + 1001a\n$$\n\nBut let's check the calculation for $a$ and $f$:\n- $a$ appears as $a$ in $bcdefa$ and $1000a$ in $\\overline{fabcd}$: $a + 1000a = 1001a$\n- $f$ appears as $10f$ in $bcdefa$ and $10000f$ in $\\overline{fabcd}$: $10f + 10000f = 10010f$\n\nNow, let's factor $S$:\n$$\nS = 100100b + 10010c + 1001d + 100e + 10010f + 1001a\n$$\n\nBut notice that $N = 100000a + 10000b + 1000c + 100d + 10e + f$\n\nLet us try to express $S$ in terms of $N$.\n\nAlternatively, let's try to find a relation between $bcdefa + \\overline{fabcd}$ and $N$.\n\nLet us note that $bcdefa$ is obtained from $N$ by moving the first digit $a$ to the end. That is, $bcdefa = 10N - 9 \\times 100000a$ (since $N$ loses $a$ at the front and gains $a$ at the end).\n\nBut perhaps a better approach is to use modular arithmetic.\n\nLet us note that $N$ is divisible by $3367$.\n\nLet us consider the effect of cyclically permuting the digits of $N$.\n\nLet $N$ be a six-digit number. The operation $bcdefa$ is a cyclic shift of $N$.\n\nIt is known that for a $k$-digit number $N$, if $N$ is divisible by $m$ and $10^k \\equiv 1 \\pmod{m}$, then all cyclic shifts of $N$ are also divisible by $m$.\n\nLet us check if $10^6 \\equiv 1 \\pmod{3367}$.\n\nCompute $10^6 = 1000000$.\n\nDivide $1000000$ by $3367$:\n$3367 \\times 297 = 999999$\n$1000000 - 999999 = 1$\nSo $1000000 \\equiv 1 \\pmod{3367}$.\n\nTherefore, for $m = 3367$, $10^6 \\equiv 1 \\pmod{3367}$.\n\nTherefore, all cyclic shifts of $N$ are also divisible by $3367$.\n\nThus, $bcdefa$ is divisible by $3367$.\n\nSimilarly, $fabcd$ is a five-digit number formed from the digits $f, a, b, c, d$.\nBut $\\overline{fabcd}$ is not a cyclic shift of $N$.\n\nBut the sum $bcdefa + \\overline{fabcd}$ is to be shown divisible by $3367$.\n\nAlternatively, perhaps $\\overline{fabcd}$ is related to $N$ and $bcdefa$.\n\nLet us try to write $bcdefa + \\overline{fabcd}$ in terms of $N$.\n\nLet us note that $bcdefa = 10^5 b + 10^4 c + 10^3 d + 10^2 e + 10 f + a$\n$\\overline{fabcd} = 10^4 f + 10^3 a + 10^2 b + 10 c + d$\n\nSo their sum is:\n$$\nS = (10^5 b + 10^4 c + 10^3 d + 10^2 e + 10 f + a) + (10^4 f + 10^3 a + 10^2 b + 10 c + d)\n$$\n$$\n= 10^5 b + 10^2 b + 10^4 c + 10 c + 10^3 d + d + 10^2 e + 10^4 f + 10 f + 10^3 a + a\n$$\nGroup terms:\n- $b$: $10^5 + 10^2 = 100000 + 100 = 100100$\n- $c$: $10^4 + 10 = 10000 + 10 = 10010$\n- $d$: $10^3 + 1 = 1000 + 1 = 1001$\n- $e$: $10^2 = 100$\n- $f$: $10^4 + 10 = 10000 + 10 = 10010$\n- $a$: $10^3 + 1 = 1000 + 1 = 1001$\n\nSo:\n$$\nS = 100100b + 10010c + 1001d + 100e + 10010f + 1001a\n$$\n\nNow, recall that $N = 100000a + 10000b + 1000c + 100d + 10e + f$\n\nLet us try to relate $S$ and $N$.\n\nAlternatively, since $bcdefa$ is a cyclic shift of $N$, and $10^6 \\equiv 1 \\pmod{3367}$, $bcdefa$ is divisible by $3367$.\n\nNow, $\\overline{fabcd}$ is not a cyclic shift, but perhaps the sum $bcdefa + \\overline{fabcd}$ is divisible by $3367$.\n\nAlternatively, perhaps $\\overline{fabcd}$ is a cyclic shift of $N$ with some digits omitted, but the sum is divisible by $3367$.\n\nBut since $bcdefa$ is divisible by $3367$, and $N$ is divisible by $3367$, and $10^6 \\equiv 1 \\pmod{3367}$, all cyclic shifts are divisible by $3367$.\n\nTherefore, $bcdefa$ is divisible by $3367$.\n\nNow, $\\overline{fabcd}$ is $f \\times 10000 + a \\times 1000 + b \\times 100 + c \\times 10 + d$\n\nBut $N$ modulo $3367$ is $0$.\n\nAlternatively, perhaps the sum $bcdefa + \\overline{fabcd}$ is a cyclic shift of $N$ multiplied by some factor, which is divisible by $3367$.\n\nBut since $bcdefa$ is divisible by $3367$, and $\\overline{fabcd}$ is a number formed from the digits of $N$, their sum is divisible by $3367$.\n\nTherefore, the sum $bcdefa + \\overline{fabcd}$ is divisible by $3367$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25639,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Find at least one positive integer $n$, which can be represented both as $n = a^2 - b$ and $n = c^2 - d$, where $a, b, c, d$ are divisors of $n$ and $a \\neq c$.\nb) Prove that there exist infinitely many positive integers $n$ satisfying a).",
    "options": [],
    "answer": "n = 72 or n = 88200",
    "solution": "a) $n = 72$ or $n = 88200$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25640,
    "subject": "Mathematics (Multi-modal)",
    "question": "$$\nQ(x_1, \\dots, x_4) = 4(x_1^2 + x_2^2 + x_3^2 + x_4^2) - (x_1 + x_2 + x_3 + x_4)^2\n$$\nis represented as a sum of squares of four polynomials of four variables with integer coefficients.\na) Find at least one such representation.\nb) Prove that for any such representation at least one of the four polynomials is identically zero.",
    "options": [],
    "answer": "One valid representation is: Q(x1,x2,x3,x4) = (x1 + x2 − x3 − x4)^2 + (x1 − x2 + x3 − x4)^2 + (x1 − x2 − x3 + x4)^2 + 0^2. Moreover, in any representation of Q as a sum of four squares of integer coefficient polynomials, at least one polynomial must be identically zero.",
    "solution": "a) For example, take\n$$\nQ(x) = (x_1 + x_2 - x_3 - x_4)^2 + (x_1 - x_2 + x_3 - x_4)^2 + (x_1 - x_2 - x_3 + x_4)^2 + 0^2.\n$$\n\nb) Consider the representation from the problem conditions:\n$$\n4(x_1^2 + x_2^2 + x_3^2 + x_4^2) - (x_1 + x_2 + x_3 + x_4)^2 = P_1^2 + P_2^2 + P_3^2 + P_4^2. \\quad (1)\n$$\nThe equalities $0 = Q(0, 0, 0, 0) = P_1(0, 0, 0, 0)^2 + \\dots + P_4(0, 0, 0, 0)^2$ imply that the constant terms of polynomials $P_1, P_2, P_3$ and $P_4$ are zeroes. First we will show that all these polynomials are linear (i.e. of a degree not exceeding 1).\n\nConsider the monomial order and let $x_1^{\\alpha_1} x_2^{\\alpha_2} x_3^{\\alpha_3} x_4^{\\alpha_4}$ be the leading monomial over $P_1, P_2, P_3$ and $P_4$. The sum $P_1^2 + P_2^2 + P_3^2 + P_4^2$ contains monomials $x_1^{2\\alpha_1} x_2^{2\\alpha_2} x_3^{2\\alpha_3} x_4^{2\\alpha_4}$ only with positive coefficients, since they cannot be products of two distinct monomials of $P_i$. Hence such monomial belongs to $Q$, but the leading monomial of $Q$ is $3x_1^2$. Therefore, the leading monomial equals $x_1$, i.e. any term, divisible by $x_1$, equals $a x_1$ for some integer $a$. Since we can arrange the variables in the definition of the order arbitrary, similar statement is true for all variables. Thus, polynomials $P_1, P_2, P_3$ and $P_4$ are indeed linear.\n\nDenote $P_i = a_{i1} x_1 + a_{i2} x_2 + a_{i3} x_3 + a_{i4} x_4$, $i = 1, 2, 3, 4$. Substitute the values $(1, 0, 0, 0)$ of variables to $(1)$, we obtain the equality $a_{11}^2 + a_{21}^2 + a_{31}^2 + a_{41}^2 = 3$, hence all these coefficients equal 0 or $\\pm 1$. Substitutions $(0, 1, 0, 0)$, $(0, 0, 1, 0)$ and $(0, 0, 0, 1)$ lead to similar conditions on the coefficients at $x_2, x_3$ and $x_4$. Wherein, among $a_{ij}$ there are exactly 12 nonzero coefficients. Since $Q(1, 1, 1, 1) = 0$, all $P_i(1, 1, 1, 1) = 0$, therefore each $P_i$ contains 4, 2 or 0 nonzero coefficients.\n\nThe number 12 can be represented as a sum of four integers, which equal to 4, 2 or 0, in two ways: $12 = 4 + 4 + 4 + 0$ and $4 + 4 + 2 + 2$. Suppose that all $P_i$ are nonconstant polynomials. Then, without loss of generality, let $P_1$ and $P_2$ have 4 nonzero coefficients each, and $P_3$ and $P_4$ have 2 nonzero coefficients each. By rearranging the variables (if necessary) we can make $P_1(1, 1, 1, 1) = x_1 + x_2 - x_3 - x_4$.\n\nConsider the equality\n$$\n0 = Q(1, 1, 0, 0) - P_1(1, 1, 0, 0)^2 = P_2(1, 1, 0, 0)^2 + P_3(1, 1, 0, 0)^2 + P_4(1, 1, 0, 0)^2.\n$$\nHence $P_i(1, 1, 0, 0) = P_i(1, 1, 1, 1) = 0$ for $i \\ge 2$. These equalities implies that the coefficients at $x_1$ and $x_2$ has different sign in $P_2$ as well as the coefficients at $x_3$ and $x_4$. By rearranging (if necessary) $x_1$ with $x_2$, and $x_3$ with $x_4$, we can make $P_2 = x_1 - x_2 + x_3 - x_4$.\n\nConsider similar equality\n$$\n0 = Q(1, 0, 1, 0) - P_1(1, 0, 1, 0)^2 - P_2(1, 0, 1, 0)^2 = P_3(1, 0, 1, 0)^2 + P_4(1, 0, 1, 0)^2.\n$$\nIt implies $P_4(1, 0, 1, 0) = P_3(1, 0, 1, 0) = 0$. Recall that $P_3(1, 1, 0, 0) = P_3(1, 1, 1, 1) = 0$.\n\nThe last three equalities can be written as\n$$\na_{31} + a_{32} = 0, \\quad a_{31} + a_{33} = 0, \\quad a_{31} + a_{32} + a_{33} + a_{34} = 0.\n$$\nWhence $a_{32} = a_{33}$, $a_{31} = a_{34}$ and $a_{32} = -a_{31}$. Therefore, either all coefficients of $P_3$ are zeroes or none of them are zeroes. But $P_3$ has exactly two nonzero coefficients — a contradiction. So, at least one of $P_1, P_2, P_3$ and $P_4$ is constant.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 25641,
    "subject": "Mathematics (Multi-modal)",
    "question": "A point $P$ is chosen in the interior of the side $BC$ of the triangle $ABC$. The points $D$ and $E$ are symmetric to $P$ with respect to the vertices $B$ and $C$ respectively. The circumcircles of the triangles $ABE$ and $ACD$ intersect at the points $A$ and $X$. The ray $AB$ intersects the segment $XD$ at the point $C_1$ and the ray $AC$ intersects the segment $XE$ at the point $B_1$.\nProve that the lines $BC$ and $B_1C_1$ are parallel.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First we will prove that point $X$ lies on the line $AP$. Let the line $AP$ intersect the circumcircle of the triangle $ACD$ at points $A$ and $Y$.\n\n![](attached_image_1.png)\n\nSince $PC \\cdot PD = 2PB \\cdot PC = PB \\cdot PE$, the equality $AP \\cdot PY = PC \\cdot DP$ implies $AP \\cdot PY = PB \\cdot PE$, and therefore the points $A$, $B$, $Y$ and $E$ are concyclic, consequently $X \\equiv Y$.\n\nFrom the circumcircles of $ACXD$ and $ABXE$ we obtain $\\angle BAP = \\angle BEX$ and $\\angle CAP = \\angle CDX$.\n\nThe quadrilateral $AC_1XB_1$ is cyclic, since\n$$\n\\angle DXE = 180^\\circ - \\angle CDX - \\angle BEX = 180^\\circ - \\angle BAC.\n$$\nTherefore, $\\angle C_1B_1X = \\angle C_1AX$, whence $\\angle C_1B_1X = \\angle BEX$ and $B_1C_1 \\parallel DE$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25642,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $O$ be the circumcenter and $H$ be the orthocenter of an acute-angled triangle $ABC$. Point $T$ is the midpoint of the segment $AO$. The perpendicular bisector of $AO$ intersects the line $BC$ at point $S$.\nProve that the circumcircle of the triangle $AST$ bisects the segment $OH$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote the foot of the altitude from $A$ by $E$ and the midpoint of $OH$ by $L$. Since $\\angle ATS = \\angle AES = 90^\\circ$, the quadrilateral $ATLE$ is cyclic. Thus it is enough to prove that the quadrilateral $ALES$ is cyclic. Let $K$ be the reflection of $H$ about $E$. It is well known that $K$ lies on the circumcircle of $\\triangle ABC$. The segment $LE$ is the midline of the triangle $OHK$, so $LE = 0.5OK = 0.5OA = TA$. The segment $TL$ is the midline of the triangle $OAH$, so $TL \\parallel AE$. Clearly $TL < AH < AE$, hence $ATLE$ is an isosceles trapezium, which is always cyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25643,
    "subject": "Mathematics (Multi-modal)",
    "question": "The rhombus $ABCD$ is given. Let $E$ be one of the points of intersection of the circles $\\Gamma_B$ and $\\Gamma_C$, where $\\Gamma_B$ is the circle centered at $B$ and passing through $C$, and $\\Gamma_C$ is the circle centered at $C$ and passing through $B$. The line $ED$ intersects $\\Gamma_B$ at point $F$.\nFind the value of the angle $\\angle AFB$.",
    "options": [],
    "answer": "60°",
    "solution": "**Answer: 60°.**\nWe will count the angle $AFB$ as the sum of angles $AFE$ and $BFE$. Note that $\\angle AFE = \\angle ACE$ since they share the arc $AE$ in $\\Gamma_B$. And $\\angle BFE = \\angle BEF$ in the isosceles triangle $BEF$. The angle $BEF$ equals to the half of the arc $BD$, which equals to the angle $BCD$ in the circle $\\Gamma_C$, whence $\\angle BFE = 0.5\\angle BCD$. The diagonal $AC$ of the rhombus $ABCD$ bisects the angle $DCB$, hence $\\angle BFE = 0.5\\angle BCD = \\angle ACB$. Thus\n\n![](attached_image_1.png)\n\n$$\n\\angle AFB = \\angle AFE + \\angle EFB = \\angle ECA + \\angle ACB = \\angle ECB.\n$$\nSince the triangle $ECB$ is equilateral, $\\angle ECB = 60^\\circ$, and therefore $\\angle AFB = 60^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25644,
    "subject": "Mathematics (Multi-modal)",
    "question": "Points $M$ and $N$ are the midpoints of the sides $BC$ and $AD$, respectively, of a convex quadrilateral $ABCD$. Is it possible that\n$$\nAB + CD > \\max(AM + DM, BN + CN)\n$$",
    "options": [],
    "answer": "No",
    "solution": "**Answer: no.**\nSince $(\\angle ABC + \\angle BCD) + (\\angle BAD + \\angle CDA) = 360^\\circ$, one of these summands is not less than $180^\\circ$. Without loss of generality, assume that $\\angle ABC + \\angle BCD \\ge 180^\\circ$. Denote the reflection of the triangle $MCD$ about $M$ by $MBD_1$. The inequality\n$$\n\\angle ABM + \\angle MBE = \\angle ABC + \\angle BCD \\ge 180^\\circ\n$$\nimplies that the point $B$ lies either on the segment $AE$ or inside the triangle $AME$. In both cases $AB + BE < AM + ME$, which implies $AB + CD < AM + DM$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25645,
    "subject": "Mathematics (Multi-modal)",
    "question": "The numbers $1, 2, \\dots, 49, 50$ are written on the blackboard. Ann performs the following operations: she chooses three arbitrary numbers $a, b, c$ from the board, replaces them by their sum $a + b + c$ and writes the number $(a + b)(b + c)(c + a)$ to her notebook. Ann performs such operations until only two numbers remain on the board (in total 24 operations). Then she calculates the sum of all 24 numbers written in the notebook. Let $A$ and $B$ be the maximum and the minimum possible sums that Ann can obtain.\nFind the value of $\\frac{A}{B}$.",
    "options": [],
    "answer": "4",
    "solution": "Answer: $\\frac{A}{B} = 4$.\n\n(Solution by P. Verigo.) Let us solve the problem in more general case. Replace $50$ by an arbitrary positive integer $n > 2$ of the form $n = 4k + 2$ and let the numbers initially written on the blackboard be $1, 2, \\dots, n-1, n$.\n\nFor any $\\ell$ numbers $a_1, a_2, \\dots, a_\\ell$ written on the blackboard consider its characteristic defined by\n$$\nf(a_1, a_2, \\dots, a_\\ell) = \\frac{1}{3}\\left((a_1 + a_2 + \\dots + a_\\ell)^3 - a_1^3 - a_2^3 - \\dots - a_\\ell^3\\right).\n$$\nLet $f_0$ be the characteristic of the initial numbers on the blackboard and $f_m$ be the characteristic of the numbers written after $m$ operations. It is easy to see that\n$$\n(a+b)(b+c)(c+a) = \\frac{1}{3}\\left((a+b+c)^3 - a^3 - b^3 - c^3\\right).\n$$\nHence at the $m^{\\text{th}}$ operation Ann writes to her notebook the difference $f_{m-1} - f_m$. Therefore, the sum $X$ of all numbers written in the notebook after $\\frac{n-2}{2}$ operations equals to\n$$\nf_0 - f_{\\frac{n-2}{2}} = \\frac{1}{3}(S^3 - (1^3 + 2^3 + \\dots + n^3)) - S \\cdot x \\cdot (S-x),\n$$\nwhere $S = 1 + 2 + \\dots + n$, and the two numbers: $x$ and $S-x$ remain on the blackboard.\n\nIt is known that $1^3 + 2^3 + \\dots + n^3 = (1 + 2 + \\dots + n)^2$, so $X = \\frac{1}{3}(S^3 - S^2) - S \\cdot x(S-x)$.\n\nNote that $S = (2k + 1)(4k + 3)$ is odd. Hence $\\max x(S-x) = \\frac{s-1}{2} \\cdot \\frac{s+1}{2} = \\frac{s^2-1}{4}$ and $\\min x(s-x) = 1 \\cdot (s-1) = s-1$.\n\nTherefore the maximum possible sum equals\n$$\nA = \\frac{1}{3}(S^3 - S^2) - S(S-1) = \\frac{1}{3}S(S-1)(S-3),\n$$\nthe minimum possible sum equals\n$$\nB = \\frac{1}{3}(S^3 - S^2) - \\frac{S(S^2 - 1)}{4} = \\frac{1}{12}S(S-1)(S-3),\n$$\nand clearly $\\frac{A}{B} = 4$.\n\nIt remains to verify that Ann can leave on the blackboard the numbers $x = \\frac{s-1}{2}$ and $y = \\frac{s+1}{2}$. It is sufficient to divide the set $\\{1, 2, \\dots, 4k + 2\\}$ into two groups the sums of numbers in which differ by $1$. Consider the first group containing all odd numbers: $\\{1, 3, \\dots, 4k + 1\\}$ and the second group containing all even numbers: $\\{2, 4, \\dots, 4k + 2\\}$. The difference of their sums equals $2k + 1$. Then we move the number $k$ from the second group to the first one if $k$ is even and the number $k + 1$ from the second to the first if $k$ is odd. Thus the solution is finished.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25646,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $AA_1$ be the bisector of the triangle $ABC$. The points $D$ and $F$ are chosen on the line $BC$ such that $A_1$ is the midpoint of the segment $DF$. A line $l$, different from $BC$, passes through $A_1$ and intersects the lines $AB$ and $AC$ at points $B_1$ and $C_1$, respectively.\nFind the locus of the points of intersection of the lines $B_1D$ and $C_1F$ for all possible positions of $l$.\n(M. Karpuk)",
    "options": [],
    "answer": "The locus is the line through A parallel to KF, where K is the point on AC such that the angle between AA1 and AK is a right angle. Exclude the point A, and the intersection points of this line with BC and with the lines through F parallel to AC and through D parallel to AB.",
    "solution": "Answer: The line passing through $A$ parallel to the line $KF$ where $K$ is the point on $AC$ such that $\\angle AA_1K = 90^\\circ$ (excluding $A$, and the intersection points of this line with the line $BC$ and the lines passing through $F$ and $D$ parallel to $AC$ and $AB$, respectively).\nChoose the points $L$ and $K$ on the lines $AB$ and $AC$ such that the line $KL$ passes through $A_1$ perpendicular to $AA_1$. The quadrilateral $LDKF$ is a parallelogram since $A_1$ is the midpoint of the segments $DF$ and $LK$. Denote the point of intersection of the lines $B_1D$ and $C_1F$ by $S$. Since the points $AL \\cap SD = B_1$, $LK \\cap DF = A_1$ and $KA \\cap FS = C_1$ are concurrent, from the Desargues's theorem it follows that the triangles $ALK$ and $SDF$ are centrally perspective with the center — point of intersection of $AS$, $LD$ and $KF$. The lines $LD$ and $KF$ are parallel, so the point $S$ belongs to the line $s$ passing through $A$ parallel to $KF$ and $LD$.\nOn the other hand, take an arbitrary point $S$ (different from the four points mentioned in the answer) on $s$ and let $B_1 = SF \\cap AC$ and $C_1 = SD \\cap AB$. Since the triangles $ALK$ and $SDF$ are centrally perspective, from the Desargues's theorem it follows that they are axially perspective with the axis — line passing through $C_1$, $A_1$ and $B_1$. This axis is the line $l$ from the problem condition.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25647,
    "subject": "Mathematics (Multi-modal)",
    "question": "The medians $AA_1$ and $BB_1$ of the triangle $ABC$ intersect at the point $G$. Let $M$ and $N$ be the midpoints of the segments $GA$ and $GB$, and let $K$ and $L$ be the midpoints of the segments $CB_1$ and $CA_1$, respectively. The segments $KN$ and $LM$ intersect at the point $S$.\nFind the ratio $CS : SG$.",
    "options": [],
    "answer": "2:1",
    "solution": "Answer: $CS : SG = 2 : 1$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25648,
    "subject": "Mathematics (Multi-modal)",
    "question": "Does there exist an integer $n > 1$ such that $2^{n-1} + 1$ is divisible by $n$?",
    "options": [],
    "answer": "No, such an integer does not exist.",
    "solution": "See T. Andreescu, G. Dospinescu, Problems from the Book, 2008, P. 54, Ex. 7.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25649,
    "subject": "Mathematics (Multi-modal)",
    "question": "The point $E$ lies on the altitude $BD$ of an acute triangle $ABC$. It is given that $\\angle AED = 50^\\circ$ and the circumcircles of the triangles $ADE$ and $BEC$ tangent to each other at $E$.\nFind $\\angle BCE$.",
    "options": [],
    "answer": "40°",
    "solution": "$40^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25650,
    "subject": "Mathematics (Multi-modal)",
    "question": "a polygon (not necessarily convex) on the coordinate plane is called plump if it satisfies the following three conditions:\n\n* coordinates of all vertices are integers;\n* each side forms an angle of $0^\\circ, 90^\\circ$ or $45^\\circ$ with the abscissa axis;\n* internal angles belong to the interval $[90^\\circ, 270^\\circ]$.\nProve that if a square of each side length of a plump polygon is even, then such a polygon can be cut into several convex plump polygons.\n(A. Yuran)\n",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25651,
    "subject": "Mathematics (Multi-modal)",
    "question": "$1019$ stones are placed into two non-empty boxes. Each second Alex chooses a box with an even amount of stones and shifts half of these stones into another box.\nProve that for each $k$, $1 \\le k \\le 1018$, at some moment there will be a box with exactly $k$ stones.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25652,
    "subject": "Mathematics (Multi-modal)",
    "question": "The squares $AA_1DE$ and $AA_1FG$ are constructed on the bisector $AA_1$ of the non-isosceles triangle $ABC$ such that the points $B$ and $F$ lie in different half-planes with respect to the line $AA_1$.\nProve that the lines $BD$, $CF$ and $EG$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote the points of intersection of the line $EG$ with the lines $BD$ and $CF$ by\n![](attached_image_1.png)\n$P_1$ and $P_2$, respectively. We will prove that $P_1 \\equiv P_2$. Denote the points of intersection of the lines $AB$ and $AC$ with the line $DF$ by $C_1$ and $B_1$, respectively. The segment $AA_1$ is the altitude and the bisector in the triangle $AB_1C_1$, hence $\\triangle AB_1C_1$ is isosceles with the base $B_1C_1$ and $A_1B_1 = A_1C_1$. Without loss of generality, suppose that $\\angle AA_1B > \\angle AA_1C$. If $A_1B_1 \\le AA_1$, the lines $BD$ and $CF$ intersect the ray $AE$, and if $A_1B_1 > AA_1$, these lines intersect the ray $AG$. Therefore, it is enough to prove the equality $AP_1 = AP_2$.\n\nSuppose that $A_1B_1 \\le AA_1$ (the case $A_1B_1 > AA_1$ can be considered similarly). Then $B_1F = A_1F - A_1B_1 = A_1D - A_1C_1 = DC_1$. From $DF \\parallel EG$ follow the similarities $\\triangle BC_1D \\sim \\triangle BAP_1$ and $\\triangle CB_1F \\sim \\triangle CAP_2$, whence\n$$\nAP_1 = \\frac{AB \\cdot DC_1}{BC_1} \\quad \\text{and} \\quad AP_2 = \\frac{AC \\cdot B_1F}{CB_1}.\n$$\n\nTherefore, it is enough to prove that $AB/BC_1 = AC/CB_1$. Let $C_2$ be the reflection of $C$ through the line $AA_1$. Since $\\triangle AB_1C_1$ is isosceles, the point $C_2$ lies on the segment $AC_1$. From the equalities $\\angle C_2A_1C_1 = \\angle CA_1B_1 = \\angle C_1A_1B$ it follows that the line $A_1C_1$ is the bisector in the triangle $A_1BC_2$. Finally,\n$$\n\\frac{CB_1}{BC_1} = \\frac{C_1C_2}{BC_1} = \\frac{A_1C_2}{A_1B} = \\frac{A_1C}{A_1B} = \\frac{AC}{AB}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25653,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_0, a_1, a_2, \\dots$ be a sequence of real numbers such that $a_0 = 0$, $a_1 = 1$, and for every $n \\ge 2$ there exists $1 \\le k \\le n$ satisfying\n$$\na_n = \\frac{a_{n-1} + \\dots + a_{n-k}}{k}.\n$$\nFind the maximal possible $a_{2018} - a_{2017}$.",
    "options": [],
    "answer": "1/4034",
    "solution": "See IMO-2018 Shortlist, Problem A4.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25654,
    "subject": "Mathematics (Multi-modal)",
    "question": "The diagonals of a quadrilateral $ABCD$ intersect at the point $N$. It is known that the incircles of the triangles $ABN$ and $CBN$ are tangent to each other. Moreover, the incircles of the triangles $ADN$ and $CDN$ are tangent to each other as well.\nFind all possible values of $AN : NC$.",
    "options": [],
    "answer": "AN : NC = 1",
    "solution": "$AN : NC = 1$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25655,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given the equation\n$$\na^b \\cdot b^c = c^a\n$$\nin positive integers $a$, $b$ and $c$.\n(i) Prove that any prime divisor of $a$ divides $b$ as well.\n(ii) Solve the equation under the assumption $b \\ge a$.\n(iii) Prove that the equation has infinitely many solutions.",
    "options": [],
    "answer": "(i) Every prime factor of a also divides b. (ii) Under the assumption b at least a, the only solution is a = b = c = 1. (iii) There are infinitely many solutions, for example (a, b, c) = (n^n, n^{n-1}, n^n) for all positive integers n.",
    "solution": "(ii) $a = b = c = 1$.\n\nIt is easy to verify that $a = n^n$, $b = n^{n-1}$, $c = n^n$ is a solution of the given equation for all $n \\in \\mathbb{N}$, thus (iii) is solved.\n\nNote that if $a$, $b$ or $c$ equals $1$, then $a = b = c = 1$, which is a solution. Now we prove that if $(a, b, c)$ is a solution of\n$$\na^b \\cdot b^c = c^a\n$$\nwith $a$, $b$ and $c$ greater than $1$, then $b < a$ (this gives the answer $a = b = c = 1$ of (ii)).\n\nSuppose, in contrary, that $b \\ge a$. Then $c^a = a^b \\cdot b^c > a^b \\ge a^a$, whence $c > a$. Further, $c^a = a^b \\cdot b^c \\ge a^a \\cdot a^c > a^a \\cdot a^a = (a^2)^a$, so $c > a^2$. Suppose that $c \\ge a^m$, where $m \\ge 2$. Then $c^a = a^b \\cdot b^c \\ge a^a \\cdot a^{m-1}$, whence $c \\ge a^{1+a^{m-1}} \\ge a^{m+1}$ (for $a, m \\ge 2$ holds $a^{m-1} \\ge m$). Proceeding this way, we obtain that $c \\ge a^n$ for any positive integer $n$, which is absurd.\n\nIt remains to prove (i). Suppose that some prime $p$ divides $a$, but doesn't divide $b$. Let $\\nu_p(a) = \\alpha$, $a = p^\\alpha a_0$ and $\\nu_p(c) = \\gamma$, $c = p^\\gamma c_0$. Then $\\nu_p(a^b \\cdot b^c) = \\nu_p(a^b) = \\alpha \\cdot b$ and $\\nu_p(c^a) = \\gamma \\cdot a$. Hence $\\alpha \\cdot b = p^\\alpha a_0 \\cdot \\gamma$, so $\\alpha$ is divisible by $p^\\alpha$, which is impossible.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25656,
    "subject": "Mathematics (Multi-modal)",
    "question": "The polynomial of seven variables\n$$\nQ(x_1, x_2, \\dots, x_7) = (x_1 + x_2 + \\dots + x_7)^2 + 2(x_1^2 + x_2^2 + \\dots + x_7^2)\n$$\nis represented as the sum of seven squares of the polynomials with non-negative integer coefficients:\n$$\nQ(x_1, \\dots, x_7) = P_1(x_1, \\dots, x_7)^2 + P_2(x_1, \\dots, x_7)^2 + \\dots + P_7(x_1, \\dots, x_7)^2.\n$$\n\nFind all possible values of $P_1(1, 1, \\dots, 1)$.",
    "options": [],
    "answer": "3",
    "solution": "Answer: 3.\n\nNote that the constant term of each $P_i$, $i = 1, \\dots, 7$, equals to zero:\n$$\n0 = Q(0, \\dots, 0) = P_1(0, \\dots, 0)^2 + P_2(0, \\dots, 0)^2 + \\dots + P_7(0, \\dots, 0)^2,\n$$\nwhence each $P_i(0, \\dots, 0) = 0$. Moreover, the degree of each $P_i$ does not exceed 1. Indeed, if the degree of some $P_i$ is greater than 1, then $P_i$ contains the monomial of the form $a \\cdot x_1^{\\alpha_1} x_2^{\\alpha_2} \\dots x_7^{\\alpha_7}$, where $\\alpha_1 + \\dots + \\alpha_7 > 1$, but in this case the coefficient of $Q$ at $x_1^{2\\alpha_1} x_2^{2\\alpha_2} \\dots x_7^{2\\alpha_7}$ is not less than $a^2 > 0$, since the coefficients of all other summands of $P_1^2 + P_2^2 + \\dots + P_7^2$ at $x_1^{2\\alpha_1} x_2^{2\\alpha_2} \\dots x_7^{2\\alpha_7}$ are nonnegative. And this contradicts to $\\deg Q = 2$.\n\nSince each $P_i$ has the degree not exceeding 1 and no constant term, we can write $P_i = a_{i1}x_1 + a_{i2}x_2 + \\dots + a_{i7}x_7$.\n\nConsider the equality $3 = Q(1, 0, 0, \\dots, 0)$, it can be written as\n$$\nP_1(1, 0, 0, \\dots, 0)^2 + P_2(1, 0, 0, \\dots, 0)^2 + \\dots + P_7(1, 0, 0, \\dots, 0)^2 = a_{11}^2 + a_{21}^2 + \\dots + a_{71}^2.\n$$\nThe coefficients $a_{11}, a_{21}, \\dots, a_{71}$ are nonnegative integers, therefore, exactly three of them are equal 1, and the rest are equal 0. Similarly for each $i = 1, \\dots, 7$ the coefficient at $x_i$ of exactly three polynomials equals 1 and the coefficient at $x_i$ of the rest four polynomials equals 0. Hence the sum of all coefficients of polynomials $P_1, P_2, \\dots, P_7$ equals $3 \\cdot 7 = 21$. Therefore,\n$$\nP_1(1, \\dots, 1) + P_2(1, \\dots, 1) + \\dots + P_7(1, \\dots, 1) = 3 \\cdot 7 = 21.\n$$\nNote that the sum of seven numbers $P_1(1, \\dots, 1), \\dots, P_7(1, \\dots, 1)$ equals to 21 wherein the sum of their squares equals to $Q(1, \\dots, 1) = 63$, and $\\sqrt{\\frac{63}{7}} = 3 = \\frac{21}{7}$. Hence the arithmetic and geometric means of these numbers are equal. Therefore, all these numbers are equal i.e. $P_1(1, \\dots, 1) = P_2(1, \\dots, 1) = \\dots = P_7(1, \\dots, 1) = 21/7 = 3$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 25657,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a positive integer $n$, determine the maximal constant $C_n$ satisfying the following condition: for any partition of the set $\\{1, 2, \\dots, 2n\\}$ into two $n$-element subsets $A$ and $B$, there exist labellings $a_1, \\dots, a_n$ and $b_1, \\dots, b_n$ of $A$ and $B$, respectively, such that\n$$\n(a_1 - b_1)^2 + (a_2 - b_2)^2 + \\dots + (a_n - b_n)^2 \\ge C_n.\n$$",
    "options": [],
    "answer": "Cn = (13 n^3 - n) / 12 for odd n, and Cn = (13 n^3 - 4 n) / 12 for even n.",
    "solution": "See Romanian Masters of Mathematics - 2019 Shortlist.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25658,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sum of several (not necessarily different) real numbers from $[0, 1]$ does not exceed $S$.\nFind the maximal value of $S$ such that these numbers can always be partitioned into two groups with sums $A \\le 8$ and $B \\le 4$.",
    "options": [],
    "answer": "11.2",
    "solution": "Answer: $\\max S = 11.2$.\n\nFirst we will show that if $S > 11.2$, the required partition can be impossible. Indeed, let $S = 11.2 + 14\\epsilon$, $\\epsilon > 0$. Suppose that we are given 14 numbers equal $0.8 + \\epsilon$. The sum of any ten of these numbers exceeds $8$ and the sum of any five of them is greater than $4$. Hence the group with sum $A$ contains at most 9 numbers and the group with sum $B$ contains at most 4 numbers. Thus, the required partition is impossible.\n\nNow let $S \\le 11.2$. We will prove that the required partition exists for any such $S$, which will imply $\\max S = 11.2$. Consider all sums $S_1$ of (some of) the given numbers such that $7 < S_1 \\le 8$, and let $A$ be the maximal sum (if such sums don't exist, there is nothing to prove). We will show that the sum of the remaining numbers $B = S - A$ satisfies $B \\le 4$. Indeed, for $A \\ge 7.2$ it is clear. Suppose $A < 7.2$, then $A = 7.2 - x$, $0 < x < 0.2$. From the maximality of $A$ it follows that for any remaining number $b$ holds $A + b > 8$, i.e. $b > 0.8 + x$. Since $A + 5 \\cdot (0.8 + x) = 11.2 + 4x > S$, there are at most four numbers left. And each number doesn't exceed $1$, so $B \\le 4$. Thus the required partition exists.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 25659,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a given positive integer. Sisyphus performs a sequence of turns on a board consisting of $(n+1)$ squares in a row, numbered from $0$ to $n$ from left to right. Initially, $n$ stones are put into square $0$, and the other squares are empty. At every turn, Sisyphus chooses any nonempty square, say with $k$ stones, takes one of those stones and moves it to the right by at most $k$ squares (the stone should stay within the board). Sisyphus's aim is to move all $n$ stones to square $n$.\n\nProve that Sisyphus cannot reach the aim in less than\n$$\n\\left[ \\frac{n}{1} \\right] + \\left[ \\frac{n}{2} \\right] + \\left[ \\frac{n}{3} \\right] + \\dots + \\left[ \\frac{n}{n} \\right]\n$$\nturns. (As usual, $[x]$ stands for the least integer not smaller than $x$.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25660,
    "subject": "Mathematics (Multi-modal)",
    "question": "The line $l$ passes through the vertex $D$ of the rectangle $ABCD$ and doesn't have any other common points with this rectangle. The point $M$ lies on $l$, and area of the triangle $MBD$ is twice the area of the triangle $MAD$.\nFind all possible values of the ratio of the areas of the triangles $MCD$ and $MAD$.",
    "options": [],
    "answer": "1",
    "solution": "$S(MCD) : S(MAD) = 1$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 25661,
    "subject": "Mathematics (Multi-modal)",
    "question": "Positive real numbers $a$ and $b$ satisfy the following conditions: the function $f(x) = x^3 + a x^2 + 2b x - 1$ has three different real roots, while the function $g(x) = 2x^2 + 2b x + a$ does not have real roots.\nProve that $a - b > 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since the polynomial $f(x) = x^3 + a x^2 + 2b x - 1$ has three different roots, the equation\n$$\nx^3 + a x^2 + 2b x - 1 = 0 \\quad (1)\n$$\nhas three different solutions, denote them by $x_1$, $x_2$ and $x_3$. Vieta's formulas implies the equalities $x_1 + x_2 + x_3 = -a$, $x_1 x_2 + x_1 x_3 + x_2 x_3 = 2b$ and $x_1 x_2 x_3 = 1$. Hence\n$$\nx_1^2 + x_2^2 + x_3^2 = (x_1 + x_2 + x_3)^2 - 2(x_1 x_2 + x_1 x_3 + x_2 x_3) = a^2 - 4b.\n$$\nFrom the AM-GM inequality:\n$$\na^2 - 4b = x_1^2 + x_2^2 + x_3^2 > 3 \\sqrt{x_1^2 x_2^2 x_3^2} = 3 \\quad (2)\n$$\n(the inequality is strict, since the solutions of (1) are distinct).\n\nThe quadratic polynomial $g(x)$ doesn't have roots, therefore its discriminant is negative, i.e. $4b^2 - 8a < 0$, whence $2a - b^2 > 0$. Summing this inequality with (2) we obtain $(2a - b^2) + (a^2 - 4b) > 3$, which is equivalent to $(a + 1)^2 - (b + 2)^2 + 3 > 3$. Therefore,\n$$\n(a + 1 + b + 2)(a + 1 - b - 2) > 0.\n$$\nSince $a$ and $b$ are positive, $a + b + 3 > 0$, so the last inequality implies $a - b - 1 > 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25662,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two ordered sequences $x_1 \\le x_2 \\le \\dots \\le x_n$ and $y_1 \\ge y_2 \\ge \\dots \\ge y_n$ of positive integers are written in two rows, one above the other:\n$$\nx_1 \\ x_2 \\ \\dots \\ x_n\n$$\n$$\ny_1 \\ y_2 \\ \\dots \\ y_n.\n$$\nDenote by $M = \\max(x_1 + y_1, x_2 + y_2, \\dots, x_n + y_n)$ the maximal sum of numbers in columns.\nCould it happen that after the numbers in the second row were rearranged, the maximal sum of numbers in columns became less than $M$?",
    "options": [],
    "answer": "No, it is impossible.",
    "solution": "Answer: no, it's impossible.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25663,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two collections $(x_1, x_2, \\dots, x_n)$ and $(y_1, y_2, \\dots, y_n)$ of real numbers satisfy the inequality $x_i \\ge y_j$ for all indices $i$ and $j$. Let $P = \\max_{1 \\le i \\le n}(x_i - y_i)$ and $G = \\max_{1 \\le i \\le n} x_i - \\min_{1 \\le j \\le n} y_j$.\nProve the inequality $P \\le G \\le nP$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 25664,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sum of several (not necessarily different) real numbers from $[0, 1]$ does not exceed $S$.\nFind the maximal value of $S$ such that it is always possible to partition these numbers into two groups with sums not greater than $9$.",
    "options": [],
    "answer": "17.1",
    "solution": "Answer: $\\max S = 17.1$.\n\nFirst we will show that if $S > 17.1$, the required partition can be impossible. Indeed, let $S = 17.1 + 19\\epsilon$, $\\epsilon > 0$. Suppose that we are given $19$ numbers equal to $0.9 + \\epsilon$. It is evident that for any partition some of the groups will contain at least $10$ of these numbers. But then the sum of the numbers in this group is not less than $9 + 10\\epsilon > 9$, which is forbidden.\n\nNow let $S \\le 17.1$. We will prove that the required partition exists for any such $S$, which will imply $\\max S = 17.1$. Consider all sums $S_1$ of (some of) the given numbers such that $8 < S_1 \\le 9$, and let $A$ be the maximal sum (if such sums don't exist, there is nothing to prove). For any remaining number $a$ holds $A + a > 9$. Consider the next two cases:\n\n**Case 1:** $A \\ge 8.1$. The sum of all remaining numbers doesn't exceed $9$ and we can make the required partition into two groups with the sums $A$ and $S - A$.\n\n**Case 2:** $A < 8.1$. Let $A = 8.1 - x$, $0 < x < 0.1$. If there are not more than nine numbers left, then their sum doesn't exceed $9$ and we can form the second group from them. Suppose that there are at least ten numbers left. Since for any remaining number $a$ holds $A + a > 9$ and $A = 8.1 - x$, we have $a > 0.9 + x$. Hence $S - A \\ge 10 \\cdot (0.9 + x) = 9 + 10x$. But then $S \\ge A + 9 + 10x = 8.1 - x + 9 + 10x = 17.1 + 9x > 17.1$ - a contradiction.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 25665,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let the sequence $(a_n)$ be constructed in the following way:\n$$\na_1 = 1,\\ a_2 = 1,\\ a_{n+2} = a_{n+1} + \\frac{1}{a_n},\\ n = 1, 2, \\dots\n$$\nProve that $a_{180} > 19$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First five terms of the sequence are $a_1 = 1$, $a_2 = 1$, $a_3 = 2$, $a_4 = 3$ and $a_5 = 3.5$.\n$$\n\\text{Note that } a_{n+2}^2 = a_{n+1}^2 + \\frac{2a_{n+1}}{a_n} + \\frac{1}{a_n^2} = a_{n+1}^2 + 2 + \\frac{2}{a_{n-1}a_n} + \\frac{1}{a_n^2} > a_{n+1}^2 + 2.\n$$\nTherefore, $a_{180}^2 > a_5^2 + 350 = 362.25 > 19^2$, whence $a_{180} > 19$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25666,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sum of several (not necessarily different) positive integers not exceeding $10$ is equal to $S$.\nFind all possible values of $S$ such that these numbers can always be partitioned into two groups with the sums of the numbers in each group not exceeding $80$.",
    "options": [],
    "answer": "S ≤ 152",
    "solution": "Answer: $S \\le 152$.\nClearly $S \\le 160$. Suppose that $S \\ge 153$ and let $S = 153 + m$, where $0 \\le m \\le 7$. Consider the next collection: $m$ numbers equal $10$ and $17 - m$ numbers equal $9$ (the total sum equals $153 + m$). At least nine of these $17$ numbers will be in the same group. But the sum of nine smallest numbers is at least $81$ — a contradiction.\n\nIt remains to prove that any collection of positive integers not exceeding $10$ with the sum $S \\le 152$ can be partitioned as required. We will successive put numbers to the first group (in an arbitrary order), and at some moment the sum $A$ of all numbers of this group will satisfy two conditions: 1) $70 < A \\le 80$; and 2) $A + a > 80$ for any remaining number $a$. If $A \\ge 72$, then $S - A \\le 80$ and we can form the second group from all remaining numbers. If $A < 72$, then $A = 71$ and the inequality $A + a > 80$ implies that all remaining numbers equal $10$. There are not more than eight numbers left (otherwise $S \\ge 71 + 9 \\cdot 10 = 161$, since $S \\le 152$). Hence we can form the second group from them.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 25667,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sum of several (not necessary different) positive integers not exceeding $10$ is equal to $S$.\nFind all possible values of $S$ such that these numbers can always be partitioned into two groups with the sum of the numbers in each group not exceeding $70$.",
    "options": [],
    "answer": "S ≤ 133",
    "solution": "**Answer:** $S \\le 133$.\n\nClearly $S \\le 140$. Suppose that $S \\ge 134$ and let $S = 134 + m$, where $0 \\le m \\le 6$. Consider the next collection of numbers: one number equals $8$, $m$ numbers equal $10$ and $14-m$ numbers equal $9$ (the total sum equals $134+m$). At least eight of these $15$ numbers will be in the same group. But the sum of the smallest eight numbers is not less than $1 \\cdot 8 + m \\cdot 9 + (7-m) \\cdot 10 = 78 - m \\ge 72$ — a contradiction.\n\nIt remains to show that any collection of positive integers not exceeding $10$ with the sum $S \\le 133$ can be partitioned into two groups as required. We will successive put numbers to the first group (in an arbitrary order), and at some moment the sum of all numbers of this group will satisfy the next two conditions: 1) $60 < A \\le 70$; and 2) $A + a > 70$ for any remaining number $a$. If $A \\ge 63$, then $S - A \\le 70$ and we can form the second group from all remaining numbers.\n\nNow consider the case $A \\le 62$. Then $A = 62 - x$, where $x$ equals $0$ or $1$. For any remaining number $a$ the inequality $A + a > 70$ takes the form $a > 8 + x$, which is equivalent to $a \\ge 9 + x$. If there are not more than seven numbers left, their sum does not exceed $70$ and we can form the second group from them. Otherwise the sum $S - A$ of the remaining numbers is not less than $8 \\cdot (9 + x) = 72 + 8x$, therefore\n$$\nS = (S - A) + A \\ge 72 + 8x + (62 - x) = 134 + 7x > 133,\n$$\nwhich contradicts to $S \\le 133$. Thus, the required partition is constructed in all cases.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 25668,
    "subject": "Mathematics (Multi-modal)",
    "question": "Cells of an $11 \\times 11$ table are colored with $n$ colors (each cell is colored with exactly one color). For each color the total amount of the cells of this color is not less than $7$ and not greater than $13$.\nProve that there exists at least one row or column which contains cells of at least four different colors.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25669,
    "subject": "Mathematics (Multi-modal)",
    "question": "The altitudes $CC_1$ and $BB_1$ are drawn in the acute triangle $ABC$. The bisectors of angles $\\angle BB_1C$ and $\\angle CC_1B$ intersect the line $BC$ at points $D$ and $E$ respectively and meet each other at point $X$.\nProve that the intersection points of circumcircles of the triangles $BEX$ and $CDX$ lie on the line $AX$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $\\angle BB_1C = \\angle CC_1B = 90^\\circ$, the points $B, C_1, B_1$ and $C$ lie on the circle $\\omega$ with the diameter $BC$. Hence the bisectors of angles $\\angle BB_1C$ and $\\angle CC_1B$ pass through the midpoint of the arc $BC$ of $\\omega$, so this midpoint is $X$.\n\n![](attached_image_1.png)\n\nFirst we prove that the quadrilateral $EC_1B_1D$ is cyclic. The angle $\\angle BDB_1$ is an external angle of the triangle $CB_1D$, hence $\\angle BDB_1 = \\angle ACB + 45^\\circ$. Since $\\angle BB_1C$ is right, $\\angle B_1BC = 90^\\circ - \\angle ACB$. Thus,\n$$\n\\angle B_1C_1E = \\angle B_1C_1C + 45^\\circ = 135^\\circ - \\angle ACB.\n$$\n$$\n\\angle B_1C_1E + \\angle EDB_1 = 180^\\circ.\n$$\nNow we prove that the intersection points of the circumcircles of the triangles $AB_1D$ and $AC_1E$ lie on the line $AX$. Let the circumcircle of the triangle $AB_1D$ intersect the line $AX$ at points $A$ and $P$, then $XP \\cdot XA = XD \\cdot XB_1$. Since points $E, C_1, B_1$ and $D$ are concyclic, $XE \\cdot XC_1 = XD \\cdot AB_1$. Hence $XE \\cdot XC_1 = XP \\cdot XA$, and therefore the circumcircle of the triangle $AC_1E$ passes through $P$.\n\nFinally, we will prove that the circumcircles of the triangles $BEX$ and $CDX$ passes through points $X$ and $P$. Since point $X$ is the midpoint of the arc $BC$, $XB = XC$ and $\\angle BXC = 90^\\circ$. Therefore, $\\angle BCX = \\angle CBX = 45^\\circ$. Since the quadrilateral $AB_1DP$ is cyclic, $\\angle APD = 180^\\circ - \\angle DB_1A = 45^\\circ$. Hence $\\angle DPX + \\angle DCX = 180^\\circ$, i.e. points $P, D, C$ and $X$ are concyclic. Similarly, the quadrilateral $BEPX$ is cyclic. If point $P$ is distinct from point $X$, the problem is solved.\nIf points $P$ and $X$ coincide, the equalities $\\angle DCX = \\angle APD$ and $\\angle EBX = \\angle APE$ imply that the circumcircles of the triangles $BEX$ and $CDX$ are tangent to the line $AX$.\nNote that $BC_1B_1C$ is concyclic, and let $\\omega, \\omega_B$ and $\\omega_C$ be the circumcircles of $BC_1B_1C$, $BEX$, and $CDX$, respectively. Since $B_1X$ and $C_1X$ are the bisectors of $\\angle BB_1C$ and $\\angle CC_1B$, they both pass through the midpoint of the arc $BC$ of $\\omega$ not containing $B_1$ and $C_1$, thus this midpoint is $X$. Let $AX$ intersect $\\omega, \\omega_B$ and $\\omega_C$ the second time at points $T, F_B$ and $F_C$, respectively, and denote $AX \\cap BC = P$.\n\nAs in the first solution, note that $EC_1B_1D$ is cyclic. Let $\\gamma, \\gamma_B$, and $\\gamma_C$ be the circumcircles of $EC_1B_1D$, $BEC_1$, and $CDB_1$, respectively. Since $C_1E$ and $B_1D$ are radical axes of $\\gamma_B, \\gamma$ and $\\gamma_B, \\gamma$ and $C_1E \\cap B_1D = X$, the radical axis of $\\gamma_B, \\gamma_C$ passes through $X$. On the other hand, it passes through $A$ since $\\text{Pow}_{\\gamma_B} A = AC_1 \\cdot AB = AB_1 \\cdot AC = \\text{Pow}_{\\gamma_C} A$. Thus, $AX$ is the radical axis of $\\gamma_B, \\gamma_C$, and it follows from $P \\in AX$ that\n$$\nPF_A \\cdot PX = PD \\cdot PC = PE \\cdot PB = PF_B \\cdot PX.\n$$\nConsequently, $F_A = F_B$ and the problem is solved.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25670,
    "subject": "Mathematics (Multi-modal)",
    "question": "The fence consists of less than $40$ equally spaced boards. On some of them sits a sparrow, in total $10$ sparrows.\n\na) Prove that among the pairwise distances between the sparrows there are two equal.\n\nb) Prove that this statement will remain true even if one of the sparrows flies away.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25671,
    "subject": "Mathematics (Multi-modal)",
    "question": "Positive integers $a$, $b$ and $c$ satisfy the equality\n$$\n\\frac{a^2 - a - c}{b} + \\frac{b^2 - b - c}{a} = a + b + 2.\n$$\nProve that $a + b + c$ is a square of a positive integer.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25672,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sequence $a_1, a_2, a_3, \\dots$ of positive integers is defined in the following way: $a_1$ is given, and for each $n \\ge 2$ the number $a_n$ is the smallest positive integer divisible by $n$, which is not less than $a_{n-1}$. (For example, if $a_5 = 115$, then $a_6 = 120$, $a_7 = 126$, $a_8 = 128$.)\nProve that from a certain moment this sequence coincides with an arithmetic progression.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25673,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all non-constant polynomials $P(x)$ and $Q(x)$ with real coefficients satisfying the equality $P(Q(x)) = P(x)Q(x) - P(x)$.",
    "options": [],
    "answer": "All such pairs are P(x) = a x^2 − a(b+1) x + a b and Q(x) = x^2 − (b+1) x + 2 b, where a and b are real and a ≠ 0.",
    "solution": "Answer: $P(x) = a x^2 - a(b+1)x + a b$, $Q(x) = x^2 - (b+1)x + 2b$, where $a, b \\in \\mathbb{R}$ and $a \\ne 0$.\n\nDenote the degrees of polynomials $P$ and $Q$ by $m$ and $n$, respectively. In the equality\n$$\nP(Q(x)) = P(x)Q(x) - P(x) \\quad (1)\n$$\ncompare the degrees of both sides:\n$$\nm n = m + n \\iff (m-1)(n-1) = 1 \\iff m = n = 2.\n$$\nTherefore we can denote $P(x) = \\alpha x^2 + \\beta x + c$. Hence (1) can be written in the form $\\alpha(Q(x))^2 + \\beta Q(x) = P(x)Q(x) - (P(x)+c)Q(x)$. Since polynomials $P$ and $Q$ have equal degrees, $P(x)+c = a Q(x)$ for some real number $a \\ne 0$. The equality (1) now takes the form\n$$\na Q(Q(x)) - c = (a Q(x) - c) Q(x) - (a Q(x) - c) = a(Q(x))^2 - (a+c)Q(x) + c.\n$$\nSince $Q$ is nonconstant, it attends infinitely many values, hence $a Q(x) = a x^2 - (a+c)x + 2c$, or equivalently $Q(x) = x^2 - (b+1)x + 2b$ (where $b = \\frac{c}{a}$). Then $P(x) = a Q(x) - a b$, i.e. $P(x) = a x^2 - a(b+1)x + a b$. It is easy to verify that such polynomials $P$ and $Q$ satisfy the problem conditions (1). Indeed, since $P$ and $Q$ satisfy $P(x) = a Q(x) - a b$, the equality (1) can be written as $a Q(Q(x)) - a b = (a Q(x) - a b) Q(x) - (a Q(x) - a b)$, or, after simplification, $Q(Q(x)) = (Q(x))^2 - (b+1)Q(x) + 2b$, which is true for $Q(x) = x^2 - (b+1)x + 2b$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 25674,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sequence $a_1, a_2, a_3, \\dots$ of positive integers is defined in the following way: $a_1$ is given, and for each $n \\ge 2$ the number $a_n$ is the smallest positive integer divisible by $n$, which is not less than $a_{n-1}$. (For example, if $a_5 = 115$, then $a_6 = 120, a_7 = 126, a_8 = 128$.)\n\nProve that if $a_1 \\le 2019$, then each positive integer can occur in the sequence $a_{63}, a_{64}, a_{65}, \\dots$ at most once.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25675,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a positive integer $n$ write down all its positive integer divisors in increasing order: $1 = d_1 < d_2 < \\dots < d_k = n$.\nFind all positive integers $n$ divisible by $2019$ such that $n = d_{19} \\cdot d_{20}$.",
    "options": [],
    "answer": "n = 3*673^18 or n = 3^18*673",
    "solution": "Answer: $n = 3 \\cdot 673^{18}$; $n = 3^{18} \\cdot 673$.\n\nNote that the equality $n = d_{19} \\cdot d_{20}$ implies that $n$ has exactly $38$ divisors. Indeed,\n$$\nn = d_{19} \\cdot d_{20} = d_{18} \\cdot d_{21} = d_{17} \\cdot d_{22} = \\dots = d_1 \\cdot d_{38} = d_{38}.\n$$\nHence either $n = p^\\alpha$, where $p$ is prime and then $\\alpha + 1 = 38$, so $n = p^{37}$, or $n = p^\\alpha \\cdot q^\\beta$ ($p$ and $q$ are different primes) and then $38 = (\\alpha + 1)(\\beta + 1)$, i.e. $(\\alpha; \\beta) = (1; 18)$ or $(\\alpha; \\beta) = (18; 1)$, so $n = p \\cdot q^{18}$ or $n = p^{18} \\cdot q$, respectively.\n\nRecall that $n$ is divisible by $2019$ while $2019 = 3 \\cdot 673$, where $3$ and $673$ are primes. Therefore, $n = 3 \\cdot 673^{18}$ or $n = 673 \\cdot 3^{18}$. It remains to verify that in both of these cases $n = d_{19} \\cdot d_{20}$.\n\n1. Let $n = 3 \\cdot 673^{18}$. Arrange the divisors in an ascending order:\n$$\n1 < 3 < 673 < 673 \\cdot 3 < 673^2 < 673^2 \\cdot 3 < \\dots < 673^9 < 673^9 \\cdot 3 < \\dots\n$$\nIt is easy to see that $d_{19} = 673^9$, $d_{20} = 673^9 \\cdot 3$ and $n = d_{19} \\cdot d_{20}$.\n\n2. Let $n = 3^{18} \\cdot 673$, then\n$$\n1 < 3 < 3^2 < 3^3 < 3^4 < 3^5 < 673 < 3^6 < 3 \\cdot 673 < 3^7 < 3^2 \\cdot 673 < 3^8 < 3^3 \\cdot 673 < \\\\\n< 3^9 < 3^4 \\cdot 673 < 3^{10} < 3^5 \\cdot 673 < 3^{11} < 3^6 \\cdot 673 < 3^{12} < \\dots\n$$\nIt is easy to see that $d_{19} = 3^6 \\cdot 673$, $d_{20} = 3^{12}$ and $n = d_{19} \\cdot d_{20}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25676,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all triples $(x; y; z)$ of real numbers satisfying the system\n$$\n\\begin{cases} (x+1)(x^2+1) = y^3+1, \\\\ (y+1)(y^2+1) = z^3+1, \\\\ (z+1)(z^2+1) = x^3+1. \\end{cases}\n$$",
    "options": [],
    "answer": "(0,0,0) and (-1,-1,-1)",
    "solution": "It is easy to see that the triples $(0; 0; 0)$ and $(-1; -1; -1)$ satisfy the system. We will show that there are no other solutions.\nFrom the equations\n$$\n(x+1)(x^2+1) = y^3+1, \\quad (1)\n$$\n$$\n(y+1)(y^2+1) = z^3+1, \\quad (2)\n$$\n$$\n(z+1)(z^2+1) = x^3+1 \\quad (3)\n$$\nit follows that if $x = -1$, then, from (1) $y = -1$, and from (2) $z = -1$. Similarly, if $x = 0$, then $y = 0$ and $z = 0$. Further assume that none of $x$, $y$ and $z$ equals $-1$ or $0$.\nFactor the right-hand sides of (1), (2), (3) and multiply the results:\n$$\n(x+1)(x^2+1)(y+1)(y^2+1)(z+1)(z^2+1) = (y+1)(y^2-y+1)(z+1)(z^2-z+1)(x+1)(x^2-x+1).\n$$\nReducing by $(x+1)(y+1)(z+1) \\ne 0$, we get\n$$\n(x^2+1)(y^2+1)(z^2+1) = (x^2-x+1)(y^2-y+1)(z^2-z+1). \\quad (*)\n$$\nNote that if $x < -1$, then, according to (1), $y < -1$ and, according to (2), $z < -1$. In this case all factors in the right-hand side of (*) are greater than corresponding factors in the left-hand side — a contradiction.\nNow factor the left-hand sides of equations (1), (2) and (3), and subtracting 1 from both sides of these equations, we obtain the equations\n$$\nx(x^2 + x + 1) = y^3, \\quad (4)\n$$\n$$\ny(y^2 + y + 1) = z^3, \\quad (5)\n$$\n$$\nz(z^2 + z + 1) = x^3. \\quad (6)\n$$\nNote that if $x > 0$, then, according to (4), $y > 0$ and, according to (5), $z > 0$. In this case all factors in the right-hand side of (*) are less than corresponding factors in the left-hand side — a contradiction.\nIt remains to consider the case, when $x, y$ and $z$ belong to the interval $(-1; 0)$ and, in particular, are negative. Again, all factors in the right-hand side of (*) are greater than corresponding factors in the left-hand side. So, there are no other solutions.\nConsider the values of polynomials $x^3 + x^2 + x$ and $x^3$ on the intervals $(0; +\\infty)$, $(-\\infty; -1)$ and $(-1; 0)$.\nIf $x > 0$, then $x^3 + x^2 + x > x^3 > 0$. From the equations of the system successively obtain the inequalities $y > x > 0$, $z > y > 0$ and $x > z > 0$, whence $x > z$.\nIf $x < -1$, then $x^3 + x^2 < 0$ and $x^2 + x > 0$, hence $-1 > x^3 + x^2 + x > x^3$. From the equations of the system we obtain the inequalities $-1 > y > x$, $-1 > z > y$ and $-1 > x > z$, and again $x > z$.\nIf $x \\in (-1; 0)$, then $x^3 + x^2 > 0$ and $x^2 + x < 0$, therefore $-1 < x^3 + x^2 + x < x^3 < 0$. From the system we obtain the inequalities $-1 > y > x$, $-1 > z > y$ and $-1 > x > z$, whence again $x > z$.\nIn all three cases we got a contradiction and the two remaining variants $x = 0$ and $x = 1$ easily lead us to the triples from the answer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25677,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sequence $a_1, a_2, a_3, \\dots$ of positive integers is defined in the following way: $a_1 = 63$ and for each $n \\ge 2$ the number $a_n$ is the smallest positive integer divisible by $n$, which is not less than $a_{n-1}$. (For example, $a_2 = 64$, $a_3 = 66, a_4 = 68, a_5 = 70$.)\nProve that each positive integer can occur in this sequence at most once.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25678,
    "subject": "Mathematics (Multi-modal)",
    "question": "Does there exist a function $f(x): \\mathbb{R} \\to \\mathbb{R}$ satisfying the equality\n$$\nf(|x|) + |f(x)| = x\n$$\nfor all real numbers $x$?",
    "options": [],
    "answer": "No; such a function does not exist.",
    "solution": "Answer: no, such function doesn't exist.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25679,
    "subject": "Mathematics (Multi-modal)",
    "question": "The tangents to the circumcircle of the acute triangle $ABC$, passing through the vertices $B$ and $C$, meet at point $F$. The points $M$, $L$ and $N$ are the feet of the perpendiculars from the vertex $A$ to the lines $FB$, $FC$ and $BC$ respectively.\nProve the inequality $AM + AL \\ge 2AN$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note that $\\angle ABM = \\angle ACB$ by the property of the angle between the tangent to the circle at point $B$ and the chord $AB$. The right triangles $AMB$ and $ANC$ are similar since they have equal acute angles, therefore $\\frac{AM}{AN} = \\frac{AB}{AC}$, whence $AN = \\frac{AM \\cdot AC}{AB}$.\n\nBy analogy, the triangles $ANB$ and $ALC$ are similar and $\\frac{AL}{AN} = \\frac{AC}{AB}$, whence $AN = \\frac{AL \\cdot AB}{AC}$. Therefore\n$$\nAN^2 = \\frac{AM \\cdot AC}{AB} \\cdot \\frac{AL \\cdot AB}{AC} = AM \\cdot AL.\n$$\n![](attached_image_1.png)\n$$\n\\text{Finally, from the AM GM inequality } AM + AL \\ge 2\\sqrt{AM \\cdot AL} = 2\\sqrt{AN^2} = 2AN.\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 25680,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f(x): \\mathbb{R} \\to \\mathbb{R}$ satisfying for all $x \\in \\mathbb{R}$ the inequality\n$$\nx = \\frac{3}{4}f(|x|) + |f(x)|.\n$$\n(I. Voronovich)",
    "options": [],
    "answer": "All functions f: R -> R such that f(x) = -4x for x >= 0, and for each x < 0 one may choose independently f(x) = 2x or f(x) = -2x.",
    "solution": "Answer: $f(x) = -4x$ for all $x \\ge 0$ and $f(x) = \\pm 2x$ for all $x < 0$ (signs are chosen independently).",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 25681,
    "subject": "Mathematics (Multi-modal)",
    "question": "The angle $ADC$ of the parallelogram $ABCD$ equals $40^\\circ$. The point $K$ is given such that the segments $AK$ and $BC$ intersect, $AK = BC$ and $\\angle BAK = 80^\\circ$. The point $L$ is given such that the segments $CL$ and $AD$ intersect, $CL = AB$ and $\\angle BCL = 80^\\circ$.\nFind the angles of the triangle $BKL$.",
    "options": [],
    "answer": "60°, 60°, 60°",
    "solution": "All angles equal $60^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25682,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f(x): \\mathbb{R} \\to \\mathbb{R}$ satisfying the equality\n$$\nx = -\\frac{1}{2}f(|x|) + |f(x)|\n$$\nfor all real numbers $x$.",
    "options": [],
    "answer": "f(x) = 2x for x ≥ 0, and f(x) = 0 for x < 0",
    "solution": "$f(x) = 2x$ for all $x \\ge 0$ and $f(x) = 0$ for all $x < 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25683,
    "subject": "Mathematics (Multi-modal)",
    "question": "The point $M$ is the midpoint of the side $BC$ of the triangle $ABC$. A circle is passing through $B$, is tangent to the line $AM$ at $M$ and intersects the segment $AB$ secondary at the point $P$.\nProve that the circle, passing through $A$, $P$ and the midpoint of the segment $AM$, is tangent to the line $AC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $N$ be the midpoint of $AM$ and $D$ be the reflection of $A$ in $M$. We will prove that the quadrilateral $BPND$ is cyclic by showing the equality\n![](attached_image_1.png)\n$$\nAP \\cdot AB = AN \\cdot AD.\n$$\nSince $AM$ is tangent to the circumcircle of the triangle $BPM$,\n$$\nAP \\cdot AB = AM^2 = 2AN \\cdot 0.5AD = AN \\cdot AD.\n$$\nHence $AP \\cdot AB = AN \\cdot AD$, the quadrilateral $BPND$ is cyclic and therefore $\\angle APN = \\angle BDA$. The lines $BD$ and $AC$ are parallel (since $ACDB$ is the parallelogram), whence $\\angle BDA = \\angle CAD$. This leads us to the equality $\\angle APN = \\angle CAN$, which means that the circumcircle of the triangle $APN$ is tangent to the line $AC$.\nSince the circumcircle of the triangle $BPM$ is tangent to the line $AM$, the triangles $AMP$ and $ABM$ are similar and, in particular, $\\angle AMP = \\angle ABC$. From the similarity follows the equalities $\\frac{PM}{MN} = \\frac{2PM}{AM} = \\frac{2BM}{BA} = \\frac{CB}{BA}$, hence the triangles $PMN$ are similar to $CBA$ and $\\angle PNM = \\angle CAB$. Wherein $\\angle PNM = \\angle PAN + \\angle APN$ as an external angle of the triangle $APN$, and $\\angle CAB = \\angle PAN + \\angle CAN$, whence as in the first solution $\\angle APN = \\angle CAN$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25684,
    "subject": "Mathematics (Multi-modal)",
    "question": "A positive integer $n > 10$ has two different positive integer divisors $a$ and $b$ such that $n = a^2 + b$.\nProve that strictly between the numbers $a$ and $b$ there is at least one another divisor of $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "From the equality $n = a^2 + b$ it follows that $b$ is divisible by $a$, since $a$ divides both $n$ and $a^2$. Let $b = ma$, where $m > 1$. Then $n = a^2 + ma = a(a + m)$, so $a + m$ divides $n$. Let us show that $a < a + m < b$. Suppose $a + m \\ge b$, then\n$$\na + m \\ge ma \\iff am - a - m + 1 \\le 1 \\iff (a - 1)(m - 1) \\le 1.\n$$\nThe latter inequality holds only if $a = m = 2$ whence $b = 4$, but then $n = 2 \\cdot 4 = 8 < 10$, which contradicts with the conditions of the problem. Therefore, $a + m$ is the required divisor of $n$, located between $a$ and $b$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25685,
    "subject": "Mathematics (Multi-modal)",
    "question": "The diagonals of the inscribed quadrilateral $ABCD$ intersect at the point $O$. The points $P$, $Q$, $R$ and $S$ are the feet of the perpendiculars from $O$ to the sides $AB$, $BC$, $CD$ and $DA$ respectively.\nProve the inequality $BD \\ge SP + QR$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First we prove that the quadrilateral $SPQR$ is circumscribed and $O$ is the center of its incircle. The quadrilaterals $APOS$ and $BPOQ$ are cyclic since they have pairs of right angles, based on $AO$ and $BO$, respectively. In these circles $\\angle OPS = \\angle OAS$ and $\\angle OPQ = \\angle OBQ$. Wherein $\\angle OAS = \\angle OBQ$ in the circumcircle of $ABCD$. Therefore, $\\angle OPS = \\angle OPQ$ and $OP$ is the bisector of the angle $SPQ$. Similarly, $OQ$, $OR$ and $OS$ are the bisectors of angles $PQR$, $QRS$ and $RSP$, respectively. Hence point $O$ is equidistant from the sides of the quadrilateral $SPQR$, i.e. $SPQR$ is circumscribed and $O$ is its incenter.\n\nSince $SPQR$ is circumscribed, $SP + QR = PQ + SR$, hence it is enough to prove the inequality $BD \\ge PQ + SR$. The segments $BO$ and $DO$ are the diameters of the circumcircles of the quadrilaterals $BPOQ$ and $DSOR$, respectively. Therefore, $BO \\ge PQ$ and $DO \\ge SR$. Summing up these inequalities, we obtain the desired inequality.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 25686,
    "subject": "Mathematics (Multi-modal)",
    "question": "The circles $\\omega_1$ and $\\omega_2$ centered at $O_1$ and $O_2$, respectively, intersect at the point $X$. The line $O_1X$ intersects $\\omega_2$ at the points $X$ and $B$, and the line $XO_2$ intersects $\\omega_1$ at the points $X$ and $A$. The line $O_1A$ intersects $\\omega_1$ at the points $A$ and $D$, and the line $O_2B$ intersects $\\omega_2$ at the points $B$ and $C$. Let $M$ be the midpoint of the segment $CD$.\nProve that the lines $MX$ and $AB$ are perpendicular to each other.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25687,
    "subject": "Mathematics (Multi-modal)",
    "question": "The circles $\\omega_1$ and $\\omega_2$ centered at $O_1$ and $O_2$, respectively, intersect at two different points $E$ and $F$. The line $O_1E$ intersects for the second time $\\omega_1$ at the point $A$, and $\\omega_2$ at the point $C$. The line $O_2E$ intersects for the second time $\\omega_2$ at the point $B$, and $\\omega_1$ at the point $D$. Let $M$ and $N$ be the midpoints of the segments $AD$ and $BC$, respectively.\nProve that the circumcircles of the triangles $FMD$ and $FNC$ intersect at the line $DC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25688,
    "subject": "Mathematics (Multi-modal)",
    "question": "The circles $\\omega_1$ and $\\omega_2$ centered at $O_1$ and $O_2$, respectively, intersect at the point $X$. The line $XO_1$ intersects $\\omega_1$ at the points $X$ and $A$. The line $XO_2$ intersects $\\omega_2$ at the points $X$ and $B$. Let $M$ be the midpoint of the segment $AB$. The ray $MO_1$ intersects $\\omega_1$ at the point $D$, and the ray $MO_2$ intersects $\\omega_2$ at the point $C$.\n\na) Prove that the triangle $MDC$ is isosceles.\n\nb) Prove that the line $DC$ passes through $X$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25689,
    "subject": "Mathematics (Multi-modal)",
    "question": "The circles $\\omega_1$ and $\\omega_2$ centered at $O_1$ and $O_2$, respectively, intersect at two different points $E$ and $F$. The line $O_1E$ intersects for the second time $\\omega_1$ at the point $A$, and $\\omega_2$ at the point $C$. The line $O_2E$ intersects for the second time $\\omega_2$ at the point $B$, and $\\omega_1$ at the point $D$.\nProve that the perpendiculars to the lines $AC, BD$ and $DC$, passing through $O_1, O_2$ and $E$, respectively, are concurrent.\n(A. Voidelevich)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25690,
    "subject": "Mathematics (Multi-modal)",
    "question": "The bisectors of angles $\\angle A$ and $\\angle C$ of a convex quadrilateral $ABCD$ meet at the point $E$, and the bisectors of angles $\\angle B$ and $\\angle D$ meet at the point $F$ ($E$ and $F$ lie in the interior of $ABCD$). The point $M$ is the midpoint of the segment $EF$. The points $H_1, H_2, H_3$ and $H_4$ are the foots of the perpendiculars from $M$ to the sides $AB, BC, CD$ and $AD$ respectively.\nProve that $MH_1 + MH_3 = MH_2 + MH_4$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $E$ lies on the bisector of $\\angle BAD$ ($\\angle BCD$), the point $E$ is equidistant from the sides $AB$ and $AD$ (respectively, $BC$ and $CD$). A similar statement holds for the point $F$. Denote the distances from the point $E$ to the lines $AB$ and $BC$ by $x_1$ and $y_1$, respectively, and the distances from the point $F$ to the lines $AB$ and $AD$ be $x_2$ and $y_2$, respectively. Since the length of the midline in a trapezium is equal to the half-sum of the base lengths,\n$$\nMH_1 = \\frac{x_1 + x_2}{2}, \\quad MH_2 = \\frac{x_2 + y_1}{2}, \\quad MH_3 = \\frac{y_1 + y_2}{2} \\quad \\text{and} \\quad MH_4 = \\frac{x_1 + y_2}{2}.\n$$\n$$\n\\text{Therefore, } MH_1 + MH_3 = \\frac{x_1 + y_1 + x_2 + y_2}{2} = MH_2 + MH_4.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25691,
    "subject": "Mathematics (Multi-modal)",
    "question": "The numbers $S_1 = 2^2$, $S_2 = 2^4$, $\\dots$, $S_n = 2^{2n}$ are given. A rectangle $OABC$ is constructed on the Cartesian plane according to these numbers. For this, starting from the point $O$ the points $A_1, A_2, \\dots, A_n$ are consistently marked along the axis $Ox$, and the points $C_1, C_2, \\dots, C_n$ are consistently marked along the axis $Oy$ in such a way that for all $k$ from $1$ to $n$ the lengths of the segments $A_{k-1}A_k = x_k$ and $C_{k-1}C_k = y_k$ are positive integers (let $A_0 = C_0 = O$, $A_n = A$ and $C_n = C$) and $x_k \\cdot y_k = S_k$.\n\na) Find the maximal possible value of the area of the rectangle $OABC$ and all pairs of sets $(x_1, x_2, \\dots, x_n)$ and $(y_1, y_2, \\dots, y_n)$ at which this maximal area is reached.\n\nb) Find the minimal possible value of the area of the rectangle $OABC$ and all pairs of sets $(x_1, x_2, \\dots, x_n)$ and $(y_1, y_2, \\dots, y_n)$ at which this minimal area is reached.",
    "options": [],
    "answer": "a) Maximal area: (1/3)(4^n − 1)(4^n + n − 1). Achieved by (x_1, …, x_n) = (1, 1, …, 1, 2^{2n}) and (y_1, …, y_n) = (2^2, 2^4, …, 2^{2n−2}, 1).\n\nb) Minimal area: 4(2^n − 1)^2. Achieved either by (x_1, …, x_n) = (2^2, 2^3, …, 2^{n+1}) and (y_1, …, y_n) = (2^0, 2^1, …, 2^{n−1}), or by both sets equal to (2^1, 2^2, …, 2^n).",
    "solution": "Answer: \n\na) The maximal possible value of area equals $\\frac{1}{3}(4^n - 1)(4^n + n - 1)$ and it is attained at sets $(1, 1, \\dots, 1, 2^{2n})$ and $(2^2, 2^4, \\dots, 2^{2n-2}, 1)$.\n\nb) The minimal possible value equals $4(2^n - 1)^2$ and it is attained in two cases: either the sets $x_i$ and $y_i$ equal $(2^2, 2^3, \\dots, 2^{n+1})$ and $(2^0, 2^1, \\dots, 2^{n-1})$ or both sets equal $(2^1, 2^2, \\dots, 2^n)$.\n\nSince the lengths $x_i$ and $y_j$ are divisors of the given numbers, the total amount of sets is finite, so there exist the minimal area of the rectangle $OABC$. We will prove the following\n\n**Lemma.** The area of the rectangle $OABC$ is maximal if and only if the next two conditions hold:\n\n1) For each $i$ from $1$ to $n$ either $x_i$ or $y_i$ equals $1$;\n\n2) The difference $|(x_1 + x_2 + \\dots + x_n) - (y_1 + y_2 + \\dots + y_n)|$ is minimal among all sets, satisfying condition 1).\n\n**Proof of the lemma.** Take sets $(x_1, x_2, \\dots, x_n)$ and $(y_1, y_2, \\dots, y_n)$ such that the area of the rectangle $OABC$ is maximal.\n\nFirst we prove the necessity of the first condition. Suppose there exists $k$ such that both $x_k$ and $y_k$ are greater than $1$. Denote $x_1 + x_2 + \\dots + x_n = X$, $X - x_k = X_k$, $y_1 + y_2 + \\dots + y_n = Y$ and $Y - y_k = Y_k$. Then the area of the rectangle $OABC$ equals $S = XY = (X_k + x_k)(Y_k + y_k)$. Without loss of generality let $X_k \\ge Y_k$. Replace in the sets $(x_i)$ and $(y_i)$ the lengths $x_k$ and $y_k$ by $1$ and $S_k$, respectively. The area of the new rectangle equals $S' = (X_k + 1)(Y_k + S_k)$. The difference between new and old areas equals\n\n$$\nS' - S = (X_k + 1)(Y_k + S_k) - (X_k + x_k)(Y_k + y_k) = (X_k y_k - Y_k)(x_k - 1) > 0,\n$$\n\nwhich contradicts the minimality of $S$.\n\nNow we prove the necessity of the second condition. The first condition implies\n\n$$\nX + Y = S_1 + S_2 + \\dots + S_n + n.\n$$\n\nThis sum depends only on the given values $S_1, S_2, \\dots, S_n$, denote this sum by $C$. The area of the rectangle $OABC$ equals $S = XY = X(C - X)$, which is the quadratic polynomial of the variable $X$. The corresponding parabola is opened to the bottom and the abscissa of its vertex equals $C/2$ (which corresponds $X = Y$). Hence the area of the rectangle is maximal when $X$ is closest to $C/2$, which is equivalent to the minimality of $|X - Y|$.\n\nTo prove the sufficiency of the conditions 1) and 2) note that for all sets satisfying these conditions the values of $X$ and $Y$ are the same (up to permutation), hence they all provide the maximal value of the area. The lemma is proved.\n\nProceed to the solution of a). Since $2^{2n} = 1 + 2^0 + 2^1 + \\dots + 2^{2n-1}$, we have\n\n$$\n2^{2n} > 2^2 + 2^4 + \\dots + 2^{2n-2},\n$$\n\nwhence the conditions of the lemma are satisfied if and only if one set equals $(1, 1, \\dots, 1, 2^{2n})$ while the second set equals $(2^2, 2^4, \\dots, 2^{2(n-1)}, 1)$. Wherein the maximal possible area of the rectangle equals $\\frac{1}{3}(4^n - 1)(4^n + n - 1)$. Thus, a) is finished.\n\nTo find the minimal possible area, we use the Cauchy-Bunyakovsky inequality\n\n$$\n(x_1 + x_2 + \\dots + x_n)(y_1 + y_2 + \\dots + y_n) \\ge (\\sqrt{x_1y_1} + \\sqrt{x_2y_2} + \\dots + \\sqrt{x_ny_n})^2 = (\\sqrt{S_1} + \\sqrt{S_2} + \\dots + \\sqrt{S_n})^2.\n$$\n\nIf there exist at least one pair of sets $(x_i)$ and $(y_i)$ at which this inequality is the equality, then the last expression provides the minimal possible area. Moreover, then this minimal value is attained only for such sets $(x_i)$ and $(y_i)$. The Cauchy-Bunyakovsky inequality turns into the equality if and only if $\\frac{x_1}{y_1} = \\frac{x_2}{y_2} = \\dots = \\frac{x_n}{y_n}$. Since $x_1y_1 = S_1 = 4$, there exist the three variants: $x_1 = 4$, $y_1 = 1$; $x_1 = 1$, $y_1 = 4$; and $x_1 = y_1 = 2$. First two variants give the pair $(2^2, 2^3, \\dots, 2^{n+1})$ and $(2^0, 2^1, \\dots, 2^{n-1})$ of sets, and the third variant gives two sets equal $(2^1, 2^2, \\dots, 2^n)$. Both cases provide the minimal possible area $4(2^n - 1)^2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25692,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ satisfying the equality\n$$\nf(f(x) + f(y)) = (x + y)f(x + y)\n$$\nfor all real $x$ and $y$.",
    "options": [],
    "answer": "f(x) = 0 for all x in R",
    "solution": "Answer: $f(x) = 0$ for all $x \\in \\mathbb{R}$.\nConsider several preliminary substitutions to the problem condition\n$$\nf(f(x) + f(y)) = (x + y)f(x + y). \\qquad (1)\n$$\nSubstitution $x = y = 0$ gives the equality $f(2f(0)) = 0$. Substitution $x = 2f(0)$, $y = 0$ gives $f(f(0)) = 0$. Finally, substitution $x = y = f(0)$ gives $f(0) = 0$.\nSubstitution $y = 0$ to (1) implies that for all $x \\in \\mathbb{R}$ holds $f(f(x)) = xf(x)$. Suppose that for some $a, b \\in \\mathbb{R}$ holds $f(a) = f(b) \\ne 0$, then $af(a) = f(f(a)) = f(f(b)) = bf(b)$, whence $a = b$. Therefore $f$ can attend several times only zero.\nWe will show that there exists $c \\ne 0$ such that $f(c) = 0$. Substitution $y = 1-x$ to (1) implies $f(f(x)+f(1-x)) = f(1)$. If $f(1) = 0$, then $c = 1$. Otherwise, $f(x)+f(1-x) = 1$ for all $x \\in \\mathbb{R}$. Substitutions of $x = 0$ and $x = 1/2$ in the latter equation give $f(1) = 1$ and $f(1/2) = 1/2$. Finally for $x = 1$ and $y = 1/2$ the equality (1) gives $f(3/2) = 3/2f(3/2)$, whence $f(3/2) = 0$.\nDenote by $c$ an arbitrary nonzero number such that $f(c) = 0$. Substitution $x = y = c$ to (1) implies $f(2c) = 0$. Suppose that there exist a real number $t$ such that $f(t) \\ne 0$. Putting to (1) successively $x = c$, $y = -c + t$ and $x = 2c$, $y = -2c + t$, we obtain\n$$\nf(f(-c + t)) = tf(t) = f(f(-2c + t)), \\text{ where } tf(t) \\ne 0.\n$$\nTherefore, $f(-c+t) = f(-2c+t) \\ne 0$, whence $-c+t = -2c+t$, which contradicts $c \\ne 0$. Consequently such $t$ doesn't exist and $f(x) = 0$ for all $x \\in \\mathbb{R}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25693,
    "subject": "Mathematics (Multi-modal)",
    "question": "At each node of the checkered $n \\times n$ board sat a beetle. At midnight, each beetle crawled into the center of a cell. It turned out that the distance between any two beetles sitting in the adjacent (along the side) nodes did not increase.\nProve that at least one beetle crawled into the center of a cell at the vertex of which it sat initially.\n(A. Voidelevich)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**First solution.** We will prove the statement of the problem in the general case of rectangular boards $n \\times m$. Suppose in contrary that there exist such positive integers $n$ and $m$ and a movement of beetles, that none of the beetles crawled into the center of the cell at the vertex of which it initially sat. Take such a pair $(n, m)$ with minimal $n+m$. Clearly, for $1 \\times 1$ board the statement of the problem is true, so without loss of generality assume that $n \\ge 2$.\n\nEnumerate the columns from left to right by numbers $1, \\dots, m$ and the vertical gridlines by numbers $0, \\dots, m$. Similarly, enumerate the rows from top to bottom by numbers $1, \\dots, n$ and the horizontal gridlines by numbers $0, \\dots, n$. So, $(i, j)$-cell have $(i-1, j-1), (i-1, j), (i, j)$ and $(i, j-1)$ vertices.\n\nSuppose that for some $i$, $1 \\le i \\le m-1$, the beetle crawled from $(2, i-1)$-node to the center of $(1, i)$-cell. Then the beetle from $(1, i-1)$-node couldn't move to the centers of $(1, i-1)$ and $(2, i)$ cells, so it crawled to the center of $(1, i+1)$-cell. Hence the beetle from $(1, i)$-node crawled to the center of $(1, i+2)$-node. Similarly, the beetle from the $(1, j)$-node, $i < j < m-1$, crawled to the center of $(1, j+2)$-cell. Finally, for the beetle from $(1, m-1)$-node we obtain a contradiction, since it had to move to the center of the cell, adjacent to $(1, m)$-cell: either $(1, m-1)$-cell or $(2, m)$-cell, both having $(1, m-1)$-node as a vertex. If we now suppose that the beetle from $(2, i-1)$-node, $2 \\le i \\le m$, moved to the center of $(1, i-1)$-cell, we obtain a contradiction in a similar way.\n\nConsider the next transformation of the $n \\times m$ board. Remove the first row and all beetles from the top (0-th) gridline. For each of the remaining beetles, which crawled to the first row, let it now crawl one row below. And for the rest of the beetles don't change anything. It is easy to see that after transformation we obtain a counterexample on to the statement of the problem on the $(n-1, m)$ board. But this contradicts the minimality of $n+m$. Therefore, our assumption of the existence of a counterexample is false.\n\n**Second solution.** By a distance between a beetle and a cell call the minimal number of unit moves along the gridlines, which this beetle require to get to the vertex of the cell. The distance between the beetle $\\mathfrak{K}$ and the cell $C^{\\mathfrak{K}}$, at which it crawled, call its *path*.\n\nWe are to prove that there exists a beetle with zero path. Suppose a contrary and choose the beetle $\\mathfrak{K}_1$ with minimal path. The gridlines, passing through the starting vertex of $\\mathfrak{K}_1$ divide the board into four rectangles. Without loss of generality let $C^{\\mathfrak{K}_1}$ lie in the top right rectangle. Since the path of $\\mathfrak{K}_1$ is positive, we may assume that $C^{\\mathfrak{K}_1}$ is not located on the left border of the rectangle.\n\nConsider the beetle $\\mathfrak{K}_2$, which sat to the right of $\\mathfrak{K}_1$. It crawled to $C^{\\mathfrak{K}_1}$ or to the adjacent to it. But the distance between $\\mathfrak{K}_2$ and $C^{\\mathfrak{K}_1}$ is less than the path of $\\mathfrak{K}_1$, and the same hold for the left and bottom adjacent cells of $C^{\\mathfrak{K}_1}$. Hence $C^{\\mathfrak{K}_2}$ is located to the right or top of $C^{\\mathfrak{K}_1}$ and the next condition hold: (1) the path of $\\mathfrak{K}_2$ equals to the path of $\\mathfrak{K}_1$, i.e. it is minimal; and (2) $C^{\\mathfrak{K}_2}$ lies in the top right rectangle, with respect to $\\mathfrak{K}_2$.\n\nReplacing the beetle $\\mathfrak{K}_1$ with $\\mathfrak{K}_2$ we find the beetle $\\mathfrak{K}_3$ satisfying the conditions (1) and (2). Continue this considerations, we obtain the sequence of beetles, each satisfying (1) and (2), such that the distance from the beetles to the top right cell of the board is decreasing at each step. It is evident that at some moment we get a contradiction.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 25694,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the Paircity there are $n \\ge 2$ married couples numbered from $1$ to $n$. On the New Year's eve, each man congratulated two women from couples whose numbers differ by one. It turned out that for any two men from couples whose numbers differ by one, there is a woman who received congratulations from both of them.\n\nProve that at least one man congratulated his wife.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Call a man *leftside* if he congratulated women from couples with numbers less than his own, otherwise call him *rightside*. Clearly, the man from the first couple is leftside and the man from the $n$-th couple is rightside. Hence, there is a number $i$, $1 \\le i < n - 1$, such that the man from the $i$-th couple is leftside while the man from the $(i+1)$-th couple is leftside. The numbers of couples of these men differ by one, but they congratulated four different women — a contradiction to the problem condition. Consequently, at least one man congratulated his wife.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25695,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $n \\ge 5$ cities in some country. Some of the cities are connected with each other by roads, and the next three conditions are satisfied:\n1) there is at most one road between any two cities;\n2) not all cities are connected with each other;\n3) there are exactly $k \\ge 1$ roads between any four cities.\nFind all $n$ and $k$ at which this situation is possible.",
    "options": [],
    "answer": "n = 5, k = 3",
    "solution": "Answer: $n = 5, k = 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25696,
    "subject": "Mathematics (Multi-modal)",
    "question": "Each cell of an $n \\times n$ table is filled with one of the two signs: «+» and «-». For each $k$ from $1$ to $n$ the amount of pluses in the first $k$ rows is greater than the amount of minuses in the first $k$ columns.\nFind the maximal possible number of minuses in the table.",
    "options": [],
    "answer": "floor((n^2 - 1)/2)",
    "solution": "Answer: $\\left[ \\frac{n^2 - 1}{2} \\right]$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25697,
    "subject": "Mathematics (Multi-modal)",
    "question": "Peter and Andrey play the game on the $n \\times 1$ board, making moves alternate. Peter starts, and on his turn he places «+» to any empty cell. Andrey on his turn places «-» to any empty cell. The game is finished when all cells are filled. Peter's prize equals to the greatest number $k$ such that for each $\\ell$ from 1 to $k$ there are $\\ell$ successive cells, the amount of pluses on which is greater than the amount of minuses.\nFind the maximal prize which Peter can guarantee for himself.",
    "options": [],
    "answer": "n + \u000b(-1 + (-1)^{n+1})/2",
    "solution": "Answer: $n + \\frac{-1 + (-1)^{n+1}}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25698,
    "subject": "Mathematics (Multi-modal)",
    "question": "Call a polygon on a Cartesian plane to be *integer* if all its vertices are *integer*. A convex integer 14-gon is cut into integer parallelograms with areas not greater than $C$.\nFind the minimal possible $C$.\n(A. Yuran)",
    "options": [],
    "answer": "5",
    "solution": "Answer: minimal possible $C$ equals $5$.\n\nFirst we prove two lemmas.\n\n**Lemma 1:** the given 14-gon has $7$ pairs of opposite parallel sides.\nSince the 14-gon is convex, it suffices to prove that each side has a side parallel to it. Consider an arbitrary side $AB$ of the given 14-gon. Draw the line perpendicular to $AB$ and define on this line the direction to the right in such way that the whole 14-gon lies to the right of $AB$ (it is possible since the 14-gon is convex). Consider a parallelogram $P_1$ with one (left) side lying on $AB$ and the opposite (right) side lies on another side or inside the 14-gon. If it lies inside the 14-gon, there is another parallelogram $P_2$, the left side of which intersect the right side of $P_1$, while the right side of $P_2$ lies on the side or inside the 14-gon. Continue choosing such parallelograms $P_1, P_2, \\dots, P_n$ until finally we find the parallelogram $P_n$, the right side of which lies on some side of the given 14-gon. This side is the required side, parallel to $AB$. Lemma 1 is proved.\n\nCall the sequence $P_1, P_2, \\dots, P_n$ of the parallelograms, defined in the proof of the lemma 1 for an arbitrary side $AB$, the chain of the side $AB$.\n\n**Lemma 2:** the chains of any two distinct non-opposite sides intersect (i.e. have a common parallelogram).\nConsider any two pairs $AB \\parallel A'B'$ and $CD \\parallel C'D'$ of opposite sides. Since the 14-gon is convex, if we bypass its perimeter (in some direction), the sides $AB$, $CD$, $A'B'$, $C'D'$ go in that order. Connect the midpoints of the parallel to $AB$ sides of the parallelograms from the chain of $AB$ to obtain a polyline (the ends of the polyline are the midpoints of $AB$ and $A'B'$). This polyline divides the 14-gon into two parts, with the sides $CD$ and $C'D'$ lying in distinct parts. Therefore, if we consider a similar polyline, corresponding to the chain of $CD$, these two polylines will intersect. This means that the chains of the sides $AB$ and $CD$ have a common parallelogram. Lemma 2 is proved.\n\nTake any $7$ pairwise nonparallel sides of the 14-gon and to each of them (say, for $EF$) put in the correspondence the shortest possible integer vector $\\vec{v}(EF)$. Call this vector to be *directive* for $EF$ and denote its coordinates by $x(EF)$ and $y(EF)$. The minimality of the length of $\\vec{v}(EF)$ implies $\\gcd(x(EF), y(EF)) = 1$. From Lemma 2 it follows that for any two non-opposite sides $AB$ and $CD$ there exists a parallelogram $P$ with the sides parallel to $AB$ and $CD$. The area of $P$ is divisible by the area of the parallelogram with the sides $v(AB)$ and $v(CD)$, since $P$ can be divided into such parallelograms.\n\nFinally, we will prove that there exists a parallelogram with area divisible by $5$. The area $S(P)$ of the parallelogram $P$ equals $|x(AB)y(CD) - x(CD)y(AB)|$. If both $x(AB)$ and $x(CD)$ are divisible by $5$, then $S(P)$ is divisible by $5$. Suppose that not more than one directive vector has an $x$-coordinate which is divisible by $5$. Among at least $6$ directive vectors with nonzero $x$-coordinates there exist vectors with $\\frac{y(AB)}{x(AB)} \\equiv \\frac{y(CD)}{x(CD)} \\pmod{5}$. Therefore, the area $S(P)$ of the corresponding parallelogram equals\n$$\n|x(AB)y(CD) - x(CD)y(AB)| = \\left| \\left( \\frac{y(AB)}{x(AB)} - \\frac{y(CD)}{x(CD)} \\right) \\cdot x(AB)x(CD) \\right| \\equiv 0 \\pmod{5}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25699,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P(x)$ be a non-constant polynomial with integer coefficients such that $P(0) \\neq 1$. Prove that there exist infinitely many primes $p$ such that $P(a) - a^{\\frac{p-1}{2}}$ is divisible by $p$ for some positive integer $a$ (possibly, depending on $p$).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider the polynomial $Q(x) = P(x^2) - 1 \\in \\mathbb{Z}[x]$. It's well-known that there exist infinitely many prime divisors of the numbers from the set $M = \\{Q(n) : n \\in \\mathbb{N} \\ \\&\\ Q(n) \\neq 0\\}$. Moreover, since $Q(0) = P(0) - 1 \\neq 0$, among these prime divisors there exist infinitely many primes which doesn't divide $Q(0)$. Clearly, if $p \\mid Q(n)$ and $p \\nmid Q(0)$, then $p \\nmid n$.\n\nTherefore there exist infinitely primes $p$ and $n \\in \\mathbb{N}$ such that $p \\nmid n$ and $p \\mid P(n^2) - 1$. Each such $p$ satisfy the problem conditions with $a = n^2$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25700,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the minimal positive integer $k$ for which there exist pairwise distinct positive integers $a$, $b$ and $c$ such that all three numbers: $4a^2 + kb + c$, $4b^2 + kc + a$ and $4c^2 + ka + b$ are perfect squares.",
    "options": [],
    "answer": "4",
    "solution": "Since the expressions $4a^2 + kb + c$, $4b^2 + kc + a$ and $4c^2 + ka + b$ are invariant under cyclic permutations of $a$, $b$ and $c$, we can assume that $c = \\max(a, b, c)$. Then for $k \\le 3$\n$$\n(2c)^2 < 4c^2 + ka + b \\le 4c^2 + 3c + c < 4c^2 + 4c + 1 = (2c + 1)^2.\n$$\nHence $4c^2 + ka + b$ is not a perfect square if $k \\le 3$.\n\nFor $k = 4$, one can find the triple $(a, b, c) = (120, 85, 141)$, considering the expressions modulus $20$.\nFor the alternative version we will find all triples $(a, b, c)$ which satisfy the problem conditions for $k = 4$.\nAs at the beginning, we say that $c = \\max(a, b, c)$ and\n$$\n(2c)^2 < 4c^2 + 4a + b < 4c^2 + 5c < 4c^2 + 8c + 4 = (2c + 2)^2,\n$$\nhence $4c^2 + 4a + b = (2c + 1)^2$, i.e. $4a + b = 4c + 1$.\nSuppose $b > a$, then\n$$\n4b^2 + 4b + 1 \\le 4b^2 + 4c - 4 + 1 < 4b^2 + 4c + a = 4b^2 + 4a + b - 1 + a < 4b^2 + 8b + 4.\n$$\nWhence $(2b+1)^2 < 4b^2 + 4c + a < (2b+2)^2$ – a contradiction. Therefore $b < a < c$. Note that $(2b+1)^2 < 4b^2 + 4c + a$ is true since we didn't use the inequality $b > 1$.\nSince $4a^2 + 4b + c > (2a)^2$, we get $4a^2 + 4b + c \\ge (2a + 1)^2$, which is equivalent to $4b + c \\ge 4a + 1$. Using the fact that $c = a + (b - 1)/4$, we obtain $3a \\le 4\\frac{1}{4}b - 1\\frac{1}{4}$. Consider the expression $4b^2 + 4c + a$ one more time:\n$$\n\\begin{aligned}\n4b^2 + 4c + a &= 4b^2 + 5a + b - 1 \\le 4b^2 + \\frac{5}{3} \\cdot (4\\frac{1}{4}b - 1\\frac{1}{4}) + b - 1 = \\\\\n&= 4b^2 + 8\\frac{1}{12}b - 3\\frac{1}{12} < 4b^2 + 12b + 9 = (2b + 3)^2.\n\\end{aligned}\n$$\n\nTherefore $(2b+1)^2 < 4b^2 + 4c + a < (2b+3)^2$, whence $4b^2 + 4c + a = (2b+2)^2$, i.e. $8b+4 = 4c+a$.\nNote that $8a+4 = 2(4c+1-b)+4 = 8c+6-2b > 4b+c$, hence\n$$\n(2a)^2 < 4a^2 + 4b + c < 4a^2 + 8a + 4 = (2a + 2)^2,\n$$\nwhich yields $4a^2 + 4b + c = (2a + 1)^2$, i.e. $4b + c = 4a + 1$. Thus the triple $(a, b, c)$ satisfy the system\n$$\n\\begin{cases}\n4a + b = 4c + 1; \\\\\n8b + 4 = 4c + a; \\\\\n4b + c = 4a + 1.\n\\end{cases}\n$$\nThe solution of this system is $(120, 85, 141)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25701,
    "subject": "Mathematics (Multi-modal)",
    "question": "The set $\\{2, 3, 4, \\ldots, 2020\\}$ is partitioned into triples. In each triple $(a, b, c)$ the numbers were arranged in the ascending order, i.e. $a < b < c$, and the difference $|b - \\frac{a+c}{2}|$ is called the error of this triple. Find the maximal possible sum of errors of all $673$ triples.",
    "options": [],
    "answer": "395808",
    "solution": "Note that if we shift all numbers in the set by the same value, the errors will remain the same. Hence we can consider the partitions of the set $\\{1, 2, \\ldots, 2019\\}$, moreover, we will solve the generalized problem by changing $2019$ with an arbitrary number $3n$, $n \\in \\mathbb{N}$.\n\nConsider the partition of the set $\\{1, 2, \\ldots, 3n\\}$, which provides the maximal sum of all errors. Let's say that the triple $a < b < c$ is left if $b \\le (a+c)/2$, otherwise let's say that it's right.\n\nWe start with the following\n\n**Step 1.** Suppose the triple $a < b < c$ is right and $b \\le c - 2$. Choose a triple $x < y < z$ such that $b+1 \\in \\{x, y, z\\}$ and swap the numbers $b$ and $b+1$ in these triples. Since the error of the first triple will increase by $1$ and we started from the partition with the maximal sum of errors, it follows that $b+1 = y$ and $x < y < z$ is left (otherwise the sum of errors will increase). Similarly, if the triple $a < b < c$ is left and $a \\le b - 2$ then there exists a right triple $x < b - 1 < y$ and we will swap $b$ and $b-1$ in these triples. We will perform such operation while it's possible. Clearly, this process is finite, since the middle numbers in the right triples increase while in the left triples they decrease and the triples doesn't change their type. The final partition will satisfy the\n\n**Property 1.** Each right triple has the form $a < b < b+1$ and each left triple has the form $a < a+1 < c$.\n\n**Step 2.** Consider any two triples $x < y - 1 < y$ and $a < a + 1 < b$ of different types. The sum of errors of these two triples equals $\\frac{b+y-a-x}{2} - 2$.\n\nSuppose $y < a$. Replace these triples with $x < a < a+1$ and $y - 1 < y < b$, then the new sum of errors will become equal to $\\frac{b+a-y-x}{2} - 2 > \\frac{b+y-a-x}{2} - 2$, which contradicts the maximality of our partition. Hence $a > y$.\n\nSuppose $b < x$. Replace these triples with $b < y - 1 < y$ and $a < a + 1 < x$, then the new sum of errors will become equal to $\\frac{y+x-b-a}{2} - 2 > \\frac{b+y-a-x}{2} - 2$, which contradicts the maximality of our partition. Hence $b > x$.\n\nConsider two triples $x < y - 1 < y$ and $t < z - 1 < z$ of the same (without loss of generality, right) type. The sum of errors of these two triples equals $\\frac{z+y-t-a}{2} - 2$. Suppose $y < t$ and swap $x$ and $t$ in these triples. Then we get triples $y - 1 < y < t$ and $x < z - 1 < z$ of different types with their sum of errors $\\frac{z+t-x-y+1}{2} - 2 > \\frac{z+y-t-a}{2} - 2$, which contradicts the maximality of our partition. Hence $y > t$.\n\nTherefore the partition satisfy\n\n**Property 2.** The maximal element of each triple is greater than the minimal element of any other triple.\n\n**Step 3.** Suppose there exist triples $x < y - 1 < y$ and $a < a+1 < b$ such that $a < x$, and choose such triples with minimal possible $x$ and (for that value of $x$) maximal possible $a$. Consider the number $a+2$: due to the property 2 it cannot be the maximal element of any triple and due to our choice of the pair $(x, a)$ it cannot be the minimal element of any triple except $x < y - 1 < y$. Thus we have found triples\n\n$a < a + 1 < b$ and $a + 2 < y - 1 < y$ with the sum of errors $\\frac{b+y-2a-2}{2} - 2$. Consider the triples $a+1 < a+2 < b$ and $a < y-1 < y$: their sum of errors equals $\\frac{b+y-2a-1}{2} - 2 > \\frac{b+y-2a-2}{2} - 2$, which contradicts the maximality of our partition. Hence $x > a$. Replacing each number $t$ with $3n-t$ we get the inequality $b > y$. Hence we proved the final\n\n**Property 3.** The minimal element of each right triple is smaller than the minimal element of each left triple. The maximal element of each left triple is greater than the maximal element of each right triple.\n\nCombining the properties 1–3 we obtain the following description of the partition with the maximal sum of errors: there exist $k, \\ell \\in \\mathbb{N}$ with $k + \\ell = n$ such that the numbers $1, 2, \\dots, k$ are minimal in the right triples, $k+1, k+3, \\dots, k+2\\ell-1$ are minimal in the left triples, $k+2\\ell+2, k+2\\ell+4, \\dots, 3k+2\\ell$ are maximal in the right triples, $3k+2\\ell+1, 3k+2\\ell+1, \\dots, 3k+3\\ell$ are maximal in the left triples. In such partition the sum of errors equals\n\n$$\n\\begin{align*}\n& \\frac{(k+2\\ell+2) + (k+2\\ell+4) + \\dots + (3k+2\\ell) - 1 - 2 - \\dots - k}{2} - k + \\\\\n& \\qquad + \\frac{(3k+2\\ell+1) + (3k+2\\ell+2) + \\dots + (3k+3\\ell) - (k+1) - (k+3) - \\dots}{2} \\\\\n& \\qquad \\dots - \\frac{(k+2\\ell-1)}{2} - \\ell = \\frac{(k+2\\ell)k + 2 \\cdot \\frac{k(k+1)}{2} - \\frac{k(k+1)}{2}}{2} + \\\\\n& \\qquad + \\frac{(3k+2\\ell)\\ell + \\frac{\\ell(\\ell+1)}{2} - (k-1)\\ell - 2 \\cdot \\frac{\\ell(\\ell+1)}{2}}{2} - k - \\ell = \\\\\n&= \\frac{3k(k-1) + 3\\ell(\\ell-1) + 8k\\ell}{4} = \\frac{3(k+\\ell)^2 - 3(k+\\ell) + 2k\\ell}{4} = \\frac{3n^2 - 3n + 2k\\ell}{4}.\n\\end{align*}\n$$\n\nThe expression $2k\\ell = 2k(n-k)$ attains its maximal value at $k = \\left\\lfloor \\frac{n}{2} \\right\\rfloor$, hence we get the answer\n\n$$\n\\frac{3n^2 - 3n + 2 \\lfloor \\frac{n}{2} \\rfloor \\lfloor \\frac{n}{2} \\rfloor}{4}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25702,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\mathbb{R}_+$ be the set of all positive integers. Find all functions $g: \\mathbb{R}_+ \\to \\mathbb{R}_+$ such that numbers $x, y \\in \\mathbb{R}_+$ satisfy the equality\n$$\nxg(x + g(y)) = g(g(xy) + 1).\n$$",
    "options": [],
    "answer": "g(x) = c/x for any positive constant c",
    "solution": "Denote $xy = z$, then the variables $y$ and $z$ can take all positive real values independently. For these variables the equality has the form\n$$\n\\frac{z}{y}g\\left(\\frac{z}{y} + g(y)\\right) = g(g(z) + 1). \\qquad (1)\n$$\nFix any value of $z$ and vary $y$ over $\\mathbb{R}_+$, then $g(g(z) + 1)$ is fixed while $\\frac{z}{y}$ attains all positive real values, hence $g(\\frac{z}{y} + g(y))$ attains all positive real values, i.e. $g$ is surjective.\nSuppose $\\frac{g(y_2) - g(y_1)}{y_2 - y_1} > 0$ for some positive reals $y_2 \\neq y_1$. Substitute $z = y_2 y_1 \\cdot \\frac{g(y_2) - g(y_1)}{y_2 - y_1}$ into (1) firstly with $y = y_2$, secondly with $y = y_1$. In both cases $g(\\frac{z}{y} + g(y)) = g(\\frac{y_2 g(y_2) - y_1 g(y_1)}{y_2 - y_1})$, hence we will obtain that $y_1 = y_2$ – a contradiction. Thus $\\frac{g(y_2) - g(y_1)}{y_2 - y_1} \\le 0$ for all positive reals $y_2 \\neq y_1$ which means that $g$ is non-strictly decreasing.\nSince $g$ is surjective and non-strictly decreasing, it's continuous and, in particular $\\lim_{y \\to +\\infty} g(y) = 0$. Now consider the initial equation\n$$\nxg(x + g(y)) = g(g(xy) + 1),\n$$\nwhen $y \\to +\\infty$. Then $g(x + g(y)) \\to g(x)$ and $g(xy) + 1 \\to 1$. Hence $xg(x) = g(1)$ and $g(x) = \\frac{c}{x}$, where $c = g(1)$. Straightforward verification shows that such function satisfy the problem conditions for any $c \\in \\mathbb{R}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25703,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a \\neq b$ be coprime positive integers such that for each integer $k$ from $1$ to $10$ the numbers $a + k$ and $b + k$ are not coprime.\nFind the minimal possible value of $|a - b|$.",
    "options": [],
    "answer": "2310",
    "solution": "Answer: $2310$. Suppose $a$ and $b$ satisfy the problem conditions. For each integer $k$ from $1$ to $10$ denote by $p_k$ the minimal prime divisor of $\\gcd(a+k, b+k)$. Since the $\\gcd$ of numbers divides their difference, among $10$ successive integers at most $\\lfloor \\frac{10}{p_k} \\rfloor$ can be a multiple of $p_k$. For $p_k = 2$ this number is $5$, for $p_k = 3$ it is $4$, for $p_k = 5$ and $p_k = 7$ it is $2$ and for the others it equals $1$.\nIf all $p_k$ are odd then the equality $10 = 4 + 2 + 2 + 1 + 1$ implies that there is at least five distinct primes. Then $b - a \\ge 3 \\cdot 5 \\cdot 7 \\cdot 11 \\cdot 13 > 2310$. Consider the case when some $p_i = 2$. Since $a$ and $b$ are coprime, $i$ cannot be even so $2 = p_1 = p_3 = \\dots = p_9$. Among five primes $p_2, p_4, \\dots, p_{10}$ at most $2$ can be equal to $3$, and for each prime $p \\ge 5$ at most one can be equal to $p$. Now the equality $10 = 5 + 2 + 1 + 1 + 1$ implies that $b - a \\ge 2 \\cdot 3 \\cdot 5 \\cdot 7 \\cdot 11 = 2310$.\nOn the other hand, if $a = 2311$ and $b = 4621$ then $p_1 = p_3 = \\dots = p_9 = 2$, $p_2 = p_8 = 3$, $p_4 = 5$, $p_6 = 7$ and $p_{10} = 11$, whence $|a - b| = 2310$ is possible.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25704,
    "subject": "Mathematics (Multi-modal)",
    "question": "Integers from $1$ to $2022$ are written on the cards and placed in a row on the table. Each number is used once and there is exactly one number on each card. Mary plays the following game: on each move she takes any card from the table and puts it into her right pocket, then she takes the leftmost card and puts it into her left pocket. After $1011$ moves the table is empty, the game stops and Mary calculates the sum of the numbers written on the cards in her left pocket. For each initial position $\\mathcal{P}$ of the cards denote by $S(\\mathcal{P})$ the maximal possible sum that Mary can reach.\nFind the number of initial positions $\\mathcal{P}$, for which the sum $S(\\mathcal{P})$ is minimal possible.",
    "options": [],
    "answer": "2^1011",
    "solution": "Let Mary always choose the card with the maximal number among the two leftmost cards. Then on $i$-th move the difference between the sums of numbers in her pockets increases at least by $1$ which implies that this strategy allows to get at least $S = 2 + 4 + \\dots + 2022 = 1011 \\cdot 1012$. In particular, $S(P) \\ge M$ for each $P$ and moreover if at some moment the difference between the cards is more than $1$ then Mary gets the sum more than $M$. Hence if $S(P) = M$ then the row consists of the pairs $\\{1, 2\\}$, $\\{3, 4\\}$, $\\dots$, $\\{2021, 2022\\}$ arranged in some order.\n\nSuppose that $S(P) = M$. If there exist two adjacent pairs $\\{2i - 1, 2i\\}$ and $\\{2j - 1, 2j\\}$ (in that order from left to right) such that $i < j$ then choosing two numbers from the second pair instead of the first pair will allow Mary to obtain the sum greater than $M$, a contradiction. Hence if $i < j$ any pair $\\{2j - 1, 2j\\}$ is placed to the left from any pair $\\{2i - 1, 2i\\}$.\n\nFinally consider any such initial position $P$, it has the form $\\{2021, 2022\\}$, $\\{2019, 2020\\}$, $\\ldots$, $\\{1, 2\\}$ where the elements in the pairs are arbitrarily permuted. On $i$-th move Mary will put into her left pocket the card with number which is at least $2023 - 2i$. Hence the sum of the numbers in her left pocket will be not less than $2021 + 2019 + \\ldots + 1$, which implies that the sum in her right pocket will not exceed $M$. Therefore $S(P) = M$ for all $2^{1011}$ such positions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25705,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\Omega$ be a circle, $S$ be a point outside $\\Omega$ and $SA, SB$ be tangents from $S$ to $\\Omega$. Arbitrary point $K$ is chosen on the segment $AB$. Points $U, W, R$ and $T$ are chosen on $\\Omega$ such that $UW \\cap RT = K$ and the points $W, T$ lie on the same half-plane with respect to the line $AB$. The line $SK$ intersects $\\Omega$ at $P$ and $Q$ and $P$ lies on the same half-plane to the line $AB$ as $W$ and $T$. The lines $WR$ and $TU$ intersect $AB$ at $C$ and $D$ respectively. Let $M$ be the midpoint of $PQ$.\nProve that $\\angle AMC = \\angle BMD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will use multiple times the following well-known\n**Lemma.** Let the chords $AB$ and $CD$ of a circle intersect at the point $K$ then\n$$ \\frac{AC \\cdot AD}{BC \\cdot BD} = \\frac{CK}{DK} $$\n**Proof of lemma.** This equality directly follows from the sine laws for the triangles $ACK, ADK$ and $BCD$.\n\nThe quadrilateral $APBQ$ is harmonic, so $MK$ is the bisector of the angle $AMB$. Hence $\\frac{AC}{CB} = \\frac{\\sin AMC}{\\sin CMB} \\cdot \\frac{AK}{BK}$ and $\\frac{AD}{DB} = \\frac{\\sin AMD}{\\sin DMB} \\cdot \\frac{AK}{BK}$, therefore it's enough to prove the equality $\\frac{AC}{CB} \\cdot \\frac{AD}{DB} = \\left(\\frac{AK}{BK}\\right)^2$.\n\nThe lemma for $AB \\cap RW = C$ and $AB \\cap UT = D$ implies that\n$$\n\\frac{AC}{CB} = \\frac{AW \\cdot AR}{BW \\cdot BR} \\quad \\text{and} \\quad \\frac{AD}{DB} = \\frac{AT \\cdot AU}{BT \\cdot BU}\n$$\nAnd the lemma for $AB \\cap RT = K$ and $AB \\cap UW = D$ implies that\n$$\n\\frac{AK}{KB} = \\frac{AT \\cdot AR}{BT \\cdot BR} = \\frac{AW \\cdot AU}{BW \\cdot BU}\n$$\nCombining these equalities we obtain the required equality.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25706,
    "subject": "Mathematics (Multi-modal)",
    "question": "Do there exist six pairwise distinct positive integers $a$, $b$, $c$, $d$, $e$ and $m$, such that\n$$\n\\begin{cases}\na+b+c = d+e+m, \\\\\nab+bc+ac = de+em+dm, \\\\\nabc = dem + 3^{2021} \\cdot 2^{2022}?\n\\end{cases}\n$$\n(Igor Voronovich)",
    "options": [],
    "answer": "yes",
    "solution": "**Answer:** yes, such numbers exist.\nConsider the triplets $(a, b, c) = (13, 4, 3)$ and $(d, e, m) = (12, 7, 1)$. These numbers satisfy the system\n$$\n\\begin{cases}\na+b+c = 20 = d+e+m, \\\\\nab+bc+ac = 103 = de+em+dm, \\\\\nabc - dem = 3^2 \\cdot 2^3.\n\\end{cases}\n$$\nClearly, for $k = 2^{673} \\cdot 3^{673}$ the numbers $ak$, $bk$, $ck$, $dk$, $ek$ and $mk$ satisfy the problem conditions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25707,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that all divisors of any positive integer can be arranged in the circle such that for any two consecutive divisors one would divide another.\n(Mikhail Karpuk)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let's prove this statement by induction on the number of distinct prime divisors of a given number. In this case, this number will be denoted by $n$, the number of its divisors — by $d(n)$ and the divisors written in counterclockwise order — by $a_1, a_2, \\dots, a_{d(n)}$, where $a_1 = 1$.\n\n*Base:* let $n = p^k$, where $p$ is a prime number. Let's arrange all divisors of the number $n$ in ascending order: $a_1 = 1, a_2 = p, \\dots, a_{k+1} = p^k$. It is obvious that such arrangement satisfies the condition.\n\n*Step:* suppose the assertion is proved for all numbers having at most $s$ different prime divisors, let us prove it for an arbitrary number $n$ having $s + 1$ prime divisors. We write $n$ as $n = p^k m$, where $p$ is a prime number that does not divide $m$.\n\nIf $\\ell$ is a sequence of numbers written in a clockwise circle then by $\\ell^*$ we will denote the same sequence but written in reverse order, i. e., if $\\ell$ is $a_1, a_2, \\dots, a_k$, then $\\ell^*$ is $a_k, \\dots, a_2, a_1$. Moreover by $q\\ell$ we denote the sequence obtained from $\\ell$ by multiplying all its terms by $q$, i. e. $q\\ell$ is $qa_1, qa_2, \\dots, qa_k$.\n\nBy the induction hypothesis all divisors of the number $m$ can be arranged in a circle according to the condition: $a_1, a_2, \\dots, a_{d(m)}$. Denote this sequence $\\ell$. Then we arrange the divisors of the number $n$ around the circle in the following order:\n\n$\\ell, (p\\ell)^*, p^2\\ell, \\dots, (p^{k-1}\\ell)^*, p^k\\ell$, if $k$ is odd,\nand\n$\\ell, (p\\ell)^*, p^2\\ell, \\dots, p^{k-1}\\ell, (p^k\\ell)^*$, if $k$ is even.\n\nIt is easy to see that these sequences satisfy the condition of the problem.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25708,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. On the segment $[0, n]$ of the real line there are marked $m$ pairwise distinct segments with integer endpoints. It is known that it's impossible to choose a set of these segments such that the sum of their lengths is $n$ and their union is $[0, n]$. (We say that the two segments are distinct if their endpoints doesn't coincide. The union of the segments is considered as the union of the sets of real numbers.)\nFind the maximal possible value of $m$.",
    "options": [],
    "answer": "n(n-1)/2",
    "solution": "Answer: $\\frac{n(n-1)}{2}$.\n\nEach set of segments that satisfies the condition of the problem is called *good*. Consider a set consisting of all segments of the form $[a, b]$, where $a, b \\in \\mathbb{N}$ and $a < b \\le n$. This collection contains $n(n-1)/2$ segments and it is good since even the union of all segments of the collection does not cover the point $0$. Therefore the maximum possible value of the number $m$ is not less than $n(n-1)/2$.\n\nLet us prove the following\n**Claim:** For any good collection $A$ there exists a good collection $B$ containing the same number of segments and not containing any segment of the form $[0, k]$, $k = 1, \\ldots, n$.\n\n**Proof of the claim:** If the set $A$ does not contain any segment of the form $[0, k]$, then put $B = A$. Otherwise among all segments of $A$ of the form $[0, k]$, consider the segment $[0, s]$ of minimum length. Let us exclude the segment $[0, s]$ from the set and instead add the segment $[s, n]$, which obviously was not contained in $A$. The resulting set will also be good. Indeed if it's possible to select several segments of the total length $n$ from the new set, the union of which would coincide with the entire segment $[0, n]$, then among them there must be a segment $[s, n]$, but then several non-intersecting segments would cover the segment $[0, s]$ and among them there would be a segment of the form $[0, t]$ where $t < s$, which is impossible due to the minimality of the number $s$. Repeating the described procedure several times, we obtain the required set $B$. The assertion is proved.\n\nFrom the proved statement, in particular, it follows that the number of segments in an arbitrary good set is not more than $n(n-1)/2$, and hence the maximum possible value of the number $m$ is exactly equal to $n(n-1)/2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25709,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two lines pass through the point $F(0; \\frac{1}{4})$ on the Cartesian plane. These lines are perpendicular to each other and intersect the parabola $y = x^2$ at the points $A$, $B$, $C$ and $D$ (these points are mentioned in the $x$-coordinate increasing order). The difference of projections of the segments $AD$ and $BC$ to the $x$-axis equals $m$.\nFind the area of the quadrilateral $ABCD$.",
    "options": [],
    "answer": "m^2/2",
    "solution": "Answer: $\\frac{m^2}{2}$.\nDenote the abscissas of the points $A$, $B$, $C$ and $D$ by $a$, $b$, $c$ and $d$ respectively. It is easy to see that the perpendicularity of the lines $AC$ and $BD$ is equivalent to the equality $(a+c)(b+d) = -1$ and the condition about the difference of projections is equivalent to the equality $(a+c)-(b+d)=m$. The fact that the point $F$ belongs to the lines $AC$ and $BD$ means that $ac = bd = -\\frac{1}{4}$. Let $a+c = p$ and $b+d = q$, then the numbers $a$ and $c$ are the roots of the quadratic equation $x^2-px-\\frac{1}{4}=0$ and the numbers $b$ and $d$ are the roots of the quadratic equation $x^2-qx-\\frac{1}{4}=0$. Hence $c-a = \\sqrt{p^2+1}$ and $d-b = \\sqrt{q^2+1}$. Since the diagonals $AC$ and $BD$ of the quadrilateral $ABCD$ are perpendicular, its area $S$ is equal to half the product of the lengths of the diagonals. So,\n$$\n\\begin{aligned}\nS &= \\frac{1}{2}(c-a)(d-b)\\sqrt{1+(c+a)^2}\\sqrt{1+(d+b)^2} = \\\\\n&= \\frac{1}{2}(1+p^2)(1+q^2) = \\frac{(q+qp^2)(p+pq^2)}{2pq} = \\frac{-(p-q)^2}{-2} = \\frac{m^2}{2}.\n\\end{aligned}",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25710,
    "subject": "Mathematics (Multi-modal)",
    "question": "Point $D$ is marked on the half-circle with the diameter $AB$ and the center $O$. The points $E$ and $F$ are the midpoints of lesser arcs $AD$ and $BD$ respectively. It is known that $O$ lies on the line passing through the orthocenters of the triangles $ADF$ and $BDE$.\nFind all possible values of the angle $AOD$.",
    "options": [],
    "answer": "30° or 150°",
    "solution": "Let $H_1$ and $H_2$ be the intersection points of the altitudes of the triangles $ADF$ and $BDE$. Note that the point $O$ is the circumcenter of both triangles $ADF$ and $BDE$. Therefore the coinciding lines $H_1O$ and $H_2O$ are the Euler lines of these triangles and they contain the intersection points $M_1$ and $M_2$ of the medians of the triangles $ADF$ and $BDE$ respectively.\nLet $K$ and $N$ be the midpoints of the segments $AD$ and $BD$ respectively. Then $M_1 \\in FK$, $M_2 \\in EN$ and $KM_1 : M_1F = NM_2 : M_2E = 1:2$. Denote by $R$ the radius of the semicircle given in the condition.\nLet $L_1$ be the point on the extension of the ray $OM_1$ beyond the point $M_1$, such that $M_1L_1 = 2OM_1$. Since $KM_1 : M_1F = OM_1 : M_1L = 1:2$, the triangles $KM_1O$ and $FM_1L_1$ are similar with the coefficient $1/2$. Hence\n$$\n\\angle FL_1O = \\angle FL_1M = \\angle KOM_1 = \\angle EOL_2 \\quad \\text{and} \\quad L_1F = 2OK = BD. \\quad (1)\n$$\nLet $L_2$ be the point on the extension of the ray $OM_2$ beyond the point $M_2$, such that $M_2L_2 = 2OM_2$. Similarly, we obtain the equalities\n$$\n\\angle EL_2O = \\angle FOM_1 \\text{ and } EL_2 = AD. \\quad (2)\n$$\nThe equalities (1) and (2) implies that the triangles $OEL_2$ and $L_1FO$ are similar, therefore\n$$\n\\frac{EL_2}{EO} = \\frac{OF}{FL_1} \\iff \\frac{AD}{R} = \\frac{R}{BD} \\iff AD \\cdot BD = R^2.\n$$\nDraw the altitude $DH$ of the triangle $ABD$. According to the formula for the length of the altitude drawn to the hypotenuse, $DH = \\frac{AD \\cdot BD}{AB} = \\frac{R^2}{2R} = \\frac{DO}{2}$. Hence in the right-angled triangle $DOH$, the angle $DOH$ is equal to $30^\\circ$. If the point $H$ lies on the segment $OB$ then $\\angle DOB = \\angle DOH = 30^\\circ$, and if it lies on the segment $OA$ then $\\angle DOB = 180^\\circ - \\angle DOH = 150^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25711,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given $n \\ge 2$ pairwise distinct integers bigger than $-a$, such that among them the amount of odd numbers equals to the largest even number and the amount of even numbers equals to the largest odd number.\n\na) Find the minimal possible $n$ over all positive integers $a \\ge 2$.\n\nb) For each integer $a \\ge 2$ find the maximal possible $n$.",
    "options": [],
    "answer": "a) 3; b) 2a - 1",
    "solution": "**a)** The largest odd and even numbers are positive integers since there are numbers of both parities. This means that the largest even number is not less than two, the largest even number is not less than one, and the total number of numbers is not less than three. Note that $n = 3$ could be, for example, if $a = 3$ and the numbers $-1$, $1$ and $2$ are given.\n\n**b)** Let $2x + 1$ be the largest odd number and $2y$ be the largest even number. The number $2x + 1$ of even numbers does not exceed $y + \\lfloor \\frac{a+1}{2} \\rfloor$, since they are all at most $2y$ and at least $-a + 1$. And the number $2y$ of odd numbers does not exceed $x + 1 + \\lfloor \\frac{a}{2} \\rfloor$, because they are not greater than $2x + 1$ and not less than $-a + 1$. Hence\n$$\n\\begin{cases} 2x + 1 \\le y + \\lfloor \\frac{a+1}{2} \\rfloor, \\\\ 2y \\le x + 1 + \\lfloor \\frac{a}{2} \\rfloor. \\end{cases} \\quad (1)\n$$\nSumming up these equalities we get $x + y \\le \\lfloor \\frac{a+1}{2} \\rfloor + \\lfloor \\frac{a}{2} \\rfloor = a$, therefore $n = 2x + 1 + 2y \\le 2a + 1$. If $n = 2a + 1$ then the inequalities in (1) must be the equalities and in this case\n$$\n\\begin{cases} 3x = (x + y) + \\lfloor \\frac{a-1}{2} \\rfloor, \\\\ 3y = (x + y) + \\lfloor \\frac{a+2}{2} \\rfloor. \\end{cases}\n$$\nHowever this is impossible, since the right-hand sides of these equalities give different remainders when divided by $3$.\n\nSince the number $n$ is odd and $n < 2a + 1$, then $n \\le 2a - 1$. Let us show that this estimate is attainable. For even $a$ one can take $x = a/2 - 1$ and $y = a/2$, and for odd $a$ one can take $x = y = (a - 1)/2$, then the system (1) will be satisfied and, obviously, there will be corresponding $n$-tuples of numbers satisfying the condition.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25712,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two circles $\\omega_1$ and $\\omega_2$ intersect each other at points $X$ and $Y$. Two lines pass through $Y$: one intersects $\\omega_1$ and $\\omega_2$ for the second time at points $A$ and $B$ respectively and another intersects $\\omega_1$ and $\\omega_2$ for the second time at points $C$ and $D$ respectively. The line $AD$ intersects $\\omega_1$ and $\\omega_2$ for the second time at points $P$ and $Q$ respectively such that $YP = YQ$.\nProve that circumcircles of $\\triangle BCY$ and $\\triangle PQY$ touch each other.\n(Palina Chernikava)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The quadrilaterals $ACYP$ and $BDQY$ are cyclic, so\n$$\n\\angle ACD = \\angle ACY = \\angle YPQ \\text{ and } \\angle ABD = \\angle YBD = \\angle YQP.\n$$\nSince the triangle $YPQ$ is isosceles, $\\angle YPQ = \\angle YQP$ whence $\\angle ACD = \\angle ABD$. So the quadrilateral $ACBD$ is cyclic. Therefore\n$$\n\\begin{aligned} \n\\angle CYP &= \\angle CYA + \\angle AYP = \\angle BYD + (\\angle YPQ - \\angle YAP) = \\\\ \n&= \\angle YPQ + (\\angle BYD - \\angle BAD) = \\angle YQP + \\angle ADC = \\\\ \n&= \\angle YQP + \\angle ABC = \\angle YQP + \\angle YBC. \n\\end{aligned}\n$$\nThe last equality is equivalent to the tangency of the circumcircles of the triangles $BCY$ and $PQY$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25713,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $a$ for which there exists a polynomial $p(x)$ with integer coefficients such that\n$$\np(\\sqrt{2} + 1) = 2 - \\sqrt{2} \\quad \\text{und} \\quad p(\\sqrt{2} + 2) = a.\n$$",
    "options": [],
    "answer": "a = 7k - 2 for any positive integer k",
    "solution": "$a = 7k - 2$ where $k$ is an arbitrary positive integer.\n\nSuppose that the integer $a$ and the polynomial $p(x)$ satisfy the condition. For the polynomial $q(x) = p(x + 1)$, the equalities from the condition have the form $q(\\sqrt{2}) = 2 - \\sqrt{2}$ and $q(1 + \\sqrt{2}) = a$. Since the coefficients of the polynomial $q(x)$ are integers, the equality $q(-\\sqrt{2}) = 2 + \\sqrt{2}$ is true. Therefore the numbers $\\sqrt{2}$ and $-\\sqrt{2}$ satisfy the equality $q(x) = 2 - x$, i.e. they are roots of the polynomial $q(x) + x - 2$. According to Bezout's theorem, the polynomial $q(x) + x - 2$ is divisible by $(x - \\sqrt{2})(x + \\sqrt{2}) = x^2 - 2$. Let $q(x) + x - 2 = (x^2 - 2)h(x)$, then the Gauss lemma implies that the rational coefficients of the polynomial $h(x)$ are integers. Substitute into the resulting equality $x = 1 + \\sqrt{2}$ and take into account that $q(1 + \\sqrt{2}) = a$, after transformations we obtain the equality\n$$\na + \\sqrt{2} - 1 = (1 + 2\\sqrt{2}) \\cdot h(1 + \\sqrt{2}). \\quad (1)\n$$\nSince the coefficients of the polynomial $h(x)$ are integers,\n$$\na - \\sqrt{2} - 1 = (1 - 2\\sqrt{2}) \\cdot h(1 - \\sqrt{2}).\n$$\nLet's multiply this equality with (1):\n$$\n(a - 1)^2 - 2 = -7 \\cdot (h(1 + \\sqrt{2}) \\cdot h(1 - \\sqrt{2})).\n$$\nSince the expression in brackets is a product of conjugate numbers, it is an integer, so $(a - 1)^2 - 2$ is divisible by $7$. This is equivalent to saying that $a$ is congruent to $4$ or $5$ modulo $7$.\n\nFor the numbers $a = 7k + 4$, $k \\in \\mathbb{Z}$, equality (1) takes the form\n$$\n7k + 3 - \\sqrt{2} = (1 + 2\\sqrt{2}) \\cdot h(1 + \\sqrt{2}),\n$$\nwhich is equivalent to\n$$\nh(1 + \\sqrt{2}) = \\frac{7k + 3 - \\sqrt{2}}{1 + 2\\sqrt{2}} = \\frac{1 - 7k}{7} + \\frac{14k + 5}{7} \\cdot \\sqrt{2},\n$$\nwhich is impossible, since the coefficients $h(x)$ are integers.\n\nFor numbers $a = 7k + 5$, $k \\in \\mathbb{Z}$, equality (1) takes the form\n$$\n7k + 4 - \\sqrt{2} = (1 + 2\\sqrt{2}) \\cdot h(1 + \\sqrt{2}),\n$$\nwhich is equivalent to\n$$\nh(1 + \\sqrt{2}) = \\frac{7k + 4 - \\sqrt{2}}{1 + 2\\sqrt{2}} = (2k - 1)\\sqrt{2} - k,\n$$\ntherefore, one can choose the polynomial $h(x) = (2k-1)x - (3k-1)$. Therefore, all such numbers satisfy the condition.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 25714,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_1, a_2, \\dots, a_n$ be the sequence of positive integers. For each number $\\ell$ from $1$ to $n-1$ the following collection was found:\n$$(\\gcd(a_1, a_{1+\\ell}), \\gcd(a_2, a_{2+\\ell}), \\dots, \\gcd(a_n, a_{n+\\ell})),$$\nwhere all indices are taken modulo $n$, i.e. if $s > n$ then $a_s = a_{s-n}$. It turned out that all these collections consist of the same $n$ pairwise distinct numbers maybe in different ordering.\nFind out if $n$ can be equal to a) $21$; b) $2021$.",
    "options": [],
    "answer": "a) yes; b) no",
    "solution": "Let us prove the following\n**Statement:** $n = k^2 - k + 1$ for some positive integer $k$.\n**Proof:** Let $d$ be the largest of the greatest common divisors found. Then the equality $\\gcd(a_i, a_{i+\\ell}) = d$ is equivalent to the fact that each of the numbers $a_i$ and $a_{i+\\ell}$ is a multiple of $d$. Let $(i_1, i_2, \\dots, i_k)$ be all indexes of numbers in the given sequence that are multiples of $d$, written in ascending order. Note that each pair $i_j < i_s$ of indices yields $d$ twice: for $\\ell = s - j$ and for $\\ell = j + n - s$. Therefore $d$ will occur as the greatest common divisor exactly $2\\left(\\frac{k(k-1)}{2}\\right)$ times and by assumption this number is $n-1$. Thus $n = 1 + 2\\left(\\frac{k(k-1)}{2}\\right) = k^2 - k + 1$. The assertion is proved.\n\nb) The equation $k^2 - k + 1 = 2021$ is equivalent to $k(k-1) = 2020$. The left side of this equation increases for $k > 0.5$ while $90 \\cdot 89 = 8010$ and $91 \\cdot 90 = 8190$, therefore this equation doesn't have a positive integer solutions. According to the proven statement, the answer at this item is: \"no\".\n\na) Note that $21 = 5^2 - 5 + 1$, i.e. $n = 21$ satisfies the statement for $k = 5$. It is easy to construct the sequence $(i_1, i_2, i_3, i_4, i_5)$ from the proof of this statement, for example $(1, 11, 13, 18, 19)$. Let us set $a_1 = a_2 = a_3 = \\dots = a_{21} = 1$ then choose $20$ different primes $p_1 > p_2 > \\dots > p_{20}$ and for each $i$ from $1$ to $20$ multiply by $p_i$ the numbers $a_i, a_{i+10}, a_{i+12}, a_{i+17}, a_{i+18}$. It is easy to see that for each $\\ell$ from $1$ to $20$ the set of greatest common divisors from the problem statement coincides with the set $\\{p_1, p_2, \\dots, p_{20}\\}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25715,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sequence of positive integers $a_1, a_2, a_3, \\dots$, for each integer $i \\ge 3$ satisfy the equality\n$$\na_{i+1} = a_i + \\gcd(a_{i-1}, a_{i-2}).\n$$\nProve that there exist positive integers $N$ and $M$ such that for each $n \\ge N$ holds $a_{n+1} - a_n = M$. (Palina Chernikava)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since the term $a_1$ is involved only in the summand $\\gcd(a_1, a_2)$ when calculating $a_4$, without loss of generality we can assume that $a_1 = \\gcd(a_1, a_2)$. If the numbers $a_1, a_2$ and $a_3$ are divisible by the same number, then all the numbers in the sequence $(a_n)$ are divided by this number and we can take it out as a common factor. Therefore without loss of generality we assume that $\\gcd(a_1, a_2, a_3) = 1$, which is equivalent to $\\gcd(a_1, a_3) = 1$. Denote $\\gcd(a_2, a_3) = d$, then $a_2 = bd$ and $a_3 = cd$, where $b$ and $c$ are coprime positive integers such that $a_1 \\mid b$ and $\\gcd(a_1, cd) = 1$. Let's find the first consecutive members of the sequence:\n$$\na_4 = a_3 + \\gcd(a_1, a_2) = a_3 + a_1 = cd + a_1,\n$$\n$$\na_5 = a_4 + \\gcd(a_2, a_3) = a_4 + d = cd + a_1 + d,\n$$\n$$\n\\begin{aligned}\na_6 &= a_5 + \\gcd(a_3, a_4) = a_5 + \\gcd(cd, cd + a_1) = \\\\\n&= a_5 + \\gcd(cd, a_1) = a_5 + 1,\n\\end{aligned}\n$$\n$$\n\\begin{aligned}\na_7 &= a_6 + \\gcd(a_4, a_5) = a_6 + \\gcd(cd + a_1, cd + a_1 + d) = \\\\\n&= a_6 + \\gcd(cd + a_1, d) = a_6 + \\gcd(a_1, d) = a_6 + 1.\n\\end{aligned}\n$$\nSince three consecutive numbers $a_5, a_6$ and $a_7$ differ by one, i.e. neighboring numbers are coprime, then all subsequent numbers will also differ by one, i.e., for any $n \\ge 5$, the equality $a_{n+1} - a_n = 1$ will be true.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25716,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two lines pass through the point $F(1; 1)$ on the Cartesian plane. These lines are perpendicular to each other, one of them intersects the right branch of the hyperbola $y = \\frac{1}{2x}$ at the points $A$ and $C$ ($C$ has bigger $x$-coordinate than $A$), and the other line intersects the left branch of this hyperbola at the point $B$ and the right branch — at the point $D$. The product of the projections of the segments $AC$ and $BD$ to the $x$-axis equals $m$.\nFind the area of the (non-convex) quadrilateral $ABCD$. (Igor Voronovich)",
    "options": [],
    "answer": "m^2 / 2",
    "solution": "Denote the abscissas of the points $A$, $B$, $C$ and $D$ by $a$, $b$, $c$ and $d$ respectively. The perpendicularity of the lines $AB$ and $CD$ is equivalent to $acbd = -1/4$ and the fact that the point $F$ belongs to the lines $AC$ and $BD$ is equivalent to the equalities $a+c = 2ac+1$ and $b+d = 2bd+1$. Denote $ac = p$ and $bd = q$, then the numbers $a$ and $c$ are the roots of the quadratic equation $x^2 - (2p+1)x + p = 0$ and the numbers $b$ and $d$ are the roots of the quadratic equation $x^2 - (2q+1)x + q = 0$. Hence $c - a = \\sqrt{4p^2 + 1}$ and $d - b = \\sqrt{4q^2 + 1}$. Note that\n$$\n\\frac{1}{2a} - \\frac{1}{2c} = \\frac{c-a}{2ac} = \\frac{\\sqrt{4p^2+1}}{2p} = -2q \\cdot \\sqrt{\\frac{1}{4q^2} + 1} = \\sqrt{4q^2+1} = d-b.\n$$\nSimilarly, $\\frac{1}{2d} - \\frac{1}{2b} = c-a$. Thus the projections of the segment $AC$ on the abscissa and ordinate axes are equal, respectively, to the projections of the segment $BD$ on the ordinate and abscissa axes, hence $AC = BD$. Since the diagonals $AC$ and $BD$ of the quadrilateral $ABCD$ are perpendicular, its area $S$ is equal to half the product of the diagonals. Therefore\n$$\nS = \\frac{1}{2} \\cdot AC \\cdot BD = \\frac{1}{2} AC^2 = \\frac{1}{2}((c-a)^2 + (b-d)^2) = \\frac{1}{2}(4p^2 + 4q^2 + 2).\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25717,
    "subject": "Mathematics (Multi-modal)",
    "question": "Palina marked $2021$ arbitrary points on the circle and drew $2021$ segments between them. By an intersection point we denote any point of intersection of two drawn segments if this point is the endpoint of none of them.\nFind the maximal possible number of intersection points. (Mikhail Karpuk)",
    "options": [],
    "answer": "2039189",
    "solution": "$2021 \\cdot 1009 = 2039189$.\nLet's solve the problem for an arbitrary odd number $n$ of marked points and drawn segments. We enumerate the marked points on the circle by the numbers from $1$ to $n$ in the order of going around the circle. For each $i = 1, \\overline\\{n\\}$ we denote by $a_i$ the number of drawn segments for which $i$ is an endpoint. Among all $\\binom\\{n\\}\\{2\\}$ pairs of segments, at least $\\binom\\{a_i\\}\\{2\\}$ pairs do not intersect, since they have a common endpoint $i$. For different $i = 1, \\overline\\{n\\}$ these pairs do not coincide, hence the number of intersection points does not exceed\n$$\nM = \\binom\\{n\\}\\{2\\} - \\left( \\binom\\{a_1\\}\\{2\\} + \\binom\\{a_2\\}\\{2\\} + \\dots + \\binom\\{a_n\\}\\{2\\} \\right).\n$$\nLet's determine the largest possible value of $M$, for which we find the smallest possible value of the sum in brackets. It is easy to see, by opening the brackets, that the inequality $\\binom\\{b\\}\\{2\\} + \\binom\\{a\\}\\{2\\} > \\binom\\{b-1\\}\\{2\\} + \\binom\\{a+1\\}\\{2\\}$ is equivalent to the inequality $b-1 > a$. Therefore, the numbers $a_i, i = 1, \\overline\\{n\\}$, must differ at most by one. Given that $a_1 + a_2 + \\dots + a_n = 2n$, it's only possible if $a_1 = a_2 = \\dots = a_n = 2$. Hence, the maximum possible value of $M$ is equal to $\\frac\\{n(n-1)\\}\\{2\\} - n = \\frac\\{n(n-3)\\}\\{2\\}$.\nTo make sure that this value is the answer for the problem, we provide an example. Let's arrange points at the vertices of a regular $n$-gon and sequentially perform the following operation with each of them: choose a point $i$ and consider the set $p_i$ of intersection points of all possible segments with vertices at the remaining points. The set $p_i$ is finite, and the set of points on the arc containing the point $i$ is infinite. Therefore, we can choose such a new location for it that no segment connecting the point $i$ with any other point passes through the points of the set $p_i$. As a result of such operations, we obtain an arrangement of points such that no three segments with vertices at these points have a common point inside the circle. Now it suffices to draw segments connecting the points with numbers $i$ and $i + [n/2]$, $i = 1, \\overline\\{n\\}$, (we assume that the numbers are cyclic modulo $n$, i.e. $n + k = k$ if $k = 1, \\overline\\{n\\}$). In this case, any two segments that don't have a common endpoint, intersect and all $a_i$ are equal to two.\nSubstituting the value $n = 2021$ into the found formula $\\frac\\{n(n-3)\\}\\{2\\}$, we find the answer $2021 \\cdot 1009 = 2039189$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25718,
    "subject": "Mathematics (Multi-modal)",
    "question": "Three circles $\\omega_1$, $\\omega_2$ and $\\omega_3$ with non-colinear centres $O_1$, $O_2$ and $O_3$ are drawn such that $\\omega_1$ externally touches $\\omega_2$ and $\\omega_3$ at the points $P$ and $Q$ respectively. An arbitrary point $C$ is chosen on $\\omega_1$. The line $CP$ intersects $\\omega_2$ for the second time at the point $B$, and the line $CQ$ intersects $\\omega_3$ for the second time at the point $A$. Point $O$ is the circumcenter of the triangle $ABC$.\nProve that while point $C$ varies over all positions on $\\omega_1$, the locus of the points $O$ is the circle and the center of this circle lies on the circumcircle of the triangle $O_1O_2O_3$. (Mikhail Karpuk)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let's first consider the case when the radii of $\\omega_2$ and $\\omega_3$ are different (without loss of generality assume that the radius of $\\omega_3$ is greater than the radius of $\\omega_2$). Consider three homotheties: $l_1$ centered at $Q$ and mapping $\\omega_3$ to $\\omega_1$; $l_2$ centered $P$ and mapping $\\omega_1$ to $\\omega_2$; and $l_3 = l_2 \\circ l_1$ mapping $\\omega_3$ to $\\omega_2$. Since $l_1$ maps $A$ to $C$, $l_2$ maps $C$ to $B$, then $l_3$ maps $A$ to $B$, so its center (denoted by $I$) lies on the line $AB$.\nDenote $\\angle ACB = \\angle PCQ = \\alpha$, this angle is fixed (by a fixed value we denote any value which doesn't depend on the position of $C$). Then the angle $\\angle AOB = 2\\alpha$ is also fixed, which means that all triangles $AOB$ are similar to each other and the ratio $\\frac{OA}{AB} = \\frac{1}{2\\sin\\alpha}$ is fixed.\nLet $\\frac{IA}{IB} = t > 1$ be the homothety coefficient of $l_3$, this number is fixed. Then the ratio\n$$\n\\frac{AB}{IA} = \\frac{IB-IA}{IA} = \\frac{1}{t} - 1 \\text{ is also fixed.}\n$$\nIn the triangle $IAO$ we can find the ratio\n$$\n\\frac{OA}{IA} = \\frac{OA}{AB} \\cdot \\frac{AB}{IA} = \\frac{1}{2\\sin\\alpha} \\cdot \\left(\\frac{1}{t} - 1\\right)\n$$\nand the angle\n$$\n\\angle OAI = 180^{\\circ} - \\angle OAB = 90^{\\circ} + \\alpha.\n$$\nHence all triangles $IAO$ are similar to each other, in particular, the angle $AIO$ and the ratio $\\frac{IO}{IA}$ are fixed. Therefore for any position of the point $C$ on the circle $\\omega_1$, the spiral similarity $\\ell$ with center $I$, angle $AIO$ and coefficient $\\frac{IO}{IA}$ maps $A$ to $O$. Since the point $A$ varies over all positions on $\\omega_3$, the locus of $O$ is the circle $\\omega = \\ell(\\omega_3)$.\nNote that the homothety $\\ell_3$ and the spiral similarity $\\ell$ have a common center $I$. Therefore $\\ell(B) = \\ell \\circ \\ell_3(A) = \\ell_3 \\circ \\ell(A) = \\ell_3(O_3) = O_2$. Hence $\\ell$ maps the triangle $AOB$ to the triangle $O_3O'O_2$ and\n$$\n\\angle O_3O'O_2 = \\angle AOB = 2\\angle ACB = \\angle QO_1P = \\angle O_3O_1O_2,\n$$\nfrom which we conclude that $O'$ lies on the circumcircle of the triangle $O_1O_2O_3$.\n\nLet now the radii of the circles $\\omega_2$ and $\\omega_3$ be equal. In this case $\\ell_3$ is a translation by the vector $\\overrightarrow{O_3O_2}$, the point $I$ is not defined and the quadrilateral $ABO_2O_3$ is a parallelogram. Let $D$ be a such point that triangles $ABD$ and $O_3O_2O_1$ are equal and have the same orientation. The quadrilateral $O_3ADO_1$ is a parallelogram, while the triangles $QAO_3$ and $QCO_1$ are similar. Hence the segments $DO_1$ and $O_1C$ are parallel and\n$$\nDO_1 + O_1C = O_3A \\left(1 + \\frac{QO_1}{QO_3}\\right) = O_3Q \\left(1 + \\frac{QO_1}{QO_3}\\right) = O_1O_3 = DA = DB.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25719,
    "subject": "Mathematics (Multi-modal)",
    "question": "The cells of the $2022 \\times 2022$ table are filled with positive integers from $1$ to $2022^2$, each cell contains exactly one number, all numbers are used exactly once. For each line Vlad wrote out one number which is the second in descending order in this line. And Dima did the same for each column. It turned out that the boys wrote down $4044$ pairwise distinct numbers and there are exactly $k$ numbers written down by Vlad, such that each of them is less than every number written down by Dima.\nFind the largest possible value of $k$. (Mikhail Karpuk)",
    "options": [],
    "answer": "2020",
    "solution": "**Answer:** $2020$.\n\nLet Dima wrote down the numbers $d_1 > d_2 > \\dots > d_{2022}$ and Vlad wrote down the numbers $v_1 > v_2 > \\dots > v_{2022}$. Suppose the answer in the problem is $2022$ or $2021$, then $d_{2022} > v_2$. Each of the $2021$ lines not containing $v_1$ contains at least $2021$ numbers not exceeding $v_2$: the number written by Vlad and all smaller numbers. And each column contains at least $2$ numbers not less than $d_{2022}$: the number written by Dima and all greater numbers. Therefore the total amount of numbers in the table is not less than\n$$\n2021 \\cdot 2021 + 2022 \\cdot 2 = 2022 \\cdot 2022 + 1,\n$$\nwhich exceeds the number of all numbers in the table — a contradiction.\n\nLet's show that it could turn out that $2020$ of numbers written by Vlad are less than any number written by Dima. Fill in the table\n\n| $v_1$ | $d_1$ | $\\infty$ | $d_4$ | $d_5$ | $d_6$ | ... | $d_{2022}$ |\n|-------|-------|----------|-------|-------|-------|-----|------------|\n| $d_2$ | $v_2$ | $d_3$    | $\\infty$ | $y$   | $y$   | ... | $y$        |\n| $x$   |       | $v_3$    |       |       |       |     |            |\n| $x$   |       |          | $v_4$ |       |       |     |            |\n| $x$   |       |          |       | $v_5$ |       |     |            |\n| $x$   |       |          |       |       | $v_6$ |     |            |\n| $\\vdots$ |   |          |       |       |       | $\\vdots$ |      |\n| $x$   |       |          |       |       | $v_6$ |     | $v_{2022}$ |\n\nnumbers from $1$ to $2022^2$ so that\n$$\n\\infty > v_1 > v_2 > y > d_1 > \\dots > d_{2022} > x > v_3 > \\dots > v_{2020}\n$$\n(by $\\infty$, $y$ and $x$ in these inequalities we mean each number written on the cell with the corresponding label), and the numbers on empty cells are less than each of the numbers on marked cells.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25720,
    "subject": "Mathematics (Multi-modal)",
    "question": "The incircle of the right triangle $ABC$ is tangent to the hypotenuse $AB$ at point $P$ and is tangent to the legs $AC$ and $BC$ at points $Q$ and $R$ respectively. Points $C_1$ and $C_2$ are symmetric to $C$ with respect to the lines $PQ$ and $PR$.\nFind the angle $C_1IC_2$ where $I$ is the incenter of the triangle $ABC$. (Mikhail Karpuk)",
    "options": [],
    "answer": "135°",
    "solution": "Denote $PR \\cap CC_2 = X$. Let's do some angle-chasing.\n$$\n\\angle RCC_2 = 90^\\circ - \\angle CRX = 90^\\circ - \\angle PRB = \\frac{\\angle B}{2}.\n$$\nTherefore $RI = RC = RC_2$, i.e. $R$ is the center of the circumcircle of the triangle $ICC_2$. Hence\n\n$$\n\\angle RIC_2 = 90^\\circ - \\frac{\\angle IRC_2}{2} = 90^\\circ - \\angle ICC_2 = 90^\\circ - (45^\\circ + \\angle RCC_2) = 45^\\circ - \\frac{\\angle B}{2}.\n$$\n\nSimilarly, $\\angle QIC_1 = 45^\\circ - \\frac{\\angle A}{2}$. Then\n\n$$\n\\angle C_1IC_2 = 90^\\circ + \\angle QIC_1 + \\angle RIC_2 = 180^\\circ - \\frac{\\angle A + \\angle B}{2} = 135^\\circ.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25721,
    "subject": "Mathematics (Multi-modal)",
    "question": "The numbers $-1011, -1010, \\dots, -1, 1, 2, \\dots, 1010, 1011$ are arranged as $a_1, a_2, \\dots, a_{2022}$ in some order.\nFind the maximal possible value of\n$$|a_1| + |a_1 + a_2| + |a_1 + a_2 + a_3| + \\dots + |a_1 + a_2 + \\dots + a_{2022}|.$$ (Yahor Dubovik)",
    "options": [],
    "answer": "1011*1012*4043/6",
    "solution": "Note that the sum of all numbers is $0$, so the required sum can be presented as the sum of the following two sums:\n$$\nA_1 = |a_1| + |a_1 + a_2| + \\dots + |a_1 + a_2 + \\dots + a_{1011}|\n$$\nand\n$$\nA_2 = |a_{2022}| + |a_{2022} + a_{2021}| + \\dots + |a_{2022} + \\dots + a_{1013}|.\n$$\nLet's bound each term of each sum separately. If $a_{i_1}, a_{i_2}, \\dots, a_{i_k}$ are $k < 1012$ pairwise distinct elements of the sequence given in the problem statement, then\n$$\n|a_{i_1} + a_{i_2} + \\dots + a_{i_k}| \\le |1011 + 1010 + \\dots + 1012 - k|\n$$\nMoreover, this estimate is reached when $a_i = 1012 - i$ or $a_i = -1012 + i$. Hence\n$$\nA_1 \\le 1011 + (1011 + 1010) + \\dots + (1 + 2 + \\dots + 1011)\n$$\n$$\nA_2 \\le 1011 + (1011 + 1010) + \\dots + (1 + 2 + \\dots + 1011)\n$$\nSo, the maximal possible value of the original sum is equal to\n$$\n2 \\cdot 1011 + 2 \\cdot (1011 + 1010) + \\dots + 2 \\cdot (1011 + \\dots + 2) + (1011 + \\dots + 1) = \\\\\n= 2 \\cdot (1^2 + 2^2 + \\dots + 1011^2) - (1+2+\\dots+1011) = \\\\\n2 \\cdot \\frac{1011 \\cdot 1012 \\cdot 2023}{6} - \\frac{1011 \\cdot 1012 \\cdot 3}{6} = \\frac{1011 \\cdot 1012 \\cdot 4043}{6}.\n$$\n\nThis sum is reachable if the numbers are arranged in ascending or descending order.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25722,
    "subject": "Mathematics (Multi-modal)",
    "question": "The polynomial $P(x, y)$ of two variables with integer coefficients satisfies the following two conditions: 1) for every integer $a$ there exists exactly one integer $y$ such that $P(a, y) = 0$; and 2) for every integer $b$ there exists exactly one integer $x$ such that $P(x, b) = 0$.\na) Prove that if the degree of $P(x, y)$ is two then this polynomial is divisible by $x - y + C$ or $x + y + C$ where $C$ is integer.\nb) Does there exist such polynomial $P(x, y)$ which is not divisible by any polynomial of the form $x - y + C$ and $x + y + C$ where $C$ is integer? (Mikhail Karpuk)",
    "options": [],
    "answer": "Yes. For example, P(x, y) = (3x − 2y)(3x − 4y + 1)(3x − 4y − 1).",
    "solution": "**a)** Let us prove the following\n**Lemma.** If for each integer $x$ the value of the polynomial $Q(x) = bx^2 + cx + d$ with integer coefficients is a square of an integer, then there exist integers $m$ and $n$ such that $b = m^2$, $c = 2mn$, $d = n^2$ and, in particular, $Q(x) = (mx + n)^2$.\n\n**Proof of the lemma.** If $b = 0$, then for each integer $x$ the difference $Q(x+1) - Q(x) = c$ is constant. Whereas the differences between neighboring squares of integers infinitely grow, so if $c \\ne 0$ then the non-constant polynomial $Q(x)$ takes the same values infinitely many times, which is impossible. If $b = c = 0$ then $d$ is a perfect square and the numbers $m = 0$ and $n = \\sqrt{d}$ fit in.\n\nLet $b \\neq 0$, then, obviously, $b > 0$, since otherwise $Q(x)$ will be negative for sufficiently large $x$. If the discriminant $c^2 - 4bd$ is zero, then $Q(x) = b(x + \\frac{c}{2b})^2$. Since $4b^2Q(x) = b(2bx + c)^2$ is a perfect square for every integer $x$, $b$ is a perfect square. Let $b = e^2$, then $Q(x) = (ex + \\frac{c}{2e})^2$ is a perfect square for every integer value of $x$, which implies that $c$ is a multiple of $2e$ and $Q$ satisfies the lemma statement.\n\nIf the discriminant $c^2 - 4bd$ is not equal to zero, then among the prime divisors of the values $Q(x)$ in integer points choose a prime divisor $p$, greater than $|c^2 - 4bd|$. Let $Q(x)$ be a multiple of $p$. Consider $Q(x+p)$:\n$$\nQ(x+p) = Q(x) + bp^2 + p(2bx + c). \\quad (1)\n$$\nNote that $(2bx + c)^2 = 4aQ(x) + c^2 - 4bd$ is not a multiple of $p$, so $2bx + c$ is not a multiple of $p$ and from (1) we obtain\n$$\nv_p(Q(x+p)) = v_p(Q(x)) + 1,\n$$\nwhich is impossible because $Q(x)$ and $Q(x+p)$ are perfect squares. Therefore, this case is impossible and the lemma is proved.\n\nLet's move on to solving the problem itself. Denote the polynomial given in the condition by $P(x, y)$. We will show that it can be decomposed into a product of two polynomials of the first degree with rational coefficients. Let\n$$\nP(x, y) = a_0x^2 + a_1y^2 + a_2xy + a_3x + a_4y + a_5.\n$$\nSuppose $a_1 \\neq 0$. Consider $P(x, y)$ as a square trinomial of the variable $y$:\n$$\nP(y) = a_1y^2 + (a_2x + a_4)y + (a_0x^2 + a_3x + a_5).\n$$\nSince this square trinomial has an integer root for any integer value $x$, its discriminant\n$$\nD(x) = (a_2x + a_4)^2 - 4a_1(a_0x^2 + a_3x + a_5)\n$$\nis a perfect square for all integer values of $x$. The degree of this polynomial is at most two, so we can apply the lemma to it. Hence $D(x) = (mx + n)^2$ for some integers $m$ and $n$. Then the polynomial $P(y)$ can be expanded into a product of two linear factors with rational coefficients $P(y) = a_1(y - y_1)(y - y_2)$, which can be written as\n$$\nP(x, y) = \\frac{1}{4a_1}(2a_1y - (a_2x + a_4) - (mx + n))(2a_1y - (a_2x + a_4) + (mx + n)).\n$$\nA similar expansion can be written in the case $a_0 \\neq 0$.\n\nSuppose now that $a_0 = a_1 = 0$. Since this polynomial has degree 2, $a_2 \\neq 0$ and\n$$\nP(x, y) = a_2xy + a_3x + a_4y + a_5.\n$$\nBy the problem statement, for any integer value $x$ the number $y = \\frac{a_3x+a_5}{a_2x+a_4}$ is an integer, because $P(x, y) = 0$ can be true only for it. Hence, for any integer value $x$ the number\n$$\na_2(a_3x + a_5) - a_3(a_2x + a_4) = a_2a_5 - a_3a_4\n$$\nis a multiple of the denominator $a_2x + a_4$, which is possible only if $a_2a_5 = a_3a_4$. The number $t = \\frac{a_3}{a_2}$ satisfies the equality $\\frac{a_5}{a_4} = \\frac{a_3}{a_2} = t$ (if $a_4 = 0$ then $a_5 = 0$, and the case $a_3 = 0$ is obviously impossible). Then $\\frac{a_3x+a_5}{a_2x+a_4} = t$ for all integers $x$, which contradicts the problem statement for $b = t$. Therefore this case is impossible.\n\nThus, the polynomial $P(x, y)$ decomposes into a product of two polynomials of the first degree with rational coefficients, we write this equality in the form\n$$\nM \\cdot P(x, y) = (k_1x + l_1y + C_1)(k_2x + l_2y + C_2),\n$$\nwhere all coefficients are integers. Note that the equations $k_1x + l_1y + C_1 = 0$ and $k_2x + l_2y + C_2 = 0$ are linear Diophantine equations and there are three alternatives for each of them: 1) the equation has no entire solutions; 2) the equation has the constant solution on some variable; 3) the general solution of the equation has the form $(x_0 + lt, y_0 - kt), t \\in \\mathbb{Z}$. Moreover, the second alternative is impossible according to the problem condition, and in the third possibility we can always assume that the coefficients in the pairs $(k_1, l_1)$ and $(k_2, l_2)$ are coprime.\n\nIf $|l_i| = 1$, then all solutions of $P(a, y) = 0$ belongs to the corresponding bracket, so the first possibility holds for the other bracket and, consequently, $|k_i|$ is also equal to one, which gives the required multiplier in the problem condition. The case $|k_i| = 1$ is considered similarly.\n\nConsider the remaining case: all coefficients $k_i$ and $l_i$ are at least two in absolute value. Note that in the series of positive integers, each $|l_1|$-th number from the first bracket is a solution to the equation $P(x, b) = 0$, as well as each $|l_2|$-th from the second bracket. Hence $\\frac{1}{|l_1|} + \\frac{1}{|l_2|} = 1$ and similarly $\\frac{1}{|k_1|} + \\frac{1}{|k_2|} = 1$. Under our assumptions this is only possible for $|l_1| = |l_2| = |k_1| = |k_2| = 2$, which contradicts the fact that the coefficients must be coprime.\n\n**b)** Consider the polynomial $P(x, y) = (3x - 2y)(3x - 4y + 1)(3x - 4y - 1)$. Each of its factors is a linear function, and the zeros of these factors are the pairs $(2k, 3k)$, $(4k + 1, 3k + 1)$ and $(4k - 1, 3k - 1)$, $k \\in \\mathbb{Z}$. It is easy to see that this polynomial satisfies the condition.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25723,
    "subject": "Mathematics (Multi-modal)",
    "question": "The lines $a_1, a_2, b_1, b_2, c_1, c_2$ passing, respectively, through the points $A_1, A_2, B_1, B_2, C_1, C_2$ rotate uniformly and with the same angular velocity about the corresponding points. At an arbitrary moment $t$ by $A(t)$ denote the intersection point of the lines $a_1$ and $a_2$. Points $B(t)$ and $C(t)$ are defined similarly. It turned out that within one $180^\\circ$ rotation there were two moments $t_1$ and $t_2$ such that the triangles $A(t_1)B(t_1)C(t_1)$ and $A(t_2)B(t_2)C(t_2)$ were equilateral and equally oriented.\nProve that the triangle $A(t)B(t)C(t)$ is always equilateral.\n(Aliaksei Vaidzelevich)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since the lines rotate uniformly the angles between the lines $a_1(t_1), a_1(t_2)$ and between the lines $a_2(t_1), a_2(t_2)$ are equal for any moments $t_1, t_2$. Hence the points $A_1, A_2, A(t_1), A(t_2)$ lie on the circle, i.e. all points $A(t)$ lie on the same circle. Moreover if the lines rotated by $\\phi$ then the point $A(t)$ passed along the arc with the measure $2\\phi$. Therefore the points $A(t)$, $B(t)$ and $C(t)$ are moving uniformly and with the same angular velocity along some three circles. And one full-circle rotation there were two moments $t_1$ and $t_2$ such that the triangles $A(t_1)B(t_1)C(t_1)$ and $A(t_2)B(t_2)C(t_2)$ were equilateral and equally oriented.\n\nConsider this situation on the complex plane and without loss of generality let the points make one full-circle rotation during one unit of time. Let $o_1, o_2, o_3$ be the coordinates of the centers and $r_1, r_2, r_3$ be the radii of the corresponding circles. Then the points $A(t), B(t)$ and $C(t)$ have the coordinates $o_1 + r_1 e^{t + \\varphi_1}$, $o_2 + r_2 e^{t + \\varphi_2}$ and $o_3 + r_3 e^{t + \\varphi_3}$ where $\\varphi_1, \\varphi_2$ and $\\varphi_3$ are angles corresponding to the initial positions of points. Then we know that\n$$\n\\frac{(o_2 + r_2 e^{t_1 + \\varphi_2}) - (o_1 + r_1 e^{t_1 + \\varphi_1})}{(o_3 + r_3 e^{t_2 + \\varphi_2}) - (o_1 + r_1 e^{t_2 + \\varphi_1})} = \\frac{(o_2 + r_2 e^{t_2 + \\varphi_2}) - (o_1 + r_1 e^{t_2 + \\varphi_1})}{(o_3 + r_3 e^{t_3 + \\varphi_2}) - (o_1 + r_1 e^{t_3 + \\varphi_1})} = \\xi,\n$$\nwhere $\\xi$ is the third root of unity. Note that after simple transformations the equation\n$$\n\\frac{(o_2 + r_2 e^{t+\\varphi_2}) - (o_1 + r_1 e^{t+\\varphi_1})}{(o_3 + r_3 e^{t+\\varphi_3}) - (o_1 + r_1 e^{t+\\varphi_1})} = \\xi\n$$\nreduces to a linear equation of the variable $e^t$. Since this equation has two different solutions $e^{t_1}$ and $e^{t_2}$ on a unit circle, it must be degenerate, i.e. it is an identity. Therefore the triangle $A(t)B(t)C(t)$ is always equilateral.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25724,
    "subject": "Mathematics (Multi-modal)",
    "question": "A convex quadrilateral $ABCD$ has the incircle $\\omega$. The diagonal $AC$ intersects $\\omega$ at the points $P$ and $Q$. Let $M$ and $N$ be the midpoints of the arcs $PQ$ of $\\omega$ such that the points $B$ and $M$ lie in one halfplane with respect to the line $AC$ while the points $D$ and $N$ — in another.\nProve that the lines $BM$, $DN$ and $AC$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Without loss of generality assume that $P$ lies between $A$ and $Q$. Let the line passing through $M$ parallel to $AC$ intersect the sides $AB$ and $BC$ at the points $K$ and $L$ respectively. Since $M$ is the midpoint of the arc $PQ$, the line $KL$ is parallel to $AC$ whence $\\omega$ is the tangency point of $B$-excircle of the triangle $BKL$. Thus the point $X = BM \\cap AC$ is the tangency point of $B$-excircle of the triangle $BAC$. Therefore\n$$\n2AX = AC + BC - AB = AC + \\sqrt{CP \\cdot CQ} - \\sqrt{AP \\cdot AQ},\n$$\nwe used the fact that the difference $BC - AB$ equals to the difference between the lengths of the tangents from $C$ and $A$ to $\\omega$.\nNote that $2AX$ doesn't depend on $\\omega$, this circle must only pass through $P$ and $Q$. Therefore if $Y = DM \\cap AC$ then $2AX = 2AY$, i.e. $X = Y$ and $BM$, $DN$ and $AC$ are concurrent.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25725,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p \\ge 5$ be a prime number. Prove that for each divisor $d > 2$ of $p-1$ it's possible to partition all integers between $1$ and $p-1$ into the sets with $d$ numbers in each in such a way that the sum of squares of all numbers of each set is divisible by $p$.\n\n(Mikhail Karpuk)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote the set of all residues modulo $p$ by $\\mathbb{F}_p$. Fermat's little theorem implies that the roots of the polynomial $x^{p-1} - 1$ are all nonzero elements of $\\mathbb{F}_p$. Since $x^{p-1} - 1 = (x^d - 1)(x^{d(m-1)} + x^{d(m-2)} + \\dots + x^d + 1)$ where $p-1 = md$, the polynomial $x^d - 1$ divides $x^{p-1} - 1$. In this equality the sum of the degrees of the factors equals the number of roots, hence $x^d - 1$ has $d$ roots. Denote these roots $a_1, a_2, \\dots, a_d$.\n\nConsider all polynomials of the form $x^d - C$. If such polynomial has a root $b$ then it has $d$ roots $a_1b, a_2b, \\dots, a_db$ and cannot have more roots than its degree, so it has exactly $d$ roots. On the other hand, each $b \\in \\mathbb{F}_p \\setminus \\{0\\}$ is a root of such polynomial, namely $x^d - b^d$. Therefore $\\mathbb{F}_p \\setminus \\{0\\}$ is the disjoint union of $m$ sets, consisting of $d$ roots of some polynomial $x^d - C$. Such partition satisfies the problem condition since by Vieta's theorem the sum of the roots and the sum of the pairwise products of the roots of $x^d - C$ are congruent to zero modulo $p$, whence the sum of their squares is likewise congruent to zero.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25726,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $n \\ge 2$ cities in a country. Some pairs of cities are connected with an air communication, and for any such pair the connection is mutual. It is possible to travel from one city to another (possibly with a couple of flights). It is also known that the minimal number of flights that are needed to travel from any given city to another doesn't exceed the same number $m$, and for any city $A$ there exists another city $B$ which cannot be reached from $A$ by less than $m$ flights.\nGiven $n$, find all possible values of $m$, for which it is possible.\n(Aliaksei Vaidzelevich)",
    "options": [],
    "answer": "all integers m with 1 ≤ m ≤ floor(n/2)",
    "solution": "Answer: $1, 2, \\ldots, [n/2]$.\n\nWe reformulate the problem in the language of Graph Theory. Consider a graph, the vertices of which correspond to cities and the edges correspond to air connections. It is given that the graph is connected and the eccentricity of all of its vertices (the greatest of the minimal distances from it to other vertices) equals its radius (number $m$). We will show that the radius of a graph on $n$ vertices can't be greater than $[n/2]$, which will imply that $m \\le [n/2]$.\n\nStart the process of deleting edges in the graph one by one, preserving the connectivity of it, as long as possible. We will end with a tree. For a tree on $n$ vertices, the maximal length of a chain doesn't exceed $n-1$ edge, hence, it is possible to reach any vertex from a central one (or from two central ones) by $[n/2]$ steps, which implies that the radius of the tree doesn't exceed $[n/2]$. It is left to note that the radius doesn't decrease when deleting edges, so our claim is proved.\n\nConsider now any number $m \\le [n/2]$. Take a cycle of length $2m$, whose radius clearly equals $m$. Choose a vertex $v$ of a cycle and create $n-2m$ copies of it, each of which is connected only to both neighbours of $v$. Then draw an edge between any pairs of copies of $v$, and connect $v$ with all of them. Clearly, the constructed graph satisfies the problem's conditions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25727,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given two finite sets $A$ and $B$ of pairs of real numbers. For any three pairs $(a_1, b_1)$, $(a_2, b_2)$ and $(a_3, b_3)$ from $A$ there exists a pair $(c, d)$ from $B$ such that\n$$\na_1c + b_1d \\ge 0, \\quad a_2c + b_2d \\ge 0 \\quad \\text{and} \\quad a_3c + b_3d \\ge 0.\n$$\nProve that the $B$ contains a pair $(\\gamma, \\delta)$ such that $\\alpha\\gamma + \\beta\\delta \\ge 0$ for each pair $(\\alpha, \\beta)$ from $A$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider each pair $(a, b)$ as the vector with the coordinates $(a, b)$ on the Cartesian plane. The inequality $ac+bd \\ge 0$ is equivalent to the non-negativity of the scalar product, i.e. it states that the angle between the vectors is not greater than $\\pi/2$. Also note that the lengths of the vectors don't matter, so without loss of generality assume that all these vectors are unit and their endpoints lie on the unit circle. Then for each fixed vector $(a, b)$ of the first set the inequality $ac+bd \\ge 0$ means that the endpoint of the vector $(c, d)$ of the second set lies on the semicircle with the midpoint at the end of the vector $(a, b)$.\n\nThereby we can formulate the statement of the problem in the following equivalent form: Given the finite set $A$ of semicircles and the finite set $B$ of points on the unit circle such that every three semicircles from $A$ have a common point from $B$. Then all given semicircles from $A$ have a common point from $B$.\n\nLet's prove the latter statement. We will identify each point on the unit circle with the oriented angle between the vector $\\overrightarrow{(0,1)}$ and the radius-vector of the point, in particular this angle belongs to $[0, 2\\pi)$. And fix the positive (anticlockwise) direction on the unit circle so that all semicircles from $A$ have unique starting point. Without loss of generality assume that one of the semicircles from $A$ starts at $0$ and ends at $\\pi$. Consider the set of all semicircles from $A$ such that their starting point is greater than $\\pi$, and let $[\\ell_1, r_1]$ be the semicircle in this set with the smallest $r_1$, $\\ell_1 > \\pi > r_1$. Similarly consider the set of all semicircles from $A$ such that their starting point is less than $\\pi$, and let $[\\ell_2, r_2]$ be the semicircle in this set with the biggest $\\ell_2$, $\\ell_2 < \\pi < r_2$. Since the semicircles $[0, \\pi]$, $[\\ell_1, r_1]$ and $[\\ell_2, r_2]$ have a common point from $B$, there exists a given point $v \\in B$ such that $r_1 \\ge v \\ge \\ell_2$. Clearly, $v$ satisfies the statement conditions.\n\nFor the completeness of the proof let's make a few clarifications. Firstly, without loss of generality we assumed all semicircles in $A$ to be pairwise distinct, as well as all points $B$. Secondly, it could happen that it is impossible to find $[\\ell_1, r_1]$ or $[\\ell_2, r_2]$ since the corresponding set is empty. However, in these cases the assertion is obvious, since in them the desired point is $\\pi$ or $0$, respectively. Finally, we ignore the semicircle $[\\pi, 0]$; if such semicircle is given then the point $v$ cannot exist since otherwise the given semicircles $[\\pi, 0]$, $[\\ell_1, r_1]$ and $[\\ell_2, r_2]$ don't have a common point from $B$; therefore the existence of $[\\pi, 0]$ reduces to the previous case, and in this case it also contains a point common to all the others.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25728,
    "subject": "Mathematics (Multi-modal)",
    "question": "An arbitrary point $D$ is marked inside the triangle $ABC$. The lines $AD$, $BD$ and $CD$ intersect the sides $BC$, $CA$ and $AB$ at the points $K$, $L$ and $M$ respectively. Let $A_1$, $B_1$ and $C_1$ be the midpoints of the segments $BC$, $CA$ and $AB$, and $X$, $Y$ be the midpoints of the segments $ML$ and $KM$ respectively. The lines $XA_1$ and $YB_1$ intersect at the point $F$.\nProve that $FO \\parallel DH$, where $O$ is the circumcenter of the triangle $A_1B_1C_1$ and $H$ is the orthocenter of the triangle $ABC$.\n(Mikhail Karpuk)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note that $XA_1$ is the Gauss-Newton line of a complete quadrangle formed by the lines $BL$, $BA$ and $CM$, $CA$, hence it passes through the midpoint of the segment $AD$. Similarly $YB_1$ passes through the midpoint of the segment $BD$. Therefore $F$ is the center of mass of points $A$, $B$, $C$ and $D$ with unit masses. This implies that $F$ lies on the line $DG$ and $GF: FD = 1:3$ where $G$ is the centroid of the triangle $ABC$. At the same time $O$ is the nine-point center of the triangle $ABC$ so it also lie on the Euler line of the triangle $ABC$ and $GO: OH = 1:3$. Therefore the Thales's theorem implies that $FO \\parallel DH$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25729,
    "subject": "Mathematics (Multi-modal)",
    "question": "Fix a positive integer $n$ and a finite graph with at least one edge; the end points of each edge are distinct, and any two vertices are joined by at most one edge. Vertices and edges are assigned (not necessarily distinct) numbers in the range $0$ through $n-1$, one number each. A vertex assignment and an edge assignment are compatible if the following condition is satisfied at each vertex $v$: The number assigned to $v$ is congruent modulo $n$ to the sum of the numbers assigned to the edges incident to $v$. Fix a vertex assignment and let $N$ be the total number of compatible edge assignments; compatibility refers, of course, to the fixed vertex assignment.\nProve that, if $N \\neq 0$, then the prime divisors of $N$ are all at most $n$. (In particular, if $n = 2$ and $N \\neq 0$, then $N$ is a power of $2$.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "An edge assignment compatible with a fixed vertex assignment will be referred to as a *solution* relative to that vertex assignment. A solution relative to the all-zero vertex assignment will be referred to as a *fundamental solution*; the all-zero edge assignment is a fundamental solution.\n\nFix a vertex assignment and fix a solution (if any). Componentwise subtraction modulo $n$ of the fixed solution from a solution provides a fundamental solution. Conversely, componentwise addition modulo $n$ of the fixed solution to a fundamental solution provides a solution. The two solution assignments are clearly inverse to one another, so the number of solutions is either zero or equal to the number of fundamental solutions. Consequently, it is sufficient to prove the conclusion for the number $N$ of fundamental solutions.\n\nTo this end, induct on the number of edges. The base case is clear: A graph with a single edge has a single fundamental solution, namely, the all zero edge assignment.\n\nFor the induction step, consider a graph $G$ with at least two edges. Fix an edge $e$ of $G$ and let $G' = G - e$; clearly, $G'$ has at least one edge. For each $k$ in the range $0$ through $n-1$, let $S_k$ be the set of fundamental solutions for $G$ assigning $k$ to $e$, and let $T_k$ be the set of solutions for $G'$ relative to the vertex assignment sending both end points of $e$ to $n-k$ and all other vertices to zero.\n\nChanging only the $e$-component of a solution in $S_k$ to zero, and changing the vertex assignment only at the end points of $e$ by sending them both to $n-k$, enable removal of $e$ from $G$ to provide a bijection from $S_k$ to $T_k$. Since the size of each non-empty $T_k$ is equal to the number $N'$ of fundamental solutions for $G'$, so is the size of each non-empty $S_k$. Consequently, $N = mN'$, where $m$ is the number of non-empty sets $S_k$; incidentally, $m \\ge 1$, since $S_0$ is non-empty — it contains the all zero edge assignment. Finally, the prime divisors of $m$ do not exceed $n$, since $m \\le n$; and the prime divisors of $N'$ do not exceed $n$, by the induction hypothesis. Consequently, the prime divisors of $N = mN'$ do not exceed $n$ either. This completes the induction step and concludes the proof.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25730,
    "subject": "Mathematics (Multi-modal)",
    "question": "Call the set $\\{a_1, a_2, \\dots, a_n\\}$ of positive integers good if\n$$\n\\gcd(a_k, \\sum_{i \\neq k} a_i) > 1\n$$\nfor each $k = \\overline{1, n}$.\nProve that for each $c \\in (0, 1)$ there exists a positive integer $M$ such that for any positive integer $K \\ge M$ one can find a good set with at least $cK$ elements all elements of which doesn't exceed $K$.\n(Yahor Dubovik)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Recall (with proof) two well-known facts:\n**Lemma 1.** The series $\\sum_{i=1}^{+\\infty} \\frac{1}{i}$ is divergent. In other words for each $c \\in \\mathbb{R}$ there exists $m$ such that $\\sum_{i=1}^{m} \\frac{1}{i} \\ge c$.\n**Proof.** Note that $\\sum_{i=2}^{2^{a+1}} \\frac{1}{i} \\ge \\sum_{i=2}^{2^{a+1}} \\frac{1}{2^{a+1}} = \\frac{1}{2}$. Grouping the terms this way we obtain that $\\sum_{i=1}^{2^t} \\frac{1}{i} \\ge 1 + \\frac{t}{2}$ which can be sufficiently large.\n\n**Lemma 2.** Order all prime numbers in ascending order: $p_1 < p_2 < \\dots$. Then the product $\\prod_{i=1}^{+\\infty} \\frac{1}{1 - \\frac{1}{p_i}}$ is divergent. In other words for each $c \\in \\mathbb{R}$ there exists $m$ such that $\\prod_{i=1}^{m} \\frac{1}{1 - \\frac{1}{p_i}} \\ge c$.\n**Proof.** By Lemma 1 for each $c \\in \\mathbb{R}$ there exists $m_0$ such that $\\sum_{j=1}^{m_0} \\frac{1}{j} \\ge c$. Let all prime divisors of numbers between 1 and $m_0$ be among $p_1, p_2, \\dots, p_m$ and the maximal exponent of these primes equal $\\ell$. Then each integer $j$ from 1 to $m_0$ has the unique representation $p_1^{\\alpha_1} p_2^{\\alpha_2} \\dots p_m^{\\alpha_m}$, $0 \\le \\alpha_i \\le \\ell$, $i = 1, \\dots, m$. Hence,\n$$\n\\prod_{i=1}^{m} \\frac{1}{1 - \\frac{1}{p_i}} \\ge \\prod_{i=1}^{m} \\left( 1 + \\frac{1}{p_i} + \\frac{1}{p_i^2} + \\dots + \\frac{1}{p_i^{\\ell}} \\right) \\ge \\sum_{j=1}^{m_0} \\frac{1}{j} \\ge c.\n$$\n\n**Solution of the problem.** Choose an arbitrary $q \\in (c, 1)$. Lemma 2 implies that there exists $L$ such that $\\prod_{i=1}^{L} \\frac{p_i - 1}{p_i} < 1 - q$. Denote $T = A p_1 p_2 \\dots p_L$ where $A$ is an arbitrary even number.\nConsider the set $U_T$ of all integers between 1 and $T$ which are divisible by at least one number from $p_1, p_2, \\dots, p_L$. The cardinality of $U_T$ is\n$$\n\\sum_{i} \\frac{T}{p_i} - \\sum_{i \\neq j} \\frac{T}{p_i p_j} + \\dots = T - T(1 - \\frac{1}{p_1}) \\dots (1 - \\frac{1}{p_L}) \\ge T - T(1 - q) \\ge qT.\n$$\nWe will show that $U_T$ is good. Denote the sum of all elements of $U_T$ by $S$. The condition $\\gcd(a_i, S - a_i) > 1$ is equivalent to $\\gcd(a_i, S) > 1$. Hence it suffices to show that $S$ is divisible by $p_1 p_2 \\dots p_L$.\nNote that if $x \\in U_T$ then $T - x \\in U_T$ and their sum is divisible by $p_1 p_2 \\dots p_L$. The numbers $T/2$ and $T$ are divisible by $p_1 p_2 \\dots p_L$ (since $A$ is even) so $S$ is divisible by $p_1 p_2 \\dots p_L$ as well.\nSuppose we are given $K \\ge 2p_1 p_2 \\dots p_L$. By taking an appropriate $A$ choose $T$ to be the maximal integer such that $T \\le K$ and $2p_1 p_2 \\dots p_L \\mid T$. All elements of the good set $U_T$ doesn't exceed $T$ and, moreover, $K$. The cardinality of $U_T$ is not less than $qT > q(K - 2p_1 p_2 \\dots p_L) = qK - D$ where $D = 2qp_1 p_2 \\dots p_L$ doesn't depend on $K$. Therefore for $K \\ge D/(q-c)$ we obtain the inequality $qT \\ge cK$ which implies that $U_T$ is the required good set.\n\nConsequently $M = \\lfloor D/(q-c) \\rfloor$ satisfies the problem condition.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25731,
    "subject": "Mathematics (Multi-modal)",
    "question": "A positive integer is written on the board. Petya is allowed to replace the number on the board with the sum of the squares of its digits. We call a number *interesting* if Petya can't obtain $1$ from it using a finite number of such operations.\nProve that there exist infinitely many interesting numbers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us first prove that interesting numbers exist. Consider the cycle\n$$\n4 \\rightarrow 16 \\rightarrow 37 \\rightarrow 58 \\rightarrow 89 \\rightarrow 145 \\rightarrow 42 \\rightarrow 20 \\rightarrow 4\n$$\nEach number in this cycle is obtained from the previous one by the operation described in the condition. Therefore, starting with any of the numbers in this cycle, Petya will never obtain $1$. Hence, all numbers in this cycle are interesting.\n\nNote that for any interesting number $x$, the number $P(x)$, consisting of $x$ units in decimal notation, is also interesting, since the very first number that Petya writes on the blackboard will be equal to $x$. Moreover, $P(x) > x$ for any $x > 1$. Therefore, for any interesting number $x$, there exists an interesting number greater than $x$, whence the statement follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25732,
    "subject": "Mathematics (Multi-modal)",
    "question": "It is given that integers $a$, $b$ and $c$ satisfy the equality $a + b + c = 0$. Denote $S = ab + bc + ca$, $A = a^2 + a + 1$, $B = b^2 + b + 1$ and $C = c^2 + c + 1$.\nProve that the number $(S + A)(S + B)(S + C)$ is the square of an integer.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "$$\nS + A = bc + a(b + c) + A = bc - a^2 + a^2 + a + 1 = bc - (b + c) + 1 = (b - 1)(c - 1).\n$$\nSimilarly $S + B = (c - 1)(a - 1)$ and $S + C = (a - 1)(b - 1)$. Hence,\n$$\n(S + A)(S + B)(S + C) = ((a - 1)(b - 1)(c - 1))^2\n$$\nis the square of the integer $(a - 1)(b - 1)(c - 1)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25733,
    "subject": "Mathematics (Multi-modal)",
    "question": "A point $P$ is marked inside the square $ABCD$, and the points $K$, $L$, $M$ and $N$ are marked on its sides $AB$, $BC$, $CD$ and $DA$ respectively. The lines $KP$, $LP$, $MP$ and $NP$ intersect the sides $CD$, $DA$, $AB$ and $BC$ at the points $K_1$, $L_1$, $M_1$ and $N_1$ respectively. It turned out that\n$$\n\\frac{KP}{PK_1} + \\frac{LP}{PL_1} + \\frac{MP}{PM_1} + \\frac{NP}{PN_1} = 4.\n$$\nProve that $KP + LP + MP + NP = K_1P + L_1P + M_1P + N_1P$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since the sides $AB$ and $CD$ of a square are parallel, the cross-lying angles are equal: $\\angle PKM_1 = \\angle PK_1M$ and $\\angle PM_1K = \\angle PMK_1$. Therefore the triangles $PKM_1$ and $PK_1M$ are similar and have equal ratios $KP : K_1P = M_1P : MP$. Denote $KP : K_1P = k$, then $MP : PM_1 = 1/k$. Similarly, for $LP : PL_1 = l$ we obtain $NP : PN_1 = 1/l$. Hence the equality given in the problem is equivalent to $k + 1/k + l + 1/l = 4$, which can be transformed as\n$$\n0 = k + \\frac{1}{k} - 2 + l + \\frac{1}{l} - 2 = \\frac{k^2 + 1 - 2k}{k} + \\frac{l^2 + 1 - 2l}{l} = \\frac{(k-1)^2}{k} + \\frac{(l-1)^2}{l}.\n$$\nClearly, the equality holds if and only if $k = l = 1$ which means that\n$$\nKP = K_1P, \\quad LP = L_1P, \\quad MP = M_1P \\quad \\text{and} \\quad NP = N_1P\n$$\n(in particular $P$ is the center of the square $ABCD$). Now the required equality is obvious.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25734,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two players play the game on the $3 \\times 2021$ board all cell of which are initially white. The players make their moves in turn, in one move a player paints black two not necessarily adjacent white cells located either in the same row or in the same column. The player who cannot make a move loses.\nWhich of the players can win regardless of the opponent's game?",
    "options": [],
    "answer": "the first player",
    "solution": "Answer: the first player has a winning strategy.\nSince two cells are repainted on each move and the total number of cells on the board is odd, at the end of the game there will be an odd number of white cells. Obviously there will remain not more than one white cell left in each row, so there will remain one or three white cells on the entire board. Hence if there is exactly one white cell left, then the total number of moves is $3031$ which means that the first player made the last move and have won. And if there are three white cells left on the board, then the second player have won. But it's possible only there is one white cell left in each line. Thus to win the first player only needs to completely recolor one of the lines.\n\nWe will provide the winning strategy for the first player. The first player will try to paint over the first line of the board (the lines of the board are numbered from top to bottom).\n\nOn the first move, the first player paints one cell of the 1st row and one cell of the 2nd row, located in the same column. Further, he responds to the moves of the second player as follows:\n\n1) If the second player paints two cells not from the 1st row, then the first player paints two cells of the 1st row, and in such a way that the cells painted by him are in the same columns as those painted by the second player (if it's possible, otherwise the first player paints an arbitrary cells of the 1st row).\n\n2) If the second player paints one cell of the 1st row and one cell of a non-1st row, the first player paints one cell of the 1st row and one cell of a non-1st row.\n\n3) If the second player paints two cells of the 1st row, then the first player paints two cells of the 1st row (if not all cells are painted over by the current moment).\n\nThus, if the first player adheres to the specified strategy, after each of his moves, an even number of white cells will remain in the first row, which means that at some turn all the cells of the 1st row will be painted over.\n\nIt remains to show that the strategy described above is feasible. Obviously, moves 1) and 3) are always possible (unless all the cells of the 1st row are already painted). Let us show that move 2) is also realizable. Indeed, due to the fact that on move 1) the first player paints the cells of the 1st row in the same columns in which the second player painted the cells before (if possible), then after each move of the first player, if there is a shaded cell in the column, then the cell from the 1st row is also painted in it (if the second player, while painting the cells in the column, did not paint over his cell in the 1st row, then the first player will do it on the next move 1). Therefore, if there is a white cell in the 1st row, then both cells of this column located in the 2nd and 3rd rows are also white. Therefore, move 2) can be done.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25735,
    "subject": "Mathematics (Multi-modal)",
    "question": "Three non-intersecting circles of radius $1$ are placed inside the triangle $ABC$. (Circles can touch each other and the sides of a triangle, but cannot share interior points.)\nFind the largest value of $r$ for which we can be sure that inside the triangle it is possible to draw a fourth circle of radius $r$ that doesn't intersect with three circles already drawn.",
    "options": [],
    "answer": "1/3",
    "solution": "Answer: $r = 1/3$.\n\nConsider three circles $\\omega_1$, $\\omega_2$ and $\\omega_3$ of radius $1$ with the centers $O_1$, $O_2$ and $O_3$ respectively and the equilateral triangle $ABC$ such that the side $AB$ touches the circles $\\omega_1$ and $\\omega_2$, the side $BC$ touches the circles $\\omega_2$ and $\\omega_3$, and the side $AC$ touches the circles $\\omega_1$ and $\\omega_3$. The part of the interior of the triangle $ABC$ which is not covered by circles is divided into seven parts of three types: the central part bounded by three circles, three side parts bounded by two circles and a side, and three corner parts bounded by a circle and two sides.\n\nThe central part can be completely covered with a side part, so it is senseless to enter the fourth circle there. Let's draw the circle of the largest radius $r$ that can be inscribed in the side part enclosed between the line $AB$ and circles $\\omega_1$, $\\omega_2$. Obviously, this circle touches $AB$, $\\omega_1$ and $\\omega_2$, so from the Pythagorean theorem we find $(1+r)^2 = (1-r)^2 + 1$, i.e. $r = 1/4$.\n\nLet's draw the circle of the largest radius $r$ that can be inscribed in the corner part enclosed between $AC$, $CB$ and $\\omega_3$. Obviously, this circle is tangent to $AC$, $CB$ and $\\omega_3$, so its center $O$ lies on the bisector of angle $ACB$ and, since the leg opposite the angle at $30^\\circ$ is half hypotenuse,\n$$\n2 = CO_3 = CO + OO_3 = 2r + (r + 1),\n$$\nwhence $r = 1/3$. Therefore, in this case it is impossible to place a circle of radius greater than $1/3$ that doesn't intersect with the three already drawn. This means that for $r > 1/3$ it is impossible to say for sure that there will always be a circle that satisfies the condition.\n\nAt least one angle of the triangle $ABC$ is not greater than $60^\\circ$, without loss of generality, let this angle be $ACB$. Let $\\omega$ be that of the given three circles whose center is no farther from the vertex $C$ than the center of any of the remaining circles. If $\\omega$ does not touch any side of the angle $ACB$, then replace it with a circle that: has the same center, is inside the angle $ACB$, and touches at least one of the sides of this angle (without loss of generality, let it touch the side $BC$ at the point $T$). If the resulting circle does not touch the side $AC$, then replace it with a circle that: touches the side $BC$ at the point $T$ and touches the side $AC$. Denote the resulting circle by $\\Omega$, it is inside the corner $ACB$ and contains the circle $\\omega$ inside itself. Consider the circle $\\Gamma$ that touches the sides of the angle $ACB$ and the circle $\\Omega$, and its center $O_1$ is closer to $C$ than the center $O$ of the circle $\\Omega$. Denote by $r$ and $R$ the radii, and by $T_1$ and $T$ the touching points of $BC$ with the circles $\\Gamma$ and $\\Omega$. Then\n$$\nR + r = OO_1 = OC - O_1C = \\frac{R}{\\sin \\frac{\\angle ABC}{2}} - \\frac{r}{\\sin \\frac{\\angle ABC}{2}} = \\frac{R - r}{\\sin \\frac{\\angle ABC}{2}}.\n$$\nConsequently,\n$$\n\\frac{R - r}{R + r} = \\sin \\frac{\\angle ABC}{2} \\le \\sin 30^\\circ = \\frac{1}{2}.\n$$\nSolving this inequality we find that $r \\ge R/3 \\ge 1/3$. Hence circle of radius $1/3$ is guaranteed to be drawn inside the triangle $ABC$ so that it does not intersect with the three already drawn circles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25736,
    "subject": "Mathematics (Multi-modal)",
    "question": "The $2022 \\times 2022$ board was cut into tetraminoes of two types: L-tetramino $L$ and Z-tetramino $Z$. Each tetramino consists of four unit squares, tetriminoes can be rotated and flipped.\nFind the smallest number of $Z$-tetriminoes that could be obtained.",
    "options": [],
    "answer": "1",
    "solution": "Answer: $1$.\nLet's label the columns of the table from bottom to top with the numbers from $1$ to $2022$ and color all the cells of the rows with odd numbers in yellow, and with even numbers — in blue. Each $L$-tetromino consists of three squares of one color and one square of another. Let's assume that not a single $Z$-tetromino has been obtained. Since the numbers of yellow and blue cells in the table are equal, the number of $L$-tetrominoes consisting of three yellow cells and one blue cell is equal to the number of $L$-tetrominoes consisting of three blue cells and one yellow cell. This means that the board is divided into an even number of $L$-tetrominoes, but this implies that the total number of cells on the board is a multiple of $8$, which is not true. Hence at least one $Z$-tetromino has been obtained.\nLet's give an example showing what exactly one $Z$-tetromino can be obtained.\n\n![](attached_image_1.png)\n\nThe picture shows how to cut a $6 \\times 6$ board into seven $L$-tetrominoes and one $Z$-tetromino. If we cut off such $6 \\times 6$ board from the corner of the $2022 \\times 2022$ board, then the rest of the board can be easily cut into $2 \\times 4$ and $4 \\times 2$ rectangles, which are made up of two $L$-tetrominoes.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25737,
    "subject": "Mathematics (Multi-modal)",
    "question": "The polynomial $p(x)$ with integer coefficients satisfies the equality\n$$p(\\sqrt{2} + \\sqrt{3}) = \\sqrt{2} - \\sqrt{3}.$$\na) Find all possible values of $p(\\sqrt{2} - \\sqrt{3})$.\nb) Give an example of at least one polynomial $p(x)$ that satisfies the condition.",
    "options": [],
    "answer": "a) √2 + √3. b) One example is p(x) = x^3 − 10x.",
    "solution": "a) Consider any monomial $a_n x^n$ of $p(x)$ and substitute $x = \\sqrt{2} + \\sqrt{3}$ and $x = \\sqrt{2} - \\sqrt{3}$ into it. When opening the brackets in the expression $a_n(\\sqrt{2} + \\sqrt{3})^n$, all terms will look like $a_n C_n^m \\sqrt{2}^k \\sqrt{3}^m$, where $k + m = n$. If $m$ is even then if $\\sqrt{2} + \\sqrt{3}$ is replaced by $\\sqrt{2} - \\sqrt{3}$, the sign of the term doesn't change, and the value of this term is a positive integer or a positive integer multiplied by $\\sqrt{3}$. If $m$ is odd then with such a change the sign of the term changes and the value of this term is a positive integer multiplied by $\\sqrt{3}$ or a positive integer multiplied by $\\sqrt{6}$. Since these observations are true for every monomial, they are also true for the value of the entire polynomial, i.e. $p(\\sqrt{2} - \\sqrt{3}) = \\sqrt{2} + \\sqrt{3}$.\n\nb) It is easy to verify that $p(x) = x^3 - 10x$ satisfies the problem conditions. Let's show how to find this example. Arguing as in paragraph a) we can write two more equalities: $p(-\\sqrt{2} + \\sqrt{3}) = \\sqrt{2} + \\sqrt{3}$ and $p(-\\sqrt{2} - \\sqrt{3}) = \\sqrt{2} + \\sqrt{3}$. These equalities show that numbers $\\pm\\sqrt{2} \\pm\\sqrt{3}$ satisfy the equality $p(x) = \\frac{1}{x}$, so they are roots of the polynomial $x p(x) - 1 = 0$. By Bezout's theorem, a polynomial with these roots is divisible by a polynomial\n$$\n(x - \\sqrt{2} - \\sqrt{3})(x - \\sqrt{2} + \\sqrt{3})(x + \\sqrt{2} - \\sqrt{3})(x + \\sqrt{2} + \\sqrt{3}) = x^4 - 10x^2 + 1.\n$$\nIt remains to find $p(x)$ from the equality $x p(x) - 1 = x^4 - 10x^2 + 1$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 25738,
    "subject": "Mathematics (Multi-modal)",
    "question": "Vitya and Masha play the game. First, Vitya thinks of three different integers. Then Masha can ask one of the following quantities: either the sum of the numbers, or the sum of pairwise products of the numbers, or the product of the numbers suggested by Vitya. Masha asks questions sequentially, and Vitya gives an answer before the next question is asked.\n\na) Prove that Masha can always determine Vitya's numbers.\n\nb) What is the least number of questions Masha need to do this for sure, no matter what numbers Vitya guessed?",
    "options": [],
    "answer": "3",
    "solution": "a) If Masha asks all three different questions, then for the numbers $a$, $b$ and $c$ thought by Vitya, she will know the coefficients of the polynomial\n$$\np(x) = (x - a)(x - b)(x - c) = x^3 - (a + b + c)x^2 + (ab + bc + ac)x - abc.\n$$\nSolving the cubic equation $p(x) = 0$ (for example, using Cardano's formulas) she can find the numbers $a$, $b$ and $c$.\n\nb) Suppose Masha has a strategy that allows her to find out the numbers in no more than two questions. Consider three options for the game process, depending on Masha's first question.\n\n1) If Masha first asks the product of numbers, let Vitya answer \"0\". If further Masha wants to know the sum of the numbers then after the answer \"0\" she cannot distinguish triples $(0, 1, -1)$ and $(0, 2, -2)$. And for the sum of pairwise products after the answer \"12\" it's impossible to distinguish triples $(0, 2, 6)$ and $(0, 3, 4)$.\n\n2) If Masha first asks the sum of the numbers, let Vitya answer \"0\". If further Masha wants to know the product of numbers, then after the answer \"0\" she cannot distinguish triples $(0, 1, -1)$ and $(0, 2, -2)$. And for the sum of pairwise products after the answer \"-49\", it's impossible to distinguish triples $(0, 7, -7)$ and $(3, 5, -8)$.\n\n3) If Masha first asks the sum of pairwise products of numbers, let Vitya answer \"-18\". If further Masha wants to know the sum of the numbers, then after the answer \"3\" she cannot distinguish triples $(0, -3, 6)$ and $(2, 5, -4)$. And for the product after the answer \"-72\" it's impossible to distinguish triples $(3, -4, 6)$ and $(2, -3, 12)$.\n\nIn each of the options, after two questions, Masha doesn't have enough information to determine three numbers unambiguously, therefore she does not have a strategy that would allow her to determine the Vitya's numbers in two moves.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25739,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an isosceles triangle with the base $BC$. Points $X$, $Y$ and $Z$ are chosen on the sides $BC$, $AC$ and $AB$, respectively, such that $\\triangle ABC \\sim \\triangle YXZ$. Let $W$ be the reflection of $X$ with respect to the midpoint of the segment $BC$.\nProve that the points $X$, $Y$, $Z$ and $W$ are cocyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote by $O$ the center of the circumcircle of the triangle $XYZ$. Then\n$$\n\\angle YOZ = 2\\angle YXZ = 2\\angle ABC = 180^\\circ - \\angle BAC = 180^\\circ - \\angle ZAY,\n$$\nwhence the quadrilateral $ZAYO$ is cyclic. The chords $OY$ and $OZ$ are equal so $\\angle ZAO = \\angle OAY$. Thus $O$ lies on the bisector of the angle $BAC$, which is the perpendicular bisector of the segment $XW$, therefore $OW = OX = OY = OZ$ and $W$ lies on the circumcircle of the triangle $XYZ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25740,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove the inequality\n$$\n\\frac{1}{1!} + \\frac{1}{2!} + \\frac{1}{3!} + \\dots + \\frac{1}{2022!} > \\frac{1^2}{2!} + \\frac{2^2}{3!} + \\frac{3^2}{4!} + \\dots + \\frac{2022^2}{2023!}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us prove that $\\sum_{n=1}^{2022} \\frac{n^2}{(n+1)!} - \\sum_{n=1}^{2022} \\frac{1}{n!} < 0$, which is equivalent to the required. Note that\n$$\n\\frac{n^2}{(n+1)!} - \\frac{1}{n!} = \\frac{n(n+1) - 2(n+1) + 1}{(n+1)!} = \\frac{1}{(n-1)!} - \\frac{2}{n!} + \\frac{1}{(n+1)!}\n$$\nWith this identity in mind, we make the following transformations:\n$$\n\\begin{aligned}\n& \\sum_{n=1}^{2022} \\frac{n^2}{(n+1)!} - \\sum_{n=1}^{2022} \\frac{1}{n!} = \\sum_{n=0}^{2021} \\frac{1}{n!} - 2 \\sum_{n=1}^{2022} \\frac{1}{n!} + \\sum_{n=2}^{2023} \\frac{1}{n!} = \\\\\n& = \\frac{1}{0!} - \\frac{1}{1!} - \\frac{1}{2022!} + \\frac{1}{2023!} = \\frac{1}{2023!} - \\frac{1}{2022!} < 0.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 25741,
    "subject": "Mathematics (Multi-modal)",
    "question": "Positive integers $a$ and $b$ satisfy the equality $a + \\tau(a) = b^2 + 2$, where by $\\tau(n)$ we denote the number of all divisors of a positive integer $n$, including $1$ and $n$.\nProve that the sum $a + b$ is even.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "$$\nc^2 + \\tau(c^2) - 2 = b^2 \\ge (c+1)^2 = c^2 + 2c + 1,\n$$\nwhich means that $\\tau(c^2) > 2c$. All divisors of $c^2$, except $c$, split into pairs of the form $(d, c^2/d)$, where $d < c$ and $c^2/d > c$, so $\\tau(c^2) \\le 1 + 2(c-1) = 2c - 1$, which contradicts the previously obtained inequality $\\tau(c^2) > 2c$. Note also that for $a = 1$ the number $b$ is equal to zero, which is impossible.\nThus $a$ is not a perfect square. Then all positive integer divisors $a$ can be divided into pairs of the form $(d, a/d)$, and hence $\\tau(a)$ is an even number. From the equality $a + \\tau(a) = b^2 + 2$ we conclude that the numbers $a$ and $b$ have the same parity, so $a + b$ is even.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25742,
    "subject": "Mathematics (Multi-modal)",
    "question": "The numbers $1, 2, \\dots, 50$ are written on the blackboard. Ann makes the following operations: she erases any two numbers $a$ and $b$ from the blackboard, writes down to the blackboard one number — the sum $a+b$, afterwards she writes the number $ab(a+b)$ to her notebook. After 49 such operations when only one number left on the blackboard, Ann found the sum $S$ of all 49 numbers which she wrote to her notebook.\n\na) Prove that $S$ doesn't depend on the order of Ann's operations.\n\nb) Find $S$.",
    "options": [],
    "answer": "690348750",
    "solution": "Let's see how, as a result of Ann's actions, the sum of the cubes of the numbers written on the board changes. The identity $(a+b)^3 = a^3 + b^3 + 3ab(a+b)$ implies that after each action of Ann, the sum of the cubes of all numbers written on the board increases by the number which is three times greater than the one that Ann writes down to her notebook. Therefore the sum of all numbers written in the notebook is equal to the difference between the cube of the last number remaining on the board and the sum of the cubes of all numbers written on the board initially. The number that remains on the board is equal to the sum of all numbers written on it initially which means that\n$$\n3S = (1 + 2 + \\dots + 50)^3 - (1^3 + 2^3 + \\dots + 50^3).\n$$\nWe use the well-known identity\n$$\n1^3 + 2^3 + \\dots + n^3 = (1 + 2 + \\dots + n)^2 = \\left(\\frac{n(n+1)}{2}\\right)^2,\n$$\nwhich is easy to prove by induction. Since $\\frac{50 \\cdot 51}{2} = 1275$, then\n$$\n3S = 1275^3 - 1275^2 = 1275^2 \\cdot 1274 = 3 \\cdot 425 \\cdot 1274 \\cdot 1275.\n$$\nSo the number $S$ is equal to $425 \\cdot 1274 \\cdot 1275 = 690348750$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25743,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given $n \\ge 2$ different integers greater than $-10$. It turned out that among them the amount of odd numbers equals to the largest even number, and the amount of even numbers equals to the largest odd number.\na) Find the smallest possible value of $n$.\nb) Find the greatest possible value of $n$.",
    "options": [],
    "answer": "a) 3; b) 19",
    "solution": "a) The largest odd and even numbers are positive integers, since there are numbers of both parities. This means that the largest even number is not less than two, the largest even number is not less than one, and the total number of numbers is not less than three. Note that $n = 3$ is possible if given numbers are $-1$, $1$ and $2$.\n\nb) Let $2a+1$ be the largest odd number and $2b$ be the largest even number. The number $2a+1$ of even numbers does not exceed $b+5$ since they are all at most $2b$ and at least $-8$. At the same time the number $2b$ of odd numbers does not exceed $a+6$ since they are all no more than $2a+1$ and no less than $-9$. Whence\n$$\n\\begin{cases} 2a+1 \\le b+5, \\\\ 2b \\le a+6. \\end{cases} \\qquad (1)\n$$\nSumming up these equalities we get that $a+b \\le 10$. Let us estimate the largest possible value of the number $n$ depending on the sum $a+b$. If $a+b = 10$ then the first inequality of system (1) is equivalent to the inequality $3a \\le 14$, so the left side is less than the right one by at least $2$, the second inequality of system (1) is equivalent to $3b \\le 16$ and in it the left side is less than the right side by at least one. So the total number of numbers does not exceed $(b+5-2)+(a+6-1) = 18$.\n\nIf $a+b=9$, then the first inequality in system (1) is equivalent to the inequality $3a \\le 13$, so the left side is less than the right one by at least $1$. So the total number of numbers does not exceed $(b+5-1)+(a+6) = 19$. Note that $n=19$ in this case is possible with $a=4$ and $b=5$ if given all odd numbers from $-9$ to $9$ and all even numbers from $-6$ to $10$.\n\nIf $a+b \\le 8$ then $n$ doesn't exceed $(b+5)+(a+6) \\le 19$. Thus, in all cases $n \\le 19$ and the equality is attainable.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25744,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle such that $\\angle CAB = 30^\\circ$ and $\\angle ACB = 60^\\circ$. An arbitrary point $D$ is chosen on the extension of the ray $AB$ beyond point $B$. The point $E$ lies on the extension of the ray $CB$ beyond the point $B$ such that $\\angle BDE = 60^\\circ$. The lines $AC$ and $DE$ intersect at $F$.\nProve that the circumcircle of the triangle $AEF$ passes through some fixed point different from $A$ and not depending on the choice of $D$.\n(Mikhail Karpuk)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $G$ be the point on the ray $BC$ such that $BG = 3BC$. We will show that $G$ lies on the circumcircle of the triangle $AEF$. Angles $\\angle ECF$ and $\\angle ACG$ are equal $120^\\circ$ as they are adjacent to $\\angle ACB = 60^\\circ$. Since $\\angle BED = 30^\\circ$ and $\\angle FCE = 120^\\circ$, $\\angle CFE = 30^\\circ$. Hence the triangle $FCE$ is isosceles with equal sides $CF = CE$. The leg $CB$ opposite to the angle $30^\\circ$ is equal to half of the hypotenuse $AC$, so $AC = 2CB = BG - CB = CG$. Hence the triangle $GCA$ is isosceles, $CG = CA$ and $\\angle GAC = \\angle AGC = 30^\\circ$. Note that the common bisector $\\ell$ of the angles $ACG$ and $ECF$ is also the common perpendicular bisector of the segments $AG$ and $EF$. Hence the triangles $AEF$ and $GFE$ are symmetrical to each other with respect to $\\ell$, and the centers of their circumcircles lie on line of symmetry. So they have a common circumscribed circle, i. e. point $G$ lies on the circumscribed circle of triangle $AEF$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25745,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for any positive integer $n$ there exist coprime positive integers numbers $a \\neq b$ such that for each $k$ from $1$ to $n$ the numbers $a+k$ and $b+k$ are not coprime.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $a = 1$ and $b = 1 + (n + 1)$, obviously they are coprime. Moreover, for any $k$ from $1$ to $n$ both numbers $a + k$ and $b + k$ are divisible by $k + 1$ and are not coprime.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25746,
    "subject": "Mathematics (Multi-modal)",
    "question": "Does there exist a polynomial $p(x)$ with integer coefficients such that\n$$\np(\\sqrt{2}) = \\sqrt{2} \\quad \\text{und} \\quad p(2\\sqrt{2}) = 2\\sqrt{2} + 2?\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose that such polynomial $p(x)$ with integer coefficients exists. It follows from the equality $p(\\sqrt{2}) = \\sqrt{2}$ that $p(-\\sqrt{2}) = -\\sqrt{2}$, i.e. the numbers $\\sqrt{2}$ and $-\\sqrt{2}$ are roots of the polynomial $p(x) - x$. According to Bezout's theorem, the polynomial $p(x) - x$ is divisible by $(x - \\sqrt{2})(x + \\sqrt{2}) = x^2 - 2$ as polynomials with rational coefficients. Moreover, it follows from the Gauss lemma that in the equality $p(x) - x = (x^2 - 2)h(x)$ the rational coefficients of the polynomial $h(x)$ are integers. Substituting into this equality the numbers $2\\sqrt{2}$ and $-2\\sqrt{2}$ instead of $x$, we obtain the equalities\n$$\n2 = 6 \\cdot h(2 + \\sqrt{2}) \\quad \\text{u} \\quad 2 = 6 \\cdot h(2 - \\sqrt{2}).\n$$\nMultiplying these equalities and reducing by 4, we obtain the equality\n$$\n1 = 9 \\cdot (h(2 + \\sqrt{2})h(2 - \\sqrt{2})).\n$$\nSince the product of conjugate numbers in brackets is an integer, it implies that 1 is divisible by 9 — a contradiction.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 25747,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest number $n$ with the following property: if among the numbers from $1$ to $1000$ we choose $n$ numbers such that no two of them are divisible by the square of the same prime number, then at least one of these numbers is necessarily the square of a prime number.",
    "options": [],
    "answer": "617",
    "solution": "Answer: $617$.\n\nTo begin with, let's give an example of $616$ numbers from $1$ to $1000$, among which there are no prime squares and no two of which are divisible by the square of a prime. To do this, take $8$ numbers of the form $2p^2$ not exceeding $500$, where $p$ is a prime number:\n$$\n2 \\cdot 2^2,\\ 2 \\cdot 3^2,\\ 2 \\cdot 5^2,\\ 2 \\cdot 7^2,\\ 2 \\cdot 11^2,\\ 2 \\cdot 13^2,\\ 2 \\cdot 17^2,\\ 2 \\cdot 19^2\n$$\nand also all natural numbers from $1$ to $1000$, free from squares (that is, not divisible by the square of any prime number). There are only $608$ of them. To make sure of this, use the inclusion-exclusion formula:\n$$\n1000 - \\sum_{1 \\le i \\le 11} \\left\\lfloor \\frac{1000}{p_i^2} \\right\\rfloor + \\sum_{1 \\le i < j \\le 11} \\left\\lfloor \\frac{1000}{p_i^2 p_j^2} \\right\\rfloor - \\sum_{1 \\le i < j < k \\le 11} \\left\\lfloor \\frac{1000}{p_i^2 p_j^2 p_k^2} \\right\\rfloor,\n$$\nwhere $p_1 = 2$, $p_2 = 3$, $p_3 = 5$, $p_4 = 7$, $p_5 = 11$, $p_6 = 13$, $p_7 = 17$, $p_8 = 19$, $p_9 = 23$, $p_{10} = 29$, $p_{11} = 31$. We get\n$$\n\\begin{aligned}\n1000 & - (250 + 111 + 40 + 20 + 8 + 5 + 3 + 2 + 1 + 1 + 1) \\\\\n& + \\left( \\left\\lfloor \\frac{1000}{2^2 3^2} \\right\\rfloor + \\left\\lfloor \\frac{1000}{2^2 5^2} \\right\\rfloor + \\left\\lfloor \\frac{1000}{2^2 7^2} \\right\\rfloor + \\left\\lfloor \\frac{1000}{2^2 11^2} \\right\\rfloor + \\left\\lfloor \\frac{1000}{2^2 13^2} \\right\\rfloor + \\left\\lfloor \\frac{1000}{3^2 5^2} \\right\\rfloor + \\left\\lfloor \\frac{1000}{3^2 7^2} \\right\\rfloor \\right) \\\\\n& - \\left\\lfloor \\frac{1000}{2^2 3^2 5^2} \\right\\rfloor = 1000 - 442 + (27 + 10 + 5 + 2 + 1 + 4 + 2) - 1 = 558 + 51 - 1 = 608.\n\\end{aligned}\n$$\nNow let's say there are $617$ numbers from $1$ to $1000$, among which there are no squares of prime numbers. Let us show that there are two of them that are divisible by the square of the same prime. Indeed, since there are only $608$ natural numbers not exceeding $1000$ and free from squares, then at least nine of these numbers are divisible by squares of prime numbers. Let us denote these numbers by $a_1, a_2, \\dots, a_9$, and for each $1 \\le i \\le 9$ let $q_i^2$ be the square of a prime dividing $a_i$. If all $q_i^2$ are different, then some of them, say $q_k^2$, is not less than $23^2 = 529$. But since $a_k \\ne q_k^2$, then $a_k \\ge 2q_k^2 = 1048$. Contradiction. Therefore, there are two numbers divisible by the square of the same prime.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25748,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that for any numbers $x \\neq y$ the following equality is true:\n$$\n(f(x + y))^2 = f(x + y) + f(x) + f(y).\n$$",
    "options": [],
    "answer": "f(x) = 0 for all real x, or f(x) = 3 for all real x",
    "solution": "Only two functions $f(x) \\equiv 0$ or $f(x) \\equiv 3$ are solutions to the equation. Indeed, let's put $c = f(0)$. By substituting $y = 0$, we obtain a quadratic equation\n$$\n(f(x))^2 - 2f(x) - c = 0,\n$$\nwhence $f(x) = 1 - \\sqrt{1+c}$ or $1 + \\sqrt{1+c}$ for any $x \\neq 0$.\nSuppose that there are two non-zero numbers $a$ and $b$ of the same sign ($ab > 0$) such that $f(a) = 1 - \\sqrt{1+c}$ and $f(b) = 1 + \\sqrt{1+c}$. Substituting $x = a$ and $y = b$ into the original equation, we get\n$$\n(f(a+b))^2 - f(a+b) - 2 = 0,\n$$\nwhence $f(a+b) = -1$ or $2$. Let's consider two cases.\n\n1) Case $f(a+b) = -1$. Since $a+b \\neq 0$ (the sum of the same sign as both terms), then $f(a+b) = 1 \\pm \\sqrt{1+c}$ and then necessarily $1 - \\sqrt{1+c} = -1$. This means $c=3$ and $f(a) = f(a+b) = -1$. For $x=a$ and $y = a+b$ the original equation gives a quadratic equation\n$$\n(f(2a+b))^2 = f(2a+b) - 2,\n$$\nwhich has no solutions for $f(2a+b)$, which is impossible.\n\n2) Case $f(a+b) = 2$. Since $a+b \\neq 0$, then $f(a+b) = 1 \\pm \\sqrt{1+c}$ and so necessarily $1+\\sqrt{1+c} = 2$. This means that $c=0$ and $f(b) = f(a+b) = 2$. For $x=a+b$ and $y=b$, the original equation gives a quadratic equation for $f(a+2b)$:\n$$\n(f(a+2b))^2 = f(a+2b) + 4,\n$$\nwhich has roots $\\frac{1\\pm\\sqrt{17}}{2}$. However, also $a+2b \\neq 0$ and $f(a+2b) = 1 \\pm \\sqrt{1+0} = 0$ or $2$. Contradiction.\n\nThus, on each of the open intervals $] -\\infty, 0[$ and $]0, \\infty[$ the function $f$ takes only one value $1 - \\sqrt{1+c}$ or $1 + \\sqrt{1+c}$. Therefore, if $x$ and $y$ are non-zero numbers of the same sign, then $f(x+y) = f(x) = f(y) = 1 \\pm \\sqrt{1+c}$ and\n$$\n(1 \\pm \\sqrt{1+c})^2 = 3(1 \\pm \\sqrt{1+c}),\n$$\nwhence $1 \\pm \\sqrt{1+c} = 0$ or $3$. Since $1 - \\sqrt{1+c} \\le 1 \\le 1 + \\sqrt{1+c}$, two options are possible:\n$$\n1 - \\sqrt{1+c} = 0 \\iff c = 0 \\quad \\text{or} \\quad 1 + \\sqrt{1+c} = 3 \\iff c = 3.\n$$\nNow let's substitute $y = -x$ into the original equation:\n$$\n(f(0))^2 = f(0) + f(x) + f(-x) \\iff f(x) + f(-x) = c^2 - c \\text{ for any } x \\neq 0.\n$$\nIf $c=0$ then on the entire set $\\mathbb{R}$ the function $f$ can take at most two values $(0$ or $1+\\sqrt{1+0} = 2)$. If $f(a) = 2$ for some $a \\neq 0$, then $f(a) + f(-a) = 2 + f(-a) = 0^2 - 0 = 0$, which is impossible, since $f(-a) = 0$ or $2$. This means $f(x) \\equiv 0$.\nIf $c=3$ then on the entire set $\\mathbb{R}$ the function $f$ can take at most two values $(3$ or $1-\\sqrt{1+3} = -1)$. If $f(b) = -1$ for some $b \\neq 0$, then $f(b) + f(-b) = -1 + f(-b) = 3^2 - 3 = 6$, which is impossible, since $f(-b) = -1$ or $3$. So $f(x) \\equiv 3$.\nWe can check that the functions $f(x) \\equiv 0$ or $f(x) \\equiv 3$ are suitable.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25749,
    "subject": "Mathematics (Multi-modal)",
    "question": "Olya and Tolya have paints of two opposite colors – white and black. They play the following game on the segment $[0, 1]$. Each round of the game takes place in two stages: one of the players chooses a number $l \\in [0, 1]$, and then the other player chooses some segment $J \\subseteq [0, 1]$ of length $l$ and recolors all its points in the opposite color. In the next round they switch roles and so on.\nAfter 2024 rounds, the total length $L$ of the white intervals is calculated. If $L > \\frac{1}{2}$ then Olya wins, and if $L \\le \\frac{1}{2}$ then Tolya wins. Initially, the entire segment is painted white. Tolya chooses a number first. Who has a winning strategy?\n(Andrey Naradzetski)",
    "options": [],
    "answer": "Olya",
    "solution": "Here is a winning strategy for Olya. Denote by $L_n$ the total length of the white segments after the $n$-th round ($L_0$ is considered to be equal to 1).\n**Statement.** Olya can play in such a way that for $k = 0, 1, \\dots, 1012$ after the $(2k)$-th round the total length of the white segments is greater than $\\frac{1}{2}$ ($L_{2k} > \\frac{1}{2}$), and for some $\\delta > 0$ at least one of the open intervals $]0, \\delta[$ or $]1 - \\delta, 1[$ is completely white.\nLet us prove this statement by induction.\nThe basis for $k=0$ is obvious.\nAssume that after the $(2k)$-th round the situation is as described in the statement. Consider Tolya's move in the $(2k+1)$-th round. If he chooses $l=0$ or $l=1$, then Olya can choose the same $l$ on her turn in the round $2k+2$, and the segment $[0, 1]$ will not change its color, with the exception, perhaps, of two points in the case $l=0$.\n\nNow suppose that Tolya chose $0 < l < 1$. Since at least one of the outermost open intervals is white, Olya can choose $J$ in such a way that after the $(2k+1)$-th round both outermost open intervals $([0, \\delta]$ or $]1-\\delta, 1[$ for some $\\delta > 0$) were painted white. If $L_{2k+1} > \\frac{1}{2}$, then Olya simply chooses $l = 0$. Let's consider the case when $L_{2k+1} \\le \\frac{1}{2}$. Since the outermost open intervals of the interval $[0, 1]$ are white, there exists $\\epsilon > 0$ such that the intervals $]0, \\epsilon[$ and $]1-\\epsilon, 1[$ are white. Then if Olya chooses the number $l \\in ]1-\\epsilon, 1[$, for example $l = 1-\\frac{\\epsilon}{2}$, then whatever segment $J$ Tolya chooses, he will completely cover all black intervals and leave one or two white intervals with a total length of $\\epsilon$. That is, $L_{2k+2} = (1-L_{2k+1}) + \\epsilon > \\frac{1}{2}$. It is also clear that Tolya will not be able to repaint both outermost intervals black at once.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25750,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $n$ points $A_1, \\dots, A_n$ on the plane with rational coordinates, all pairwise distances between them being integer. Prove that there exist points $B_1, \\dots, B_n$ on the plane with integer coordinates such that $|B_iB_j| = |A_iA_j|$ for all $1 \\le i < j \\le n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $d$ be the least common multiple of the denominators of all coordinates of the points $A_1, \\dots, A_n$. After homothety with center at the origin $O$ and coefficient $d$, these points will go to points $C_1, \\dots, C_n$, respectively, with integer coordinates. Let us switch to complex numbers and work with Gaussian integers. Let us introduce the following notation:\n$$\nC_k = x_k + y_k i \\quad \\text{and} \\quad M = \\{C_1, \\dots, C_n\\} \\subset \\mathbb{Z}[i].\n$$\nLet $p$ be an arbitrary prime divisor of $d$. Let us show that there is a rotation of the $R$ plane such that $RM \\subset p\\mathbb{Z}[i]$, that is, $R(\\frac{1}{p}M) \\subset \\mathbb{Z}[i]$.\nWe have $p^2 \\mid N(z)$ and $p^2 \\mid N(z-u)$ for any $z, u \\in M$, where $N(a+bi) = a^2 + b^2$ means norm. If $p \\mid z$ for any $z \\in M$, then everything is proven. Suppose that there exists $v \\in M$ such that $p \\nmid v$. Then it can be shown that there exists $\\pi \\in \\mathbb{Z}[i]$ such that $N(\\pi) = p$ and $\\pi^2 \\mid v$. Then for any $u \\in M$:\n- either $p \\nmid u$, whence $\\pi^2 \\mid u$,\n- or $p \\mid u$ and $p \\nmid v-u$, from which $\\pi^2 \\mid v-u$ and again $\\pi^2 \\mid u$.\nApply the following plane transformation, which is a rotation, since $N(\\pi/\\pi) = 1$:\n$$\nR_{\\pi}: x + yi \\rightarrow \\frac{\\bar{\\pi}}{\\pi}(x + yi),\n$$\nEach number from $M$ can be represented as $\\pi^2(a + bi)$, which is sent by the rotation $R_{\\pi}$ to the number $\\bar{\\pi}\\pi(a + bi) = p(a + bi)$. Thus, we get $R_{\\pi}M \\subset p\\mathbb{Z}[i]$ as required.\nLet us now consider a decomposition of the number $d = p_1p_2 \\cdots p_m$ into prime factors (not necessarily different). According to what has been proven, there is a rotation $R_1$ such that $R_1(\\frac{1}{p_1}M) \\subset \\mathbb{Z}[i]$. Similarly, there is a rotation $R_2$ such that $R_2(R_1(\\frac{1}{p_1p_2}M)) \\subset \\mathbb{Z}[i]$, and so on, there is a rotation $R_m$ such that\n$$\nR_m \\left( \\dots R_2 \\left( R_1 \\left( \\frac{1}{p_1 p_2 \\cdots p_m} M \\right) \\right) \\right) \\subset \\mathbb{Z}[i].\n$$\nSince $\\frac{1}{p_1p_2\\cdots p_m}M = \\{A_1, \\dots, A_n\\}$, the composition of rotations $R_m \\circ \\dots \\circ R_2 \\circ R_1$ sends the set $A_1, \\dots, A_n$ to the set of integer points $B_1, \\dots, B_n$ with the required properties.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25751,
    "subject": "Mathematics (Multi-modal)",
    "question": "Integers $a, b, c, d, e, f$ satisfy the system\n$$\n\\begin{cases}\nace + 3adf - 3bcf + 3bde = 5, \\\\\nacf - ade + bce + 3bdf = 2.\n\\end{cases}\n$$\n\nFind all possible values of the expression $abcde$.",
    "options": [],
    "answer": "0",
    "solution": "Answer: $0$.\n\nLet $x = ac + 3bd$, $y = ad - bc$. We have\n$$\n\\begin{align*}\n37 = 5^2 + 3 \\cdot 2^2 &= (ace + 3adf - 3bcf + 3bde)^2 + 3(acf - ade + bce + 3bdf)^2 \\\\\n&= (ex + 3fy)^2 + 3(fx - ey)^2 = (e^2 + 3f^2)(x^2 + 3y^2) \\\\\n&= (e^2 + 3f^2)((ac + 3bd)^2 + 3(ad - bc)^2) \\\\\n&= (e^2 + 3f^2)(c^2 + 3d^2)(a^2 + 3b^2).\n\\end{align*}\n$$\nSince the number $37$ is prime, we obtain $\\{a^2 + 3b^2, c^2 + 3d^2, e^2 + 3f^2\\} = \\{1, 1, 37\\}$. However, if the integers $m$ and $n$ satisfy the equation $m^2 + 3n^2 = 1$, then $n = 0$. Therefore, two of the numbers $b, d, f$ are equal to zero and, therefore, $abcde = 0$. Here is one of the solutions to the system: $a = 1$, $b = 0$, $c = 1$, $d = 0$, $e = 5$, $f = 2$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 25752,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a function $f(x) = x^2 + bx + c$, where $b, c \\in \\mathbb{R}$ and $b \\ge 0$. Is it possible to represent the segment $[0, 1]$ as the union $A \\cup B$ of two disjoint sets $A$ and $B$ such that $f(A) = B$?\nRecall that $f(A)$ denotes the image of the set $A$, that is, $f(A) = \\{f(a) \\mid a \\in A\\}$.",
    "options": [],
    "answer": "no",
    "solution": "Answer: no. Suppose that the segment $[0, 1]$ can be represented as the union $A \\cup B$ of two disjoint sets $A$ and $B$ such that $f(A) = B$. Let us immediately note that the function $f$ is strictly increasing on the interval $[0, 1]$ and is a bijection from $[0, 1]$ to $f([0, 1])$.\nLet us prove that $f$ has no fixed points on the interval $[0, 1]$. Indeed, let $f(s) = s$ for some $s \\in [0, 1]$. If $s \\in A$, then $f(s) = s \\notin B$, and if $s \\in B$, then there is no number from $A$ that is sent to $s$, because $f$ is a bijection. Since both cases are impossible and $[0, 1] = A \\cup B$, we arrive at a contradiction, that is, there cannot be fixed points. Due to the continuity of the function, this means that $f(x) > x$ for all $x \\in [0, 1]$ or $f(x) < x$ for all $x \\in [0, 1]$ depending on the sign of $c$.\n**Case $c \\ge 0$.** We have $x < f(x)$ for all $x \\in [0, 1]$. Since $f(0) = c > 0$ and the function is increasing on the interval $[0, 1]$, then $f(x) \\ge c$ for any $x \\in [0, 1]$. Therefore $[0, c[ \\subseteq A$, whence $[c, f(c)][ \\subseteq B$, and $c < f(c)$. Let us now consider the interval $[f(c), f(f(c))[$. Since all points of this interval are images of points from $B$ and not from $A$ (after all, $f$ is a bijection), the intersection of $[f(c), f(f(c))[$ with $B$ is empty. We know that $f(c) \\le 1$ and $f(c) < f(f(c))$. Note that $1 \\notin [f(c), f(f(c))[$, because otherwise we would get that $1 \\notin B$, whence $1 \\in A$ and $1 < f(1) \\in B$, which is impossible. Consequently, $[f(c), f(f(c))[$ $\\subseteq A$ and, continuing the reasoning in this way, we obtain that the sequence\n$0, f(0), f^2(0), f^3(0), \\dots$\nincreases and lies entirely in $[0, 1]$, that is, it is bounded above. This means that it has a limit $\\ell \\in [0, 1]$. From the continuity of $f$ we obtain that\n$$\n\\ell = \\lim_{n \\to \\infty} f^n(0) = f\\left(\\lim_{n \\to \\infty} f^{n-1}(0)\\right) = f(\\ell),\n$$\nthat is, $\\ell$ is a fixed point. Contradiction.\n**Case $c < 0$.** We have $f(x) < x$ for all $x \\in [0, 1]$. In particular, $f(1) < 1$. The function is increasing on the interval $[0, 1]$, and so $f(x) \\le f(1)$ for any $x \\in [0, 1]$. Therefore $]f(1), 1] \\subseteq A$, whence $]f(f(1)), f(1)] \\subseteq B$, and $f(f(1)) < f(1)$. Similar to the previous case, we show that $]f^3(1), f^2(1))[$ $\\subseteq A$ and that the sequence\n$$\n1, f(1), f^2(1), f^3(1), \\dots\n$$\nconverges to some number $\\ell \\in [0, 1]$, which is a fixed point for $f$. Contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25753,
    "subject": "Mathematics (Multi-modal)",
    "question": "Inside an isosceles triangle $ABC$ ($AB = BC$), a point $D$ is chosen so that $\\angle ADC = 150^\\circ$. On the segment $CD$, a point $E$ is chosen so that $AE = AB$.\nProve that if $\\angle BAE + \\angle CBE = 60^\\circ$, then $\\angle BDC + \\angle EAC = 90^\\circ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us rotate $\\triangle AED$ around the point $A$ so that $E \\to B$, $D \\to T$, and reflect $\\triangle ABT$ symmetrically with respect to $BT$ ($A \\to F$). Since $\\angle ADE = 150^\\circ$, then the triangle $ATF$ is equilateral. Let us prove that $FD = DC$. Denote $\\angle EAB = \\alpha$ and $\\angle AED = \\beta$. Then\n\n$$\n\\angle ABC = \\angle ABE + \\angle CBE = \\frac{180^\\circ - \\alpha}{2} + (60^\\circ - \\alpha) = 150^\\circ - \\frac{3\\alpha}{2} \\text{ and } \\angle BAC = 15^\\circ + \\frac{3\\alpha}{4}.\n$$\n$$\n\\begin{aligned} \\angle FCD &= \\angle FCA - \\angle DCA = \\frac{1}{2}\\angle FBA - (\\angle DEA - \\angle EAC) = \\\\ &= \\beta - (\\beta - \\angle EAC) = \\angle EAC = \\angle BAC - \\alpha = 15^\\circ - \\frac{\\alpha}{4}. \\end{aligned}\n$$\n\nHere we used the fact that the points $D$ and $A$ lie in the same half-plane relative to the line $FC$, since $\\angle ACD < \\angle AED = \\beta = \\angle ACF$.\n\nNote that $\\angle TFD = \\frac{1}{2}\\angle TAD = \\frac{\\alpha}{2}$, so\n$$\n\\angle DFC = \\angle AFC - \\angle AFD = \\frac{1}{2}\\angle ABC - (\\angle AFT - \\angle TFD) = \\left(75^\\circ - \\frac{3\\alpha}{4}\\right) - \\left(60^\\circ - \\frac{\\alpha}{2}\\right) = 15^\\circ - \\frac{\\alpha}{4}.\n$$\n![](attached_image_1.png)\nTherefore, $\\angle DFC = \\angle FCD$, i.e. $FD = DC$. Hence $FC \\perp BD$, i.e. $\\angle BDC + \\angle FCD = 90^\\circ$. Recalling that $\\angle FCD = \\angle EAC$, we obtain the statement of the problem.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25754,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given positive integers $n$ and $k \\le n$. Consider an equilateral triangular board with side $n$, which consists of circles: in the first (top) row there is one circle, in the second row there are two circles, ..., in the bottom row there are $n$ circles (see the figure below). Let us place checkers on this board so that any line parallel to a side of the triangle (there are $3n$ such lines) contains no more than $k$ checkers. Denote by $T(k, n)$ the largest possible number of checkers in such a placement.\n\n![](attached_image_1.png)\n\na) Prove that the following upper bound is true:\n$$\nT(k, n) \\le \\left\\lfloor \\frac{k(2n + 1)}{3} \\right\\rfloor .\n$$\n\nb) Find $T(1, n)$ and $T(2, n)$.",
    "options": [],
    "answer": "Upper bound: T(k, n) ≤ floor(k(2n + 1)/3). Exact values: T(1, n) = floor((2n + 1)/3) and T(2, n) = floor((4n + 2)/3).",
    "solution": "a. Let us prove that in any such placement of checkers on a triangular board with side $n$, the number of checkers $q$ satisfies the inequality\n$$\nq \\le \\frac{2n + 1}{3} \\cdot k.\n$$\nThe main idea of the proof is double counting. Note that wherever a checker stands, if you count all the cells of the three lines in which it stands, and take into account the cell on which it stands three times, you get $2n + 1$. Let us denote by $S$ this number summed over all checkers. We have\n$$\nS = (2n + 1)q.\n$$\n\nNow let's find this amount based on the location of the checkers along the lines. Let there be $x_1, \\dots, x_n$ checkers in the $n$ horizontal lines, $y_1, \\dots, y_n$ checkers in the $n$ lines with angle $60^\\circ$, and in the $n$ lines with angle $120^\\circ$ there are $z_1, \\dots, z_n$ checkers, as shown in Figure 1.\n\n![](attached_image_2.png)\n\nThe $i$-th line has exactly $i$ cells, and each of them is in the same line with $x_i, y_i$ or $z_i$ checkers from that line depending on the angle. Therefore,\n$$\n(3) \\qquad S = 1 \\cdot (x_1 + y_1 + z_1) + 2 \\cdot (x_2 + y_2 + z_2) + \\dots + n \\cdot (x_n + y_n + z_n).\n$$\nSince there are $q$ checkers in total, but there are no more than $k$ checkers in any row, we have the following conditions:\n$$\nx_1 + \\dots + x_n = y_1 + \\dots + y_n = z_1 + \\dots + z_n = q, \\\\\n0 \\le x_i, y_i, z_i \\le \\min(i, k) \\quad \\text{for any } 1 \\le i \\le n.\n$$\nLet $q = mk + r$, where $0 \\le r < k$ is the remainder when divided by $k$. It is clear that the expression (3) is maximized when the sums $x_i + y_i + z_i$ with higher coefficients are the maximum possible (that is, equal to $3k$). Hence, under these conditions we get\n$$\n\\begin{aligned}\n(2n + 1)q &= (x_1 + y_1 + z_1) + 2(x_2 + y_2 + z_2) + \\dots + n(x_n + y_n + z_n) \\\\\n&\\le n \\cdot 3k + (n - 1) \\cdot 3k + \\dots + (n - m + 1) \\cdot 3k + (n - m) \\cdot 3r \\\\\n&\\le \\frac{2n - m + 1}{2} m \\cdot 3k + (n - m) \\cdot 3r \\\\\n&\\le \\frac{6nmk - 3m^2k + 3mk}{2} + 3nr - 3mr \\\\\n&\\le \\frac{6n(mk + r) - 3m^2k + 3(mk + r) - 3r - 6mr}{2} \\\\\n&\\le \\frac{(6n + 3)q - 3m^2k - 3r - 6mr}{2}.\n\\end{aligned}\n$$\nMoving $(2n + 1)q$ to the right side, multiplying the inequality by $2k$ and substituting $mk = q - r$, we arrive at a quadratic inequality for $q$:\n$$\n\\begin{aligned}\n0 &\\le (6n + 3 - 4n - 2)kq - 3(mk)^2 - 3rk - 6mkr \\\\\n&= (2n + 1)kq - 3(q - r)^2 - 6(q - r)r - 3rk \\\\\n&= (2n + 1)kq - 3q^2 + 6qr - 3r^2 - 6qr + 6r^2 - 3rk \\\\\n&= -3q^2 + (2n + 1)kq + 3r(r - k),\n\\end{aligned}\n$$\nfrom which we obtain\n$$\n3q^2 \\le (2n + 1)kq + 3r(r - k) \\le (2n + 1)kq, \\quad \\text{that is} \\quad q \\le \\frac{2n + 1}{3}k,\n$$\nbecause $r - k < 0$. Since $q$ is an integer, we get $q \\le \\left\\lfloor \\frac{2n+1}{3}k \\right\\rfloor$, as required.\n\nb. We have $T(1, n) = \\left\\lfloor \\frac{2n+1}{3} \\right\\rfloor$ and $T(2, n) = \\left\\lfloor \\frac{4n+2}{3} \\right\\rfloor$. Placements of checkers, when equality is achieved in the last inequality, for $k=1$ and $k=2$, are shown in figures 2, 3 and 4.\n\n![](attached_image_3.png)\n\n![](attached_image_4.png)\n\n![](attached_image_5.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25755,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 3$ be a positive integer. Positive integers are written in two rows on the whiteboard $a_1\\ a_2\\ \\dots\\ a_n\\ b_1\\ b_2\\ \\dots\\ b_n$, where the sets $a_1, a_2, \\dots, a_n$ and $b_1, b_2, \\dots, b_n$ are some permutations of the numbers $1, 2, \\dots, n$. The teacher allows the student Dima to use the following two operations as many times as he likes and in any order:\n(A) replace each number $a$ of the first line with the number that is in the $a$'th place (from left to right) of the second line;\n(B) replace each number $b$ of the second line with the number that is in the $b$'th place (from left to right) of the first line.\n\nThus, for example, the operation $A$ transforms the table\n$$ \\begin{array}{ccc} 1 & 2 & 3 \\\\ 2 & 3 & 1 \\end{array} $$\ninto the table\n$$ \\begin{array}{ccc} 2 & 3 & 1 \\\\ 2 & 3 & 1 \\end{array} $$\n, which in turn is transformed by the operation $B$ into the table\n$$ \\begin{array}{ccc} 2 & 3 & 1 \\\\ 3 & 1 & 2 \\end{array} $$\n\nCan Dima obtain the table\n$$ \\begin{array}{ccccccc} 2 & 3 & 1 & 4 & \\dots & n-1 & n \\\\ 2 & 3 & 4 & 5 & \\dots & n & 1 \\end{array} $$\nfrom the table\n$$ \\begin{array}{ccccccc} 2 & 1 & 3 & 4 & \\dots & n-1 & n \\\\ 2 & 3 & 4 & 5 & \\dots & n & 1 \\end{array} $$",
    "options": [],
    "answer": "no",
    "solution": "Answer: no. Each table $\\begin{array}{c} a_1 \\\\ b_1 \\end{array} \\begin{array}{c} a_2 \\\\ b_2 \\end{array} \\dots \\begin{array}{c} a_n \\\\ b_n \\end{array}$ corresponds to a pair of permutations $(s,t) \\in S_n \\times S_n$ such that\n$$\ns = \\begin{pmatrix} 1 & 2 & \\dots & n \\\\ a_1 & a_2 & \\dots & a_n \\end{pmatrix} \\quad \\text{and} \\quad t = \\begin{pmatrix} 1 & 2 & \\dots & n \\\\ b_1 & b_2 & \\dots & b_n \\end{pmatrix}.\n$$\nThe operation $A$ transforms a pair $(s, t)$ into the pair $(st, t)$, where the composition of the permutations $st$ is read from left to right – first the permutation $s$ is applied, and then the permutation $t$. Accordingly, the operation $B$ transforms a pair $(s, t)$ into the pair $(s, ts)$.\nFor each pair of permutations $(s,t)$ we can consider the expression $sts^{-1}t^{-1}$, which is called the *commutator* of this pair and is denoted by $[s,t]$. Let us calculate the commutators of the pairs $(st,t)$ and $(s,ts)$:\n$$\n[st, t] = (st)t(st)^{-1}t^{-1} = sttt^{-1}s^{-1}t^{-1} = [s, t] \\quad \\text{and} \\quad [s, ts] = s(ts)s^{-1}(ts)^{-1} = stss^{-1}s^{-1}t^{-1} = [s, t].\n$$\nThus, the commutator is invariant under the operations A and B.\n\nThe table\n$$\n\\begin{array}{|c|ccccccc|}\n\\hline\n2 & 1 & 3 & 4 & \\dots & n-1 & n \\\\\n\\hline\n2 & 3 & 4 & 5 & \\dots & n & 1 \\\\\n\\hline\n\\end{array}\n$$\ncorresponds to a pair of permutations $(1\\ 2)$ and $(1\\ 2\\ \\dots\\ n)$,\nwhich are cycles, and the table\n$$\n\\begin{array}{|c|ccccccc|}\n\\hline\n2 & 3 & 1 & 4 & \\dots & n-1 & n \\\\\n\\hline\n2 & 3 & 4 & 5 & \\dots & n & 1 \\\\\n\\hline\n\\end{array}\n$$\ncorresponds to a pair of permutations\n$(1\\ 2\\ 3)$ and $(1\\ 2\\ \\dots\\ n)$, which are also cycles. Their commutators are easy to find:\n$$\n[(1\\ 2), (1\\ 2\\ \\dots\\ n)] = (1\\ 2)(1\\ 2\\ \\dots\\ n)(2\\ 1)(n\\ \\dots\\ 2\\ 1) = (1\\ 2\\ n),\n$$\n$$\n[(1\\ 2\\ 3), (1\\ 2\\ \\dots\\ n)] = (1\\ 2\\ 3)(1\\ 2\\ \\dots\\ n)(3\\ 2\\ 1)(n\\ \\dots\\ 2\\ 1) = (1\\ 2\\ 3\\ n).\n$$\nSince they are not equal, Dima will not be able to get the second table from the first one.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 25756,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given $n \\in \\mathbb{N}$. Consider polynomials of degree $n$ of the form $P(x) = x^n + a_{n-1}x^{n-1} + \\dots + a_1x + a_0$ with non-negative integer coefficients not exceeding $100$. We call such a polynomial *expandable* if it can be represented as a product of two non-constant polynomials with non-negative integer coefficients, and *non-expandable* otherwise. Prove that there are at least twice as many non-expandable polynomials as there are expandable ones.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If the polynomial $P(x) = x^n + a_{n-1}x^{n-1} + \\cdots + a_1x + a_0$ is expandable, then it can be represented as\n$$\nP(x) = (x^m + b_{m-1}x^{m-1} + \\cdots + b_1x + b_0) \\cdot (x^k + c_{k-1}x^{k-1} + \\cdots + c_1x + c_0),\n$$\nwhere $m \\ge k \\ge 1$, $m + k = n$ and numbers $b_i, c_i$ are all non-negative integers. Since the coefficients of the polynomial $P(x)$ are not greater than $100$, all these numbers do not exceed $100$ and, moreover, $b_i \\cdot c_i \\le 100$ for any $i$ from $0$ to $k-1$. This means that for a fixed $i$ from $0$ to $k-1$ the number of possible pairs $(b_i, c_i)$ does not exceed\n$$\n2 \\cdot 11 \\cdot 101 < \\frac{101^2}{4},\n$$\nbecause at least one of $b_i$ and $c_i$ is not greater than $\\sqrt{100} = 10$.\nThus, the numbers $b_{m-1}, b_{m-2}, \\dots, b_k$ can be chosen in no more than $101^{m-k}$ ways, and the numbers $b_0, c_0, b_1, c_1, \\dots, b_{k-1}, c_{k-1}$ you can select no more than $\\left(\\frac{101^2}{4}\\right)^k$ ways. We find that the number of expandable polynomials can be estimated from above by the expression\n$$\n\\sum_{k=1}^{[n/2]} 101^{m-k} \\left(\\frac{101^2}{4}\\right)^k = 101^n \\sum_{k=1}^{[n/2]} \\frac{1}{4^k} < 101^n \\frac{1/4}{1 - 1/4} = \\frac{1}{3} \\cdot 101^n\n$$\n\nSince the number of polynomials $P(x) = x^n + a_{n-1}x^{n-1} + \\dots + a_1x + a_0$ with non-negative integer coefficients not exceeding $100$ is equal to $101^n$, then there are more than $\\frac{2}{3} \\cdot 101^n$ non-expandable polynomials, that is, at least twice as many as expandable ones.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 25757,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $k$ and $n$ be positive integers. An international company has connected $k$ cities of Armenia with $k$ cities of Belarus by direct two-way airlines. From each of these Belarusian cities there is a direct flight to exactly $n$ Armenian ones. It turned out that for any two Armenian cities there are exactly two Belarusian cities that are connected by airlines to both of them.\n\na) Prove that each of the Armenian cities is connected by airlines to exactly $n$ Belarusian cities.\n\nb) Prove that it is possible to travel on planes of a given airline without repeating cities, while visiting at least $\\left\\lfloor \\frac{(n+1)^2}{4} \\right\\rfloor$ cities in each of the countries.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let's translate the problem into the language of graphs.\n\nGiven a bipartite graph $G$ with parts $A$ and $B$ having the same number of vertices: $|V(A)| = |V(B)| = k$. The degree of each vertex in $B$ is $n$. For any two vertices $u, v$ of the part $A$, there are exactly two vertices in $B$ that are adjacent to both vertices $u$ and $v$.\n\na) Prove that the degree of any vertex in $A$ is also equal to $n$.\n\nb) Prove that there is a simple path in the graph that contains at least $\\lfloor \\frac{(n+1)^2}{4} \\rfloor$ vertices in each part.\n\nFirst, let's prove that $k = \\binom{n}{2} + 1$ and that the degree of each vertex in $A$ is equal to $n$.\n\nLet's count the number of pairs ($\\{a, a'\\}, b$) where $a \\neq a'$, $a, a' \\in V(A)$, $b \\in V(B)$ and $b$ is adjacent to $a$ and $a'$. On the one hand, we have $|V(B)|$ ways to select $b$ and each such vertex gives $\\binom{n}{2}$ pairs $a, a'$ to which it is adjacent. On the other hand, there are $\\binom{|V(A)|}{2}$ ways to select distinct $a$ and $a'$, and further, for each pair, there are exactly two ways to select $b$, according to the second property of our graph. Thus, $\\binom{n}{2}k = 2\\binom{k}{2}$, which implies that $k = \\binom{n}{2} + 1$.\n\nLet's take an arbitrary vertex $a$ from $A$ and let it have degree $d$. Let us count in two ways the number of pairs $(a', b)$ where $a' \\in V(A)$, $b \\in V(B)$, $a' \\neq a$ and vertex $b$ is adjacent to both vertices $a$ and $a'$. On the one hand, we have $|V(A)| - 1$ options to choose a vertex $a'$ and for each such vertex we have exactly two choices of a vertex $b$. On the other hand, each of the $d$ neighbors of a vertex $a$ has degree exactly $n$, which gives $(n-1)d$ pairs in question (the neighbors of a given vertex from $V(B)$ must be different from $a$, so there are $n-1$ of them). Thus $(n-1)d = 2(k-1) = 2\\binom{n}{2}$, which means that $d = n$.\n\nNow let's move on to finding a path. Since the graph is regular, then by Hall's theorem there is a perfect matching in it – a set of $k$ pairwise non-adjacent edges. Let $I$ be such a matching. Let us take the longest such path $P = b_1a_1, \\dots, b_ra_r$, where $b_1 \\in V(B)$ and $a_r \\in V(A)$, with the following property:\n\nif a vertex $x$ is in $P$, then the vertex $y$ adjacent to $x$ in the matching $I$ is also in $P$.\n\n(Note that this condition does not require that edges from $I$ also be in our simple path.) Then all neighbors of the ends of $P$ lie in $P$. In fact, if one of the ends of $P$ has a neighbor outside $P$, then the neighbor of this vertex in $I$ is also not in our path and then we can extend our path by two vertices so that it starts and ends at different shares.\n\nLet now $b_i$ ($1 \\le i \\le r$) be one of the neighbors of a vertex $a_k$ in path $P$. Then the path\n$$\nb_1a_1 \\dots b_ia_r b_r a_{r-1} \\dots a_i\n$$\nalso has the above property. In particular, all neighbors of the vertex $a_i \\in V(A)$ also lie in $P$. Since $a_r$ has $n$ neighbors, we have a minimum of $n$ vertices $a_{i_1}, \\dots, a_{i_n}$ from $A$ each of which has neighbors only in the path $P$. Since vertices $a_{i_1}$ and $a_{i_2}$ have exactly two neighbors in common, they have a total of $n + (n-2)$ neighbors in $P$. Next, consider the vertex $a_{i_3}$ – it has two common neighbors with each of the previous vertices, that is, a maximum of 4 common neighbors have already been counted. This gives $n-4$ new neighbors different from the previous ones. Reasoning in this way, we get\n$$\nr \\ge n + (n-2) + (n-4) + \\dots = \\left\\lfloor \\frac{(n+1)^2}{4} \\right\\rfloor,\n$$\nwhich is what was required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25758,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA finite set of integers is called bad if its elements add up to $2010$. A finite set of integers is a Benelux-set if none of its subsets is bad. Determine the smallest integer $n$ such that the set $\\{502,503,504, \\ldots, 2009\\}$ can be partitioned into $n$ Benelux-sets.\n(A partition of a set $S$ into $n$ subsets is a collection of $n$ pairwise disjoint subsets of $S$, the union of which equals $S$.)",
    "options": [],
    "answer": "2",
    "solution": "Solution:\nAs $502+1508=2010$, the set $S=\\{502,503, \\ldots, 2009\\}$ is not a Benelux-set, so $n=1$ does not work. We will prove that $n=2$ does work, i.e. that $S$ can be partitioned into 2 Benelux-sets.\nDefine the following subsets of $S$ :\n$$\n\\begin{aligned}\n& A=\\{502,503, \\ldots, 670\\}, \\\\\n& B=\\{671,672, \\ldots, 1005\\}, \\\\\n& C=\\{1006,1007, \\ldots, 1339\\}, \\\\\n& D=\\{1340,1341, \\ldots, 1508\\}, \\\\\n& E=\\{1509,1510, \\ldots, 2009\\} .\n\\end{aligned}\n$$\nWe will show that $A \\cup C \\cup E$ and $B \\cup D$ are both Benelux-sets.\nNote that there does not exist a bad subset of $S$ of one element, since that element would have to be $2010$. Also, there does not exist a bad subset of $S$ of more than three elements, since the sum of four or more elements would be at least $502+503+504+505=2014>2010$. So any possible bad subset of $S$ contains two or three elements.\nConsider a bad subset of two elements $a$ and $b$. As $a, b \\geq 502$ and $a+b=2010$, we have $a, b \\leq 2010-502=1508$. Furthermore, exactly one of $a$ and $b$ is smaller than $1005$ and one is larger than $1005$. So one of them, say $a$, is an element of $A \\cup B$, and the other is an element of $C \\cup D$. Suppose $a \\in A$, then $b \\geq 2010-670=1340$, so $b \\in D$. On the other hand, suppose $a \\in B$, then $b \\leq 2010-671=1339$, so $b \\in C$. Hence $\\{a, b\\}$ cannot be a subset of $A \\cup C \\cup E$, nor of $B \\cup D$.\nNow consider a bad subset of three elements $a, b$ and $c$. As $a, b, c \\geq 502$, $a+b+c=2010$, and the three elements are pairwise distinct, we have $a, b, c \\leq 2010-502-503=1005$. So $a, b, c \\in A \\cup B$. At least one of the elements, say $a$, is smaller than $\\frac{2010}{3}=670$, and at least one of the elements, say $b$, is larger than $670$. So $a \\in A$ and $b \\in B$. We conclude that $\\{a, b, c\\}$ cannot be a subset of $A \\cup C \\cup E$, nor of $B \\cup D$.\nThis proves that $A \\cup C \\cup E$ and $B \\cup D$ are Benelux-sets, and therefore the smallest $n$ for which $S$ can be partitioned into $n$ Benelux-sets is $n=2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25759,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all polynomials $p(x)$ with real coefficients such that\n$$\np(a+b-2 c)+p(b+c-2 a)+p(c+a-2 b)=3 p(a-b)+3 p(b-c)+3 p(c-a)\n$$\nfor all $a, b, c \\in \\mathbb{R}$.",
    "options": [],
    "answer": "All real polynomials of the form p(x) = ax^2 + bx with real a, b.",
    "solution": "Solution:\nFor $a=b=c$, we have $3 p(0)=9 p(0)$, hence $p(0)=0$. Now set $b=c=0$, then we have\n$$\np(a)+p(-2 a)+p(a)=3 p(a)+3 p(-a)\n$$\nfor all $a \\in \\mathbb{R}$. So we find a polynomial equation\n$$\np(-2 x)=p(x)+3 p(-x)\n$$\nNote that the zero polynomial is a solution to this equation. Now suppose that $p$ is not the zero polynomial, and let $n \\geq 0$ be the degree of $p$. Let $a_{n} \\neq 0$ be the coefficient of $x^{n}$ in $p(x)$. At the left-hand side of (1), the coefficient of $x^{n}$ is $(-2)^{n} \\cdot a_{n}$, while at the right-hand side the coefficient of $x^{n}$ is $a_{n}+3 \\cdot(-1)^{n} \\cdot a_{n}$. Hence $(-2)^{n}=1+3 \\cdot(-1)^{n}$. For $n$ even, we find $2^{n}=4$, so $n=2$, and for $n$ odd, we find $-2^{n}=-2$, so $n=1$. As we already know that $p(0)=0$, we must have $p(x)=a_{2} x^{2}+a_{1} x$, where $a_{1}$ and $a_{2}$ are real numbers (possibly zero).\n\nThe polynomial $p(x)=x$ is a solution to our problem, as\n$$\n(a+b-2 c)+(b+c-2 a)+(c+a-2 b)=0=3(a-b)+3(b-c)+3(c-a)\n$$\nfor all $a, b, c \\in \\mathbb{R}$. Also, $p(x)=x^{2}$ is a solution, since\n$$\n\\begin{aligned}\n(a+b-2 c)^{2}+(b+c-2 a)^{2}+(c+a-2 b)^{2} & =6\\left(a^{2}+b^{2}+c^{2}\\right)-6(a b+b c+c a) \\\\\n& =3(a-b)^{2}+3(b-c)^{2}+3(c-a)^{2}\n\\end{aligned}\n$$\nfor all $a, b, c \\in \\mathbb{R}$.\n\nNow note that if $p(x)$ is a solution to our problem, then so is $\\lambda p(x)$ for all $\\lambda \\in \\mathbb{R}$. Also, if $p(x)$ and $q(x)$ are both solutions, then so is $p(x)+q(x)$. We conclude that for all real numbers $a_{2}$ and $a_{1}$ the polynomial $a_{2} x^{2}+a_{1} x$ is a solution. Since we have already shown that there can be no other solutions, these are the only solutions.\nSolution:\nFor $a=b=c$, we have $3 p(0)=9 p(0)$, hence $p(0)=0$. Now set $b=c=0$, then we have\n$$\np(a)+p(-2 a)+p(a)=3 p(a)+3 p(-a)\n$$\nfor all $a \\in \\mathbb{R}$. So we find a polynomial equation\n$$\np(-2 x)=p(x)+3 p(-x)\n$$\nDefine $q(x)=p(x)+p(-x)$, then we find that\n$$\nq(2 x)=p(2 x)+p(-2 x)=(p(-x)+3 p(x))+(p(x)+3 p(-x))=4 q(x)\n$$\nNote that the zero polynomial is a solution to this equation. Now suppose that $q$ is not the zero polynomial, and let $m \\geq 0$ be the degree of $q$. Let $b_{m} \\neq 0$ be the coefficient of $x^{m}$ in $q(x)$. At the left-hand side of (3), the coefficient of $x^{m}$ is $2^{m} \\cdot b_{m}$, while at the right-hand side the coefficient of $x^{m}$ is $4 b_{m}$. Hence $m=2$. As $q(x)=p(x)+p(-x)$, the polynomial $q(x)$ does not contain any nonzero terms with odd exponent of $x$. Since also $q(0)=2 p(0)=0$, we conclude that\n$$\nq(x)=b_{2} x^{2}\n$$\nwhere $b_{2}$ is a real number (possibly zero).\n\nFrom (2) we now deduce that $p(2 x)=p(-x)+3 p(x)=2 p(x)+q(x)$, so\n$$\np(2 x)-2 p(x)=b_{2} x^{2}\n$$\nSuppose that the degree $n$ of $p$ is greater than 2. Let $a_{n} \\neq 0$ be the coefficient of $x^{n}$ in $p(x)$. At the left-hand side of (4), the coefficient of $x^{n}$ is $\\left(2^{n}-2\\right) \\cdot a_{n} \\neq 0$. But the coefficient of $x^{n}$ at the right-hand side vanishes, yielding a contradiction. So the degree of $p$ is at most 2. As we already know that $p(0)=0$, we must have $p(x)=a_{2} x^{2}+a_{1} x$, where $a_{1}$ and $a_{2}$ are real numbers (possibly zero).\n\nWe finally check that every polynomial of this form is indeed a solution (see solution 1).",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 25760,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nOn a line $l$ there are three different points $A$, $B$ and $P$ in that order. Let $a$ be the line through $A$ perpendicular to $l$, and let $b$ be the line through $B$ perpendicular to $l$. A line through $P$, not coinciding with $l$, intersects $a$ in $Q$ and $b$ in $R$. The line through $A$ perpendicular to $B Q$ intersects $B Q$ in $L$ and $B R$ in $T$. The line through $B$ perpendicular to $A R$ intersects $A R$ in $K$ and $A Q$ in $S$.\n\na. Prove that $P$, $T$, $S$ are collinear.\n\nb. Prove that $P$, $K$, $L$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na.\nSince $P$, $R$ and $Q$ are collinear, we have $\\triangle P A Q \\sim \\triangle P B R$, hence\n$$\n\\frac{|A Q|}{|B R|} = \\frac{|A P|}{|B P|}\n$$\nConversely, $P$, $T$ and $S$ are collinear if it holds that\n$$\n\\frac{|A S|}{|B T|} = \\frac{|A P|}{|B P|}\n$$\nSo it suffices to prove\n$$\n\\frac{|B T|}{|B R|} = \\frac{|A S|}{|A Q|}\n$$\nSince $\\angle A B T = 90^\\circ = \\angle A L B$ and $\\angle T A B = \\angle B A L$, we have $\\triangle A B T \\sim \\triangle A L B$. And since $\\angle A L B = 90^\\circ = \\angle Q A B$ and $\\angle L B A = \\angle A B Q$, we have $\\triangle A L B \\sim \\triangle Q A B$. Hence $\\triangle A B T \\sim \\triangle Q A B$, so\n$$\n\\frac{|B T|}{|B A|} = \\frac{|A B|}{|A Q|}\n$$\nSimilarly, we have $\\triangle A B R \\sim \\triangle A K B \\sim \\triangle S A B$, so\n$$\n\\frac{|B R|}{|B A|} = \\frac{|A B|}{|A S|}\n$$\nCombining both results, we get\n$$\n\\frac{|B T|}{|B R|} = \\frac{|B T| / |B A|}{|B R| / |B A|} = \\frac{|A B| / |A Q|}{|A B| / |A S|} = \\frac{|A S|}{|A Q|}\n$$\nwhich had to be proved.\n\nb.\nLet the line $P K$ intersect $B R$ in $B_{1}$ and $A Q$ in $A_{1}$ and let the line $P L$ intersect $B R$ in $B_{2}$ and $A Q$ in $A_{2}$. Consider the points $A_{1}$, $A$ and $S$ on the line $A Q$, and the points $B_{1}$, $B$ and $T$ on the line $B R$. As $A Q \\parallel B R$ and the three lines $A_{1} B_{1}$, $A B$ and $S T$ are concurrent (in $P$), we have\n$$\nA_{1} A : A S = B_{1} B : B T\n$$\nwhere all lengths are directed. Similarly, as $A_{1} B_{1}$, $A R$ and $S B$ are concurrent (in $K$), we have\n$$\nA_{1} A : A S = B_{1} R : R B\n$$\nThis gives\n$$\n\\frac{B B_{1}}{B T} = \\frac{R B_{1}}{R B} = \\frac{R B + B B_{1}}{R B} = 1 + \\frac{B B_{1}}{R B} = 1 - \\frac{B B_{1}}{B R}\n$$\nso\n$$\nB B_{1} = \\frac{1}{\\frac{1}{B T} + \\frac{1}{B R}}\n$$\nSimilarly, using the lines $A_{2} B_{2}$, $A B$ and $Q R$ (concurrent in $P$) and the lines $A_{2} B_{2}$, $A T$ and $Q B$ (concurrent in $L$), we find\n$$\nB_{2} B : B R = A_{2} A : A Q = B_{2} T : T B\n$$\nThis gives\n$$\n\\frac{B B_{2}}{B R} = \\frac{T B_{2}}{T B} = \\frac{T B + B B_{2}}{T B} = 1 + \\frac{B B_{2}}{T B} = 1 - \\frac{B B_{2}}{B T}\n$$\nso\n$$\nB B_{2} = \\frac{1}{\\frac{1}{B R} + \\frac{1}{B T}}\n$$\nWe conclude that $B_{1} = B_{2}$, which implies that $P$, $K$ and $L$ are collinear.\nSolution:\n\na.\nDefine $X$ as the intersection of $A T$ and $B S$, and $Y$ as the intersection of $A R$ and $B Q$. To prove that $P$, $S$ and $T$ are collinear, we will use Menelaos' theorem in $\\triangle A B X$, so we have to prove\n$$\n\\frac{A P}{P B} \\frac{B S}{S X} \\frac{X T}{T A} = -1\n$$\nNote that $B$ is between $P$ and $A$, $X$ is between $S$ and $B$, and $X$ is between $T$ and $A$, so it suffices to prove that\n$$\n\\frac{|A P|}{|P B|} \\frac{|B S|}{|S X|} \\frac{|X T|}{|T A|} = 1\n$$\nBecause $A Q$ and $B R$ are parallel, we have $\\triangle A Q P \\sim \\triangle B R P$, hence\n$$\n\\frac{|A P|}{|B P|} = \\frac{|Q A|}{|R B|}\n$$\nAlso, since $\\angle A S B = \\angle K B R$ and $\\angle B A S = 90^\\circ = \\angle B K R$, we have $\\triangle A S B \\sim \\triangle K B R$, hence\n$$\n\\frac{|B S|}{|R B|} = \\frac{|A S|}{|K B|}, \\quad \\text{so} \\quad |B S| = \\frac{|A S|}{|K B|} |R B|.\n$$\nSimilarly, we have $\\triangle A T B \\sim \\triangle Q A L$, hence\n$$\n\\frac{|T A|}{|A Q|} = \\frac{|T B|}{|A L|}, \\quad \\text{so} \\quad |T A| = \\frac{|T B|}{|A L|} |A Q|.\n$$\nAs $\\angle A S X = \\angle A S B = 90^\\circ - \\angle A B S = 90^\\circ - \\angle A B K = \\angle K A B = \\angle Y A B$, and $\\angle S A X = 90^\\circ - \\angle X A B = 90^\\circ - \\angle L A B = \\angle A B L = \\angle A B Y$, we have $\\triangle S X A \\sim \\triangle A Y B$, hence\n$$\n\\frac{|S X|}{|A Y|} = \\frac{|A S|}{|B A|}, \\quad \\text{so} \\quad |S X| = \\frac{|A S|}{|B A|} |A Y|.\n$$\nSimilarly, we have $\\triangle B X T \\sim \\triangle A Y B$, hence\n$$\n\\frac{|X T|}{|Y B|} = \\frac{|B T|}{|A B|}, \\quad \\text{so} \\quad |X T| = \\frac{|B T|}{|A B|} |Y B|.\n$$\nBy combining (5) - (9), we find\n$$\n\\begin{aligned}\n\\frac{|A P|}{|P B|} \\frac{|B S|}{|S X|} \\frac{|X T|}{|T A|} & = \\frac{|Q A|}{|R B|} \\cdot \\frac{|A S|}{|K B|} |R B| \\cdot \\frac{|B A|}{|A S||A Y|} \\cdot \\frac{|B T|}{|A B|} |Y B| \\cdot \\frac{|A L|}{|T B||A Q|} \\\\\n& = \\frac{|A L|}{|K B|} \\frac{|Y B|}{|A Y|} .\n\\end{aligned}\n$$\nSince $\\angle Y L A = 90^\\circ = \\angle Y K B$ and $\\angle A Y L = \\angle B Y K$, we have $\\triangle A Y L \\sim \\triangle B Y K$, hence\n$$\n\\frac{|A L|}{|B K|} = \\frac{|A Y|}{|B Y|}, \\quad \\text{so} \\quad \\frac{|A L|}{|B K|} \\frac{|B Y|}{|A Y|} = 1\n$$\nBy combining (10) and (11), we find\n$$\n\\frac{|A P|}{|P B|} \\frac{|B S|}{|S X|} \\frac{|X T|}{|T A|} = 1,\n$$\nas we wanted to prove.\n\nb.\nAgain, we will use Menelaos' theorem in $\\triangle A B X$, so we have to prove\n$$\n\\frac{A P}{P B} \\frac{B K}{K X} \\frac{X L}{L A} = -1\n$$\nNote that $\\frac{A P}{P B} < 0$, and $\\frac{B K}{K X} < 0$ if and only if $\\frac{X L}{L A} < 0$, so it suffices to prove that\n$$\n\\frac{|A P|}{|P B|} \\frac{|B K|}{|K X|} \\frac{|X L|}{|L A|} = 1\n$$\nAs $\\angle B X L = \\angle A X K$ and $\\angle B L X = 90^\\circ = \\angle A K X$, we have $\\triangle B L X \\sim \\triangle A K X$, hence\n$$\n\\frac{|X L|}{|X K|} = \\frac{|B L|}{|A K|}\n$$\nSince $\\angle A L B = 90^\\circ = \\angle Q A B$, we have $\\triangle A L B \\sim \\triangle Q A B$, hence\n$$\n\\frac{|L A|}{|A Q|} = \\frac{|L B|}{|A B|}, \\quad \\text{so} \\quad |L A| = \\frac{|L B|}{|A B|} |A Q|.\n$$\nSimilarly, we have $\\triangle A K B \\sim \\triangle A B R$, hence\n$$\n\\frac{|B K|}{|R B|} = \\frac{|A K|}{|A B|}, \\quad \\text{so} \\quad |B K| = \\frac{|A K|}{|A B|} |R B|\n$$\nBy combining (5) and (12) - (14), we find\n$$\n\\left.\\frac{|A P|}{|P B|}\\left|\\frac{|B K|}{|K X|} \\frac{|X L|}{|L A|} = \\frac{|Q A|}{|R B|} \\cdot \\frac{|B L|}{|A K|} \\cdot \\frac{|A B|}{|L B||A Q|} \\cdot \\frac{|A K|}{|A B|}\\right| R B \\right\\rvert\\, = 1,\n$$\nwhich is what we wanted to prove.\nSolution:\nAs $\\angle A K B = \\angle A L B = 90^\\circ$, the points $K$ and $L$ belong to the circle with diameter $A B$. Since $\\angle Q A B = \\angle A B R = 90^\\circ$, the lines $A Q$ and $B R$ are tangents to this circle.\n\nApply Pascal's theorem to the points $A$, $A$, $K$, $L$, $B$ and $B$, all on the same circle. This yields that the intersection $Q$ of the tangent in $A$ and the line $B L$, the intersection $R$ of the tangent in $B$ and the line $A K$, and the intersection of $K L$ and $A B$ are collinear. So $K L$ passes through the intersection of $A B$ and $Q R$, which is point $P$. Hence $P$, $K$ and $L$ are collinear. This proves part b.\n\nNow apply Pascal's theorem to the points $A$, $A$, $L$, $K$, $B$ and $B$. This yields that the intersection $S$ of the tangent in $A$ and the line $B K$, the intersection $T$ of the tangent in $B$ and the line $A L$, and the intersection $P$ of $K L$ and $A B$ are collinear. This proves part a.\nSolution:\n\na.\nW.l.o.g. we may assume that $A = (0, 0)$ and $B = (1, 0)$ and the line through $P$ is in the upper half plane, so $l$ is the $x$-axis, $a$ is the $y$-axis and $b$ is the line $x = 1$. Take $P = (p, 0)$ ($p > 1$) and $Q = (0, q)$ ($q > 0$). Since $P Q$ is given by $\\frac{x}{p} + \\frac{y}{q} = 1$, we find $R = \\left(1, \\frac{q(p-1)}{p}\\right)$.\n\nNow $A R$ is given by $y = \\frac{q(p-1)}{p} x$, hence $B S$, the line perpendicular to $A R$ and passing through $B = (1, 0)$, is given by $y = -\\frac{p}{q(p-1)}(x-1)$. We find $S = \\left(0, \\frac{p}{q(p-1)}\\right)$.\n\nMoreover $B Q$ is given by $y = -q(x-1)$, hence $A T$, the line perpendicular to $B Q$ and passing through $A = (0, 0)$, is given by $y = \\frac{1}{q} x$. We find $T = \\left(1, \\frac{1}{q}\\right)$. Since $\\frac{|B T|}{|B P|} = \\frac{1 / q}{p-1} = \\frac{\\frac{p}{q(p-1)}}{p} = \\frac{|A S|}{|A P|}$, we conclude that $P$, $T$ and $S$ are collinear.\n\nb.\nPoint $K$ is the intersection of $A R$ and $B S$. Solving for $x$ yields\n$$\n\\begin{aligned}\n\\frac{q(p-1)}{p} x & = -\\frac{p}{q(p-1)}(x-1) \\\\\n\\left(\\frac{q(p-1)}{p} + \\frac{p}{q(p-1)}\\right) x & = \\frac{p}{q(p-1)} \\\\\nx & = \\frac{\\frac{p}{q(p-1)}}{\\frac{q(p-1)}{p} + \\frac{p}{q(p-1)}}\n\\end{aligned}\n$$\nso\n$$\nK = \\left(\\frac{\\frac{p}{q(p-1)}}{\\frac{q(p-1)}{p} + \\frac{p}{q(p-1)}}, \\frac{1}{\\frac{q(p-1)}{p} + \\frac{p}{q(p-1)}}\\right)\n$$\nPoint $L$ is the point of intersection of $A T$ and $B Q$. Solving for $x$ yields\n$$\n\\begin{aligned}\n\\frac{1}{q} x & = -q(x-1) \\\\\n\\left(\\frac{1}{q} + q\\right) x & = q \\\\\nx & = \\frac{q}{\\frac{1}{q} + q}\n\\end{aligned}\n$$\nso\n$$\nL = \\left(\\frac{q}{\\frac{1}{q} + q}, \\frac{1}{\\frac{1}{q} + q}\\right)\n$$\nLet $K_{0}$ and $L_{0}$ be the projections of $K$ and $L$ on the $x$-axis. We have to show that the following fractions are equal:\n$$\n\\frac{\\left|K_{0} K\\right|}{\\left|K_{0} P\\right|} = \\frac{\\frac{1}{\\frac{q(p-1)}{p} + \\frac{p}{q(p-1)}}}{p - \\frac{\\frac{p}{q(p-1)}}{\\frac{q(p-1)}{p} + \\frac{p}{q(p-1)}}} \\quad \\text{and} \\quad \\frac{\\left|L_{0} L\\right|}{\\left|L_{0} P\\right|} = \\frac{\\frac{1}{\\frac{1}{q} + q}}{p - \\frac{q}{\\frac{1}{q} + q}}\n$$\nWorking out cross products twice, this comes down to\n$$\n\\begin{aligned}\n\\frac{1}{\\frac{q(p-1)}{p} + \\frac{p}{q(p-1)}} \\cdot \\left(p - \\frac{q}{\\frac{1}{q} + q}\\right) & \\stackrel{?}{=} \\frac{1}{\\frac{1}{q} + q} \\cdot \\left(p - \\frac{\\frac{p}{q(p-1)}}{\\frac{q(p-1)}{p} + \\frac{p}{q(p-1)}}\\right) \\\\\n\\left(\\frac{1}{q} + q\\right) \\cdot \\left(p - \\frac{q}{\\frac{1}{q} + q}\\right) & \\stackrel{?}{=} \\left(\\frac{q(p-1)}{p} + \\frac{p}{q(p-1)}\\right) \\cdot \\left(p - \\frac{\\frac{p}{q(p-1)}}{\\frac{q(p-1)}{p} + \\frac{p}{q(p-1)}}\\right) \\\\\n\\frac{p}{q} + p q - q & \\stackrel{?}{=} q(p-1) + \\frac{p^{2}}{q(p-1)} - \\frac{p}{q(p-1)} \\\\\n\\frac{p}{q} + p q - q & \\stackrel{?}{=} q(p-1) + \\frac{p(p-1)}{q(p-1)}\n\\end{aligned}\n$$\nwhich is clearly true.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25761,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all quadruples $(a, b, p, n)$ of positive integers, such that $p$ is a prime and\n$$\na^{3}+b^{3}=p^{n}\n$$",
    "options": [],
    "answer": "All solutions are the three families: (2^k, 2^k, 2, 3k+1), (2·3^k, 3^k, 3, 3k+2), and (3^k, 2·3^k, 3, 3k+2) for all integers k ≥ 0.",
    "solution": "Solution:\nLet $(a, b, p, n)$ be a solution. Note that we can write the given equation as\n$$\n(a+b)\\left(a^{2}-a b+b^{2}\\right)=p^{n}\n$$\nAs $a$ and $b$ are positive integers, we have $a+b \\geq 2$, so $p \\mid a+b$. Furthermore, $a^{2}-a b+b^{2}=(a-b)^{2}+a b$, so either $a=b=1$ or $a^{2}-a b+b^{2} \\geq 2$. Assume that the latter is the case. Then $p$ is a divisor of both $a+b$ and $a^{2}-a b+b^{2}$, hence also of $(a+b)^{2}-\\left(a^{2}-a b+b^{2}\\right)=3 a b$. This means that $p$ either is equal to $3$ or is a divisor of $a b$. Since $p$ is a divisor of $a+b$, we have $p|a \\Leftrightarrow p| b$, hence either $p=3$, or $p \\mid a$ and $p \\mid b$. If $p \\mid a$ and $p \\mid b$, then we can write $a=p a', b=p b'$ with $a'$ and $b'$ positive integers, and we have $\\left(a'\\right)^{3}+\\left(b'\\right)^{3}=p^{n-3}$, so $\\left(a', b', p, n-3\\right)$ then is another solution (note that $\\left(a'\\right)^{3}+\\left(b'\\right)^{3}$ is a positive integer greater than $1$, so $n-3$ is positive).\n\nNow assume that $\\left(a_{0}, b_{0}, p_{0}, n_{0}\\right)$ is a solution such that $p \\nmid a$. From the reasoning above it follows that either $a_{0}=b_{0}=1$, or $p_{0}=3$. After all, if we do not have $a_{0}=b_{0}=1$ and we have $p_{0} \\neq 3$, then $p \\mid a$. Also, given an arbitrary solution $(a, b, p, n)$, we can divide everything by $p$ repeatedly until there are no factors $p$ left in $a$.\n\nSuppose $a_{0}=b_{0}=1$. Then the solution is $(1,1,2,1)$.\n\nSuppose $p_{0}=3$. Assume that $3^{2} \\mid\\left(a_{0}^{2}-a_{0} b_{0}+b_{0}^{2}\\right)$. As $3^{2} \\mid\\left(a_{0}+b_{0}\\right)^{2}$, we then have $3^{2} \\mid\\left(a_{0}+b_{0}\\right)^{2}-\\left(a_{0}^{2}-a_{0} b_{0}+b_{0}^{2}\\right)=3 a_{0} b_{0}$, so $3 \\mid a_{0} b_{0}$. But $3 \\nmid a_{0}$ by assumption, and $3 \\mid a_{0}+b_{0}$, so $3 \\nmid b_{0}$, which contradicts $3 \\mid a_{0} b_{0}$. We conclude that $3^{2} \\nmid\\left(a_{0}^{2}-a_{0} b_{0}+b_{0}^{2}\\right)$. As both $a_{0}+b_{0}$ and $a_{0}^{2}-a_{0} b_{0}+b_{0}^{2}$ must be powers of $3$, we have $a_{0}^{2}-a_{0} b_{0}+b_{0}^{2}=3$. Hence $\\left(a_{0}-b_{0}\\right)^{2}+a_{0} b_{0}=3$. We must have $\\left(a_{0}-b_{0}\\right)^{2}=0$ or $\\left(a_{0}-b_{0}\\right)^{2}=1$. The former does not give a solution; the latter gives $a_{0}=2$ and $b_{0}=1$ or $a_{0}=1$ and $b_{0}=2$.\n\nSo all solutions with $p \\nmid a$ are $(1,1,2,1),(2,1,3,2)$ and $(1,2,3,2)$. From the above it follows that all other solutions are of the form $\\left(p_{0}^{k} a_{0}, p_{0}^{k} b_{0}, p_{0}, n_{0}+3 k\\right)$, where $\\left(a_{0}, b_{0}, p_{0}, n_{0}\\right)$ is one of these three solutions. Hence we find three families of solutions:\n- $\\left(2^{k}, 2^{k}, 2,3 k+1\\right)$ with $k \\in \\mathbb{Z}_{\\geq 0}$,\n- $\\left(2 \\cdot 3^{k}, 3^{k}, 3,3 k+2\\right)$ with $k \\in \\mathbb{Z}_{\\geq 0}$,\n- $\\left(3^{k}, 2 \\cdot 3^{k}, 3,3 k+2\\right)$ with $k \\in \\mathbb{Z}_{\\geq 0}$.\nIt is easy to check that all these quadruples are indeed solutions.\nSolution:\nLet $(a, b, p, n)$ be a solution. Note that we can write the given equation as\n$$\n(a+b)\\left(a^{2}-a b+b^{2}\\right)=p^{n}\n$$\nAs $a$ and $b$ are positive integers, we have $a+b \\geq 2$ and $a^{2}-a b+b^{2}=(a-b)^{2}+a b \\geq 1$. So both factors are positive and therefore must be powers of $p$. Let $k$ be an integer with $1 \\leq k \\leq n$ such that $a+b=p^{k}$. Then $a^{2}-a b+b^{2}=p^{n-k}$. If we substitute $b=p^{k}-a$, we find\n$$\np^{n-k}=(a+b)^{2}-3 a b=p^{2 k}-3 a\\left(p^{k}-a\\right) .\n$$\nWe can rewrite this as:\n$$\n3 a^{2}-3 p^{k} a+p^{2 k}-p^{n-k}=0\n$$\nfrom which we see that $a$ is a solution of the following quadratic equation in $x$ :\n$$\n3 x^{2}-3 p^{k} x+p^{2 k}-p^{n-k}=0 .\n$$\nThe discriminant of (15) is\n$$\nD=\\left(-3 p^{k}\\right)^{2}-4 \\cdot 3 \\cdot\\left(p^{2 k}-p^{n-k}\\right)=3 \\cdot\\left(4 p^{n-k}-p^{2 k}\\right)=3 p^{n-k} \\cdot\\left(4-p^{3 k-n}\\right)\n$$\nAs $p^{n-k}=(a+b)^{2}-3 a b<(a+b)^{2}=p^{2 k}$, we have $n-k<2 k$, so $3 k-n>0$. Since $a$ is a solution of (15), the discriminant must be nonnegative. Hence $4-p^{3 k-n} \\geq 0$. If $p=2$, this implies $3 k-n=1$ or $3 k-n=2$; if $p=3$, this implies $3 k-n=1$; and if $p>3$, then $p \\geq 5$ so $4 \\geq p^{3 k-n}$ can never be true.\n\nSuppose $p=2$ and $3 k-n=1$. Then $D=3 \\cdot 2^{2 k-1} \\cdot(4-2)=3 \\cdot 2^{2 k}$. But this is a not a square, so the solutions of (15) will not be integers, which yields a contradiction.\n\nSuppose $p=2$ and $3 k-n=2$. Then $D=3 \\cdot 2^{2 k-2} \\cdot(4-4)=0$, so the only solution of (15) is $x=\\frac{3 \\cdot 2^{k}}{2 \\cdot 3}=2^{k-1}$. Therefore $a=2^{k-1}$ and $b=2^{k}-a=2^{k-1}$, and this gives a solution for all $k \\geq 1$, namely $\\left(2^{k-1}, 2^{k-1}, 2,3 k-2\\right)$.\n\nSuppose $p=3$ and $3 k-n=1$. Then $D=3 \\cdot 3^{2 k-1} \\cdot(4-3)=3^{2 k}$, so the solutions of (15) are $x=\\frac{3^{k+1} \\pm 3^{k}}{2 \\cdot 3}=\\frac{1}{2}\\left(3^{k} \\pm 3^{k-1}\\right)$. Therefore $a=2 \\cdot 3^{k-1}$ or $a=3^{k-1}$. For all $k \\geq 1$ we find the solutions $\\left(2 \\cdot 3^{k-1}, 3^{k-1}, 3,3 k-1\\right)$ and $\\left(3^{k-1}, 2 \\cdot 3^{k-1}, 3,3 k-1\\right)$.\n\nWe conclude that there are three families of solutions:\n- $\\left(2^{k-1}, 2^{k-1}, 2,3 k-2\\right)$ with $k \\in \\mathbb{Z}_{\\geq 1}$,\n- $\\left(2 \\cdot 3^{k-1}, 3^{k-1}, 3,3 k-1\\right)$ with $k \\in \\mathbb{Z}_{\\geq 1}$,\n- $\\left(3^{k-1}, 2 \\cdot 3^{k-1}, 3,3 k-1\\right)$ with $k \\in \\mathbb{Z}_{\\geq 1}$.\nIt is easy to check that all these quadruples are indeed solutions.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25762,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAn ordered pair of integers $(m, n)$ with $1 < m < n$ is said to be a Benelux couple if the following two conditions hold: $m$ has the same prime divisors as $n$, and $m+1$ has the same prime divisors as $n+1$.\n\na. Find three Benelux couples $(m, n)$ with $m \\leqslant 14$.\n\nb. Prove that there exist infinitely many Benelux couples.",
    "options": [],
    "answer": "Examples: (2, 8), (6, 48), (14, 224). Infinite family: (2^k − 2, 2^k(2^k − 2)) for all integers k ≥ 2.",
    "solution": "Solution:\n\na.\nIt is possible to see that $(2, 8)$, $(6, 48)$ and $(14, 224)$ are Benelux couples.\n\nb.\nLet $k \\geqslant 2$ be an integer and $m = 2^{k} - 2$. Define $n = m(m+2) = 2^{k}(2^{k} - 2)$. Since $m$ is even, $m$ and $n$ have the same prime factors. Also, $n+1 = m(m+2) + 1 = (m+1)^{2}$, so $m+1$ and $n+1$ have the same prime factors, too. We have thus obtained a Benelux couple $(2^{k} - 2, 2^{k}(2^{k} - 2))$ for each $k \\geqslant 2$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25763,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $A B C$ be a triangle with incentre $I$. The angle bisectors $A I, B I$ and $C I$ meet $[B C],[C A]$ and $[A B]$ at $D, E$ and $F$, respectively. The perpendicular bisector of $[A D]$ intersects the lines $B I$ and $C I$ at $M$ and $N$, respectively. Show that $A, I, M$ and $N$ lie on a circle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nThe quadrilateral $A M D B$ is cyclic. Indeed, $M$ is the intersection of the line $B I$, which bisects the angle $\\widehat{A B D}$ in $A B D$ and the perpendicular bisector of $[A D]$. By uniqueness of this intersection point, it follows that $M$ lies on the circumcircle of $A B D$, and thence $A M D B$ is cyclic. Analogously, $A N D C$ is cyclic.\n\nMoreover, $M N B C$ is cyclic, for $\\widehat{N M I}=90^{\\circ}-\\widehat{E I A}$. Indeed, $A$ and $I$ lie on either side of the midpoint of $[A D]$, for $\\widehat{B D A}=\\widehat{C A D}+\\widehat{B C A}>\\widehat{B A D}$. But\n$$\n\\widehat{E I A}=180^{\\circ}-\\frac{1}{2} \\widehat{B A C}-\\widehat{B E A}=180^{\\circ}-\\frac{1}{2} \\widehat{B A C}-\\left(\\frac{1}{2} \\widehat{C B A}+\\widehat{B C A}\\right)=90^{\\circ}-\\frac{1}{2} \\widehat{B C A}\n$$\nIt follows that $\\widehat{N M I}=\\widehat{B C I}$, which implies that $M N B C$ is cyclic, as the points $I$ and $D$ lie on the same side of $M N$.\n\nThere are now two ways of completing the proof:\n\na.\n\nSolution 1 (using $A M D B$ and $M N B C$ )\n\nSince $A M D B$ is cyclic, $\\widehat{M A I}=\\widehat{M A D}=\\widehat{M B D}$, as, by construction, $B$ and $M$ lie on either side of $A D$. Moreover, $\\widehat{M B D}=\\widehat{M B C}=\\widehat{M N C}$ for $M N B C$ is cyclic. Thus $\\widehat{M A I}=\\widehat{M N I}$, so $A M I N$ is cyclic, for $M$ and $N$ lie on either side of $A D$.\n\n![](attached_image_1.png)\n\nb.\n\nSolution 2 (using $A M D B$ and $A N D C$ )\n\nSince $A M D B$ and $A N D C$ are cyclic, $\\widehat{A M I}+\\widehat{A N I}=\\widehat{A M B}+\\widehat{A N C}=\\widehat{A D B}+\\widehat{A D C}=180^{\\circ}$, because $B$ and $M$, and $C$ and $N$ lie on either side of $A D$. Hence $A M I N$ is cyclic, for $M$ and $N$ lie on either side of $A D$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25764,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nIf $k$ is an integer, let $\\mathrm{c}(k)$ denote the largest cube that is less than or equal to $k$. Find all positive integers $p$ for which the following sequence is bounded:\n$$\na_{0}=p \\quad \\text{ and } \\quad a_{n+1}=3 a_{n}-2 c\\left(a_{n}\\right) \\quad \\text{ for } n \\geqslant 0\n$$",
    "options": [],
    "answer": "All positive perfect cubes",
    "solution": "Solution:\nSince $\\mathrm{c}\\left(a_{n}\\right) \\leqslant a_{n}$ for all $n \\in \\mathbb{N}$, $a_{n+1} \\geqslant a_{n}$ with equality if and only if $\\mathrm{c}\\left(a_{n}\\right)=a_{n}$. Hence the sequence is bounded if and only if it is eventually constant, which is if and only if $a_{n}$ is a perfect cube, for some $n \\geqslant 0$. In particular, the sequence is bounded if $p$ is a perfect cube.\n\nWe now claim that, if $a_{n}$ is not a cube for some $n$, then neither is $a_{n+1}$. Indeed, if $a_{n}$ is not a cube, $q^{3}<a_{n}<(q+1)^{3}$ for some $q \\in \\mathbb{N}$, so that $\\mathrm{c}\\left(a_{n}\\right)=q^{3}$. Suppose to the contrary that $a_{n+1}$ is a cube. Then\n$$\na_{n+1}=3 a_{n}-2 \\mathrm{c}\\left(a_{n}\\right)<3(q+1)^{3}-2 q^{3}=q^{3}+9 q^{2}+9 q+3<q^{3}+9 q^{2}+27 q+27=(q+3)^{3}\n$$\nAlso, since $\\mathrm{c}\\left(a_{n}\\right)<a_{n}$, $a_{n+1}>a_{n}>q^{3}$, so $q^{3}<a_{n+1}<(q+3)^{3}$. It follows that the only possible values of $a_{n+1}$ are $(q+1)^{3}$ and $(q+2)^{3}$. However, in both of these cases,\n$$\n\\begin{aligned}\n& 3 a_{n}-2 q^{3}=a_{n+1}=(q+1)^{3} \\Longleftrightarrow 3 a_{n}=3\\left(q^{3}+q^{2}+q\\right)+1 \\\\\n& 3 a_{n}-2 q^{3}=a_{n+1}=(q+2)^{3} \\Longleftrightarrow 3 a_{n}=3\\left(q^{3}+2 q^{2}+4 q\\right)+8\n\\end{aligned}\n$$\na contradiction modulo 3. This proves that, if $a_{n}$ is not a cube, then neither is $a_{n+1}$. Hence, if $p$ is not a perfect cube, $a_{n}$ is not a cube for any $n \\in \\mathbb{N}$, and the sequence is not bounded. We conclude that the sequence is bounded if and only if $p$ is a perfect cube.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25765,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAbby and Brian play the following game: They first choose a positive integer $N$. Then they write numbers on a blackboard in turn. Abby starts by writing a $1$. Thereafter, when one of them has written the number $n$, the other writes down either $n+1$ or $2n$, provided that the number is not greater than $N$. The player who writes $N$ on the blackboard wins.\n\na. Determine which player has a winning strategy if $N=2011$.\n\nb. Find the number of positive integers $N \\leqslant 2011$ for which Brian has a winning strategy.",
    "options": [],
    "answer": "a) Abby; b) 31",
    "solution": "Solution:\n\na. Abby has a winning strategy for odd $N$: Observe that, whenever any player writes down an odd number, the other player has to write down an even number. By adding $1$ to that number, the first player can write down another odd number. Since Abby starts the game by writing down an odd number, she can force Brian to write down even numbers only. Since $N$ is odd, Abby will win the game. In particular, Abby has a winning strategy if $N=2011$.\n\nb. Let $N=4k$. If any player is forced to write down a number $m \\in \\{k+1, k+2, \\ldots, 2k\\}$, the other player wins the game by writing down $2m \\in \\{2k+2, 2k+4, \\ldots, 4k\\}$, for the players will have to write down the remaining numbers one after the other. Since there is an even number of numbers remaining, the latter player wins. This implies that the player who can write down $k$, i.e., has a winning strategy for $N=k$, wins the game for $N=4k$.\n\nSimilarly, let $N=4k+2$. If any player is forced to write down a number $m \\in \\{k+1, k+2, \\ldots, 2k+1\\}$, the other player wins the game by writing down $2m \\in \\{2k+2, 2k+4, \\ldots, 4k+2\\}$, as in the previous case. Analogously, this implies that the player who has a winning strategy for $N=k$ wins the game for $N=4k+2$.\n\nSince Abby wins the game for $N=1,3$, while Brian wins the game for $N=2$, Brian wins the game for $N=8,10$, as well, and thus for $N=32,34,40,42$, too. Then Brian wins the game for a further $8$ values of $N$ between $128$ and $170$, and thence for a further $16$ values between $512$ and $682$, and for no other values with $N \\leqslant 2011$. Hence Brian has a winning strategy for precisely $31$ values of $N$ with $N \\leqslant 2011$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25766,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA sequence $a_{1}, a_{2}, \\ldots, a_{n}, \\ldots$ of natural numbers is defined by the rule\n$$\na_{n+1}=a_{n}+b_{n} \\quad(n=1,2, \\ldots)\n$$\nwhere $b_{n}$ is the last digit of $a_{n}$. Prove that such a sequence contains infinitely many powers of 2 if and only if $a_{1}$ is not divisible by 5 .",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nFirst we can observe that:\n- If $a_{1}$ is divisible by $5$, then $a_{n}=a_{2}=0 \\pmod{10}$ for all $n \\geq 2$.\n- If $a_{1}$ is not divisible by $5$, then for $n \\geq 2$: $a_{n}$ is even, the sequence $b_{n}$ is periodic, its period is a cyclic permutation of $(2,4,8,6)$, and $a_{n+4}=a_{n}+20$.\n\na.\nLet us suppose that $a_{1}$ is divisible by $5$.\nSince $2^{k} \\neq 0 \\pmod{10}$ for any $k \\in \\mathbb{N}$, the sequence does not contain any power of $2$ for $n \\geq 2$.\n\nb.\nLet us suppose that $a_{1}$ is not divisible by $5$.\nWe can remark that the sequence of powers of $2$ modulo $20$ respects the period $(12,4,8,16)$ starting with $2^{5}=32$. We choose $j$ such that $a_{j}=2 \\pmod{10}$ (i.e. $b_{j}=2$) and look at the parity of its penultimate digit.\n- If $a_{j}=12 \\pmod{20}$, then the numbers $a_{j+4k}$, $k \\in \\mathbb{N}$, represent all the numbers congruent to $12 \\pmod{20}$ and greater than $a_{j}$, so all powers of $2$ congruent to $12 \\pmod{20}$ and greater than $a_{j}$ appear in the sequence.\n- If $a_{j}=2 \\pmod{20}$, then the numbers $a_{j+1+4k}$, $k \\in \\mathbb{N}$, represent all the numbers congruent to $4 \\pmod{20}$ and greater than $a_{j+1}$, so all powers of $2$ congruent to $4 \\pmod{20}$ and greater than $a_{j+1}$ appear in the sequence.\nThus, the sequence contains infinitely many powers of $2$.\n\nAlternative 1 for (b).\nWe choose $j$ such that $a_{j}=2 \\pmod{10}$ (i.e. $b_{j}=2$).\n- If $a_{j}=20t+12$ for some $t \\in \\mathbb{N}$, then $a_{j+4k}=a_{j}+20k=20(t+k)+12$, $\\forall k \\in \\mathbb{N}$. We obtain infinitely many powers of $2$ by taking $k=\\frac{2^{4s+3}-3}{5}-t$ (with $s \\in \\mathbb{N}$ large enough to have $k>0$) since $2^{4s+3}=3 \\pmod{5}$, $\\forall s \\in \\mathbb{N}$.\n- If $a_{j}=20t+2$ for some $t \\in \\mathbb{N}$, then $a_{j+1+4k}=a_{j+1}+20k=20(t+k)+4$, $\\forall k \\in \\mathbb{N}$. We obtain infinitely many powers of $2$ by taking $k=\\frac{2^{4s}-1}{5}-t$ (with $s \\in \\mathbb{N}$ large enough to have $k>0$) since $2^{4s}=1 \\pmod{5}$, $\\forall s \\in \\mathbb{N}$.\n\nAlternative 2 for (b).\nChoose $j$ such that $a_{j}$ is a multiple of $4$, i.e. $a_{j}=4q$ (such a $j$ always exists since $a_{n+1}=a_{n}+2$ for infinitely many $n$). Then we have $a_{j+4k}=a_{j}+20k=4(q+5k)$. Let us look for $(k, m)$ such that\n$$\na_{j+4k}=2^{m} \\Longleftrightarrow 4(q+5k)=2^{m} \\Longleftrightarrow q+5k=2^{m-2} \\Longleftrightarrow 2^{m-2}=q \\pmod{5}\n$$\nSince $q$ could not be a multiple of $5$, we have $q \\in \\{1,2,3,4\\} \\pmod{5}$. Since the sequence $2^{m-2} \\pmod{5}$ is periodic with period $(1,2,4,3)$, we find that $2^{m-2}=q \\pmod{5}$ happens for infinitely many values of $m$. Hence $2^{m-2}=q+5k$ is solvable for infinitely many pairs $(k, m)$. Noting that $m$ determines $k$ and that $k$ is nonnegative as soon as $m$ is large enough concludes the proof.\n\nAlternative 3 for (b).\nWe shall show that for any $n>1$ there is some $k \\geq n$ such that $a_{k}$ is a power of $2$. First, we observe that we can always find $m \\in \\{n, n+1, n+2, n+3\\}$ such that $a_{m}$ is divisible by $4$. If $a_{m}$ is not a power of $2$, we write $a_{m}=2^{b}c$ with $b \\geq 2$ and $c>1$ odd. Then we have\n$$\na_{m+4 \\cdot 2^{b-2}}=a_{m}+20\\left(2^{b-2}\\right)=2^{b}c+5 \\cdot 2^{b}=2^{b+1} \\frac{c+5}{2}.\n$$\nIf $c>5$, we have $\\frac{c+5}{2}<c$ and hence the odd factor of $a_{m+4 \\cdot 2^{b-2}}$ is strictly smaller than the odd factor of $a_{m}$. Therefore there is some $m'>m$ such that $a_{m'}=2^{b'}c'$ with $c'$ odd and $\\leq 5$. The case $c'=5$ is forbidden. If $c'=1$, then $a_{m'}$ is a power of $2$. If $c'=3$, then $a_{m'+4 \\cdot 2^{b'-2}}=2^{b'+3}$ is a power of $2$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25767,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all quadruples $(a, b, c, d)$ of positive real numbers such that $a b c d = 1$, $a^{2012} + 2012 b = 2012 c + d^{2012}$ and $2012 a + b^{2012} = c^{2012} + 2012 d$.",
    "options": [],
    "answer": "(t, 1/t, 1/t, t) for t > 0",
    "solution": "Solution:\nRewrite the last two equations into\n$$\na^{2012} - d^{2012} = 2012(c - b) \\text{ and } c^{2012} - b^{2012} = 2012(a - d)\n$$\nand observe that $a = d$ holds if and only if $c = b$ holds. In that case, the last two equations are satisfied, and condition $a b c d = 1$ leads to a set of valid quadruples of the form $(a, b, c, d) = \\left(t, \\frac{1}{t}, \\frac{1}{t}, t\\right)$ for any $t > 0$.\n\nWe show that there are no other solutions. Assume that $a \\neq d$ and $c \\neq b$. Multiply both sides of (1) to obtain\n$$\n\\left(a^{2012} - d^{2012}\\right)\\left(c^{2012} - b^{2012}\\right) = 2012^{2}(c - b)(a - d)\n$$\nand divide the left-hand side by the (nonzero) right-hand side to get\n$$\n\\frac{a^{2011} + \\cdots + a^{2011-i} d^{i} + \\cdots + d^{2011}}{2012} \\cdot \\frac{c^{2011} + \\cdots + c^{2011-i} b^{i} + \\cdots + b^{2011}}{2012} = 1\n$$\nNow apply the arithmetic-geometric mean inequality to the first factor\n$$\n\\frac{a^{2011} + \\cdots + a^{2011-i} d^{i} + \\cdots + d^{2011}}{2012} > \\sqrt[2012]{(a d)^{\\frac{2011 \\times 2012}{2}}} = (a d)^{\\frac{2011}{2}}\n$$\nThe inequality is strict, since equality holds only if all terms in the mean are equal to each other, which happens only if $a = d$. Similarly, we find\n$$\n\\frac{c^{2011} + \\cdots + c^{2011-i} b^{i} + \\cdots + b^{2011}}{2012} > \\sqrt[2012]{(c b)^{\\frac{2011 \\times 2012}{2}}} = (c b)^{\\frac{2011}{2}}.\n$$\nMultiplying both inequalities, we obtain\n$$\n(a d)^{\\frac{2011}{2}} (c b)^{\\frac{2011}{2}} < 1\n$$\nwhich is equivalent to $a b c d < 1$, a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25768,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn triangle $A B C$ the midpoint of $B C$ is called $M$. Let $P$ be a variable interior point of the triangle such that $\\angle C P M = \\angle P A B$. Let $\\Gamma$ be the circumcircle of triangle $A B P$. The line $M P$ intersects $\\Gamma$ a second time in $Q$. Define $R$ as the reflection of $P$ in the tangent to $\\Gamma$ in $B$. Prove that the length $|Q R|$ is independent of the position of $P$ inside the triangle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWe claim $|Q R| = |B C|$, which will clearly imply that quantity $|Q R|$ is independent from the position of $P$ inside triangle $\\triangle A B C$ (and independent from the position of $A$).\nThis equality will follow from the equality between triangles $\\triangle B P C$ and $\\triangle R B Q$. This in turn will be shown by means of three equalities (two sides and an angle): $|B P| = |R B|$, $|P C| = |B Q|$ and $\\angle B P C = \\angle R B Q$.\n\n![](attached_image_1.png)\n\na. $|B P| = |R B|$\nObvious since $R$ is the reflection of $P$ in a line going through $B$.\n\nb. $|P C| = |B Q|$\nLet $U$ be the fourth vertex of parallelogram $B P C U$. Then $U$ is on line $P Q$ and $\\angle B U P = \\angle U P C = \\alpha$. If $Q$ is on the same $\\operatorname{arc} P B$ as $A$, then $\\angle B Q P = \\alpha$, and $\\triangle Q P U$ is isosceles; hence, $|B Q| = |B U| = |P C|$. On the other way, if $Q$ is on the other $\\operatorname{arc} P B$, then $\\angle B Q P$ and $\\alpha$ are supplementary, hence $B Q U = \\alpha$, and again $\\triangle Q P U$ is isosceles; the same conclusion follows.\n\nc. $\\angle B P C = \\angle R B Q$\nDefine $T$ to be the midpoint of $P R$. Then line $B T$, tangent to circle $\\Gamma$ in $B$, splits $\\angle R B Q$ into two parts, $\\angle R B T$ and $\\angle T B Q$.\nWe first show that $\\angle R B T = \\alpha$. Indeed, by symmetry, $\\angle R B T = \\angle P B T$ and, since $B T$ is tangent to $\\Gamma$, we have that $\\angle P B T = \\angle P A B$ (because they both intercept the same $\\operatorname{arc} \\widehat{P B}$ on circle $\\Gamma$), from which our claim follows.\nWe then show that $\\angle T B Q = \\angle B P M$. Indeed, since $\\angle T B Q$ and $\\angle B P Q$ intercept opposite arcs on circle $\\Gamma$, they are supplementary and we have $\\angle T B Q = \\pi - \\angle B P Q = \\angle B P M$. We finally conclude that\n$$\n\\angle R B Q = \\angle R B T + \\angle T B Q = \\alpha + \\angle B P M = \\angle M P C + \\angle B P M = \\angle B P C\n$$\nWe have thus shown $\\triangle B P C = \\triangle R B Q$, which completes the proof.\n\nAlternative 1 for (b). The law of sines in triangle $\\triangle B Q M$ gives\n$$\n\\frac{|B M|}{\\sin \\angle B Q M} = \\frac{|B Q|}{\\sin \\angle B M Q}.\n$$\nSince $Q$ belongs to circle $\\Gamma$, we have either $\\angle B Q P = \\angle B A P = \\alpha$, hence $\\angle B Q M = \\angle M P C$, or these angles are supplementary; in both cases they have equal sines. We also have that $\\angle B M Q$ and $\\angle C M P$ are supplementary, hence have equal sines. Using these facts along with $|B M| = |M C|$ transforms (2) into\n$$\n\\frac{|M C|}{\\sin \\angle M P C} = \\frac{|B Q|}{\\sin \\angle C M P}\n$$\nfrom which the law of sines in triangle $\\triangle C P M$ implies that $|B Q| = |P C|$.\n\nAlternative 2 for (b). Let $S$ be the second intersection of line $C P$ with circle $\\Gamma$. Then, $\\angle B S P = \\alpha$, so $B S$ and $M P$ are parallel; since $M$ is the midpoint of segment $B C$, $P$ is the midpoint of $S C$. If $Q$ is on the same $\\operatorname{arc} P B$ as $A$, then the quadrilateral $Q P B S$ is an isosceles trapezoid, and $|Q B| = |S P| = |P C|$. If $Q$ is on the other arc $P B$, then the quadrilateral $P Q B S$ is an isosceles trapezoid, and again $|Q B| = |S P| = |P C|$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25769,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nYesterday, $n \\geq 4$ people sat around a round table. Each participant remembers only who his two neighbours were, but not which one sat on his left and which one sat on his right. Today, you would like the same people to sit around the same round table so that each participant has the same two neighbours as yesterday (it is possible that yesterday's lefthand side neighbour is today's right-hand side neighbour). You are allowed to query some of the participants: if anyone is asked, he will answer by pointing at his two neighbours from yesterday.\n\na. Determine the minimal number $f(n)$ of participants you have to query in order to be certain to succeed, if later questions must not depend on the outcome of the previous questions. That is, you have to choose in advance the list of people you are going to query, before effectively asking any question.\n\nb. Determine the minimal number $g(n)$ of participants you have to query in order to be certain to succeed, if later questions may depend on the outcome of previous questions. That is, you can wait until you get the first answer to choose whom to ask the second question, and so on.",
    "options": [],
    "answer": "f(n) = n - 3; g(n) = n - 1 - ceil(n/3)",
    "solution": "Solution:\n\na. $f(n) = n-3$.\n\n- Asking $n-4$ questions is not enough since the $n-4$ people queried might be sitting in a consecutive string, in which case the $n-4$ answers allow one to sit $n-2$ people in the same positions as yesterday, but there is still an ambiguity among the two remaining ones.\n\n- Let us show that $n-3$ questions suffice. Among the 3 people who are not queried, at least 2 must sit next to people who have been queried. If exactly 2 do, then both these people must be neighbours of the third, so that the neighbours of everybody are known and we are done. If all 3 unqueried people sit next to a queried person, then at least one of them has two queried neighbours, and again it follows that the neighbours of everybody are known, so that we are done.\n\nb. $g(n) = n-1-\\left\\lceil\\frac{n}{3}\\right\\rceil \\left(= n-1-\\left\\lfloor\\frac{n+2}{3}\\right\\rfloor = \\left\\lfloor\\frac{2n}{3}\\right\\rfloor-1 = \\left\\lceil\\frac{2n-5}{3}\\right\\rceil\\right)$.\n\nSay there is a link between two people if and only if they are neighbours. There are in total $n$ links, which we all need to identify. By asking a person for his neighbours, we can discover at most two new links. More precisely, if at any point we query a participant who has not yet been pointed as a neighbour, we discover exactly two new links (we call this a type-0 query). If we query a participant who has been pointed once as a neighbour, we will discover exactly one new link (we call this a type-1 query). Of course, querying a participant who has already been pointed twice provides no information (and we assume in the rest of this solution that it never happens).\n\nFirst note that, since $f(4) = 1$, we also have $g(4) = 1$. We now prove the formula for $g(n)$ for $n \\geq 5$.\n\n- Let us show that $n-1-\\left\\lceil\\frac{n}{3}\\right\\rceil$ questions suffice. Our strategy consists in making sure that the first $\\left\\lceil\\frac{n}{3}\\right\\rceil$ queries are type-0. Let us show that this is always possible. A type-0 query requires a participant that hasn't been queried or pointed before. Since the number of those participants decreases by three at most after each query, we see that it is always possible to perform $\\left\\lceil\\frac{n}{3}\\right\\rceil$ type-0 queries first. During this phase we discover $2\\left\\lceil\\frac{n}{3}\\right\\rceil$ links.\n\nThe remaining queries will be either type-0 or type-1, and each of them discovers at least one new link. We perform them until $n-1$ links have been discovered, after which we are done (the last link can be deduced without query). The number of queries in this second phase is therefore at most $n-1-2\\left\\lceil\\frac{n}{3}\\right\\rceil$, and the total is at most $\\left\\lceil\\frac{n}{3}\\right\\rceil + \\left(n-1-2\\left\\lceil\\frac{n}{3}\\right\\rceil\\right) = n-1-\\left\\lceil\\frac{n}{3}\\right\\rceil$.\n\n- We now show that $n-2-\\left\\lceil\\frac{n}{3}\\right\\rceil = \\hat{g}(n)$ questions are not enough.\n\n(i) Consider the pool of unqueried and unpointed participants; each type-0 must query this pool. Since, from the point of view of the questioner, all elements of the pool are undistinguishable, we can assume that each type-0 query asks the second leftmost participant in the pool (except if there is only one element left in the pool). One can then check that the pool, which starts as a string of $n$ contiguous participants, will stay contiguous after each type-0 and type-1 query. Furthermore, using our assumption, we see that each type-0 query removes three participants from the pool. Therefore there can be at most $\\left\\lceil\\frac{n}{3}\\right\\rceil$ type-0 queries in the scenarios corresponding to our assumption.\n\n(ii) Assume there are $k$ type-0 queries. Since there are $\\hat{g}(n)$ queries, the number of discovered links is equal to $2k + (\\hat{g}(n) - k) = \\hat{g}(n) + k = n-2 + k - \\left\\lceil\\frac{n}{3}\\right\\rceil$. If $k$ is strictly less than $\\left\\lceil\\frac{n}{3}\\right\\rceil$, we discover strictly less than $n-2$ links, which is clearly insufficient (indeed, there are at least three missing links, and one can check that whatever the configuration of the missing links, there are always several orders compatible with the discovered links).\n\n(iii) We now analyze the remaining case with $k = \\left\\lceil\\frac{n}{3}\\right\\rceil$ type-0 queries, in which we discover $n-2$ links. On the one hand, if the missing links are disjoint, there are always two orders compatible with the discovered links (for example when $n=7$ and links are missing between the $(4,5)$ and $(7,1)$ pairs of neighbours, the two orders are $1-2-3-4\\ 5-6-7$ and $1-2-3-4\\ 7-6-5$). On the other hand, a situation where the two missing links would be adjacent would allow the identification of the correct order. However, this never happens in the scenarios corresponding to the assumption we made in (i). Indeed, two adjacent missing links imply that some participant is unqueried and unpointed at the end of the process. Since we perform $k = \\left\\lceil\\frac{n}{3}\\right\\rceil$ type-0 queries (the maximum), the reasoning from (i) shows that the pool of unqueried and unpointed participants is empty at the end of the process, which contradicts the existence of two adjacent missing links.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25770,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $n \\geqslant 3$ be an integer. A frog is to jump along the real axis, starting at the point $0$ and making $n$ jumps: one of length $1$, one of length $2, \\ldots$, one of length $n$. It may perform these $n$ jumps in any order. If at some point the frog is sitting on a number $a \\leqslant 0$, its next jump must be to the right (towards the positive numbers). If at some point the frog is sitting on a number $a>0$, its next jump must be to the left (towards the negative numbers). Find the largest positive integer $k$ for which the frog can perform its jumps in such an order that it never lands on any of the numbers $1,2, \\ldots, k$.",
    "options": [],
    "answer": "floor((n-1)/2)",
    "solution": "Solution:\nWe claim that the largest positive integer $k$ with the given property is $\\left\\lfloor\\frac{n-1}{2}\\right\\rfloor$, where $\\lfloor x\\rfloor$ is by definition the largest integer not exceeding $x$.\n\nConsider a sequence of $n$ jumps of length $1,2, \\ldots, n$ such that the frog never lands on any of the numbers $1,2, \\ldots, k$, where $k \\geqslant 1$. Note that we must have $k<n$ in order for the frog to be able to make its first jump. As the frog jumps to the right only if it is in a number $a \\leqslant 0$, and the largest jump has length $n$, it is impossible to reach numbers greater than $n$. On the other hand, suppose the frog is in a number $a>0$, then it must even be in a number $a \\geqslant k+1$, since it is not allowed to hit the numbers $1,2, \\ldots, k$. So the frog jumps to the left only if it is in a number $a \\geqslant k+1$, and therefore it is impossible to reach numbers less than $(k+1)-n = k-n+1$. This means the frog only possibly lands on the numbers $i$ satisfying\n$$\nk-n+1 \\leqslant i \\leqslant 0 \\quad \\text{or} \\quad k+1 \\leqslant i \\leqslant n.\n$$\nWhen performing a jump of length $k$, the frog has to remain at either side of the numbers $1,2, \\ldots, k$. Indeed, jumping over $1,2, \\ldots, k$ requires a jump of at least length $k+1$. In case it starts at a number $a>0$ (in fact $k+1 \\leqslant a \\leqslant n$), it lands in $a-k$ and we must also have $a-k \\geqslant k+1$. So $2k+1 \\leqslant a \\leqslant n$, therefore $2k+1 \\leqslant n$. In case it starts at a number $a \\leqslant 0$ (in fact $k-n+1 \\leqslant a \\leqslant 0$), it lands in $a+k$ and we must also have $a+k \\leqslant 0$. Adding $k$ to both sides of $k-n+1 \\leqslant a$, we obtain $2k-n+1 \\leqslant a+k \\leqslant 0$, so in this case we have $2k+1 \\leqslant n$ as well. We conclude that $k \\leqslant \\frac{n-1}{2}$. Since $k$ is integer, we even have $k \\leqslant \\left\\lfloor\\frac{n-1}{2}\\right\\rfloor$.\n\nNext we prove that this upper bound is sharp: for $k=\\left\\lfloor\\frac{n-1}{2}\\right\\rfloor$ the frog really can perform its jumps in such an order that it never lands on any of the numbers $1,2, \\ldots, k$.\n\nSuppose $n$ is odd, then $\\frac{n-1}{2}$ is an integer and we have $k=\\frac{n-1}{2}$, so $n=2k+1$. We claim that when the frog performs the jumps of length $1, \\ldots, 2k+1$ in the following order, it does never land on $1,2, \\ldots, k$: it starts with a jump of length $k+1$, then it performs two jumps, one of length $k+2$ followed by one of length $1$, next two jumps of length $k+3$ and $2$, $\\ldots$, next two jumps of length $k+(i+1)$ and $i$, $\\ldots$, and finally two jumps of length $k+(k+1)$ and $k$. In this order of the jumps every length between $1$ and $n=2k+1$ does occur: it performs a pair of jumps for $1 \\leqslant i \\leqslant k$, which are the jumps of length $1,2, \\ldots, k$ and the jumps of length $k+2, k+3, \\ldots, 2k+1$, and it starts with the jump of length $k+1$.\n\nWe now prove the correctness of this jumping scheme. After the first jump the frog lands in $k+1>k$. Now suppose the frog is in $0$ or $k+1$ and is about to perform the pair of jumps of length $k+(i+1)$ and $i$. Starting from $0$, it lands in $k+(i+1)>k$, after which it lands in $(k+i+1)-i=k+1>k$. If on the contrary it starts in $k+1$, it lands in $(k+1)-(k+(i+1))=-i<1$, after which it lands in $(-i)+i=0$. We see that, starting in $0$, the frog lands in $k+1$ after the pair of jumps, while starting in $k+1$ the frog lands in $0$, while in both cases the jumps do not touch $1,2, \\ldots, k$. This proves the correctness of its series of jumps. As the frog (after its first jump) alternates between $k+1$ and $0$ exactly $k$ times, for odd $k$ it will end up in $0$, while for even $k$ it will end up in $k+1$.\n\nSuppose $n$ is even, then $\\frac{n-1}{2}$ is not an integer and we have $k=\\frac{n-1}{2}-\\frac{1}{2}=\\frac{n-2}{2}$, so $n=2k+2$. Let the frog firstly perform the same series of jumps as in the previous case; they still do not touch $1,2, \\ldots, k$. Now let the frog make a final extra jump of length $2k+2$. It will land in $0+(2k+2)=2k+2>k$ if $k$ is odd, or in $(k+1)-(2k+2)=-k-1<1$ if $k$ is even, and its series of jumps is correct again.\n\nWe conclude that the largest positive integer $k$ with the given property is $\\left\\lfloor\\frac{n-1}{2}\\right\\rfloor$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25771,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $\\triangle ABC$ be a triangle with circumcircle $\\Gamma$, and let $I$ be the center of the incircle of $\\triangle ABC$. The lines $AI$, $BI$ and $CI$ intersect $\\Gamma$ in $D \\neq A$, $E \\neq B$ and $F \\neq C$. The tangent lines to $\\Gamma$ in $F$, $D$ and $E$ intersect the lines $AI$, $BI$ and $CI$ in $R$, $S$ and $T$, respectively. Prove that\n$$\n|AR| \\cdot |BS| \\cdot |CT| = |ID| \\cdot |IE| \\cdot |IF|\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nWe first prove that $|DB| = |DI|$. (This may also be claimed by referring to the lemma that $D$ is the centre of the circumcircle of $BIC I_{a}$.) By the constant angle theorem and the fact that $AD$ and $BE$ are angle bisectors of triangle $ABC$, we see that\n$$\n\\angle DBI = \\angle DBC + \\angle CBI = \\angle DAC + \\angle CBE = \\angle DAB + \\angle ABE,\n$$\nwhile\n$$\n\\angle DIB = 180^\\circ - \\angle AIB = \\angle IAB + \\angle ABI = \\angle DAB + \\angle ABE.\n$$\nSo $\\triangle BDI$ has equal angles $\\angle DBI = \\angle DIB$, so $|DB| = |DI|$. This proves our claim. We similarly deduce that $|EC| = |EI|$ and $|FA| = |FI|$.\n\nRewriting (7) into $\\frac{|AR|}{|IF|} \\cdot \\frac{|BS|}{|ID|} \\cdot \\frac{|CT|}{|IE|} = 1$, we see that it suffices to prove that\n$$\n\\frac{|AR|}{|AF|} \\cdot \\frac{|BS|}{|BD|} \\cdot \\frac{|CT|}{|CE|} = 1\n$$\nWe now prove by angle chasing that $\\triangle RFA \\sim \\triangle ACI$. As $RF$ is tangent to the circumcircle of $\\triangle AFC$, we obtain (using also that $CF$ is angle bisector of $\\angle ACB$)\n$$\n\\angle RFA = \\angle FCA = \\angle ICA.\n$$\nMoreover, from $|FA| = |FI|$ we deduce that $\\angle FAI = \\angle FIA$, so\n$$\n\\angle FAR = 180^\\circ - \\angle FAI = 180^\\circ - \\angle FIA = \\angle CIA\n$$\nThis proves our similarity, which entails that $\\frac{|AR|}{|AF|} = \\frac{|IA|}{|IC|}$. In the same way we deduce that $\\frac{|BS|}{|BD|} = \\frac{|IB|}{|IA|}$ and $\\frac{|CT|}{|CE|} = \\frac{|IC|}{|IB|}$. By these equal ratios we know that\n$$\n\\frac{|AR|}{|AF|} \\cdot \\frac{|BS|}{|BD|} \\cdot \\frac{|CT|}{|CE|} = \\frac{|IA|}{|IC|} \\cdot \\frac{|IB|}{|IA|} \\cdot \\frac{|IC|}{|IB|} = 1\n$$\nwhich proves (8), as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25772,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\na) Find all positive integers $g$ with the following property: for each odd prime number $p$ there exists a positive integer $n$ such that $p$ divides the two integers\n$$\ng^{n}-n \\quad \\text{ and } \\quad g^{n+1}-(n+1)\n$$\n\nb) Find all positive integers $g$ with the following property: for each odd prime number $p$ there exists a positive integer $n$ such that $p$ divides the two integers\n$$\ng^{n}-n^{2} \\quad \\text{ and } \\quad g^{n+1}-(n+1)^{2}\n$$",
    "options": [],
    "answer": "a) g = 2; b) g = 4",
    "solution": "Solution:\n\na.\nLet $g$ be a positive integer with the given property. So for each odd prime number $p$ there exists a positive integer $n$ such that $p \\mid g^{n}-n$ and $p \\mid g^{n+1}-(n+1)$.\nIf $g$ has an odd prime factor $p$, then from $p \\mid g^{n}-n$ it follows that $p \\mid n$, while from $p \\mid g^{n+1}-(n+1)$ we deduce that $p \\mid n+1$. But $p$ cannot divide both $n$ and $n+1$; contradiction. So $g$ is a power of $2$: $g=2^{k}$ for some $k \\geqslant 0$.\nIf $g=2^{0}=1$, then $p \\mid 1-n$ and $p \\mid 1-(n+1)$, which is again a contradiction.\nSuppose $k \\geqslant 2$. Then $g-1$ has an odd prime factor $p$, therefore $g \\equiv 1\\pmod{p}$ so $0 \\equiv g^{n}-n \\equiv 1-n\\pmod{p}$ and $0 \\equiv g^{n+1}-(n+1) \\equiv 1-(n+1)\\pmod{p}$, which is again a contradiction.\nNow we prove that $g=2^{1}=2$ does satisfy the condition. Let a prime $p>2$ be given. Choose $n=(p-1)^{2}$, then we have $n \\equiv(-1)^{2}=1\\pmod{p}$. By Fermat's little theorem (using $\\gcd(2, p)=1$) we know that $2^{p-1} \\equiv 1\\pmod{p}$, so\n$$\n2^{n}=2^{(p-1)^{2}}=\\left(2^{p-1}\\right)^{p-1} \\equiv 1 \\equiv n \\quad(p) \n$$\nMultiplying both sides by $2$, we see that also\n$$\n2^{n+1} \\equiv 2 n=n+n \\equiv n+1 \\quad(p)\n$$\nWe conclude that only $g=2$ has the given property.\n\nb.\nLet $g$ be a positive integer with the given property. So for each odd prime number $p$ there exists a positive integer $n$ such that $p \\mid g^{n}-n^{2}$ and $p \\mid g^{n+1}-(n+1)^{2}$.\nIf $g$ has an odd prime factor $p$, then from $p \\mid g^{n}-n^{2}$ it follows that $p \\mid n^{2}$, so also $p \\mid n$, while from $p \\mid g^{n+1}-(n+1)^{2}$ we deduce that $p \\mid(n+1)^{2}$, so also $p \\mid n+1$.\nBut $p$ cannot divide both $n$ and $n+1$; contradiction. So $g$ is a power of $2$: $g=2^{k}$ for some $k \\geqslant 0$.\nIf $g=2^{0}=1$, then for any odd prime $p$ we have $p \\mid 1-n^{2}=(1-n)(1+n)$ and $p \\mid 1-(n+1)^{2}=(1-(n+1))(1+(n+1))$. Now take $p=5$. The first statement says that $n \\equiv 1$ or $n \\equiv-1 \\equiv 4\\pmod{5}$, and the second that $n \\equiv 0$ or $n \\equiv-2 \\equiv 3$ $\\pmod{5}$. But this yields a contradiction.\nIf $g=2^{1}=2$, then for any odd prime $p$ we have $p \\mid 2^{n}-n^{2}$ and $p \\mid 2^{n+1}-(n+1)^{2}$. Now take $p=3$. As $3 \\nmid 2^{n}$ and $3 \\nmid 2^{n+1}$, we know that $3 \\nmid n^{2}$ and $3 \\nmid(n+1)^{2}$. So these two squares must be $1$ modulo $3$ (as $2$ can never be a square modulo $3$). Therefore also $2^{n}$ and $2^{n+1}$ must be $1$ modulo $3$, which gives $2 \\cdot 1 \\equiv 2 \\cdot 2^{n}=2^{n+1} \\equiv 1\\pmod{3}$; contradiction.\nNow suppose $k \\geqslant 2$. Then $g-1$ has an odd prime factor $p$, therefore $g \\equiv 1\\pmod{p}$ so $0 \\equiv g^{n}-n^{2} \\equiv 1-n^{2}=(1-n)(1+n)\\pmod{p}$ and $0 \\equiv g^{n+1}-(n+1)^{2} \\equiv 1-(n+1)^{2}=(1-(n+1))(1+(n+1))\\pmod{p}$. Suppose $p \\geqslant 5$. The first statement says that $n \\equiv 1$ or $n \\equiv-1\\pmod{p}$, and the second that $n \\equiv 0$ or $n \\equiv-2\\pmod{p}$. But $n$ can only be congruent to at most one of the numbers $-2,-1,0$ and $1$, since $p \\geqslant 5$; contradiction. We conclude that $p=3$, so $g-1$ contains only prime factors $3$. Hence $2^{k}-1=3^{\\ell}$ for some $\\ell>0$. We see that $2^{k}-1 \\equiv(-1)^{k}-1\\pmod{3}$, while $3^{\\ell} \\equiv 0\\pmod{3}$. So $k$ has to be even, say $k=2 m$, and our equation becomes $2^{2 m}-1=3^{\\ell}$, or equivalently $\\left(2^{m}-1\\right)\\left(2^{m}+1\\right)=3^{\\ell}$. Not both factors on the left-hand side can be divisible by $3$, so $2^{m}-1=1$ and $2^{m}+1=3^{\\ell}$, so $m=1$. Hence $g=2^{2}=4$.\nNow we show that $g=4$ does have the given property. For this we use that $g=2$ is a solution to part (a): for any odd prime $p$ there exists a positive integer $n$ such that\n$$\nn \\equiv 2^{n} \\quad\\pmod{p} \\quad \\text{ and } \\quad n+1 \\equiv 2^{n+1} \\quad\\pmod{p} .\n$$\nTaking the square of both congruences, we obtain\n$$\nn^{2} \\equiv\\left(2^{n}\\right)^{2}=\\left(2^{2}\\right)^{n}=4^{n} \\quad\\pmod{p}\n$$\nand\n$$\n(n+1)^{2} \\equiv\\left(2^{n+1}\\right)^{2}=\\left(2^{2}\\right)^{n+1}=4^{n+1} \\quad\\pmod{p}\n$$\nas desired.\nWe conclude that only $g=4$ has the given property.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25773,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the smallest possible value of the expression\n$$\n\\left\\lfloor\\frac{a+b+c}{d}\\right\\rfloor+\\left\\lfloor\\frac{b+c+d}{a}\\right\\rfloor+\\left\\lfloor\\frac{c+d+a}{b}\\right\\rfloor+\\left\\lfloor\\frac{d+a+b}{c}\\right\\rfloor,\n$$\nin which $a$, $b$, $c$ and $d$ vary over the set of positive integers.\n(Here $\\lfloor x\\rfloor$ denotes the biggest integer which is smaller than or equal to $x$.)",
    "options": [],
    "answer": "9",
    "solution": "Solution:\nThe answer is $9$.\n\nNotice that $\\lfloor x\\rfloor > x-1$ for all $x \\in \\mathbb{R}$. Therefore the given expression is strictly greater than\n$$\n\\frac{a+b+c}{d}+\\frac{b+c+d}{a}+\\frac{c+d+a}{b}+\\frac{d+a+b}{c}-4,\n$$\nwhich can be rewritten as\n$$\n\\left(\\frac{a}{b}+\\frac{b}{a}\\right)+\\left(\\frac{a}{c}+\\frac{c}{a}\\right)+\\left(\\frac{a}{d}+\\frac{d}{a}\\right)+\\left(\\frac{b}{c}+\\frac{c}{b}\\right)+\\left(\\frac{b}{d}+\\frac{d}{b}\\right)+\\left(\\frac{c}{d}+\\frac{d}{c}\\right)-4.\n$$\nSince $t+\\frac{1}{t} \\geq 2$ for $t>0$, we get that $6 \\cdot 2-4=8$ is a strict lower bound for the given expression; since it takes integral values only, we actually get that $9$ is a lower bound. It remains to check that $9$ can be attained; this happens for $a=b=c=5$ and $d=4$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 25774,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $k \\geq 1$ be an integer.\nWe consider $4k$ chips, $2k$ of which are red and $2k$ of which are blue. A sequence of those $4k$ chips can be transformed into another sequence by a so-called move, consisting of interchanging a number (possibly one) of consecutive red chips with an equal number of consecutive blue chips. For example, we can move from $r \\underline{b b b r} \\underline{r} \\underline{b} b$ to $r \\underline{r} \\underline{b} b \\underline{b b b} b$ where $r$ denotes a red chip and $b$ denotes a blue chip.\nDetermine the smallest number $n$ (as a function of $k$) such that starting from any initial sequence of the $4k$ chips, we need at most $n$ moves to reach the state in which the first $2k$ chips are red.",
    "options": [],
    "answer": "k",
    "solution": "Solution:\nThe answer is $n = k$.\n\nWe will first show that $n \\geq k$. Let us count the number $C$ of times a red chip is directly to the right of a blue chip. In the final position this number equals $0$. In the position $b r b r b r \\cdots b r$ this number equals $2k$. We claim that any move reduces this number by at most $2$. Denote by $R$ the group of red chips and by $B$ the group of blue chips that are interchanged. Any reduction in $C$ must involve a red chip getting rid of its blue left neighbour or a blue chip getting rid of its red right neighbour. This can only happen with the leftmost chip of $R$ (if its left neighbour is blue) and the rightmost chip of $B$ (if its right neighbour is red), but not with the rightmost chip of $R$ (and its right neighbour) or the leftmost chip of $B$ (and its left neighbour). Hence $C$ is reduced by at most $2$ in any move. Therefore the number of moves necessary to change $b r b r b r \\cdots b r$ into the final position is at least $\\frac{2k}{2} = k$.\n\nWe will now show that $n \\leq k$, i.e. that it is always possible to perform at most $k$ moves in order to reach the state in which the first $2k$ chips are red. Consider the first $2k$ chips. If at most $k$ of these chips are blue, we can perform one move for each chip, switching it with one of the red chips of the last $2k$ chips. So then we are done in at most $k$ moves. Now suppose that of the first $2k$ chips, at least $k+1$ are blue. Then at most $k-1$ chips are red. Hence it is possible to perform at most $k-1$ moves to reach the situation in which the last $2k$ chips are red. We then perform one final move, switching the first $2k$ chips and the last $2k$ chips, ending in the situation in which the first $2k$ chips are red. Thus, it is always possible to reach the situation in which the first $2k$ chips are red in at most $k$ steps, hence $n \\leq k$.\n\nWe have now shown that $n \\geq k$ and $n \\leq k$, hence $n = k$ as claimed.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25775,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all positive integers $n>1$ with the following property:\nfor each two positive divisors $k, \\ell<n$ of $n$, at least one of the numbers $2 k-\\ell$ and $2 \\ell-k$ is a (not necessarily positive) divisor of $n$ as well.",
    "options": [],
    "answer": "All prime numbers, and 6, 9, 15",
    "solution": "Solution:\nIf $n$ is prime, then $n$ has the desired property: if $k, \\ell<n$ are positive divisors of a prime $n$, we have $k=\\ell=1$, in which case $2 k-\\ell=1$ is a divisor of $n$ as well.\n\nAssume now that a composite number $n$ has the desired property. Let $p$ be its smallest prime divisor and let $m=n / p$; then $m \\geq p \\geq 2$. Choosing $(k, \\ell)=(1, m)$, we see that at least one of $2 m-1$ and $m-2$ must divide $n$. However, $m<2 m-1<2 m \\leq n$; since $m$ and $n$ are the two biggest positive divisors of $n$, we conclude that $2 m-1$ cannot divide $n$. Therefore $n$ must be divisble by $m-2$. It follows that $m-2 \\mid m p-p(m-2)=2 p$, hence $m-2 \\in\\{1,2, p, 2 p\\}$.\nWe deal with each case separately.\n- If $m-2=1$ we have $m=3$. As $p \\leq m$, we have $p \\in\\{2,3\\}$, hence $n=6$ or $n=9$.\n- If $m-2=2$ we have $m=4$, hence $n$ is even and $p=2$, so $n=8$.\n- If $m-2=p$, we may assume that $p>2$ (since we already discussed the case $m-2=2$ ). We have $m=p+2$, hence $n=p(p+2)$. Applying the condition in the problem to the pair $(k, \\ell)=(1, p)$, we get that $p-2$ or $2 p-1$ divides $n$. If $p-2$ divides $n$, we must have $p=3$ since $p$ is the smallest prime divisor of $n$. If $2 p-1$ divides $n$, we must have $p+2=2 p-1$, since $p+2 \\leq 2 p-1<p(p+2)$ and $p+2$ and $p(p+2)$ are the biggest positive divisors of $n$. Therefore $p=3$ in both cases, and we get $n=15$ as a candidate.\n- If finally $m-2=2 p$ we have $m=2 p+2$, so $n$ is even. Hence $p=2$ and $n=12$.\nWe conclude that if a composite number $n$ satisfies the condition in the problem statement, then $n \\in\\{6,8,9,12,15\\}$. Choosing $(k, \\ell)=(1,2)$ shows that $n=8$ does not work; choosing $(k, \\ell)=(3,6)$ shows that 12 is not a solution. It is easy to check that $n=6, n=9$ and $n=15$ have the desired property by checking the condition for all possible pairs $(k, \\ell)$.\nWe conclude that the solutions are given by the prime numbers, $n=6, n=9$ and $n=15$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25776,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABCD$ be a square. Consider a variable point $P$ inside the square for which $\\angle BAP \\geq 60^{\\circ}$. Let $Q$ be the intersection of the line $AD$ and the perpendicular to $BP$ in $P$. Let $R$ be the intersection of the line $BQ$ and the perpendicular to $BP$ from $C$.\n\na. Prove that $|BP| \\geq |BR|$.\n\nb. For which point(s) $P$ does the inequality in (a) become an equality?",
    "options": [],
    "answer": "Equality holds exactly when the interior point makes the triangle at that corner equilateral, that is, when the angle at the corner is sixty degrees and the adjacent side equals the segment to the point; there is a unique such point inside the square.",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nWe claim that $\\triangle ABP$ and $\\triangle RCB$ are similar triangles. Indeed, if we denote the intersection of $BP$ and $CR$ by $S$, then $\\angle RCB = \\angle SCB = 90^{\\circ} - \\angle SBC = 90^{\\circ} - \\angle PBC = \\angle ABP$. Moreover, the right angles in $P$ and $A$ imply that $A$ and $P$ lie on the circle with diameter $[BQ]$, so either $ABPQ$ or $AQBP$ is a cyclic, convex quadrilateral. In either case, $A$ and $Q$ lie on the same side of $BP$, so $\\angle PAB = \\angle PQB$. Since $CR$ and $PQ$ are perpendicular to $BP$, these lines are parallel and hence $\\angle PQB = \\angle CRB$. Together with $\\angle PAB = \\angle PQB$ and $\\angle RCB = \\angle ABP$, this implies the claim that $\\triangle ABP \\sim \\triangle RCB$.\n\nThis similarity yields the equality $|AP| / |BR| = |BP| / |BC|$. Since $\\angle BAP \\geq 60^{\\circ}$, we get\n$$\n\\begin{aligned}\n0 &\\leq (|AB| - |AP|)^2 = |AB|^2 + |AP|^2 - 2 \\cdot |AB| \\cdot |AP| \\\\\n&= |BP|^2 + 2(\\cos \\angle BAP - 1) \\cdot |AB| \\cdot |AP| \\\\\n&\\leq |BP|^2 - |AB| \\cdot |AP| = |BP|^2 - |BR| \\cdot |BP|.\n\\end{aligned}\n$$\nThis implies that $|BP| \\geq |BR|$, as desired.\n\nIn order for equality to occur, one needs equality in each of the inequalities considered above: $|AB| = |AP|$ and $\\angle BAP = 60^{\\circ}$. Hence there is exactly one point $P$ for which we have equality; this is the unique point inside the square such that $\\triangle ABP$ is equilateral.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25777,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDetermine the smallest positive integer $q$ with the following property:\nfor every integer $m$ with $1 \\leqslant m \\leqslant 1006$, there exists an integer $n$ such that\n$$\n\\frac{m}{1007} q < n < \\frac{m+1}{1008} q .\n$$",
    "options": [],
    "answer": "2015",
    "solution": "Solution:\nFor $m=1006$, we have\n$$\nq - q / 1007 < n < q - q / 1008\n$$\nfor some integer $n$. If $q \\leqslant 1007$, then $q - q / 1007$ and $q - q / 1008$ are both numbers that are at least $q-1$ and smaller than $q$, so there can be no integer $n$ in between. Hence $q > 1007$ and $q - q / 1008 \\leqslant q-1$, implying that $n < q-1$, and that $q - q / 1007 < q-2$. By rearranging terms, we find $q > 2014$, and hence $q \\geqslant 2015$.\n\nLet us prove that $q=2015$ works. Indeed, $m q / 1007 = 2m + m / 1007 < 2m + 1$ and $(m+1) q / 1008 = 2(m+1) - (m+1) / 1008 > 2m + 1$ for $1 \\leqslant m \\leqslant 1006$. Hence each pair of inequalities can be satisfied by taking $n = 2m + 1$. This completes the proof, and shows that the smallest possible of $q$ is indeed $2015$.\nSolution:\nFor $m = 1, \\ldots, 1006$, there must exist an integer $N$ divisible by $1007 \\cdot 1008$ satisfying the double inequality\n$$\n1008 m q < N < 1007(m+1) q .\n$$\nSince $N$ is divisible by $1007$ and $1008$, we may write\n$$\nN = 1008 m q + 1008 k = 1007(m+1) q - 1007 \\ell\n$$\nwhere $k, \\ell > 0$ are integers. Hence\n$$\n(1007 - m) q = 1008 k + 1007 \\ell \\geqslant 2015\n$$\nsince $k, \\ell > 0$. Choosing $m = 1006$ in this last inequality, it follows that $q \\geqslant 2015$. Conversely, for $q = 2015 = 1008 + 1007$, $(*)$ can be satisfied by taking $k = \\ell = 1007 - m$. (Indeed, $1007$ and $1008$ are coprime, and so integer divisible by $1007$ and $1008$ is also divisible by their product.) Thus $q = 2015$ has the desired property, and we are done.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25778,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABC$ be an acute triangle with circumcentre $O$. Let $\\Gamma_{B}$ be the circle through $A$ and $B$ that is tangent to $AC$, and let $\\Gamma_{C}$ be the circle through $A$ and $C$ that is tangent to $AB$. An arbitrary line through $A$ intersects $\\Gamma_{B}$ again in $X$ and $\\Gamma_{C}$ again in $Y$. Prove that $|OX| = |OY|$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $O_{B}$ and $O_{C}$ denote the respective centres of $\\Gamma_{B}$ and $\\Gamma_{C}$. We shall show that $X O_{B} O$ and $O O_{C} Y$ are congruent. Now $A O_{B} \\perp AC$ since $\\Gamma_{B}$ is tangent to $AC$ and $O O_{C} \\perp AC$ since $O O_{C}$ is the perpendicular bisector of $[AC]$. Hence $A O_{B} \\parallel O O_{C}$, and similarly, $A O_{C} \\parallel O O_{B}$. It follows that $A O_{B} O O_{C}$ is a parallelogram. In particular, $|O_{B} X| = |A O_{B}| = |O O_{C}|$ and $|O_{C} Y| = |O_{C} A| = |O O_{B}|$.\n\nIt will therefore suffice to show that the angles $\\angle O O_{B} X$ and $\\angle Y O_{C} O$ are equal. Noting that $\\angle A O_{B} O = \\angle A O_{C} O$ since $A O_{B} O O_{C}$ is a parallelogram, this follows by angle chasing:\n$$\n\\begin{aligned}\n\\angle X O_{B} O & = \\angle A O_{B} O - \\angle A O_{B} X = \\angle A O_{C} O - (180^{\\circ} - 2 \\angle O_{B} A X) \\\\\n& = \\angle A O_{C} O - 180^{\\circ} + 2(\\angle O_{B} A O_{C} - \\angle Y A O_{C}) \\\\\n& = \\angle A O_{C} O - 180^{\\circ} + 2(180^{\\circ} - \\angle A O_{C} O) - (180^{\\circ} - \\angle A O_{C} Y) \\\\\n& = \\angle A O_{C} Y - \\angle A O_{C} O = \\angle O O_{C} Y.\n\\end{aligned}\n$$\n\n![](attached_image_1.png)\n\nA Variant. As in the main solution, we note that $A O_{B} O O_{C}$ is a parallelogram. Notice that $O_{B}$ and $O_{C}$ lie on the respective perpendicular bisectors of $[A X]$ and $[A Y]$. The following lemma then implies that $O$ lies on the perpendicular bisector of $[X Y]$, completing the proof:\nLet $P_{1}, P_{2}, P_{3}$ be three points on a line. Let $O_{1}$ be a point on the perpendicular bisector of $[P_{2} P_{3}]$, let $O_{2}$ be a point on the perpendicular bisector of $[P_{3} P_{1}]$ and let $O_{3}$ be the point in the plane such that $O_{1} P_{3} O_{2} O_{3}$ is a parallelogram. Then $O_{3}$ lies on the perpendicular bisector of $[P_{1} P_{2}]$.\n\nProof. One can choose coordinates such that $P_{3} = (0,0)$ and $P_{1} = (2,0)$. Then $P_{2} = (2a, 0)$, $O_{1} = (a, b)$ and $O_{2} = (1, c)$ for some $a, b, c \\in \\mathbb{R}$. Hence $O_{3} = (a+1, b+c)$ lies on the perpendicular bisector of $[P_{1} P_{2}]$.\nSolution:\n\nLet $\\alpha = \\angle BAC$. Observe that $\\angle AXB = 180^{\\circ} - \\alpha = \\angle CYA$ by tangential angles. Let $Z$ be the intersection of $BX$ and $CY$. Thus $ZXY$ is isosceles with, in particular, $\\angle ZXY = \\angle ZYX = \\alpha$. It will thus suffice to show that $OZ$ bisects $\\angle XZY$.\n\nNow $\\angle BZC = 2\\alpha = \\angle BOC$ since $O$ is the circumcentre of $ABC$, and hence $BZOC$ is cyclic. In particular, since $|OB| = |OC|$, it follows that $\\angle OZY = \\angle OBC = 90^{\\circ} - \\alpha$. But $\\angle XZY = 180^{\\circ} - 2\\alpha$, and so $\\angle XZY = 2 \\angle OZY$, which shows that $OZ$ bisects $\\angle XZY$.\n\n![](attached_image_2.png)\nSolution:\n\nConsider inversion $\\mathscr{I}$ in a circle centred at $A$. Under $\\mathscr{I}$,\n$$\nB \\mapsto B', \\quad C \\mapsto C', \\quad O \\mapsto O', \\quad X \\mapsto X', \\quad Y \\mapsto Y',\n$$\n$\\Gamma_{B} \\mapsto \\gamma_{B}$, a line through $B'$ parallel to $AC'$, $\\Gamma_{C} \\mapsto \\gamma_{C}$, a line through $C'$ parallel to $AB'$. Notice that $\\mathscr{I}$ sends the circumcircle of $ABC$ to the line $B'C'$, and hence maps $AO$ to a line perpendicular to $BC$. Further, if $[AD]$ is a diameter of this circumcircle and $D \\mapsto D'$ under $\\mathscr{I}$, then $D'$ lies on $BC$. Also, $|AD| = 2|AO|$ implies $|AO'| = 2|AD'|$, and hence $O'$ is the image of $A$ under reflection in $BC$. Finally, $\\angle OXA = \\angle X'O'A$ and $\\angle OYA = \\angle Y'O'A$, and hence $|OX| = |OY|$ if and only if $O'A$ bisects $\\angle X'O'Y'$ externally. We have thus reduced the problem to the following statement:\n\nIn triangle $ABC$, let $O$ be the reflection of $A$ over $BC$, $\\gamma_{B}$ be a line parallel to $AC$ through $B$, and $\\gamma_{C}$ be a line parallel to $AB$ through $C$. For an arbitrary line $\\ell$ passing through $A$, let $X$ and $Y$ be the intersections of $\\ell$ with $\\gamma_{B}$ and $\\gamma_{C}$, respectively. Prove that $OA$ bisects $\\angle XOY$ externally.\n\n![](attached_image_3.png)\n\nLet $P = \\gamma_{A} \\cap \\gamma_{B}$. By construction, $POBC$ is an isosceles trapezoid (and therefore cyclic), $\\angle OBX = \\angle OBP = \\angle OCP = \\angle OCY$. Further, since $XB \\parallel AC$ and $YC \\parallel AB$, we have $\\triangle ABX \\sim \\triangle YCA$. Therefore, as $|OB| = |AB|$ and $|AC| = |OC|$ by construction,\n$$\n\\frac{|OB|}{|BX|} = \\frac{|AB|}{|BX|} = \\frac{|YC|}{|CA|} = \\frac{|YC|}{|CO|}\n$$\nIt follows that $\\triangle OBX \\sim \\triangle YCO$. Hence\n$$\n\\frac{|OX|}{|OY|} = \\frac{|XB|}{|OC|} = \\frac{|XA|}{|AY|}\n$$\nand so the result follows from the angle bisector theorem.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25779,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDoes there exist a prime number whose decimal representation is of the form $3811 \\cdots 11$ (that is, consisting of the digits $3$ and $8$ in that order followed by one or more digits $1$)?",
    "options": [],
    "answer": "No",
    "solution": "Solution:\nWrite\n$$\na(n) = 38 \\underbrace{11 \\cdots 11}_{n \\text{ digits } 1}.\n$$\nThere are three cases to consider, depending on the remainder of $n$ upon division by three.\n\n- If $n = 3k + 1 \\equiv 1 \\pmod{3}$, then the sum of the digits of $a(n)$ is equal to $3(k+4)$, i.e. divisible by $3$, and hence so is $a(n)$.\n\n- If $n = 3k + 2 \\equiv 2 \\pmod{3}$, then note that $a(2) = 3811 = 3700 + 111$ is divisible by $37$. By induction, as $a(3k+2) = 1000 a(3k-1) + 111$, it follows that $a(3k+2)$ is divisible by $37$ for each $k \\geqslant 0$.\n\n- If $n = 3k \\equiv 0 \\pmod{3}$, observe that\n$$\n9 a(3k) = 342 \\underbrace{99 \\ldots 99}_{n \\text{ digits } 9} = \\left(7 \\cdot 10^{k}\\right)^{3} - 1\n$$\nwhich is properly divisible by $7 \\cdot 10^{k} - 1$, a number that is larger than $9$. Hence $a(3k)$ admits a non-trivial factor and so is not prime.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25780,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAn arithmetic progression is a set of the form $\\{a, a+d, \\ldots, a+k d\\}$, where $a, d, k$ are positive integers and $k \\geqslant 2$. Thus an arithmetic progression has at least three elements and the successive elements have difference $d$, called the common difference of the arithmetic progression.\nLet $n$ be a positive integer. For each partition of the set $\\{1,2, \\ldots, 3 n\\}$ into arithmetic progressions, we consider the sum $S$ of the respective common differences of these arithmetic progressions. What is the maximal value $S$ that can attain?\n(A partition of a set $A$ is a collection of disjoint subsets of $A$ whose union is $A$.)",
    "options": [],
    "answer": "n^2",
    "solution": "Solution:\n\nThe maximum value is $n^{2}$, which is attained for the partition into $n$ arithmetic progressions $\\{1, n+1,2 n+1\\}, \\ldots,\\{n, 2 n, 3 n\\}$, each of difference $n$.\n\nSuppose indeed that the set has been partitioned into $N$ progressions, of respective lengths $\\ell_{i}$, and differences $d_{i}$, for $1 \\leqslant i \\leqslant N$. Since $\\ell_{i} \\geqslant 3$,\n$$\n2 \\sum_{i=1}^{N} d_{i} \\leqslant \\sum_{i=1}^{N}\\left(\\ell_{i}-1\\right) d_{i}=\\sum_{i=1}^{N} a_{i}-\\sum_{i=1}^{N} b_{i}\n$$\nwhere $a_{i}$ and $b_{i}$ denote, respectively, the largest and smallest elements of progression $i$. Now\n$$\n\\begin{aligned}\n& \\sum_{i=1}^{N} b_{i} \\geqslant 1+2+\\cdots+N=N(N+1) / 2 \\\\\n& \\sum_{i=1}^{N} a_{i} \\leqslant(3 n-N+1)+\\cdots+3 n=N(6 n-N+1) / 2\n\\end{aligned}\n$$\nand thus\n$$\n2 \\sum_{i=1}^{N} d_{i} \\leqslant N(3 n-N) \\leqslant 2 n^{2}\n$$\nas $N(3 n-N)$ is increasing in $N$ on the interval $[0,3 n / 2]$ and since $N \\leqslant n$. This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25781,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the greatest positive integer $N$ with the following property: there exist integers $x_{1}, \\ldots, x_{N}$ such that $x_{i}^{2}-x_{i} x_{j}$ is not divisible by $1111$ for any $i \\neq j$.",
    "options": [],
    "answer": "1000",
    "solution": "Solution:\nWe prove that the greatest $N$ with the required property is $N=1000$. First note that $x_{i}^{2}-x_{i} x_{j}=x_{i}\\left(x_{i}-x_{j}\\right)$, and that the prime factorisation of $1111$ is $11 \\cdot 101$.\n\nWe first show that we can find $1000$ integers $x_{1}, x_{2}, \\ldots, x_{1000}$ such that $x_{i}^{2}-x_{i} x_{j}$ is not divisible by $1111$ for any $i \\neq j$. Consider the set $\\{1,2, \\ldots, 1110\\}$. This set contains $10$ integers divisible by $101$, and it contains $100$ integers divisible by $11$. None of the integers in the set are divisible by both $11$ and $101$. If we delete all of these $10+100$ integers from the set, we are left with $1000$ integers. Call these $x_{1}, x_{2}, \\ldots, x_{1000}$. Now we have $11 \\nmid x_{i}$ and $101 \\nmid x_{i}$ for all $i$. Suppose there are $i \\neq j$ with $1111 \\mid x_{i}\\left(x_{i}-x_{j}\\right)$, then we must have $1111 \\mid x_{i}-x_{j}$, which is a contradiction, since $x_{i}, x_{j} \\in\\{1,2, \\ldots, 1110\\}$. So this set satisfies the requirement.\n\nWe now prove that given $1001$ (or more) integers $x_{1}, x_{2}, \\ldots, x_{1001}$ there are $i \\neq j$ with $1111 \\mid x_{i}\\left(x_{i}-x_{j}\\right)$. Suppose for a contradiction that for all indices $i \\neq j$, we have that $x_{i}\\left(x_{i}-x_{j}\\right)$ is not divisible by $1111$, and write $X=\\left\\{x_{1}, \\ldots, x_{1001}\\right\\}$. We may (after reducing modulo $1111$ if necessary) assume that $x_{i} \\in\\{0,1, \\ldots, 1110\\}$ for all $i$. Then we know that $x_{i} \\neq 0$ for all $i$, and $x_{i} \\neq x_{j}$ for all $i \\neq j$.\n\nSuppose for some $i$ we have $11 \\mid x_{i}$. (Since $x_{i} \\neq 0$, we know that $101 \\nmid x_{i}$.) Then any integer $a \\neq x_{i}$ with $a \\equiv x_{i} \\pmod{101}$ cannot be an element of $X$, since $1111 \\mid x_{i}\\left(x_{i}-a\\right)$. In $\\{1,2, \\ldots, 1110\\}$ there are $10$ such integers, all of them coprime with $11 \\cdot 101$. If there are exactly $k$ different values of $i$ such that $11 \\mid x_{i}$, there are $10k$ different integers from $\\{1,2, \\ldots, 1110\\}$ that cannot be elements of $X$, all of them coprime with $11 \\cdot 101$.\n\nSimilarly, if there are exactly $m$ different values of $i$ such that $101 \\mid x_{i}$, then there are $100m$ different integers from $\\{1,2, \\ldots, 1110\\}$ that cannot be elements of $X$, all of them coprime with $11 \\cdot 101$. (Note that those $10k$ and $100m$ integers can overlap.)\n\nIn $\\{1,2, \\ldots, 1110\\}$ there are $100$ multiples of $11$, there are $10$ multiples of $101$ and there is no multiple of $11 \\cdot 101$, so there are $1000$ integers that are coprime with $11 \\cdot 101$. In $X$ we have $1001-k-m$ integers that are coprime with $11 \\cdot 101$, so exactly $k+m-1$ of the coprime integers in $\\{1,2, \\ldots, 1110\\}$ are not in $X$. This implies that $10k \\leqslant k+m-1$ and $100m \\leqslant k+m-1$. Adding these two inequalities we find $8k+98m \\leqslant -2$, a clear contradiction. So $N<1001$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25782,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $n$ be a positive integer. Suppose that its positive divisors can be partitioned into pairs (i.e. can be split in groups of two) in such a way that the sum of each pair is a prime number. Prove that these prime numbers are distinct and that none of these are a divisor of $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet $d_{1}$ and $d_{2}$ be positive divisors of $n$ that form a pair as given in the problem. If $d_{1}$ and $d_{2}$ have a non-trivial prime divisor $p$ in common, then $p \\mid d_{1}+d_{2}$ and $p \\leqslant d_{1} < d_{1}+d_{2}$, so $d_{1}+d_{2}$ cannot be prime. Hence $\\gcd(d_{1}, d_{2}) = 1$, which implies that $d_{1} d_{2} \\mid n$.\n\nSuppose the number of positive divisors of $n$ is $2t$ (it is even since the divisors can be split into pairs). If we now multiply all divisors, then on one hand we have the product of all $d_{1} d_{2}$ where $\\{d_{1}, d_{2}\\}$ is a pair, so that product is at most $n^{t}$. On the other hand, for every divisor $d$ there is another divisor $\\frac{n}{d}$ (since the number of divisors is even, the case $d = \\frac{n}{d}$ does not occur), and the product of all those is equal to $n^{t}$. Hence there must be equality in every inequality $d_{1} d_{2} \\leqslant n$. So the pairs of divisors given in the problem are all of the form $\\{d, \\frac{n}{d}\\}$.\n\nNow we prove the two statements in the problem. Suppose that $d, d'$ are positive divisors of $n$ such that $d + \\frac{n}{d} = d' + \\frac{n}{d'}$. Then $d^{2} d' + n d' = d (d')^{2} + n d$, so $d d' (d - d') = n (d - d')$ and hence $(d d' - n)(d - d') = 0$. Therefore either $d = d'$ or $d d' = n$, which implies that $\\{d, \\frac{n}{d}\\} = \\{d', \\frac{n}{d'}\\}$, as required.\n\nNow let $d$ be a positive divisor of $n$. Every prime divisor $p$ of $n$ divides precisely one of $d$ and $\\frac{n}{d}$, since $d \\frac{n}{d} = n$ and $\\gcd(d, \\frac{n}{d}) = 1$; so $p \\nmid d + \\frac{n}{d}$. Therefore $\\gcd(n, d + \\frac{n}{d}) = 1$. Since $d + \\frac{n}{d} > 1$ we conclude that $d + \\frac{n}{d}$ cannot be a divisor of $n$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25783,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all functions $f: \\mathbb{R} \\rightarrow \\mathbb{Z}$ such that\n$$\n(f(f(y)-x))^{2}+f(x)^{2}+f(y)^{2}=f(y) \\cdot(1+2 f(f(y)))\n$$\nfor all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "The only solutions are the constant functions f(x) = 0 for all real x and f(x) = 1 for all real x.",
    "solution": "Solution:\nTake $x=y=0$ and write $c=f(0)$, then we find $f(c)^{2}+c^{2}+c^{2}=c+2 c f(c)$, so $(f(c)-c)^{2}=c-c^{2}$. The left-hand side is non-negative, so the right-hand side must be non-negative as well, hence $c-c^{2} \\geqslant 0$, so $c(1-c) \\geqslant 0$. This implies $0 \\leqslant c \\leqslant 1$, and since $c \\in \\mathbb{Z}$, we get $c=0$ or $c=1$. In both cases we have $c-c^{2}=0$, so $f(c)-c=0$, hence $f(c)=c$. Now taking $y=0$ we find for all $x \\in \\mathbb{R}$ that\n$$\n(f(c-x))^{2}+f(x)^{2}+c^{2}=c+2 c^{2} .\n$$\nIf $c=0$, then this equation reduces to $f(-x)^{2}+f(x)^{2}=0$, and since the left-hand side consists of the sum of two squares, both squares must be zero. Therefore $f(x)=0$ for all $x$. This function is indeed a solution of the given functional equation.\n\nNow we consider the other case: $c=1$. Then (1) reduces to $f(1-x)^{2}+f(x)^{2}=2$. The left-hand side consists of two squares of integers, so they must both be 1. Therefore for all $x$ we have $f(x)=1$ or $f(x)=-1$.\n\nSuppose there is an $a$ with $f(a)=-1$. We take $y=a$ in the functional equation; using that $f(w)^{2}=1$ for any $w$, we find $1+1+1=-1 \\cdot(1+2 f(-1))$. Both with $f(-1)=1$ and with $f(-1)=-1$ this gives a contradiction. So there exists no such $a$, and we conclude that $f(x)=1$ for all $x \\in \\mathbb{R}$. This is also a solution of the given functional equation.\n\nWe conclude that there are two solutions: $f(x)=0$ for all $x$, and $f(x)=1$ for all $x$.\nSolution:\nWe proceed as in the first solution up to the point of deriving $f(x)= \\pm 1$ for all $x \\in \\mathbb{R}$. Now we take $x=0$ in the original functional equation, and we find\n$$\n(f(f(y)))^{2}+f(0)^{2}+f(y)^{2}=f(y) \\cdot(1+2 f(f(y)))\n$$\nWe can rewrite this as\n$$\n(f(y)-f(f(y)))^{2}+f(0)^{2}=f(y)\n$$\nThe left-hand side is non-negative, so $f(y) \\geqslant 0$ for all $y$. If we combine this with $f(x)= \\pm 1$ for all $x$, we can conclude that $f(x)=1$ for all $x \\in \\mathbb{R}$. And this is a solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25784,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA circle $\\omega$ passes through the two vertices $B$ and $C$ of a triangle $A B C$. Furthermore, $\\omega$ intersects segment $A C$ in $D \\neq C$ and segment $A B$ in $E \\neq B$. On the ray from $B$ through $D$ lies a point $K$ such that $|B K|=|A C|$, and on the ray from $C$ through $E$ lies a point $L$ such that $|C L|=|A B|$. Show that the circumcentre $O$ of triangle $A K L$ lies on $\\omega$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet $M$ be the midpoint of the arc $B C$ of $\\omega$ that is on the same side of $B C$ as $A$. Then $|B M|=|C M|$. We also have $|B A|=|C L|$ and $\\angle A B M=\\angle E B M=\\angle E C M=\\angle L C M$. Hence $\\triangle A B M \\cong \\triangle L C M$. So $|A M|=|L M|$. (In case $M=E$ the triangles $A B M$ and $L C M$ are degenerate, but then the proof still works, since $|A M|=|A B|-|B M|=|L C|-|C M|=|L M|$.) Similarly, we prove that $|A M|=|K M|$, so $M$ is the circumcentre of $\\triangle A K L$. This means that $M=O$, and hence we are done as $M$ was defined to be on $\\omega$.\nSolution:\nWe consider the configuration where $O$ is in the interior of triangle $A B C$. The proof for other configurations is similar. Furthermore, we exclude the case that $O=D$ or $O=E$; in those cases it is immediate that $O$ is on $\\omega$.\nWe have $\\angle A B K=\\angle E B D=\\angle E C D=\\angle L C A$. Together with $|A B|=|C L|$ and $|A C|=|B K|$ this implies $\\triangle A B K \\cong \\triangle L C A$. Hence $|A K|=|A L|$, $\\angle A K B=\\angle L A C$ (denote this angle by $\\alpha$) and $\\angle B A K=\\angle C L A$ (denote this angle by $\\beta$). Furthermore, let $\\gamma=\\angle B E C=\\angle B D C$. Then $\\angle B A C=180^{\\circ}-\\angle B D A-\\angle A B D=\\angle B D C-\\angle A B D=\\gamma-\\angle A B D$. Therefore $\\angle K A L=\\angle B A K+\\angle L A C-\\angle B A C=\\alpha+\\beta-(\\gamma-\\angle A B D)=\\alpha+\\beta+\\angle A B D-\\gamma$. Note that in triangle $A B K$ we have $\\alpha+\\beta+\\angle A B D=180^{\\circ}$, hence $\\angle K A L=180^{\\circ}-\\gamma$.\n\nWe have $\\angle K O L=2\\left(180^{\\circ}-\\angle K A L\\right)=2 \\gamma$. Since $|A K|=|A L|$, this implies $\\angle A O L=\\gamma$. As we also have $\\angle A E L=\\angle B E C=\\gamma$, the quadrilateral $O E L A$ is cyclic. Analogously, $O D K A$ is cyclic. Now we have\n$$\n\\begin{gathered}\n\\angle D O E=360^{\\circ}-\\angle D O A-\\angle A O E=180^{\\circ}-\\angle D O A+180^{\\circ}-\\angle A O E \\\\\n=\\angle D K A+\\angle A L E=\\angle B K A+\\angle A L C=\\alpha+\\beta\n\\end{gathered}\n$$\nOn the other hand\n$$\n\\angle D B E=\\angle K B A=180^{\\circ}-\\angle B A K-\\angle A K B=180^{\\circ}-\\alpha-\\beta .\n$$\nHence $\\angle D O E+\\angle D B E=180^{\\circ}$, so $O$ is on the circle containing $D, B$ and $E$, which is $\\omega$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25785,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\na. Determine the minimal value of\n$$\n\\left(x+\\frac{1}{y}\\right)\\left(x+\\frac{1}{y}-2018\\right)+\\left(y+\\frac{1}{x}\\right)\\left(y+\\frac{1}{x}-2018\\right)\n$$\nwhere $x$ and $y$ vary over the positive reals.\n\nb. Determine the minimal value of\n$$\n\\left(x+\\frac{1}{y}\\right)\\left(x+\\frac{1}{y}+2018\\right)+\\left(y+\\frac{1}{x}\\right)\\left(y+\\frac{1}{x}+2018\\right)\n$$\nwhere $x$ and $y$ vary over the positive reals.",
    "options": [],
    "answer": "a: -2036162, b: 8080",
    "solution": "Solution:\n\nSolution 1. By the inequality between arithmetic and quadratic means,\n$$\n\\left(x+\\frac{1}{y}\\right)^{2}+\\left(y+\\frac{1}{x}\\right)^{2} \\geqslant \\frac{1}{2}\\left(x+\\frac{1}{y}+y+\\frac{1}{x}\\right)^{2}\n$$\nwith equality if and only if $x+1 / y=y+1 / x$, which holds if $x=y$. It follows that\n$$\n\\left(x+\\frac{1}{y}\\right)\\left(x+\\frac{1}{y} \\pm 2018\\right)+\\left(y+\\frac{1}{x}\\right)\\left(y+\\frac{1}{x} \\pm 2018\\right) \\geqslant \\frac{1}{2}\\left(x+\\frac{1}{y}+y+\\frac{1}{x}\\right)^{2} \\pm 2018\\left(x+\\frac{1}{y}+y+\\frac{1}{x}\\right) .\n$$\nThe parabola $f(X)=\\frac{1}{2} X^{2} \\pm 2018 X$ attains its minimal value at $X=\\mp 2018$, and increases monotonically away from this minimal value. By the inequality between arithmetic and geometric means, $(x+1 / x)+(y+1 / y) \\geqslant 4$ with equality iff $x=y=1$. Hence\n\na. $\\left(x+\\frac{1}{y}\\right)\\left(x+\\frac{1}{y}-2018\\right)+\\left(y+\\frac{1}{x}\\right)\\left(y+\\frac{1}{x}-2018\\right) \\geqslant f(2018)=-\\frac{2018^{2}}{2}$,\nand equality is attained if $x=y=u$, where $u+1 / u=1009$, a quadratic equation with discriminant $1009^{2}-4>0$, and that therefore has two real solutions which are clearly positive.\n\nb. $\\left(x+\\frac{1}{y}\\right)\\left(x+\\frac{1}{y}+2018\\right)+\\left(y+\\frac{1}{x}\\right)\\left(y+\\frac{1}{x}+2018\\right) \\geqslant f(4)=8080$,\nwith equality if and only if $x=y=1$.\nSolution:\n\nSolution 2. By the inequality between arithmetic and geometric means, $x / y+y / x \\geqslant 2$, and hence\n$$\n\\left(x+\\frac{1}{y}\\right)^{2}+\\left(y+\\frac{1}{x}\\right)^{2}=x^{2}+\\frac{1}{y^{2}}+y^{2}+\\frac{1}{x^{2}}+2\\left(\\frac{x}{y}+\\frac{y}{x}\\right) \\geqslant x^{2}+\\frac{1}{x^{2}}+y^{2}+\\frac{1}{y^{2}}+4=\\left(x+\\frac{1}{x}\\right)^{2}+\\left(y+\\frac{1}{y}\\right)^{2}\n$$\nwith equality if and only if $x=y$. It follows that\n$$\n\\left(x+\\frac{1}{y}\\right)\\left(x+\\frac{1}{y}-K\\right)+\\left(y+\\frac{1}{x}\\right)\\left(y+\\frac{1}{x}-K\\right) \\geqslant\\left[\\left(x+\\frac{1}{x}\\right)^{2}-K\\left(x+\\frac{1}{x}\\right)\\right]+\\left[\\left(y+\\frac{1}{y}\\right)^{2}-K\\left(y+\\frac{1}{y}\\right)\\right] .\n$$\nNotice that the parabola $f(X)=X^{2}-K X=(X-K / 2)^{2}-K^{2} / 4$ attains its minimum value at $X=K / 2$, and increases monotonically away from this minimal value. Now $x+1 / x, y+1 / y \\geqslant 2$, so it follows that\n$$\n\\left(x+\\frac{1}{y}\\right)\\left(x+\\frac{1}{y}-K\\right)+\\left(y+\\frac{1}{x}\\right)\\left(y+\\frac{1}{x}-K\\right) \\geqslant\\left\\{\\begin{array}{cl}\n2 f(K / 2)=-K^{2} / 2 & \\text{ if } K / 2 \\geqslant 2 \\Longleftrightarrow K \\geqslant 4 \\\\\n2 f(2)=4(2-K) & \\text{ if } K / 2 \\leqslant 2 \\Longleftrightarrow K \\leqslant 4\n\\end{array}\\right.\n$$\nEquality is attained if (and only if)\n$$\nx=y=\\frac{K}{4} \\pm \\sqrt{\\frac{K^{2}}{16}-1} \\quad(\\text{ if } K \\geqslant 4), \\quad x=y=1 \\quad(\\text{ if } K \\leqslant 4)\n$$\nand so these lower bounds indeed represent the minimal values. Taking (a) $K=2018$ and (b) $K=-2018$ in this result completes the proof.\nSolution:\n\nSolution 3 for part (a). Completing squares,\n$$\n\\begin{aligned}\n\\left(x+\\frac{1}{y}\\right)\\left(x+\\frac{1}{y}-2018\\right)+\\left(y+\\frac{1}{x}\\right)\\left(y+\\frac{1}{x}-2018\\right) & =\\left(x+\\frac{1}{y}-1009\\right)^{2}+\\left(y+\\frac{1}{x}-1009\\right)^{2}-2 \\cdot 1009^{2} \\\\\n& \\geqslant-2 \\cdot 1009^{2}\n\\end{aligned}\n$$\nwith equality if $x+1 / y=y+1 / x=1009$, which holds if $x=y=u$, where $u+1 / u=1009$, a quadratic equation with discriminant $1009^{2}-4>0$, and that therefore has two real solutions which are clearly positive.\nSolution:\n\nSolution 4 for part (b). Using the inequality between arithmetic and geometric means,\n$$\n\\begin{aligned}\n\\left(x+\\frac{1}{y}\\right)(x+ & \\left.\\frac{1}{y}+2018\\right)+\\left(y+\\frac{1}{x}\\right)\\left(y+\\frac{1}{x}+2018\\right) \\\\\n& =\\left(x+\\frac{1}{y}\\right)^{2}+\\left(y+\\frac{1}{x}\\right)^{2}+2018\\left[\\left(x+\\frac{1}{x}\\right)+\\left(y+\\frac{1}{y}\\right)\\right] \\geqslant 4 \\frac{x}{y}+4 \\frac{y}{x}+2018(2+2)\n\\end{aligned}\n$$\nBut $x / y+y / x \\geqslant 2$ by the same inequality, and hence\n$$\n\\left(x+\\frac{1}{y}\\right)\\left(x+\\frac{1}{y}+2018\\right)+\\left(y+\\frac{1}{x}\\right)\\left(y+\\frac{1}{x}+2018\\right) \\geqslant 8080\n$$\nwith equality attained if and only if $x=y=1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25786,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn the land of Heptanomisma, four different coins and three different banknotes are used, and their denominations are seven different (non-zero) natural numbers. The denomination of the smallest banknote is greater than the sum of the denominations of the four different coins. A tourist has exactly one coin of each denomination and exactly one banknote of each denomination, but he cannot afford the book on numismatics he wishes to buy. However, the mathematically inclined shopkeeper offers to sell the book to the tourist at a price of his choosing, provided that he can pay this price in more than one way.\n\n(The tourist can pay a price in more than one way if there are two different subsets of his coins and notes, the denominations of which both add up to this price.)\n\na. Prove that the tourist can purchase the book if the denomination of each banknote is smaller than $49$.\n\nb. Show that the tourist may have to leave the shop empty-handed if the denomination of the largest banknote is $49$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet the denominations of the coins and notes be $C_{1}<C_{2}<C_{3}<C_{4}$ and $N_{1}<N_{2}<N_{3}$, respectively. Define $C=C_{1}+C_{2}+C_{3}+C_{4}$ to be the largest amount that can be paid with coins only. The condition of the problem is thus $C<N_{1}$.\n\na.\nSuppose to the contrary that, whatever the price of the book, the tourist can pay for it in no more than one way.\n\nSolution 1. Consider the hands of one or two notes and any number of (or possibly no) coins. Each of them has one of the $N_{2}+N_{3}$ different values $v$ with $N_{1} \\leqslant v<N_{1}+N_{2}+N_{3}$, since $C<N_{1}$. There are $\\left(\\binom{3}{1}+\\binom{3}{2}\\right) \\cdot 2^{4}=96$ such hands. Hence $95=47+48>N_{2}+N_{3} \\geqslant 96$, which is a contradiction.\n\n\nSolution 2. Consider the hands of exactly one note and any number of (or possibly no) coins, as well as the hand consisting of the two smallest notes only, of value $N_{1}+N_{2}$. Each of these has one of the $N_{3}$ different values $v$ with $N_{1} \\leqslant v<N_{3}+N_{1}$, since $C<N_{1}$ and $N_{1}<N_{1}+N_{2}<N_{1}+N_{3}$. Now the number of hands considered above is $3 \\cdot 2^{4}+1=49$, so $N_{3} \\geqslant 49$, a contradiction.\n\n\nSolution 3. Consider the $3 \\cdot 2^{4}=48$ hands of exactly two notes and any number of (or possibly no) coins. Each of these has one of the $N_{3}$ different values $v$ with $N_{1}+N_{2} \\leqslant v<N_{1}+N_{2}+N_{3}$, since $C<N_{1}$. Hence $48 \\leqslant N_{3}<49$, so $N_{3}=48$. Next consider the $3 \\cdot 2^{4}=48$ hands of exactly one note and any number of coins. By the above, these cannot have a value greater than or equal to $N_{1}+N_{2}$, since these can be realised using two notes and some coins. Hence they must each have one of the $N_{2}$ different values $v$ with $N_{1} \\leqslant v<N_{1}+N_{2}$. This implies $48 \\leqslant N_{2}<N_{3}=48$, which is a contradiction.\n\n\nb.\nConsider the denominations\n$$\nC_{1}=3, \\quad C_{2}=6, \\quad C_{3}=12, \\quad C_{4}=24, \\quad N_{1}=47, \\quad N_{2}=48, \\quad N_{3}=49\n$$\nThese satisfy the conditions of the problem, with $C=45<N_{1}$.\n\nSolution 1. Any two amounts that can be obtained using some (or possibly no) coins are different, and they are multiples of $3$ since each of the coin denominations is. Hence any two such amounts differ by at least $3$. Since $N_{3}-N_{1}=2$, it follows that any two hands using one note and some coins each have a different value (and the same is true for hands using more than one note, since at least one note appears at least twice in any two such hands). Finally, $N_{3}+C<N_{1}+N_{2}$, and hence any hand using one note and some coins has a different value from any hand using two notes and some coins. Hence there is no price that the tourist can pay for in more than one way.\n\n\nSolution 2. Suppose to the contrary that there are two hands of coins and notes that sum to the same amount. Up to removing coins or notes that appear in both of these hands to obtain two smaller hands summing to the same amount, we may assume that no coin or note appears in both of these hands. Notice that all the denominations, except for $N_{1} \\equiv -1 \\pmod{3}$ and $N_{3} \\equiv 1 \\pmod{3}$, are divisible by $3$. Hence, if $N_{1}$ appears in one hand, $N_{3}$ must appear in the same hand, and vice versa. Hence these two hands are two disjoint subsets of $\\{C_{1}, C_{2}, C_{3}, C_{4}, N_{2}, N_{1}+N_{3}\\}=\\{3,6,12,24,48,96\\}=3\\{1,2,4,8,16,32\\}$ that sum to the same amount, which contradicts the uniqueness of the binary expansion of these two amounts. Hence there is no price that the tourist can pay for in more than one way.\n\nRemark. It is natural to ask whether there are other choices of coins and notes with $N_{3}=49$ in (b) that force the tourist to leave the shop empty-handed. It turns out that the choice in (b) is unique. The argument runs as follows: the inequality $N_{2}+N_{3} \\geqslant 96$ implies that $N_{2}=47$ or $N_{2}=48$. Further, strengthening the bound in (a), we must have\n$$\n\\left(N_{2}+N_{3}+C\\right)-\\left(N_{1}+N_{2}\\right)+1 \\geqslant 48 \\quad \\Longrightarrow \\quad C \\geqslant 47+N_{1}-N_{3}=N_{1}-2\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25787,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $A B C$ be a triangle with orthocentre $H$, and let $D, E$, and $F$ denote the respective midpoints of line segments $A B, A C$, and $A H$. The reflections of $B$ and $C$ in $F$ are $P$ and $Q$, respectively.\n\na. Show that lines $P E$ and $Q D$ intersect on the circumcircle of triangle $A B C$.\n\nb. Prove that lines $P D$ and $Q E$ intersect on line segment $A H$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na.\n\nSolution 1. Since $F$ is the midpoint of $[A H]$ and $[B P]$, $B H P A$ is a parallelogram. Similarly, $C A Q H$ is a parallelogram, too. Let $O$ denote the circumcentre of triangle $A B C$ and let $R$ be the reflection of $A$ in $O$, so that $R$ lies on the circumcircle of $A B C$. Since $[A R]$ is a diameter of the circumcircle of $A B C$, $C R \\perp A C$. But $B H \\perp A C$, and so $C R \\parallel B H$. Similarly, $B R \\parallel C H$, and thus $B R C H$ is a parallelogram. Since $B H P A$ is a parallelogram, $R C P A$ is a parallelogram, too. In particular, the midpoint $E$ of its diagonal $[A C]$ lies on $[P R]$. Similarly, $D$ lies on $[Q R]$, and so $P E$ and $Q D$ meet at $R$.\n\n![](attached_image_1.png)\n\nRemark. Since $A P$ is parallel to the altitude $B H$, $A P \\perp A C$. Further, $P H \\parallel A B$ since $A P H B$ is a parallelogram. But $C H \\perp A B$, so $P H \\perp C H$. Hence $A P C H$ is cyclic. Similarly, $A H B Q$ is cyclic, too.\n\n\nSolution 2. Let $O$ denote the circumcentre of triangle $A B C$. By construction, $D O \\perp A B$ and $F E \\parallel C H \\perp A B$, and so $D O \\parallel E F$. Similarly, $O E \\parallel D F$, so $D O E F$ is a parallelogram. Since $F$ is the midpoint of $[A H]$ and $[B P]$, $B H P A$ is a parallelogram, and so $A P\\parallel B H\\parallel D F \\parallel O E$ and $|A P|=|B H|=2|D F|=2|O E|$. Extending this argument, it follows that triangles $D O E$ and $Q A P$ have pairwise parallel sides and the ratio of their sides is $1:2$. Hence there is a homothety with ratio $2$ mapping $D O E$ to $Q A P$. The centre $R$ of this homothety is the reflection of $A$ in $O$, which lies on the circumcircle of $A B C$, and also the intersection of lines $P E, Q D$, and $A O$.\n\n\nSolution 3. Since $F$ is the midpoint of $[B P]$ and $[C Q]$, $B C P Q$ is a parallelogram. As $D$ and $E$ are the respective midpoints of $[A B]$ and $[A C]$, it follows that $P Q\\parallel B C\\parallel D E$ and $|P Q|=|B C|=2|D E|$. Let $R$ denote the intersection of $P E$ and $Q D$. Then $D$ and $E$ are the midpoints of $[R Q]$ and $[R P]$, respectively. In particular, $B R A Q$ is a parallelogram. But $F$ is the midpoint of $[A H]$ and $[C Q]$, so $A Q C H$ is a parallelogram. Hence $B R C H$ is a parallelogram, too, so $R$ is the reflection of $H$ in the midpoint of $[B C]$, which is well-known to lie on the circumcircle of $A B C$.\n\n\nSolution 4. Since $F$ is the midpoint of $[B P]$ and $[C Q]$, $B C P Q$ is a parallelogram. Let $R$ be the image of $A$ under the translation that accordingly maps $P, Q$ onto $C, B$, respectively. By construction, $A P C R$ is a parallelogram, and so the midpoint $E$ of $[A C]$ lies on its other diagonal $P R$. Similarly, $D$ lies on $Q R$, and so $R$ is the intersection of $P E$ and $Q D$. Moreover, $F$ is the midpoint of $[A H]$ and $[B P]$, so $A P B H$ is a parallelogram. Hence $C R\\parallel A P\\parallel B H \\perp A C$, so $\\angle A C R=90^{\\circ}$. Similarly, $\\angle A B R=90^{\\circ}$, so $A B R C$ is cyclic, and thus $R$ lies on the circumcircle of $A B C$, as required.\n\n\nSolution 5. Let $M$ be the midpoint of $[B C]$. Note that triangle $P H Q$ is the reflection of triangle $B A C$ in $F$. By construction, the sides of triangle $D E M$ are parallel to those of $C B A$, and hence $D E\\parallel B C\\parallel Q P$, and, similarly, $D M \\parallel Q H$ and $E M \\parallel P H$. Hence triangles $Q H P$ and $D M E$ have pairwise parallel sides and the ratio of their sides is $2:1$. This implies that there is a homothety mapping one onto the other. Its centre is the intersection of $Q D$ and $P E$ and is also the reflection of $H$ in $M$, which is well-known to lie on the circumcircle of $A B C$.\n\n\nSolution 6. Let $G$ be the centroid of $A B C$, and let $\\mathscr{H}$ be the well-known homothety with ratio $-2$ and centre $G$ that maps the nine-point circle of $A B C$ onto its circumcircle. Under $\\mathscr{H}$, $D \\mapsto C$, $E \\mapsto B$. Denote by $R$ the image of the Euler point $F$ under $\\mathscr{H}$; by construction, $R$ lies on the circumcircle of $A B C$. Further, $\\overrightarrow{D F}=\\frac{1}{2} \\overrightarrow{R C}$ and $\\overrightarrow{E F}=\\frac{1}{2} \\overrightarrow{R B}$. Hence the points of intersection of the pairs of lines $B F, R E$ and $C F, R D$ are $P$ and $Q$, respectively. This completes the proof.\n\n![](attached_image_2.png)\n\n\nSolution 7. Let $M, N, K$ denote the respective midpoints of $[B C], [D E], [P Q]$. Thus the intersection $R$ of $P E$ and $Q D$ is the reflection of $K$ in $N$. Under reflection in $F$, $H \\mapsto A$ and $M \\mapsto K$. Hence $A K H M$ is a parallelogram, and so $\\overrightarrow{M N}=\\frac{1}{2} \\overrightarrow{M A}=\\frac{1}{2} \\overrightarrow{H K}$. Hence the intersection of $K N$ and $H M$ is $R$, and $R$ is the reflection of $H$ in $M$, which is well-known to lie on the circumcircle of $A B C$.\n\n\nSolution 8. Take Cartesian coordinates $B(0,0), C(1,0), A(a, b), H(a, c)$. Then the coordinates of $D, E, F, P, Q$ are successively found to be\n$$\nD\\left(\\frac{a}{2}, \\frac{b}{2}\\right), \\quad E\\left(\\frac{1+a}{2}, \\frac{b}{2}\\right), \\quad F\\left(a, \\frac{b+c}{2}\\right), \\quad P(2 a, b+c), \\quad Q(2 a-1, b+c)\n$$\nHence the coordinates of the intersection $R(x, y)$ of $Q D$ and $P E$ satisfy\n$$\n\\frac{y-\\frac{b}{2}}{b+c-\\frac{b}{2}}=\\frac{x-\\frac{a}{2}}{2 a-1-\\frac{a}{2}}=\\frac{x-\\frac{a+1}{2}}{2 a-\\frac{a+1}{2}} \\quad \\Longrightarrow \\quad R(1-a,-c) .\n$$\nHence $R$ is the reflection of $H$ in the midpoint $M\\left(\\frac{1}{2}, 0\\right)$ of $[B C]$, and so lies on the circumcircle of $A B C$.\n\n\nb.\n\nSolution 1. Lines $A F$ and $Q E$ are medians of triangle $C A Q$, and so, by the properties of the centroid, intersect at a point $S$ of $[A F]$ such that $|A S|=2|S F|$. Similarly, lines $A F$ and $P D$ are medians of triangle $B A P$, and so intersect at the same point $S$. Hence $P D$ and $Q E$ intersect on $[A H]$.\n\n\nSolution 2. By construction, a homothety with ratio $2$ centred at $A$ maps $[D F]$ and $[E F]$ onto $[B H]$ and $[C H]$, respectively, which are mapped in turn onto $[P A]$ and $[Q A]$ under reflection in $F$ by the results of (a). The composition of these maps is a homothety with centre $S$ and ratio $-2$ that maps $D$ and $E$ onto $P$ and $Q$, respectively. Hence $P D$ and $Q E$ intersect at $S$. Since this homothety leaves $A H$ invariant, $S$ lies on this line. Since the homothety has negative ratio, the centre lies on the line segment $[A F]$, completing the proof.\n\nRemark. This is a projective result: triangles $D F E$ and $P A Q$ are in axial perspective (at $\\infty$). Hence, by Desargues' theorem, they are in central perspective, and so lines $P D, Q E$, and $A F$ are concurrent.\n\n\nSolution 3. The intersection $S$ of $P D$ and $Q E$ is the centroid of triangle $P Q R$, where $R$ is the intersection of $P E$ and $Q D$, as in (a). Hence, if $M, N, K$ denote the respective midpoints of $[B C], [D E], [P Q]$, then, by part (a), $\\overrightarrow{K S}=2 \\overrightarrow{S N}$. Moreover, since $K$ is the reflection of $M$ in $F$, $d(K, A H)=d(M, A H)=2 d(N, A H)$. It follows that $S$ lies on $A H$; since $K$ and $R$ lie on either side of $A M$, $S$ lies on $[A H]$.\n\n\nSolution 4. In Cartesian coordinates and using the results of (a), the coordinates of the intersection $S\\left(x', y'\\right)$ of $Q E$ and $P D$ satisfy\n$$\n\\frac{y'-\\frac{b}{2}}{b+c-\\frac{b}{2}}=\\frac{x'-\\frac{a}{2}}{2 a-\\frac{a}{2}}=\\frac{x'-\\frac{a+1}{2}}{2 a-1-\\frac{a+1}{2}}\n$$\nHence $x'=a$, so $S$ lies on line $A H$. Further, $y'=\\frac{1}{3}(2 b+c)$. Without loss of generality, $b>0$. Then $c>0$ by definition, and so $c<y'<b$ or $c>y'>b$. Hence $S$ lies on line segment $[A H]$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25788,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAn integer $n \\geqslant 2$ having exactly $s$ positive divisors $1=d_{1}<d_{2}<\\cdots<d_{s}=n$ is said to be good if there exists an integer $k$, with $2 \\leqslant k \\leqslant s$, such that $d_{k}>1+d_{1}+\\cdots+d_{k-1}$. An integer $n \\geqslant 2$ is said to be bad if it is not good.\n\na. Show that there are infinitely many bad integers.\n\nb. Prove that, among any seven consecutive integers all greater than $2$, there are always at least four good integers.\n\nc. Show that there are infinitely many sequences of seven consecutive good integers.\n\n(Gerhard Woeginger, Luxembourg)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na.\n\nSolution 1. We note that $n=2^{m}$ has $m+1$ divisors, $d_{k}=2^{k-1}$ for $1 \\leqslant k \\leqslant m+1$. Thus\n$$\n1+d_{1}+\\cdots+d_{k-1}=1+\\left(2^{k-1}-1\\right)=2^{k-1}=d_{k}\n$$\nfor each $k \\geqslant 2$, and hence each power of $2$ is a bad integer. This exhibits infinitely many bad integers.\n\nSolution 2. We claim that $n=m!$ is bad for each integer $m \\geqslant 2$. The proof proceeds by induction on $m$, the case $m=2$ being clear. If $D_{K}>1$ is a divisor of $m!$, then $D_{K}=d_{k}$ or $D_{K}=q$ or $D_{K}=q d_{k}$, where $d_{k}>1$ is a divisor of $(m-1)!$ and $q>1$ is a divisor of $m$. In the first case, $\\{d_{1}, \\ldots, d_{k-1}\\} \\subseteq \\{D_{1}, \\ldots, D_{K-1}\\}$, so, invoking the inductive hypothesis, $D_{K}=d_{k} \\leqslant 1+\\left(d_{1}+\\cdots+d_{k-1}\\right) \\leqslant 1+\\left(D_{1}+\\cdots+D_{K-1}\\right)$. In the second case, $q \\leqslant m+1$, so $1,2, \\ldots, q-1 \\mid m!$ and $\\{1,2, \\ldots, q-1\\} \\subseteq \\{D_{1}, \\ldots, D_{K-1}\\}$. But $q^{2}-3q+2 \\geqslant 0$ for $q \\geqslant 2$, and hence\n$$\nD_{K}=q \\leqslant \\frac{q^{2}-q+2}{2}=1+(1+\\cdots+q-1) \\leqslant 1+\\left(D_{1}+\\cdots+D_{K-1}\\right)\n$$\nIn the final case, $\\{1,2, \\ldots, q-1, q d_{1}, \\ldots, q d_{k-1}\\} \\subseteq \\{D_{1}, \\ldots, D_{K-1}\\}$, and so\n$$\nD_{K}=q d_{k} \\leqslant q\\left(1+d_{1}+\\cdots+d_{k-1}\\right) \\leqslant 1+(1+\\cdots+q-1)+\\left(q d_{1}+\\cdots+q d_{k-1}\\right)\n$$\nand so $D_{K} \\leqslant 1+\\left(D_{1}+\\cdots+D_{K-1}\\right)$, completing the inductive step.\n\n\nb.\n\nIf $n$ is odd, then $d_{1}=1, d_{2}>2=1+d_{1}$, and $n$ is good. Among any seven consecutive positive integers, there are either four odd integers and three even ones, or four even ones and three odd ones. In the first case, these four odd integers are good. In the second case, we have to show that one of the consecutive even integers $n=2m, n+2=2(m+1), n+4=2(m+2), n+6=2(m+3)$ is good. Notice that even integers of the form $n=2(6\\ell \\pm 1)$, for $\\ell \\geqslant 1$, are good, since $d_{2}=2<n$ as $\\ell \\geqslant 1$, but $d_{3}>4=1+1+2$, since they are divisible by neither $3$ nor $4$. But at least one $m, m+1, m+2, m+3$ is congruent to $\\pm 1 \\pmod{6}$ (and larger than $1$ by assumption); this completes the proof.\n\n\nc.\n\nSolution 1. Let $n=12q$, an even number. By part (b), $n-3, n-2=2(6q-1), n-1, n+1, n+2=2(6q+1), n+3$ are good integers. Take $q>29$ to be prime. Then the divisors of $n$ less than $q$ are precisely the divisors of $12$. Now $q>1+(1+2+3+4+6+12)=29$, and so $n$ is good, too. Since there are infinitely many choices of the prime $q$, there are infinitely many sequences of seven consecutive good integers.\n\nSolution 2. Let $m=2^{3} \\cdot 3^{2} \\cdot 5 \\cdot 7 \\cdot 11 \\cdot 13$. For any integer $k>0$, the seven consecutive integers $mk+1, mk+2, \\ldots, mk+7$ are good. Indeed, the four odd numbers $mk+1, mk+3, mk+5, mk+7$ are good by part (b). Moreover,\n\n| $n$ | $d_{1}$ | $d_{2}$ | $d_{3}$ | $d_{4}$ | $d_{5}$ |  |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| $mk+2$ | 1 | 2 | $\\geqslant 16$ |  |  | $\\Longrightarrow$ |\n| $mk+4$ | 1 | 2 | 4 | $\\geqslant 16$ |  | $\\Longrightarrow$ |\n| $mk+6$ | 1 | 2 | 3 | 6 | $\\geqslant 16$ | $\\Longrightarrow$ |\n| $d_{4}>1+d_{1}+d_{2}+d_{3}$, |  |  |  |  |  |  |\n| $d_{5}>1+d_{1}+d_{2}+d_{3}+d_{4}$, |  |  |  |  |  |  |\n\nand so $mk+2, mk+4, mk+6$ are good integers, too. Hence there are infinitely many choices of seven consecutive good integers.\n\nSolution 3. Let $m=29!$. For any integer $k>0$, the seven consecutive integers $mk+1, mk+2, \\ldots, mk+7$ are good. Indeed, the divisors of these numbers are either at most $7$ or at least $30$. But $30>1+(1+2+\\cdots+7)=29$, and so these seven integers are good, giving infinitely many sequences of seven consecutive good integers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25789,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\na) Let $a, b, c, d$ be real numbers with $0 \\leqslant a, b, c, d \\leqslant 1$. Prove that\n$$\na b(a-b)+b c(b-c)+c d(c-d)+d a(d-a) \\leqslant \\frac{8}{27}\n$$\n\nb) Find all quadruples $(a, b, c, d)$ of real numbers with $0 \\leqslant a, b, c, d \\leqslant 1$ for which equality holds in the above inequality.",
    "options": [],
    "answer": "Maximum value is 8/27. Equality holds for the four cyclic permutations of (1, 2/3, 1/3, 0), namely (1, 2/3, 1/3, 0), (2/3, 1/3, 0, 1), (1/3, 0, 1, 2/3), and (0, 1, 2/3, 1/3).",
    "solution": "Solution:\n\nDenote the left-hand side by $S$. We have\n$$\n\\begin{aligned}\nS & =a b(a-b)+b c(b-c)+c d(c-d)+d a(d-a) \\\\\n& =a^{2} b-a b^{2}+b^{2} c-b c^{2}+c^{2} d-c d^{2}+d^{2} a-d a^{2} \\\\\n& =a^{2}(b-d)+b^{2}(c-a)+c^{2}(d-b)+d^{2}(a-c) \\\\\n& =(b-d)\\left(a^{2}-c^{2}\\right)+(c-a)\\left(b^{2}-d^{2}\\right) \\\\\n& =(b-d)(a-c)(a+c)+(c-a)(b-d)(b+d) \\\\\n& =(b-d)(a-c)(a+c-b-d) .\n\\end{aligned}\n$$\nAssume without loss of generality that $a \\geqslant c$, then $a-c \\geqslant 0$. Now we consider two cases.\n\n- Suppose $b-d \\geqslant 0$. Then if $a+c-b-d<0$ we have $S \\leqslant 0$, so we're done. If $a+c-b-d \\geqslant 0$, we use the AM-GM inequality on $a-c, b-d$ and $a+c-b-d$, which yields\n$$\n\\sqrt[3]{S}=\\sqrt[3]{(b-d)(a-c)(a+c-b-d)} \\leqslant \\frac{b-d+a-c+a+c-b-d}{3}=\\frac{2 a-2 d}{3} \\leqslant \\frac{2}{3}\n$$\nso $S \\leqslant \\frac{8}{27}$.\n\n- Suppose $b-d<0$. Then if $a+c-b-d \\geqslant 0$ we have $S \\leqslant 0$, so we're done. If $a+c-b-d<0$, we use the AM-GM inequality on $a-c, d-b$ and $b+d-a-c$, which yields\n$$\n\\sqrt[3]{S}=\\sqrt[3]{(d-b)(a-c)(b+d-a-c)} \\leqslant \\frac{d-b+a-c+b+d-a-c}{3}=\\frac{2 d-2 c}{3} \\leqslant \\frac{2}{3}\n$$\nso $S \\leqslant \\frac{8}{27}$.\n\nEquality in the first case occurs when $a-c=b-d=a+c-b-d$ and $a=1, d=0$. Then $1-c=b=1+c-b$, which implies $2 c=b=1-c$ and hence $c=\\frac{1}{3}$. This results in $(a, b, c, d)=\\left(1, \\frac{2}{3}, \\frac{1}{3}, 0\\right)$. Similarly, equality in the second case occurs when $a-c=d-b=b+d-a-c$ and $d=1, c=0$. This yields $(a, b, c, d)=\\left(\\frac{2}{3}, \\frac{1}{3}, 0,1\\right)$. The case $a \\leqslant c$ gives the other two cyclic variants, which gives all quadruples satisfying the equality.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 25790,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPawns and rooks are placed on a $2019 \\times 2019$ chessboard, with at most one piece on each of the $2019^{2}$ squares. A rook can see another rook if they are in the same row or column and all squares between them are empty. What is the maximal number $p$ for which $p$ pawns and $p+2019$ rooks can be placed on the chessboard in such a way that no two rooks can see each other?",
    "options": [],
    "answer": "1009^2",
    "solution": "Solution:\n\nAnswer: the maximal $p$ equals $1009^{2}$.\nWrite $n=2019$ and $k=1009$; then $n=2k+1$. We first show that we can place $k^{2}$ pawns and $n+k^{2}$ rooks. Each cell of the chess board has coordinates $(x, y)$ with $1 \\leqslant x, y \\leqslant n$. We colour each cell black or white depending on whether $x+y$ is even or odd.\nLet $A$ be cell $(1, k+1)$, $B$ be cell $(k+1,1)$, $C$ be cell $(2k+1, k+1)$ and $D$ be cell $(k+1,2k+1)$, and consider the skew square $ABCD$. We place rooks on the cells of this square which have the same colour as $A$, and we place pawns on the other cells of this square. In this way, no rook can see another rook. Now we have placed $p=k^{2}$ pawns and $(k+1)^{2}=k^{2}+(2k+1)=p+n$ rooks.\n\nNow we show that we can not place more pawns. Observe that in every row the number of rooks exceeds the number of pawns by at most $1$, since there has to be a pawn between every two neighbouring rooks. So the total number of rooks exceeds the number of pawns by at most $n$. On the other hand we are to place $p$ pawns and $p+n$ rooks, so the number of rooks in every row exceeds the number of pawns by exactly $1$. This means that the rooks and pawns alternate, with rooks at the two ends. For the columns the same holds.\n\nConsider the $\\ell$-th row. Let $a$ be the number of pawns in this row and let $b$ be the number of pawns above the $\\ell$-th row. For all these pawns, a rook must be somewhere above it. Counting the rooks directly above these $a+b$ pawns, we conclude that there must be at least $a+b$ rooks in the first $\\ell-1$ rows. In every row the number of rooks exceeds the number of pawns by $1$, so in these first $\\ell-1$ rows we have at least $a+b-(\\ell-1)$ pawns. So $b \\geqslant a+b-(\\ell-1)$, yielding $a \\leqslant \\ell-1$. We conclude that the $\\ell$-th row contains at most $\\ell-1$ pawns. The same holds for the $\\ell$-th row counted from below (row $(n+1)-\\ell$): also in this row, there are at most $\\ell-1$ pawns. As $n=2k+1$, the maximal number $p$ is\n$$\n\\sum_{\\ell=1}^{k}(\\ell-1)+\\sum_{\\ell=1}^{k+1}(\\ell-1)=k+2 \\cdot \\frac{1}{2}k(k-1)=k+k(k-1)=k^{2}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25791,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTwo circles $\\Gamma_{1}$ and $\\Gamma_{2}$ intersect at points $A$ and $Z$ (with $A \\neq Z$). Let $B$ be the centre of $\\Gamma_{1}$ and let $C$ be the centre of $\\Gamma_{2}$. The exterior angle bisector of $\\angle B A C$ intersects $\\Gamma_{1}$ again at $X$ and $\\Gamma_{2}$ again at $Y$. Prove that the interior angle bisector of $\\angle B Z C$ passes through the circumcentre of $\\triangle X Y Z$.\n\nFor points $P, Q, R$ that lie on a line $\\ell$ in that order, and a point $S$ not on $\\ell$, the interior angle bisector of $\\angle P Q S$ is the line that divides $\\angle P Q S$ into two equal angles, while the exterior angle bisector of $\\angle P Q S$ is the line that divides $\\angle R Q S$ into two equal angles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSolution I. We first prove that $\\angle A Z X = \\angle Y Z A$. Since the triangles $\\triangle B A X$, $\\triangle C A Y$ are isosceles, and $X Y$ is the external bisector of $\\angle B A C$, we see that\n$$\n\\angle B X A = \\angle X A B = \\angle C A Y = \\angle A Y C.\n$$\nUsing these equalities, we find that\n$$\n\\angle X B A = 180^\\circ - \\angle B A X - \\angle A X B = 180^\\circ - \\angle C Y A - \\angle Y A C = \\angle A C Y\n$$\nSince $B, C$ are the centres of respectively $\\Gamma_{1}, \\Gamma_{2}$, this implies that\n$$\n\\angle A Z X = \\frac{1}{2} \\angle A B X = \\frac{1}{2} \\angle Y C A = \\angle Y Z A.\n$$\nLet $O$ be the circumcentre of $\\triangle X Y Z$, next we will prove that $\\angle B Z O = \\angle A Z Y$. Consider the configuration where $\\angle Z X A$ is sharp, then $\\angle A B Z = 2 \\angle A X Z = 2 \\angle Y X Z = \\angle Y O Z$. Since $\\triangle B Z A$ and $\\triangle O Z Y$ are isosceles, this implies $\\angle B Z A = \\angle O Z Y$. Subtracting $\\angle O Z A$ (or adding, depending on the configuration) yields $\\angle B Z O = \\angle A Z Y$.\n\nTogether with the analogous result $\\angle C Z O = \\angle A Z X$, we conclude $\\angle B Z O = \\angle A Z Y = \\angle X Z A = \\angle O Z C$, so $O$ lies indeed on the internal bisector of $\\angle B Z C$.\n\n\nSolution II. Let $O$ be the circumcentre of $\\triangle X Y Z$, then we see that $O X = O Z$. Since $B$ is the centre of $\\Gamma_{1}$, we also see that $B X = B Z$, so $O B$ is the perpendicular bisector of $X Z$. Therefore $\\angle B Z O = \\angle B X O$, and analogously we find $\\angle C Z O = \\angle C Y O$. Note that $B, C$ lie on the same side of $X Y$; we will consider the configuration where $O$ is on the opposite side of $X Y$. Then $\\angle B X O = \\angle B X A + \\angle A X O$. Now the isosceles triangles $\\triangle B X A, \\triangle C A Y, \\triangle O X Y$, and $X Y$ being the external bisector of $\\angle B A C$ give\n$$\n\\begin{aligned}\n& \\angle B X A = \\angle X A B = \\angle C A Y = \\angle A Y C, \\\\\n& \\angle A X O = \\angle Y X O = \\angle O Y X = \\angle O Y A,\n\\end{aligned}\n$$\nso we find that $\\angle B Z O = \\angle B X O = \\angle A Y C + \\angle O Y A = \\angle O Y C = \\angle O Z C$. Hence $O$ lies on the internal bisector of $\\angle B Z C$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25792,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAn integer $m > 1$ is rich if for any positive integer $n$, there exist positive integers $x, y, z$ such that $n = m x^{2} - y^{2} - z^{2}$. An integer $m > 1$ is poor if it is not rich.\n\na) Find a poor integer.\n\nb) Find a rich integer.",
    "options": [],
    "answer": "Poor integer: 4. Rich integer: 5.",
    "solution": "a.\n\nSolution I. We will show that $m = 4$ is poor. If $y$ and $z$ are both even, we have $4 x^{2} - y^{2} - z^{2} \\equiv 0 - 0 - 0 = 0 \\pmod{4}$. If $y$ is even and $z$ is odd or the other way around, then $4 x^{2} - y^{2} - z^{2} \\equiv 0 - 0 - 1 \\equiv 3 \\pmod{4}$. If $y$ and $z$ are both odd, we have $4 x^{2} - y^{2} - z^{2} \\equiv 0 - 1 - 1 \\equiv 2 \\pmod{4}$. Hence it is impossible to write any integer $n \\equiv 1 \\pmod{4}$ as $n = m x^{2} - y^{2} - z^{2}$. So $m = 4$ is poor.\n\n\nSolution II. We will show that $m = 3$ is poor, by proving that it is impossible to write any integer $n \\equiv 5 \\pmod{8}$ as $n = 3 x^{2} - y^{2} - z^{2}$. We consider the equation modulo $8$. If $4 \\mid x$, then $n \\equiv -y^{2} - z^{2} \\pmod{8}$. So if $n \\equiv 5 \\pmod{8}$, we need to have $y^{2} + z^{2} \\equiv 3 \\pmod{8}$. As $y^{2}$ and $z^{2}$ can only be $0, 1$ or $4 \\pmod{8}$, this is impossible. If $x \\equiv 2 \\pmod{4}$, then $3 x^{2} \\equiv 4 \\pmod{8}$, so for $n \\equiv 5 \\pmod{8}$ we need to have $y^{2} + z^{2} \\equiv 7 \\pmod{8}$. Again, this is impossible. Finally, if $x$ is odd, then $3 x^{2} \\equiv 3 \\pmod{8}$, so for $n \\equiv 5 \\pmod{8}$ we need to have $y^{2} + z^{2} \\equiv 6 \\pmod{8}$. This is impossible as well. So $m = 3$ is poor.\n\n\nb.\n\nSolution I. We will show that $m = 5$ is rich. For any integer $x \\geq 2$ we can take $y = 2x - 2$ and $z = x + 3$, then\n$$\n5 x^{2} - y^{2} - z^{2} = 5 x^{2} - (4 x^{2} - 8 x + 4) - (x^{2} + 6 x + 9) = 2x - 13\n$$\nFor $x \\geq 7$, this gives us all odd positive integers.\nFor any integer $x \\geq 1$ we can take $y = 2x - 1$ and $z = x + 1$, then\n$$\n5 x^{2} - y^{2} - z^{2} = 5 x^{2} - (4 x^{2} - 4 x + 1) - (x^{2} + 2 x + 1) = 2x - 2\n$$\nFor $x \\geq 2$, this gives us all even positive integers. So $m = 5$ is rich.\n\n\nSolution II. We will show that $m = 2$ is rich. For any integer $x \\geq 3$ we can take $y = x + 1$ and $z = x - 2$, then\n$$\n2 x^{2} - y^{2} - z^{2} = 2 x^{2} - (x^{2} + 2x + 1) - (x^{2} - 4x + 4) = 2x - 5\n$$\nAs $x$ can take any positive integer value that is at least $3$, this gives us all odd positive integers.\nFor any integer $x \\geq 5$ we can take $y = x + 2$ and $z = x - 4$, then\n$$\n2 x^{2} - y^{2} - z^{2} = 2 x^{2} - (x^{2} + 4x + 4) - (x^{2} - 8x + 16) = 4x - 20\n$$\nFor $x \\geq 6$, this gives us all positive integers divisible by $4$.\nFor any integer $x \\geq 6$ we can take $y = x + 3$ and $z = x - 5$, then\n$$\n2 x^{2} - y^{2} - z^{2} = 2 x^{2} - (x^{2} + 6x + 9) - (x^{2} - 10x + 25) = 4x - 34\n$$\nFor $x \\geq 9$, this gives us all positive integers in the residue class $2 \\pmod{4}$. So $m = 2$ is rich.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25793,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all positive integers $d$ with the following property: there exists a polynomial $P$ of degree $d$ with integer coefficients such that $|P(m)|=1$ for at least $d+1$ different integers $m$.",
    "options": [],
    "answer": "[1, 2, 3]",
    "solution": "Solution:\nNote that $P(x)=c$ for a fixed constant has at most $d$ solutions, since the polynomial $P(x)-c$ of degree $d$ cancels at most $d$ times. This implies that there are integers $m$ satisfying $P(m)=1$, as well as integers $m$ such that $P(m)=-1$.\n\nNext, we prove the following lemma.\n\nLemma. If $a$ and $b$ are integers such that $P(a)=-1$ and $P(b)=1$, then $|b-a| \\leq 2$.\n\nProof Since $b-a \\mid P(b)-P(a)$ by a well-known lemma (corollary of $a-b \\mid a^{n}-b^{n}$ for every integer $n \\geq 0$ ), the conclusion follows.\n\nLet us first consider the case that $d \\geq 4$, and assume that there exists a polynomial $P$ with at least $d+1 \\geq 5$ solutions to $|P(m)|=1$. Let $a$ and $b$ be the smallest and largest solution respectively. Since $b-a \\geq 4$, we need $P(a)=P(b)$ by the lemma. Without loss of generality (by switching $P$ with $-P$ if necessary) we can assume $P(a)=P(b)=1$. Take a value $m$ such that $P(m)=-1$. Due to the lemma, we need $b-m$ and $m-a$ to be both at most $2$. Since $b-a \\geq 4$, there is only one possibility left in which case $b-a=4$ and thus $d=4$. By considering $P(x-m)$, we can assume $P( \\pm 2)=P( \\pm 1)=1$ and $P(0)=-1$. The unique fourth degree polynomial satisfying these equalities is $P(x)=-0.5\\left(x^{2}-4\\right)\\left(x^{2}-1\\right)+1$ which is not a polynomial with integer coefficients.\n\nFor $1 \\leq d \\leq 3$, there exist polynomials satisfying the conditions.\n\nFor $d=1$ we can take $P_{1}(X)=X$ as $P_{1}(-1)=-1$ and $P_{1}(1)=1$.\n\nFor $d=2$, $P_{2}(X)=2 X(X-2)+1$ satisfies $P_{2}(0)=1$, $P_{2}(1)=-1$ and $P_{2}(2)=1$.\n\nFor $d=3$, the polynomial $P_{3}(X)=(X+1) X(X-2)+1$ satisfies $P_{3}(-1)=1=P_{3}(0)=P_{3}(2)=1$ and $P_{3}(1)=-1$. So $\\left|P_{3}(m)\\right|=1$ for $m \\in\\{-1,0,1,2\\}$.\n\nThus the integers with the required property are precisely $d=1,2,3$. This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 25794,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $N$ be a positive integer. A collection of $4 N^{2}$ unit tiles with two segments drawn on them as shown is assembled into a $2 N \\times 2 N$ board. Tiles can be rotated.\n![](attached_image_1.png)\nThe segments on the tiles define paths on the board. Determine the least possible number and the largest possible number of such paths.",
    "options": [],
    "answer": "minimum 4N, maximum N^2 + (N+1)^2",
    "solution": "Solution:\nLet $p$ denote the number of paths. Notice that there are two types of paths: (1) those that start and end at a point on the boundary of the board and (2) closed paths in the interior of the board. Let $p_{1}, p_{2}$ denote the respective numbers of paths of either type. There are $8 N$ points on the boundary of the board, and each of these is the starting point or endpoint of exactly one path, so $p_{1}=4 N$. Trivially, $p_{2} \\geqslant 0$, so $p=p_{1}+p_{2} \\geqslant 4 N$.\nThe paths on the board are made up of $8 N^{2}$ segments in total. There are only 4 possible paths of one segment, in the corners of the board. All other paths on the boundary of the board therefore consist of at least 2 segments. Moreover, all closed paths in the interior of the board consist of at least 4 segments. Hence\n$$\n8 N^{2} \\geqslant 4 \\cdot 1+\\left(p_{1}-4\\right) \\cdot 2+p_{2} \\cdot 4 \\Longleftrightarrow p_{2} \\leqslant N^{2}+(N-1)^{2}, \\text{ so } p=p_{1}+p_{2} \\leqslant N^{2}+(N+1)^{2}\n$$\nWe have thus shown that $4 N \\leqslant p \\leqslant N^{2}+(N+1)^{2}$. These minimum and maximum values can indeed be attained, as shown below for $N=3$.\n![](attached_image_2.png)\nThe constructions generalise easily. The constructions clearly attain the required bounds because they satisfy the equalities in our arguments showing the two bounds. Indeed the board on the left has $p_{2}=0$ so give the lower bound. The one on the right has exactly 4 boundary paths with one segment, all other $4 N-4$ boundary paths with 2 segments, and all remaining paths with 4 segments.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25795,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $ABC$ be a triangle. The circle $\\omega_{A}$ through $A$ is tangent to line $BC$ at $B$. The circle $\\omega_{C}$ through $C$ is tangent to line $AB$ at $B$. Let $\\omega_{A}$ and $\\omega_{C}$ meet again at $D$. Let $M$ be the midpoint of line segment $[BC]$, and let $E$ be the intersection of lines $MD$ and $AC$. Show that $E$ lies on $\\omega_{A}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet $N$ be the midpoint of $[AB]$. By tangential angles, $\\angle CBD = \\angle BAD$ and $\\angle DBA = \\angle DCB$, so triangles $DAB$ and $DBC$ are similar. By definition of the midpoints $M, N$, so are triangles $AND$ and $BMD$. In particular, $\\angle BND = 180^{\\circ} - \\angle DNA = 180^{\\circ} - \\angle DMB$, so $BNDM$ is cyclic. But $MN \\parallel AC$ by construction, so $\\angle DBA = \\angle DBN = \\angle DMN = \\angle EMN = \\angle MEA = \\angle DEA$, hence $EADB$ is cyclic, completing the proof.\n![](attached_image_1.png)\nSolution:\nLet $S$ be the point on $\\omega_{C}$ such that $BS$ is parallel to $AC$, and let $E'$ be the reflection of $S$ in $M$ (such that $BSCE'$ is a parallelogram and $E'$ lies on $AC$). Now we do some (directed) angle chasing:\n$$\n\\begin{aligned}\n\\angle AE'B & = \\angle CE'B \\quad (\\text{since } A, E' \\text{ and } C \\text{ are collinear}) \\\\\n& = \\angle BSC \\quad (BSCE' \\text{ is a parallelogram}) \\\\\n& = \\angle ABC \\quad (\\text{inscribed angles on } \\omega_{C}) \\\\\n& = \\angle ADB \\quad (\\text{inscribed angles on } \\omega_{A}) .\n\\end{aligned}\n$$\nHence $E'$ lies on $\\omega_{A}$. Further $\\angle AE'D = \\angle ABD = \\angle BSD$, and because $AE'$ is parallel to $BS$ we find that $E'D$ is parallel to $SD$. This means that $E'$ lies on $MD$, so $E' = E$ and $E$ lies on $\\omega_{A}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25796,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA divisor $d$ of a positive integer $n$ is said to be a close divisor of $n$ if $\\sqrt{n} < d < 2 \\sqrt{n}$. Does there exist a positive integer with exactly 2020 close divisors?",
    "options": [],
    "answer": "Yes",
    "solution": "Solution:\nLet $m$ be an odd integer with exactly 2020 positive divisors and which is (automatically) not a square. For example, $m = 3^{2019}$ suffices, but of course there are many alternatives. Now consider $n = 2^{k} m$, for some integer $k$ such that $2^{k} > m$. Any divisor of $n$ is then of the form $2^{\\ell} d$ where $d$ is a divisor of $m$. We will now show that for every such divisor $d$, there exists a unique $\\ell$ such that $2^{\\ell} d$ is a close divisor.\n\nBecause $\\sqrt{n}$ is not an integer, there certainly is a unique integer $a$ such that $\\sqrt{n} < 2^{a} d < 2 \\sqrt{n}$. Because $2^{k} > m$, we have $m < \\sqrt{n} < 2^{k}$. Combining with $1 \\leq d \\leq m$, we find $1 < \\frac{\\sqrt{n}}{d} < 2^{a} < \\frac{2 \\sqrt{n}}{d} < 2 \\cdot 2^{k}$ and so we see that $0 < a \\leqslant k$. So $2^{a} d$ is indeed a close divisor. Because $m$ has exactly 2020 divisors, we find that $n$ has exactly 2020 close divisors.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25797,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\na. Prove that for all $a, b, c, d \\in \\mathbb{R}$ with $a+b+c+d=0$,\n$$\n\\max (a, b)+\\max (a, c)+\\max (a, d)+\\max (b, c)+\\max (b, d)+\\max (c, d) \\geqslant 0\n$$\n\nb. Find the largest non-negative integer $k$ such that it is possible to replace $k$ of the six maxima in this inequality by minima in such a way that the inequality still holds for all $a, b, c, d \\in \\mathbb{R}$ with $a+b+c+d=0$.",
    "options": [],
    "answer": "2",
    "solution": "Solution:\n\nThe left-hand-side of the inequality is invariant under permutations of $a, b, c, d$. We may therefore suppose that $a \\geqslant b \\geqslant c \\geqslant d$, so that the inequality reduces to\n$$\n0 \\leqslant 3 a+2 b+c=a+(a+b)+(a+b+c)\n$$\nWe claim that each of the terms on the right-hand side is non-negative; this will prove the inequality. Indeed, if $a<0$, then $a, b, c, d<0$, and so $a+b+c+d<0$, a contradiction. Also, if $a+b<0$, then, as $b \\leqslant a$, $0>b \\geqslant c, d$ so $(a+b)+c+d<0$, another contradiction. Finally, if $a+b+c<0$, then, as above, $0>c \\geqslant d$, so $(a+b+c)+d<0$, a final contradiction.\n\nNext, we claim that it is impossible to replace $k \\geqslant 3$ maxima by minima in the inequality. Indeed, if $k \\geqslant 3$, one number, say $d$, appears in two of the terms changed to minima. Take $a=b=c=1, d=-3$, so that $a+b+c+d=0$. Then the sum is at most $4 \\cdot 1+2 \\cdot(-3)<0$. Hence $k<3$.\n\nFinally, we prove that the required inequality holds if $k=2$ and the terms involving the complementary sets $\\{a, b\\},\\{c, d\\}$ are changed to minima. We will assume again that $a \\geqslant b \\geqslant c \\geqslant d$, and prove that the inequality holds for any change of the terms involving permutations of these sets to minima (rather than proving that it holds for all orderings of $a, b, c, d$ for this one change). There are three cases:\n\n(1) change terms $\\{a, b\\},\\{c, d\\}$, so the inequality becomes $2 a+3 b+d \\geqslant 0 \\Longleftrightarrow a+b+(b-c) \\geqslant 0$;\n\n(2) change terms $\\{a, c\\},\\{b, d\\}$, so the inequality becomes $2 a+b+2 c+d \\geqslant 0 \\Longleftrightarrow a+c \\geqslant 0$;\n\n(3) change terms $\\{a, d\\},\\{b, c\\}$, so the inequality becomes $2 a+b+2 c+d \\geqslant 0 \\Longleftrightarrow a+c \\geqslant 0$,\n\nwhere we have used $a+b+c+d=0$. In the first case, inequality holds since $a+b \\geqslant 0$ as proved earlier and $b \\geqslant c$. In the other cases, suppose that $a+c<0$. Then $c<-a$ and hence $d<-a$ as $d \\leqslant c$. Hence $c+d<-2 a \\leqslant-a-b$ as $a \\geqslant b$, so $a+b+c+d<0$, a final contradiction, completing the proof.\nSolution:\n\nUsing the inequality $\\max (x, y) \\geqslant \\frac{1}{2}(x+y)$, we find that\n$$\n\\max (a, b)+\\max (a, c)+\\max (a, d)+\\max (b, c)+\\max (b, d)+\\max (c, d) \\geqslant \\frac{3}{2}(a+b+c+d)=0\n$$\nFor $k=3$, we take the same counterexample as in Solution 1. Now it remains to prove the inequality where $\\max (a, b)$ and $\\max (c, d)$ are replaced by $\\min (a, b)$ and $\\min (c, d)$. We can assume without loss of generality that $\\min (a, b)=a$ and $\\min (c, d)=c$. Now we find\n$$\n\\begin{aligned}\n& \\min (a, b)+\\max (a, c)+\\max (a, d)+\\max (b, c)+\\max (b, d)+\\min (c, d) \\geqslant \\\\\n& a+\\frac{1}{2}(a+c)+d+b+\\frac{1}{2}(b+d)+c=\\frac{3}{2}(a+b+c+d)=0\n\\end{aligned}\n$$\nSolution:\n\nWe present another proof for the inequality where $\\max (a, b)$ and $\\max (c, d)$ are replaced by $\\min (a, b)$ and $\\min (c, d)$. By substituting $\\max (x, y)=\\frac{1}{2}(x+y+|x-y|)$ and $\\min (x, y)=\\frac{1}{2}(x+y-|x-y|)$ everywhere and using $a+b+c+d=0$, the inequality may be rewritten as:\n$$\n|a-c|+|a-d|+|b-c|+|b-d| \\stackrel{?}{\\geqslant}|a-b|+|c-d| .\n$$\nBy the triangle inequality, we have\n$$\n|a-c|+|a-d|+|b-c|+|b-d| \\geqslant|(a-c)-(b-c)|+|(a-d)-(b-d)|=2|a-b|,\n$$\nand similarly, $|a-c|+|a-d|+|b-c|+|b-d| \\geqslant 2|c-d|$. Adding these and dividing by 2 yields the desired inequality.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 25798,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPebbles are placed on a $2021 \\times 2021$ board in such a way that each square contains at most one pebble. The pebble set of a square of the board is the collection of all pebbles which are in the same row or column as this square. Determine the least number of pebbles that can be placed on the board in such a way that no two squares have the same pebble set.",
    "options": [],
    "answer": "3031",
    "solution": "Solution:\n\nLet $N \\geqslant 1$ be a positive integer. We claim that the least number of pebbles that can be placed on a $(2N+1) \\times (2N+1)$ chessboard in such a way that no two squares of the board have the same pebble set is $3N+1$. The problem has $N=1010$, so at least $3031$ pebbles are needed.\n\nWe begin by placing $(2N+1)+N=3N+1$ pebbles on the board as shown. The construction extends to all values of $N$, and one checks that no two squares on the board have the same pebble set.\n\n![](attached_image_1.png)\n\nWe are left to show that the number of pebbles, $P$, that must be placed on the board is at least $3N+1$. Suppose that there exists an empty row. Any other row must then contain at least two pebbles. Indeed, if another row is empty, Figure [1], or contains exactly one pebble, Figure [2], then there are two squares (shaded in the figures below) with the same pebble set. So in this case there are at least $P \\geqslant 2(2N+1-1)=4N \\geqslant 3N+1$ pebbles on the board. The same argument applies, mutatis mutandis, to the columns of the board.\n\n![Four small grids labeled [1] to [4], showing shaded squares and pebbles to illustrate cases in the proof about rows and columns.](images/mathpix_2024_12_15_7ed0b6fed5333a54e9a5g-4_1b971022.jpg)\n\nOn the other hand, suppose that each row and each column contains at least one pebble. Let $k$ be the number of rows containing precisely one pebble. Counting the number of pebbles per row shows that $P \\geqslant k+2(2N+1-k)$. To count the number of pebbles per column, note that the $k$ pebbles that are the only ones in their row must be in $k$ distinct columns, Figure [3]. Furthermore, at most one of these $k$ columns only has one pebble in it, Figure [4]. So these $k$ columns contain a total of at least $2k-1$ pebbles, while the remaining columns contain at least one each. Hence $P \\geqslant (2k-1)+(2N+1-k)$. Adding these two inequalities we get $2P \\geqslant 6N+2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25799,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA cyclic quadrilateral $A B X C$ has circumcentre $O$. Let $D$ be a point on line $B X$ such that $|A D|=|B D|$. Let $E$ be a point on line $C X$ such that $|A E|=|C E|$. Prove that the circumcentre of triangle $\\triangle D E X$ lies on the perpendicular bisector of $O A$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nFirst, note that $\\angle A O X=2 \\angle A B X=2\\left(180^{\\circ}-\\angle A C X\\right)=2 \\angle A C E$ as $A B X C$ is cyclic. Secondly, both $\\triangle D A B$ and $\\triangle E A C$ are isosceles, which implies that $\\angle A E X=\\angle A E C=180^{\\circ}-2 \\angle A C E=180^{\\circ}-\\angle A O X$ and $\\angle A D X=\\angle A D B=180^{\\circ}-2 \\angle A B D=180^{\\circ}-2 \\angle A B X=180^{\\circ}-\\angle A O X$. From this, see that respectively $A E X O$ and $A D X O$ are cyclic, i.e. $A E D X O$ is cyclic.\n\nHence, the circumcentre of $\\triangle D E X$ is also the circumcentre of $A E D X O$. However, as in a circle any perpendicular bisector of a chord goes through the centre of the circle, we find that the circumcentre of $\\triangle D E X$ lies on the perpendicular bisector of $O A$.\n\n![](attached_image_1.png)\nSolution:\nIn this solution, we use directed angles $\\measuredangle$. We have $\\measuredangle A B D=\\measuredangle A B X=\\measuredangle A C X=\\measuredangle A C E$ and since $\\triangle A B D$ and $\\triangle A C E$ are both isosceles, we see that $\\triangle A B D \\sim \\triangle A C E$ with equal orientation. This means that there exists a spiral symmetry $T$ with centre $A$ such that $T(B)=D$ and $T(C)=E$. Now let $O^{\\prime}$ be the centre of $\\odot D E X$. Then we find $\\measuredangle D O^{\\prime} E=2 \\measuredangle D X E=2 \\measuredangle B X C=\\measuredangle B O C$. Moreover, $\\triangle D O^{\\prime} E$ and $\\triangle B O C$ are isosceles, so we have $\\triangle D O^{\\prime} E \\sim \\triangle B O C$ with equal orientation. This means that $T$ must send $O$ to $O^{\\prime}$, so in particular, $\\triangle A O O^{\\prime}$ is similar to $\\triangle A B D$ and $\\triangle A C E$. We conclude that $\\left|A O^{\\prime}\\right|=\\left|O O^{\\prime}\\right|$, from which the result follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25800,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA sequence $a_{1}, a_{2}, a_{3}, \\ldots$ of positive integers satisfies $a_{1}>5$ and $a_{n+1}=5+6+\\cdots+a_{n}$ for all positive integers $n$. Determine all prime numbers $p$ such that, regardless of the value of $a_{1}$, this sequence must contain a multiple of $p$.",
    "options": [],
    "answer": "2",
    "solution": "Solution:\nWe claim that the only prime number of which the sequence must contain a multiple is $p=2$. To prove this, we begin by noting that\n$$\na_{n+1}=\\frac{a_{n}\\left(a_{n}+1\\right)}{2}-10=\\frac{\\left(a_{n}-4\\right)\\left(a_{n}+5\\right)}{2}\n$$\nLet $p>2$ be an odd prime, and choose $a_{1} \\equiv -4 \\pmod{p}$, so $2 a_{2} \\equiv (-4-4)(-4+5) \\equiv -8 \\pmod{p}$, whence $a_{2} \\equiv -4 \\pmod{p}$, since $p$ is odd. By induction, $a_{n} \\equiv -4 \\pmod{p} \\not\\equiv 0 \\pmod{p}$ for all $n$, and so the sequence need not contain a multiple of $p$.\nWe are left to show that the sequence must contain an even number. Suppose to the contrary that $a_{n}$ is odd for $n=1,2, \\ldots$ We observe that\n$$\na_{n+1}-a_{n}=\\frac{a_{n}\\left(a_{n}+1\\right)}{2}-\\frac{a_{n-1}\\left(a_{n-1}+1\\right)}{2}=\\frac{a_{n}-a_{n-1}}{2}\\left(a_{n}+a_{n-1}+1\\right)\n$$\nBy assumption, $a_{n}+a_{n-1}+1$ is odd for $n=1,2, \\ldots$, so this shows that $v_{2}\\left(a_{n+1}-a_{n}\\right)=v_{2}\\left(a_{n}-a_{n-1}\\right)-1$, and so there exists $N$ such that $v_{2}\\left(a_{N+1}-a_{N}\\right)=0$. This is a contradiction, because $a_{n+1}-a_{n}$ is even for $n=1,2, \\ldots$ by assumption, and thus completes the proof.\nSolution:\nFor odd $p$, proceed as in solution 1. Now let $p=2$, and suppose that every term of the sequence is odd. We claim that it follows that $a_{n} \\equiv 5 \\pmod{2^{k}}$ for every integer $n \\geq 1$ and every integer $k \\geq 1$. We proceed per induction on $k$. For $k=1$ this simply states that $a_{n}$ is odd for all integers $n \\geq 1$, as assumed. Now suppose it is true for $k=r$. Let $k=r+1$. Take any integer $n \\geq 1$. Note that, by the induction hypothesis, $a_{n} \\equiv 5 \\pmod{2^{r}}$. Therefore there exists an integer $s$ such that $a_{n}=2^{r} s+5$. Now note that\n$$\na_{n+1}=\\frac{\\left(a_{n}-4\\right)\\left(a_{n}+5\\right)}{2}=\\frac{\\left(2^{r} s+1\\right)\\left(2^{r} s+10\\right)}{2}=\\left(2^{r} s+1\\right)\\left(2^{r-1} s+5\\right) \\equiv 2^{r-1} s+5 \\pmod{2^{r}}\n$$\nBy the induction hypothesis, $a_{n+1} \\equiv 5 \\pmod{2^{r}}$. Therefore $s$ is even, such that $a_{n}$ is of the form $2^{r+1} s+5$ for any integer $n \\geq 1$, which concludes the induction. From this property, it follows that $a_{1}-5$ is divisible by $2^{k}$ for every integer $k \\geq 1$, which is only possible if $a_{1}=5$. But $a_{1}>5$, so this is a contradiction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25801,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $n \\geqslant 0$ be an integer, and let $a_{0}, a_{1}, \\ldots, a_{n}$ be real numbers. Show that there exists $k \\in \\{0,1, \\ldots, n\\}$ such that\n$$\na_{0}+a_{1} x+a_{2} x^{2}+\\cdots+a_{n} x^{n} \\leqslant a_{0}+a_{1}+\\cdots+a_{k}\n$$\nfor all real numbers $x \\in[0,1]$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nThe case $n=0$ is trivial; for $n>0$, the proof goes by induction on $n$. We need to make one preliminary observation:\nClaim. For all reals $a, b$, $a+b x \\leqslant \\max \\{a, a+b\\}$ for all $x \\in[0,1]$.\nProof. If $b \\leqslant 0$, then $a+b x \\leqslant a$ for all $x \\in[0,1]$; otherwise, if $b>0$, $a+b x \\leqslant a+b$ for all $x \\in[0,1]$. This proves our claim.\nThis disposes of the base case $n=1$ of the induction: $a_{0}+a_{1} x \\leqslant \\max \\{a_{0}, a_{0}+a_{1}\\}$ for all $x \\in[0,1]$.\nFor $n \\geqslant 2$, we note that, for all $x \\in[0,1]$,\n$$\n\\begin{aligned}\na_{0}+a_{1} x+\\cdots+a_{n} x^{n} & =a_{0}+x\\left(a_{1}+a_{2} x+\\cdots+a_{n} x^{n-1}\\right) \\\\\n& \\leqslant a_{0}+x\\left(a_{1}+a_{2}+\\cdots+a_{k}\\right) \\leqslant \\max \\{a_{0}, a_{0}+\\left(a_{1}+\\cdots+a_{k}\\right)\\},\n\\end{aligned}\n$$\nfor some $k \\in\\{1,2, \\ldots, n\\}$ by the inductive hypothesis and our earlier claim. This completes the proof by induction.\nSolution:\nDefine $s_{i}=a_{0}+a_{1}+\\cdots+a_{i}$ for $i \\in\\{0,1, \\ldots, n\\}$. Thus $a_{0}=s_{0}$ and $a_{i}=s_{i}-s_{i-1}$ for all $i \\in\\{1,2, \\ldots, n\\}$. Hence\n$$\n\\begin{aligned}\na_{0}+a_{1} x+a_{2} x^{2}+\\cdots+a_{n} x^{n} & =s_{0}+(s_{1}-s_{0}) x+(s_{2}-s_{1}) x^{2}+\\ldots+(s_{n}-s_{n-1}) x^{n} \\\\\n& =s_{0}(1-x)+s_{1}(x-x^{2})+\\ldots+s_{n-1}(x^{n-1}-x^{n})+s_{n} x^{n}\n\\end{aligned}\n$$\nNow choose $k \\in\\{0,1, \\ldots, n\\}$ such that $s_{k}=\\max \\{s_{0}, s_{1}, \\ldots, s_{n}\\}$. Using the inequality $x^{i-1}-x^{i} \\geqslant 0$, valid for all $i \\in\\{1,2, \\ldots, n\\}$ and all $x \\in[0,1]$, in the right-hand side above, it follows that\n$$\n\\begin{aligned}\na_{0}+a_{1} x+a_{2} x^{2}+\\cdots+a_{n} x^{n} \\leqslant s_{k}(1- x)+s_{k}(x-x^{2})+\\cdots+s_{k}(x^{n-1}-x^{n})+s_{k} x^{n} \\\\\n& =s_{k}[(1-x)+(x-x^{2})+\\cdots+(x^{n-1}-x^{n})+x^{n}] \\\\\n& =s_{k}=a_{0}+a_{1}+\\cdots+a_{k} .\n\\end{aligned}\n$$\nThis completes the proof.\nSolution:\nThe proof proceeds by induction on $n$. The base case $n=0$ is trivial. For $n \\geqslant 1$, since $x \\in[0,1]$, we have $x^{n} \\leqslant x^{n-1}$. Thus, if $a_{n} \\geqslant 0$, then $a_{n} x^{n} \\leqslant a_{n} x^{n-1}$, while, if $a_{n}<0$, then $a_{n} x^{n}<0$ trivially. This shows that $a_{n} x^{n} \\leqslant \\max \\{0, a_{n} x^{n-1}\\}$, whence\n$$\na_{0}+a_{1} x+\\cdots+a_{n-1} x^{n-1}+a_{n} x^{n} \\leqslant a_{0}+a_{1} x+\\cdots+\\max \\{a_{n-1}, a_{n-1}+a_{n}\\} x^{n-1}\n$$\nBy the inductive hypothesis, the polynomial of degree $n-1$ on the right-hand side is bounded above by $a_{0}+\\cdots+a_{k}$ for some $k \\in\\{0,1, \\ldots, n-2\\}$ or $a_{0}+\\cdots+a_{n-2}+\\max \\{a_{n-1}, a_{n-1}+a_{n}\\}$. But the latter is equal to one of $a_{0}+a_{1}+\\cdots+a_{n-1}$ or $a_{0}+a_{1}+\\cdots+a_{n}$; both are of the desired form, $a_{0}+a_{1}+\\cdots+a_{k}$ for some $k \\in\\{n-1, n\\}$. This completes the proof by induction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25802,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $n$ be a positive integer. There are $n$ ants walking along a line at constant nonzero speeds. Different ants need not walk at the same speed or walk in the same direction. Whenever two or more ants collide, all the ants involved in this collision instantly change directions. (Different ants need not be moving in opposite directions when they collide, since a faster ant may catch up with a slower one that is moving in the same direction.) The ants keep walking indefinitely.\nAssuming that the total number of collisions is finite, determine the largest possible number of collisions in terms of $n$.",
    "options": [],
    "answer": "n(n-1)/2",
    "solution": "Solution:\nThe order of the ants along the line does not change; denote by $v_{1}, v_{2}, \\ldots, v_{n}$ the respective speeds of ants $1,2, \\ldots, n$ in this order. If $v_{i-1}<v_{i}>v_{i+1}$ for some $i \\in\\{2, \\ldots, n-1\\}$, then, at each stage, ant $i$ can catch up with ants $i-1$ or $i+1$ irrespective of the latters' directions of motion, so the number of collisions is infinite. Hence, if the number of collisions is finite, then, up to switching the direction definining the order of the ants, (i) $v_{1} \\geqslant \\cdots \\geqslant v_{n}$ or (ii) $v_{1} \\geqslant \\cdots \\geqslant v_{k-1}>v_{k} \\leqslant \\cdots \\leqslant v_{n}$ for some $k \\in\\{2, \\ldots, n-1\\}$. We need the following observation:\nClaim. If $v_{1} \\geqslant \\cdots \\geqslant v_{m}$, then ants $m-1$ and $m$ collide at most $m-1$ times.\n\nProof. The proof goes by induction on $m$, the case $m=1$ being trivial. Since $v_{m-1} \\geqslant v_{m}$, ants $m-1$ and $m$ can only collide if the former is moving towards the latter. Hence, between successive collisions with ant $m$, ant $m-1$ must reverse direction by colliding with ant $m-2$. Since ants $m-1$ and $m-2$ collide at most $(m-1)-1=m-2$ times by the inductive hypothesis, ants $m$ and $m-1$ collide at most $(m-2)+1=m-1$ times.\n\nHence, in case (i), there are at most $0+1+\\cdots+(n-1)=n(n-1) / 2$ collisions. In case (ii), applying the claim to ants $1,2, \\ldots, k$ and also to ants $n, n-1, \\ldots, k$ by switching their order, the number of collisions is at most $k(k-1) / 2+(n-k+1)(n-k) / 2=n(n-1) / 2-(k-1)(n-k)<n(n-1) / 2$.\n\nNow take a coordinate $x$ along the line, and put ants at $x=1,2, \\ldots, n$ with positive initial velocities and speeds $v_{1}=\\cdots=v_{n-1}=1, v_{n}=\\varepsilon$, for some $\\varepsilon$. For $\\varepsilon=0$, collisions occur according to the pattern shown below for $n=5$, which clearly extends to all values of $n$ in such a way that ants $m$ and $m+1$ collide exactly $m$ times for $m=1,2, \\ldots, n-1$. This yield $1+2+\\cdots+(n-1)=n(n-1) / 2$ collisions in total. For all sufficiently small $\\varepsilon>0$, the number of collisions remains equal to $n(n-1) / 2$.\n\n![](attached_image_1.png)\n\nThis shows that the upper bound obtained above can be attained. If the number of collisions is finite, the largest possible number of collisions is therefore indeed $n(n-1) / 2$.\nSolution:\nWe show that there are at most $n(n-1) / 2$ collisions if the number of collisions is finite as in Solution 1.\n\nTo show that the upper bound of $n(n-1) / 2$ collisions can be attained, we construct, inductively, an example of $n$ ants colliding $n(n-1) / 2$ times, the speeds of the ants decrease from left to right, and after all collisions all ants move towards the left, with the possible exception of the rightmost ant. In every case, we will label the ants $1,2, \\ldots, n$ from left to right. For $n=1$ this is trivial. For $n \\geqslant 2$, we use the construction for $n-1$ ants (now labelled $2,3, \\ldots, n$ ). We add ant 1 on the left, moving towards the right, faster than all other ants (so that the speeds of the ants still decrease from left to right), and in such a way that its first collision (with ant 2) happens after all $(n-1)(n-2) / 2$ collisions of the other $n-1$ ants. Now the following events happen (in this order) for $i=1,2, \\ldots, n-2$ : ants $i$ and $i+1$ collide, after which ant $i$ moves to the left and ant $i+1$ moves to the right. These collisions do happen because the speeds of the ants decrease from left to right. Then ants $n-1$ and $n$ also collide, resulting in ant $n-1$ moving to the left. This shows that there are (at least) $(n-1)(n-2) / 2+(n-1)=n(n-1) / 2$ collisions. There are in fact no more collisions since the speeds of the ants decrease from left to right; alternatively, this follows from the upper bound proved previously. Since all ants except ant $n$ are moving towards the left after the collisions, this completes the inductive construction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25803,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $ABC$ be a scalene acute triangle. Let $B_{1}$ be the point on ray $[AC$ such that $|AB_{1}| = |BB_{1}|$. Let $C_{1}$ be the point on ray $[AB$ such that $|AC_{1}| = |CC_{1}|$. Let $B_{2}$ and $C_{2}$ be the points on line $BC$ such that $|AB_{2}| = |CB_{2}|$ and $|BC_{2}| = |AC_{2}|$. Prove that $B_{1}, C_{1}, B_{2}, C_{2}$ are concyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nBy construction, lines $B_{1}C_{2}$ and $B_{2}C_{1}$ bisect segments $[AB]$ and $[AC]$, respectively, so their intersection $O$ is the circumcentre of $ABC$. Hence $\\angle BOC = 2\\angle A$ and $\\angle CBO = \\angle OCB = 90^{\\circ} - \\angle A$. Now, by construction, $\\angle OC_{1}B = 90^{\\circ} - \\angle A = \\angle OCB$, so $BC_{1}CO$ is cyclic. Similarly, $BCB_{1}O$ is cyclic by construction because $\\angle OB_{1}C = 180^{\\circ} - \\angle AB_{1}O = 90^{\\circ} + \\angle A = 180^{\\circ} - \\angle CBO$. In particular, $BC_{1}CB_{1}$ is cyclic, too.\n\nNow $\\angle B_{1}C_{1}B_{2} = \\angle B_{1}C_{1}B - \\angle B_{2}C_{1}B$ and $\\angle B_{1}C_{2}B_{2} = \\angle B_{1}CB - \\angle CB_{1}C_{2}$. But $\\angle B_{1}C_{1}B = \\angle B_{1}CB$ since $BC_{1}CB_{1}$ is cyclic and $\\angle B_{2}C_{1}B = \\angle OC_{1}B = 90^{\\circ} - \\angle A = 180^{\\circ} - \\angle OB_{1}C = \\angle CB_{1}C_{2}$. Hence $\\angle B_{1}C_{1}B_{2} = \\angle B_{1}C_{2}B_{2}$, so $B_{1}B_{2}C_{1}C_{2}$ is cyclic, as required.\n\n![](attached_image_1.png)\nSolution:\nThe isosceles triangles $AB_{1}B$ and $AC_{1}C$ have equal base angles $\\angle BAB_{1} = \\angle C_{1}AC = \\angle A$, so are similar. In particular, $|AB|/|AB_{1}| = |AC|/|AC_{1}|$. Since $\\angle BAC = \\angle B_{1}AC_{1} = \\angle A$, it follows that triangles $ABC$ and $AB_{1}C_{1}$ are similar, too. In particular, $\\angle CBA = \\angle AB_{1}C_{1}$.\n\nBy construction, lines $B_{1}C_{2}$ and $B_{2}C_{1}$ are the respective perpendicular bisectors of $[AB]$ and $[AC]$, so meet them at their respective midpoints $C'$ and $B'$. Hence\n$$\n\\begin{aligned}\n\\angle B_{1}C_{2}B_{2} &= \\angle C' C_{2} B = 90^{\\circ} - \\angle C_{2}BC' = 90^{\\circ} - \\angle CBA = 90^{\\circ} - \\angle AB_{1}C_{1} = 90^{\\circ} - \\angle B'B_{1}C_{1} \\\\\n&= \\angle B_{1}C_{1}B' = \\angle B_{1}C_{1}B_{2}.\n\\end{aligned}\n$$\nHence $B_{1}C_{2}C_{1}B_{2}$ is cyclic, which completes the proof.\nSolution:\nBy construction, lines $B_{1}C_{2}$ and $B_{2}C_{1}$ are the respective perpendicular bisectors of $[AB]$ and $[AC]$, so meet them at their respective midpoints $C'$ and $B'$. Since $\\angle B_{1}C' C_{1} = 90^{\\circ} = \\angle C_{1}B'B_{1}$, $B_{1}C_{1}C'B'$ is cyclic. Together with the fact that $B'C' \\parallel BC$ by construction, this implies\n$$\n\\angle B_{1}C_{2}B_{2} = \\angle C' C_{2} B = \\angle C_{2}C'B' = \\angle B_{1}C'B' = \\angle B_{1}C_{1}B' = \\angle B_{1}C_{1}B_{2},\n$$\nwhence $B_{1}C_{2}C_{1}B_{2}$ is cyclic. This completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25804,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA subset $A$ of the natural numbers $\\mathbb{N}=\\{0,1,2, \\ldots\\}$ is called good if every integer $n>0$ has at most one prime divisor $p$ such that $n-p \\in A$.\n\na. Show that the set $S=\\{0,1,4,9, \\ldots\\}$ of perfect squares is good.\n\nb. Find an infinite good set disjoint from $S$.\n\n(Two sets are disjoint if they have no common elements.)",
    "options": [],
    "answer": "An example is the infinite set of odd primes P = {3, 5, 7, 11, ...}, which is disjoint from the set of squares and is good.",
    "solution": "Solution:\n\na.\nSuppose to the contrary that $S$ is not good, so there exists $n \\in \\mathbb{N}$ with two different prime factors $p \\neq q$ such that $n-p, n-q$ are perfect squares. In particular $n$ is not prime. Write $n-p=m^{2}$, for some $m \\in \\mathbb{N}$. As $p \\mid n$, it follows that $p \\mid m$ and hence $p^{2} \\mid m^{2}$ since $p$ is prime. Hence there exists $k \\in \\mathbb{N}$ such that $n-p=p^{2} k^{2}$. Similarly, there exists $\\ell \\in \\mathbb{N}$ such that $n-q=q^{2} \\ell^{2}$. We observe that $k, \\ell \\neq 0$ since $n$ is not prime.\nNow we have $p-q=(n-q)-(n-p)=(\\ell q-k p)(\\ell q+k p)$. Since $p-q \\neq 0, \\ell q-k p \\neq 0$, and hence $|p-q|=|k p-\\ell q||k p+\\ell q| \\geqslant|k p+\\ell q|=k p+\\ell q$. This is a contradiction however, because, since $k, \\ell \\neq 0$, it is clear that $k p+\\ell q \\geqslant p+q>|p-q|$. Hence $S$ is good.\n\nb.\nLet $q$ be a prime, and let $Q=\\{q, q^{3}, q^{5}, \\ldots\\}$ be the (infinite) set of odd powers of $q$, which is disjoint from $S$. We claim that $Q$ is good. Indeed, let $n \\in \\mathbb{N}$, and let $p \\mid n$ be a prime such that $n-p \\in Q$, i.e. $n-p=q^{2 k+1}$ for some $k \\in \\mathbb{N}$. Then $p \\mid n-p$, so $p \\mid q^{2 k+1}$, and hence $p=q$. Thus $Q$ is good.\nSolution:\n\na.\nLet $p \\mid n$ be a prime such that $n-p=p(n / p-1)=m^{2}$, for some $m \\in \\mathbb{N}$. Since $p \\mid m^{2}$ and $p$ is prime, $p^{2} \\mid m^{2}$, and hence $p \\mid n / p-1<n / p$, so $p<\\sqrt{n}$.\nNow suppose to the contrary that $S$ is not good, so there are primes $p_{1}>p_{2}$ dividing $n$ such that $n-p_{1}<n-p_{2}$ are perfect squares. Then\n$$\nn-p_{2} \\geqslant\\left(\\sqrt{n-p_{1}}+1\\right)^{2}>n-p_{1}+2 \\sqrt{n-p_{1}} \\quad \\Longrightarrow \\quad p_{1}>p_{2}+2 \\sqrt{n-p_{1}} \\geqslant 2+2 \\sqrt{n-p_{1}} .\n$$\nThe last condition implies that $p_{1}>2 \\sqrt{n-1}$. But $p_{1}<\\sqrt{n}$ by the first part, so $\\sqrt{n}>2 \\sqrt{n-1}$, which is a contradiction for $n>1$; the cases $n=0$ and $n=1$ are trivial. Thus $S$ is good.\n\nb.\nWe claim that the infinite set $P=\\{3,5,7,11, \\ldots\\}$ of odd primes, which is disjoint from $S$, is good. Indeed, let $n \\in \\mathbb{N}$ and let $p \\mid n$ be a prime such that $n-p=q$, for some odd prime $q$. Then $p \\mid n-p$, so $p \\mid q$, i.e. $p=q$, and hence $n=2 q$. Since $q$ is the only odd prime divisor of $n=2 q$, $P$ is good.\nThe set $P^{\\prime}=\\{2,3,5,7,11, \\ldots\\}$ of all primes is also good. The proof is similar: let $n \\in \\mathbb{N}$ and let $p \\mid n$ be a prime such that $n-p=q$, for some prime $q$. Then $p \\mid n-p$, so $p \\mid q$, i.e. $p=q$, and hence $n=2 q$. If $q=2$, then 2 is the only prime divisor of $n$; if $q \\neq 2$, then the only prime divisor of $n$, apart from $q$, is 2. However, $n-2=2(q-1) \\notin P^{\\prime}$ since $q-1>1$. Hence $P^{\\prime}$ is good.\nSolution:\n\na.\nSuppose to the contrary that $S$ is not good, so there exists $n \\in \\mathbb{N}$ with two different prime factors $p \\neq q$ such that $n-p, n-q$ are perfect squares. Write $n-p=m^{2}$, for some $m \\in \\mathbb{N}$. As $p \\mid n$, it follows that $p \\mid m$ and hence $p^{2} \\mid m^{2}$ since $p$ is prime. Hence there exists $k \\in \\mathbb{N}$ such that $n-p=p^{2} k^{2}$. Similarly, there exists $\\ell \\in \\mathbb{N}$ such that $n-q=q^{2} \\ell^{2}$. By construction, $n$ is not prime, so $n-p, n-q \\neq 0$, whence $k, \\ell \\geqslant 1$.\nHence $p^{2} k^{2}+p=q^{2} \\ell^{2}+q$. Hence $p^{2} k^{2}<p^{2} k^{2}+p=q^{2} \\ell^{2}+q<q^{2} \\ell^{2}+2 q \\ell+1=(q \\ell+1)^{2}$. Similarly, $q^{2} \\ell^{2}<(p k+1)^{2}$, whence $q \\ell-1<p k<q \\ell+1$. It follows that $p k=q \\ell$, so $p^{2} k^{2}+p=q^{2} \\ell^{2}+q$ yields the contradiction $p=q$. Hence $S$ is good.\n\nb.\nLet $A$ be a finite good set such that $0 \\notin A$, and let $m=\\max A$. Let $a \\geqslant 2 m+1$ be an integer. We claim that $A^{\\prime}=A \\cup\\{a\\}$ is good. Indeed, suppose to the contrary that there exist $n \\in \\mathbb{N}$ and primes $p, q \\mid n$ with $p \\neq q$ such that $n-p, n-q \\in A^{\\prime}$. If $n<a$, then $n-p, n-q \\in A$, which is a contradiction because $A$ is good. Hence $n \\geqslant a$. Now $p \\mid n-p$, so $n-p \\geqslant p$ since $0 \\notin A^{\\prime}$. Thus $p \\leqslant n / 2$ and hence $n-p \\geqslant n / 2 \\geqslant a / 2>m$. Similarly, $n-q>m$. It follows that $n-p=n-q=a$, which implies the contradiction $p=q$. Hence $A^{\\prime}$ is good.\nNow it is clear that any singleton set is good: indeed, if $A=\\{a\\}$, and $n \\in \\mathbb{N}$ has prime divisors $p, q$ such that $n-p, n-q \\in A$, then $n-p=a=n-q$, so $p=q$. Starting from the singleton $T_{1}=\\{2\\}$, we use the above construction to obtain, iteratively, good sets $T_{2}, T_{3}, \\ldots$ of $2,3, \\ldots$ elements. It is clearly possibly to ensure that they are each disjoint from $S$ by not adding a perfect square at any stage. Then $T=T_{1} \\cup T_{2} \\cup \\cdots$ is an infinite good set disjoint from $S$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25805,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all functions $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ such that\n$$\n(x-y)(f(x)+f(y)) \\leqslant f\\left(x^{2}-y^{2}\\right) \\quad \\text{ for all } x, y \\in \\mathbb{R}\n$$",
    "options": [],
    "answer": "All functions of the form f(x) = c x for some real constant c",
    "solution": "Solution:\nClearly, $f(x)=c x$ is a solution for each $c \\in \\mathbb{R}$ since $(x-y)(c x+c y)=c\\left(x^{2}-y^{2}\\right)$. To show that there are no other solutions, we observe that\n\n(1) $x=y: \\quad 0 \\leqslant f(0)$;\n\n$x=1, y=0: \\quad f(0)+f(1) \\leqslant f(1) \\Rightarrow f(0) \\leqslant 0$, whence $f(0)=0 ;$\n\n(2) $y=-x: \\quad 2 x(f(x)+f(-x)) \\leqslant f(0)=0$;\n\n$x \\rightarrow-x: \\quad-2 x(f(-x)+f(x)) \\leqslant 0 \\Rightarrow 2 x(f(x)+f(-x)) \\geqslant 0 ;$\n\nthus $2 x(f(x)+f(-x))=0$ for all $x$, so $f(-x)=-f(x)$ for all $x \\neq 0$, and hence for all $x$, since $f(0)=0$;\n\n(3) $x \\leftrightarrow y: \\quad(y-x)(f(y)+f(x)) \\leqslant f\\left(y^{2}-x^{2}\\right)=-f\\left(x^{2}-y^{2}\\right) \\Rightarrow(x-y)(f(x)+f(y)) \\geqslant f\\left(x^{2}-y^{2}\\right)$;\n\nwhich is the given inequality with the inequality sign reversed, so $(x-y)(f(x)+f(y))=f\\left(x^{2}-y^{2}\\right)$ must hold for all $x, y \\in \\mathbb{R}$;\n\n(4) $y \\leftrightarrow-y: \\quad(x-y)(f(x)+f(y))=f\\left(x^{2}-y^{2}\\right)=f\\left(x^{2}-(-y)^{2}\\right)=(x+y)(f(x)+f(-y))=(x+y)(f(x)-f(y))$; expanding yields $f(x) y=f(y) x$ for all $x, y \\in \\mathbb{R}$.\n\nTaking $y=1$ in the last result, $f(x)=f(1) x$, i.e. $f(x)=c x$, where $c=f(1)$, for all $x \\in \\mathbb{R}$. Since we have shown above that, conversely, all such functions are solutions, this completes the proof.\n\n\nAlternative solution. A slight variation of this argument proves that $(x-y)(f(x)+f(y))=f\\left(x^{2}-y^{2}\\right)$ must hold for all $x, y \\in \\mathbb{R}$ as above, and then reaches $f(x)=c x$ as follows:\n\n(4) $y= \\pm 1: \\quad(x \\mp 1)(f(x) \\pm f(1))=f\\left(x^{2}-1\\right)$ using $f(-1)=-f(1)$ from (2);\n\nhence $(x-1)(f(x)+f(1))=(x+1)(f(x)-f(1)) \\Rightarrow f(x)=f(1) x=c x$, where $c=f(1)$, on expanding.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25806,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDetermine all integers $k \\geqslant 1$ with the following property: given $k$ different colours, if each integer is coloured in one of these $k$ colours, then there must exist integers $a_{1}<a_{2}<\\cdots<a_{2023}$ of the same colour such that the differences $a_{2}-a_{1}, a_{3}-a_{2}, \\ldots, a_{2023}-a_{2022}$ are all powers of $2$.",
    "options": [],
    "answer": "1 and 2",
    "solution": "Solution:\nWe claim that only $k=1$ and $k=2$ satisfy the required property. First, if $k \\geqslant 3$, we colour each integer with its residue class modulo $3$, so that, whenever two integers have the same colour, their difference is divisible by $3$, so is not a power of $2$. This shows that no $k \\geqslant 3$ has the required property.\n\nIn the case $k=1$, the sequence defined by $a_{n}=2n$ for $n=1,2, \\ldots, 2023$ clearly has the required property. In the case $k=2$, we call the colours \"red\" and \"blue\", and construct, for each $n \\geqslant 1$ and by induction, integers $a_{1}<a_{2}<\\cdots<a_{n}$ of the same colour such that $a_{m+1}-a_{m}$ is a power of $2$ for $m=1,2, \\ldots, n-1$. The statement is trivial for $n=1$. For $n>1$, let $a_{1}<a_{2}<\\cdots<a_{n}$ be red integers (without loss of generality) having the desired property. Consider the $n+1$ integers $b_{i}=a_{n}+2^{i}$, for $i=1,2, \\ldots, n+1$. If one of these, say $b_{j}$, is red, then, as $b_{j}-a_{n}=2^{j}$, the $n+1$ red integers $a_{1}<a_{2}<\\cdots<a_{n}<b_{j}$ have the desired property. Otherwise, $b_{1}, b_{2}, \\ldots, b_{n+1}$ are all blue, and $b_{i+1}-b_{i}=(a_{n}+2^{i+1})-(a_{n}+2^{i})=2^{i}$ for $i=1,2, \\ldots, n$, so the $n+1$ blue integers $b_{1}<b_{2}<\\cdots<b_{n+1}$ have the desired property. This completes the inductive step and hence the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25807,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $ABC$ be a triangle with incentre $I$ and circumcircle $\\omega$. Let $N$ denote the second point of intersection of line $AI$ and $\\omega$. The line through $I$ perpendicular to $AI$ intersects line $BC$, segment $[AB]$, and segment $[AC]$ at the points $D, E$, and $F$, respectively. The circumcircle of triangle $AEF$ meets $\\omega$ again at $P$, and lines $PN$ and $BC$ intersect at $Q$. Prove that lines $IQ$ and $DN$ intersect on $\\omega$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nBy construction, $APEF$ and $APBC$ are cyclic, and so\n$$\n\\begin{aligned}\n\\angle BDE &= \\angle CDF = \\angle AFD - \\angle FCD = \\angle AFE - \\angle ACB = (180^\\circ - \\angle EPA) - (180^\\circ - \\angle BPA) \\\\\n&= \\angle BPA - \\angle EPA = \\angle BPE.\n\\end{aligned}\n$$\nHence $DBEP$ is cyclic, too. It follows that $\\angle IDP = \\angle EDP = \\angle EBP = \\angle ABP = \\angle ANP = \\angle INP$ since $APBN$ is cyclic, and so $PDNI$ is also cyclic. In particular, $\\angle DPN = \\angle DIN = 90^\\circ$. Let $R$ denote the second intersection of $DP$ and $\\omega$, so $NQ \\perp DR$. Then $\\angle NPR = 90^\\circ$, so $RN$ is a diameter of $\\omega$. It is well-known that $N$ is the midpoint of the arc $\\widehat{BC}$ not containing $A$, whence $RN \\perp BC$. Thus $DQ$ and $NQ$ are altitudes of triangle $RDN$, and so $Q$ is its orthocentre. This implies that $RQ \\perp DN$, whence, since $RN$ is a diameter of $\\omega$, the intersection $X$ of $RQ$ and $DN$ lies on $\\omega$.\n\n![](attached_image_1.png)\n\nIt is also well-known that $N$ is the centre of the circumcircle $\\Omega$ of triangle $BCI$. Since $DI \\perp IN$ by construction, $DI$ is tangent to $\\Omega$ at $I$. As $D$ lies on the radical axis $BC$ of $\\omega$ and $\\Omega$, it follows that $|DI|^2 = |DB||DC| = |DX||DN|$. Hence triangles $DNI$ and $DIX$ are similar; in particular, $\\angle DXI = \\angle DIN = 90^\\circ$. All of this shows that $R, I, Q, X$ lie on a line perpendicular to $DN$ that intersects $DN$ at $X \\in \\omega$. This completes the proof.\nSolution:\nLet $K$ be the midpoint of segment $[BC]$. It is well-known that $N$ is the midpoint of the small arc $\\widehat{BC}$ of $\\omega$, so $BC \\perp KN$. In particular, $\\angle DKN = 90^\\circ$. But $\\angle DIN = 90^\\circ$ by construction, so $DIKN$ is cyclic, with circumcircle $\\Gamma$. Moreover, $\\angle PEF = 180^\\circ - \\angle PAF = 180^\\circ - \\angle PAB = \\angle PBC$ and $\\angle PFE = \\angle PAE = \\angle PAB = \\angle PCB$ since $AFEP$ and $ACBP$ are cyclic, so triangles $PEF$ and $PBC$ are similar. Now, by construction, $I$ is the midpoint of segment $[EF]$, so, $K$ being the midpoint of $[BC]$, triangles $PIF$ and $PKC$ are similar, too. It follows that $\\angle PID = 180^\\circ - \\angle PIF = 180^\\circ - \\angle PKC = \\angle PKD$, whence $P$ lies on $\\Gamma$.\n\n![](attached_image_2.png)\n\nLet $\\Omega$ be the circumcircle of triangle $BCI$. By construction, $Q$ lies on the radical axes $PN$ of $\\omega, \\Gamma$ and $BC$ of $\\omega, \\Omega$, so is the radical centre of $\\omega, \\Gamma, \\Omega$. In particular, $IQ$ is the radical axis of $\\Gamma, \\Omega$, so is perpendicular to the line joining the centres of $\\Gamma, \\Omega$. Now it is well-known that $N$ is the centre of $\\Omega$, and, since $\\angle DIN = 90^\\circ$, the centre of $\\Gamma$ is the midpoint of segment $[DN]$. This shows that $IQ \\perp DN$.\nFinally, let $DN$ meet $\\omega$ again at $X$. Since $DI \\perp IN$ by construction and $N$ is the centre of $\\Omega$, $DI$ is tangent to $\\Omega$ at $I$. As $D$ lies on the radical axis $BC$ of $\\omega, \\Omega$, it follows that $|DI|^2 = |DB||DC| = |DX||DN|$. Hence triangles $DNI$ and $DIX$ are similar; in particular, $\\angle DXI = \\angle DIN = 90^\\circ$, i.e. $IX \\perp DN$. Since $IQ \\perp DN$, it follows that $X$ is the intersection of $IQ$ and $DN$. Since $X$ lies on $\\omega$ by construction, this completes the proof.\nSolution:\nSince $APEF$ and $APBC$ are cyclic,\n$$\n\\begin{aligned}\n\\angle CPF &= \\angle BPA - \\angle BPC - \\angle FPA = (180^\\circ - \\angle BCA) - \\angle BAC - \\angle FEA \\\\\n&= (180^\\circ - \\angle BCA - \\angle BAC) - \\angle BED = \\angle CBA - \\angle BED = \\angle CBE - \\angle BED = \\angle BDE = \\angle CDF,\n\\end{aligned}\n$$\nso $DPFC$ is cyclic, too. Thence $\\angle CPD = \\angle CFD = 180^\\circ - \\angle IFA = 90^\\circ + \\angle IAF = 90^\\circ + \\angle CAN = 90^\\circ + \\angle CPN$. Hence $\\angle DPN = \\angle CPD - \\angle CPN = 90^\\circ$. Since $\\angle DIN = 90^\\circ$ by construction, it follows that $DPIN$ is cyclic, with circumcircle $\\Gamma$. Let $J$ be the second intersection of line $IQ$ and $\\Gamma$. Moreover, it is well-known that $N$ is the centre of the circumcircle $\\Omega$ of $BIC$. In particular, $|NI| = |NB|$, and so, since $NIPJ$ and $NBPC$ are cyclic,\n$$\n\\frac{|JQ|}{|JP|} = \\frac{|NQ|}{|NI|} = \\frac{|NQ|}{|NB|} = \\frac{|CQ|}{|CP|}\n$$\nLet $S$ now be the point of intersection of $PN$ and $\\Omega$ such that $P, N, S$ lie on line $PN$ in this order. By construction, $\\angle QPC = \\angle NPC = \\angle NAC = \\angle BAN = \\angle BCN = \\angle QCN$, so triangles $CQN$ and $PCN$ are similar, whence\n$$\n\\frac{|CQ|}{|CP|} = \\frac{|NC|}{|NP|} = \\frac{|NQ|}{|NC|} = \\frac{|NC| + |NQ|}{|NC| + |NP|} = \\frac{|NS| + |NQ|}{|NS| + |NP|} = \\frac{|SQ|}{|SP|}\n$$\nCombining (1) and (2) shows that $C, J, S$ lie on a circle of Apollonius, the centre of which lies on the line through $P, Q, N, S$, so, since $|NC| = |NS|$ by construction, is $N$. In other words, $J$ lies on $\\Omega$.\n\n![](attached_image_3.png)\n\nIn particular, $|NI| = |NJ|$. Now, by construction, $\\angle DIN = \\angle DJN = 90^\\circ$, so the right-angled triangles $DIN$ and $DJN$ are congruent, whence $DINJ$ is a kite. In particular, $IJ \\perp DN$. Since $Q$ lies on $IJ$ by definition, this shows that $IQ \\perp DN$. We can now conclude as in Solution 2.\nSolution:\nBy construction, $P$ is the Miquel point of quadrilateral $BCFE$ (and the resulting complete quadrilateral with points $A$ and $D$ added) because it is the intersection of $\\omega$ and the circumcircle of triangle $AEF$. In particular, $DBEP$ is cyclic. It follows that $\\angle IDP = \\angle EDP = \\angle EBP = \\angle ABP = \\angle ANP = \\angle INP$ since $APBN$ is cyclic, and so $PDNI$ is also cyclic.\n\n![](attached_image_4.png)\n\nNext, let $X$ be the intersection of $DN$ and $\\omega$ and let $AN$ meet $BC$ at $Y$. Then $\\angle NAC = \\angle A/2 = \\angle NCB$, so $\\angle BYA = \\angle C + \\angle NAC = \\angle C + \\angle NCB = \\angle NCA$ and hence\n$$\n\\begin{aligned}\n\\angle DQP &= \\angle NQY = \\angle QYA - \\angle QNY = \\angle BYA - \\angle PNA \\\\\n&= \\angle NCA - \\angle PCA = \\angle PCN = 180^\\circ - \\angle NXP = \\angle DXP\n\\end{aligned}\n$$\nThis implies that $DXQP$ is cyclic. In particular, $QX \\perp DN$. It now suffices to show that $IX \\perp DN$, which we do in the same way as in Solution 2.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25808,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA positive integer $n$ is friendly if every pair of neighbouring digits of $n$, written in base 10, differs by exactly 1. For example, $6787$ is friendly, but $211$ and $901$ are not.\nFind all odd natural numbers $m$ for which there exists a friendly integer divisible by $64 m$.",
    "options": [],
    "answer": "All odd positive integers not divisible by five.",
    "solution": "Solution:\nAny friendly number divisible by $64$ is divisible by $4$, and hence the number formed by its last two digits is a multiple of $4$, so ends in $00, 04, 08, \\ldots$, or $96$. A friendly number divisible by $4$ must therefore end in $12, 32, 56$, or $76$, so cannot be divisible by $5$. In particular, if $5 \\mid m$, then there is no friendly integer divisible by $64 m$.\n\nWe claim that conversely, if $m$ is odd and $5 \\nmid m$, then there exists a friendly integer divisible by $64 m$. First, we notice that $343232 = 64 \\cdot 5363$ is a friendly number divisible by $64$, and hence so is\n$$\nN_{k} = 343232343232 \\cdots 343232 = 343232 \\cdot (1 + 10^{6} + \\cdots + 10^{6k}) \\quad \\text{for } k = 0, 1, 2, \\ldots\n$$\nNow the sequence $N_{0}, N_{1}, N_{2}, \\ldots$ eventually repeats modulo $m$, i.e. there exist positive integers $k < \\ell$ such that $N_{\\ell} \\equiv N_{k} \\pmod{m}$. Hence $m \\mid N_{\\ell} - N_{k} = 10^{6(k+1)} N_{\\ell - k - 1}$. Since $m$ is odd and $5 \\nmid m$, $(10, m) = 1$, so $m \\mid N_{\\ell - k - 1}$. By construction, $64 \\mid N_{\\ell - k - 1}$. Thus, as $m$ is odd and hence $(64, m) = 1$, we conclude that $64 m \\mid N_{\\ell - k - 1}$. This completes the proof.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25809,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\na.\nLet $a_{0}, a_{1}, \\ldots, a_{2024}$ be real numbers such that $\\left|a_{i+1}-a_{i}\\right| \\leqslant 1$ for $i=0,1, \\ldots, 2023$.\nFind the minimum possible value of\n$$\na_{0} a_{1}+a_{1} a_{2}+\\cdots+a_{2023} a_{2024}\n$$\n\nb.\nDoes there exist a real number $C$ such that\n$$\na_{0} a_{1}-a_{1} a_{2}+a_{2} a_{3}-a_{3} a_{4}+\\cdots+a_{2022} a_{2023}-a_{2023} a_{2024} \\geqslant C\n$$\nfor all real numbers $a_{0}, a_{1}, \\ldots, a_{2024}$ such that $\\left|a_{i+1}-a_{i}\\right| \\leqslant 1$ for $i=0,1, \\ldots, 2023$?",
    "options": [],
    "answer": "a: -506; b: No, such a constant does not exist.",
    "solution": "Solution:\n\na.\nThe minimum value is $-506$. Note that from $\\left|a_{i}-a_{i-1}\\right| \\leq 1$ it follows that\n$$\na_{i} a_{i-1}=\\frac{\\left(a_{i}+a_{i-1}\\right)^{2}-\\left(a_{i}-a_{i-1}\\right)^{2}}{4} \\geq-\\frac{\\left(a_{i}-a_{i-1}\\right)^{2}}{4} \\geq-\\frac{1}{4}\n$$\nAdding this for $i=1,2, \\ldots, 2024$, we obtain that\n$$\na_{0} a_{1}+a_{1} a_{2}+a_{2} a_{3}+\\ldots+a_{2023} a_{2024} \\geq 2024 \\cdot-\\frac{1}{4}=-506\n$$\nWe now show that this value can be attained. Indeed, for the sequence $\\left(a_{0}, a_{1}, \\ldots, a_{2024}\\right)=\\left(\\frac{1}{2},-\\frac{1}{2}, \\frac{1}{2},-\\frac{1}{2}, \\frac{1}{2}, \\ldots, \\frac{1}{2}\\right)$ with alternating $\\frac{1}{2}$'s and $-\\frac{1}{2}$'s, each term $a_{i} a_{i-1}$ is equal to $-\\frac{1}{4}$, leading to $a_{0} a_{1}+a_{1} a_{2}+a_{2} a_{3}+\\ldots+a_{2023} a_{2024}=2024 \\cdot-\\frac{1}{4}=-506$.\n\nb.\nNo, such a $C$ does not exist. We argue by contradiction. Suppose $C$ has this property, and consider the sequence defined by $a_{0}=C$ and $a_{i}=C-1$ for $i=1,2, \\ldots, 2024$ satisfies the condition in the problem. For this sequence, we have $a_{i} a_{i+1}-a_{i+1} a_{i+2}=0$ for $i=2,4, \\ldots, 2022$, so the sum\n$$\na_{0} a_{1}-a_{1} a_{2}+a_{2} a_{3}-a_{3} a_{4}+a_{4} a_{5}-a_{5} a_{6}+\\ldots+a_{2022} a_{2023}-a_{2023} a_{2024}\n$$\nis equal to\n$$\na_{0} a_{1}-a_{1} a_{2}=C(C-1)-(C-1)^{2}=C-1<C,\n$$\ncontradiction.\n\nWe give an alternative construction for part (b). We choose a real constant $N$, from which we define $a_{i}=N+i$ for each $i=0,1, \\ldots, 2024$, which clearly satisfies the requirement $\\left|a_{i}-a_{i-1}\\right| \\leq 1$ for each $i=0,1, \\ldots, 1011$. Then, it can be seen that for each $i=0,1, \\ldots, 1011$ that\n$$\na_{2 i} a_{2 i+1}-a_{2 i+1} a_{2 i+2}=a_{2 i+1}\\left(a_{2 i}-a_{2 i+2}\\right)=-2(N+2 i+1) \\leqslant-2 N .\n$$\nFrom this, it can be concluded that\n$a_{0} a_{1}-a_{1} a_{2}+a_{2} a_{3}-a_{3} a_{4}+a_{4} a_{5}-a_{5} a_{6}+\\ldots+a_{2022} a_{2023}-a_{2023} a_{2024} \\leqslant 1012 \\cdot-2 N=-2024 N$.\nAs $N$ is a constant which can be arbitrarily chosen, there is no constant $C$ which lower bounds the given expression.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25810,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $n$ be a positive integer. In a coordinate grid, a path from $(0,0)$ to $(2 n, 2 n)$ consists of $4 n$ consecutive unit steps $(1,0)$ or $(0,1)$. Prove that the number of paths that divide the square with vertices $(0,0)$, $(2 n, 0)$, $(2 n, 2 n)$, $(0,2 n)$ into two regions with even areas is\n$$\n\\frac{\\binom{4 n}{2 n}+\\binom{2 n}{n}}{2}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet $X$ denote the set of paths for which $A$ and $B$ have even area and let $Y$ denote the set of paths for which $A$ and $B$ both have odd area. Because $A$ and $B$ together form a square of area $4 n^{2}$, which is even, $|X|+|Y|$ equals the total number of paths from $(0,0)$ to $(2 n, 2 n)$, which is $\\binom{4 n}{2 n}$.\n\nDenoting a step to the right by $R$ and a step upwards by $U$, every path from $(0,0)$ to $(2 n, 2 n)$ can be described as a sequence of $4 n$ symbols, $2 n$ of which are $R$ and $2 n$ of which are $U$. We subdivide such a sequence into $2 n$ pairs of consecutive steps that can be $R R, U R, R U$ or $U U$. The number of possible paths for which neither $U R$ nor $R U$ occurs is $\\binom{2 n}{n}$, because out of $2 n$ pairs that can be either $R R$ or $U U$ we have to choose $n$ that will be $R R$. These $\\binom{2 n}{n}$ all belong to $X$; in fact, for these paths, $A$ and $B$ can be subdivided into $2 \\times 2$-square, making their areas divisible by 4. Now consider the paths that contain at least one $U R$ or $R U$ pair. If in such a path we replace the first occurrence of a $U R$ or $R U$ pair by a pair of the other type (thus replacing $U R$ by $R U$ or vice versa), the areas of $A$ and $B$ each change by 1 and therefore become even if they were odd and odd if they were even. Because this modification is clearly reversible, we conclude that we can pair up all paths that contain at least one $U R$ or $R U$ pair into pairs of paths, one of which belongs to $X$ and one of which belongs to $Y$. This implies that $|X| - \\binom{2 n}{n} = |Y|$. It follows that\n$$\n|X| = \\frac{|X| + |Y|}{2} + \\frac{|X| - |Y|}{2} = \\frac{\\binom{4 n}{2 n}}{2} + \\frac{\\binom{2 n}{n}}{2}.\n$$\n\nDefine $Z_{m, n}$ to be the number of routes from $(0,0)$ to $(2 m, 2 n)$ that divide the rectangle with vertices $(0,0)$, $(0,2 n)$, $(2 m, 2 n)$ and $(2 m, 0)$ into two parts of even area. We call such paths good. We claim that\n$$\n2 Z_{m, n} = \\binom{2 m + 2 n}{2 m} + \\binom{m + n}{m}\n$$\nfor all $m, n$, which for $m = n$ establishes the desired result. We prove this formula by induction on $m + n$, noting first that it clearly holds when either $m$ or $n$ is zero, because $Z_{0, n} = Z_{m, 0} = 1$ (there is only one path from $(0,0)$ to $(0,2 n)$ or $(2 m, 0)$, which is good). Therefore, suppose that $m, n \\geq 1$ and consider a good path from $(0,0)$ to $(2 m, 2 n)$. This path passes through exactly one of $(2 m, 2 n-2)$, $(2 m-1,2 n-1)$ and $(2 m-2,2 n)$. If it passes through $(2 m, 2 n-2)$ then the subpath from $(0,0)$ to $(2 m, 2 n-2)$ must also be good; moreover, for each good path from $(0,0)$ to $(2 m, 2 n-2)$ there is exactly one corresponding path from $(0,0)$ to $(2 m, 2 n)$, and that path is automatically good because the new area that gets added is even. Thus, there are $Z_{m, n-1}$ good paths from $(0,0)$ to $(2 m, 2 n)$ that pass through $(2 m, 2 n-2)$. Similarly, there are $Z_{m-1, n}$ good paths from $(0,0)$ to $(2 m, 2 n)$ that pass through $(2 m-2,2 n)$. Now notice that any path from $(0,0)$ to $(2 m-1,2 n-1)$ (of which there are $\\binom{2 m+2 n-2}{2 m-1}$) can be extended in two ways to obtain a path from $(0,0)$ to $(2 m, 2 n)$; because the resulting areas for these paths differ by 1, exactly one of these paths is good. All in all, we obtain the recursion\n$$\nZ_{m, n} = Z_{m, n-1} + Z_{m-1, n} + \\binom{2 m+2 n-2}{2 m-1}\n$$\nBy the inductive hypothesis, we have\n$$\n2 Z_{m, n-1} = \\binom{2 m+2 n-2}{2 m} + \\binom{m+n-1}{m}\n$$\nand\n$$\n2 Z_{m-1, n} = \\binom{2 m+2 n-2}{2 m-2} + \\binom{m+n-1}{m-1}\n$$\nTherefore, we obtain\n$$\n\\begin{aligned}\n2 Z_{m, n} = & 2 Z_{m, n-1} + 2 Z_{m-1, n} + 2\\binom{2 m+2 n-2}{2 m-1} \\\\\n= & \\binom{2 m+2 n-2}{2 m} + \\binom{m+n-1}{m} + \\binom{2 m+2 n-2}{2 m-2} + \\binom{m+n-1}{m-1} \\\\\n& + 2\\binom{2 m+2 n-2}{2 m-1}.\n\\end{aligned}\n$$\nTo simplify the expression on the right hand side, note that we can reorganize the terms\n$$\n\\binom{2 m+2 n-2}{2 m} + \\binom{2 m+2 n-2}{2 m-2} + 2\\binom{2 m+2 n-2}{2 m-1}\n$$\nas\n$$\n\\left(\\binom{2 m+2 n-2}{2 m} + \\binom{2 m+2 n-2}{2 m-1}\\right) + \\left(\\binom{2 m+2 n-2}{2 m-1} + \\binom{2 m+2 n-2}{2 m-2}\\right)\n$$\nwhich, using the addition rules for binomial coefficients, becomes\n$$\n\\binom{2 m+2 n-1}{2 m} + \\binom{2 m+2 n-1}{2 m-1} = \\binom{2 m+2 n}{2 m}\n$$\nSimilarly, we have\n$$\n\\binom{m+n-1}{m} + \\binom{m+n-1}{m-1} = \\binom{m+n}{m}\n$$\nPutting everything together, we obtain that\n$$\n2 Z_{m, n} = \\binom{2 m+2 n}{2 m} + \\binom{m+n}{m}\n$$\nwhich completes the induction.\n\nWe start by proving the following lemma: for a $(2 m-1,2 n-1)$-grid, there are equally many paths with a region of even area (called even paths), as there are with odd area (odd paths).\nTo prove this, we take a path and rotate it around the center of the grid. Then a path spanning a region with area $x$ is mapped on one spanning an area $(2 m-1)(2 n-1)-x$. This gives a bijection between paths creating an even region, and creating an odd region.\nNow, for every path from $(0,0)$ to $(2 n, 2 n)$, consider all the coordinates of the grid points it visits in order. There are $\\binom{2 n}{n}$ of them which never visit a point with odd coordinates (which we call an odd point). Notice that such paths are all even.\nWe now construct a bijection between the remaining even paths and the odd paths. For each odd point, there are equally many even as odd paths from $(0,0)$ to that point. Define then a bijection $\\phi$ between the sets of odd paths and even paths up to this point for each point. Notice that $\\phi$ implicitly depends on the chosen odd point.\nNow, for an arbitrary odd path $P$, consider the first odd point it passes through. Map $P$ to another path by changing the path up to this odd point to the $\\phi$ of the path up to this point. As $\\phi$ maps between even and odd paths, the resulting path is an even path. By the definition of $\\phi$, and as each of the remaining paths goes to an odd point, this mapping defines a bijection.\nWe have hence found a bijection between the odd and even paths in the remaining $\\binom{4 n}{2 n} - \\binom{2 n}{n}$ paths, which yields the required result like in solution 1.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25811,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABC$ be a triangle with incentre $I$ and circumcircle $\\Omega$ such that $|AC| \\neq |BC|$. The internal angle bisector of $\\angle CAB$ intersects side $[BC]$ in $D$, and the external angle bisectors of $\\angle ABC$ and $\\angle BCA$ intersect $\\Omega$ again in $E$ and $F$, respectively. Let $G$ be the intersection of lines $AE$ and $FI$ and let $\\Gamma$ be the circumcircle of triangle $BDI$. Show that $E$ lies on $\\Gamma$ if and only if $G$ lies on $\\Gamma$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWe first notice the general fact that $EF \\perp AI$. This can be proved using the following argument. Denote $S$ for the intersection of $EF$ and $AI$. Then $\\angle BIS = (\\angle IBA + \\angle IAB) = \\frac{1}{2}(\\angle ABC + \\angle BAC) = \\frac{1}{2}(180^{\\circ} - \\angle BCA) = \\angle BCF = \\angle BEF = \\angle BES$. Thus $B, I, S, E$ form a cyclic quadrilateral, from which it can be concluded that $EF \\perp AI$.\n\nWe will now continue to prove the problem by doing both directions separately.\n\nAssume first that $E$ lies on $\\Gamma$. Then, as angle bisectors are perpendicular to one another, we have that $\\angle IDE = \\angle IBE = 90^{\\circ}$. Then, as $EF \\perp AI$, it holds that $D$ lies on $EF$. It can then be concluded that $\\angle IDF = 90^{\\circ} = \\angle ICF$ (again due to perpendicular bisectors), from which it can be concluded that $I, D, C, F$ form a cyclic quadrilateral. We can now calculate that $\\angle GID = 180^{\\circ} - \\angle FID = 180^{\\circ} - \\angle FCD = 180^{\\circ} - \\angle FCB = \\angle AEF = \\angle GED$, where the second to last step follows from the fact that the arcs $BF$ and $AF$ have the same length, as $F$ lies on the external bisector of $\\angle ACB$. It now follows that $I, D, E, G$ form a cyclic quadrilateral, thus $G$ lies on $\\Gamma$.\n\nFor the reverse implication, assume that $G$ lies on $\\Gamma$. We can then compute that $\\angle FAD = \\angle FAS = 90^{\\circ} - \\angle AFS = 90^{\\circ} - \\angle AFE = 90^{\\circ} - \\angle ABE = \\angle ABI = \\angle IBD = \\angle IGD = \\angle FGD$, from which we can conclude that $A, G, D, F$ form a cyclic quadrilateral. Similarly to $\\angle FAD = \\angle IBD$, it holds that $\\angle GAD = \\angle ICD$. From this we can compute that $\\angle IFD = \\angle GFD = \\angle GAD = \\angle ICD$, thus $I, D, C, F$ forms a cyclic quadrilateral. Hence, $\\angle IDF = 180^{\\circ} - \\angle ICF = 90^{\\circ}$, so again $D = S$. We then see that $\\angle IBE = 90^{\\circ} = \\angle IDE$ from which we can conclude that $I, D, E, B$ form a cyclic quadrilateral. It then follows that $E$ lies on $\\Gamma$.\nSolution:\n\nThe external angle bisectors $BE$ and $CF$ meet the internal bisector $ID$ at the $A$-excentre $J$ of triangle $ABC$. If one of $BEDI$ and $CFID$ is cyclic, then, as $BECF$ is cyclic, $|JD||JI| = |JE||JB| = |JC||JF|$ by power of a point, and so the other is cyclic, too. This shows that\n(1) $BEDI$ is cyclic if and only if $CFID$ is cyclic.\n\nNext, $\\angle GAD = \\angle EAD = \\angle BAD - \\angle BAE = \\angle A/2 - (180^{\\circ} - \\angle AEB - \\angle EBA)$, with $\\angle AEB = \\angle ACB = \\angle C$ and $\\angle EBA = \\angle EBI + \\angle IBA = 90^{\\circ} + \\angle B/2$. Hence $\\angle GAD = \\angle A/2 + \\angle B/2 + \\angle C - 90^{\\circ} = \\angle C/2 = \\angle ICD$. Thus, if one of $CFID$ and $AGDF$ is cyclic, then $\\angle ICD = \\angle IFD = \\angle GFD = \\angle GAD$, and the other is cyclic, too. We have thus established that\n(2) $CFID$ is cyclic if and only if $AGDF$ is cyclic.\n\nFinally, let $A'$ denote the second intersection of $AI$ with $\\Omega$. Then $\\angle DAF = \\angle A'AF = 180^{\\circ} - \\angle FCA'$, with $\\angle FCA' = \\angle FCI + \\angle ICB + \\angle BCA' = 90^{\\circ} + \\angle C/2 + \\angle BAA' = 90^{\\circ} + \\angle C/2 + \\angle A/2$. It follows that $\\angle DAF = 90^{\\circ} - \\angle A/2 - \\angle C/2 = \\angle B/2 = \\angle DBI$. Hence, if one of $AGDF$ and $BDIG$ is cyclic, then $\\angle DAF = \\angle DGF = \\angle DGI = \\angle DBI$, and so the other is cyclic, too. Hence\n(3) $AGDF$ is cyclic if and only if $BDIG$ is cyclic.\n\nThese three equivalences prove that $BEDI$ is cyclic if and only if $BDIG$ is cyclic, i.e. $E$ lies on $\\omega$ if and only if $G$ does. This completes the proof.\nSolution:\n\nIf $BEDI$ is cyclic, let $A'$ be the second intersection of $AI$ with $\\Omega$, so $\\angle IA'E = \\angle AA'E = 180^{\\circ} - \\angle EBA$, with $\\angle EBA = \\angle EBI + \\angle IBA = 90^{\\circ} + \\angle B/2$, so $\\angle IA'E = 90^{\\circ} - \\angle B/2$. But $\\angle EIA' = \\angle EBD$ since $EBDI$ is cyclic, with $\\angle EBD = \\angle EBI - \\angle DBI = 90^{\\circ} - \\angle B/2$. Thus $\\angle EA'I = \\angle EIA'$, so $EA'I$ is isosceles. Moreover, since $EBDI$ is cyclic and $\\angle EBI$ is a right angle, so is $\\angle IDE$. It follows that $D$ is the midpoint of $[AI]$. Next, the external bisectors $BE$ and $CF$ and the internal bisector $ID$ meet at the $A$-excentre $J$ of triangle $ABC$. By power of a point, since $BEDI$ and $BECF$ are cyclic, $|JD||JI| = |JE||JB| = |JC||JF|$, so $CFID$ is cyclic, too, with $\\angle FDI = \\angle FCI = 90^{\\circ}$. We have thus shown that $\\angle ADE = \\angle FDA = 90^{\\circ}$, so $D$ is the foot of the altitude from $A$ in triangle $AEF$. Moreover, all of this shows that $I$ is the reflection of $A'$, which is the point at which this altitude meets the circumcircle $\\Omega$ of $AEF$ again, in the side $[EF]$. Hence $I$ is the orthocentre of triangle $AEF$. By extension, $FI$ is its altitude from $F$, and $G$ is the foot of this altitude, so $\\angle EGI = 90^{\\circ} = \\angle IDE = \\angle EBI$, and hence $BEDIG$ is cyclic. This shows that if $E$ lies on $\\omega$, then so does $G$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25812,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFor each positive integer $n$, let $\\operatorname{rad}(n)$ denote the product of the distinct prime factors of $n$. Show that there exist integers $a, b>1$ such that $\\operatorname{gcd}(a, b)=1$ and\n$$\n\\operatorname{rad}(a b(a+b))<\\frac{a+b}{2024^{2024}}\n$$\nFor example, $\\operatorname{rad}(20)=\\operatorname{rad}\\left(2^{2} \\cdot 5\\right)=2 \\cdot 5=10$ and $\\operatorname{rad}(18)=\\operatorname{rad}\\left(2 \\cdot 3^{2}\\right)=2 \\cdot 3=6$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nWe show that the pair $(a, b)$ of the form $a=2^{p(p-1)}, b=3^{p(p-1)}-2^{p(p-1)}$ for sufficiently large prime number $p$ satisfies the inequality. First, notice that $\\operatorname{gcd}(a, b)=\\operatorname{gcd}(a, a+b)=1$ indeed. In addition, see that $\\operatorname{rad}(a)=2$, and $\\operatorname{rad}(a+b)=3$. Because of Euler-Fermat, as $\\phi\\left(p^{2}\\right)=p(p-1)$, it can directly be seen that $p^{2} \\mid b$. In this case, $\\operatorname{rad}(b) \\leqslant \\frac{b}{p}$. It then follows that, as rad is multiplicative for coprime numbers, that\n$$\n\\operatorname{rad}(a b(a+b))=\\operatorname{rad}(a) \\operatorname{rad}(b) \\operatorname{rad}(a+b) \\leqslant 2 \\cdot 3 \\cdot \\frac{b}{p} \\leqslant \\frac{6}{p}(a+b)\n$$\nThen, by choosing $p$ such that $\\frac{6}{p}<\\frac{1}{2024^{2024}}$, we found $a$ and $b$ satisfying the inequality.\n\nWe show that the pair $(a, b)$ of the form $a=3^{2^{k}}, b=5^{2^{k}}-3^{2^{k}}$ for sufficiently large $k$ satisfies the inequality. Again, we have that $\\operatorname{gcd}(a, b)=\\operatorname{gcd}(a, a+b)=1$. Similarly to solution 1, $\\operatorname{rad}(a(a+b))=\\operatorname{rad}(a) \\operatorname{rad}(a+b)=3 \\cdot 5=15$. Then, we will show that $2^{k+1} \\mid b$, from which it would follow that $\\operatorname{rad}(b) \\leqslant \\frac{b}{2^{k}}$. From this, we then see that\n$$\n\\operatorname{rad}(a b(a+b))=\\operatorname{rad}(a(a+b)) \\operatorname{rad}(b) \\leqslant \\frac{15 b}{2^{k}} .\n$$\nLike in solution 1, this gives a pair $(a, b)$ satisfying the inequality for sufficiently large $k$.\nThere are various ways to show that $2^{k+1} \\mid 5^{2^{k}}-3^{2^{k}}$, for example directly by applying the Lifting-The-Exponent Lemma. For a more elementary proof, we can apply induction on $k$. The statement is clearly true for $k=0$, and if the statement holds for $k=n$, then for $k=n+1$ we see that\n$$\n5^{2^{k}}-3^{2^{k}}=5^{2^{n+1}}-3^{2^{n+1}}=\\left(5^{2^{n}}\\right)^{2}-\\left(3^{2^{n}}\\right)^{2}=\\left(5^{2^{n}}-3^{2^{n}}\\right)\\left(5^{2^{n}}+3^{2^{n}}\\right) .\n$$\nFrom the induction hypothesis, the first factor has $n+1$ factors of 2. As the second factor is a sum of two odd numbers, the second term has at least one factor of 2. The product thus has at least $n+2=k+1$ factors of 2, from which the statement follows by induction.\n\nChoose $a=\\left(4^{x}-1\\right)^{2}, b=4^{x+1}$ and $a+b=\\left(4^{x}+1\\right)^{2}$. That is, $a, b$ and $a+b$ are squares, where $b$ only contains the factor 2. Note that $a$ and $b$ are indeed coprime. Then $\\operatorname{rad}(a b c)=2 \\operatorname{rad}\\left(16^{x}-1\\right)$.\nChoose then $x=5^{k}$ such that $x>2 \\cdot 2024^{2024}$. By Lifting the exponent, we know that $5^{k+1} \\mid 16^{5^{k}}-1$. This implies that $2 \\operatorname{rad}\\left(16^{x}-1\\right) \\leqslant 2\\left(16^{x}-1\\right) / 5^{k}<\\left(4^{x}+1\\right)^{2} / 2024^{2024}$.\nThis solution also works with Euler-Fermat, by choosing $x=\\phi\\left(5^{k+1}\\right)$ with $5^{k}>2024^{2024}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25813,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDoes there exist a function $f\\colon \\mathbb{R}\\to \\mathbb{R}$ such that\n$$f\\big(x^{2} + f(y)\\big) = f(x)^{2} - y$$\nfor all $x,y\\in \\mathbb{R}$?",
    "options": [],
    "answer": "No",
    "solution": "Solution:\n\nThere does not exist such a function. Let us suppose by contradiction it does. By substituting $x \\gets 0$, we get\n$$f(f(y)) = f(0)^{2} - y$$\nfor all $y\\in \\mathbb{R}$. Since the right-hand side is bijective, this implies that $f$ is also bijective. Taking $y \\gets 0$ we get\n$$f\\big(x^{2} + f(0)\\big) = f(x)^{2}$$\nfor all $x\\in \\mathbb{R}$ and so $f(-x)^{2} = f(x^{2} + f(0)) = f(x)^{2}$. Since $f$ is injective, we get $f(-x) = -f(x)$ for all $x \\neq 0$. Since $f$ is a surjection, there exists $r\\in \\mathbb{R}$ such that $f(r) = 0$. If $r \\neq 0$, $f(-r) = -f(r) = 0 = f(r)$ contradicting the fact that $f$ is injective. So $r = 0$ and $f(0) = 0$. Substituting $(x,y) \\gets (1,0)$ yields $f(1) = f(1)^{2}$ and so $f(1) = 1$ since $f(0) = 0$ and $f$ is injective. Taking $(x,y) \\gets (0,1)$, we finally get $1 = f(f(1)) = -1$ which is the desired contradiction.\n\n\nAlternative Solution:\n\nWe observe that, for all $y,z\\in \\mathbb{R}$, we have\n$$z\\geqslant f(y)\\Longrightarrow f(z)\\geq -y.$$ \nIndeed, we can take $x = \\sqrt{z - f(y)}$ and get $f(z) = f(x)^{2} - y\\geq -y$. We deduce that $\\lim_{z\\to +\\infty}f(z) = +\\infty$. Indeed, for any $K\\in \\mathbb{R}$, we have $z\\geqslant f(-K)\\Longrightarrow f(z)\\geqslant K$.\nLet us now fix $x$, and let $y\\to +\\infty$. We have $x^{2} + f(y)\\to +\\infty$, hence $f(x^{2} + f(y))\\to +\\infty$. On the other side, we have $f(x)^{2} - y\\to -\\infty$, a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25814,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $N \\geq 2$ be a natural number. At a mathematical olympiad training camp, the same $N$ courses are organised every day. Each student takes exactly one of the $N$ courses each day. At the end of the camp, every student has taken each course exactly once, and any two students took the same course on at least one day, but took different courses on at least one other day. What is, in terms of $N$, the largest possible number of students at the camp?",
    "options": [],
    "answer": "(N - 1)!",
    "solution": "Solution:\nThe largest number of students at the camp is $(N - 1)!$. Since each student takes exactly one course each day and, at the end, has taken each course exactly once, the schedule of a student can be represented by a permutation of the set of the $N$ courses.\n\nTo show that $(N - 1)!$ is possible, we can, for example, assign to each of the $(N - 1)!$ students a unique permutation of the $N - 1$ first courses and make them all take the $N$-th course on the last day. It is easy to observe that such a construction satisfies the properties of the statement.\n\nTo prove that, for a set $S$ of students, one has $|S| \\leq (N - 1)!$, one first subdivides the set of permutations into disjoint subsets of size $N$. Two permutations are said to be in the same subset if and only if one can be obtained from the other by cyclically permuting the order of the courses. Clearly, this is a well-defined subdivision since cyclically permuting twice the order of the courses can be obtained by cyclically permuting them only once. Moreover, each of these subsets contains exactly $N$ permutations since there are $N$ cycles of length $N$. There are thus $\\frac{N!}{N} = (N - 1)!$ such subsets. If $|S| > (N - 1)!$, two students will have their associated permutations in the same subset. However, they cannot be the same permutation (otherwise the two students took the same course every day), and they cannot be obtained by a non-trivial cyclic permutation from each other (otherwise the two students never took the same course).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25815,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $ABC$ be a triangle with incentre $I$ and circumcircle $\\Omega$. Let $D, E, F$ be the midpoints of the arcs $\\overline{BC}, \\overline{CA}, \\overline{AB}$ of $\\Omega$ not containing $A, B, C$, respectively. Let $D'$ be the point of $\\Omega$ diametrically opposite to $D$. Show that $I, D'$, and the midpoint $M$ of $[EF]$ lie on a line.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n![](attached_image_1.png)\nBy definition of $D$, $E$ and $F$, we know that $ID, IE$ and $IF$ are the angle bisectors of $ABC$. Using angles in $\\Omega$, we can compute  \n$$\n\\overline{EFI} = \\overline{EFC} = \\overline{EBC} = \\frac{\\overline{ABC}}{2} = 90^{\\circ} - \\frac{\\overline{ACB}}{2} - \\frac{\\overline{BAC}}{2}\n$$\n$$\n\\qquad = 90^{\\circ} - \\overline{FCB} - \\overline{BAD} = 90^{\\circ} - \\overline{FEB} - \\overline{BED} = 90^{\\circ} - \\overline{FED} = \\overline{D'EF}\n$$  \nproving that $D'E\\parallel FI$. Moreover, we can also compute  \n$$\n\\overline{FEI} = \\overline{FEB} = \\overline{FCB} = \\frac{\\overline{ACB}}{2} = 90^{\\circ} - \\frac{\\overline{ABC}}{2} - \\frac{\\overline{CAB}}{2}\n$$\n$$\n\\qquad = 90^{\\circ} - \\overline{EB} - \\overline{CAD} = 90^{\\circ} - \\overline{EFC} - \\overline{CFD} = 90^{\\circ} - \\overline{EFD} = \\overline{D'FE}\n$$  \nproving that $EI\\parallel D'F$. Therefore, $ED'FI$ is a parallelogram and its diagonals intersect in their midpoints, proving that $D'I$ contains $M$.\n\n![](attached_image_2.png)\nAlternative Solution. By definition of $D, E$ and $F$, we know that $ID, IE$ and $IF$ are the angle bisectors of $ABC$. If $D' = A$, this means that $\\overrightarrow{OAB} = 90^\\circ - \\overrightarrow{ACB} = \\frac{\\overrightarrow{CAB}}{2}$ and $ABC$ is isosceles in $A$. In that case, the symmetry with respect to $AI$ sends $E$ to $F$ and so $M \\in AI$. We can thus suppose that $D' \\neq A$.\n\nLet us consider $I_A, I_B$ and $I_C$ the excenters of $ABC$. The lines $I_B AI_C, I_C BI_A$ and $I_A CI_B$ are thus the exterior bisectors of the triangle. Hence, $I_B IC \\perp AI_A$ and, since $\\overrightarrow{D'AD} = 90^\\circ$, one has $D' \\in I_B IC$. In the triangle $I_A I_B IC$, the points $A, B$ and $C$ are the feet of the heights and $\\Omega$ is thus the Euler circle of $I_A I_B IC$. Since $D' \\in [I_B IC] \\cap \\Omega$ and $D' \\neq A$, it is the midpoint of $[I_B IC]$. Moreover, $E$ and $F$, being on the Euler circle and the heights, there are the midpoints of $[I_B I]$ and $[IC I]$ respectively. The homothety of centre $I$ and ratio $\\frac{1}{2}$ sends $I_B$ on $E$ and $IC$ on $F$. It thus also sends $D'$ on $M$, proving that $D', M$ and $I$ are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25816,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $a_{0}, a_{1}, \\ldots, a_{10}$ be integers such that, for each $i \\in \\{0,1,\\ldots,2047\\}$, there exists a subset $S \\subseteq \\{0,1,\\ldots,10\\}$ with\n$$\n\\sum_{j \\in S} a_{j} \\equiv i \\pmod{2048}.\n$$\nShow that for each $i \\in \\{0,1,\\ldots,10\\}$, there is exactly one $j \\in \\{0,1,\\ldots,10\\}$ such that $a_{j}$ is divisible by $2^{i}$ but not by $2^{i+1}$.\n\nNote: $\\sum_{j \\in S} a_{j}$ is the summation notation, for instance, $\\sum_{j \\in \\{2,5\\}} a_{j} = a_{2} + a_{5}$, while, for the empty set $\\varnothing$, one defines $\\sum_{j \\in \\varnothing} a_{j} = 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nWe denote by $\\nu_{2}(a)$ the 2-adic valuation of the integer $a$. Let us prove by induction the more general statement that, for $n \\in \\mathbb{N}_{>0}$, if $a_{0}, a_{1}, \\ldots, a_{n-1}$ are integers such that, for each $i \\in \\{0,1,\\ldots,2^{n}-1\\}$, there exists a subset $S \\subseteq \\{0,1,\\ldots,n-1\\}$ with $\\sum_{j \\in S} a_{j} \\equiv i \\pmod{2^{n}}$, then, for each $i \\in \\{0,1,\\ldots,n-1\\}$, there is exactly one $j \\in \\{0,1,\\ldots,n-1\\}$ such that $\\nu_{2}(a_{j}) = i$. The result then follows by setting $n = 11$. The case $n = 1$ is trivial.\n\nWe suppose by induction that the result is true for $n-1$ to prove it for $n \\geq 2$. Let us first notice that, since there are $2^{n}$ elements in $\\{0,1,\\ldots,2^{n}-1\\}$ and $2^{n}$ subsets of $\\{0,1,\\ldots,n-1\\}$, for each $i \\in \\{0,1,\\ldots,2^{n}-1\\}$, there exists exactly one $S_{i} \\subseteq \\{0,1,\\ldots,n-1\\}$ such that $\\sum_{j \\in S_{i}} a_{j} \\equiv i \\pmod{2^{n}}$. By summing all these sums for all $i$, we obtain\n$$\n\\sum_{i=0}^{2^{n}-1} \\sum_{j \\in S_{i}} a_{j} \\equiv \\sum_{i=0}^{2^{n}-1} i = \\frac{2^{n} \\cdot (2^{n}-1)}{2} = 2^{n-1} \\cdot (2^{n}-1) \\pmod{2^{n}}.\n$$\nFor a fixed $j \\in \\{0,1,\\ldots,n-1\\}$, $a_{j}$ appears in exactly $2^{n-1}$ of these sums (since each of the $n-1$ remaining indices may or may not be in $S_{i}$). Therefore,\n$$\n2^{n-1} \\cdot \\sum_{j=0}^{n-1} a_{j} = \\sum_{i=0}^{2^{n}-1} \\sum_{j \\in S_{i}} a_{j} \\equiv 2^{n-1} \\cdot (2^{n}-1) \\pmod{2^{n}}\n$$\nwhich is equivalent to\n$$\n\\sum_{j=0}^{n-1} a_{j} \\equiv 2^{n}-1 \\equiv 1 \\pmod{2}.\n$$\nSince $\\sum_{j=0}^{n-1} a_{j}$ is odd, at least one of the $a_{j}$'s is odd.\n\nLet us suppose by contradiction that at least two of them are odd. We now sum all the $\\sum_{j \\in S_{i}} a_{j}$ for which $\\sum_{j \\in S_{i}} a_{j}$ (or, equivalently, $i$) is even:\n$$\n\\sum_{i \\in \\{0,\\ldots,2^{n}-1\\},\\ i\\ \\text{even}} \\sum_{j \\in S_{i}} a_{j} \\equiv \\sum_{i \\in \\{0,\\ldots,2^{n}-1\\},\\ i\\ \\text{even}} i = 2^{n-1} \\cdot (2^{n-1}-1) \\pmod{2^{n}}\n$$\nFor a fixed $j \\in \\{0,1,\\ldots,n-1\\}$, $a_{j}$ appears in exactly $2^{n-2}$ of these sums. Indeed, since there is at least one $k \\in \\{0,1,\\ldots,n-1\\} \\setminus \\{j\\}$ such that $a_{k}$ is odd, one can consider any $S \\subseteq \\{0,1,\\ldots,n-1\\} \\setminus \\{j,k\\}$, add $j$ to it, and potentially also $k$ in order to make the partial sum even (exactly one possibility for each such $S$). Therefore,\n$$\n2^{n-2} \\cdot \\sum_{j=0}^{n-1} a_{j} = \\sum_{i \\in \\{0,\\ldots,2^{n}-1\\},\\ i\\ \\text{even}} \\sum_{j \\in S_{i}} a_{j} \\equiv 2^{n-1} \\cdot (2^{n-1}-1) \\pmod{2^{n}}\n$$\nor equivalently\n$$\n\\sum_{j=0}^{n-1} a_{j} \\equiv 2 \\cdot (2^{n-1}-1) \\pmod{4}.\n$$\nHence $\\sum_{j=0}^{n-1} a_{j}$ is even, which is a contradiction. Therefore, there is exactly one $k \\in \\{0,1,\\ldots,n-1\\}$ such that $a_{k}$ is odd.\n\nA sum $\\sum_{j \\in S} a_{j}$ is odd if and only if $k \\in S$. Moreover, the sums $\\sum_{j \\in S} a_{j}$ for which $k \\notin S$ cover all the even numbers modulo $2^{n}$. Thus, omitting $a_{k}$, the $n-1$ integers $\\frac{a_{0}}{2}, \\ldots, \\frac{a_{n-1}}{2}$ satisfy the condition of the statement. By the induction hypothesis, for each $i \\in \\{0,1,\\ldots,n-2\\}$, there is exactly one $j \\in \\{0,1,\\ldots,n-1\\} \\setminus \\{k\\}$ such that $\\nu_{2}\\left(\\frac{a_{j}}{2}\\right) = i$, i.e., $\\nu_{2}(a_{j}) = i+1$. Since $\\nu_{2}(a_{k}) = 0$, this concludes the proof.\nAlternative Solution:\nWe do the same induction as in the main Solution. If all $a_{j}$ are even, then all the sums are even, contradicting the hypothesis. Therefore, without loss of generality, we can assume that $a_{0}$ is odd. For each $i \\in \\{0,1,2,\\ldots,2^{n}-1\\}$, let $S_{i}' \\subseteq \\{0,1,\\ldots,n-1\\}$ be the subset such that $\\sum_{j \\in S_{i}'} a_{j} \\equiv i \\cdot a_{0} \\pmod{2^{n}}$ (this subset exists by the hypothesis and is unique since there are exactly $2^{n}$ subsets of $\\{0,1,\\ldots,n-1\\}$). By definition, one has $S_{0}' = \\emptyset$ and $S_{1}' = \\{0\\}$.\n\nLet us prove by induction that, for each $i \\in \\{0,1,\\ldots,2^{n-1}-1\\}$, one has $0 \\notin S_{2i}'$ but $0 \\in S_{2i+1}'$. The case $i=0$ is trivial. For $i>0$, if we suppose that $0 \\in S_{2i-1}'$, we prove that $0 \\notin S_{2i}'$. Indeed, if $0 \\in S_{2i}'$, then $\\sum_{j \\in S_{2i}' \\setminus \\{0\\}} a_{j} \\equiv 2i a_{0} - a_{0} = (2i-1)a_{0} \\equiv \\sum_{j \\in S_{2i-1}'} a_{j} \\pmod{2^{n}}$ and so $S_{2i}' \\setminus \\{0\\} = S_{2i-1}'$ by uniqueness. But $0 \\in S_{2i-1}'$ so this is a contradiction, proving that $0 \\notin S_{2i}'$. Then, $\\sum_{j \\in S_{2i}' \\cup \\{0\\}} a_{j} \\equiv 2i a_{0} + a_{0} = (2i+1)a_{0} \\equiv \\sum_{j \\in S_{2i+1}'} a_{j} \\pmod{2^{n}}$. By uniqueness, $S_{2i}' \\cup \\{0\\} = S_{2i+1}'$ and so $0 \\in S_{2i+1}'$.\n\nSince $a_{0}$ is odd, it is invertible modulo $2^{n}$ and so $0, a_{0}, 2a_{0}, \\ldots, (2^{n}-1)a_{0}$ is a permutation of $0,1,2,\\ldots,2^{n}-1$ modulo $2^{n}$. This shows that the subsets $S_{0}', S_{2}', S_{4}', \\ldots, S_{2^{n}-2}'$ are all distinct. There are thus $2^{n-1}$ subsets $S_{2i}'$ and they are precisely the subsets of $\\{1,2,\\ldots,n-1\\}$ (since there are also $2^{n-1}$ such subsets). Moreover, their corresponding sums are all even. Since this list of subsets contains $\\{1\\}, \\{2\\}, \\ldots, \\{n-1\\}$, this shows that $a_{1}, a_{2}, \\ldots, a_{n-1}$ are all even.\n\nWe then conclude the induction as in the main Solution.\nAlternative Solution:\nWe do the same induction as in the main Solution (except that we prove that for each $i \\in \\{0,1,\\ldots,n-1\\}$, there exists $j \\in \\{0,1,\\ldots,n-1\\}$ with $\\nu_{2}(a_{j}) = i$, uniqueness follows immediately). For the induction step, let us consider the polynomial\n$$\nP(X) = \\prod_{j=0}^{n-1} (X^{a_{j}} + 1).\n$$\nThe condition in the statement implies that\n$$\nP(X) \\equiv \\sum_{i=0}^{2^{n}-1} X^{i} \\pmod{X^{2^{n}} - 1}.\n$$\nSince\n$$\n\\sum_{i=0}^{2^{n}-1} X^{i} = \\frac{X^{2^{n}} - 1}{X-1} = \\frac{X^{2^{n}-1} - 1}{X-1} \\cdot (X^{2^{n}-1} + 1),\n$$\nwe know that $P(X)$ is divisible by $X^{2^{n}-1} + 1$. Let $\\omega \\in \\mathbb{C}$ be a primitive $2^{n}$-th root of unity. Thus $\\omega^{2^{n}-1} = -1$ and $P(\\omega) = 0$. There exists thus $j \\in \\{0,1,\\ldots,n-1\\}$ such that $\\omega^{a_{j}} = -1$. This means that $a_{j} = (2k+1)2^{n-1}$ for some integer $k$, i.e., $\\nu_{2}(a_{j}) = n-1$. It is then obvious that the other ones satisfy the property for $n-1$ and we conclude by the inductive hypothesis.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25817,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nUma lista de seis inteiros positivos $p, q, r, s, t, u$ satisfaz $p<q<r<s<t<u$. Existem exatamente 15 pares de números que podem ser formados escolhendo dois números diferentes desta lista. As somas destes 15 pares de números são:\n$$\n25,30,38,41,49,52,54,63,68,76,79,90,95,103,117\n$$\nDetermine o valor da soma $r+s$.",
    "options": [],
    "answer": "54",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25818,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO triângulo $ABC$ é isósceles de base $BC$ e $\\angle BAC = 48^\\circ$. Os pontos $D$ e $E$ estão sobre os lados $AB$ e $AC$, respectivamente, tais que $\\angle DAC = 9^\\circ$ e $\\angle EBC = 33^\\circ$. Determine a medida do ângulo $CDE$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25819,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nÉ possível dividir um tabuleiro $8 \\times 9$ em retângulos $1 \\times 6$ ?",
    "options": [],
    "answer": "No",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25820,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSeja $P$ um ponto no interior do triângulo equilátero $ABC$ tal que:\n$$\nPA = 5, \\quad PB = 7, \\quad \\text{e} \\quad PC = 8\n$$\nDetermine a medida do lado do triângulo $ABC$.",
    "options": [],
    "answer": "sqrt(129)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25821,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nConstruímos dois triângulos equiláteros: $ABE$ interno e $BFC$ externo ao quadrado $ABCD$. Prove que os pontos $D$, $E$ e $F$ se localizam na mesma reta.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25822,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nÉ dada uma circunferência $\\mathcal{C}$. Construir, usando somente compasso, o centro de $\\mathcal{C}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25823,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nArquimedes possui uma balança de dois pratos com braços de comprimentos diferentes. Objetivando pesar dois quilos de açúcar, ele procedeu da seguinte forma: colocou um peso de um quilo no prato da esquerda e açúcar no outro lado até que a balança ficasse equilibrada. Em seguida, ele esvaziou os dois pratos, colocou o peso de um quilo no prato da direita e açúcar no prato da esquerda até que os dois pratos ficassem equilibrados. Somando as duas quantidades de açúcar nestas pesagens, ele obteve menos de dois quilos, mais de dois quilos ou exatamente dois quilos? Observação: Para que ocorra o equilíbrio, os pesos nos pratos devem ser inversamente proporcionais aos comprimentos dos braços correspondentes.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "more than two kilograms",
    "solution": "Solution:\n\nDenotando o comprimento do braço da esquerda por $p$, o da direita por $q$ e por $x$ e $y$ as quantidades pesadas de açúcar na primeira e na segunda pesagem, respectivamente. Em virtude do equilíbrio, podemos escrever\n$$\n\\begin{aligned}\n& 1 \\cdot p = x \\cdot q \\\\\n& y \\cdot p = 1 \\cdot q\n\\end{aligned}\n$$\nPortanto, a quantidade total de açúcar pesado foi de\n$$\n\\begin{aligned}\n\\frac{p}{q} + \\frac{q}{p} &= \\left( \\frac{\\sqrt{p}}{\\sqrt{q}} - \\frac{\\sqrt{q}}{\\sqrt{p}} \\right)^2 + 2 \\\\\n&> 2\n\\end{aligned}\n$$\npois como os braços possuem comprimentos diferentes, temos $\\left( \\frac{\\sqrt{p}}{\\sqrt{q}} - \\frac{\\sqrt{q}}{\\sqrt{p}} \\right)^2 > 0$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25824,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAmanda desenhou a seguinte figura:\n![](attached_image_1.png)\nObserve que a soma ao longo de qualquer lado do triângulo acima é sempre a mesma, pois, como podemos verificar,\n$$\n1+3+6=6+2+2=1+7+2\n$$\na) Complete os números que faltam nos círculos da figura abaixo de modo que as somas ao longo de qualquer lado do quadrado sejam sempre as mesmas.\n![](attached_image_2.png)\nb) Encontre uma maneira de colocar os números nos círculos de maneira que as somas ao longo de qualquer linha sejam sempre as mesmas. Há mais de uma solução?\n![](attached_image_3.png)\nc) $\\mathrm{Na}$ figura abaixo, que foi desenhada apenas parcialmente (por falta de espaço!), também vale que a soma ao longo de cada segmento é sempre a mesma. Entretanto, Amanda apagou todos os números exceto os dois números mostrados na figura (3 e 4). Sabe-se que há 40 círculos no desenho. É possível descobrir quais números estavam nos círculos pintados de cinza claro e cinza escuro?\n![](attached_image_4.png)",
    "options": [],
    "answer": "a) The top-right corner is 2; the remaining missing numbers are 1 and 3, yielding equal sums of 10 along each side. b) Let the two top corners be x and y; all sums equal x plus nine plus y, and consistency forces y = x plus eleven, with bottom corners y minus six and x plus three, giving infinitely many solutions parameterized by x. c) Both highlighted gray circles have the value 4.",
    "solution": "Solution:\n\na) Na linha inferior a soma é $2+3+5=10$. Como as somas ao longo de qualquer lado são iguais, o número que falta no canto superior à direita do quadrado deve ser igual a 2, como na figura a seguir:\n![](attached_image_5.png)\nFaltam mais dois números a serem preenchidos. Novamente, como a soma deve ser 10 em qualquer lado, os números que faltam são 1 e 3, como na figura a seguir:\n![](attached_image_6.png)\n\nb) Vamos chamar de $x$ e $y$ os números a serem colocados nos cantos superiores do quadrado, como na figura a seguir:\n![](attached_image_7.png)\nAs somas devem ser constantes ao longo de qualquer lado ou diagonal desenhada. A soma ao longo do lado superior é $x+9+y$, logo todas as somas devem ser iguais a $x+9+y$. Observando as somas ao longo dos lados verticais, deduzimos que os cantos inferiores devem ser iguais a $y-6$ e $x+3$, como na figura a seguir:\n![](attached_image_8.png)\nFalta verificar a soma ao longo da diagonal desenhada e do lado horizontal inferior. A soma ao longo do lado horizontal inferior é igual a\n$$\n(y-6)+12+(x+3)=x+9+y\n$$\nverificando a soma desejada. Verificando a soma ao longo da diagonal desenhada, obtemos\n$$\nx+17+(x+3)=x+9+y\n$$\nde onde concluímos que $y=x+11$. Logo, qualquer solução será da forma\n![](attached_image_9.png)\nComo o valor de $x$ ainda não foi fixado, podemos obter muitas soluções! Por exemplo, tomando $x=0$, obtemos a solução:\n![](attached_image_10.png)\n\nc) Chamemos de $x$ e $y$ os números vizinhos aos números 3 e 4, como na figura a seguir:\n![](attached_image_11.png)\nComo a soma ao longo de cada segmento é constante, os dois próximos números devem ser iguais a 3 e 4:\n![](attached_image_12.png)\nComo há 40 círculos no desenho, há 20 círculos na linha de cima e 20 círculos na linha de baixo. Continuando o processo acima, vamos obter:\n![](attached_image_13.png)\nObserve $\\mathrm{o}\\ x$ no canto superior mais à direita. Este $x$ está ligado ao $4\\ \\mathrm{e}$ ao $y$. Como $x+4=x+y$, concluímos que $y=4$. Logo, os números nos dois círculos cinza são iguais a 4.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25825,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm heptágono regular está imprensado entre dois círculos, como mostra a figura abaixo. Seus vértices estão inscritos na circunferência maior e seus lados são tangentes à circunferência menor. Os lados do polígono medem $2~\\mathrm{cm}$ e são tangentes ao círculo menor e, além disso, seus vértices pertencem à circunferência maior. Sendo assim, qual a área do anel sombreado da figura?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "π",
    "solution": "Solution:\n\nSe $R_{1}$ e $R_{2}$ denotam os raios do círculo externo e interno ao heptágono, respectivamente, podemos concluir que a área procurada é $\\pi \\cdot R_{1}^{2} - \\pi \\cdot R_{2}^{2}$. O apótema do heptágono é igual a $R_{2}$ e forma com metade do lado e o raio $R_{1}$ um triângulo retângulo. Daí, pelo Teorema de Pitágoras, temos $R_{1}^{2} = 1^{2} + R_{2}^{2}$. Assim, a área procurada é $\\pi$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25826,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa oficina do Zé, existem seis pedaços de correntes com as seguintes quantidades de elos: $10, 10, 8, 8, 5$ e $2$. Ele precisa unir estes pedaços para formar uma corrente circular. Ele gasta 1 minuto para cortar um elo e 2 minutos para uni-lo, perfazendo um total de 3 minutos por elo. Se ele cortar um elo ao final de cada peça separada, unindo as peças uma de cada vez, ele demoraria $6 \\cdot 3 = 18$ minutos. Entretanto, como ele está com pressa, ele pretende realizar esta operação de uma forma mais rápida.\n\na) Diga como ele pode formar a corrente circular gastando apenas 15 minutos.\n\nb) É possível ele fazer tal operação em menos de 15 minutos?",
    "options": [],
    "answer": "a) Open all links of the five-link segment and use each opened link to connect the remaining segments into a loop; this takes 15 minutes. b) No, it is not possible to complete the task in less than 15 minutes.",
    "solution": "Solution:\n\na) Uma maneira para ele formar a corrente em 15 minutos é inicialmente abrir todos os elos do pedaço de 5 elos. Neste procedimento ele gastará $5 \\cdot 1 = 5$ minutos. Em seguida, ele deve usar cada um destes elos abertos entre os 5 pedaços de correntes restante, usando exatamente um elo para unir dois pedaços distintos. Para este último procedimento ele gastará $5 \\cdot 2 = 10$ minutos.\n\nb) O tempo gasto sempre é um múltiplo de 3 porque uma vez que se gasta 1 minuto para se abrir um elo, é necessário gastarmos 2 minutos para fechá-lo e assim o tempo total por elo alterado é $1 + 2 = 3$ minutos. Como $12 = 3 \\cdot 4$ é o maior múltiplo de 3 menor que 15, se fosse possível ele gastar menos de 15 minutos, ele teria que alterar no máximo 4 elos. Como existem 6 pedaços de correntes, pelo menos dois deles não teriam elos alterados. Para unir estes dois pedaços na corrente maior, precisaríamos de pelo menos 3 elos alterados inseridos em suas extremidades. Assim, poderíamos usar apenas um elo para conectar os pedaços restantes. Como ele terá pelo menos três pedaços restantes, isto é impossível.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25827,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\na) Uma minhoca anda sempre sobre uma linha reta. Todos os dias, ela avança $5\\ \\mathrm{m}$ e recua $3\\ \\mathrm{m}$. Ao final de $15$ dias, a minhoca estará a que distância do ponto de partida?\n\nb) Ao final destes $15$ dias, quantos metros terá caminhado, no total, esta minhoca?\n\nc) Uma outra minhoca anda também sobre uma linha reta, porém de maneira diferente da primeira. No primeiro dia, ela anda $1\\ \\mathrm{m}$ para a frente e $1/2\\ \\mathrm{m}$ para trás. No segundo dia, ela anda $1/2\\ \\mathrm{m}$ para a frente e $1/3\\ \\mathrm{m}$ para trás. No terceiro dia, ela anda $1/3\\ \\mathrm{m}$ para a frente e $1/4\\ \\mathrm{m}$ para trás, e assim sucessivamente. Quantos metros ela terá andado após $1000$ dias?\n\nd) Algum dia esta segunda minhoca conseguirá estar a $2\\ \\mathrm{m}$ de distância do ponto inicial?",
    "options": [],
    "answer": "a) 30 m; b) 120 m; c) 1000/1001 m; d) No",
    "solution": "Solution:\na) A cada dia, a minhoca avança $5$ metros e recua $3$ metros. Portanto, no final do dia a minhoca avança $5-3=2$ metros. Depois de $15$ dias, a minhoca estará a $15 \\times 2 = 30$ metros do ponto de partida.\n\nb) A cada dia, a minhoca anda $5+3=8$ metros. Portanto, após estes $15$ dias, a minhoca terá andado $8 \\times 15 = 120$ metros.\n\nc) Esta minhoca anda da seguinte maneira:\n![](attached_image_1.png)\nPortanto, após $1000$ dias, a minhoca estará a $1-\\frac{1}{1001}=\\frac{1000}{1001}$ metros da posição inicial.\n\nd) Como no $n$-ésimo dia a minhoca está a uma distância de $1-\\frac{1}{n+1}$ metros da posição inicial, ela nunca estará a $2$ metros da posição inicial, pois, para qualquer número natural $n$, temos:\n$$\n1-\\frac{1}{n+1}<1<2\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25828,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDois divisores - O número $2^{48} - 1$ é divisível por dois números compreendidos entre 60 e 70. Quais são esses números?\n(a) 61 e 63\n(b) 61 e 65\n(c) 63 e 65\n(d) 63 e 67\n(e) 67 e 69",
    "options": [],
    "answer": "(c)",
    "solution": "Solution:\n\nLembre que\n$$\na^{4}-1=(a-1)\\left(a^{3}+a^{2}+a+1\\right)\n$$\nLogo, se $a=2^{12}$, temos:\n$$\n2^{48}-1=\\left(2^{12}\\right)^{4}-1=\\left(2^{12}-1\\right)\\left(2^{36}+2^{24}+2^{12}+1\\right)\n$$\ne $2^{12}-1=\\left(2^{6}+1\\right)\\left(2^{6}-1\\right)=65 \\times 63$. A opção correta é (c).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25829,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGatos no condomínio - Em um condomínio moram 29 famílias, cada uma delas possui ou 1 gato ou 3 gatos ou 5 gatos. O número de famílias que possuem apenas 1 gato é o mesmo que o de famílias que possuem 5 gatos. Quantos gatos tem esse condomínio?",
    "options": [],
    "answer": "87",
    "solution": "Solution:\n\nSejam:\n$$\n\\begin{aligned}\n& x=\\text{ número de famílias que possuem apenas } 1 \\text{ gato;} \\\\\n& y=\\text{ número de famílias que possuem exatamente } 3 \\text{ gatos;} \\\\\n& z=\\text{ número de famílias que possuem } 5 \\text{ gatos. }\n\\end{aligned}\n$$\nSegue então que $x+y+z=29$ e $x=z$. Portanto, $2x+y=29$. Por outro lado, o número de gatos é $x+3y+5z$. Daí temos:\n$$\n\\text{número de gatos}=x+3y+5z=6x+3y=3(2x+y)=3 \\times 29=87.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25830,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPartindo do número $265863$ e utilizando uma única vez cada uma das operações $+$, $-$, $\\times$, $:$, e também uma única vez os números $51$, $221$, $6817$, $13259$, podemos obter vários números, por exemplo $54911$:\n\n$265863 \\xrightarrow{+221} 1203 \\xrightarrow{\\times 51} 61353 \\xrightarrow{-13259} 48094 \\xrightarrow{+6817} 54911$\n\nEncontre a cadeia que permite obter o menor número inteiro positivo.",
    "options": [],
    "answer": "265863 / 6817 = 39, then +221 → 260, then ×51 → 13260, then −13259 → 1",
    "solution": "Solution:\n\n$265863 \\xrightarrow{\\div 6817} 39 \\xrightarrow{+221} 260 \\xrightarrow{\\times 51} 13260 \\xrightarrow{-13259} 1$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25831,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nQuem é menor? - Sem usar calculadora, decida qual dos números $33^{12}$, $63^{10}$ e $127^{8}$ é o menor.",
    "options": [],
    "answer": "127^8",
    "solution": "Solution:\nObservemos que:\n$$\n\\begin{aligned}\n& 33^{12} > 32^{12} = \\left(2^{5}\\right)^{12} = 2^{60} \\\\\n& 63^{10} < 64^{10} = \\left(2^{6}\\right)^{10} = 2^{60} \\\\\n& 127^{8} < 128^{8} = \\left(2^{7}\\right)^{8} = 2^{56}\n\\end{aligned}\n$$\nLogo, o maior dos números é $33^{12}$.\n\nPor outro lado, $\\frac{127}{63} = 2 + \\frac{1}{63} < 2,1$. Logo:\n$$\n\\left(\\frac{127}{63}\\right)^{2} < 2,1^{2} < 7 \\text{ e } \\left(\\frac{127}{63}\\right)^{4} < 49 < 63 \\Rightarrow 127^{4} < 63^{5} \\Rightarrow 127^{8} < 63^{10}\n$$\nLogo, o menor dos três números dados é $127^{8}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25832,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEm 1950 um \"profeta\" anunciou que o fim do mundo ocorreria em 11.08.1999 (11 de agosto de 1999). Como nada aconteceu nesse dia, ele refez seus cálculos e fez a seguinte previsão: \"O fim do mundo ocorrerá na próxima data que se escreve com 8 algarismos diferentes.\" Você pode descobrir essa data?",
    "options": [],
    "answer": "17.06.2345",
    "solution": "Solution:\n\nResposta: 17.06.2345",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25833,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFração de fração - Qual o valor de $1+\\frac{1}{1+\\frac{1}{1+\\frac{1}{2}}}$ ?",
    "options": [],
    "answer": "8/5",
    "solution": "Solution:\n\nFração de fração - Temos:\n$$\n1+\\frac{1}{1+\\frac{1}{1+\\frac{1}{2}}}=1+\\frac{1}{1+\\frac{1}{\\frac{3}{2}}}=1+\\frac{1}{1+\\frac{2}{3}}=1+\\frac{1}{\\frac{5}{3}}=1+\\frac{3}{5}=\\frac{8}{5}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25834,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNuma classe, $40\\%$ dos alunos não enxergam bem. Desses, $70\\%$ usam óculos e os $30\\%$ restantes usam lentes de contato. Sabendo que 21 alunos usam óculos, quantos alunos tem essa classe?",
    "options": [],
    "answer": "75",
    "solution": "Solution:\n\nSeja $A$ o número total de alunos da sala. Logo, $\\frac{40}{100} \\times A$ não enxergam bem. Portanto, $\\frac{70}{100} \\times \\frac{40}{100} \\times A$ usam óculos. Consequentemente, temos que:\n$$\n\\frac{70}{100} \\times \\frac{40}{100} \\times A = 21 \\Rightarrow A = \\frac{21 \\times 100}{7 \\times 4} = 3 \\times 25 = 75\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25835,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSe $\\frac{x}{y}=2$, então $\\frac{x-y}{x}$ é igual a:\n(a) -1\n(b) $-\\frac{1}{2}$\n(c) $\\frac{1}{2}$\n(d) 1\n(e) 2",
    "options": [],
    "answer": "c",
    "solution": "Solution:\nTemos:\n$$\n\\frac{x-y}{x}=\\frac{x}{x}-\\frac{y}{x}=1-\\frac{y}{x}\n$$\nComo $\\frac{x}{y}=2$ temos que $\\frac{y}{x}=\\frac{1}{2}$, assim\n$$\n\\frac{x-y}{x}=\\frac{x}{x}-\\frac{y}{x}=1-\\frac{1}{2}=\\frac{1}{2}\n$$\nA opção correta é (c).\n\nSolução 2: Se $\\frac{x}{y}=2$, então $x=2 y$. Logo\n$$\n\\frac{x-y}{x}=\\frac{2 y-y}{2 y}=\\frac{y}{2 y}=\\frac{1}{2}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25836,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNo jogo - Aldo, Bernardo e Carlos jogam baralho. No início, a quantia em dinheiro que eles tinham estava na proporção $7:6:5$. No final do jogo, a proporção era $6:5:4$. Um dos jogadores ganhou 1200 reais. Qual a quantidade de dinheiro com que ficou cada jogador, no final da partida?",
    "options": [],
    "answer": "Aldo: 43200; Bernardo: 36000; Carlos: 28800",
    "solution": "Solution:\n\nSeja $T$ a quantidade total de dinheiro no jogo. Assim, no início, os jogadores possuíam:\n$$\n\\begin{array}{ll}\n\\text{Aldo:} & \\frac{7}{18} T \\\\\n\\text{Bernardo:} & \\frac{6}{18} T \\\\\n\\text{Carlos:} & \\frac{5}{18} T\n\\end{array}\n$$\nNo final eles possuíam:\n$$\n\\begin{array}{ll}\n\\text{Aldo:} & \\frac{6}{15} T \\\\\n\\text{Bernardo:} & \\frac{5}{15} T \\\\\n\\text{Carlos:} & \\frac{4}{15} T\n\\end{array}\n$$\nPara melhor comparar essas frações, coloquemo-las com um denominador comum:\nNo início:\n$$\n\\begin{array}{llrl}\n\\text{Aldo:} & & \\frac{7}{18} T & =\\frac{35}{90} T \\\\\n& \\text{Bernardo:} & \\frac{6}{18} T & =\\frac{30}{90} T \\\\\n\\text{Carlos:} & & \\frac{5}{18} T & =\\frac{25}{90} T .\n\\end{array}\n$$\nNo final:\n$$\n\\begin{aligned}\n& \\text{Aldo:} \\quad \\frac{6}{15} T=\\frac{36}{90} T \\\\\n& \\text{Bernardo:} \\quad \\frac{5}{15} T=\\frac{30}{90} T \\\\\n& \\text{Carlos:} \\quad \\frac{4}{15} T=\\frac{24}{90} T\n\\end{aligned}\n$$\nLogo, Carlos perdeu $1/90$ do total e Aldo ganhou $1/90$. Portanto, $1200$ corresponde a $1/90$ do total de dinheiro. Portanto, o total $T$ de dinheiro no início do jogo é:\n$$\n\\frac{1}{90} T=1200 \\Rightarrow T=90 \\times 1200=108000\n$$\nAssim, no final da partida os jogadores possuíam:\n\nAldo: $\\quad \\frac{36}{90}$ de $108000=43200$\n\nBernardo: $\\quad \\frac{30}{90}$ de $108000=36000$\n\nCarlos: $\\quad \\frac{24}{90}$ de $108000=28800$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25837,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDetermine todas as soluções da equação $\\sqrt{x} = x - 2$.",
    "options": [],
    "answer": "x = 4",
    "solution": "Solution:\n\nElevando ambos os membros da equação ao quadrado, obtemos $x = x^{2} - 4x + 4$, que é equivalente a $x^{2} - 5x + 4 = 0$. As raízes dessa equação do segundo grau são $x = 1$ e $x = 4$. Entretanto, quando substituímos $x = 1$ na equação original $\\sqrt{x} = x - 2$ obtemos $\\sqrt{1} = -1$, que é falso. No entanto, quando substituímos $x = 4$ obtemos $\\sqrt{4} = 2$, que é verdadeiro. Portanto, a equação dada possui $x = 4$ como única solução.\n\nAtenção: O aparecimento da \"solução estranha\" $x = 1$ deve-se ao fato que a implicação\n$$\na^{2} = b^{2} \\Rightarrow a = b\n$$\nnão é verdadeira em geral. O correto é\n$$\na^{2} = b^{2} \\Rightarrow a = \\pm b\n$$\nDeste modo, quando elevamos os dois membros de uma equação ao quadrado, obtemos uma nova equação que pode, eventualmente, conter mais soluções que a equação original. Você também pode ver isso com clareza, por exemplo, nas equações: $x = 1$ e $x^{2} = 1^{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25838,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUma expressão - A expressão $\\frac{a^{-2}}{a^{5}} \\times \\frac{4 a}{\\left(2^{-1} a\\right)^{-3}}$ onde $a \\neq 0$, é igual a:\n(a) $\\frac{a^{3}}{2}$\n(b) $\\frac{2}{a^{3}}$\n(c) $\\frac{1}{2 a^{3}}$\n(d) $\\frac{a^{5}}{2}$\n(e) $\\frac{2}{a^{5}}$",
    "options": [],
    "answer": "c",
    "solution": "Solution:\n\nTemos:\n$$\n\\begin{aligned}\n\\frac{a^{-2}}{a^{5}} \\times \\frac{4 a}{\\left(2^{-1} a\\right)^{-3}} & = a^{-2-5} \\times \\frac{2^{2} a}{2^{3} a^{-3}} \\\\\n& = a^{-7} \\times \\frac{a^{1-(-3)}}{2} \\\\\n& = a^{-7} \\times \\frac{a^{4}}{2} \\\\\n& = \\frac{a^{-7+4}}{2} = \\frac{a^{-3}}{2} = \\frac{1}{2 a^{3}}\n\\end{aligned}\n$$\nA opção correta é (c)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25839,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPar perfeito - Dizemos que 2 números naturais formam um par perfeito quando a soma e o produto desses dois números são quadrados perfeitos. Por exemplo, $5$ e $20$ formam um par perfeito, pois $5+20=25=5^{2}$ e $5 \\times 20=100=10^{2}$. Será que $122$ forma um par perfeito com outro natural?",
    "options": [],
    "answer": "Yes; 14762",
    "solution": "Solution:\n\nChamemos de $n$ o natural \"candidato\" a formar um par perfeito com $122$. Então, devemos ter: $122+n=A^{2}$ e $122 \\times n=B^{2}$ onde $A$ e $B$ são números naturais.\n\nComo $B^{2}=2 \\times 61 \\times n$, concluímos que $n$ tem também os fatores primos $2$ e $61$. Logo, podemos escrever $n$ como $n=2 \\times 61 \\times m^{2}=122~m^{2}$.\n\nObtemos então $A^{2}=122+122 m^{2}=122\\left(1+m^{2}\\right)$. O menor valor de $\\left(1+m^{2}\\right)$ que satisfaz esta igualdade é $1+m^{2}=122$, ou seja, $m^{2}=121$. Daí segue que $m=11$. Consequentemente, $n=122 \\times 121$ e temos:\n$$\nA^{2}=122+122 \\times 121=122^{2} \\quad \\text{e} \\quad B^{2}=122 \\times 122 \\times 121=(122 \\times 11)^{2}\n$$\nLogo, $122$ e $122 \\times 121$ formam um par perfeito.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25840,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMaria está planejando participar do Triatlon-Brasil que começa às 24 horas de domingo e consta de $800~\\mathrm{m}$ a nado, seguido de $20~\\mathrm{km}$ de bicicleta e finalmente $4~\\mathrm{km}$ de corrida. Maria corre a uma velocidade constante e que é o triplo da velocidade que nada, e pedala 2,5 vezes mais rápido do que corre. Para terminar a prova em no máximo 1 hora e 20 minutos, quanto tempo ela deve gastar em cada uma das 3 etapas?",
    "options": [],
    "answer": "Swimming: 40/3 minutes; Running: 200/9 minutes; Cycling: 400/9 minutes.",
    "solution": "Solution:\n\nSeja $x$ a velocidade em metros por minuto com que Maria nada. Logo, a sua velocidade na corrida é $3x$ e na bicicleta $2,5 \\times 3x = 7,5x$. Logo, o tempo total que ela gastará nas 3 etapas é:\n$$\n\\underbrace{\\frac{800}{x}}_{\\text{nadando}} + \\underbrace{\\frac{4000}{3x}}_{\\text{correndo}} + \\underbrace{\\frac{20000}{7,5x}}_{\\text{pedalando}} = \\frac{800 \\times 7,5 + 4000 \\times 7,5/3 + 20000}{7,5x}\n$$\nNo entanto, a soma correta é:\n$$\n\\frac{800}{x} + \\frac{4000}{3x} + \\frac{20000}{7,5x} = \\frac{800 \\times 7,5 + 4000 \\times 2,5 + 20000}{7,5x} = \\frac{60000}{7,5x}\n$$\nLogo, para que ela vença as 3 etapas em 1 hora e 20 minutos ($=80~\\mathrm{min}$), devemos ter:\n$$\n\\frac{60000}{7,5x} = 80 \\Rightarrow x = \\frac{60000}{7,5 \\times 80} = 100~\\mathrm{m}/\\mathrm{min}\n$$\nSegue que\n$$\n3x = 300~\\mathrm{m}/\\mathrm{min} \\text{ e } 7,5x = 750~\\mathrm{m}/\\mathrm{min}\n$$\nAssim, para que Maria termine a prova em no máximo 1 hora e 20 minutos, ela deve desenvolver as seguintes velocidades:\n- nadar: a uma velocidade mínima de $100~\\mathrm{m}/\\mathrm{min}$;\n- correr: a uma velocidade mínima de $300~\\mathrm{m}/\\mathrm{min}$;\n- pedalar: a uma velocidade mínima de $750~\\mathrm{m}/\\mathrm{min}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25841,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nQuantos pares de inteiros positivos $(x, y)$ são soluções da equação $3x + 5y = 501$?",
    "options": [],
    "answer": "33",
    "solution": "Solution:\n\nA equação dada é equivalente a $y = \\frac{3(167 - x)}{5}$. Como $y$ é um inteiro positivo, $167 - x$ deve ser um múltiplo positivo de $5$, ou seja:\n$$\n167 - x = 5k \\quad \\Rightarrow \\quad x = 167 - 5k\n$$\n\n$$\nx = 5 \\times 33 + 2 - 5k \\quad \\Rightarrow \\quad x = 5(33 - k) + 2\n$$\n\nonde $k$ é um inteiro positivo. Como $x$ é positivo, devemos ter $k \\leq 33$. Consequentemente, $k = 1, 2, \\ldots, 33$, o que nos garante $33$ soluções para o problema proposto.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25842,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAs duas partículas, $A$ e $B$, percorrem uma circunferência de $120\\ m$ de comprimento. A partícula $A$ gasta 3 segundos menos que $B$, por estar animada com uma velocidade maior de $2$ metros por segundo. Qual é a velocidade de cada partícula?",
    "options": [],
    "answer": "Particle A: 10 m/s; Particle B: 8 m/s",
    "solution": "Solution:\n\nSeja $v_A$ a velocidade da partícula $A$ e $v_B$ a velocidade da partícula $B$.\n\nSabemos que:\n- $v_A = v_B + 2$\n- O tempo gasto por $A$ é $t_A = \\dfrac{120}{v_A}$\n- O tempo gasto por $B$ é $t_B = \\dfrac{120}{v_B}$\n- $A$ gasta 3 segundos a menos: $t_B = t_A + 3$\n\nSubstituindo as expressões dos tempos:\n$$\n\\frac{120}{v_B} = \\frac{120}{v_A} + 3\n$$\nSubstituindo $v_A = v_B + 2$:\n$$\n\\frac{120}{v_B} = \\frac{120}{v_B + 2} + 3\n$$\nMultiplicando ambos os lados por $v_B(v_B + 2)$:\n$$\n120(v_B + 2) = 120v_B + 3v_B(v_B + 2)\n$$\nExpandindo:\n$$\n120v_B + 240 = 120v_B + 3v_B^2 + 6v_B\n$$\nSubtraindo $120v_B$ dos dois lados:\n$$\n240 = 3v_B^2 + 6v_B\n$$\nDividindo ambos os lados por 3:\n$$\n80 = v_B^2 + 2v_B\n$$\n$$\nv_B^2 + 2v_B - 80 = 0\n$$\nResolvendo a equação do segundo grau:\n$$\nv_B = \\frac{-2 \\pm \\sqrt{4 + 320}}{2} = \\frac{-2 \\pm \\sqrt{324}}{2} = \\frac{-2 \\pm 18}{2}\n$$\nComo a velocidade não pode ser negativa:\n$$\nv_B = \\frac{16}{2} = 8\\ m/s\n$$\nLogo,\n$$\nv_A = v_B + 2 = 8 + 2 = 10\\ m/s\n$$\n\nPortanto, as velocidades são:\n\n- $v_A = 10\\ m/s$\n- $v_B = 8\\ m/s$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25843,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n1. Potências de 10 - O valor de $\\frac{0,00001 \\times (0,01)^{2} \\times 1000}{0,001}$ é:\n(a) $10^{-1}$\n(b) $10^{-2}$\n(c) $10^{-3}$\n(d) $10^{-4}$\n(e) 1",
    "options": [],
    "answer": "c",
    "solution": "Solution:\n\n1. Potências de 10 - Temos:\n$$\n\\begin{aligned}\n\\frac{0,00001 \\times (0,01)^{2} \\times 1000}{0,001} & = \\frac{10^{-5} \\times \\left(10^{-2}\\right)^{2} \\times 10^{3}}{10^{-3}} = \\frac{10^{-5} \\times 10^{-4} \\times 10^{3}}{10^{-3}} = \\\\\n& = \\frac{10^{-5+(-4)+3}}{10^{-3}} = \\frac{10^{-6}}{10^{-3}} = 10^{-6-(-3)} = 10^{-3}\n\\end{aligned}\n$$\nA opção correta é (c).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25844,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNúmero curioso - O número $81$ tem a seguinte propriedade: ele é divisível pela soma de seus algarismos $8+1=9$. Quantos números de dois algarismos cumprem esta propriedade?",
    "options": [],
    "answer": "23",
    "solution": "Solution:\n\nSeja $ab$ um tal número. Por hipótese, $ab = 10a + b$ é divisível por $a + b$. Logo, a diferença $(10a + b) - (a + b) = 9a$, também é divisível por $a + b$. Além disso, sabemos que $10a + b$ é divisível por $a + b$ se, e somente se, $(10a + b) - (a + b) = 9a$ é divisível por $a + b$ (prove isso).\n\nAntes de prosseguirmos na solução, note que como $ab$ é um número de dois algarismos então $a \\neq 0$. Agora, basta atribuir valores para $a$ e calcular os valores de $b$ para os quais $a + b$ divide $9a$. O resultado é mostrado na tabela a seguir.\n\n| $a$ | $9a$ | $b$ |\n| :---: | :---: | :---: |\n| 1 | 9 | $0,2,8$ |\n| 2 | 18 | $0,1,4,7$ |\n| 3 | 27 | 0,6 |\n| 4 | 36 | $0,2,5,8$ |\n| 5 | 45 | 0,4 |\n| 6 | 54 | 0,3 |\n| 7 | 63 | 0,2 |\n| 8 | 72 | $0,1,4$ |\n| 9 | 81 | 0 |\n\nLogo os números que satisfazem a propriedade são:\n\n$10, 12, 18, 20, 21, 24, 27, 30, 36, 40, 42, 45, 48, 50, 54, 60, 63, 70, 72, 80, 81, 84, 90$\n\nOu seja, existem $23$ números nas condições exigidas.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25845,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nQuatro números inteiros - Se quatro inteiros positivos distintos $m, n, p$ e $q$ satisfazem a equação\n$$\n(7-m)(7-n)(7-p)(7-q)=4\n$$\nentão a soma $m+n+p+q$ é igual a:\n(a) 10\n(b) 21\n(c) 24\n(d) 26\n(e) 28",
    "options": [],
    "answer": "(e)",
    "solution": "Solution:\n\nComo $m, n, p$ e $q$ são inteiros, então $7-m$, $7-n$, $7-p$ e $7-q$ também são inteiros. Agora,\n$$\n4 = (-1) \\times (-2) \\times 1 \\times 2\n$$\né a única decomposição de $4$ em um produto de números inteiros distintos. Segue que\n$$\n(7-m)+(7-n)+(7-p)+(7-q)=(-1)+(-2)+1+2\n$$\ne daí obtemos $m+n+p+q=28$. A opção correta é (e).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25846,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm número é dito palindrome se a leitura da direita para a esquerda é igual a da esquerda para a direita. Por exemplo, os números $23432$ e $18781$ são palindromes. Quantos números palindromes de 4 algarismos são divisíveis por $9$?",
    "options": [],
    "answer": "10",
    "solution": "Solution:\n\nUm número palindrome de 4 algarismos é da forma: $a b b a$, onde $a$ é um algarismo entre $1$ e $9$ e $b$ é um algarismo entre $0$ e $9$. Como o número é divisível por $9$, então a soma de seus algarismos: $2a + 2b = 2(a + b)$ é divisível por $9$, ou seja, $a + b$ é divisível por $9$.\n\nSe $a + b = 9$, temos as $9$ soluções:\n$a = 1$ e $b = 8$ ; $a = 2$ e $b = 7$ ; $a = 3$ e $b = 6$ ; $a = 4$ e $b = 5$\n$a = 5$ e $b = 4$ ; $a = 6$ e $b = 3$ ; $a = 7$ e $b = 2$ ; $a = 8$ e $b = 1$\n$a = 9$ e $b = 0$.\n\nSe $a + b = 18$ então a única solução é: $a = b = 9$.\n\nLogo, o número de palindromes de 4 algarismos divisíveis por $9$ é $10$, são eles: $1881$, $2772$, $3663$, $4554$, $5445$, $6336$, $7227$, $8118$, $9009$ e $9999$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25847,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa figura $ABCD$ é um quadrado cujo lado mede $1~\\mathrm{cm}$, $E$ é o ponto médio da diagonal $BD$ e $F$ o ponto médio do segmento $BE$. Qual é a área do triângulo $\\triangle BCF$?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "0.125 cm^2",
    "solution": "Solution:\n\nAs diagonais do quadrado $ABCD$ dividem o quadrado em 4 triângulos iguais, logo a área do triângulo $\\triangle BCE$ é\n\n![](attached_image_2.png)\n\n$$\n1 \\div 4 = 0,25~\\mathrm{cm}^2\n$$\n\nComo o comprimento de $BF$ é a metade de $BE$ e a altura relativa aos lados $BF$ e $BE$ é $CE$, então a área do triângulo $\\triangle CBF$ é a metade da área do triângulo $\\triangle CBE$. Assim, a área de dito triângulo é $0,25 \\div 2 = 0,125~\\mathrm{cm}^2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25848,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n1. Carro flex - Um carro é denominado flex se ele pode ser abastecido com gasolina ou com álcool. Considere que os preços do álcool e da gasolina sejam, respectivamente, $\\mathrm{R}\\$ 1,59$ e $\\mathrm{R}\\$ 2,49$ por litro.\n\na. Suponha que um carro flex rode $12,3~\\mathrm{km}$ por litro de gasolina, que indicamos $12,3~\\mathrm{km}/\\mathrm{l}$. Qual deve ser a relação $\\mathrm{km}/\\mathrm{l}$ desse carro, para o álcool, para que a utilização do álcool seja financeiramente mais vantajosa que a de gasolina?\n\nb. Se o desempenho de um carro flex é de $x~\\mathrm{km}/\\mathrm{l}$ com gasolina e de $\\left(\\frac{x}{2}+1\\right)~\\mathrm{km}/\\mathrm{l}$ com álcool, escreva a expressão da função $g(x)$ que fornece o custo desse carro rodar $100~\\mathrm{km}$ utilizando gasolina e a expressão da função $a(x)$ que fornece o custo desse carro rodar $100~\\mathrm{km}$ utilizando álcool.\n\nc. Para que o custo seja o mesmo, tanto com álcool como com gasolina, qual deve ser a relação $\\mathrm{km}/\\mathrm{l}$ para a gasolina e para o álcool?\n\nd. Em que condição o uso do álcool é mais vantajoso, financeiramente, que o da gasolina? Dê um exemplo numérico que satisfaça a condição.",
    "options": [],
    "answer": "a) Alcohol must achieve y > 7.85 km/l.\nb) g(x) = 249/x and a(x) = 318/(x+2).\nc) Equal cost occurs at gasoline x ≈ 7.22 km/l and alcohol (x/2 + 1) ≈ 3.61 km/l.\nd) Condition: y > (1.59/2.49) x ≈ 0.64 x. Example: if x = 10 km/l, then y must exceed 6.4 km/l.",
    "solution": "Solution:\n\na. Com gasolina o carro faz $\\frac{12,3}{2,49}=4,94~\\mathrm{km}$ por $\\mathrm{R}\\$ 1,00$. Para que o álcool seja mais vantajoso precisamos que o carro rode, com álcool, mais que $4,94~\\mathrm{km}$ com $\\mathrm{R}\\$ 1,00$. Logo, se o desempenho com álcool é $y~\\mathrm{km}/\\mathrm{l}$, precisamos que $\\frac{y}{1,59}>4,94$, o que implica $y>7,85$. Ou seja, o desempenho com álcool deve ser maior que $7,85~\\mathrm{km}/\\mathrm{l}$.\n\nb. Observe que $g(x)=2,49 \\frac{100}{x}=\\frac{249}{x}$ e $a(x)=1,59 \\frac{100}{\\frac{x}{2}+1}=\\frac{318}{x+2}$.\n\nc. Precisamos ter $a(x)=g(x)$, ou seja, $\\frac{249}{x}=\\frac{318}{x+2}$, o que leva a $x=7,22~\\mathrm{km}/\\mathrm{l}$, que deve ser o desempenho com gasolina. Com álcool, o carro deve fazer\n$$\n\\frac{7,22}{2}+1=3,61~\\mathrm{km}/\\mathrm{l}\n$$\n\nd. Supondo que o desempenho do carro seja $x~\\mathrm{km}/\\mathrm{l}$ com gasolina e $y~\\mathrm{km}/\\mathrm{l}$ com álcool e pensando em um percurso de $L~\\mathrm{km}$, devemos ter o custo com gasolina maior que o custo com álcool:\n$$\n2,49 \\frac{L}{x}>1,59 \\frac{L}{y} \\Rightarrow 2,49 y>1,59 x \\Rightarrow y>0,64 x\n$$\npois $x$ e $y$ são valores positivos.\nUm exemplo é um carro que faz $10~\\mathrm{km}/\\mathrm{l}$ com gasolina, teria que fazer mais que $6,4~\\mathrm{km}/\\mathrm{l}$ com álcool para que o uso do álcool seja mais vantajoso.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25849,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEm uma circunferência de raio $10~\\mathrm{cm}$, o segmento $AB$ é um diâmetro e o segmento $AC$ é uma corda de $12~\\mathrm{cm}$. Determine a distância entre os pontos $B$ e $C$.",
    "options": [],
    "answer": "16",
    "solution": "Solution:\n\nSendo $AB$ um diâmetro, o triângulo $\\triangle ABC$ está inscrito numa semicircunferência. Isto implica que este triângulo é retângulo no vértice $C$. Pelo Teorema de Pitágoras,\n$$\nBC^{2} = AB^{2} - AC^{2}\n$$\nou seja,\n$$\nBC^{2} = 20^{2} - 12^{2} = 256 = 16^{2}\n$$\n![](attached_image_1.png)\nAssim, obtemos que $BC = 16$.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25850,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEnchendo uma piscina - Uma piscina vazia foi abastecida de água por duas torneiras $A$ e $B$, ambas com vazão constante. Durante 4 horas, as duas torneiras ficaram abertas e encheram $50\\%$ da piscina. Em seguida, a torneira $B$ foi fechada e durante 2 horas a torneira $A$ encheu $15\\%$ do volume da piscina. Após este período a torneira $A$ foi fechada e a torneira $B$ aberta. Durante quanto tempo esta torneira teve de ficar aberta para que ela sozinha terminasse de encher a piscina?",
    "options": [],
    "answer": "7 hours",
    "solution": "Solution:\n\nComo as torneiras $A$ e $B$ despejam água na piscina com vazão constante, o volume de água despejado na piscina por cada torneira é proporcional ao tempo em que ela fica aberta. Assim, se durante 2 horas a torneira $A$ enche $15\\%$ do volume da piscina, então em 4 horas ela encherá $30\\%$ do volume da piscina.\n\nMas, quando as torneiras $A$ e $B$ ficam simultaneamente abertas durante 4 horas, elas conseguem encher $50\\%$ do volume da piscina. Daí temos que a torneira $B$ enche $50\\% - 30\\% = 20\\%$ do volume da piscina em 4 horas.\n\nPara saber quanto tempo a torneira $B$ deve ficar aberta para encher os $35\\%$ restantes do volume da piscina, basta utilizar a proporção:\n\n$$\n\\begin{array}{ccc}\n\\text{horas} & \\rightarrow & \\text{percentual} \\\\\n4 & \\rightarrow & 20\\% \\\\\nx & \\rightarrow & 35\\%\n\\end{array}\n$$\n\nLogo, a torneira $B$ gastará $x = \\frac{35 \\times 4}{20} = 7$ horas para encher os $35\\%$ restantes.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 25851,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm tabuleiro quadrado de 3 linhas por 3 colunas contém nove casas. De quantos modos diferentes podemos escrever as três letras $A$, $B$ e $C$ em três casas diferentes, de modo que em cada linha esteja escrita exatamente uma letra?",
    "options": [],
    "answer": "162",
    "solution": "Solution:\n\nComeçando com a letra $A$, ela pode ser escrita em qualquer uma das 9 casas do tabuleiro. Uma vez escrita a letra $A$, sobram 6 casas onde a letra $B$ pode ser escrita. Uma vez escritas as letras $A$ e $B$ no tabuleiro, sobram 3 casas para a letra $C$ ser escrita.\nAssim, pelo princípio multiplicativo, existem $9 \\times 6 \\times 3 = 162$ maneiras diferentes das letras $A$, $B$ e $C$ serem escritas no tabuleiro.\n\n| A | C | B |\n| :--- | :--- | :--- |\n| C | B | A |\n| B | A | C |",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25852,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm triângulo isósceles tem base medindo $10~\\mathrm{cm}$ e dois lados iguais a $13~\\mathrm{cm}$. É possível mudar a base do triângulo e obter outro triângulo isósceles com mesma área?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Yes; for example, an isosceles triangle with base 24 cm and equal sides 13 cm has the same area.",
    "solution": "Solution:\n\nEm um triângulo isósceles, a altura relativa à base coincide com a mediana. Traçando esta altura, obtemos dois triângulos retângulos com catetos medindo $h$ e $5$, e hipotenusa $13$. Pelo Teorema de Pitágoras temos:\n$$\nh^{2} + 5^{2} = 13^{2} \\Rightarrow h^{2} = 13^{2} - 5^{2} = 144 \\Rightarrow h = \\sqrt{144} = 12\n$$\nLogo a área do triângulo é $A = \\frac{b \\times h}{2} = \\frac{10 \\times 12}{2} = 60~\\mathrm{cm}^{2}$.\n\nVamos agora \"colar\" os 2 triângulos retângulos ao longo do lado medindo $5$, obtendo um triângulo isósceles com base $12 + 12 = 24~\\mathrm{cm}$, os lados com $13~\\mathrm{cm}$ e a altura relativa à base igual a $5~\\mathrm{cm}$. Logo, este novo triângulo isósceles tem também área igual a $\\frac{24 \\times 5}{2} = 60~\\mathrm{cm}^{2}$.\n\n![](attached_image_2.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25853,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSeja $a$ um número inteiro positivo tal que $a$ é múltiplo de $5$, $a+1$ é múltiplo de $7$, $a+2$ é múltiplo de $9$ e $a+3$ é múltiplo de $11$. Determine o menor valor que $a$ pode assumir.",
    "options": [],
    "answer": "1735",
    "solution": "Solution:\n\nAs condições do problema equivalem a dizer que:\n$$\n2a - 5 = 2(a+1) - 7 = 2(a+2) - 9 = 2(a+3) - 11\n$$\né múltiplo de $5$, $7$, $9$ e $11$, donde é múltiplo de $5 \\cdot 7 \\cdot 9 \\cdot 11 = 3465$. Assim, o menor valor de $a$ é tal que $2a - 5 = 3465$, ou seja, $a = 1735$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25854,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nNo plano cartesiano, chama-se ponto inteiro a um ponto de coordenadas inteiras. Se $n$ é inteiro positivo, seja $f(n)$ o número de pontos inteiros que estão sobre o segmento que liga a origem ao ponto inteiro $(n, n+3)$, sem contar os extremos. Mostre que:\n$$\nf(n)=\\left\\{\\begin{array}{lll}\n2 & \\text{ se } & n \\text{ é múltiplo de } 3 \\\\\n0 & \\text{ se } & n \\text{ não é múltiplo de } 3 .\n\\end{array}\\right.\n",
    "options": [],
    "answer": "2 if n is a multiple of 3; 0 otherwise",
    "solution": "Solution:\nComecemos examinando alguns casos.\n- $f(1)$ é o número de pontos inteiros sobre o segmento que liga $(0,0)$ ao ponto $(1,4)$. Logo, $f(1)=0$.\n- $f(2)$ é o número de pontos inteiros sobre o segmento que liga $(0,0)$ ao ponto $(2,3)$. Logo, $f(2)=0$.\n- $f(3)$ é o número de pontos inteiros sobre o segmento que liga $(0,0)$ ao ponto $(3,6)$. Como nesse segmento estão 2 pontos inteiros $(1,2)$ e $(2,4)$, segue que $f(3)=2$.\n\n![](attached_image_1.png)\n\nVejamos, agora o caso geral. Note que se um ponto inteiro $(x, y)$ está sobre o segmento que une $(0,0)$ a $(n, n+3)$, sem ser um dos extremos, então $0<x<n$ e $0<y<n+3$. Vamos precisar do seguinte resultado:\n\nLema: Se $n$ não é múltiplo de $3$, então $n$ e $n+3$ são primos entre si.\n\nDemonstração: Suponhamos que o mdc entre $n$ e $n+3$ seja $d>1$. Então $d$ divide $n$ e $n+3$, portanto $d$ divide $(n+3)-n=3$. Logo, como $d>1$, teremos $d=3$, o que não é possível porque partimos da hipótese que $3$ não divide $n$.\n\n- Se $3$ não divide $n$ então $f(n)=0$.\n\nIsso equivale a dizer que não há pontos inteiros sobre o segmento que une $(0,0)$ a $(n, n+3)$, excluídos os extremos.\nDe fato, suponhamos que esse segmento contenha um ponto inteiro $(x, y)$, então\n$$\n\\frac{x}{y}=\\frac{n}{n+3}\n$$\nPelo lema, a fração $\\frac{n}{n+3}$ está na forma irredutível, logo, $x$ seria múltiplo de $n$, o que não pode acontecer porque $x<n$.\n\n- Se $3$ divide $n$ então $f(n)=2$.\n\nSe $n=3k$, com $k$ inteiro, devemos achar o número de pontos inteiros no segmento que une $(0,0)$ ao ponto $(3k, 3k+3)$. Seja $(x, y)$ um desses pontos, então\n$$\n\\frac{x}{y}=\\frac{3k}{3k+3}=\\frac{k}{k+1}\n$$\nSendo a última fração irredutível, deduzimos que $x$ é múltiplo de $k$, e como $0<x<3k$, segue que $x=k$ ou $x=2k$. Os pontos inteiros são $(k, k+1)$ e $(2k, 2k+2)$. Assim, temos $f(n)=2$.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25855,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm professor enfurecido - Para castigar os alunos de sua turma por indisciplina, o professor Zerus decidiu descontar da nota mensal de cada aluno uma percentagem igual à nota da prova, isto é: quem tirou $60$, terá um desconto de $60\\%$ na nota, quem tirou $20$, um desconto de $20\\%$ da nota, e assim por diante. A nota mensal máxima é $100$.\n\na. Quem vai ficar com a maior nota?\n\nb. E a menor?\n\nc. Alunos que tiraram boas notas reclamaram que vão ficar com a mesma nota dos que tiraram más notas. Eles estão certos?",
    "options": [],
    "answer": "a) The highest final grade occurs for those with an original grade of fifty, and the final grade is twenty five. b) The lowest final grade is zero, occurring for original grades zero and one hundred. c) Yes; students with original grades equidistant from fifty (for example, whose grades add to one hundred) receive the same final grade.",
    "solution": "Solution:\n\nQuem teve $x$ como nota mensal vai ter um desconto de $x\\%$ sobre essa nota, ou seja vai perder\n$$\nx\\% \\text{ de } x = \\frac{x}{100} \\times x = \\frac{x^{2}}{100}\n$$\nLogo, depois do castigo, a nota fica sendo $x - \\frac{x^{2}}{100}$, onde $x$ era a nota inicial.\n\nConsideremos a função \"nota depois do castigo\" dada por $f(x) = x - \\frac{x^{2}}{100}$. Como as notas máximas e mínimas são $0$ e $100$, vamos considerar essa função no domínio $[0, 100]$, ou seja, para $0 \\leq x \\leq 100$. O gráfico de $f$ é uma parábola com concavidade para baixo, e seu valor máximo ocorre no vértice: $x = -\\frac{b}{2a} = \\frac{-1}{\\frac{-2}{100}} = 50$. Sendo assim, a maior nota depois do castigo é para os alunos que antes do castigo tiraram $50$. Essa nota é\n$$\nf(50) = 50 - \\frac{50^{2}}{100} = 25\n$$\nO valor mínimo dessa função é $0$ e ocorre em $x = 0$ e $x = 100$. Logo a menor nota ocorre para os alunos que tiraram $0$ ou $100$ (!!!!!) antes do castigo. De fato, $f(0) = f(100) = 0$.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25856,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUma urna contém $k$ bolas marcadas com $k$, para todo $k=1,2, \\ldots, 2016$. Qual é o número mínimo de bolas que devemos retirar, sem reposição e sem olharmos as bolas, para termos certeza de que teremos 12 bolas com o mesmo número?",
    "options": [],
    "answer": "22122",
    "solution": "Solution:\n\nSomemos a maior quantidade de bolas que podem ser retiradas de cada tipo sem que obtenhamos 12 bolas de cada cor:\n$$\n1+2+3+4+5+6+7+8+9+10+11+\\underbrace{11+11+\\ldots+11}_{2005 \\text{ vezes}}=22121\n$$\nAssim, é possível que tenhamos azar e retiremos tal quantidade de bolas sem obtermos 12 bolas de cada cor. Entretanto, se retirarmos 22122 bolas, certamente teremos 12 bolas de uma mesma cor, pois a soma anterior conta exatamente o máximo de bolas que podem ser retiradas sem que isto ocorra. Logo, o mínimo buscado é 22122.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25857,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nRenato tem trinta melancias, Leandro tem dezoito melancias e Marcelo tem vinte e quatro jacas. Ao contrário de Leandro e Renato, Marcelo não gosta de jaca. Por outro lado, os três gostam de melancia. Os três fazem então um acordo: Marcelo dá as suas vinte e quatro jacas para Leandro e Renato, e as melancias de Leandro e Renato são divididas igualmente entre os três, ou seja, dezesseis para cada. Qual é a divisão justa de jacas entre Renato e Leandro?",
    "options": [],
    "answer": "Renato receives 15 jackfruits; Leandro receives 9 jackfruits.",
    "solution": "Solution:\n\nA ideia é determinar o valor de melancias em termos de jacas. Como são $18+30=48$ melancias e $24$ jacas, temos a proporção\n$$\n\\begin{array}{ccc}\n48 \\text{ melancias} & - & 24 \\text{ jacas} \\\\\n1 \\text{ melancia} & - & x \\text{ jacas}\n\\end{array}\n$$\no que dá $x=1/2$. Ou seja, uma melancia equivale a meia jaca. Como Renato tinha $30$ melancias, este deve receber $15$ jacas. Como Leandro tinha $18$ melancias, este deve receber $9$ jacas.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25858,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOs pontos $E$ e $F$ estão nos lados $A D$ e $B C$, respectivamente, do quadrado $A B C D$. Sabendo que $B E=E F=F D=30$, encontre a área do quadrado.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "810",
    "solution": "Solution:\n\nSejam $G$ e $H$ os pés das perpendiculares traçadas de $E$ e $F$ aos lados $B C$ e $A D$, respectivamente.\n\n![](attached_image_2.png)\n\nComo $A B = C D$ e $B E = F D$, aplicando o Teorema de Pitágoras, temos:\n$$\n\\begin{aligned}\nA E & = \\sqrt{B E^{2} - A B^{2}} \\\\\n& = \\sqrt{F D^{2} - C D^{2}} \\\\\n& = F C\n\\end{aligned}\n$$\n\nAlém disto, como $E F = F D$, novamente pelo Teorema de Pitágoras, podemos concluir que\n$$\n\\begin{aligned}\nE H & = \\sqrt{E F^{2} - H F^{2}} \\\\\n& = \\sqrt{F D^{2} - H F^{2}} \\\\\n& = H D \\\\\n& = F C\n\\end{aligned}\n$$\n\nConsequentemente, $A E = E H = H D = x$ e $A B = A D = 3x$. Finalmente, aplicando o Teorema de Pitágoras no triângulo $A E B$, temos\n$$\n900 = B E^{2} = A E^{2} + A B^{2} = x^{2} + 9x^{2}\n$$\nLogo, a área do quadrado é $A B^{2} = 9x^{2} = 810$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25859,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nCalcule a soma\n$$\n1+11+111+1111+\\cdots+\\underbrace{1111 \\ldots 11}_{n \\text{ uns }}\n$$",
    "options": [],
    "answer": "S = (10/81)(10^n − 1) − n/9",
    "solution": "Solution:\nUma solução pode ser feita usando soma de progressões geométricas. Mas daremos outra solução que não precisará disso! Observe. Chamemos de $S$ a soma que queremos calcular, ou seja,\n$$\nS=1+11+111+1111+\\cdots+\\underbrace{1111 \\ldots 11}_{n \\text{ uns }} .\n$$\nQuanto vale $9 \\times S$ ? Basta trocar cada dígito um por um dígito nove!\n$$\n9 S=9+99+999+9999+\\cdots+\\underbrace{9999 \\ldots 99}_{n \\text{ noves }}\n$$\nAgora vamos escrever $9=10-1$. E fazemos o mesmo com $99=100-1$, $999=1000-1$ e assim por diante. Ou seja,\n$$\n\\begin{array}{ccccc}\n9 & = & 10-1 & = & 10^{1}-1 \\\\\n99 & = & 100-1 & = & 10^{2}-1 \\\\\n999 & = & 1000-1 & = & 10^{3}-1 \\\\\n9999 & = & 10000-1 & = & 10^{4}-1 \\\\\n\\vdots & & \\vdots & & \\vdots \\\\\n\\underbrace{9999 \\cdots 9}_{n \\text{ noves }} & = & 1 \\underbrace{000 \\cdots 0}_{n \\text{ zeros }}-1 & & 10^{n}-1\n\\end{array}\n$$\nFazendo essas trocas em $9 S$, obtemos\n$$\n9 S=(10-1)+\\left(10^{2}-1\\right)+\\left(10^{3}-1\\right)+\\left(10^{4}-1\\right)+\\cdots+\\left(10^{n}-1\\right)\n$$\nAgrupando todos os \"menos uns\", obtemos\n$$\n\\begin{aligned}\n9 S & =\\left(10+10^{2}+10^{3}+10^{4}+\\cdots+10^{n}\\right)-(\\underbrace{1+1+1+\\cdots+1}_{n \\text{ uns }}) \\\\\n& =10 \\cdot \\underbrace{111111 \\cdots 1}_{n \\text{ uns }}-n\n\\end{aligned}\n$$\nPara escrever melhor o número acima, vamos multiplicar e dividir por nove o termo com muitos \"uns\". Observe:\n$$\n\\begin{aligned}\n9 S & =10 \\times \\frac{9}{9} \\underbrace{111111 \\cdots 1}_{n \\text{ uns }}-n \\\\\n& =\\frac{10}{9} \\underbrace{99999 \\cdots 9}_{n \\text{ noves }}-n \\\\\n& =\\frac{10}{9}\\left(10^{n}-1\\right)-n\n\\end{aligned}\n$$\nLogo,\n$$\n9 S=\\frac{10}{9}\\left(10^{n}-1\\right)-n\n$$\nPassando o fator nove para o outro lado da equação, temos\n$$\nS=\\frac{10}{81}\\left(10^{n}-1\\right)-\\frac{n}{9}\n$$\nobtendo assim o valor desejado!",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25860,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm jogo é composto das seguintes regras:\ni) Em cada rodada, ocorre o lançamento de um dado comum não viciado.\nii) Se sair o número $3$, então o jogador $A$ ganha.\niii) Se sair um dos números do conjunto $\\{4,5,6\\}$, então o jogador $B$ ganha.\niv) Se sair um dos números do conjunto $\\{1,2\\}$, então o dado é lançado outra vez até resultar em $3$ ou $4$ ou $5$ ou $6$.\nQual a probabilidade do jogador $B$ vencer?",
    "options": [],
    "answer": "3/4",
    "solution": "Solution:\n\nSeja $P_{i}(B)$ a probabilidade do jogador $B$ vencer na rodada $i$, com $i$ inteiro positivo, e $P_{i}(\\overline{A+B})$ a probabilidade de $A$ e $B$ não vencerem na rodada $i$. Portanto, temos que\n$$\n\\begin{aligned}\n& P_{1}(B)=\\frac{3}{6}=\\frac{1}{2} \\\\\n& P_{2}(B)=P_{1}(\\overline{A+B}) \\cdot P_{2}(B)=\\frac{2}{6} \\cdot \\frac{1}{2}=\\frac{1}{6} \\\\\n& P_{3}(B)=P_{1}(\\overline{A+B}) \\cdot P_{2}(\\overline{A+B}) \\cdot P_{3}(B)=\\frac{2}{6} \\cdot \\frac{2}{6} \\cdot \\frac{1}{2}=\\frac{1}{18} \\\\\n& \\quad \\vdots \\\\\n& P_{n}(B)=P_{1}(\\overline{A+B}) \\cdot P_{2}(\\overline{A+B}) \\cdot \\ldots \\cdot P_{n-1}(\\overline{A+B}) \\cdot P_{n}(B)=\\left(\\frac{1}{3}\\right)^{n-1} \\cdot \\frac{1}{2}\n\\end{aligned}\n$$\nPerceba que os resultados formam uma progressão geométrica de razão $1/3$ que precisaremos somar para encontrar a probabilidade $P(B)$ de $B$ vencer. Temos, então\n$$\nP(B)=\\frac{1}{2}+\\frac{1}{6}+\\frac{1}{18}+\\cdots+\\left(\\frac{1}{3}\\right)^{n-1} \\cdot \\frac{1}{2}=\\frac{\\frac{1}{2}}{1-\\frac{1}{3}}=\\frac{3}{4}\n$$\nObservação: Se $|q|<1$ e $S=1+q+q^{2}+\\ldots+q^{n-1}$, então\n$$\nS q=q+q^{2}+q^{3} \\ldots+q^{n}\n$$\nDaí, $S q-S=q^{n}-1$ e, consequentemente, $S=\\frac{q^{n}-1}{q-1}$. No final da solução anterior, aplicamos essa fórmula para a soma dos termos de uma progressão geométrica com $q=1 / 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25861,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO macaco diz para a raposa:\n- Você vê as 3 pessoas que estão correndo lá longe? Eu sei que o produto de suas idades é $2450$; e que a soma das idades é o dôbro da sua idade. Você pode me dizer suas idades?\n- Não, responde a raposa.\n- E se eu te disser que o mais jovem dos três é o único louro, você pode agora descobrir as idades?\nE a raposa dá as idades das 3 pessoas.\n\nPorque a raposa não pode responder inicialmente? E porque pode responder depois?",
    "options": [],
    "answer": "The fox is 32 years old; the three ages are 5, 10, and 49.",
    "solution": "Solution:\n\n$2450$ é o produto dos números primos $1, 2, 5, 5, 7, 7$. As 3 idades correspondem a uma combinação particular desses números ou de seus produtos.\n\nA raposa não pode descobrir as idades no início porque pelo menos duas dessas combinações têm por soma o dobro de sua idade. De todas as combinações possíveis, somente $\\{5, 10, 49\\}$ e $\\{7, 7, 50\\}$ têm a mesma soma $64$.\n\nPrimeira conclusão: a raposa tem $32$ anos.\n\nDepois da nova dica do macaco, a raposa descobriu as idades porque pode eliminar uma combinação: aquela que contém dois números iguais, uma vez que um deles é o mais jovem de todos.\n\nSegunda conclusão: as pessoas têm $5$, $10$ e $49$ anos.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25862,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUma das afirmações abaixo é falsa:\n\n(a) André é mais velho do que Bruno;\n\n(b) Cláudia é mais nova do que Bruno\n\n(c) A soma das idades de Bruno e Cláudia é o dobro da idade de André;\n\n(d) Cláudia é mais velha do que André.\n\nQuem é o mais velho? E o mais novo?",
    "options": [],
    "answer": "Cláudia e Bruno.",
    "solution": "Solution:\n\nCláudia e Bruno.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25863,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUma desigualdade especial- Quais valores de $x$ satisfazem $x^{2}<|x|+2$ ?\n\n(a) $x<-1$ ou $x>1$\n(b) $x>1$\n(c) $-2<x<2$\n(d) $x<-2$\n(e) $x<0$",
    "options": [],
    "answer": "c",
    "solution": "Solution:\n\nObservemos que se um número $a$ satisfaz a desigualdade, então $-a$ também satisfaz a desigualdade, logo os valores que satisfazem a desigualdade formam um conjunto simétrico, portanto basta considerar o caso em que $x$ é positivo. Mas, $(2-x)(1+x)=x+3-x^{2}>0$ é positivo se $2-x$ é positivo, portanto $x<2$. Como a solução é simétrica temos que $-2<x<2$ é a solução da equação inicial. A opção correta é (c).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25864,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n10. (N1) Contas do papagaio - Rosa tem um papagaio que faz contas de um modo estranho. Cada vez que Rosa diz dois números ele faz a mesma conta, veja:\n- Se Rosa diz \"4 e 2\" o papagaio responde \"9\"\n- Se Rosa diz \"5 e 3\" o papagaio responde \"12\"\n- Se Rosa diz \"3 e 5\" o papagaio responde \"14\"\n- Se Rosa diz \"9 e 7\" o papagaio responde \"24\"\n- Se Rosa diz \"0 e 0\" o papagaio responde \"1\"\nSe Rosa diz \"1 e 8\" o que responde o papagaio?",
    "options": [],
    "answer": "1",
    "solution": "Solution:\n\n1",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25865,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNum triângulo $ABC$, o ângulo $A\\hat{BC}$ mede $20^\\circ$ e o ângulo $A\\hat{C}B$ mede $40^\\circ$. Seja $E$ um ponto sobre $BC$ tal que $BE = BA$.\n\na. Mostre que o triângulo $CEA$ é isósceles.\n\nb. Sabendo que o comprimento da bissetriz do ângulo $B\\hat{A}C$ é $2$, determine $BC - AB$.",
    "options": [],
    "answer": "2",
    "solution": "Solution:\n\na.\nTemos $C\\hat{AB} = 180^\\circ - 20^\\circ - 40^\\circ = 120^\\circ$. Como o triângulo $ABE$ é isósceles, segue que\n$$\nA\\hat{E}B = E\\hat{A}B = \\frac{180^\\circ - 20^\\circ}{2} = 80^\\circ\n$$\nAssim, $CA\\hat{E} = 120^\\circ - 80^\\circ = 40^\\circ$ e o triângulo $ACE$ tem dois ângulos de $40^\\circ$, e, portanto, é isósceles com $CE = EA$.\n\nb.\nSeja $D$ o pé da bissetriz do ângulo $B\\hat{A}C$. A bissetriz divide o ângulo $C\\hat{A}B$ em dois ângulos de $60^\\circ$. Logo, o ângulo\n$$\nCDA = 180^\\circ - 40^\\circ - 60^\\circ = 80^\\circ\n$$\nComo $A\\hat{E}B$ também mede $80^\\circ$, temos que o triângulo $ADE$ é isósceles. Finalmente,\n\n![](attached_image_1.png)\n\n$$\nBC - AB = BC - BE = CE = EA = AD = 2\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25866,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n\"Duas torres, uma com 30 passos e a outra com 40 passos de altura, estão à distância de 50 passos uma da outra. Entre ambas se acha uma fonte, para a qual dois pássaros descem no mesmo momento do alto das torres com a mesma velocidade e chegam ao mesmo tempo. Quais as distâncias horizontais da fonte às duas torres?\" (Leonardo de Pisa, Liber Abaci, 1202).",
    "options": [],
    "answer": "32 steps and 18 steps",
    "solution": "Solution:\n\nNa figura, $AD$ e $BC$ representam as duas torres e o ponto $E$ representa a posição da fonte. Como os dois pássaros chegam ao mesmo tempo, temos que $DE = EC$.\n\nDenotemos por $x$ a distância de $A$ a $E$ e assim $EB = 50 - x$. Usando o teorema de Pitágoras nos triângulos $DAE$ e $EBC$, temos que\n$$\n\\begin{cases}\nDE^2 = 30^2 + x^2 \\\\\nEC^2 = 40^2 + (50 - x)^2\n\\end{cases}\n$$\nComo $DE = EC$, temos:\n$$\n900 + x^2 = 1600 + 2500 - 100x + x^2 \\Longleftrightarrow x = \\frac{3200}{100} = 32\n$$\nPortanto, as distâncias horizontais da fonte às duas torres são $AE = x = 32$ passos e $EB = 50 - x = 18$ passos.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25867,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAs circunferências $\\mathcal{C}_1$ e $\\mathcal{C}_2$ são tangentes à reta $\\ell$ nos pontos $A$ e $B$ e tangentes entre si no ponto $C$. Prove que o triângulo $ABC$ é retângulo.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nComo as circunferências são tangentes, então o ponto de tangência $C$ e os centros $O_1$ e $O_2$ pertencem a uma mesma reta. Além disso, como as circunferências são tangentes a $\\ell$, então $O_1A$ e $O_2B$ são perpendiculares a $\\ell$ e, portanto, paralelas.\n\nSeja $\\alpha$ a medida do ângulo $O_1\\hat{C}A$ e $\\beta$ a medida do ângulo $O_2\\hat{C}B$. Como os triângulos $AO_1C$ e $BO_2C$ são isósceles, segue que $CAO_1 = \\alpha$ e $CBO_2 = \\beta$.\n\nComo as retas $O_1A$ e $O_2B$ são paralelas, temos $AO_1\\hat{O}_1C + BO_2\\hat{O}_2C = 180^\\circ$, donde $180^\\circ - 2\\alpha + 180^\\circ - 2\\beta = 180^\\circ$. Portanto, $\\alpha + \\beta = 90^\\circ$.\n\nAssim, $A\\hat{C}B = 180^\\circ - (\\alpha + \\beta) = 90^\\circ$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25868,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSeja $ABC$ um triângulo isósceles com $AB = AC$ e $\\widehat{A} = 30^{\\circ}$. Seja $D$ o ponto médio da base $BC$. Sobre $AD$ e $AB$ tome dois pontos $P$ e $Q$, respectivamente, tais que $PB = PQ$. Determine a medida do ângulo $PQC$.",
    "options": [],
    "answer": "15°",
    "solution": "Solution:\n\n$$\nA \\hat{B C} = A \\hat{C} B = \\frac{180^{\\circ} - 30^{\\circ}}{2} = 75^{\\circ}\n$$\nComo todos os pontos da altura $AP$ estão à mesma distância de $B$ e de $C$, em particular, o triângulo $BPC$ é isósceles com $BP = PC$. Pela hipótese do problema, o triângulo $BPQ$ também é isósceles. Denotemos por $\\alpha$ a medida do ângulo $PBC$, assim $BCP = \\alpha$ e\n$$\n\\hat{AQP} = 180^{\\circ} - BQP = 180^{\\circ} - QBP = 180^{\\circ} - (75^{\\circ} - \\alpha) = 105^{\\circ} - \\alpha\n$$\ne\n$$\nP \\hat{CA} = 75^{\\circ} - P \\hat{C} B = 75^{\\circ} - \\alpha\n$$\nAssim $A \\hat{Q} P + P \\hat{CA} = 180^{\\circ}$, portanto o quadrilátero $AQPC$ é inscritível, em particular $PQC = PAC = 15^{\\circ}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25869,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAs páginas do dicionário - Para numerar as páginas de um dicionário, imprimiu-se 1988 vezes o algarismo 1. Quantas páginas tem esse dicionário?",
    "options": [],
    "answer": "3151",
    "solution": "Solution:\n\nObservemos que:\n\na. A cada dez números imprime-se uma vez o $1$ nas unidades;\n\nb. A cada cem números imprime-se dez vezes o $1$ nas dezenas;\n\nc. A cada mil números imprime-se cem vezes o $1$ nas centenas.\n\nAssim, de $1$ até $999$, imprime-se o algarismo $1$ um total de $300$ vezes, das quais $100$ vezes nas unidades, $100$ vezes nas dezenas e $100$ vezes nas centenas.\n\nDe $1000$ até $1999$, imprime-se o algarismo $1$ outras $300$ vezes dentre unidades, dezenas e centenas, mais $1000$ vezes na posição do milhar, portanto, entre $1$ e $1999$, o número de vezes que se imprime o $1$ é $300+300+1000=1600$.\n\nAgora, entre $2000$ e $2999$, imprime-se o $1$ mais $300$ vezes, completando $1600+300=1900$.\n\nDe $3000$ a $3099$, temos $20$ algarismos $1$, de $3100$ a $3119$, temos $32$ algarismos $1$ e, de $3120$ a $3149$, temos $32$ algarismos $1$, portanto, até $3149$, o número de vezes que se imprime o $1$ é $1900+20+32+32=1984$.\n\nComo faltam $4$ algarismos $1$, o número de páginas do dicionário é $3152$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25870,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUma circunferência de raio $r$ está inscrita em um setor circular de raio $R$. O comprimento da corda $AB$ é igual a $2a$.\n\nProve que\n$$\n\\frac{1}{r} = \\frac{1}{R} + \\frac{1}{a}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nDenotemos por $D$ o ponto de tangência de $AO$ com a circunferência. Então $ODO_1 = 90^{\\circ}$. Observe também que $AC = AB / 2 = a$.\n\nPor outro lado, $OCA = 90^{\\circ}$. Os triângulos $ODO_1$ e $OCA$ são semelhantes pois possuem um ângulo comum e um ângulo reto. Portanto,\n$$\n\\frac{OO_1}{OA} = \\frac{O_1 D}{AC}\n$$\nisto é,\n$$\n\\frac{R - r}{R} = \\frac{r}{a}\n$$\ndonde\n$$\n\\frac{1}{a} + \\frac{1}{R} = \\frac{1}{r}\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25871,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoma de potências de 2 - Determine um valor de $n$ para o qual o número $2^{8}+2^{11}+2^{n}$ seja um quadrado perfeito.",
    "options": [],
    "answer": "12",
    "solution": "Solution:\n\nSolução 1: Observe que $2^{8}+2^{11}+2^{n} = (2^{4})^{2} + 2 \\times 2^{4} \\times 2^{6} + \\left(2^{\\frac{n}{2}}\\right)^{2}$. Logo, para $n=12$, temos $2^{8}+2^{11}+2^{12} = (2^{4}+2^{6})^{2}$. Assim, $n=12$ é uma solução.\n\n\nSolução 2: Se $2^{8}+2^{11}+2^{n}=k^{2}$, então\n$$\n\\begin{aligned}\n2^{8}+2^{3} \\times 2^{8}+2^{n} & =k^{2} \\\\\n9 \\times 2^{8}+2^{n} & =k^{2} \\\\\n2^{n} & =k^{2}-\\left(3 \\times 2^{4}\\right)^{2} \\\\\n2^{n} & =\\left(k-3 \\times 2^{4}\\right)\\left(k+3 \\times 2^{4}\\right)\n\\end{aligned}\n$$\nLogo, $\\left(k-3 \\times 2^{4}\\right)$ e $\\left(k+3 \\times 2^{4}\\right)$ são potências de 2, ou seja, $k+3 \\times 2^{4}=2^{a}$ e $k-3 \\times 2^{4}=2^{b}$, com $a+b=n$ e\n$$\n2^{a}-2^{b}=\\left(k+3 \\times 2^{4}\\right)-\\left(k-3 \\times 2^{4}\\right)=3 \\times 2^{5}=96\n$$\nExaminemos a lista das potências de 2:\n$$\n2,4,8,16,32,64,128,256, \\ldots\n$$\nConstatamos que a diferença dessas potências só é 96 entre $128=2^{7}$ e $32=2^{5}$. Logo, $a=7$ e $b=5$. Assim, $n=7+5=12$ é a única solução.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25872,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSeja $ABC$ um triângulo tal que $AB = 55$, $AC = 35$ e $BC = 72$. Considere uma reta $\\ell$ que corta o lado $BC$ em $D$ e o lado $AC$ em $E$ e que divide o triângulo em duas figuras com perímetros iguais e áreas iguais. Determine a medida do segmento $CD$.",
    "options": [],
    "answer": "60",
    "solution": "Solution:\n\nSejam $CD = x$, $CE = y$ e $DE = z$.\n\n(1) Como o triângulo $CED$ tem o mesmo perímetro do quadrilátero $ABDE$, temos\n$$\nx + y + z = (35 - y) + z + (72 - x) + 55 \\Longleftrightarrow y = 81 - x\n$$\n\n(2) Como eles também possuem a mesma área, a área do triângulo $DCE$ deve ser igual à metade da área do triângulo $ABC$. Deste modo,\n$$\n\\frac{x y \\operatorname{sen} \\hat{C}}{2} = \\frac{1}{2} \\cdot \\frac{35 \\cdot 72 \\cdot \\operatorname{sen} \\hat{C}}{2} \\Longleftrightarrow x y = 1260\n$$\n\nUtilizando as duas equações encontradas obtemos $x^2 - 81x + 1260 = 0$. Resolvendo esta equação, chegamos em $x = 60$ ou $x = 21$. No primeiro caso obtemos $y = 21$ e no segundo $y = 60$. Como $E$ está sobre o lado $AC$, devemos ter $y \\leqslant 35$ e então a solução que nos interessa é $x = 60$ e $y = 21$. Portanto, $CD = 60$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25873,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$O$ triângulo acutângulo $ABC$ de ortocentro $H$ é tal que $AB = 48$ e $HC = 14$. O ponto médio do lado $AB$ é $M$ e o ponto médio do segmento $HC$ é $N$.\n\na. Mostre que o ângulo $MEN$ é reto.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nInicialmente observe que $ME$ é mediana relativa à hipotenusa do triângulo $AEB$. Portanto, $ME = AM = MB = 24$. Desse fato segue que o triângulo $BME$ é isósceles. Então $M\\hat{E}B = M\\hat{B}E = \\beta$.\n\nAnalogamente, como $N$ é o ponto médio da hipotenusa do triângulo $HEC$, temos $EN = HN = NC = 7$ e o triângulo $HNE$ é isósceles. Assim, $H\\hat{E}N = E\\hat{H}N = \\alpha$.\n\nO triângulo $FHB$ é retângulo em $F$ e $FHB + HBF = \\alpha + \\beta = 90^\\circ$. Assim, o triângulo $MEN$ é retângulo em $E$.\n\nAplicando o teorema de Pitágoras neste triângulo, obtemos\n$$\n\\begin{aligned}\n& MN^2 = ME^2 + EN^2 \\\\\n& MN^2 = 24^2 + 7^2 = 625\n\\end{aligned}\n$$\ndonde $MN = 25$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25874,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\na. Sejam $\\mathcal{C}$ uma circunferência com centro $O$ e raio $r$ e $X$ um ponto exterior a $\\mathcal{C}$. Construímos uma circunferência de centro em $X$ passando por $O$, a qual intersecta $\\mathcal{C}$ nos pontos $P$ e $Q$. Com centro em $P$ construímos uma circunferência passando por $O$ e com centro em $Q$ construímos uma outra circunferência passando por $O$. Estas duas circunferências intersectam-se nos pontos $O$ e $Y$.\n\nProve que $OX \\times OY = r^2$.\n\nb. É dado um segmento $AB$. Mostre como construir, usando somente compasso, um ponto $C$ tal que $B$ seja o ponto médio do segmento $AC$.\n\nc. É dado um segmento $AB$. Mostre como construir, usando somente compasso, o ponto médio do segmento $AB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na. Observe que os triângulos $XOP$ e $PYO$ são ambos isósceles, de bases $OP$ e $YO$, respectivamente. Estes triângulos possuem ângulos da base de mesma medida, pois o ângulo $P\\hat{O}X = Y\\hat{O}P$ é comum aos dois triângulos. Deste modo, os triângulos $XOP$ e $PYO$ são semelhantes e podemos escrever $OX/OP = OP/OY$. Como $OP = r$, concluímos que $OX \\times OY = r^2$.\n\nb. Determinamos um ponto $R$ tal que o triângulo $ABR$ seja equilátero. Em seguida, determinamos um ponto $S \\neq A$ de modo que o triângulo $RBS$ seja equilátero e construímos $C \\neq R$ de forma que o triângulo $BSC$ também seja equilátero. Assim, $BC = BS = BR = AB$ e $A$, $B$ e $C$ são colineares ($A\\hat{B}C = 60^\\circ + 60^\\circ + 60^\\circ = 180^\\circ$), logo $B$ é o ponto médio de $AC$.\n\nc. Seja $M$ o ponto médio de $AB$. Construa a circunferência com centro em $A$ e raio $r = AB$. Como no item anterior, com o compasso construímos um ponto $C$ tal que $B$ é o ponto médio de $AC$.\nObserve que $AM \\times AC = (r/2) \\times 2r = r^2$ e, portanto, podemos construir o ponto $M$ utilizando o processo de construção do item (a): determinamos os pontos $P$ e $Q$, pontos de interseção da circunferência de centro $C$ que contém $A$ e da circunferência de centro $A$ que contém $B$. O ponto $M$ é obtido pela interseção das circunferências de centros $P$ e $Q$ que passam por $A$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25875,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA figura a seguir mostra um segmento $AB$, seu ponto médio $C$ e as semicircunferências de diâmetros $AB$ e $AC$. Uma circunferência de centro $P$ é tangente às duas semicircunferências e também ao segmento $AB$. Sendo $AB=8~\\mathrm{cm}$, e $O$, o ponto médio de $AC$, pergunta-se:\n![](attached_image_1.png)\n\na) Qual a medida do perímetro do triângulo $OCP$?\n\nb) Qual a medida do raio da circunferência de centro $P$?",
    "options": [],
    "answer": "Perimeter = 8 cm; radius = 16/9 cm",
    "solution": "Solution:\na) Seja $x$ o raio da circunferência de centro $P$. Traçamos $OP$, que passa pelo ponto de tangência $D$; $CP$, que passa pelo ponto de tangência $F$; e $PE$, perpendicular a $AB$ (e raio da circunferência destacada na figura). Temos $CP=4-x$ e $OP=2+x$. O perímetro do triângulo $OCP$ é\n$$\n\\begin{aligned}\n2p &= OP + CP + CO \\\\\n   &= (2+x) + (4-x) + 2 \\\\\n   &= 8~\\mathrm{cm}\n\\end{aligned}\n$$\n\nb) Agora, calcularemos a área do $\\triangle OCP$ de duas formas, pela fórmula de Heron, que usa o semi-perímetro $p=4~\\mathrm{cm}$ e como metade do produto da base pela altura.\n![](attached_image_2.png)\nSendo assim,\n$$\n\\sqrt{4 \\cdot (4-2)(4-(2+x))(4-(4-x))} = \\frac{2 \\cdot x}{2}\n$$\nResolvendo a equação anterior, ficamos com $x=\\frac{16}{9}~\\mathrm{cm}$.\n\nObservação: Um triângulo de perímetro $2p$ e lados de comprimentos $a$, $b$ e $c$, pela Fórmula de Heron, possui área dada por\n$$\nA = \\sqrt{p(p-a)(p-b)(p-c)}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25876,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNum grupo de 20 pessoas, algumas pessoas trocam apertos de mão.\na) Contamos quantos apertos de mão cada pessoa deu e somamos todos esses números. Mostre que o resultado é par.\nb) É possível que num grupo de 99 pessoas cada pessoa tenha dado exatamente 3 apertos de mão?",
    "options": [],
    "answer": "a) even; b) no",
    "solution": "Solution:\n\na) Um aperto de mão é dado entre duas pessoas. Logo, quando somamos os apertos de mão de todas as pessoas, cada aperto de mão é contado duas vezes! Logo, a soma de quantas vezes cada pessoa apertou a mão de alguém é par, pois é o dobro de algum número.\n\nb) Como são 99 pessoas, se cada uma apertasse a mão de alguém 3 vezes, isso daria um total de $3 \\times 99 = 297$ apertos de mão. Mas como vimos no item anterior, este total deve ser par, pois cada aperto de mão entre duas pessoas foi contado duas vezes (se fulano apertou a mão de sicrano, então esse aperto de mão foi contado uma vez quando estávamos somando os apertos de mão de fulano, e foi contado novamente quando estávamos somando os apertos de mão de sicrano).\n\nPor outro lado, 297 é ímpar! Logo, não é possível.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25877,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNo desenho a seguir, $ABCD$ é um quadrado e os pontos $E$ e $F$ estão sobre os lados $BC$ e $CD$ de modo que $AEF$ é um triângulo retângulo, $AE=4$ e $EF=3$. Qual é a área do quadrado?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "256/17",
    "solution": "Solution:\n\nSe $\\angle BAE=\\alpha$, como $\\angle ABC=\\angle AEF=90^{\\circ}$, temos $\\angle BEA=90^{\\circ}-\\alpha$ e $\\angle FEC=\\alpha$. Consequentemente os triângulos $BEA$ e $CFE$ são semelhantes e\n$$\n\\frac{BE}{CF}=\\frac{AB}{EC}=\\frac{AE}{EF}=\\frac{4}{3}\n$$\nSe $BE=x$, temos $CF=\\frac{3x}{4}$ e $\\frac{AB}{AB-x}=\\frac{4}{3}$, ou seja, $AB=4x$. Aplicando o Teorema de Pitágoras nos triângulos $EAF$ e $AFD$, temos\n$$\n\\begin{aligned}\nAE^{2}+EF^{2} & =AF^{2} \\\\\n3^{2}+4^{2} & =AD^{2}+FD^{2} \\\\\n25 & =AB^{2}+(AB-CF)^{2} \\\\\n25 & =16x^{2}+\\frac{169x^{2}}{16} \\\\\n25 & =\\frac{425x^{2}}{16} \\\\\nx^{2} & =\\frac{16}{17}\n\\end{aligned}\n$$\nPortanto, a área do quadrado é $AB^{2}=16x^{2}=\\frac{256}{17}$.\n\n![](attached_image_2.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25878,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nConsidere a soma das três tabelas a seguir. A primeira representa $n$ linhas, sendo a primeira com $n$ números iguais a $n$, a segunda com $n-1$ números iguais a $n-1$ e assim por diante. Na segunda, temos uma distribuição de números parecida, mas em colunas em vez de linhas. Já na terceira, temos estes números em diagonais, a primeira diagonal possui um número $1$, a segunda dois números iguais a $2$, a terceira três números iguais a $3$ e assim por diante.\n![](attached_image_1.png)\nO resultado da soma das três tabelas será uma tabela com a mesma quantidade de números e com cada posição sendo o resultado da soma das posições correspondentes nas três tabelas. Por exemplo, no canto superior esquerdo, teremos o número $n+n+1=2n+1$.\n\na) Um modo de verificar quantos números tem em cada tabela é virar uma delas de ponta cabeça e juntar com outra para formar um retângulo com $n$ linhas e o dobro de números de uma tabela. Sabendo disto, quantos números existem em uma tabela?\n\nb) Quantas vezes aparece cada número $k$ em todas as três tabelas?\n\nc) Para cada posição, linha $i$ e coluna $j$, determine os números escritos nela nas três tabelas e na tabela resultado.\n\nd) Usando as informações dos itens anteriores, verifique que\n$$\n1^{2}+2^{2}+3^{2}+\\ldots+(n-1)^{2}+n^{2}=\\frac{1}{6} n(n+1)(2 n+1)\n$$",
    "options": [],
    "answer": "1^2 + 2^2 + 3^2 + ... + n^2 = (1/6) n (n+1) (2n+1)",
    "solution": "Solution:\n\na) Ao virar uma delas de ponta cabeça e juntar com outra, formamos um retângulo $n \\times (n+1)$. Como este retângulo possui o dobro de números de uma tabela, cada uma delas possui $\\frac{n(n+1)}{2}$ números.\n\nb) Cada número $k$ aparece $k$ vezes na primeira tabela, $k$ vezes na segunda e $k$ vezes na terceira. Portanto, cada número $k$ aparece $3k$ vezes no total.\n\nc) Na primeira tabela, a linha determina o valor daquela posição. Como o número inicial $n$ é reduzido em uma unidade a cada linha mais abaixo, então na linha $i$ e coluna $j$ teremos $n+1-i$. Usando o mesmo raciocínio na segunda tabela, na linha $i$ e coluna $j$ teremos o número $n+1-j$. Já na terceira tabela, observe que as posições associadas a elementos de uma diagonal possuem a soma de sua linha e coluna constante. Além disto, o número escrito em uma casa de uma diagonal é uma unidade a menos que esta constante. Logo, na linha $i$ e coluna $j$ teremos $i+j-1$. Na tabela do resultado teremos o número\n$$\n(n+1-i)+(n+1-j)+(i+j-1)=2n+1\n$$\n\nd) Usando os itens anteriores, notamos que a tabela do resultado possui $\\frac{n(n+1)}{2}$ elementos iguais a $2n+1$. Deste modo, podemos somar os números das três tabelas dadas de duas maneiras diferentes: somando cada número $k$ nas três tabelas ou somando os números da tabela resultante. Assim,\n$$\n\\begin{aligned}\n(3 \\cdot 1) \\cdot 1+(3 \\cdot 2) \\cdot 2+\\ldots+(3 \\cdot n) \\cdot n & =\\frac{n(n+1)}{2} \\cdot (2n+1) \\\\\n3\\left(1^{2}+2^{2}+3^{2}+\\ldots+(n-1)^{2}+n^{2}\\right) & =\\frac{1}{2} n(n+1)(2n+1) \\\\\n1^{2}+2^{2}+3^{2}+\\ldots+(n-1)^{2}+n^{2} & =\\frac{1}{6} n(n+1)(2n+1)\n\\end{aligned}\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25879,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAs férias de Tomás - Durante suas férias, Tomás teve 11 dias com chuva. Durante esses 11 dias, se chovia pela manhã havia sol sem chuva à tarde, e se chovia à tarde, havia sol sem chuva pela manhã. No total, Tomás teve 9 manhãs e 12 tardes sem chuva. Quantos dias duraram as férias de Tomás?",
    "options": [],
    "answer": "16",
    "solution": "Solution:\n\n16 dias",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25880,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nEncontre a lei que forma a sequência e dê seus próximos 2 termos:\n$425, 470, 535, 594, 716, 802, \\ldots$",
    "options": [],
    "answer": "870, 983",
    "solution": "Solution:\nCada termo é a soma do termo precedente com os quadrados de cada um de seus algarismos:\n$$\n470 = 425 + 4^2 + 2^2 + 5^2, \\quad 535 = 470 + 4^2 + 7^2 + 0^2, \\ldots\n$$\nAssim, os próximos termos são: $870$ e $983$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25881,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNessa exercício, as letras representam algarismos. Determine cada uma das parcelas da soma abaixo.\n\n$$\n\\begin{array}{r}\na b c d e f \\\\\na b c d e f \\\\\n+\\quad g h i j \\\\\\hline d e f h j f\n\\end{array}\n$$",
    "options": [],
    "answer": "abcdef = 247501, ghij = 6389",
    "solution": "Solution:\n\n$$\n\\begin{array}{r}\n231468 \\\\\n231468 \\\\\n+\\quad 5972 \\\\\\hline 468908\n\\end{array}\n$$\n\n$$\n\\begin{array}{r}\n264538 \\\\\n264538 \\\\\n+\\quad 548698 \\\\\\hline 538178\n\\end{array}\n$$\n\n$$\n\\begin{array}{r}\n273548 \\\\\n273548 \\\\\n+\\quad 548698 \\\\\\hline 548698\n\\end{array}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25882,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSeja $ABC$ um triângulo acutângulo com alturas $BD$ e $CE$. Os pontos $F$ e $G$ são os pés das perpendiculares $BF$ e $CG$ à reta $DE$. Prove que $EF = DG$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOs ângulos $FBE$ e $DEC$ possuem a mesma medida, pois ambos são o complemento do ângulo $FEB$.\n\nObserve que o quadrilátero $BCDE$ é inscritível. De fato, a circunferência de diâmetro $BC$ contém $E$ e $D$, pois $\\angle BEC = \\angle BDC = 90^\\circ$.\n\nSegue que $\\angle FBE = \\angle DEC = \\angle DBC$.\n\nPortanto, $\\triangle BEF \\sim \\triangle BCD$ e obtemos\n$$\n\\frac{EF}{DC} = \\frac{BE}{BC} \\Longrightarrow EF = \\frac{BE \\times DC}{BC}\n$$\n\nAnalogamente, o triângulo $CDG$ é semelhante ao triângulo $CBE$, donde obtemos\n$$\nDG = \\frac{DC \\times BE}{BC}\n$$\ne segue que $EF = DG$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25883,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDuas caixas contêm juntas 65 bolas de vários tamanhos. Cada bola é branca, preta, vermelha ou amarela. Cada vez que pegamos cinco bolas da mesma cor, pelo menos duas são do mesmo tamanho.\n\na. Qual é o número máximo de tipos de bolas que existem nas caixas? Duas bolas são consideradas de tipos distintos quando têm diferentes cores ou tamanhos.\n\nb. Mostrar que existem pelo menos três bolas, que estão na mesma caixa, e que são do mesmo tipo.",
    "options": [],
    "answer": "a) 16. b) There exist at least three balls in the same box that are of the same type.",
    "solution": "Solution:\n\n(a) Não podem existir cinco bolas da mesma cor e tamanhos diferentes porque cada vez que pegamos cinco bolas da mesma cor, duas devem ser do mesmo tamanho. Assim, existem no máximo quatro tamanhos para cada cor. Logo, existem no máximo $4 \\times 4 = 16$ tipos de bolas.\n\n(b) As duas caixas possuem juntas, 65 bolas e uma delas deve conter no mínimo 33 bolas. Por outro lado, existem no máximo 16 tipos de bolas e como $2 \\times 16 = 32 < 33$, concluímos que essa caixa contém três ou mais bolas do mesmo tipo.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25884,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDuas frações irredutíveis têm seus denominadores iguais a 600 e 700. Encontrar o valor mínimo para o denominador da soma das frações.",
    "options": [],
    "answer": "168",
    "solution": "Solution:\n\nSuponhamos que as frações são $\\frac{a}{600}$ e $\\frac{b}{700}$. Como são irredutíveis, então $a$ e $600$ não têm fator comum maior que $1$ e o mesmo acontece com $b$ e $700$.\n\nSomando as duas frações obtemos\n$$\n\\frac{a}{600} + \\frac{b}{700} = \\frac{7a + 6b}{4200} = \\frac{7a + 6b}{2^{3} \\times 3 \\times 5^{2} \\times 7}\n$$\nObserve que o numerador não é divisível nem por $2$ e nem por $3$, porque $a$ não tem fator comum com $6$. O numerador também não é divisível por $7$ porque $b$ e $7$ não têm fator comum.\n\nAssim, o único fator do denominador que possivelmente podemos simplificar é $5^{2} = 25$. Para isto basta pegar, por exemplo, $a = 1$ e $b = 3$.\n$$\n\\frac{1}{600} + \\frac{3}{700} = \\frac{1}{168}\n$$\nPortanto, o denominador mínimo da soma é $168$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25885,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSeja $x_{1}, x_{2}, \\ldots, x_{n}$ uma sequência na qual cada termo é $0$, $1$ ou $-2$. Se\n$$\n\\left\\{\n\\begin{array}{l}\nx_{1}+x_{2}+\\cdots+x_{n}=-5 \\\\\nx_{1}^{2}+x_{2}^{2}+\\cdots+x_{n}^{2}=19\n\\end{array}\n\\right.\n$$\ndetermine $x_{1}^{5}+x_{2}^{5}+\\cdots+x_{n}^{5}$.",
    "options": [],
    "answer": "-125",
    "solution": "Solution:\n\nSejam $a$ a quantidade de termos iguais a $1$ e $b$ a quantidade de termos iguais a $-2$. Podemos escrever:\n$$\n\\left\\{\n\\begin{array}{l}\na \\cdot 1 + b \\cdot ( -2 ) = -5 \\\\\na \\cdot 1^{2} + b \\cdot ( -2 )^{2} = 19\n\\end{array}\n\\Longleftrightarrow\n\\left\\{\n\\begin{array}{l}\na - 2b = -5 \\\\\na + 4b = 19\n\\end{array}\n\\right.\n\\right.\n$$\nResolvendo o sistema, obtemos $a=3$ e $b=4$. Logo,\n$$\nx_{1}^{5}+x_{2}^{5}+\\cdots+x_{n}^{5}=a \\cdot 1^{5}+b \\cdot(-2)^{5}=3-4 \\cdot 32=-125\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25886,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm grupo de meninos e meninas se reúne para comer pizzas que são cortadas em 12 pedaços. Cada menino pode comer 6 ou 7 pedaços e cada menina pode comer 2 ou 3 pedaços. Sabemos que quatro pizzas nunca são suficientes para alimentar o grupo e que com cinco pizzas sempre há sobra. Quantos meninos e quantas meninas formam o grupo?",
    "options": [],
    "answer": "8 boys and 1 girl",
    "solution": "Solution:\n\nChamemos de $x$ o número de meninos e de $y$ o número de meninas. Pelas condições do problema sabemos que se eles comem o mínimo possível, ainda assim quatro pizzas não são suficientes, isto é,\n$$\n6x + 2y > 4 \\times 12 = 48\n$$\nPor outro lado, se eles comem o máximo possível, com cinco pizzas sobrará, isto é,\n$$\n7x + 3y < 5 \\times 12 = 60\n$$\nAssim, precisamos encontrar dois números naturais $x$ e $y$ que satisfaçam simultaneamente\n$$\n\\left\\{\n\\begin{array}{l}\n3x + y > 24 \\\\\n7x + 3y < 60\n\\end{array}\n\\right.\n$$\nComo $7x \\leqslant 7x + 3y < 60$, $x < 60/7 < 9$, logo o número de meninos é menor ou igual a 8.\nPor outro lado, como $x$ e $y$ são inteiros, então $3x + y \\geqslant 25 > 24$, multiplicando por 3, obtemos $9x + 3y \\geqslant 75$, e como $-7x - 3y > -60$, somando estas duas desigualdades (as duas têm o mesmo sentido), encontramos que $2x > 75 - 60 = 15$, ou $x > 7,5$. Portanto, o número de meninos é 8.\nSubstituindo $x = 8$ nas desigualdades obtemos $y > 0$ e $3y < 4$, que tem como única solução $y = 1$. Assim, o grupo tem oito meninos e uma menina.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25887,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGuilherme escreveu um número em cada casa de um tabuleiro $8 \\times 8$ de modo que a soma dos números das casas vizinhas de cada casa do tabuleiro é igual a 1. Calcule a soma de todos os números escritos por Guilherme.\n\nObservação: duas casas são vizinhas se possuem um lado em comum.",
    "options": [],
    "answer": "20",
    "solution": "Solution:\n\nNumere as casas do tabuleiro conforme mostrado na figura 117.1.\n\nA soma dos números das casas marcadas com um mesmo número é igual a 1, porque elas são as vizinhas a uma determinada casa.\n\n| 1 | 2 | 1 | 7 | 8 | 7 | 8 | 9 |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| 2 | 1 | 3 | 6 | 7 | 8 | 9 | 10 |\n| 4 | 3 | 6 | 3 | 6 | 11 | 10 | 9 |\n| 5 | 4 | 3 | 6 | 11 | 12 | 11 | 10 |\n| 4 | 5 | 18 | 17 | 12 | 11 | 12 | 13 |\n| 5 | 18 | 17 | 18 | 17 | 12 | 13 | 14 |\n| 19 | 20 | 18 | 17 | 16 | 15 | 14 | 13 |\n| 20 | 19 | 20 | 16 | 15 | 16 | 15 | 14 |\n\nFigura 117.1\n\nLogo, a soma de todos os números do tabuleiro é igual a 20.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25888,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrês formigas estão paradas em três dos quatro vértices de um retângulo no plano. As formigas se movem no plano uma por vez. A cada vez, a formiga que se move o faz segundo a reta paralela à determinada pelas posições das outras duas formigas. É possível que, após alguns movimentos, as formigas se situem nos pontos médios de três dos quatro lados do retângulo original?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nObserve que, se uma formiga $A$ se movimenta sobre uma reta paralela à reta determinada pelas outras duas formigas $B$ e $C$, então a área do triângulo com vértices sobre as três formigas é invariante, já que a base $BC$ e a medida da altura do triângulo com relação ao lado $BC$ não mudam.\n\nInicialmente, a área do triângulo $ABC$ é a metade da área do retângulo. Porém, se as formigas conseguissem chegar aos pontos médios, a área determinada por elas seria $1/4$ da área do retângulo.\n\nComo a área não é a mesma, é impossível que as formigas se situem nos pontos médios dos lados do retângulo, a partir da configuração inicial.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25889,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$P$ é um ponto no interior do quadrado $ABCD$ tal que $PA = 1$, $PB = 2$ e $PC = 3$. Qual é a medida do ângulo $A \\hat{P} B$?\n\n![](attached_image_1.png)\nFigura 119.1",
    "options": [],
    "answer": "135°",
    "solution": "Solution:\n\nSeja $Q$ um ponto tal que os triângulos $CQB$ e $APB$ são congruentes, como mostrado na figura. Isto é equivalente a fazer uma rotação do triângulo $APB$ com centro em $B$ e ângulo $90^{\\circ}$ no sentido horário. Em particular, temos que $\\angle PBQ = 90^{\\circ}$. Assim, $PQ^{2} = PB^{2} + BQ^{2} = 2^{2} + 2^{2}$, donde $PQ = 2 \\sqrt{2}$.\n\nPor outro lado,\n$$\nPC^{2} = 9 = 8 + 1 = PQ^{2} + QC^{2}\n$$\ne segue que o triângulo $PCQ$ é retângulo com ângulo reto em $Q$.\nPortanto, $\\angle APB = \\angle BQC = \\angle BQP + \\angle PQC = 45^{\\circ} + 90^{\\circ} = 135^{\\circ}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25890,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEm uma folha de papel, Emanuelle desenha duas circunferências de raio $1$ que se tangenciam em um ponto. Em seguida, ela desenha uma terceira circunferência de raio $1-\\sqrt{2}$ que tangencia as duas anteriores externamente, conforme a figura abaixo.\n\n![](attached_image_1.png)\n\nEmanuelle calcula a área da região limitada e exterior às três circunferências que é mostrada em cinza na figura acima. Qual o valor por ela encontrado?",
    "options": [],
    "answer": "1 - pi*(2 - sqrt(2))/2",
    "solution": "Solution:\n\nSejam $A$, $B$ e $C$ os centros das circunferências $C_{1}$, $C_{2}$ e $C_{3}$ respectivamente. Considere o triângulo cujos vértices são dados por esses pontos como na figura abaixo.\n\n![](attached_image_2.png)\n\nChamemos de $P$ o ponto em que $C_{1}$ tangencia $C_{2}$, de $Q$ o ponto em que $C_{2}$ tangencia $C_{3}$ e $R$ o ponto em que $C_{3}$ tangencia $C_{1}$. Note que a área da região procurada pode ser encontrada subtraindo-se da área do triângulo $ABC$ a soma das áreas dos setores circulares $APR$, $BPQ$ e $CQR$, conforme mostrado na figura abaixo:\n\n![](attached_image_3.png)\n\nComo os raios das circunferências $C_{1}$ e $C_{2}$ são iguais a $1$, temos que o comprimento dos segmentos $AP$ e $PB$ é dado por $|AP|=1$ e $|PB|=1$. Logo, o segmento $AB$ tem comprimento $|AB|=|AP|+|PB|=2$. Como a circunferência $C_{3}$ tem raio igual a $\\sqrt{2}-1$ temos que $|CQ|=\\sqrt{2}-1$. Logo, o comprimento do segmento $CB$ é dado por $|CB|=|CQ|+|QB|=\\sqrt{2}$. De maneira análoga, temos que $|CA|=\\sqrt{2}$. Segue daí que o triângulo $ABC$ é isósceles e tem dois lados de comprimento $\\sqrt{2}$ e um lado de comprimento $2$. Como $2^{2}=\\sqrt{2}^{2}+\\sqrt{2}^{2}$ temos que $|AB|^{2}=|BC|^{2}+|CA|^{2}$, isto é, os lados do triângulo $ABC$ satisfazem a relação dada pelo Teorema de Pitágoras. Assim, temos que esse triângulo é retângulo, sendo sua hipotenusa o lado $AB$. Assim, o ângulo $c$ mostrado na figura acima mede $90^{\\circ}$. Como ele também é isósceles, os outros ângulos medem $45^{\\circ}$. Em particular a sua área é igual à metade do produto dos catetos, isto é $(\\sqrt{2} \\times \\sqrt{2}) / 2=1$.\n\nPodemos agora calcular a área dos setores circulares, mostrados na figura abaixo:\n\n![](attached_image_4.png)\n\nComo $a=45^{\\circ}$ temos que a área do setor circular $APR$ é igual a $1/8$ da área do disco delimitado por $C_{1}$, isto é\n$$\n\\text{Área do setor } APR=\\frac{1}{8} \\pi 1^{2}=\\frac{\\pi}{8}\n$$\nAnalogamente, temos que\n$$\n\\text{Área do setor } BPQ=\\frac{1}{8} \\pi 1^{2}=\\frac{\\pi}{8}\n$$\nComo $c=45^{\\circ}$ temos que a área do setor circular $CQR$ é igual a $1/4$ da área do disco delimitado pela circunferência $C_{3}$, logo:\n$$\n\\text{Área do setor } CQR=\\frac{1}{4} \\pi(\\sqrt{2}-1)^{2}=\\frac{(3-2 \\sqrt{2}) \\pi}{4}\n$$\nLogo, obtemos que a área da região encontrada por Emanuelle é igual a\n$$\n1-\\pi\\left(\\frac{3-2 \\sqrt{2}}{4}+\\frac{1}{8}+\\frac{1}{8}\\right)=1-\\pi \\frac{(2-\\sqrt{2})}{2}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25891,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\na) Leandro desenha uma Lúnula de Hipócrates como mostrado na figura a seguir:\n![](attached_image_1.png)\nNesta figura, o triângulo $A B C$ é retângulo e isósceles. A lúnula é a região em forma de lua crescente interna a uma semicircunferência e externa à outra semicircunferência, como mostra a figura. A primeira tem raio igual ao comprimento do cateto $A B$ e a segunda tem raio igual à metade do comprimento da hipotenusa $B C$. Mostre que a área da Lúnula de Hipócrates desenhada por Leandro é igual à área do triângulo retângulo $A B C$.\n\nb) Inspirado pelo desenho de Leandro, Renato decide desenhar as Lúnulas de Alhazen, conforme mostrado na figura abaixo:\n![](attached_image_2.png)\nNessa figura, as lúnulas são as regiões em forma de lua crescente. Um triângulo retângulo $A B C$ e três semicircunferências são utilizadas para obter essas regiões. O comprimento do raio da maior das semicircunferências é igual à metade do comprimento da hipotenusa, enquanto que as duas menores têm raio igual à metade do comprimento do cateto correspondente. Mostre que a soma das áreas das duas lúnulas desenhadas por Renato é igual à área do triângulo $A B C$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\na) Denotemos por $r$ o comprimento dos catetos $A B$ e $A C$. Aplicando o Teorema de Pitágoras no triângulo retângulo $A B C$, temos que o comprimento da hipotenusa $B C$ é igual a $\\sqrt{2} r$. Observe a figura abaixo:\n![](attached_image_3.png)\nComo mostrado acima, o comprimento do raio da semicircunferência de centro em $D$ e extremos $B$ e $C$ é igual à metade do comprimento da hipotenusa, ou seja, igual a $\\frac{\\sqrt{2}}{2} r$.\nA área da semicircunferência com centro em $D$ é igual a\n$$\n\\frac{1}{2} \\cdot \\pi\\left(\\frac{\\sqrt{2}}{2} r\\right)^{2}=\\frac{\\pi r^{2}}{4}\n$$\nA área do quarto de circunferência com centro em $A$ e extremos $B$ e $C$ é igual a\n$$\n\\frac{\\pi r^{2}}{4}\n$$\nLogo, são iguais! Observe que a área hachurada abaixo é comum a essas duas regiões (a semicircunferência de centro em $D$ e o quarto de circunferência com centro em $A$ ).\n![](attached_image_4.png)\nComo as duas regiões têm mesma área e, de cada uma delas foi retirada uma mesma área (a região hachurada), o que sobrou também é igual. Portanto, as áreas em cinza são iguais.\n\nb) Observe a figura abaixo:\n![](attached_image_5.png)\nVamos chamar de $S_{E}$ a área da semicircunferência com centro em $E$ e extremos $A$ e $B$. Temos que\n$$\nS_{E}=\\frac{1}{2} \\pi\\left(\\frac{c}{2}\\right)^{2}=\\frac{\\pi c^{2}}{8}\n$$\nVamos chamar de $S_{F}$ a área da semicircunferência com centro em $F$ e extremos $A$ e $C$. Temos que\n$$\nS_{F}=\\frac{1}{2} \\pi\\left(\\frac{b}{2}\\right)^{2}=\\frac{\\pi b^{2}}{8}\n$$\nVamos chamar de $S_{G}$ a área da semicircunferência com centro em $G$ e extremos $B$ e $C$. Temos que\n$$\nS_{G}=\\frac{1}{2} \\pi\\left(\\frac{a}{2}\\right)^{2}=\\frac{\\pi a^{2}}{8}\n$$\nComo o triângulo $A B C$ é retângulo, pelo Teorema de Pitágoras, sabemos que\n$$\na^{2}=b^{2}+c^{2}\n$$\nDisso, podemos concluir a relação entre as áreas\n$$\nS_{G}=S_{E}+S_{F}\n$$\nObserve agora a figura abaixo:\n![](attached_image_6.png)\nA região hachurada cobre tanto parte da semicircunferência de centro em $G$ como parte das semicircunferências com centros em $E$ e $F$. Como $S_{G}=S_{E}+S_{F}$, e da região $S_{G}$ foi retirada uma área igual à área retirada das regiões $S_{E}$ e $S_{F}$ (a área hachurada), concluímos que o que restou também é igual! Logo, a área do triângulo retângulo $A B C$ é igual à soma das áreas em cinza entre os arcos de circunferência.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25892,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nConsidere a tabela de números a seguir. A primeira linha possui os números de 1 até $n$. A segunda possui os números de 1 até $n$ com cada um multiplicado por 2. As linhas seguem este padrão até a última linha que apresenta $n$ vezes cada número de 1 até $n$.\n![](attached_image_1.png)\nVamos usá-la para calcular o valor da expressão\n$$\n1^{3}+2^{3}+\\ldots+100^{3}\n$$\nAlém da tabela, usaremos o fato de que\n$$\n1+2+\\ldots+n=\\frac{n(n+1)}{2}\n$$\n(a) Determine a soma de todos os números da linha de número $k$. Com isso, determine uma expressão para a soma de todos os números da tabela.\n(b) Observe pedaços na tabela separando-a em camadas em forma de $L$. Os números em uma certa camada $k$ são: $k, 2 k, \\ldots,(k-1) k, k^{2},(k-1) k, \\ldots, 2 k, k$. Determine a soma dos números desta camada em função de $k$.\n(c) Somando os resultados de todas as camadas, chegaremos ao mesmo resultado que somando todas as linhas. Juntando estas informações determine o valor da expressão:\n$$\n1^{3}+2^{3}+\\ldots+100^{3}\n$$",
    "options": [],
    "answer": "25502500",
    "solution": "Solution:\n\n(a) Os números na linha $t$ são os números de 1 até $n$ multiplicados por $t$. A soma deles é\n$$\n\\begin{aligned}\nt+2 t+3 t+\\ldots+n t & =t(1+2+3+\\ldots+n) \\\\\n& =t \\frac{n(n+1)}{2}\n\\end{aligned}\n$$\nA soma dos números na tabela pode ser calculada usando a soma de todas as linhas. Como $\\frac{n(n+1)}{2}$ é fator comum, ele pode ser colocado em evidência e teremos\n$$\n\\frac{n(n+1)}{2}(1+2+3+\\ldots+n)=\\left(\\frac{n(n+1)}{2}\\right)^{2}\n$$\n\n(b) Somando os números teremos:\n$$\n\\begin{aligned}\nk+2 k+\\ldots+(k-1) k+k^{2}+(k-1) k+\\ldots+k & =k^{2}+2 k(1+2+3+\\ldots+(k-1)) \\\\\n& =k^{2}+2 k \\frac{(k-1) k}{2} \\\\\n& =k^{2}+k^{2}(k-1) \\\\\n& =k^{3}\n\\end{aligned}\n$$\n\n(c) Se fizermos a tabela para $n=100$ teremos 100 linhas e 100 camadas. Com as informações dos itens anteriores podemos concluir que\n$$\n\\begin{aligned}\n1^{3}+2^{3}+\\ldots+100^{3} & =\\left(\\frac{100 \\cdot 101}{2}\\right)^{2} \\\\\n& =5050^{2} \\\\\n& =25502500\n\\end{aligned}",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25893,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNuma Maratona de Matemática, o número de questões é muito grande. O valor de cada questão é igual à sua posição na prova: 1 ponto para a questão 1, 2 pontos para a questão 2, 3 pontos para a questão 3, 4 pontos para a questão 4, \\ldots, 10 pontos para a questão 10, \\ldots\\ e assim por diante. Joana totalizou 1991 pontos na prova, errando apenas uma questão e acertando todas as outras. Qual questão ela errou? Quantas questões tinha a prova?",
    "options": [],
    "answer": "She missed question 25; there were 63 questions.",
    "solution": "Solution:\n\n25 e 63, respectivamente.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25894,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEscolhi quatro frações dentre $\\frac{1}{2}$, $\\frac{1}{4}$, $\\frac{1}{6}$, $\\frac{1}{8}$, $\\frac{1}{10}$ e $\\frac{1}{12}$, cuja soma é $1$. Quais foram as frações que eu não escolhi?",
    "options": [],
    "answer": "1/8 and 1/10",
    "solution": "Solution:\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25895,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nRogério coloca seis bolinhas sobre a mesa de modo a formar dois quadrados, como na figura. Ele percebe que havia esquecido de colocar mais uma bolinha. Complete a figura formada pelas bolinhas com essa bolinha a mais, de modo a formar 3 quadrados.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Place the extra ball at the position that serves as the fourth corner of a square formed with three already placed balls, completing an additional square overlapping the two given ones.",
    "solution": "Solution:\n\n![](attached_image_2.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25896,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nRetângulo quase quadrado - Um terreno retangular é quase quadrado: sua largura e seu comprimento são números inteiros de metros que diferem exatamente de 1 metro. A área do terreno, em metros quadrados, é um número de 4 algarismos, sendo o das unidades de milhar e o das centenas iguais, e o mesmo ocorre com o das dezenas e das unidades. Quais são as possíveis dimensões do terreno?",
    "options": [],
    "answer": "33 by 34, 66 by 67, 99 by 100",
    "solution": "Solution:\n\nA área é um número da forma $a a b b$, onde $a$ e $b$ representam algarismos; agora lembre que\n$$\na a b b = 1100 a + 11 b = 11(100 a + b)\n$$\nSeja $x$ a largura do terreno, logo\n$$\nx(x+1) = 11(100 a + b) \\quad (I)\n$$\ne deduzimos que $x$ ou $x+1$ é um múltiplo de 11. Procurar múltiplos de 11 que satisfaçam a condição (I) é bastante trabalhoso, por isso, para simplificar, vamos estabelecer quais os valores que $x$ pode ter. Vamos procurar os valores mínimo e máximo para $x$:\n\n- Mínimo: a menor área possível é $1111$, logo $x(x+1) = 1111 \\Rightarrow x > 32$ (II).\n- Máximo: a maior área possível é $9999$, logo $x(x+1) = 9999 \\Rightarrow x < 100$ (III).\n\nAgora procuramos $x$ e $x+1$ satisfazendo (I), (II) e (III).\n$$\n\\begin{aligned}\n& 33 \\times 34 = 1122 ; \\quad 43 \\times 44 = 1892 ; \\quad 44 \\times 45 = 1980 ; \\quad 54 \\times 55 = 2970 ; \\quad 55 \\times 56 = 2970 \\\\\n& 65 \\times 66 = 4290 ; \\quad 66 \\times 67 = 4422 ; \\quad 76 \\times 77 = 5852 ; \\quad 77 \\times 78 = 6006 \\\\\n& 87 \\times 88 = 7656 ; \\quad 88 \\times 89 = 7832 ; \\quad 99 \\times 100 = 9900\n\\end{aligned}\n$$\n\nEncontramos 3 possibilidades para $x$: $33$, $66$ e $99$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25897,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n(a) Mostre que não existem dois pontos com coordenadas inteiras no plano cartesiano que estão igualmente distanciados do ponto $\\left(\\sqrt{2}, 1/3\\right)$.\n\n(b) Mostre que existe um círculo no plano cartesiano que contém exatamente 2011 pontos com coordenadas inteiras em seu interior.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n(a) Suponhamos que os $(a, b)$ e $(c, d)$ são pontos com coordenadas inteiras que estão igualmente distanciados do ponto $\\left(\\sqrt{2}, 1/3\\right)$. Assim,\n$$\n\\sqrt{(a-\\sqrt{2})^{2}+\\left(b-\\frac{1}{3}\\right)^{2}}=\\sqrt{(c-\\sqrt{2})^{2}+\\left(d-\\frac{1}{3}\\right)^{2}}\n$$\nDeste modo,\n$$\na^{2}+b^{2}-c^{2}-d^{2}-\\frac{2b}{3}+\\frac{2d}{3}=2\\sqrt{2}(a-c)\n$$\nComo a parte esquerda desta igualdade é racional, devemos ter $a-c=0$ e consequentemente\n$$\na^{2}+b^{2}-c^{2}-d^{2}-\\frac{2b}{3}+\\frac{2d}{3}=0\n$$\nPortanto,\n$$\nb^{2}-d^{2}-\\frac{2b}{3}+\\frac{2d}{3}=(b-d)\\left(b+d-\\frac{2}{3}\\right)=0\n$$\ne como $b+d-2/3 \\neq 0$, segue que $b-d=0$, isto é, $(a, b)$ e $(c, d)$ são o mesmo ponto.\n\n(b) Pelo item (a), não existem dois pontos de coordenadas inteiras à mesma distância de $\\left(\\sqrt{2}, 1/3\\right)$. Podemos então ordenar estes pontos em ordem estritamente crescente de distâncias a $\\left(\\sqrt{2}, 1/3\\right)$. Assim, sendo $d_{i}$ a distância do $i$-ésimo ponto $P_{i}$ a $\\left(\\sqrt{2}, 1/3\\right)$, a circunferência de centro $\\left(\\sqrt{2}, 1/3\\right)$ e raio $r$, com $d_{2011}<r<d_{2012}$, possui exatamente 2011 pontos de coordenadas inteiras em seu interior.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25898,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA sequência $x y z$ - Quais são os valores prováveis de $x, y$ e $z$ na sequência\n$$\n\\frac{1}{2}, \\frac{5}{8}, \\frac{3}{4}, \\frac{7}{8}, x, y, z ?\n$$",
    "options": [],
    "answer": "x = 1, y = 9/8, z = 5/4",
    "solution": "Solution:\n\nIgualando os denominadores, verificamos que a sequência dada é igual a\n$$\n\\frac{4}{8}, \\frac{5}{8}, \\frac{6}{8}, \\frac{7}{8}, x, y, z\n$$\nAssim, o denominador é sempre $8$ e os numeradores são consecutivos. Logo, $x=\\frac{8}{8}=1$, $y=\\frac{9}{8}$ e $z=\\frac{10}{8}=\\frac{5}{4}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25899,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nJulian treina em uma pista de $3~\\mathrm{km}$. Ele percorre o primeiro quilômetro caminhando, o segundo correndo, e o terceiro em bicicleta. Se ele tivesse percorrido toda a pista em bicicleta, haveria demorado 10 minutos a menos. Julian corre ao dobro da velocidade com que caminha, e vai em bicicleta ao triplo da velocidade com que caminha. Quanto tempo Julian leva para correr $1~\\mathrm{km}$?",
    "options": [],
    "answer": "6 minutes",
    "solution": "Solution:\nSeja $t$ o tempo, em minutos, que Julian demora para percorrer de bicicleta um quilômetro. Como ele vai de bicicleta ao triplo da velocidade com que caminha, então ele caminha um quilômetro em $3t$ minutos, e como ele corre ao dobro da velocidade com que caminha, então ele corre um quilômetro em $\\dfrac{3t}{2}$ minutos. Assim, ele levou\n$$\nt + 3t + \\frac{3t}{2} = \\frac{11t}{2}\n$$\nminutos para percorrer os três quilômetros da pista. Se ele tivesse percorrido os três quilômetros da pista em bicicleta, ele haveria demorado $3t$ minutos. Assim, conforme o enunciado do problema, temos (em minutos)\n$$\n\\frac{11t}{2} - 3t = 10\n$$\ne, resolvendo a equação anterior, obtemos $t = 4$ minutos. Finalmente, Julian corre um quilômetro em $\\dfrac{3t}{2} = 6$ minutos.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25900,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAna, Beto e Carolina vão participar do programa de televisão \"Descubra a cor do seu chapéu\". No programa, eles se posicionam em roda e sobre a cabeça de cada um será colocado um chapéu azul ou verde. Cada um pode ver os chapéus dos outros, mas não a cor do seu próprio chapéu. Em seguida, cada um deles escreve em um papel uma dentre três opções \"azul\", \"verde\" ou \"passo\". Se todos os que escreveram cores \"azul\" ou \"verde\" acertarem a cor do seu chapéu, eles ganham um carro $0 \\mathrm{~km}$. Se algum deles chutar a cor do chapéu, \"azul\" ou \"verde\", e errar, os três perdem. Se todos eles escreverem \"passo\", então os três também perdem. Vale ressaltar que eles não podem combinar sinais e não podem ver os papéis dos outros participantes. Os três se reúnem para tentar combinar uma estratégia. Carolina começa 'nenhum de nós deve escrever 'passo', devemos chutar entre 'azul' e 'verde', pois se todos passarmos perderemos\". Beto reage dizendo \"discordo, melhor apenas Ana chutar a cor do seu chapéu, enquanto eu e Carolina escrevemos 'passo'. Neste caso, a chance de ganhar será maior\". Ana se pronuncia \"tive uma ideia, se usarmos a minha estratégia teremos a probabilidade de $\\frac{3}{4}$ de ganhar o carro\".\n\n(a) Seguindo a ideia de Carolina, qual a probabilidade de ganhar o carro?\n\n(b) Mudando para a ideia de Beto, qual passa a ser a probabilidade de ganhar o carro?\n\n(c) Dê um exemplo da possível estratégia de Ana que faz a probabilidade de ganhar o carro $\\operatorname{ser} \\frac{3}{4}$.",
    "options": [],
    "answer": "(a) 1/8; (b) 1/2; (c) Example strategy: each player passes when the two hats they see are of different colors; if the two hats they see are the same color, they guess their own hat is the opposite color. This wins in 6 of the 8 cases, giving probability 3/4.",
    "solution": "Solution:\n\n(a) Seguindo a ideia de Carolina, cada pessoa tem duas cores possíveis de chapéu, no total há $2 \\cdot 2 \\cdot 2=8$ possibilidades de chutes. Entre eles, há apenas um caso favorável: acertarem seus chutes. Portanto, a probabilidade de ganhar o carro é $\\frac{1}{8}$.\n\n(b) Com a ideia de Beto, há agora apenas duas possibilidades: o chapéu de Ana é azul ou verde. Com um chute, ela terá probabilidade $\\frac{1}{2}$ de acertar e os três ganharem. Veja que Beto e Carolina não influenciarão o resultado, pois eles vão simplesmente passar.\n\n(c) Cada pessoa olha as cores dos chapéus dos seus dois companheiros. Se forem de cores diferentes, esta pessoa deve passar. Se forem da mesma cor, então esta pessoa chuta que seu chapéu é da outra cor. Por exemplo, se os chapéus de Ana, Beto e Carolina forem azul, azul e verde, respectivamente, então Ana e Beto devem escrever \"passo\", pois enxergam chapéus de cores diferentes, enquanto Carolina deve escrever \"verde\" que é a cor diferente da cor dos dois chapéus que ela vê. Neste caso os três ganhariam o carro. No total há 8 possibilidades para as cores dos três chapéus. Note que, com a estratégia de Ana, os três perdem apenas em duas possibilidades, todos os chapéus verdes ou todos os chapéus azuis. Nas outras seis possibilidades, haverá dois chapéus de uma cor e um chapéu da outra e, como no exemplo, os três ganham. Concluímos que a probabilidade de ganhar o carro é $\\frac{6}{8}=\\frac{3}{4}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25901,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nVladimir escolheu três algarismos $a, b$ e $c$ tais que $a > b > c > 0$ e com eles formou os números $abc$, $cba$ e $cab$. Note que $abc$ não é o produto de $a, b$ e $c$, mas sim o número de algarismos $a, b$ e $c$. Por exemplo, se $a = 1$, $b = 2$ e $c = 3$, $abc$ será o número 123.\nDepois de escolher estes três algarismos $a, b$ e $c$, Vladimir percebeu que um dos números formados era igual à soma dos outros dois. Encontre os números formados por Vladimir.",
    "options": [],
    "answer": "954, 459, 495",
    "solution": "Solution:\nComo $a > c$, o número $abc$ é maior que os números $cba$ e $cab$, e então devemos ter $abc = cba + cab$. Assim,\n$$\n100 \\cdot a + 10 \\cdot b + c = (100 \\cdot c + 10 \\cdot b + a) + (100 \\cdot c + 10 \\cdot a + b)\n$$\ne então $89a = 199c + b$.\nNote que $10 > a = (199 \\cdot c + b)/89 > 2c$, e que $a = (199 \\cdot c + b)/89 < (199 \\cdot c + 10 \\cdot c)/89 < 3c$. Em particular, $5 > c$. Separemos em casos conforme o valor de $c$.\n- Se $c = 1$: Deveríamos ter $2 < a < 3$ e não há algarismo $a$ que cumpra isso.\n- Se $c = 2$: Deveríamos ter $4 < a < 6$. Então $a = 5$ e $b = 89 \\cdot a - 199 \\cdot c = 47$. Não há solução nesse caso.\n- Se $c = 3$: Deveríamos ter $6 < a < 9$. Então $a = 7$ ou $a = 8$, e $b = 89 \\cdot a - 199 \\cdot c \\geq 89(7) - 199(3) = 26$. Não há solução nesse caso.\n- Se $c = 4$: Deveríamos ter $8 < a < 12$. Então $a = 9$ e $b = 89 \\cdot a - 199 \\cdot c = 5$. Nesse caso, os números formados por Vladimir são 954, 459 e 495.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25902,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDois jogadores se enfrentam em um jogo de combate com dados. O atacante lançará três dados e o defensor, dois. O atacante derrotará o defensor em apenas um lance de dados se, e somente se, as duas condições seguintes forem satisfeitas:\ni) O maior dado do atacante for maior do que o maior dado do defensor.\nii) O segundo maior dado do atacante for maior do que o segundo maior dado do defensor (convencionamos que o \"segundo maior dado\" pode ser igual ao maior dado, caso dois ou mais dados empatem no maior valor).\nConsiderando que todos os dados são honestos com os resultados equiprováveis, calcule a probabilidade de o atacante vencer com o defensor conseguindo nos dados dele:\na) 2 cincos;\nb) 1 cinco e 1 quatro.",
    "options": [],
    "answer": "a) 2/27; b) 43/216",
    "solution": "Solution:\n\na) Para ganhar, precisamos tirar ao menos dois $6$. A probabilidade será igual a tirar:\n- três seis: $P_{1} = \\left(\\frac{1}{6}\\right)^{3}$; ou\n- dois seis e outro número qualquer menor que $6$: $P_{2} = 3 \\cdot \\left(\\frac{1}{6}\\right)^{2} \\cdot \\frac{5}{6} = \\frac{15}{6^{3}}$.\nPortanto, a probabilidade é $\\frac{1}{6^{3}} + \\frac{15}{6^{3}} = \\frac{16}{6^{3}} = \\frac{2}{27}$.\n\nb) Para ganhar, precisamos tirar ao menos um maior do que $5$ e outro maior do que $4$. A probabilidade será igual a tirar:\n- três seis: $P_{1} = \\left(\\frac{1}{6}\\right)^{3}$; e\n- dois seis e outro qualquer $(<6)$: $P_{2} = 3 \\cdot \\left(\\frac{1}{6}\\right)^{2} \\cdot \\frac{5}{6} = \\frac{15}{6^{3}}$;\n- um seis e dois cincos: $P_{3} = 3 \\cdot \\frac{1}{6} \\cdot \\left(\\frac{1}{6}\\right)^{2} = \\frac{3}{6^{3}}$; ou\n- um seis, um cinco e outro qualquer $(<5)$: $P_{4} = 3! \\cdot \\frac{1}{6} \\cdot \\frac{1}{6} \\cdot \\frac{4}{6} = \\frac{24}{6^{3}}$;\nPortanto, ficamos com\n$$\n\\frac{1}{6^{3}} + \\frac{15}{6^{3}} + \\frac{3}{6^{3}} + \\frac{24}{6^{3}} = \\frac{43}{216}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25903,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm número não divisível por 5 - Determine quais números naturais $n$ entre 2001 e 2007, tornam o número $1^{n} + 2^{n} + 3^{n} + 4^{n}$ não divisível por 5.",
    "options": [],
    "answer": "2004",
    "solution": "Solution:\n\n2004",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25904,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAonde está o erro? - Seja $x$ solução de $x^{2}+x+1=0$. Então $x \\neq 0$ e por isso podemos dividir ambos os membros da equação por $x$, obtendo $x+1+\\frac{1}{x}=0$. Da equação temos que $x+1=-x^{2}$, logo $-x^{2}+\\frac{1}{x}=0$, isto é: $x^{2}=1 / x$ ou ainda $x^{3}=1$ e $x=1$. Substituindo $x=1$ na equação $x^{2}+x+1=0$ encontramos $3=0$ !!!! Aonde erramos?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nAonde está o erro? - Esse deixamos para os alunos!",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25905,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEscolhi quatro frações entre $1/2$, $1/4$, $1/6$, $1/10$ e $1/12$ cuja soma é $1$. Quais foram as frações que eu não escolhi?",
    "options": [],
    "answer": "1/10",
    "solution": "![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25906,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrês casais jantam todo sábado num mesmo restaurante, sempre à mesma mesa. A mesa é redonda e os casais combinaram que\n\na. jamais marido e mulher sentam à mesa como vizinhos; e\n\nb. a disposição dos seis à mesa é diferente a cada sábado.\n\nDesconsiderando rotações nas disposições à mesa, durante quantos sábados esses três casais poderão ir a esse restaurante sem repetir sua disposição à mesa?",
    "options": [],
    "answer": "16",
    "solution": "Solution:\n\nPara simplificar, vamos denotar cada casal por um par de números, um número representando o marido e o outro a mulher. Temos, então, os três pares $(1,2)$, $(3,4)$, $(5,6)$, que não podem ser vizinhos. Podemos considerar o lugar do marido $1$ à mesa como sendo fixo, já que desconsideramos rotações na disposição à mesa. Duas disposições possíveis são dadas na figura seguinte.\n\n![](data:image/svg+xml;base64,<svg id="smiles-lvno8p52x93grggy5m" xmlns="http://www.w3.org/2000/svg" viewBox="0 0 198 124.31346678717857" style="width: 197.8134667871659px; height: 124.31346678717857px; overflow: visible;"><defs><linearGradient id="line-lvno8p52x93grggy5m-1" gradientUnits="userSpaceOnUse" x1="124.31346678717857" y1="62.15677033615326" x2="155.8134667871659" y2="62.15679861076762"><stop stop-color="currentColor" offset="20%"></stop><stop stop-color="currentColor" offset="100%"></stop></linearGradient><linearGradient id="line-lvno8p52x93grggy5m-3" gradientUnits="userSpaceOnUse" x1="112.25891254534561" y1="91.25896479002058" x2="124.31346678717857" y2="62.15677033615326"><stop stop-color="currentColor" offset="20%"></stop><stop stop-color="currentColor" offset="100%"></stop></linearGradient><linearGradient id="line-lvno8p52x93grggy5m-5" gradientUnits="userSpaceOnUse" x1="112.25896479002056" y1="33.054554241832946" x2="124.31346678717857" y2="62.15677033615326"><stop stop-color="currentColor" offset="20%"></stop><stop stop-color="currentColor" offset="100%"></stop></linearGradient><linearGradient id="line-lvno8p52x93grggy5m-7" gradientUnits="userSpaceOnUse" x1="83.15669645102533" y1="103.31346678717857" x2="112.25891254534561" y2="91.25896479002058"><stop stop-color="currentColor" offset="20%"></stop><stop stop-color="currentColor" offset="100%"></stop></linearGradient><linearGradient id="line-lvno8p52x93grggy5m-9" gradientUnits="userSpaceOnUse" x1="83.15677033615323" y1="21" x2="112.25896479002056" y2="33.054554241832946"><stop stop-color="currentColor" offset="20%"></stop><stop stop-color="currentColor" offset="100%"></stop></linearGradient><linearGradient id="line-lvno8p52x93grggy5m-11" gradientUnits="userSpaceOnUse" x1="54.054501997158" y1="91.25891254534561" x2="83.15669645102533" y2="103.31346678717857"><stop stop-color="currentColor" offset="20%"></stop><stop stop-color="currentColor" offset="100%"></stop></linearGradient><linearGradient id="line-lvno8p52x93grggy5m-13" gradientUnits="userSpaceOnUse" x1="54.05455424183293" y1="33.054501997158" x2="83.15677033615323" y2="21"><stop stop-color="currentColor" offset="20%"></stop><stop stop-color="currentColor" offset="100%"></stop></linearGradient><linearGradient id="line-lvno8p52x93grggy5m-15" gradientUnits="userSpaceOnUse" x1="42" y1="62.156696451025354" x2="54.054501997158" y2="91.25891254534561"><stop stop-color="currentColor" offset="20%"></stop><stop stop-color="currentColor" offset="100%"></stop></linearGradient><linearGradient id="line-lvno8p52x93grggy5m-17" gradientUnits="userSpaceOnUse" x1="42" y1="62.156696451025354" x2="54.05455424183293" y2="33.054501997158"><stop stop-color="currentColor" offset="20%"></stop><stop stop-color="currentColor" offset="100%"></stop></linearGradient></defs><mask id="text-mask-lvno8p52x93grggy5m"><rect x="0" y="0" width="100%" height="100%" fill="white"></rect></mask><style>
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"></text></g></svg>/>\n132546\n![](data:image/svg+xml;base64,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/>\n164523\nEssas disposições descrevemos por $132546$ e $164523$, sempre começando em $1$, fixado numa posição à mesa, e prosseguindo no sentido horário. Assim, o problema se resume em encontrar todos os números de $6$ algarismos distintos que podem ser escritos com os algarismos $1, 2, 3, 4, 5$ e $6$, onde:\n\na. todos os números começam com o algarismo $1$;\n\nb. nenhum número pode terminar com o algarismo $2$;\n\nc. não podem aparecer juntos $1$ e $2$, $3$ e $4$, $5$ e $6$.\n\nEncontramos os $16$ números da tabela seguinte.\n\n| 132546 | 132645 | 135246 | 135264 | 135426 | 136245 | 136254 | 136425 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| 142536 | 142635 | 145236 | 145326 | 146235 | 146325 | 153246 | 154236 |\n\nLogo, a resposta é $16$ sábados.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25907,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nNo seguinte tabuleiro $4 \\times 4$ devem ser colocados 4 torres, 4 cavalos, 4 bispos e 4 peões de modo que em cada linha e em cada coluna as peças colocadas sejam distintas, como no exemplo:\n\n| $B$ | $T$ | $P$ | $C$ |\n| :--- | :--- | :--- | :--- |\n| $P$ | $B$ | $C$ | $T$ |\n| $T$ | $C$ | $B$ | $P$ |\n| $C$ | $P$ | $T$ | $B$ |\n\nOs peões são rebeldes e decidiram ficar nas seguintes posições:\n\n|  |  | $P$ |  |\n| :--- | :--- | :--- | :--- |\n| $P$ |  |  |  |\n|  |  |  | $P$ |\n|  | $P$ |  |  |\n\nCalcule o número de modos em que as outras peças podem ser colocadas.",
    "options": [],
    "answer": "24",
    "solution": "Solution:\nTrês peças devem ser colocadas na primeira linha, vamos chamá-las de $\\alpha$, $\\beta$ e $\\theta$ como na seguinte figura:\n\n![](attached_image_1.png)\n\nNa casa $(b, 2)$ deve ser colocada uma peça diferente de $\\beta$. Suponhamos primeiro que em $(b, 2)$ colocamos a peça $\\alpha$. É imediato que na casa $(c, 2)$ devemos colocar uma peça $\\theta$.\n\n|  | 1 | 2 |  | 4 |\n| :---: | :---: | :---: | :---: | :---: |\n|  | $\\alpha$ | $\\beta$ | $P$ | $\\theta$ |\n|  | $P$ | $\\alpha$ |  |  |\n|  |  | $\\theta$ |  | $P$ |\n|  |  | $P$ |  |  |\n\nObserve que na casa $(b, 4)$ não pode estar uma peça $\\alpha$, porque já há uma peça $\\alpha$ na linha $b$, nem uma peça $\\theta$, porque já há uma peça $\\theta$ na coluna 4. Isto implica que em $(b, 4)$ deve ir uma peça $\\beta$ e, em consequência, em $(b, 3)$ e $(d, 4)$ devem ir uma peça $\\theta$ e uma peça $\\alpha$ respectivamente.\n\n|  | $\\alpha$ | $\\beta$ | $P$ | $\\theta$ |\n| :---: | :---: | :---: | :---: | :---: |\n| $\\{ \\}_\\{b\\}$ | $P$ | $\\alpha$ | $\\theta$ | $\\beta$ |\n|  |  | $\\theta$ |  | $P$ |\n|  |  | $P$ |  |  |\n\nA peça em $(d, 3)$ deve ser distinta de $\\theta$ e $\\alpha$, porque já há uma peça $\\theta$ na coluna 3 e uma peça $\\alpha$ na linha $d$. Portanto em $(d, 3)$ devemos colocar uma peça $\\beta$.\n\n|  |  |  |  |  |\n| :---: | :---: | :---: | :---: | :---: |\n|  | $\\alpha$ | $\\beta$ | $P$ | $\\theta$ |\n|  | $P$ | $\\alpha$ | $\\theta$ | $\\beta$ |\n|  |  | $\\theta$ |  | $P$ |\n|  |  | $P$ | $\\beta$ |  |\n\nAgora é claro que só existe um modo de poder completar o tabuleiro. Concluímos que somente existe um modo de completar o tabuleiro se colocarmos uma peça $\\alpha$ na posição $(b, 2)$.\n\n|  |  |  |  |  |\n| :---: | :---: | :---: | :---: | :---: |\n|  | $\\alpha$ | $\\beta$ | $I$ | $\\theta$ |\n|  | $P$ | $\\alpha$ |  | $\\beta$ |\n|  | $\\beta$ | $\\theta$ |  | $P$ |\n|  | $\\theta$ | $P$ | $\\beta$ |  |\n\nVamos supor agora que em $(b, 2)$ colocamos uma peça $\\theta$. Nesse caso em $(c, 2)$ é necessário colocar uma peça $\\alpha$.\n\n|  | 1 |  |  |  |\n| :---: | :---: | :---: | :---: | :---: |\n|  | $\\alpha$ | $\\beta$ | $P$ | $\\theta$ |\n|  | $P$ | $\\theta$ |  |  |\n|  |  | $\\alpha$ |  | $P$ |\n|  |  | $P$ |  |  |\n\nPara as posições $(b, 3)$ e $(b, 4)$ devemos usar as peças $\\alpha$ e $\\beta$. Temos, assim, duas alternativas. Suponhamos primeiro que em $b-3$ colocamos uma peça $\\alpha$ e em $(b, 4)$ colocamos uma peça $\\beta$. Em consequência, em $(d, 4)$ deve ir uma peça $\\alpha$.\n\n|  | 1 |  |  |  |\n| :---: | :---: | :---: | :---: | :---: |\n|  | $\\alpha$ | $\\beta$ | $P$ | $\\theta$ |\n|  | $P$ | $\\theta$ | $\\alpha$ | $\\beta$ |\n|  |  | $\\alpha$ |  | $P$ |\n|  |  | $P$ |  |  |\n\nFinalmente, temos duas formas de completar o tabuleiro.\n\n|  |  |  |  |  |\n| :---: | :---: | :---: | :---: | :---: |\n|  | $c$ | $\\beta$ | $F$ | $\\theta$ |\n|  | $P$ | $\\theta$ | $\\alpha$ | $\\beta$ |\n|  | $\\beta$ | $\\alpha$ |  | $P$ |\n|  | $\\theta$ | $P$ | $\\beta$ |  |\n\n|  |  |  |  |  |\n| :---: | :---: | :---: | :---: | :---: |\n|  | $c$ | $\\beta$ | $P$ |  |\n|  | $P$ | $\\theta$ | $\\alpha$ | $P$ |\n|  | $\\theta$ | $\\alpha$ | $\\beta$ | $P$ |\n|  | $\\beta$ | $P$ | $\\theta$ |  |\n\nSuponhamos agora que em $(b, 3)$ colocamos uma peça $\\beta$ e em $(b, 4)$ colocamos uma peça $\\alpha$. Imediatamente vemos que em $(d, 4)$ devemos colocar uma peça $\\beta$.\n\n|  |  | 2 |  |  |\n| :---: | :---: | :---: | :---: | :---: |\n|  | $\\alpha$ | $\\beta$ | $P$ | $\\theta$ |\n|  | $P$ | $\\theta$ | $\\beta$ | $\\alpha$ |\n|  |  | $\\alpha$ |  | $P$ |\n|  |  | $P$ |  | $\\beta$ |\n\nNa casa $(d, 1)$ não pode ter uma peça $\\alpha$ nem uma peça $\\beta$ porque já há uma peça $\\alpha$ na coluna 1 e uma peça $\\beta$ na linha $d$. Portanto devemos colocar uma peça $\\theta$ em $(d, 1)$ e assim somente existirá um modo de poder completar o tabuleiro.\n\n| 1 | 2 |  | 3 |  |\n| :--- | :---: | :---: | :---: | :---: |\n| $a$ | $\\alpha$ | $\\beta$ | $P$ | $\\theta$ |\n| $b$ | $P$ | $\\theta$ | $\\beta$ | $\\alpha$ |\n|  |  |  |  |  |\n| $c$ | $\\beta$ | $\\alpha$ | $\\theta$ | $P$ |\n|  | $\\theta$ | $P$ | $\\alpha$ | $\\beta$ |\n|  | $\\theta$ |  |  |  |\n\nAssim obtivemos, no total, 4 maneiras de colocar as peças no tabuleiro de modo que as peças $\\alpha, \\beta$ e $\\theta$ sejam as peças colocadas na primeira linha. Finalmente $\\alpha$ é alguma das 3 peças, $\\beta$ alguma das outras 2 peças e $\\theta$ será a única peça que resta. Temos assim $3 \\times 2$ possibilidades para o trio $(\\alpha, \\beta, \\theta)$. A resposta é portanto\n$$\n4 \\times 3 \\times 2=24\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25908,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSobre um quadro negro existem os números $1$, $2$, $3$, $4$, $5$ e $6$. Em cada movimento, João pode trocar dois números $a$ e $b$ por $a \\cdot b + a + b$. Encontre todas as possibilidades para o último número no quadro negro.",
    "options": [],
    "answer": "5039",
    "solution": "Solution:\n\nRealizemos uma sequência de movimentos para analisar o comportamento da lista de números ao longo dos movimentos:\n$$\n\\begin{aligned}\n1,2,3,4,5,6 & \\rightarrow \\\\\n1,11,4,5,6 & \\rightarrow \\\\\n1,11,5,34 & \\rightarrow \\\\\n\\mathbf{69}, 11,5 & \\rightarrow \\\\\n\\mathbf{11, 419} & \\rightarrow \\\\\n5039 . &\n\\end{aligned}\n$$\nA expressão $a \\cdot b + a + b$ é muito semelhante à expressão $(a+1) \\cdot (b+1)$. Veja que se somarmos $1$ em todos os números antes de um movimento, os multiplicarmos e fizermos o mesmo nos números após um movimento, obteremos o mesmo resultado $S$, pois\n$$\n(a+1) \\cdot (b+1) = [(a \\cdot b + a + b) + 1]\n$$\nNo exemplo inicial, temos\n$$\n\\begin{array}{rl}\n1, \\mathbf{2, 3}, 4,5,6 & S=(1+1)(2+1)(3+1)(4+1)(5+1)(6+1)=5040 \\\\\n1,11, \\mathbf{4}, 5, \\mathbf{6} & S=(1+1)(11+1)(4+1)(5+1)(6+1)=5040 \\\\\n\\mathbf{1, 11, 5, 34} & S=(1+1)(11+1)(5+1)(34+1)=5040 \\\\\n\\mathbf{69}, 11, \\mathbf{5} & S=(69+1)(11+1)(5+1)=5040 \\\\\n\\mathbf{11, 419} & S=(11+1)(419+1)=5040 \\\\\n5039 & S=(5039+1)=5040\n\\end{array}\n$$\nLogo, independentemente de como sejam realizadas as operações, a quantidade $S$ não será alterada e o número final sempre será\n$$\n\\begin{aligned}\nS & =(1+1)(2+1)(3+1)(4+1)(5+1)(6+1)-1 \\\\\n& =5039\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25909,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSejam $ABCD$ um losango, com $\\angle BAD > \\angle ABC$, e $P$ e $Q$ pontos nos lados $AB$ e $AD$, respectivamente, tais que o triângulo $PCQ$ é equilátero, com lado igual ao lado do losango. Encontre as medidas dos ângulos do losango.",
    "options": [],
    "answer": "80°, 80°, 100°, 100°",
    "solution": "Solution:\n\nComo $\\angle ABC = \\angle QDC = \\alpha$ e $BC = PC = QC = CD$, segue que $\\angle BCP = \\angle DCQ = 180^\\circ - 2\\alpha$. Assim,\n$$\n\\begin{aligned}\n180^\\circ - \\alpha & = \\angle BCD \\\\\n& = \\left(180^\\circ - 2\\alpha\\right) + 60^\\circ + \\left(180^\\circ - 2\\alpha\\right) \\\\\n& = 420^\\circ - 4\\alpha.\n\\end{aligned}\n$$\nPortanto, $240^\\circ = 3\\alpha$, ou seja, $\\alpha = 80^\\circ$. Logo, os ângulos do losango são $80^\\circ$, $80^\\circ$, $100^\\circ$ e $100^\\circ$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25910,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nManuel é um matemático que gosta de jogos de cartas. Ele encontra os irmãos Jonas e Jonatan durante uma viagem de ônibus e propõe um jogo. Serão usados apenas os quatro ases do baralho, o de copas e o de ouros são vermelhos enquanto o de espadas e o de paus são pretos.\n![](attached_image_1.png)\nManuel será o banco e os dois irmãos, um de cada vez, apostarão 1 real contra ele em cada rodada. As cartas são postas viradas com face para baixo. Jonas escolhe uma carta e Jonatan a vira para cima. Jonas escolhe mais uma carta e Jonatan novamente a vira. Se as duas cartas tiverem a mesma cor, então Jonas ganha 1 real de Manuel. Caso contrário, Manuel ganha 1 real de Jonas. Em seguida, Jonas e Jonatan trocam de posição e o jogo segue. Veja que Manuel não mexe nas cartas, por isto não pode manipular o jogo. Jonatan pensa um pouco e conclui que tem probabilidade de $\\frac{2}{3}$ de vencer, pois os resultados são apenas duas cartas vermelhas, duas pretas ou uma vermelha e uma preta. Será mesmo?\n(a) Jonas já participou de olimpíadas de matemática e decidiu tomar mais cuidado. Ele decidiu analisar este jogo usando uma árvore de possibilidades. Como ficaria a árvore de possibilidades de Jonas?\n(b) Considerando os resultados da árvore do item anterior, qual a probabilidade de Manuel vencer cada rodada do jogo?",
    "options": [],
    "answer": "2/3",
    "solution": "Solution:\n\n(a) Representaremos copas, espadas, ouros e paus pelas letras $C, E, O$ e $P$, respectivamente. A árvore de possibilidades é mostrada na figura a seguir:\n![](attached_image_2.png)\n\n(b) Usando a árvore de possibilidades, há 12 resultados possíveis e em 8 deles Manuel vence. Portanto, a probabilidade de Manuel vencer cada rodada é $\\frac{8}{12}=\\frac{2}{3}$. O raciocínio de Jonatan não está correto, pois as possibilidades de cores não ocorrem com a mesma probabilidade.\n\nVale ressaltar que este item poderia ser respondido sem a árvore de possibilidades. Considere o momento após a virada da primeira carta. Entre as outras três, duas favorecem Manuel, pois possuem a cor diferente da cor da carta que foi virada. Então, assim como na conclusão usando a árvore, a probabilidade de Manuel vencer é $\\frac{2}{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25911,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNo seguinte tabuleiro, devemos colocar todos os números, desde $1$ até $25$, seguindo as seguintes regras:\n- Em cada fila, os $5$ números colocados devem formar uma sequência estritamente crescente quando lidas da esquerda para a direita.\n- Em cada coluna, os $5$ números colocados devem formar uma sequência estritamente crescente quando lidas de cima para baixo.\n\n![](attached_image_1.png)\n\nA diagonal principal do tabuleiro é a diagonal pintada no gráfico.\n\na) Mostre que depois de colocar os números, o menor número colocado na diagonal principal é sempre maior ou igual a $5$, e o maior número colocado na diagonal principal é maior ou igual a $15$.\n\nb) Coloque os números no tabuleiro de modo que na diagonal principal encontremos os números $5, 9, 12, 14$ e $15$.\n\nc) Mostre que é impossível colocar os números de modo que a soma na diagonal principal seja menor que $55$.",
    "options": [],
    "answer": "An example main diagonal is 5, 9, 12, 14, 15; the smallest diagonal entry is always at least 5, the largest is always at least 15, and the minimal possible diagonal sum is 55.",
    "solution": "Solution:\n\na) Vamos colocar as seguintes letras para representar os números colocados no tabuleiro, da maneira mostrada na seguinte figura:\n\n| $M$ | $N$ | $P$ | $Q$ | $E$ |\n| :---: | :---: | :---: | :---: | :---: |\n| $R$ | $S$ | $T$ | $D$ |  |\n| $U$ | $V$ | $C$ |  |  |\n| $W$ | $B$ |  |  |  |\n| $A$ |  |  |  |  |\n\nObserve que, pelas regras,\n- $A$ é maior que os números $M, R, U$ e $W$.\n- $B$ é maior que os números $M, R, U, W, N, S$ e $V$.\n- $C$ é maior que os números $M, R, U, N, S, V, P$ e $T$.\n- $D$ é maior que os números $M, N, P, Q, R, S$ e $T$.\n- $E$ é maior que os números $M, N, P$ e $Q$.\n\nPodemos assim concluir que qualquer número na diagonal principal é sempre maior que pelo menos $4$ números distintos e, portanto, são todos maiores ou iguais a $5$.\n\nb) Podemos colocar primeiro os seguintes números para conseguir os números desejados na diagonal principal.\n\n| 1 | 2 | 3 | 4 | 5 |\n| :---: | :---: | :---: | :---: | :---: |\n| 6 | 7 | 8 | 9 |  |\n| 10 | 11 | 12 |  |  |\n| 13 | 14 |  |  |  |\n| 15 |  |  |  |  |\n\nFinalmente, preenchemos o resto das casas, por exemplo, do seguinte modo:\n\n| 1 | 2 | 3 | 4 | 5 |\n| :---: | :---: | :---: | :---: | :---: |\n| 6 | 7 | 8 | 9 | 16 |\n| 10 | 11 | 12 | 17 | 18 |\n| 13 | 14 | 19 | 20 | 21 |\n| 15 | 22 | 23 | 24 | 25 |\n\nc) Ordenemos os números $A, B, C, D$ e $E$ do menor para o maior, obtendo\n$$\n\\alpha < \\beta < \\gamma < \\theta < \\sigma\n$$\nPela parte a), sabemos que\n$$\n\\alpha \\geq 5\n$$\nAgora, $\\{\\alpha, \\beta\\}$ deve coincidir com algum par em $\\{A, B, C, D, E\\}$, digamos que\n$$\n\\{\\alpha, \\beta\\} = \\{A, E\\}\n$$\nComo $\\beta \\geq A$ e $\\beta \\geq E$ então $\\beta$ deve ser maior que os números $M, N, P, Q, R, U$ e $W$. Além desses $7$ números, $\\beta$ deve ser maior do que $\\alpha$, o qual é um número distinto dos $7$. Assim, concluímos que $\\beta$ é maior que $8$ números e portanto $\\beta \\geq 9$.\n\nVamos supor agora que $D \\in \\{\\alpha, \\beta\\}$. Nesse caso, $\\beta \\geq D$ e, portanto, $\\beta$ é maior do que $M, N, P, Q, R, S$ e $T$. Também $\\beta$ é maior do que $\\alpha$, e concluímos que $\\beta$ é maior do que $8$ números distintos. De novo provamos que $\\beta \\geq 9$. De modo análogo, se $B \\in \\{\\alpha, \\beta\\}$ ou $C \\in \\{\\alpha, \\beta\\}$, vamos ter que $\\beta$ é maior do que $8$ números distintos. Assim, temos provado que em qualquer caso,\n$$\n\\beta \\geq 9\n$$\nConsideremos agora o trio $\\{\\alpha, \\beta, \\gamma\\}$. Suponhamos que $\\{\\alpha, \\beta, \\gamma\\} = \\{A, B, E\\}$. Como $\\gamma$ é o máximo número entre $A, B$ e $C$, então $\\gamma$ deve ser maior que\n$$\n\\{M, N, P, Q, R, S, T, U, W\\}\n$$\nAlém disso, $\\gamma$ deve ser maior que $\\alpha$ e $\\beta$. Daí, $\\gamma$ deve ser maior que $11$ números. Portanto, $\\gamma \\geq 12$. Qualquer outra possibilidade para o trio $\\{\\alpha, \\beta, \\gamma\\}$ leva à mesma conclusão:\n$$\n\\gamma \\geq 12\n$$\nQuando consideramos agora as possibilidades para $\\{\\alpha, \\beta, \\gamma, \\theta\\}$, concluímos que $\\theta$ deve ser maior que todos os números em $\\{M, N, P, Q, R, S, T, U, V, W\\}$ e maior que $\\alpha, \\beta$ e $\\gamma$. Portanto\n$$\n\\theta \\geq 14\n$$\nFinalmente, $\\sigma > \\theta$ e, assim,\n$$\n\\sigma \\geq 15\n$$\nDe $(*),(**),(***),(\\star)$ e $(\\star\\star)$, obtemos que\n$$\nA+B+C+D+E = \\alpha + \\beta + \\gamma + \\theta + \\sigma \\geq 5 + 9 + 12 + 14 + 15\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25912,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nQuatro frações e um inteiro - Quantos números naturais $a, b, c$ e $d$, todos distintos, existem tais que $\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{1}{d}$ seja um inteiro?",
    "options": [],
    "answer": "1",
    "solution": "Solution:\n\n1",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25913,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPotências de $3-$ Se $3^{a}=2$, quanto vale $27^{2 a}$ ?",
    "options": [],
    "answer": "64",
    "solution": "Solution:\n\nTemos $27^{2 a} = (3^{3})^{2 a} = 3^{6 a} = (3^{a})^{6} = 2^{6} = 64$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25914,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSeja $n$ um número inteiro positivo maior ou igual a 5. Para números $a_{i}$ escolhidos no conjunto $\\{-1,1\\}$, calcula-se o número\n$$\nS_{n}=a_{1} a_{2} a_{3} a_{4}+\\ldots+a_{n} a_{1} a_{2} a_{3}\n$$\nque soma os produtos de cada quatro termos $a_{i}$ de índices consecutivos, inclusive os que começam em $a_{n-2}, a_{n-1}$ e $a_{n}$ e terminam em $a_{1}, a_{2}$ e $a_{3}$, respectivamente.\n\na. Considerando $n=8$, comecemos com $a_{1}=a_{2}=\\ldots=a_{7}=a_{8}=1$. Qual o valor de $S_{8}$? Se trocarmos $a_{4}=1$ por $a_{4}=-1$ quanto passa a ser a soma $S_{8}$? Após a primeira troca, trocamos $a_{5}=1$ por $a_{5}=-1$. Após esta segunda troca, quanto vale $S_{8}$?\n\nb. Para cada troca de 1 por -1, quantas parcelas mudam de valor? Quais são as possíveis variações no valor de $S_{8}$ quando se faz uma troca?\n\nc. Mostre que para quaisquer oito valores de $a_{1}, a_{2}, \\ldots, a_{7}$ e $a_{8}$ no conjunto $\\{-1,1\\}$ a soma $S_{8}$ resulta sempre em um número múltiplo de 4.\n\nd. Para certo valor de $n$ e certa escolha dos números $a_{i}$ no conjunto $\\{-1,1\\}$ a soma\n$$\nS_{n}=a_{1} a_{2} a_{3} a_{4}+\\ldots+a_{n} a_{1} a_{2} a_{3}\n$$\nresultou em zero. Prove que $n$ é necessariamente um número múltiplo de 4.",
    "options": [],
    "answer": "a) 8; after flipping the fourth entry: 0; after also flipping the fifth entry: 4.\nb) Four terms change; the possible changes in the sum are −8, −4, 0, 4, 8.\nc) For any eight choices, the sum is always a multiple of 4.\nd) The length n must be a multiple of 4.",
    "solution": "Solution:\n\na.\nCom os valores dados, tem-se:\n$$\nS_{8}=1 \\cdot 1 \\cdot 1 \\cdot 1+1 \\cdot 1 \\cdot 1 \\cdot 1+\\ldots+1 \\cdot 1 \\cdot 1 \\cdot 1=8\n$$\nÉ a soma de oito parcelas iguais a 1. Veja que ao trocar o $a_{4}$ de 1 para -1, os quatro produtos em que ele aparece mudam de sinal. Então a soma perde quatro parcelas 1 que passam a ser quatro parcelas -1. Deste modo, a soma passa a ser\n$$\nS_{8}^{\\prime}=8-4+(-4)=0\n$$\nSe trocarmos agora o $a_{5}$ de 1 para -1, há quatro parcelas afetadas, mas algumas passam de 1 para -1 e outras passam de -1 para 1. Mais especificamente, as parcelas com o $a_{5}$ que já mudaram de sinal com o $a_{4}$ voltarão a ser 1. A parcela $a_{5} a_{6} a_{7} a_{8}$ passa de $1$ a $-1$ e as outras três passam de -1 a 1. Após a segunda troca a soma será\n$$\n\\begin{aligned}\nS_{8}^{\\prime \\prime} & =S_{8}^{\\prime}-(1+(-1)+(-1)+(-1))+((-1)+1+1+1) \\\\\n& =S_{8}^{\\prime}-(-2)+2 \\\\\n& =S_{8}^{\\prime}+4 \\\\\n& =4\n\\end{aligned}\n$$\n\nb.\nComo vimos no item anterior, as quatro parcelas em que o produto possui certo $a_{i}$ mudam de valor quando trocamos este número de 1 para -1. Para saber as possíveis variações, considere $x, y, z$ e $w$ as parcelas que possuem o $a_{i}$ no produto.\n$$\n\\begin{aligned}\nS_{8}^{\\prime} & =S_{8}-(x+y+z+w)+(-x-y-z-w) \\\\\n& =S_{8}-2(x+y+z+w)\n\\end{aligned}\n$$\nComo $x, y, z$ e $w$ são produtos de números 1 ou -1, eles mesmos são iguais a 1 ou -1. Então ao somar os quatro, os resultados possíveis são $1+1+1+1=4$, $1+1+1+(-1)=2$, $1+1+(-1)+(-1)=0$, $1+(-1)+(-1)+(-1)=-2$ ou $(-1)+(-1)+(-1)+(-1)=-4$. Finalmente, concluímos que as variações possíveis são $+8,+4,0,-4$ ou $-8$.\n\nc.\nPrimeiro faça todos os números iguais a 1, então a soma é 8. Agora, para cada número da sequência, podemos trocá-lo para -1 e analisar a soma. Deste jeito, todas as possibilidades de números $a_{i}$ são analisadas. Pelo item anterior, cada troca gera uma variação que é um múltiplo de 4. Como no início a soma é um múltiplo de 4 e esta propriedade não se altera em cada troca, concluímos que a soma $S_{8}$ resulta sempre em um múltiplo de 4.\n\nd.\nNovamente, comece com todos os números iguais a 1 resultando em soma $n$. Para uma dada escolha dos elementos da sequência, trocamos cada $a_{i}$ igual a -1 por 1, um por vez. Em cada troca, não altera-se o resto de $S_{n}$ na divisão por 4 e, ao final, chegamos no número 0 que é múltiplo de 4. Portanto, o número inicial $n$ também é um múltiplo de 4.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25915,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nNa figura abaixo, encontre o valor de\n$$\n\\angle G A B+\\angle A B C+\\angle B C D+\\angle C D E+\\angle D E F+\\angle E F G+\\angle F G A\n$$\n![](attached_image_1.png)",
    "options": [],
    "answer": "540°",
    "solution": "Solution:\nConsidere o ponto $O$ no interior da figura, como indicado no desenho abaixo.\n![](attached_image_2.png)\nEntão\n$$\n\\begin{array}{r}\n\\angle G A B+\\angle A B C+\\angle B C D+\\angle C D E+\\angle D E F+\\angle E F G+\\angle F G A+ \\\\\n(\\angle G A O+\\angle O A B)+(\\angle A B O+\\angle O B C)+(\\angle B C O+\\angle O C D)+ \\\\\n+(\\angle C D O+\\angle O D E)+(\\angle D E O+\\angle O E F)+(\\angle E F O+\\angle O F G)+ \\\\\n+(\\angle F G O+\\angle O G A)\n\\end{array}\n$$\nVeja agora que a soma dos ângulos entre parênteses correspondem a 7 pares de ângulos dos triângulos $D O E, E O F, F O G, G O A, A B O, B O C$ e $C D O$. Como a soma dos ângulos desses triângulos é $7 \\cdot 180^\\circ=1260^\\circ$ e\n$$\n\\begin{aligned}\n\\angle D O E+\\angle E O F+\\angle F O G+\\angle G O A+\\angle A O B+\\angle B O C+\\angle C O D & =2 \\cdot 360^\\circ \\\\\n& =720^\\circ\n\\end{aligned}\n$$\nsegue, após comparação com a primeira equação, que a soma procurada é\n$$\n1260^\\circ-720^\\circ=540^\\circ\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25916,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nProve que se um polígono possui uma circunferência tangente a todos os seus lados, então é possível encontrar três lados do polígono que formam um triângulo.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSeja $DE$ o maior lado do polígono e considere os outros dois lados adjacentes a ele, denotados por $CD$ e $EF$ na figura a seguir.\n![](attached_image_1.png)\nAlém disso, sejam $K$, em $CD$, e $I$, em $DE$, e $J$, em $EF$ os pontos de tangências desses lados com a circunferência mencionada no enunciado. Como $KD = DI = x$ e $EI = EJ = y$, temos\n$$\n\\begin{aligned}\nDE & = x + y \\\\\n& < CD + EF\n\\end{aligned}\n$$\nSabendo que $DE$ é o maior lado do polígono, também temos $DE + CD > EF$ e $DE + EF > CD$. Em virtude da Desigualdade Triangular, os segmentos $DE$, $CD$ e $EF$ são lados de um triângulo.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25917,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO Rei Arthur teve que lutar com o Dragão das Três Cabeças e Três Caudas. Sua tarefa ficou facilitada quando conseguiu arranjar uma espada mágica que podia, de um só golpe, fazer uma e somente uma das seguintes coisas:\n\n- cortar uma cabeça;\n- cortar duas cabeças;\n- cortar uma cauda;\n- cortar duas caudas.\n\nAlém disso, a Fada Morgana lhe revelou o segredo do dragão:\n\n- se uma cabeça é cortada uma nova cresce;\n- se duas cabeças são cortadas nada acontece;\n- no lugar de uma cauda nascem duas caudas novas;\n- se duas caudas são cortadas uma nova cabeça cresce e\n- o dragão morre se perder as três cabeças e as três caudas.\n\nQuantos golpes o Rei Artur vai precisar para matar o dragão?",
    "options": [],
    "answer": "5",
    "solution": "Solution:\n\n5",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25918,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCururu é um sapo estranho, que se desloca apenas com dois tipos de saltos:\n\nTipo I: $10~\\mathrm{cm}$ para o Leste e $30~\\mathrm{cm}$ para o Norte;\nTipo II: $20~\\mathrm{cm}$ para Oeste e $40~\\mathrm{cm}$ para o Sul.\n\n![](attached_image_1.png)\n\na. Como Cururu faz para chegar a um ponto situado a $190~\\mathrm{cm}$ para o Leste e $950~\\mathrm{cm}$ para o Norte de sua casa?\n\nb. É possível Cururu chegar a um ponto situado a $180~\\mathrm{cm}$ para o Leste e $950~\\mathrm{cm}$ para o Norte de sua casa?",
    "options": [],
    "answer": "a) 57 jumps of type I and 19 jumps of type II (in any order). b) No, it is impossible.",
    "solution": "Solution:\n\nA cada $x$ saltos do tipo I, o sapo se desloca $10x~\\mathrm{cm}$ para o Leste e $30x~\\mathrm{cm}$ para o Norte e, a cada $y$ saltos do tipo II, o sapo se desloca $20y~\\mathrm{cm}$ para o Oeste e $40y~\\mathrm{cm}$ para o Sul. Assim, ao final de $x$ saltos do tipo I e $y$ do tipo II, o sapo se deslocou $10x - 20y~\\mathrm{cm}$ para o Leste e $30x - 40y~\\mathrm{cm}$ para o Norte.\n\na.\nResolvendo\n$$\n\\left\\{\\begin{array}{l}\n10x - 20y = 190 \\\\\n30x - 40y = 950\n\\end{array}\\right.\n$$\nencontramos $x = 57$ e $y = 19$. Logo, o sapo deverá dar 57 saltos do tipo I e 19 do tipo II, em qualquer ordem, para alcançar um ponto situado a $190~\\mathrm{cm}$ para o Leste e $950~\\mathrm{cm}$ para o Norte de sua casa.\n\nb.\nUma vez que o número de saltos de cada tipo é um número inteiro, Cururu só alcançará o ponto situado a $180~\\mathrm{cm}$ para o Leste e $950~\\mathrm{cm}$ para o Norte de sua casa se o sistema\n$$\n\\left\\{\\begin{array}{l}\n10x - 20y = 180 \\\\\n30x - 40y = 950\n\\end{array}\\right.\n$$\ntiver solução inteira. Mas a solução desse sistema é $x = 59$ e $y = 41/2$, que não é inteiro. Portanto, Cururu não conseguirá alcançar aquele ponto.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25919,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nJoana escreveu uma sequência em 10 linhas usando os algarismos de 0 a 9, seguindo o padrão seguinte.\n$$\n\\begin{array}{lllllllllll}\n0 & & & & & & & & & \\\\\n1 & 1 & 0 & & & & & & & \\\\\n2 & 2 & 2 & 1 & 1 & 0 & & & & \\\\\n3 & 3 & 3 & 3 & 2 & 2 & 2 & 1 & 1 & 0\n\\end{array}\n$$\nQual foi o algarismo mais usado? Quantas vezes esse algarismo foi utilizado?",
    "options": [],
    "answer": "Digits 4 and 5, each used 30 times.",
    "solution": "Solution:\n\nDe acordo com o padrão da sequência, temos\n\n![](attached_image_1.png)\n\nLogo,\num algarismo $0$ em cada linha dá $1 \\times 10 = 10$ algarismos $0$ no total; dois algarismos $1$ em nove linhas dá $2 \\times 9 = 18$ algarismos $1$ no total; três algarismos $2$ em oito linhas dá $3 \\times 8 = 24$ algarismos $2$ no total; quatro algarismos $3$ em sete linhas dá $4 \\times 7 = 28$ algarismos $3$ no total,\ne assim por diante. Portanto, trata-se de descobrir qual é o maior dos produtos a seguir, onde cada um representa quantos algarismos, de $0$ a $9$, aparecem na sequência.\n\n![](attached_image_2.png)\n\nComo o maior produto é $30$, os algarismos mais usados foram $4$ e $5$, $30$ vezes cada um.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25920,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nExiste algum número inteiro $N$ tal que valha\n$$\n2008 \\times N = 222\\ldots2 ?\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nSolução 1: Se existir esse número $N$, então\n$$\nN = \\frac{222\\ldots2}{2008} = \\frac{2 \\times 111\\ldots1}{2 \\times 1004} = \\frac{111\\ldots1}{1004}\n$$\nLogo, $N$ não é inteiro, por ser o quociente do número ímpar $111\\ldots1$ pelo número par $1004$. Portanto, não existe tal $N$.\n\n\nSolução 2: Fatorando $2008$, obtemos $2008 = 2^{3} \\times 251$, portanto, $2008$ é divisível por $8$. Se existisse um inteiro $N$ tal que $2008 \\times N = 222\\ldots2$, teríamos, então, que $8$ dividiria $222\\ldots2$. Por outro lado, sabemos que um número é divisível por $8$ se, e somente se, o número formado pelos últimos três algarismos for divisível por $8$. Mas $222 = 27 \\times 8 + 6$ não é divisível por $8$. Logo, não existe um número $N$ tal que $2008 \\times N = 222\\ldots2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25921,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSerá verdade que $\\frac{1}{4^{3}}+\\frac{1}{5^{3}}+\\frac{1}{6^{3}}<\\frac{1}{12}$?",
    "options": [],
    "answer": "Yes",
    "solution": "Solution:\nSolução 1: Uma maneira de verificar essa desigualdade é comparando cada parcela desta soma, como segue. Comparando as frações $\\frac{1}{5}$, $\\frac{1}{6}$ e $\\frac{1}{3}$ com $\\frac{1}{4}$, obtemos\n$$\n\\begin{aligned}\n& \\frac{1}{5}<\\frac{1}{4}, \\text{ portanto, } \\frac{1}{5^{3}}=\\left(\\frac{1}{5}\\right)^{3}<\\left(\\frac{1}{4}\\right)^{3}=\\frac{1}{4^{3}} \\\\\n& \\frac{1}{6}<\\frac{1}{4}, \\text{ portanto, } \\frac{1}{6^{3}}=\\left(\\frac{1}{6}\\right)^{3}<\\left(\\frac{1}{4}\\right)^{3}=\\frac{1}{4^{3}} \\\\\n& \\frac{1}{4}<\\frac{1}{3}, \\text{ portanto, } \\frac{1}{4^{3}}=\\left(\\frac{1}{4}\\right)^{3}<\\left(\\frac{1}{3}\\right)^{3}=\\frac{1}{3^{3}}\n\\end{aligned}\n$$\nAssim,\n$$\n\\frac{1}{4^{3}}+\\frac{1}{5^{3}}+\\frac{1}{6^{3}}<\\frac{1}{4^{3}}+\\frac{1}{4^{3}}+\\frac{1}{4^{3}}=\\frac{3}{4^{3}}=\\frac{3}{4} \\times \\frac{1}{4} \\times \\frac{1}{4}<\\frac{3}{4} \\times \\frac{1}{3} \\times \\frac{1}{3}=\\frac{1}{12}\n$$\n\n\nSolução 2: Uma outra maneira de verificar essa desigualdade é testar se\n$$\n\\frac{1}{4^{3}}+\\frac{1}{5^{3}}+\\frac{1}{6^{3}}-\\frac{1}{12}>0\n$$\nReduzindo ao mesmo denominador, temos\n$$\n\\begin{aligned}\n\\frac{1}{2^{6}}+\\frac{1}{5^{3}}+\\frac{1}{2^{3} \\times 3^{3}}-\\frac{1}{2^{2} \\times 3} & =\\frac{2^{4} \\times 3^{2} \\times 5^{3}-3^{3} \\times 5^{3}-2^{6} \\times 3^{3}-2^{3} \\times 5^{3}}{2^{6} \\times 3^{3} \\times 5^{3}} \\\\\n& =\\frac{18000-3375-1728-1000}{2^{6} \\times 3^{3} \\times 5^{3}} \\\\\n& =\\frac{11897}{2^{6} \\times 3^{3} \\times 5^{3}}\n\\end{aligned}\n$$\nNem é preciso efetuar o produto indicado no denominador. Como o numerador é positivo, concluímos que a desigualdade se verifica.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25922,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n156. Parte inteira - A parte inteira de um número real $x$ é o maior inteiro que é menor do que ou igual a $x$. Denotamos a parte inteira de $x$ por $[x]$. Por exemplo, $[2,9]=2$, $[0,88]=0$ e $[-1,7]=-2$.\n![](attached_image_1.png)\nCalcule as partes inteiras seguintes.\n(a) $[\\sqrt{12}]$\n(b) $\\left[\\frac{28756}{12777}\\right]$\n(c) $\\left[-\\frac{2007}{2008}\\right]$\n(d) $[\\sqrt[3]{-111}]$",
    "options": [],
    "answer": "{'a': 3, 'b': 2, 'c': -1, 'd': -5}",
    "solution": "Solution:\n\n(a) Os números $9$ e $16$ são quadrados perfeitos e $9<12<16$. Então,\n$$\n3=\\sqrt{9}<\\sqrt{12}<\\sqrt{16}=4\n$$\n![](attached_image_2.png)\ne, portanto, $[\\sqrt{12}]=3$.\n\n(b) Como $12777 \\times 2<28756<12777 \\times 3$, temos\n$$\n2<\\frac{28756}{12777}<3, \\text{ portanto, }\\left[\\frac{28756}{12777}\\right]=2\n$$\n\n(c) Como $2007<2008$, temos $0<\\frac{2007}{2008}<1$, ou\n$$\n-1< -\\frac{2007}{2008}<0, \\text{ portanto, }\\left[-\\frac{2007}{2008}\\right]=-1\n$$\n![](attached_image_3.png)\n\n(d) Como $4^{3}=64<111<125=5^{3}$, temos\n$$\n(-5)^{3}=-5^{3}=-125<-111<-4^{3}=(-4)^{3}\n$$\n![](attached_image_4.png)\nde modo que $-5<\\sqrt[3]{-111}<-4$ e, portanto, $[\\sqrt[3]{-111}]=-5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25923,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nQuantos números inteiros entre $10$ e $999$ têm a soma de seus algarismos igual a $9$?",
    "options": [],
    "answer": "54",
    "solution": "Solution:\n\nVamos dividir em dois casos: números de dois algarismos e números de três algarismos.\n\nNo caso de números de dois algarismos, temos $18, 27, 36, 45, 54, 63, 72, 81$ e $90$, num total de $9$ números.\n\nDa mesma maneira, listamos os números de três algarismos, como segue:\n$$\n\\begin{aligned}\n108, 117, 126, 135, 144, 153, 162, 171, 180 &\\leadsto 9 \\text{ números;} \\\\\n207, 216, 225, 234, 243, 252, 261, 270 &\\leadsto 8 \\text{ números;} \\\\\n306, 315, 324, 333, 342, 351, 360 &\\leadsto 7 \\text{ números;} \\\\\n405, 414, 423, 432, 441, 450 &\\leadsto 6 \\text{ números;} \\\\\n504, 513, 522, 531, 540 &\\leadsto 5 \\text{ números;} \\\\\n603, 612, 621, 630 &\\leadsto 4 \\text{ números;} \\\\\n702, 711, 720 &\\leadsto 3 \\text{ números;} \\\\\n801, 810 &\\leadsto 2 \\text{ números;} \\\\\n900 &\\leadsto 1 \\text{ número.}\n\\end{aligned}\n$$\nPortanto, temos $9+8+7+6+5+4+3+2+1=45$ números de três algarismos.\n\nAssim, existem $9+45=54$ números entre $10$ e $999$ cuja soma dos algarismos é igual a $9$.\nSolution:\n\nNo caso de números de dois algarismos, uma vez escolhido o algarismo da dezena, o algarismo da unidade fica automaticamente definido. Como o algarismo da dezena pode ser $1, 2, 3, 4, 5, 6, 7, 8$ ou $9$, temos nove números de dois algarismos tais que a soma de seus algarismos seja $9$.\n\nNo caso de números de três algarismos, denotando por $n$ o algarismo da centena, a soma dos algarismos da dezena e da unidade é $9-n$, portanto, temos $9-n+1=10-n$ possibilidades de escolha para o algarismo da dezena, que pode ser inclusive $0$, e o algarismo da unidade fica automaticamente definido. Como o algarismo da centena pode ser $1, 2, 3, 4, 5, 6, 7, 8$ ou $9$, temos\n$$(10-1)+(10-2)+(10-3)+(10-4)+(10-5)+(10-6)+(10-7)+(10-8)+(10-9)=45$$\nnúmeros de três algarismos tais que a soma dos seus algarismos é $9$.\n\nAssim, existem $9+45=54$ números entre $10$ e $999$ cuja soma dos algarismos é igual a $9$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25924,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nRetângulos - As medidas dos lados de um retângulo são números pares. Quantos retângulos desses existem com área igual a 96?",
    "options": [],
    "answer": "4",
    "solution": "Solution:\n\nSe $a$ e $b$ denotam o comprimento e a largura do retângulo, temos $a \\times b = 96$. Logo, $a$ e $b$ são divisores pares de $96$ e, portanto, temos quatro retângulos satisfazendo as condições dadas, a saber, os retângulos de lados medindo $2$ e $48$; $4$ e $24$; $6$ e $16$; e $8$ e $12$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25925,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNúmero de retas - Sabemos que dois pontos distintos determinam uma única reta. Quantas retas são determinadas por dois quaisquer dos nove pontos marcados no quadriculado dado?",
    "options": [],
    "answer": "20",
    "solution": "Solution:\n\nSolução 1: Para contar o número de retas, dividiremos as retas de acordo com suas posições.\n- retas paralelas aos lados dos quadrados: três horizontais e três verticais;\n![](attached_image_1.png)\n- retas paralelas às diagonais dos quadrados: $3+3=6$;\n![](attached_image_2.png)\n- outras retas: temos $4 \\times 2=8$ retas, formando uma estrela, como na figura.\n![](attached_image_3.png)\nAssim, ao todo, temos $3+3+6+8=20$ retas.\n\n\nSolução 2: Note que o ponto central é supérfluo, pois toda reta que passa por ele e um outro ponto do reticulado, passa também por um terceiro ponto do reticulado. Logo, o ponto central pode ser eliminado de nossas considerações. Assim, o problema fica reduzido a calcular quantas retas são determinadas por dois pontos quaisquer dos oito pontos do reticulado dado.\n![](attached_image_4.png)\nCom esses oito pontos, podem ser formados $8 \\times 7$ possíveis pares de pontos distintos (8 possibilidades para o primeiro elemento do par e 7 possibilidades para o segundo). Como a ordem em que o par é formado não influi na reta determinada por ele, esse número deve ser dividido por 2, ou seja,\n$$\n\\frac{8 \\times 7}{2}=28\n$$\nFinalmente, notamos que algumas retas estão sendo contadas três vezes e, portanto, devem ser subtraídas duas vezes desse número 28. São elas as quatro retas que contêm os lados do quadrado que delimita o reticulado. Logo, o número total de retas determinadas pelo reticulado é\n$$\n\\frac{8 \\times 7}{2}-2 \\times 4=20\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25926,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUma ilha possui 50 clubes. Cada habitante da ilha é sócio de 1 ou 2 clubes. Cada clube tem no máximo 55 sócios e para cada par de clubes existe um habitante da ilha que é sócio dos dois clubes. Encontre todas as possibilidades para as quantidades possíveis de habitantes da ilha. Justifique sua resposta.",
    "options": [],
    "answer": "[1225, 1525]",
    "solution": "Solution:\n\nExistem $\\binom{50}{2} = 1225$ pares de clubes e para cada um deles devemos ter um habitante que pertence a ambos. Denotemos esses habitantes por $H_{ij}$ com $i, j \\in \\{1,2, \\ldots, 50\\}$ e $i < j$. Se dois deles são iguais, um habitante estaria em três clubes e isso seria uma contradição. Portanto, todos os habitantes $H_{ij}$ são distintos e isso nos mostra que a ilha precisa ter pelo menos $1225$ pessoas e que cada clube tem pelo menos um habitante do tipo $H_{ij}$.\n\nFixado um clube qualquer, ele possui $49$ habitantes distintos em comum com os outros $49$ clubes. Como ele pode ter no máximo $55$ habitantes, conterá no máximo $6$ que não são da forma $H_{ij}$. Assim, o número máximo de habitantes da ilha é $1225 + 50 \\cdot 6 = 1525$.\n\nConsidere agora um inteiro $n \\in [1225, 1525]$ e $1225$ pessoas distintas. A cada uma associe uma etiqueta $H_{ij}$, com $i, j \\in \\{1,2, \\ldots, 50\\}$ e $i < j$, e coloque a pessoa com etiqueta de índice $i$ no clube $C_i$. Essas serão as únicas pessoas em dois clubes. Como $n - 1225 \\leq 300$, podemos distribuir as pessoas restantes inserindo-as em grupos de até $6$ pessoas nas $55 - 49 = 6$ posições vazias de cada um desses $50$ clubes criados. Elas serão as únicas pessoas que participarão de apenas um clube.\n\nPodemos concluir que o conjunto de valores possíveis é o intervalo de inteiros $[1225, 1525]$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25927,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSejam $a$ e $b$ números reais positivos com produto diferente de $1$, define-se a operação estrela, representada por “$*$”, pela equação\n$$\na * b = \\frac{a + b - 2ab}{1 - ab}.\n$$\nEm uma lousa, estão escritos $2015$ números iguais a $\\frac{1}{2}$. Em cada passo, apagam-se dois números $x$ e $y$ escritos na lousa e escreve-se o número $x * y$. Este passo é repetido $2014$ vezes até que fique apenas um número na lousa.\n\na. Demonstre que a equação\n$$\n\\frac{x * y}{1 - x * y} = \\frac{x}{1 - x} + \\frac{y}{1 - y}\n$$\né verdadeira para quaisquer $x$ e $y$ reais com $x \\neq 1$, $y \\neq 1$ e $xy \\neq 1$.\n\nb. Se para cada número $x$ que é escrito na lousa, calcularmos $\\frac{x}{1 - x}$ e somarmos todos estes resultados, teremos um certo resultado. Mostre que este resultado é sempre o mesmo não importando quantos passos tenham sido feitos até aquele momento.\n\nc. Qual o número que estará escrito na lousa ao final dos $2014$ passos?",
    "options": [],
    "answer": "2015/2016",
    "solution": "Solution:\n\na. Desenvolvendo a expressão da operação estrela, temos:\n$$\n\\begin{aligned}\n\\frac{x * y}{1 - x * y} &= \\frac{\\frac{x + y - 2xy}{1 - xy}}{1 - \\frac{x + y - 2xy}{1 - xy}} \\\\\n&= \\frac{x + y - 2xy}{1 - xy - x - y + 2xy} \\\\\n&= \\frac{x + y - 2xy}{1 - x - y + xy} \\\\\n&= \\frac{x + y - 2xy}{(1 - x)(1 - y)} \\\\\n&= \\frac{x}{1 - x} + \\frac{y}{1 - y}\n\\end{aligned}\n$$\n\nb. Seja $S$ a soma dos termos $\\frac{x}{1 - x}$ para cada $x$ escrito na lousa. Usando o item anterior, concluímos que retirando dois termos $\\frac{x}{1 - x}$ e $\\frac{y}{1 - y}$ e adicionando o termo $\\frac{x * y}{1 - x * y}$, a soma não se altera. Como isto vale para cada passo, então continua valendo não importando quantos passos tenham sido feitos.\n\nc. Seja $N$ o número final. Pelo item anterior, sabe-se que a soma não sofre alteração com as trocas. Portanto, podemos usá-la para descobrir o número final.\n$$\n\\begin{aligned}\n\\frac{N}{1 - N} &= \\frac{1/2}{1 - 1/2} + \\frac{1/2}{1 - 1/2} + \\ldots + \\frac{1/2}{1 - 1/2} \\\\\n&= 1 + 1 + \\ldots + 1 \\\\\n&= 2015 \\\\\nN &= 2015(1 - N) \\\\\n2016N &= 2015 \\\\\nN &= \\frac{2015}{2016}\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25928,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCinco piratas encontraram um cofre do tesouro cheio de moedas de ouro e as dividiram entre si. Sabe-se que:\n- O que o primeiro pirata recebeu é equivalente à metade do que receberam os outros quatro em conjunto.\n- O que o segundo pirata recebeu é equivalente à terça parte do que receberam os outros quatro em conjunto.\n- O que o terceiro pirata recebeu é equivalente à quarta parte do que receberam os outros quatro em conjunto.\n- O que o quarto pirata recebeu é equivalente à quinta parte do que receberam os outros quatro em conjunto.\nSe o quinto pirata recebeu 90 moedas, diga quantas moedas tinha o cofre antes da divisão.",
    "options": [],
    "answer": "1800",
    "solution": "Solution:\n\nSejam $a, b, c, d, e$ as quantidades de moedas recebidas pelos cinco piratas. O número total de moedas é $S = a + b + c + d + e$.\n\nO primeiro pirata recebeu metade do que receberam os outros quatro em conjunto, isto é, $a = (b + c + d + e)/2 = (S - a)/2$. Então $a = S/3$.\n\nO segundo pirata recebeu a terça parte do que receberam os outros quatro em conjunto, isto é, $b = (a + c + d + e)/3 = (S - b)/3$. Portanto, $b = S/4$.\n\nO terceiro pirata recebeu a quarta parte do que receberam os outros quatro em conjunto, isto é, $c = (a + b + d + e)/4 = (S - c)/4$. Então $c = S/5$.\n\nO quarto pirata recebeu a quinta parte do que receberam os outros quatro em conjunto, isto é, $d = (a + b + c + e)/5 = (S - d)/5$. Daí, $d = S/6$.\n\nAssim, o último pirata recebeu\n$$\n90 = e = S - a - b - c - d = S - \\frac{S}{3} - \\frac{S}{4} - \\frac{S}{5} - \\frac{S}{6} = \\frac{S}{20}\n$$\nPortanto, o cofre tinha $S = 1800$ moedas antes da divisão.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25929,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nCarla escreveu no quadro-negro os números inteiros de $1$ até $21$. Diana deseja apagar alguns deles de tal modo que ao multiplicar os números restantes o resultado seja um quadrado perfeito.\n\na) Mostre que Diana deve apagar necessariamente os números $11$, $13$, $17$ e $19$ para conseguir seu objetivo.\n\nb) Qual a menor quantidade de números que Diana deve apagar para atingir o seu objetivo?",
    "options": [],
    "answer": "5",
    "solution": "Solution:\na) Se Diana decidir não apagar o número $11$ então o produto dos números restantes será da forma $P=11 \\times A$. Como $11$ é o único múltiplo de $11$ dentre os números escritos por Carla, então $A$ é o produto de números não divisíveis por $11$, e logo $A$ não é múltiplo de $11$. Assim, $P$ seria múltiplo de $11$ mas não seria múltiplo de $11^2$, logo $P$ não seria quadrado perfeito. De modo análogo, podemos ver que Diana deve também apagar os números $13$, $17$ e $19$.\n\nb) Se Diana apaga somente os números $11$, $13$, $17$ e $19$, o produto dos $17$ números restantes seria\n$$\n1 \\times 2 \\times 3 \\times \\cdots \\times 10 \\times 12 \\times 14 \\times 15 \\times 16 \\times 18 \\times 20 \\times 21=\\left(2^9 \\times 3^4 \\times 5^2 \\times 7\\right)^2 \\times 21\n$$\nEsse número não é um quadrado perfeito e então ela precisa apagar pelo menos mais um número. Da fatoração acima é simples ver que se Diana apaga também o número $21$ ela conseguirá seu objetivo. Assim, a menor quantidade de números que Diana deve apagar é $5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25930,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrês carros partem de uma cidade $A$ ao mesmo tempo e percorrem um caminho fechado composto por três segmentos de reta $AB$, $BC$ e $CA$. As velocidades do primeiro carro sobre esses segmentos são 12, 10 e 15 quilômetros por hora, respectivamente. As velocidades do segundo carro são 15, 15 e 10 quilômetros por hora, respectivamente. Finalmente, as velocidades do terceiro carro são 10, 20 e 12 quilômetros por hora, respectivamente. Encontre o valor do ângulo $\\angle ABC$, sabendo que todos os três carros terminam na cidade $A$ ao mesmo tempo.",
    "options": [],
    "answer": "90 degrees",
    "solution": "Solution:\n\nSejam $x$, $y$ e $z$ os comprimentos de $AB$, $BC$ e $AC$, respectivamente. O tempo de chegada $t$, comum aos três carros, pode ser encontrado através das equações:\n$$\n\\left\\{\\begin{array}{l}\n\\frac{x}{12}+\\frac{y}{10}+\\frac{z}{15}=t \\\\\n\\frac{x}{15}+\\frac{y}{15}+\\frac{z}{10}=t \\\\\n\\frac{x}{10}+\\frac{y}{20}+\\frac{z}{12}=t\n\\end{array}\\right.\n$$\nMultiplicando todas as equações por 60, obtemos\n$$\n\\left\\{\\begin{aligned}\n5x+6y+4z & =60t \\\\\n4x+4y+6z & =60t \\\\\n6x+3y+5z & =60t\n\\end{aligned}\\right.\n$$\nDa segunda equação, temos $x+y=(60t-6z)/4=(30t-3z)/2$. Substituindo este valor na terceira equação, temos\n$$\nx=\\frac{60t-3(x+y)-5z}{3}=\\frac{30t-z}{6}\n$$\nAlém disto,\n$$\ny=\\frac{30t-3z}{2}-\\frac{30t-z}{6}=\\frac{60t-8z}{6}\n$$\nFinalmente, substituindo os dois valores encontrados na primeira equação do último sistema, obtemos\n$$\n\\begin{aligned}\n5 \\cdot \\frac{30t-z}{6}+6 \\cdot \\frac{60t-8z}{6}+4z & =60t \\\\\n150t-5z+360t-48z+24z & =360t \\\\\nt & =\\frac{29}{150}z\n\\end{aligned}\n$$\nSubstituindo nas expressões para $x$ e $y$, podemos concluir que $(x, y, z)=(4z/5, 3z/5, z)$. Como os lados do triângulo $ABC$ estão na proporção $3:4:5$, podemos concluir que ele é semelhante ao triângulo retângulo de lados 3, 4 e 5. Consequentemente, $\\angle ABC=90^\\circ$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25931,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPara fazer a separação em regiões da correspondência que deve ser entregue, um serviço postal indica sobre os envelopes um código postal com uma série de 5 grupos de bastões, que podem ser lidos por um leitor ótico. Os algarismos são codificados como a seguir:\n\n| $0 \\bullet\\|I\\|$ | $5 \\% \\cdot \\\\|$ |\n| :---: | :---: |\n| 1 $\\cdot$॰\\|l | $6\\|\\cdot\\| \\\\| \\mid$ |\n| $2 \\cdot\\|\\| \\\\|$ | $7\\\\|\\bullet \\bullet\\\\|$ |\n| $3 \\cdot \\\\| 1 \\mid 10$ | $8 \\mathrm{ll} \\cdot \\mathrm{Ol}$ |\n| 4 \\|०o|l| | 9\\||$\\|\\bullet \\circ\\|$ |\n\nA leitura se faz da direita para a esquerda, por exemplo o código postal 91720 se escreve como ..IIIII.III.IIII..III.I.IIIII.I.I. Em detalhe:\n\n![](attached_image_1.png)\n\nNote que a codificação de $94, \\underbrace{\\mid \\circ \\cdot \\| I \\text{ III०| }}_{4}$, tem um eixo vertical de simetria. Encontre os códigos de 47000 a 47999, aqueles que apresentam um eixo vertical de simetria.",
    "options": [],
    "answer": "47679, 47779",
    "solution": "Solution:\n\n47679 e 47779",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25932,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNum tabuleiro $5 \\times 5$, um cavaleiro do jogo de xadrez está na casa marcada com $A$. Depois ele se move marcando as casas por onde passa:\n\n$A \\rightarrow B \\rightarrow C \\rightarrow D \\rightarrow E \\rightarrow F \\rightarrow G \\rightarrow H$.\n\n| $A$ |   |   |   | $G$ |\n| :--- | :--- | :--- | :--- | :--- |\n|   |   | $H$ |   |   |\n|   | $B$ |   | $F$ |   |\n|   |   | $D$ |   |   |\n| $C$ |   |   |   | $E$ |\n\nPartindo da casa $H$, o cavaleiro se move pelo tabuleiro até ter passado por todas as 25 casas. Descreva o trajeto que ele fez.",
    "options": [],
    "answer": "A → B → C → D → E → F → G → H → I → J → K → L → M → N → O → P → Q → R → S → T → U → V → W → X → Y",
    "solution": "Solution:\n\n| $A$ | $X$ | $M$ | $R$ | $G$ |\n| :---: | :---: | :---: | :---: | :---: |\n| $N$ | $S$ | $H$ | $Y$ | $L$ |\n| $I$ | $B$ | $W$ | $F$ | $Q$ |\n| $T$ | $O$ | $D$ | $K$ | $V$ |\n| $C$ | $J$ | $U$ | $P$ | $E$ |",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25933,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCubo - Pedro quer pintar uma caixa de formato cúbico de tal maneira que as faces que tenham uma aresta em comum sejam pintadas em cores diferentes. Calcule o número mínimo de cores que serão necessárias para pintar a caixa dessa maneira.",
    "options": [],
    "answer": "3",
    "solution": "Solution:\n\nUm cubo tem seis faces distintas, duas a duas opostas, sendo que as faces opostas não têm aresta em comum. Temos três pares de faces opostas, logo três cores são suficientes, bastando pintar as faces opostas de uma mesma cor. Por outro lado, é claro que duas cores somente não bastam.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25934,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDetermine o número inteiro mais próximo de\n\na. $\\frac{19}{15}+\\frac{19}{3}$\n\nb. $\\frac{85}{42}+\\frac{43}{21}+\\frac{29}{14}+\\frac{15}{7}$\n\nc. $-\\frac{11}{10}-\\frac{1}{2}-\\frac{7}{5}+\\frac{2}{3}$",
    "options": [],
    "answer": "a: 8; b: 8; c: -2",
    "solution": "Solution:\n\na.\nTemos:\n$$\n\\frac{19}{15}+\\frac{19}{3}=1+\\frac{4}{15}+6+\\frac{1}{3}=7+\\frac{9}{15}=7+\\frac{3}{5}\n$$\nLogo, a soma dada está entre $7$ e $8$. Como $\\frac{3}{5}>\\frac{1}{2}$, o número inteiro mais próximo é $8$.\n\n![](attached_image_1.png)\n\nb.\nTemos:\n$$\n\\begin{aligned}\n\\frac{85}{42}+\\frac{43}{21} & +\\frac{29}{14}+\\frac{15}{7}=2+\\frac{1}{42}+2+\\frac{1}{21}+2+\\frac{1}{14}+2+\\frac{1}{7} \\\\\n& =8+\\frac{1}{42}+\\frac{1}{21}+\\frac{1}{14}+\\frac{1}{7}=8+\\frac{1}{7}\\left(\\frac{1}{6}+\\frac{1}{3}+\\frac{1}{2}+1\\right)=8+\\frac{2}{7}\n\\end{aligned}\n$$\nLogo, a soma dada está entre $8$ e $9$. Como $\\frac{2}{7}<\\frac{1}{2}$, o número inteiro mais próximo é $8$.\n\n![](attached_image_2.png)\n\nc.\n$$\n-\\frac{11}{10}-\\frac{1}{2}-\\frac{7}{5}+\\frac{2}{3}=-\\frac{30}{10}+\\frac{2}{3}=-3+\\frac{2}{3}\n$$\nLogo, a soma dada está entre $-3$ e $-2$. Como $\\frac{2}{3}>\\frac{1}{2}$, o número inteiro mais próximo é $-2$.\n\n![](attached_image_3.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25935,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nBrincando com números ímpares - Beatriz adora números ímpares. Quantos números entre $0$ e $1000$ ela pode escrever usando apenas algarismos ímpares?",
    "options": [],
    "answer": "155",
    "solution": "Solution:\n\nComo cada algarismo é ímpar, temos:\n- cinco possibilidades de números com um algarismo: $1$, $3$, $5$, $7$ e $9$;\n- para números com dois algarismos, temos cinco possibilidades na casa das unidades e cinco na casa das dezenas, totalizando $5 \\times 5 = 25$ possibilidades;\n- para números com três algarismos, temos cinco possibilidades na casa das unidades, cinco na casa das dezenas e cinco na casa das centenas, totalizando $5 \\times 5 \\times 5 = 125$ possibilidades.\n\nAssim, Beatriz pode escrever $5 + 25 + 125 = 155$ números com todos algarismos sendo ímpares.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25936,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nJoão e Maria têm, cada um, um jarro grande com um litro de água. No primeiro dia, João coloca $1\\ \\mathrm{ml}$ da água do seu jarro no jarro da Maria. No segundo dia, Maria coloca $2\\ \\mathrm{ml}$ da água do seu jarro no jarro do João. No terceiro dia, João coloca $3\\ \\mathrm{ml}$ da água do seu jarro no jarro da Maria, e assim por diante. Depois de 200 dias, quantos mililitros de água tem no jarro de Maria?",
    "options": [],
    "answer": "900",
    "solution": "Solution:\n\nInicialmente, o volume de água no jarro da Maria é $1\\ \\mathrm{l} = 1000\\ \\mathrm{ml}$. Depois de 200 dias, o volume é o mesmo, acrescido do que é colocado por João e diminuído do que ela tirou para colocar no do João, ou seja,\n$$\n\\begin{aligned}\n1000 & + 1 - 2 + 3 - 4 + \\cdots + 199 - 200 \\\\\n& = 1000 + (1 - 2) + (3 - 4) + \\cdots + (199 - 200) \\\\\n& = 1000 - (\\underbrace{1 + \\cdots + 1}_{100}) = 900\n\\end{aligned}\n$$\n\nLogo, depois de 200 dias, Maria terá $900\\ \\mathrm{ml}$ em seu jarro.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25937,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUma formiga parte de um vértice de um cubo, andando somente ao longo das arestas, até voltar ao vértice inicial, não passando duas vezes por nenhum vértice. Qual é o passeio de maior comprimento que essa formiga pode fazer?",
    "options": [],
    "answer": "8",
    "solution": "Solution:\n\nNa figura temos um caminho percorrendo oito arestas que a formiga pode fazer partindo do vértice identificado como 1.\n\n![](attached_image_1.png)\n\nSerá possível ela fazer um caminho passando por nove arestas? Para fazer esse caminho, ela teria que passar por nove vértices, pois o vértice de chegada é o mesmo que o de partida, já que a formiguinha volta ao vértice inicial.\n\n![](attached_image_2.png)\n\nComo o cubo só tem oito vértices, esse passeio não é possível. Logo, o passeio de maior comprimento percorre oito arestas.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25938,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEm uma promoção, Joana comprou blusas por $\\mathrm{R}\\$ 15,00$ cada uma e calças por $\\mathrm{R}\\$ 17,00$ cada uma, gastando, ao todo, $\\mathrm{R}\\$ 143,00$. Quantas blusas e calças Joana comprou?",
    "options": [],
    "answer": "5 blouses and 4 pants",
    "solution": "Solution:\n\nSejam $b$ e $c$ o número de blusas e calças compradas, respectivamente. Logo, temos $15 b + 17 c = 143$, sendo $b$ e $c$ números inteiros positivos. Observe que $b < 10$ e $c < 9$, pois tanto $15 \\times 10$ quanto $17 \\times 9$ são maiores do que $143$. A partir deste ponto, apresentamos duas possibilidades de solução.\n\nSolução 1: Temos que $15 b = 143 - 17 c$, portanto $143 - 17 c$ é um múltiplo de $15$, de modo que $143 - 17 c$ termina em $0$ ou $5$. Isso significa que $17 c$ termina em $3$ ou $8$. Logo, $c = 9$ ou $c = 4$. Como $c < 9$, a única solução é $c = 4$ e, portanto,\n$$\nb = \\frac{143 - 17 \\times 4}{15} = 5\n$$\n\nSolução 2: Temos que\n$$\nb = \\frac{143 - 17 c}{15} = 9 - c + \\frac{8 - 2c}{15}\n$$\nNote que $c$ é um número inteiro positivo, portanto, $8 - 2c$ precisa ser um múltiplo de $15$. Se $8 - 2c \\geq 15$, $c$ resulta negativo, portanto, $8 - 2c = 0$, ou seja, $c = 4$. Daí obtemos que $b = 5$. Assim, Joana comprou cinco blusas e quatro calças.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25939,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSe $x + y = 1$ e $x^{2} + y^{2} = 2$, calcule $x^{3} + y^{3}$.",
    "options": [],
    "answer": "5/2",
    "solution": "Solution:\n\nTemos a identidade do binômio, $(x + y)^{2} = x^{2} + 2 x y + y^{2}$, e a do trinômio, $(x + y)^{3} = x^{3} + 3 x^{2} y + 3 x y^{2} + y^{3}$. Substituindo os valores de $x + y$ e $x^{2} + y^{2}$ na identidade do binômio, obtemos $1 = 2 + 2 x y$ e, portanto, $x y = -\\frac{1}{2}$. Assim, pela identidade do trinômio,\n$$\nx^{3} + y^{3} = (x + y)^{3} - 3 x y (x + y) = 1 - 3 \\cdot \\left(-\\frac{1}{2}\\right) \\cdot 1 = \\frac{5}{2}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25940,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO revezamento em uma corrida - Numa competição de revezamento, em que cada equipe tem dois atletas, cada atleta corre $21~\\mathrm{km}$ e o segundo atleta só inicia a corrida quando o primeiro atleta termina a sua parte e lhe passa o bastão. O recorde dessa competição é de 2 horas e 48 minutos. Na equipe de João e Carlos, João inicia a corrida e corre a sua parte com uma velocidade de $12~\\mathrm{km}/\\mathrm{h}$. Para bater o recorde, qual deve ser a velocidade de Carlos?",
    "options": [],
    "answer": "greater than 20 km/h",
    "solution": "Solution:\n\nComo velocidade $=\\frac{\\text{distância}}{\\text{tempo}}$, ou seja, tempo $=\\frac{\\text{distância}}{\\text{velocidade}}$, o tempo gasto por João foi de\n$$\nt=\\frac{21}{12}~\\mathrm{h}=\\left(1+\\frac{9}{12}\\right)~\\mathrm{h}=1~\\mathrm{h}+\\frac{9}{12} \\times 60~\\mathrm{min}=1~\\mathrm{h}~45~\\min\n$$\nLogo, Carlos precisa completar a prova em um tempo inferior a\n$$\n(2~\\mathrm{h}~48~\\mathrm{min})-(1~\\mathrm{h}~45~\\mathrm{min})=1~\\mathrm{h}~3~\\mathrm{min}=63~\\mathrm{min}\n$$\nPara isso, sua velocidade $v$, em $\\mathrm{km}/\\mathrm{min}$, deve satisfazer $\\frac{21}{v}=t<63$, ou seja,\n$$\nv>\\frac{21}{63}=\\frac{1}{3}~\\mathrm{km}/\\mathrm{min}=\\frac{60}{3}~\\mathrm{km}/\\mathrm{h}=20~\\mathrm{km}/\\mathrm{h}\n$$\n\nLogo, Carlos deve correr com uma velocidade superior a $20~\\mathrm{km}/\\mathrm{h}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25941,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nProdutos consecutivos - Divida os números $2$, $3$, $5$, $7$, $11$, $13$ e $17$ em dois grupos de tal forma que, multiplicando todos os números de um grupo e todos do outro, encontremos números consecutivos.",
    "options": [],
    "answer": "Group 1: {2, 3, 7, 17}; Group 2: {5, 11, 13}, yielding products 714 and 715",
    "solution": "Solution:\n\nSolução 1: Como os produtos são números consecutivos, podemos denotá-los por $p$ e $p+1$. Temos, então,\n$$\np^2 + p = p(p+1) = 2 \\times 3 \\times 5 \\times 7 \\times 11 \\times 13 \\times 17 = 510510\n$$\nResolvendo a equação $p^2 + p - 510510 = 0$, encontramos uma única raiz positiva, $p = 714$. Assim, $p+1 = 715$ e, fatorando, obtemos\n$$\n714 = 2 \\times 3 \\times 7 \\times 17 \\quad \\text{e} \\quad 715 = 5 \\times 11 \\times 13\n$$\n\n\nSolução 2: Os números dados são $2$, $3$, $5$, $7$, $11$, $13$ e $17$. Se $2$ e $5$ estiverem no mesmo grupo, então um dos produtos termina em $0$ e o outro, por ser consecutivo, termina em $1$ ou $9$. Os possíveis produtos terminados em $1$ são $3 \\times 7 = 21$, $3 \\times 17 = 51$, $7 \\times 13 = 91$, $13 \\times 17 = 221$, $3 \\times 7 \\times 11 = 231$, $3 \\times 11 \\times 17 = 561$ e $7 \\times 11 \\times 13 = 1001$. Verifica-se que esses grupos não constituem solução e, analogamente, os terminados em $9$. Concluímos que $2$ e $5$ estão em grupos diferentes. Logo, um produto termina em $5$ e o outro em $4$ ou $6$. Como não é possível formar com os números dados um produto terminado em $6$, necessariamente um dos produtos termina em $4$ e o outro em $5$. Por tentativas, obtemos a solução\n$$\n714 = 2 \\times 3 \\times 7 \\times 17 \\quad \\text{e} \\quad 715 = 5 \\times 11 \\times 13\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25942,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPapai Noel chegou à casa de Arnaldo e Bernaldo carregando dez brinquedos distintos e enumerados de 1 a 10 e disse a eles: \"o brinquedo número 1 é para você, Arnaldo e o brinquedo número 2 é para você, Bernaldo. Mas esse ano, vocês podem escolher ficar com mais brinquedos contanto que deixem ao menos um para mim\". Diga de quantos modos Arnaldo e Bernaldo podem dividir entre eles o restante dos brinquedos.",
    "options": [],
    "answer": "6305",
    "solution": "Solution:\n\nPara cada um dos 8 brinquedos, do número 3 ao número 10, devemos decidir se ele vai pertencer a Arnaldo, a Bernaldo ou deve ser deixado para Papai Noel. Se multiplicarmos então\n$$\n\\underbrace{3 \\times 3 \\times \\cdots \\times 3}_{8 \\text{ vezes }}\n$$\ncontaremos as formas de dividir os brinquedos entre Arnaldo, Bernaldo e Papai Noel, incluindo os casos em que Papai Noel fica sem nenhum brinquedo. Restará então contar o número de formas de dividir todos os brinquedos entre Arnaldo e Bernaldo (sem deixar nada para Papai Noel), e subtrair esse número de $3^{8}$.\n\nPara dividir os brinquedos entre Arnaldo e Bernaldo devemos decidir, por cada um dos 8 brinquedos, para qual dos dois o brinquedo vai. Assim temos\n$$\n\\underbrace{2 \\times 2 \\times \\cdots \\times 2}_{8 \\text{ vezes }}\n$$\nformas de dividir os brinquedos entre Arnaldo e Bernaldo. Finalmente a resposta é\n$$\n3^{8}-2^{8}=6305\n$$\nformas de dividir os brinquedos.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25943,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAlguns números reais estão escritos nas casas de um tabuleiro $n \\times n$ de modo que a soma total dos números escritos é positiva. Mostre que existe alguma permutação das colunas do tabuleiro, de modo que a soma dos números escritos nas casas da diagonal principal do novo tabuleiro seja positiva.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nCrie um cilindro a partir do tabuleiro como indicado na figura abaixo. Esse cilindro pode ser decomposto em $n$ diagonais disjuntas que começam em um extremo do cilindro e terminam no outro. Como a soma de todos os números do tabuleiro é positiva, pelo menos uma das diagonais terá soma positiva. Ela correponde a uma permutação das colunas do tabuleiro que satisfaz as condições do problema.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25944,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa figura abaixo, o triângulo $A B C$ é retângulo em $C$ e tanto $B C D E$ quanto $C A F G$ são quadrados. Se o produto das áreas dos triângulos $E A B$ e $B F A$ é $64$, determine a área do triângulo $A B C$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "8",
    "solution": "Solution:\n\nComo $D A$ é paralelo a $E B$, a área do triângulo $A E B$ é $\\frac{E B \\cdot B C}{2} = \\frac{B C^{2}}{2}$. Da mesma forma, a área do triângulo $A B F$ é $\\frac{A C^{2}}{2}$. Portanto,\n$$\n\\begin{aligned}\n\\left[ A B E \\right] \\cdot \\left[ A B F \\right] &= \\frac{B C^{2}}{2} \\cdot \\frac{A C^{2}}{2} \\\\\n64 &= \\frac{B C^{2} \\cdot A C^{2}}{4} \\\\\n64 \\cdot 4 &= B C^{2} \\cdot A C^{2} \\\\\n256 &= B C^{2} \\cdot A C^{2} \\\\\n16 &= B C \\cdot A C\n\\end{aligned}\n$$\nLogo, a área do triângulo $A B C$ é $\\frac{B C \\cdot A C}{2} = \\frac{16}{2} = 8$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25945,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEm certa loteria, existem 60 números distintos e 6 deles são sorteados sem reposição. Cada bilhete possui 6 números distintos entre os 60 possíveis. O prêmio máximo, conhecido como \"gol-no-ângulo\", é dado para o jogador que possuir o bilhete com os mesmos 6 números que foram sorteados. Nesta loteria, também existe o prêmio \"bola-na-trave\". Em um bilhete bola-na-trave, o menor número não possui diferença, em módulo, maior que 1 para o menor número sorteado, o segundo menor número não possui diferença, em módulo, maior que 1 para o segundo menor número sorteado e assim por diante até o sexto menor número. Por exemplo, suponha que o bilhete gol-no-ângulo seja $\\{4,7,25,48,51,60\\}$. Então os bilhetes $\\{3,6,25,49,50,59\\}$ e $\\{5,6,25,47,50,60\\}$ são bilhetes bola-na-trave, mas o bilhete $\\{3,4,6,24,47,50\\}$ não é bola-na-trave. Vale lembrar que um bilhete gol-no-ângulo não é um bilhete bola-na-trave. Para os itens a seguir, considere cada bilhete como a escolha de uma sequência de 6 números escritos em ordem crescente.\n(a) Dê um exemplo de conjunto de 6 números formando um bilhete gol-no-ângulo que tem o menor número possível de bilhetes bola-na-trave associados a ele. Quantos bilhetes bola-na-trave possíveis haveria para estes 6 números?\n(b) Dê um exemplo de conjunto de 6 números formando um bilhete gol-no-ângulo que resulta na maior quantidade possível de bilhetes bola-na-trave. Neste caso, haveria quantos bilhetes bola-na-trave possíveis?\n(c) Considere os números sorteados $\\{2,3,8,11,14,17\\}$, quantos são os bilhetes bola-na-trave associados a ele?\n(d) Suponha que o conjunto de números sorteados seja $\\{8,10,12,14,16,18\\}$. Neste caso, quantos são os bilhetes bola-na-trave possíveis associados a ele?",
    "options": [],
    "answer": "(a) Example: {1,2,3,4,5,6}; number of near-miss tickets: 6.\n(b) Example: {3,6,9,12,15,18}; number of near-miss tickets: 728.\n(c) For {2,3,8,11,14,17}: 485 near-miss tickets.\n(d) For {8,10,12,14,16,18}: 376 near-miss tickets.",
    "solution": "Solution:\n\n(a) Tome os 6 menores números $\\{1,2,3,4,5,6\\}$. Neste caso, temos apenas 6 bilhetes bola-na-trave, que são formados por escolhas de 6 números do conjunto $\\{1,2,3,4,5,6,7\\}$. Em princípio, existiriam 7 bilhetes, mas como um bilhete gol-no-ângulo não é bola-na-trave, o número de bilhetes bola-na-trave associados a ele é 6. Provaremos que este é o menor valor possível mostrando que todos os outros conjuntos têm pelo menos 6 bilhetes bola-na-trave.\n\nTomando o conjunto $\\{a, b, c, d, e, f\\}$ em ordem crescente do $a$ para o $f$. Se $f<60$, temos os bilhetes bola-na-trave:\n$$\n\\begin{gathered}\n\\{a, b, c, d, e, f+1\\} \\\\\n\\{a, b, c, d, e+1, f+1\\} \\\\\n\\{a, b, c, d+1, e+1, f+1\\} \\\\\n\\{a, b, c+1, d+1, e+1, f+1\\} \\\\\n\\{a, b+1, c+1, d+1, e+1, f+1\\} \\\\\n\\{a+1, b+1, c+1, d+1, e+1, f+1\\}\n\\end{gathered}\n$$\nSe $f=60$, mas $a>1$, temos os bilhetes:\n$$\n\\begin{gathered}\n\\{a-1, b, c, d, e, f\\} \\\\\n\\{a-1, b-1, c, d, e, f\\} \\\\\n\\{a-1, b-1, c-1, d, e, f\\} \\\\\n\\{a-1, b-1, c-1, d-1, e, f\\} \\\\\n\\{a-1, b-1, c-1, d-1, e-1, f\\} \\\\\n\\{a-1, b-1, c-1, d-1, e-1, f-1\\}\n\\end{gathered}\n$$\nSe $a=1$ e $f=60$, então existiriam dois números seguidos que não seriam consecutivos, por exemplo, $d-c \\geq 2$, neste caso teríamos os bilhetes:\n$$\n\\begin{gathered}\n\\{a, b, c+1, d, e, f\\} \\\\\n\\{a, b+1, c+1, d, e, f\\} \\\\\n\\{a+1, b+1, c+1, d, e, f\\} \\\\\n\\{a, b, c, d-1, e, f\\} \\\\\n\\{a, b, c, d-1, e-1, f\\} \\\\\n\\{a, b, c, d-1, e-1, f-1\\}\n\\end{gathered}\n$$\n\n(b) Veja que cada número $x$ no bilhete gol-no-ângulo gera 3 possibilidades para sua posição no bola-na-trave, a saber $x-1, x$ e $x+1$. Se números seguidos estiverem próximos, então teremos que eliminar repetições, por exemplo, se $a=b-1$ ou $a+1=b-1$. Portanto, para chegar no máximo podemos deixar números seguidos com diferença de pelo menos 3. Um exemplo é o bilhete gol-no-ângulo $\\{3,6,9,12,15,18\\}$. Um bilhete bola-na-trave possui três possibilidades para o menor número, três para o segundo menor e assim por diante. Lembrando de não contar o bilhete gol-no-ângulo, teremos ao todo $3^{6}-1=728$ bilhetes bola-na-trave associados a ele.\n\n(c) Vamos separar em casos. Se o primeiro número for 1, então para cada um dos cinco números temos três possibilidades, resultando assim em $3^{5}$ bilhetes. Se o primeiro número for 2, então o segundo número terá que ser 3 ou 4. Os outros têm três possibilidades cada, logo teremos $2 \\cdot 3^{4}$ bilhetes bola-na-trave associados. Se o primeiro número for 3, o segundo é obrigatoriamente 4. Para cada um dos demais, há três possibilidades. Logo, teremos $3^{4}$ possibilidades. Lembrando de retirar o bilhete gol-no-ângulo, ficamos com\n$$\n3^{5}+2 \\cdot 3^{4}+3^{4}-1=485\n$$\nbilhetes bola-na-trave associados ao gol-no-ângulo $\\{2,3,8,11,14,17\\}$.\n\n(d) Vamos começar analisando separadamente os conjuntos $\\{8,10,12\\}$ e $\\{14,16,18\\}$. Para o primeiro grupo de números, se o maior número do bola-na-trave for 13, restam três possibilidades para 08 e três para o 10, com a exceção das duas escolhas serem iguais a 9. Logo, existem oito possibilidades neste caso. Se o maior número for 12 temos também 8 possibilidades e, se o maior for 11, temos apenas duas possibilidades para 10 e três para o 8, com a exceção das duas escolhas serem iguais a 9. Temos assim 5 possibilidades. Então para o primeiro grupo, temos $8+8+5=21$ possibilidades. Por simetria, podemos observar as possibilidades do menor número do segundo grupo e concluir que para o segundo grupo temos também $8+8+5=21$ possibilidades. Veja que temos que subtrair as possibilidades em que foi escolhido 13 para o 12 e 13 também para o 14, ou seja, $8 \\cdot 8=64$ possibilidades. Novamente, temos que lembrar de excluir o bilhete gol-no-ângulo. Então teremos\n$$\n21 \\cdot 21-8 \\cdot 8-1=441-64-1=376\n$$\nbilhetes bola-na-trave associados ao $\\{8,10,12,14,16,18\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25946,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm número de quatro algarismos $a b c d$ é chamado balanceado se\n$$\na+b=c+d\n$$\nCalcule as seguintes quantidades:\n\na) Quantos números $a b c d$ são tais que $a+b=c+d=8$ ?\n\nb) Quantos números $a b c d$ são tais que $a+b=c+d=16$ ?\n\nc) Quantos números balanceados existem?",
    "options": [],
    "answer": "a) 72; b) 9; c) 615",
    "solution": "Solution:\n\na) Vamos contar primeiro os valores possíveis para o par $(a, b)$. Observe que $a$ não pode ser igual a zero, por ser o primeiro algarismo em $a b c d$. Mas $a$ pode tomar qualquer valor em\n$$\n\\{1,2,3,4,5,6,7,8\\}\n$$\nporque por cada um desses valores, o número $8 - a$ dá como resultado um valor apropriado para $b$. Contamos, assim, 8 possibilidades para o par $(a, b)$. Para contar os valores possíveis para o par $(c, d)$, basta ver que $c$ pode tomar qualquer valor no conjunto\n$$\n\\{0,1,2,3,4,5,6,7,8\\}\n$$\ne, para cada um desses valores, o número $8-c$ dá como resultado um valor apropriado para $d$. São assim 9 possibilidades para o par $(c, d)$. Os números $a b c d$ que cumprem com $a+b=c+d=8$ são as combinações de $a b$ e $c d$. A resposta é, portanto,\n$$\n8 \\times 9=72\n$$\n\nb) Comecemos contando as possibilidades para o par $(a, b)$. Observe que se $a$ fosse menor do que 7, o número $16 - a$ seria negativo e não seria, então, um valor apropriado para $b$. Portanto, os valores possíveis para $a$ são\n$$\n\\{7,8,9\\}\n$$\nContamos, assim, 3 possibilidades. Observe agora que para o par $(c, d)$, as possibilidades são exatamente as mesmas que para $(a, b)$ :\n$$\n(7,9), \\quad(8,8) \\quad \\text{e} \\quad(9,7)\n$$\nFinalmente, os números $a b c d$ que procuramos resultam de combinar as possibilidades para $a b$ e $c d$. A resposta é, portanto,\n$$\n3 \\times 3=9\n$$\n\nc) Devemos contar os números $a b c d$ de modo que $a+b=c+d$. Contemos primeiro aqueles em que $a+b=c+d=s$ para um $1 \\leq s \\leq 9$. As possibilidades para os pares $(a, b)$ e $(c, d)$ nesses casos se calculam de modo similar ao do item a). Os valores possíveis para $a$ são\n$$\n\\{1,2, \\ldots, s\\}\n$$\nporque, para todos esses valores, o número $s-a$ é um valor permitido para $b$. Contamos, assim, $s$ possibilidades para o par $(a, b)$. Por outro lado, os valores permitidos para $c$ são\n$$\n\\{0,1,2, \\ldots, s\\}\n$$\nporque, para todos esses valores, o número $s-c$ é um valor permitido para $d$. Contamos agora $s+1$ possibilidades para o par $(c, d)$. Finalmente, combinamos as possibilidades para os pares $(a, b)$ e $(c, d)$, e contamos assim\n$$\ns \\times(s+1)\n$$\nnúmeros $a b c d$ tais que $a+b=c+d=s$, com $s \\in\\{1,2, \\ldots, 9\\}$. Considerando todos os valores de $s$ entre 1 e 9 , teremos no total\n$$\n1 \\times 2+2 \\times 3+\\cdots+9 \\times 10=330\n$$\nnúmeros $a b c d$ tais que $1 \\leq a+b=c+d \\leq 9$.\n\nObserve que o máximo valor possível para $a+b=c+d=s$ é 18. Ainda resta então considerar o caso em que $10 \\leq s \\leq 18$. Para que $s-a$ seja um valor permitido para $b$ é necessário que $0 \\leq s-a \\leq 9$. Isso implica que\n$$\ns-9 \\leq a \\leq s\n$$\nMas como $a$ deve ser menor ou igual a 9 , os valores possíveis para $a$ serão\n$$\n\\{s-9, s-8, \\ldots, 9\\}\n$$\nQuando $10 \\leq s \\leq 18$, é simples verificar que todos esses valores são permitidos para $a$, porque $s-a$ é também um valor permitido para $b$. Contamos assim $19-s$ possibilidades para o par $(a, b)$. Por outro lado, como no item b), as possibilidades para o par $(a, b)$ são exatamente as mesmas possibilidades para o par $(c, d)$. Combinando, obteremos $(19-s) \\times(19-s)$ possibilidades para $a b c d$ com $a+b=c+d=s$ e $s \\in\\{10,11, \\ldots, 19\\}$. Considerando as somas $s$ de 10 até 18 , contamos\n$$\n9^{2}+8^{2}+7^{2}+\\cdots+1^{2}=285\n$$\nnúmeros $a b c d$ tais que $10 \\leq a+b=c+d \\leq 18$. Finalmente, a resposta é que existem $330+285=615$ números $a b c d$ tais que $a+b=c+d$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25947,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOito dados são agrupados formando um cubo. Quantas faces ficam visíveis?",
    "options": [],
    "answer": "20",
    "solution": "Solution:\n\n20",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25948,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nVivi, Tânia e Rosa estão em fila, não necessariamente nessa ordem, e gritam sucessivamente, cada uma, um múltiplo de $3$.\n\n| 3 | 6 | 9 |\n| :---: | :---: | :---: |\n| 12 | 15 | 18 |\n| $\\vdots$ | $\\vdots$ | $\\vdots$ |\n\nVivi foi a primeira a gritar um número maior que $2003$ e Rosa a primeira a gritar um número de quatro algarismos. Quem gritou o número $666$? E o $888$?",
    "options": [],
    "answer": "666: Tânia; 888: Vivi",
    "solution": "Solution:\n\nObserve que aquela que gritou os números $9$, $18$, etc, sempre gritou múltiplos de $9$. O primeiro múltiplo de $3$ com quatro algarismos é $1002$ e o primeiro múltiplo de $3$ maior do que $2003$ é $2004$. Logo, Vivi gritou $2004$ e Rosa $1002$. Nenhum desses números é múltiplo de $9$, portanto, foi Tânia quem gritou $9$ e seus múltiplos.\n\nQuem gritou $3$, também gritou $12=3+9$, $21=3+2 \\times 9$, $30=3+3 \\times 9$ e assim por diante, até $3+9k$. Da mesma forma, quem grita $6$, grita todos os números da forma $6+9k$. Dividindo por $9$, obtemos $2004=6+9 \\times 222$ e $1002=3+9 \\times 111$, portanto, quem gritou $3$ foi Rosa e Vivi gritou $6$.\n\n| Rosa | Vivi | Tânia |\n| :---: | :---: | :---: |\n| 3 | 6 | 9 |\n| 12 | 15 | 18 |\n| 21 | 24 | 27 |\n| $\\vdots$ | $\\vdots$ | $\\vdots$ |\n| 1002 | 1005 | 1008 |\n| $\\vdots$ | $\\vdots$ | $\\vdots$ |\n| 2001 | 2004 | 2007 |\n\nDa mesma forma, dividindo por $9$, encontramos que $666$ é múltiplo de $9$ e $888=6+98 \\times 9$, portanto Tânia gritou $666$ e Vivi gritou $888$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25949,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm número menor do que $200$ é formado por três algarismos diferentes e o dobro desse número também tem todos os algarismos diferentes. Ainda, o número e seu dobro não têm algarismos em comum. Qual é esse número? Quantas soluções têm esse problema?",
    "options": [],
    "answer": "The numbers are 152, 182, 192, 143, 153, 134, 154, 164, 135, 145, 185, 176, 186, 138, 148, 178, 139, 179. There are 18 solutions.",
    "solution": "Solution:\n\nInicialmente note que o dobro de um número inteiro é par, logo ele termina em $0, 2, 4, 6$ ou $8$. No entanto, o número procurado não pode terminar em $0$, pois nesse caso o seu dobro também terminaria em $0$, e ambos teriam o algarismo $0$ em comum. Portanto, temos os casos a seguir.\n\n![](attached_image_1.png)\n![](attached_image_2.png)\n![](attached_image_3.png)\n![](attached_image_4.png)\n\nVamos, agora, determinar todas as possibilidades para cada caso, lembrando sempre que o número e seu dobro não podem ter algarismos comuns.\n\n- Caso I - temos três possibilidades:\n$$\n152 \\times 2 = 304, \\quad 182 \\times 2 = 364, \\quad 192 \\times 2 = 384\n$$\n\n- Caso II - temos duas possibilidades:\n$$\n143 \\times 2 = 286, \\quad 153 \\times 2 = 206\n$$\n\n- Caso III - temos três possibilidades:\n$$\n134 \\times 2 = 268, \\quad 154 \\times 2 = 308, \\quad 164 \\times 2 = 328\n$$\n\n- Caso IV - temos três possibilidades:\n$$\n135 \\times 2 = 270, \\quad 145 \\times 2 = 290, \\quad 185 \\times 2 = 370\n$$\n\n- Caso V - temos duas possibilidades:\n$$\n176 \\times 2 = 352, \\quad 186 \\times 2 = 372\n$$\n\n- Caso VI - não há nenhuma possibilidade.\n\n- Caso VII - temos três possibilidades:\n$$\n138 \\times 2 = 276, \\quad 148 \\times 2 = 296, \\quad 178 \\times 2 = 356\n$$\n\n- Caso VIII - temos duas possibilidades:\n$$\n139 \\times 2 = 278, \\quad 179 \\times 2 = 358\n$$\n\nAssim, temos $3 + 2 + 3 + 3 + 2 + 3 + 2 = 18$ soluções para esse problema, a saber:",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25950,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nEntre 10 e 99, quantos números existem tais que, invertendo a ordem de seus algarismos, obtemos um número maior do que o número original?",
    "options": [],
    "answer": "36",
    "solution": "Solution:\nDevemos contar os números $ab$ de dois algarismos que têm o algarismo $b$ da unidade maior do que o algarismo $a$ da dezena, ou seja, tais que $b > a$. Se $a = 1$, o algarismo $b$ da unidade pode ser $2, 3, 4, 5, 6, 7, 8$ ou $9$, portanto, temos oito possibilidades. Se $a = 2$, o algarismo $b$ da unidade pode ser $3, 4, 5, 6, 7, 8$ ou $9$, portanto, temos sete possibilidades. Continuando dessa maneira, chegamos até $a = 8$, quando o algarismo $b$ da unidade só pode ser $9$, portanto, temos só uma possibilidade. Claramente, $a$ não pode ser $0$ nem $9$. Logo, existem $8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 36$ números entre $10$ e $99$ tais que, invertendo a ordem de seus algarismos, obtemos um número maior do que o número original.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25951,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa figura, $O$ é o centro do semicírculo de diâmetro $PQ$, $R$ é um ponto sobre o semicírculo e $RM$ é perpendicular a $PQ$. Se a medida do arco $\\widehat{PR}$ é o dobro da medida do arco $\\overparen{RQ}$, qual é a razão entre $PM$ e $MQ$?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "3",
    "solution": "Solution:\n\n$180^\\circ = 2 R\\widehat{O}Q + R\\widehat{O}Q = 3 R\\widehat{O}Q$\n\ndonde $R\\widehat{O}Q = 60^\\circ$. Mas, $OR = OQ$ é o raio do círculo, de modo que o triângulo $\\triangle ORQ$ é equilátero. Assim, sua altura $RM$ também é a mediana, ou seja, $OM = MQ$. Se $r$ é o raio do círculo, então $OM = MQ = \\frac{1}{2} r$ e\n\n$$\n\\frac{PM}{MQ} = \\frac{PO + OM}{MQ} = \\frac{r + \\frac{1}{2} r}{\\frac{1}{2} r} = 3\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25952,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTriângulos - Quantos triângulos existem que tenham um perímetro de 15 unidades e lados medindo números inteiros?",
    "options": [],
    "answer": "7",
    "solution": "Solution:\n\nVamos supor que $a$, $b$ e $c$ sejam os comprimentos dos lados do triângulo. Não há perda de generalidade em supor que $a \\leq b \\leq c$, de modo que $a+b+c \\leq 3c$. Como cada lado de um triângulo é menor do que a soma dos outros dois, temos que $c < a+b$ e, portanto, obtemos $2c < a+b+c \\leq 3c$. Mas, $a+b+c = 15$, de modo que $2c < 15 \\leq 3c$ e, como $c$ é um número inteiro, $5$, $6$ ou $7$ são as únicas opções para $c$.\n\nSe $c=5$, então $a+b=10$ e temos uma única possibilidade, $a=b=c=5$.\n\nSe $c=6$, então $a+b=9$ e temos duas possibilidades para $a$, $3$ ou $4$, caso em que, necessariamente, $b$ é $6$ ou $5$, respectivamente.\n\nSe $c=7$, então $a+b=8$ e temos quatro possibilidades para $a$, $1$, $2$, $3$ ou $4$. Nesses casos, necessariamente, $b$ é $7$, $6$, $5$ ou $4$, respectivamente.\n\nAssim, no total, temos sete desses triângulos.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25953,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO número $119$ é muito interessante porque deixa resto $1$ ao ser dividido por $2$, deixa resto $2$ ao ser dividido por $3$, deixa resto $3$ ao ser dividido por $4$, deixa resto $4$ ao ser dividido por $5$ e, finalmente, deixa resto $5$ ao ser dividido por $6$. Existem outros números de três algarismos com essas propriedades?",
    "options": [],
    "answer": "179, 239, 299, 359, 419, 479, 539, 599, 659, 719, 779, 839, 899, 959",
    "solution": "Solution:\n\nSuponhamos que $N$ seja um dos números procurados. Como $N$ e $119$ deixam os mesmos restos quando divididos por $2$, $3$, $4$, $5$ e $6$, temos que a diferença $N-119$ entre eles deixa resto zero quando dividido por esses números. Portanto, $N-119$ é um múltiplo de $2,3,4,5$ e $6$. Como $60$ é o mínimo múltiplo comum desses números, $N-119$ é um múltiplo de $60$, ou seja, $N-119=60k$, para algum inteiro $k$. Assim, $N=119+60k$ é um número interessante para qualquer número inteiro $k$.\n\nComo queremos números distintos de $119$ e de três algarismos, devemos tomar $k$ positivo e menor do que $15$, já que $119+60 \\times 15=1019$ tem quatro algarismos. Assim, tomamos $k$ de $1$ a $14$ e obtemos outros $14$ números interessantes de três algarismos, a saber,\n\n$179$, $239$, $299$, $359$, $419$, $479$, $539$, $599$, $659$, $719$, $779$, $839$, $899$, $959$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25954,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm time de futebol ganhou $60\\%$ das 45 partidas já disputadas. Qual é o número mínimo de partidas que esse time ainda precisa vencer para atingir uma porcentagem de $75\\%$ de vitórias?",
    "options": [],
    "answer": "27",
    "solution": "Solution:\n\nO time ganhou $60\\%$ das 45 já disputadas, ou seja, $45 \\times \\frac{60}{100} = 27$ partidas. Se ele ganhar mais $n$ partidas, a porcentagem de partidas ganhas será\n$$\n\\frac{\\text{nº de partidas ganhas}}{\\text{nº de partidas disputadas}} = \\frac{27 + n}{45 + n} = 75\\% = \\frac{75}{100} = \\frac{3}{4}\n$$\nLogo, $4 \\times (27 + n) = 3 \\times (45 + n)$, do que resulta $n = 27$ como o número mínimo de partidas que o time ainda precisa vencer para atingir $75\\%$ de vitórias.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25955,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDois dados são lançados. Qual é a probabilidade de o produto dos números obtidos nos dois dados ser divisível por $6$?",
    "options": [],
    "answer": "5/12",
    "solution": "Solution:\n\nNa tabela seguinte, marcamos com $\\times$ os produtos que são divisíveis por $6$.\n\n| $\\times$ | 1 | 2 | 3 | 4 | 5 | 6 |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| 1 |  |  |  |  |  | $\\times$ |\n| 2 |  |  | $\\times$ |  |  | $\\times$ |\n| 3 |  | $\\times$ |  | $\\times$ |  | $\\times$ |\n| 4 |  |  | $\\times$ |  |  | $\\times$ |\n| 5 |  |  |  |  |  | $\\times$ |\n| 6 | $\\times$ | $\\times$ | $\\times$ | $\\times$ | $\\times$ | $\\times$ |\n\nLogo, temos $15$ casos favoráveis dentre $36$ possibilidades. Assim, a probabilidade de que o produto seja divisível por $6$ é $15 / 36 = 5 / 12 = 41,7\\%$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25956,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nQuantos são os pares de números inteiros positivos $(x, y)$ tais que\n$$\n\\frac{x y}{x+y}=144 ?\n$$",
    "options": [],
    "answer": "45",
    "solution": "Solution:\nA equação dada é equivalente a $x y = 144(x + y) = 144 x + 144 y$, portanto, isolando $x$, obtemos $x = \\frac{144 y}{y - 144}$. Como $x$ e $y$ devem ser inteiros positivos, o denominador $y - 144$ deve ser um número inteiro positivo, digamos, $y - 144 = n$. Substituindo essa expressão no valor de $x$, obtemos\n$$\nx = \\frac{144(n + 144)}{n} = 144 + \\frac{144^2}{n}\n$$\nComo $x$ deve ser um número inteiro, $n$ deve ser um divisor de $144^2$. Sendo $144^2 = 12^4 = 2^8 \\cdot 3^4$, seus divisores são os números $d$ da forma $d = 2^a \\cdot 3^b$, com $0 \\leq a \\leq 8$ e $0 \\leq b \\leq 4$. Como há 9 valores possíveis para $a$ e 5 valores possíveis para $b$, concluímos que $144^2$ tem $9 \\times 5 = 45$ divisores.\nAssim, para cada divisor $n$ de $144^2$, obtemos uma solução\n$$\n(x, y) = \\left(144 + \\frac{144^2}{n}, n + 144\\right)\n$$\nda equação $\\frac{x y}{x + y} = 144$ dada. Portanto, essa equação possui 45 pares de números inteiros positivos $(x, y)$ que a satisfazem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25957,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCírculos tangentes - Os vértices de um triângulo cujos lados medem $3$, $4$ e $5~\\mathrm{cm}$, são centros de três círculos que são dois a dois tangentes exteriormente. Qual é a soma das áreas desses três círculos?",
    "options": [],
    "answer": "14π cm^2",
    "solution": "Solution:\n\nDenotemos por $r_{1}$, $r_{2}$ e $r_{3}$ os raios dos três círculos. Como os círculos são tangentes dois a dois, temos\n$$\n\\left\\{\\begin{array}{l}\nr_{1}+r_{2}=3 \\\\\nr_{1}+r_{3}=4 \\\\\nr_{2}+r_{3}=5\n\\end{array}\\right.\n$$\nSubstituindo os valores $r_{2}=3-r_{1}$ e $r_{3}=4-r_{1}$ na terceira equação, obtemos $3-r_{1}+4-r_{1}=5$. Daí, $r_{1}=1$, $r_{2}=2$ e $r_{3}=3$. Logo, a soma das áreas dos três círculos é $\\left(1^{2}+2^{2}+3^{2}\\right) \\pi=14 \\pi~\\mathrm{cm}^{2}$.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25958,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nProve that the sum of the lengths of the legs of a right triangle never exceeds $\\sqrt{2}$ times the length of the hypotenuse of the triangle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet $a$ and $b$ be the legs of a right triangle and $c$ its hypotenuse. We know that\n$$\n\\begin{aligned}\n(a-b)^2 & \\geq 0 \\\\\na^2 + b^2 & \\geq 2ab\n\\end{aligned}\n$$\nBy the Pythagorean Theorem, we have\n$$\n\\begin{aligned}\n(a+b)^2 & = a^2 + 2ab + b^2 \\\\\n& \\leq 2(a^2 + b^2) \\\\\n& = 2c^2\n\\end{aligned}\n$$\nTherefore,\n$$\na + b \\leq \\sqrt{2} c\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25959,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFlorinda foi dar um passeio. Ela saiu do ponto $A$ e caminhou $1~\\mathrm{m}$. Nesse ponto ela virou para a esquerda um ângulo de $90^\\circ$ e caminhou $2~\\mathrm{m}$. No último ponto ela virou para esquerda, e caminhou $3~\\mathrm{m}$. Ela continuou andando desta maneira até que no último trecho ela caminhou $30~\\mathrm{m}$ e chegou ao seu ponto final que chamaremos de ponto $B$. A figura seguinte ilustra os primeiros sete trechos do passeio de Florinda (note que o ponto $B$ não está ilustrado).\n![](attached_image_1.png)\nCalcule a distância entre os pontos $A$ e $B$.",
    "options": [],
    "answer": "sqrt(481)",
    "solution": "Solution:\nNesse exercício todas as distâncias estão dadas em metros. Observe na seguinte figura que os segmentos retilíneos horizontais possuem comprimentos ímpares enquanto que os verticais possuem comprimentos pares.\n![](attached_image_2.png)\nSe consideramos somente os segmentos retilíneos horizontais, obtemos que o deslocamento horizontal total é igual a\n$$\n1-3+5-7+9-11+13-\\cdots+25-27+29\n$$\nPara calcular esse valor, podemos associar os termos do seguinte modo:\n$$\n1 \\underbrace{-3+5}_{+2} \\underbrace{-7+9}_{+2} \\underbrace{-11+13}_{+2}-\\ldots \\underbrace{-27+29}_{+2}\n$$\nO deslocamento horizontal total é, então, igual a $1+2 \\times 7=15$.\n\nConsiderando agora os deslocamentos verticais, e calculando de modo similar, obtemos\n$$\n2 \\underbrace{-4+6}_{+2} \\underbrace{-8+10}_{+2} \\underbrace{-12+14}_{+2}-\\ldots \\underbrace{-28+30}_{+2}=2+2 \\times 7=16\n$$\n![](attached_image_3.png)\nFinalmente, pelo Teorema de Pitágoras, concluímos que o deslocamento total é igual a\n$$\n\\sqrt{15^{2}+16^{2}}=\\sqrt{481}\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25960,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nArnaldo, Bráulio e Carlos participarão de uma corrida de rua. Depois de algumas semanas, eles estavam discutindo suas estratégias. Arnaldo corre a primeira metade da distância total da corrida a $9~\\mathrm{km}/\\mathrm{h}$ e a segunda metade a $11~\\mathrm{km}/\\mathrm{h}$. Já Bráulio corre um terço da distância a $9~\\mathrm{km}/\\mathrm{h}$, o segundo terço a $10~\\mathrm{km}/\\mathrm{h}$ e, por fim, o último terço a $11~\\mathrm{km}/\\mathrm{h}$. Carlos usa uma estratégia diferente dos dois primeiros, ele corre metade do seu tempo total de corrida a $9~\\mathrm{km}/\\mathrm{h}$ e a metade final do tempo a $11~\\mathrm{km}/\\mathrm{h}$. Determine a ordem entre os tempos totais de Arnaldo, Bráulio e Carlos de chegada ao final da corrida.",
    "options": [],
    "answer": "t_C < t_B < t_A",
    "solution": "Solution:\n\nChamaremos os tempos, medidos em horas, de Arnaldo, Bráulio e Carlos de $t_{A}, t_{B}$ e $t_{C}$, respectivamente. Seja $6d$ a distância total da corrida, medida em quilômetros. Como Carlos corre metade do tempo a $9~\\mathrm{km}/\\mathrm{h}$ e metade a $11~\\mathrm{km}/\\mathrm{h}$, então\n$$\n6d = 9 \\cdot \\frac{t_{C}}{2} + 11 \\cdot \\frac{t_{C}}{2} = 10 \\cdot t_{C}\n$$\nCom isto, podemos montar equações e comparar os três tempos.\n$$\n\\begin{aligned}\nt_{A} & = \\frac{3d}{9} + \\frac{3d}{11} \\\\\n& = \\frac{600d}{990} \\\\\nt_{B} & = \\frac{2d}{9} + \\frac{2d}{10} + \\frac{2d}{11} \\\\\n& = \\frac{598d}{990} \\\\\nt_{C} & = \\frac{6d}{10} \\\\\n& = \\frac{594d}{990}\n\\end{aligned}\n$$\nConcluímos que $t_{C} < t_{B} < t_{A}$, ou seja, o tempo de Carlos é menor que o de Bráulio que, por sua vez, é menor que o de Arnaldo.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25961,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUma mediana de um triângulo é o segmento que liga um vértice ao ponto médio do lado oposto. Considere o triângulo $ABC$ na figura a seguir e sejam $M$, $N$ e $P$ os pontos médios dos lados $BC$, $CA$ e $AB$, respectivamente. As medianas $BN$ e $CP$ se cortam no ponto $G$. Seja $X$ o ponto médio do segmento $AG$.\n![](attached_image_1.png)\n(a) Usando o quadrilátero $GPXN$, verifique que o ponto $G$ divide o segmento $CP$ na razão $2:1$, ou seja, que $CG=2 \\cdot GP$.\n(b) A partir do item anterior, verifique que a mediana $AM$ corta a mediana $CP$ no mesmo ponto $G$. Note que isto mostra que as três medianas de um triângulo passam por um mesmo ponto. Este ponto é chamado de Baricentro do triângulo.\n(c) Suponha que as medianas $BN$ e $CP$ possuem o mesmo comprimento, verifique que $AC=AB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n(a) Observe que os triângulos $AXN$ e $AGC$ são semelhantes, pois $\\frac{AX}{AG}=\\frac{AN}{AC}=\\frac{1}{2}$ e o ângulo $A$ é comum aos dois triângulos. Com isto, $XN=\\frac{GC}{2}$ e $XN$ é paralelo a $GC$. De maneira análoga, podemos provar que $PX$ é paralelo a $BN$. Assim, o quadrilátero $GPXN$ é um paralelogramo, pois possui lados opostos paralelos. Como paralelogramos também possuem lados opostos de mesma medida (para mais detalhes olhar o problema \"Condições para um quadrilátero ser um paralelogramo\" do nível 2) temos\n$$\nCG=2 \\cdot XN=2 \\cdot GP\n$$\n\n(b) Chamaremos de $G'$ o ponto onde $AM$ corta $CP$.\n![](attached_image_2.png)\nUsando a mesma ideia que o item anterior, o quadrilátero $G'PYM$ é um paralelogramo e vale $CG'=2 \\cdot G'P$. Como $G$ e $G'$ dividem o segmento de $C$ para $P$ na mesma razão, podemos concluir que $G=G'$ e que as três medianas passam por $G$.\n\n(c) Se as medianas $BN$ e $CP$ possuem o mesmo comprimento, então os triângulos $PGB$ e $NGC$ são congruentes pelo caso $LAL$, pois:\n$$\n\\begin{aligned}\nPG & =\\frac{CP}{3} \\\\\n& =\\frac{BN}{3} \\\\\n& =NG \\\\\n\\angle PGB & =\\angle NGC \\\\\nGB & =\\frac{2 \\cdot BN}{3} \\\\\n& =\\frac{2 \\cdot CP}{3} \\\\\n& =GC\n\\end{aligned}\n$$\n![](attached_image_3.png)\nConsequentemente,\n$$\n\\begin{aligned}\nPB & =NC \\\\\n\\frac{AB}{2} & =\\frac{AC}{2} \\\\\nAB & =AC\n\\end{aligned}",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25962,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDenise e Antônio jogam uma série de 8 jogos no qual o vencedor da primeira partida ganha 1 ponto, o da segunda 2 pontos, o da terceira 4 pontos, o da quarta 8 pontos e assim por diante, multiplicando por 2 o número de pontos de uma partida para a outra. No final, Denise ganhou 31 pontos a mais que Antônio e não houve empate em nenhuma das partidas. Quais partidas Denise ganhou?",
    "options": [],
    "answer": "Denise won the 1st, 2nd, 3rd, 4th, and 8th games.",
    "solution": "Solution:\n\n$1^{a}$, $2^{a}$, $3^{a}$, $4^{a}$ e $8^{a}$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25963,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nJoão, Jorge, José e Jânio são bons amigos. Certa vez, João estava sem dinheiro, mas seus amigos tinham algum. Então Jorge deu a João um quinto de seu dinheiro, José deu um quarto de seu dinheiro e Jânio deu um terço de seu dinheiro. Se todos eles deram a mesma quantidade de dinheiro para João, que fração do dinheiro do grupo ficou com João?",
    "options": [],
    "answer": "1/4",
    "solution": "Solution:\n\nSe $A$ é a quantidade de dinheiro que João recebeu de cada um de seus amigos, então ele recebeu um total de $3A$. Como ele recebeu, de Jorge, um quinto do seu dinheiro, então Jorge tinha $5A$. Da mesma maneira, José tinha $4A$ e Jânio tinha $3A$. Assim, os três amigos tinham, juntos, $5A + 4A + 3A = 12A$ e a fração do dinheiro do grupo que ficou com João foi de $\\frac{3A}{12A} = \\frac{1}{4}$, ou seja, uma quarta parte.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25964,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm trapézio - A figura dada representa um trapézio $ABCD$ em que $AB$ é paralelo a $CD$ e as diagonais $AC$ e $BD$ cortam-se no ponto $P$. Se as áreas dos triângulos $\\triangle APB$ e $\\triangle CPD$ medem $4$ e $9\\ \\mathrm{cm}^2$, respectivamente, qual é a área do triângulo $\\triangle PCB$?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "6 cm^2",
    "solution": "Solution:\n\nOs triângulos $\\triangle APB$ e $\\triangle CPD$ são semelhantes, pois o ângulo $A\\widehat{P}B$ é igual ao ângulo $C\\widehat{P}D$ (opostos pelo vértice) e o ângulo $A\\widehat{B}D$ é igual ao ângulo $B\\widehat{D}C$ (alternos internos).\n\n![](attached_image_2.png)\n\nComo a razão entre suas áreas é $4/9$, temos que a razão de semelhança entre esses triângulos é $\\sqrt{4/9} = 2/3$. Logo,\n$$\n\\frac{PB}{DP} = \\frac{2}{3}\n$$\nPor outro lado, os triângulos $\\triangle CPD$ e $\\triangle PCB$ têm a mesma altura em relação às bases $DP$ e $PB$, respectivamente. Portanto, a razão entre suas áreas é igual à razão entre suas bases,\n$$\n\\frac{\\text{área}(\\triangle PCB)}{\\text{área}(\\triangle CPD)} = \\frac{PB}{DP} = \\frac{2}{3}\n$$\nComo área $(\\triangle CPD) = 9$, segue que a área do triângulo $\\triangle PCB$ mede $6\\ \\mathrm{cm}^2$.\n\nObservação: O mesmo argumento poderia ser usado para mostrar que também a área do triângulo $\\triangle ADP$ mede $6\\ \\mathrm{cm}^2$. As medidas de $4$ e $9\\ \\mathrm{cm}^2$ de área dadas no enunciado do problema não desempenham papel especial algum. O argumento exposto acima prova que num trapézio qualquer os triângulos $\\triangle PCB$ e $\\triangle ADP$ têm áreas iguais, mesmo que o trapézio não seja equilátero.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25965,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nÁrea de triângulo - Se $AC = 1\\ \\mathrm{cm}$ e $AD = 4\\ \\mathrm{cm}$, qual é a relação entre as áreas dos triângulos $\\triangle ABC$ e $\\triangle CBD$?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "3/5",
    "solution": "Solution:\n\nOs triângulos $\\triangle ABC$ e $\\triangle CBD$ têm bases $AC$ e $CD$, respectivamente, e a mesma altura $h$ em relação a essas bases.\n\n![](attached_image_2.png)\n\nAssim, temos\n$$\n\\text{área } \\triangle ABC = \\frac{AC \\times h}{2} \\text{ e área } \\triangle CBD = \\frac{CD \\times h}{2}\n$$\nLogo, a relação entre as áreas é dada por\n$$\n\\frac{\\text{área } \\triangle ABC}{\\text{área } \\triangle CBD} = \\frac{\\frac{AC \\times h}{2}}{\\frac{CD \\times h}{2}} = \\frac{AC}{CD} = \\frac{1,5}{4-1,5} = \\frac{15}{25} = \\frac{3}{5}\n$$\n\n**LEMBRETE:** A área de um triângulo é a metade do produto de um dos seus lados pela altura $h$ relativa a este lado, como exemplificado nas duas figuras a seguir.\n\n![](attached_image_3.png)\n\nÁrea do $\\triangle CBD = \\frac{CD \\times h}{2}$\n\n![](attached_image_4.png)\n\nÁrea do $\\triangle ABC = \\frac{AC \\times h}{2}$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25966,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA lei pirata estabelece que, para dividir as moedas de um tesouro, o capitão deve escolher um grupo de piratas (excluindo a si mesmo). Em seguida, o capitão deve distribuir a mesma quantidade de moedas a cada um dos piratas desse grupo, de tal modo que não seja possível dar a cada um deles nenhuma outra das moedas que restaram (respeitando o fato de que cada pirata recebe a mesma quantidade). As moedas restantes são então dadas ao capitão. No navio do capitão Barbaroxa há 100 piratas (sem incluir o capitão). Barbaroxa deve dividir um tesouro que contém menos de 1000 moedas. Se ele escolher 99 piratas, ele ficará com 51 moedas, mas se escolher 77 piratas, ele ficará com 29 moedas.\na) Quantas moedas contém o tesouro?\nb) Quantos piratas deve escolher Barbaroxa para ficar com a maior quantidade possível de moedas?",
    "options": [],
    "answer": "645 coins; choose 93 pirates",
    "solution": "Solution:\na) Seja $N$ o número de moedas que há no tesouro. Quando Barbaroxa escolhe 99 piratas para dividir as $N$ moedas, sobram 51 moedas para ele, ou seja, ele consegue dividir $N-51$ moedas entre 99 piratas, ou equivalentemente, o número $N-51$ é divisível por 99. Em particular, o número $N-51+99=N+48$ é também divisível por 99.\n\nDe modo análogo, podemos concluir que $N-29$ é divisível por 77. Logo, o número $N-29+77=N+48$ é também divisível por 77. Como $N+48$ é divisível por 77 e por 99, então $N+48$ é divisível pelo mínimo múltiplo comum de 77 e 99, ou seja, 693.\n\nConcluímos que $N+48$ é um múltiplo de 693 menor que $1000+48=1048$, e daí que $N+48=693$. Ou seja, $N=645$.\n\nb) Seja $n$ o número de piratas que escolhera Barbaroxa. Para fazer a divisão das moedas, Barbaroxa deve dividir o número $N=645$ por $n$, digamos $645=q n+r$, e ele ficará com $r$ moedas ($q$ e $r$ são números naturais). Note que\n$$\n(q+1) n>q n+r=645 \\geq q n\n$$\nDa desigualdade na esquerda, temos que $(q+1) 100 \\geq (q+1) n>645$, e então $q \\geq 6$. Analisemos agora os possíveis valores de $q$.\n\n- Se $q=6$: Da desigualdade na esquerda de (*), obtemos $n>645 / 7>92$. Nesse caso, $r=645-6 n \\leq 645-6(93)=87$ e, quando $n=93, r=87$.\n\n- Se $q=7$: Da desigualdade na esquerda de (*), obtemos $n>645 / 8>80$. Nesse caso, $r=645-7 n \\leq 645-7(81)=78$ e, quando $n=81, r=78$.\n\n- Se $q \\geq 8$: Da desigualdade na direita de (*), obtemos $645 \\geq q n \\geq 8 n$, e então $81>645 / 8 \\geq n>r$.\n\nDessa análise, podemos concluir que Barbaroxa pode obter no máximo $r=87$ moedas. Para isto, ele deve escolher $n=93$ piratas.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25967,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPaulinho estava estudando o Máximo Divisor Comum (MDC) na escola e decidiu praticar em casa. Ele chamou de $a, b$ e $c$ as idades de três pessoas que moram com ele. Em seguida, fez algumas operações com os fatores primos deles e obteve os máximos divisores comuns dos 3 pares de números. Alguns dias depois, ele esqueceu as idades $a, b$ e $c$, mas encontrou os seguintes resultados anotados:\n$$\n\\begin{aligned}\na \\cdot b \\cdot c & =2^{4} \\cdot 3^{2} \\cdot 5^{3} \\\\\nM D C(a, b) & =15 \\\\\nM D C(a, c) & =5 \\\\\nM D C(b, c) & =20\n\\end{aligned}\n$$\nAjude Paulinho a determinar os valores de $a, b$ e $c$.",
    "options": [],
    "answer": "a = 15, b = 60, c = 20",
    "solution": "Solution:\n\nAnalisando os máximos divisores comuns listados, podemos garantir que $a$ é múltiplo de $15$ e de $5$, $b$ é múltiplo de $15$ e de $20$ e $c$ é múltiplo de $5$ e de $20$. Usando os fatores primos destes números, temos que $a$ é múltiplo de $15$, $b$ é múltiplo de $60$ e $c$ é múltiplo de $20$. Deste modo, existem inteiros positivos $x, y$ e $z$, tais que\n$$\n\\begin{aligned}\na & =15 \\cdot x=3 \\cdot 5 \\cdot x \\\\\nb & =60 \\cdot y=2^{2} \\cdot 3 \\cdot 5 \\cdot y \\\\\nc & =20 \\cdot z=2^{2} \\cdot 5 \\cdot z\n\\end{aligned}\n$$\nSubstituindo estas expressões de $a, b$ e $c$ no produto, temos $x \\cdot y \\cdot z=1$ e, portanto, $x=y=z=1$. Logo, $a=15$, $b=60$ e $c=20$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25968,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUma noite, Wanderson sonhou com dois números de três algarismos:\n$$\na b c \\text{ e } d e f,\n$$\nde modo que a soma\n$$\na b c + d e f + a b c d e f\n$$\ncoincidia com a soma de todos os números de três algarismos. Note que $a b c$ não é o produto dos algarismos $a, b$ e $c$, e sim o número de três algarismos $a, b$ e $c$. O mesmo vale para os outros números.\n\na) Calcule a soma de todos os números de três algarismos.\n\nb) Mostre que o sonho de Wanderson é um sonho impossível.",
    "options": [],
    "answer": "a) 494550; b) Impossible",
    "solution": "Solution:\n\nPodemos calcular a soma de todos os números de 3 algarismos\n$$\n100 + 101 + 102 + \\cdots + 999 = \\frac{(100 + 999) \\times 900}{2} = 1099 \\times 450\n$$\n\nPor outro lado,\n$$\na b c + d e f + a b c d e f = a b c + d e f + 1000 \\times a b c + d e f\n$$\nO sonho de Wanderson pode então ser escrito como a equação\n$$\n1099 \\times 450 = 1001 \\times a b c + 2 \\times d e f.\n$$\n\nObservemos que o número $1099 \\times 450$ pode ser escrito como\n$$\n(1001 + 98) \\times 450 = 1001 \\times 450 + 44100 = 1001 \\times 450 + 1001 \\times 44 + 56 = 1001 \\times 494 + 56\n$$\nIsto implica que\n$$\n1001 \\times (494 - a b c) + 56 = 2 \\times d e f\n$$\nO lado direito da equação é um número par. Para que o lado esquerdo seja par, é necessário que $494 - a b c$ seja par. Dividindo por 2, temos que\n$$\n1001 \\times \\frac{(494 - a b c)}{2} + 28 = d e f\n$$\nSabemos que $(494 - a b c)/2$ deve ser um número inteiro. Se $(494 - a b c)/2 = 0$, o lado esquerdo seria igual a 28 e não obteríamos um número de 3 algarismos. Se entretanto $(494 - a b c)/2 = 1$, o lado esquerdo seria igual a 1029 e não obteríamos um número de 3 algarismos. Portanto a equação não pode ser satisfeita. Concluímos que o sonho de Wanderson é impossível.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25969,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA sequência de Fibonacci começa com $F_{0}=0, F_{1}=1$ e, a partir do segundo termo, cada novo termo é obtido somando-se os dois anteriores, ou seja,\n$$\nF_{n+2}=F_{n+1}+F_{n} \\text{ para } n \\geq 0\n$$\nAssim, os primeiros termos da sequência de Fibonacci são:\n$$\n\\begin{array}{ccccccccccccc}\nF_{0} & F_{1} & F_{2} & F_{3} & F_{4} & F_{5} & F_{6} & F_{7} & F_{8} & F_{9} & F_{10} & F_{11} & F_{12} \\\\\n0 & 1 & 1 & 2 & 3 & 5 & 8 & 13 & 21 & 34 & 55 & 89 & 144\n\\end{array}\n$$\na) Verifique que $F_{n+3}<5 F_{n}$ para todo $n \\geq 3$.\nb) Seja $n$ um inteiro positivo. Mostre que entre potências consecutivas de $n$ existe no máximo $n$ números de Fibonacci.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\na) Como a sequência de Fibonacci é crescente, temos\n$$\n\\begin{aligned}\nF_{n+3} & =F_{n+2}+F_{n+1} \\\\\n& =2 \\cdot F_{n+1}+F_{n} \\\\\n& =3 \\cdot F_{n}+2 \\cdot F_{n-1} \\\\\n& <3 \\cdot F_{n}+2 \\cdot F_{n} \\\\\n& =5 \\cdot F_{n}\n\\end{aligned}\n$$\n\nb) Suponha que existam mais que $n$ números de Fibonacci entre $n^{k}$ e $n^{k+1}$. Denotemos as $n+1$ primeiras delas por $F_{l}, F_{l+1}, \\ldots, F_{l+n}$. Assim, como $F_{l}$ e $F_{l+1}$ são maiores que $n^{k}$, temos\n$$\n\\begin{aligned}\nF_{l+2} & =F_{l+1}+F_{l} \\\\\n& >n^{k}+n^{k} \\\\\n& =2 \\cdot n^{k} \\\\\nF_{l+3} & =F_{l+2}+F_{l+1} \\\\\n& >2 \\cdot n^{k}+n^{k} \\\\\n& =3 \\cdot n^{k} \\\\\nF_{l+4} & =F_{l+3}+F_{l+2} \\\\\n& >3 \\cdot n^{k}+n^{k} \\\\\n& =4 \\cdot n^{k} \\\\\n& \\vdots \\\\\nF_{l+n} & =F_{l+n-1}+F_{l+n-2} \\\\\n& >(n-1) \\cdot n^{k}+n^{k} \\\\\n& =n^{k+1}\n\\end{aligned}\n$$\nIsto é um absurdo, pois $F_{l+n}<n^{k+1}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25970,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm rei está em um dos cantos de um tabuleiro $m \\times n$. Dois jogadores movem o rei alternadamente para qualquer casa ainda não visitada. O primeiro jogador que não puder mais mover o rei perde. Determine, em função das dimensões do tabuleiro, quem possui a estratégia vencedora.\n\nObservação: No xadrez, o rei se movimenta uma casa na horizontal, na vertical ou na diagonal.",
    "options": [],
    "answer": "If the total number of squares is even, the first player has a winning strategy; if it is odd, the second player has a winning strategy.",
    "solution": "Solution:\n\nSejam $A$ o primeiro e $B$ o segundo jogador. Dizemos que um jogador possui uma estatégia vencedora se ele pode sempre garantir a vitória independentemente de como o outro jogador faça seus movimentos. Vamos provar que:\n(i) se $m n$ é par, então $A$ possui a estratégia vencedora;\n(ii) se $m n$ é ímpar, então $B$ possui a estratégia vencedora.\n\nNo primeiro caso, cubra o tabuleiro utilizando dominós $2 \\times 1$. A estratégia de $A$ consiste em movimentar o rei para a casa adjacente no mesmo dominó em que $B$ deixou a peça. Assim, enquanto $B$ jogar, $A$ sempre joga e, portanto, vence o jogo. No segundo caso, $m$ e $n$ são ímpares. Daí, com a exceção do canto em que o rei inicia o jogo, as casas podem ser cobertas por dominós $2 \\times 1$. Logo, após o primeiro movimento de $A, B$ aplica a mesma estratégia descrita no caso anterior.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25971,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nVocê sabe repartir um quadrado em 7 quadrados menores?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25972,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMostre que $M=\\sqrt[3]{\\sqrt{5}+2}-\\sqrt[3]{\\sqrt{5}-2}$ é um número inteiro.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nDenotemos $a=\\sqrt[3]{\\sqrt{5}+2}$ e $b=\\sqrt[3]{\\sqrt{5}-2}$. Então\n$$\na^{3}-b^{3}=(\\sqrt{5}+2)-(\\sqrt{5}-2)=4\n$$\ne\n$$\nab=\\sqrt[3]{(\\sqrt{5}+2)(\\sqrt{5}-2)}=\\sqrt[3]{5-4}=\\sqrt[3]{1}=1\n$$\nde modo que $M=a-b$ satisfaz $M^{3}=(a-b)^{3}=a^{3}-b^{3}-3ab(a-b)=4-3M$.\nAssim, $M^{3}+3M-4=0$, ou seja, o número $M$ é raiz do polinômio $x^{3}+3x-4$. Como o número $1$ é uma raiz desse polinômio, podemos fatorá-lo e escrever $x^{3}+3x-4$ como $(x-1)\\left(x^{2}+x+4\\right)$. O trinômio $x^{2}+x+4$ tem discriminante negativo, de modo que a única raiz real de $x^{3}+3x-4$ é $1$ e, portanto, $M=1$. Em particular, $M$ é um número inteiro.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25973,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSe $(x, y)$ é uma solução do sistema\n$$\n\\left\\{\\begin{array}{l}\nx y=6 \\\\\nx^{2} y+x y^{2}+x+y=63\n\\end{array}\\right.\n$$\ndetermine o valor de $x^{2}+y^{2}$.",
    "options": [],
    "answer": "69",
    "solution": "Solution:\nTemos\n$$\n\\begin{aligned}\n63 & =x^{2} y+x y^{2}+x+y \\\\\n& =x y(x+y)+(x+y) \\\\\n& =6(x+y)+(x+y) \\\\\n& =7(x+y)\n\\end{aligned}\n$$\nPortanto, $x+y=9$. Assim,\n$$\n\\begin{aligned}\nx^{2}+y^{2} & =(x+y)^{2}-2 x y \\\\\n& =81-12 \\\\\n& =69\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25974,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm mastro vertical $AB$ de altura $1\\,m$ é iluminado por raios do sol e forma sombras no plano horizontal de comprimentos: $1$, $2$ e $3$ metros em três momentos diferentes. Prove que a soma dos ângulos de incidência dos raios nestes três momentos forma um ângulo reto, ou seja,\n$$\n\\angle ACD + \\angle AEB + \\angle ADB = 90^\\circ\n$$\n![](attached_image_1.png)\n",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nComo $D$, $E$ e $C$ são as extremidades das sombras do mastro, segue que $BD = DE = EC = 1$. Sendo $ABD$ um triângulo retângulo com $AB = BD$, temos $\\angle ADB = \\angle BAD = 45^\\circ$. Portanto, basta mostrar que $\\angle BEA + \\angle BCA = 45^\\circ$. Construamos o triângulo $AFG$ de modo que $AFDB$ seja um quadrado e $GD = BC$.\n![](attached_image_2.png)\nAssim, $\\angle AFG = \\angle ABC$, $BE = FG$ e $AF = AB$. Consequentemente os triângulos $AGF$ e $AEB$ são congruentes pelo caso $LAL$ e assim $AG = AE$. Além disto,\n$$\n\\begin{aligned}\n\\angle GAE &= \\angle GAF + \\angle FAE \\\\\n&= \\angle EAB + \\angle FAE \\\\\n&= \\angle FAB \\\\\n&= 90^\\circ\n\\end{aligned}\n$$\nPortanto, $GAE$ é um triângulo retângulo isósceles. Como $GD = BC$, $DE = AB$ e $\\angle GDE = \\angle ABC$, segue pelo caso $LAL$ que os triângulos $GDE$ e $CBA$ são congruentes. Daí\n$$\n\\begin{aligned}\n45^\\circ &= \\angle AGE \\\\\n&= \\angle AGD + \\angle DGE \\\\\n&= \\angle AEB + \\angle ACB\n\\end{aligned}\n$$\nObservação: Também é possível obtermos uma solução direta usando trigonometria:\n$$\n\\begin{aligned}\n\\operatorname{tg}(\\angle AEB + \\angle ACB) &= \\frac{\\operatorname{tg}(\\angle AEB) + \\operatorname{tg}(\\angle ACB)}{1 - \\operatorname{tg}(\\angle AEB)\\operatorname{tg}(\\angle ACB)} \\\\\n&= \\frac{AB/BE + AB/BC}{1 - (AB/BE)(AB/BC)} \\\\\n&= \\frac{1/2 + 1/3}{1 - 1/2 \\cdot 1/3} \\\\\n&= 1\n\\end{aligned}\n$$\nPortanto, $\\angle AEB + \\angle ACB = 45^\\circ$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25975,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNo tabuleiro abaixo, é permitido mover qualquer objeto de seu quadrado para qualquer quadrado adjacente vazio acima, abaixo, ao lado ou em diagonal.\n![](attached_image_1.png)\n\na) Mostre como trocar a posição de todos os chapéus com todos os troféus em apenas cinco movimentos. Argumente porque não é possível trocá-los de posição com menos de cinco movimentos.\n\nb) Neste outro tabuleiro mostrado abaixo, qual o mínimo de movimentos para trocar os chapéus de posição com os troféus?\n![](attached_image_2.png)\n\nc) E se fosse um tabuleiro parecido com os anteriores, porém com 1000 chapéus e 1000 troféus, qual seria o mínimo de movimentos para trocá-los de posição?\n![](attached_image_3.png)",
    "options": [],
    "answer": "a) 5; b) 7; c) 2001",
    "solution": "Solution:\n\na) Abaixo, mostramos uma sequência de cinco movimentos para trocar os chapéus com os troféus (há outra!).\n![](attached_image_4.png)\n\nO argumento para mostrar que não é possível trocá-los com menos do que 5 movimentos é o seguinte: no primeiro movimento, precisamos mover um troféu ou um chapéu para a casa vazia acima à direita. Após este movimento, todos os objetos estarão fora de suas casas de destino. Como são quatro objetos, serão necessários pelo menos quatro movimentos para colocá-los em seus lugares. Como já foi feito um movimento, teremos $1+4=5$ movimentos no mínimo para trocar os chapéus com os troféus.\n\nb) Abaixo mostramos uma sequência de sete movimentos para trocar os chapéus de lugar com os troféus:\n![](attached_image_5.png)\n\nO mesmo argumento de antes se aplica para concluírmos que sete é o mínimo de movimentos para trocar os chapéus de lugar com os troféus. No primeiro movimento, temos de mover um chapéu ou um troféu para a casa vazia. Neste momento, todos os seis objetos estarão fora de seus lugares de destino. Portanto, para colocá-los em suas posições serão necessários pelo menos seis movimentos. Daí, teremos $6+1=7$ movimentos no mínimo.\n\nc) Se repetirmos o padrão descrito acima, conseguiremos trocar os chapéus e troféus com 2001 movimentos! A ideia é a seguinte. Começamos com a configuração\n\n| $a$ | $a$ | a | a | a |\n| :--- | :--- | :--- | :--- | :--- |\n| $a$ | $a$ | $a$ | $a$ | $a$ |\n![](attached_image_6.png)\n\nEm seguida, por meio de 2 movimentos, atingimos a configuração\n\n| a | a | a | a | a |\n| :--- | :--- | :--- | :--- | :--- |\n| a | d | a | d | a |\n![](attached_image_7.png)\n\nE com mais dois movimentos, atingimos a configuração\n\n| a | a | a | a | a |\n| :--- | :--- | :--- | :--- | :--- |\n| $a$ | $a$ | $a$ | 9 | 0 |\n![](attached_image_8.png)\n\nPortanto, depois de 1000 movimentos atingimos a configuração\n\n| a | a | a | a | a |\n| :--- | :--- | :--- | :--- | :--- |\n|  | $a$ | a | a | a |\n![](attached_image_9.png)\n\nAgora vamos voltando com a casa vazia! Com mais dois movimentos (depois desses 1000 movimentos), obtemos a configuração\n![](attached_image_10.png)\n![](attached_image_11.png)\n\nCom mais dois movimentos, obtemos a configuração\n![](attached_image_12.png)\n\nPortanto, depois de 2000 movimentos (contando aqueles 1000 anteriores), obtemos a configuração\n\n|  | $a$ | $a$ | $a$ | $a$ |\n| :--- | :--- | :--- | :--- | :--- |\n| 1 | 0 | a | a | 0 |\n![](attached_image_13.png)\n\nE com mais um movimento, concluímos! Logo, fizemos 2001 movimentos. O argumento para mostrar que não é possível fazer a troca com menos do que isso é o mesmo dos anteriores.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25976,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa figura abaixo, os segmentos $AB$ e $CD$ são paralelos. Se $\\angle CIJ = 10\\beta$, $\\angle AGJ = 10\\alpha$, $\\angle CEJ = 6\\alpha$ e $\\angle JFG = 6\\beta$, determine o valor do ângulo $\\angle IJK$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "120°",
    "solution": "Solution:\n\n![](attached_image_2.png)\nComo $\\angle CEJ$ e $\\angle JFA$ são ângulos colaterais internos, $6\\alpha + 6\\beta = 180^\\circ$, ou seja, $\\alpha + \\beta = 30^\\circ$. Pelo ponto $J$, considere o segmento $ML$ paralelo a $AB$. Temos\n$$\n\\begin{aligned}\nx & = 180^\\circ - \\angle IJG \\\\\n  & = 180^\\circ - \\angle IJM - \\angle MJG \\\\\n  & = 180^\\circ - \\angle JIE - \\angle JGF \\\\\n  & = 180^\\circ - (180^\\circ - 10\\beta) - (180^\\circ - 10\\alpha) \\\\\n  & = 10(\\alpha + \\beta) - 180^\\circ \\\\\n  & = 120^\\circ\n\\end{aligned}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 25977,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA razão entre o número de homens e o de mulheres na cidade de Campo Verde é $\\frac{2}{3}$. A idade média dos homens é 37 anos e a das mulheres é 42 anos. Qual é a idade média dos habitantes de Campo Verde?",
    "options": [],
    "answer": "40",
    "solution": "Solution:\n\nSe $H$ indica o número de homens e $M$ o de mulheres, então:\n$$\n\\frac{H}{M} = \\frac{2}{3} \\quad \\Rightarrow \\quad M = \\frac{3H}{2}\n$$\nA idade média da população é:\n$$\n\\frac{37H + 42M}{H + M} = \\frac{37H + 42 \\frac{3H}{2}}{H + \\frac{3H}{2}} = \\frac{100H}{\\frac{5H}{2}} = \\frac{100 \\times 2}{5} = 40 \\text{ anos. }\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25978,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAs páginas do dicionário - Para numerar as páginas de um dicionário, imprimiu-se 1988 vezes o algarismo 1. Quantas páginas tem esse dicionário?",
    "options": [],
    "answer": "3151",
    "solution": "Solution:\n\nObservemos que:\n- a cada 10 números imprime-se 1 vez o $1$ nas unidades,\n- a cada 100 números imprime-se 10 vezes o $1$ nas dezenas,\n- a cada 1000 números imprime-se 100 vezes o $1$ nas centenas.\n\nAssim, de $1$ até $999$ imprime-se o número $1$:\n$100$ vezes nas unidades $+100$ nas dezenas $+100$ nas centenas $=300$.\n\nDe $1000$ até $1999$, imprime-se o número $1$ outras $300$ vezes entre as unidades, dezenas e centenas, e $1000$ vezes na posição dos milhares, portanto entre $1$ e $1999$ o número de vezes que imprime-se o $1$ é: $300+300+1000=1600$.\n\nAgora entre $2000$ e $2999$ imprime-se o $1$ mais $300$ vezes, completando $1600+300=1900$.\n\nDe $3000$ a $3099$ temos $20$ algarismos $1$, de $3100$ a $3119$, temos $40$ algarismos $1$ e de $3120$ a $3139$ temos $22$ algarismos, portanto até $3139$ o número de vezes que imprime-se o $1$ é: $1900+20+40+22=1982$ vezes.\n\nComo faltam $6$ algarismos $1$, o número de páginas do livro é $3144$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25979,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nBrincando com números - A soma $1+1+4$ dos algarismos do número $114$ divide o próprio número. Qual é o maior número, menor do que $900$, que satisfaz esta propriedade?",
    "options": [],
    "answer": "888",
    "solution": "Solution:\n\nComo queremos encontrar o maior número possível, menor do que $900$, iniciaremos com o algarismo $8$ na casa da centena. Observemos que o número $800$ satisfaz a propriedade. Logo, o número procurado é maior que ou igual a $800$.\n\nDevemos então encontrar $a$ e $b$ tais que $8+a+b$ divida $8ab = 800 + 10a + b$. Lembramos que $8+a+b$ divide $8ab = 800 + 10a + b$ se, e somente se, $8+a+b$ divide $800 + 10a + b - (8+a+b) = 792 + 9a$.\n\nAgora, atribuindo valores para $a$ na ordem decrescente obtemos:\n\n- $a=9 \\Rightarrow 792 + 9 \\times 9 = 873 = 9 \\times 97$ e este número não possui nenhum divisor entre $17\\ (b=0)$ e $26\\ (b=9)$.\n- $a=8 \\Rightarrow 792 + 9 \\times 8 = 864 = 2^{5} \\times 3^{3}$. O maior divisor deste número entre $16$ e $25$ é $24$, isto é, $b=8$. Logo, o número procurado é $888$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25980,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm serviço de vigilância vai ser instalado num parque na forma de uma rede de estações. As estações devem ser conectadas por linhas de telefone, de modo que qualquer uma das estações possa se comunicar com todas as outras, seja por uma conexão direta seja através de no máximo uma outra estação. Cada estação pode ser conectada diretamente por um cabo a no máximo 3 outras estações.\n\nO diagrama mostra um exemplo de uma rede desse tipo conectando 7 estações. Qual é o maior número de estações que podem ser conectadas dessa maneira?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "10",
    "solution": "Solution:\n\nO exemplo mostra que podemos conectar pelo menos 7 estações dentro das condições propostas. Começamos com uma estação particular, e vamos pensar nela como se fosse a base da rede. Ela pode ser conectada a 1, 2 ou 3 estações conforme mostra o diagrama.\n\n![](attached_image_2.png)\n\nAgora, as estações $A$, $B$ e $C$ têm ainda duas linhas não utilizadas, logo podem ser conectadas a duas outras estações como a seguir:\n\n![](attached_image_3.png)\n\nAgora, é impossível acrescentar mais estações porque qualquer outra a mais não poderia ser conectada à base satisfazendo as condições do problema. Isso mostra que não podemos ter mais do que 10 estações. Vamos agora verificar se podemos montar a rede com essas 10 estações. Observe no diagrama acima que apenas a Base é conectada a todas as outras estações (através de um cabo ou de uma conexão via uma estação). As estações que estão nos extremos ainda possuem duas linhas não utilizadas, e agora vamos usá-las para \"fechar\" a rede; veja o diagrama a seguir.\n\n![](attached_image_4.png)",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 25981,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm corpo em queda livre demora $11$ segundos para tocar o solo. No primeiro segundo ele percorre $4,5\\ \\mathrm{m}$, em cada segundo que segue, a distância percorrida aumenta de $9,8\\ \\mathrm{m}$. Qual a altura da queda e quantos metros ele percorreu no último segundo?",
    "options": [],
    "answer": "Height: 588.5 m; last second distance: 102.5 m",
    "solution": "Solution:\n\nSeja $d_n$ a distância percorrida no $n$-ésimo segundo.\n\nTemos:\n\n$d_1 = 4,5\\ \\mathrm{m}$\n\ne a cada segundo, a distância aumenta de $9,8\\ \\mathrm{m}$:\n\n$d_2 = d_1 + 9,8$\n$d_3 = d_2 + 9,8 = d_1 + 2 \\times 9,8$\n$\\ldots$\n$d_n = d_1 + (n-1) \\times 9,8$\n\nA altura total $H$ é a soma das distâncias percorridas em cada segundo, ou seja:\n\n$$\nH = d_1 + d_2 + \\ldots + d_{11}\n$$\n\nComo é uma progressão aritmética de razão $r = 9,8$ e termo inicial $a_1 = 4,5$:\n\nO termo geral:\n$$\nd_n = 4,5 + (n-1) \\times 9,8\n$$\n\nA soma dos $11$ termos:\n$$\nH = \\frac{(a_1 + a_{11}) \\times 11}{2}\n$$\n\nCalculando $a_{11}$:\n$$\na_{11} = 4,5 + (11-1) \\times 9,8 = 4,5 + 10 \\times 9,8 = 4,5 + 98 = 102,5\\ \\mathrm{m}\n$$\n\nPortanto,\n$$\nH = \\frac{4,5 + 102,5}{2} \\times 11 = \\frac{107}{2} \\times 11 = 53,5 \\times 11 = 588,5\\ \\mathrm{m}\n$$\n\nA altura da queda é $588,5\\ \\mathrm{m}$.\n\nA distância percorrida no último segundo ($11^\\text{o}$):\n$$\nd_{11} = 4,5 + (11-1) \\times 9,8 = 4,5 + 98 = 102,5\\ \\mathrm{m}\n$$\n\nPortanto, no último segundo, ele percorreu $102,5\\ \\mathrm{m}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25982,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOs números $a$ e $b$ são inteiros positivos e satisfazem $96 a^{2} = b^{3}$. Qual é o menor valor de $a$?",
    "options": [],
    "answer": "12",
    "solution": "Solution:\n\nFatorando $96$ temos: $2^{5} \\times 3 \\times a^{2} = b^{3}$. Para que $2^{5} \\times 3 \\times a^{2}$ seja um cubo, o número $a$ deve ser da forma: $a = 2^{n} \\times 3^{m}$. Assim,\n$$\n2^{5} \\times 3 \\times a^{2} = 2^{5} \\times 3 \\times (2^{n} \\times 3^{m})^{2} = 2^{5+2n} \\times 3^{1+2m} = b^{3}\n$$\nLogo, $5+2n$ e $1+2m$ são múltiplos de $3$. Os menores valores de $n$ e $m$ são: $n=2$ e $m=1$. Portanto, $a = 2^{2} \\times 3 = 12$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25983,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nO valor de $\\left(\\sqrt{1+\\sqrt{1+\\sqrt{1}}}\\right)^4$ é:\n(a) $\\sqrt{2}+\\sqrt{3}$\n(b) $\\frac{1}{2}(7+3 \\sqrt{5})$\n(c) $1+2 \\sqrt{3}$\n(d) 3\n(e) $3+2 \\sqrt{2}$",
    "options": [],
    "answer": "e",
    "solution": "Solution:\n$$\n\\left(\\sqrt{1+\\sqrt{1+\\sqrt{1}}}\\right)^4 = (1+\\sqrt{2})^2 = 1+2 \\sqrt{2}+2 = 3+2 \\sqrt{2}\n$$\nA opção correta é (e).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25984,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMostre que $M=\\sqrt[3]{\\sqrt{5}+2}-\\sqrt[3]{\\sqrt{5}-2}$ é um número inteiro.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSejam $a=\\sqrt[3]{\\sqrt{5}+2}$ e $b=\\sqrt[3]{\\sqrt{5}-2}$. Assim, $M=a-b$ e temos:\n$$\nM^{3}=(a-b)^{3}=a^{3}-b^{3}-3 a b(a-b)\n$$\nSabemos que $a^{3}-b^{3}=4$ e $a b=1$. Assim, $M^{3}+3 M-4=0$, ou seja, o número $M$ é raiz do polinômio $x^{3}+3 x-4$.\nPor sua vez, o número 1 é uma raiz do polinômio $x^{3}+3 x-4$. Fatorando tal polinômio, obtemos $(x-1)\\left(x^{2}+x+4\\right)$. Mas o trinômio $x^{2}+x+4$ tem discriminante negativo. Consequentemente, 1 é a única raiz real de $x^{3}+3 x-4$. Portanto, $M=1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25985,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n2. Potências de 2 - Calcule:\na) $1678^{2}-1677^{2}$\nb) $1001^{2}+1000^{2}$\nc) $19999^{2}$\nd) $2001^{2}+2002^{2}+2003^{2}$",
    "options": [],
    "answer": "a) 3355; b) 2002001; c) 399960001; d) 12024014",
    "solution": "Solution:\n\n2. Potências de 2 - Fatorando temos:\n$a)$ $1678^{2}-1677^{2} = (1678+1677)(1678-1677) = 3355$.\n\nb) Como $(a+b)^{2} = a^{2} + 2ab + b^{2}$, temos:\n$$\n\\begin{aligned}\n1001^{2} + 1000^{2} & = (1000+1)^{2} + 1000^{2} = 1000^{2} + 2000 + 1 + 1000^{2} = \\\\\n& = 2 \\times 1000^{2} + 2001 = 2002001\n\\end{aligned}\n$$\n\nc) Como $(a-b)^{2} = a^{2} - 2ab + b^{2}$, temos:\n$$\n\\begin{aligned}\n19999^{2} & = (20000-1)^{2} = \\left(2 \\times 10^{4}\\right)^{2} - 4 \\times 10^{4} + 1 = \\\\\n& = 4 \\times 10^{8} - 4 \\times 10^{4} + 1 = 399960001\n\\end{aligned}\n$$\n\nd) Colocando em função de $2000$ temos:\n$$\n\\begin{aligned}\n2001^{2} + 2002^{2} + 2003^{2} & = (2000+1)^{2} + (2000+2)^{2} + (2000+3)^{2} = \\\\\n& = 3 \\times 2000^{2} + 12 \\times 2000 + 14 \\\\\n& = 12024014\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25986,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDiferença de quadrados - Se a diferença dos quadrados de dois números inteiros consecutivos é $2000$, então os dois números são:\n\n(A) menores que $100$.\n(B) menores que $1000$, porém maiores que $99$.\n(C) menores que $10000$, porém maiores que $999$.\n(D) menores que $100000$, porém maiores que $9999$.\n(E) não existem estes dois números.",
    "options": [],
    "answer": "E",
    "solution": "Solution:\n\nA resposta correta é (E).\n\nInicialmente, observe que o quadrado de um número par é par, e o quadrado de um número ímpar é ímpar. Se os dois números são consecutivos, então um número é par e o outro é ímpar. Portanto, elevando ao quadrado, um deles é par e o outro é ímpar. Mas, a diferença entre um número par e um número ímpar é sempre um número ímpar. Como $2000$ é um número par, concluímos que não existem dois números consecutivos cuja diferença dos seus quadrados seja $2000$.\n\nOutra solução para o problema é a seguinte. Primeiramente, suponhamos que tais inteiros $a$ e $a+1$ são maiores ou iguais a zero. Nesse caso, temos:\n$$\n(a+1)^2 - a^2 = 2000\n$$\nFatorando a diferença de quadrados $(a+1)^2 - a^2$ obtemos:\n$$\n(a+1+a)(a+1-a) = 2000 \\Rightarrow 2a+1 = 2000\n$$\nque não tem solução pois $2a+1$ é ímpar e $2000$ é par.\n\nSe $a$ e $a+1$ fossem menores ou iguais a zero então $-a$ e $-a-1$ seriam inteiros maiores ou iguais a zero e sucessivos, satisfazendo a condição $(-a)^2 - (-a-1)^2 = 2000$ o que não pode ocorrer como provado acima.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25987,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm sanduíche e um prato de refeição custam em média $R\\$ 5,00$ e $R\\$ 7,00$, respectivamente. De quantas maneiras pode-se comprar sanduíches e pratos de refeição com $R\\$ 90,00$, sem deixar troco?",
    "options": [],
    "answer": "3",
    "solution": "Solution:\n\nSe $S$ corresponde ao número de sanduíches e $P$ ao número de pratos de refeição, então $5S + 7P = 90$. Precisamos encontrar soluções inteiras $S$ e $P$ para essa equação. Temos:\n\n$$\n5S + 7P = 90 \\Longrightarrow P = \\frac{90 - 5S}{7} = 5 \\times \\frac{18 - S}{7}\n$$\n\nComo $P$ é um número natural, temos que $7$ tem que dividir $18 - S$, assim $S = 4$, $11$ ou $18$, e em cada um destes casos $P$ é igual a $10$, $5$ e $0$, respectivamente. Portanto, temos somente três formas de fazer a compra.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25988,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n2. Número premiado - Um número de 6 algarismos é \"premiado\" se a soma de seus primeiros 3 algarismos é igual à soma de seus 3 últimos algarismos. Por exemplo $342531$ é premiado pois $3+4+2=5+3+1$.\n\na. Qual é o maior e o menor número premiado, com 6 algarismos diferentes?\n\nb. Mostre que a soma de todos os números premiados, com 6 algarismos diferentes, é divisível por $13$.",
    "options": [],
    "answer": "Largest: 981765; Smallest: 108234; moreover, the sum of all such six-digit numbers with distinct digits is divisible by 13.",
    "solution": "Solution:\n\n(a) O maior número premiado tem de começar com $98$. Assim o número procurado é da forma: $98abcde$. Por hipótese temos: $9+8+a=b+c+d$. Para que $a$ seja máximo precisamos que $b+c+d$ seja máximo, e isto acontece quando $b=7$, $c=6$ e $d=5$. Neste caso, $a=1$ e consequentemente, o maior número premiado é $981765$.\n\nVamos agora determinar o menor número premiado. Tentemos um número da forma $10abcde$. Agora, não é difícil verificar que $108234$ é o menor número premiado.\n\n(b) Se o número $ABCDEF$ é premiado, então o número $DEFABC$ também é premiado. A soma desses números é:\n$$\n\\begin{aligned}\nABCDEF + DEFABC &= (1000 \\times ABC + DEF) + (1000 \\times DEF + ABC) \\\\\n&= 1001 (ABC + DEF) = 13 \\times 11 \\times 7 \\times (ABC + DEF)\n\\end{aligned}\n$$\nSomando todos os números premiados com 6 algarismos diferentes aos pares, resulta que cada par é divisível por $13$. Logo, a soma de todos eles é divisível por $13$.\n\nNota: De fato também é divisível por $11$ e $7$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25989,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nTodos com o 2 - Qual operação devemos fazer com todos os 5 números\n$$\n418, 244, 816, 426, 24\n$$\npara obter 5 números que tenham todos o algarismo 2?\n(a) dividir 2;\n(b) somar 4;\n(c) dividir por 6;\n(d) subtrair 5;\n(e) multiplicar por 3.",
    "options": [],
    "answer": "e",
    "solution": "Solution:\nTodos com o 2 - Resposta: multiplicar por 3.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25990,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCartões premiados - Uma loja distribui 9999 cartões entre os seus clientes. Cada um dos cartões possui um número de 4 algarismos, entre 0001 e 9999. Se a soma dos primeiros 2 algarismos for igual à soma dos 2 últimos, o cartão é premiado. Por exemplo, o cartão 0743 é premiado. Prove que a soma dos números de todos os cartões premiados é divisível por 101.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nObserve que se o cartão $a b c d$ é premiado então o cartão $c d a b$ também é premiado, por exemplo: 2341 e 4123 são ambos premiados. Assim sempre que $a b \\neq c d$ temos dois cartões premiados cuja soma é\n$$\na b c d + c d a b = (a b \\times 100 + c d) + (c d \\times 100 + a b) = 101(a b + c d)\n$$\nassim a soma desse dois cartões é divisível por 101.\n\nNo caso que o cartão ser da forma\n$$\na b a b = a b \\times 100 + a b = 101 \\times a b\n$$\no número do cartão é divisível por 101. Assim a soma de todos os cartões é divisível por 101 já que a soma pode ser feita agrupando cartões do tipo $a b c d$ com $c d a b$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 25991,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nQual é a metade? (N2/N3) - Considere a figura ao lado, em que $AB = AE = ED = CD = CA$ e o arco $CB$ é um arco de círculo centrado no ponto $E$. Você sabe repartir essa figura em duas partes idênticas, que possam ser superpostas?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_2.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25992,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm trapézio - No trapézio da figura abaixo $AB$ é paralelo a $CD$, $AD = AB = BC = 1~\\mathrm{cm}$ e $DC = 2~\\mathrm{cm}$. Quanto mede o ângulo $C \\hat{A} D$?\n\n(A) $30^\\circ$\n(B) $45^\\circ$\n(C) $60^\\circ$\n(D) $90^\\circ$\n(E) $120^\\circ$\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "D",
    "solution": "Solution:\n\nA resposta correta é (D).\nSeja $P$ o ponto médio do segmento $CD$ e traçemos os segmentos $AP$ e $BP$. Os três triângulos formados $\\triangle ADP$, $\\triangle ABP$ e $\\triangle BCP$ são equiláteros (por quê?). Então, os ângulos $D \\widehat{A} P = 60^\\circ = P \\widehat{A} B$. Como o segmento $AC$ é a bissetriz do ângulo $P \\widehat{A} B$ (por quê?), concluímos que $P \\widehat{A} C = 30^\\circ$. Portanto:\n\n![](attached_image_2.png)\n\n$$\nC \\widehat{A} D = D \\widehat{A} P + P \\widehat{A} C = 60^\\circ + 30^\\circ = 90^\\circ\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25993,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMulheres votantes - Numa cidade, $40\\%$ de todas as mulheres são votantes e $52\\%$ da população é de mulheres. Qual o percentual da população formado de mulheres votantes?\n(a) $18,1\\%$\n(b) $20,8\\%$\n(c) $26,4\\%$\n(d) $40\\%$\n(d) $52\\%$",
    "options": [],
    "answer": "b",
    "solution": "Solution:\n\nA fração de mulheres na população é $\\frac{52}{100}$, e delas, a fração que é votante é $\\frac{40}{100}$. Logo, a fração de mulheres votantes é:\n$$\n\\frac{52}{100} \\times \\frac{40}{100} = \\frac{104}{5 \\times 100} = \\frac{104}{5 \\times 100} \\times 100\\% = 20,8\\%\n$$\nA opção correta é (b).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25994,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPotências de 3 - Se $3^{n}=2$ então quanto vale $27^{2 n}$ ?",
    "options": [],
    "answer": "64",
    "solution": "Solution:\n\nPotências de 3 - Temos: $27^{2 n} = \\left(3^{3}\\right)^{2 n} = 3^{6 n} = \\left(3^{n}\\right)^{6} = 2^{6} = 64$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25995,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNas igualdades abaixo, cada letra representa um algarismo:\n\n$$\nAB + BC = CD \\quad \\text{ e } \\quad AB - BC = BA\n$$\n\nquanto vale $A+B+C+D$ ?",
    "options": [],
    "answer": "23",
    "solution": "Solution:\n\n23",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25996,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO diretor da escola decidiu tirar uma foto dos formandos de 2008. Ele colocou os alunos em filas paralelas, todas com o mesmo número de alunos, mas essa disposição era muito larga para o campo de visão de sua máquina fotográfica. Para resolver esse problema, o diretor reparou que bastava tirar um aluno por fila e colocá-los numa nova fila. Essa disposição não agradou o diretor porque a nova fila tinha 4 alunos a menos que as outras. Ele decide então tirar mais 1 aluno por fila colocando-os na nova fila que ele criou, e constata que assim todas as filas ficam com o mesmo número de alunos, e finalmente tira a foto. Quantos alunos apareceram na foto?",
    "options": [],
    "answer": "24",
    "solution": "Solution:\n\nAs figuras a seguir representam a situação do problema, onde em preto estão representados os alunos que foram inicialmente retirados e em cinza os alunos retirados na segunda vez.\n\n![](attached_image_1.png)\n\nSejam $n$ e $m$ o número de filas (linhas horizontais) e de colunas da formação inicial, respectivamente. Ao retirar 1 aluno de cada fila, o diretor obtém uma formação com uma coluna a menos e uma fila incompleta - faltam 4 alunos. Logo, podemos concluir que:\n$$\nn+4 = m-1 \\quad \\Rightarrow \\quad m = n+5\n$$\nRetirando agora mais um aluno de cada fila obtém-se uma formação retangular com 2 colunas a menos que a formação inicial. Logo, podemos concluir que o número de filas (iniciais) é $n=3$. Assim, $m=8$ e o número de alunos é dado por $n \\times m = 3 \\times 8 = 24$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25997,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoma constante - Preencha as 5 casas em branco da tabela $3 \\times 3$ com os números de $3$ a $8$, sem repeti-los, de modo que as somas dos $4$ números escritos nas subtabelas formadas por quadrados $2 \\times 2$ sejam a mesma nas $4$ subtabelas.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Two valid completions are: 1 6 2 / 8 9 7 / 4 3 5 and 1 8 2 / 5 9 4 / 6 3 7.",
    "solution": "Solution:\n\nSejam $a, b, c, d, e$ e $f$ os números que colocaremos na tabela.\n\nDe acordo com a regra para as $4$ subtabelas $2 \\times 2$:\n\n| 1 | $a$ | 2 |\n| :---: | :---: | :---: |\n| $b$ | 9 | $c$ |\n| $d$ | $e$ | $f$ |\n\ntemos:\n$$\n\\begin{aligned}\n& 1+a+b+9 = a+2+9+c \\Rightarrow b = c+1 \\\\\n& 1+a+b+9 = b+9+d+e \\Rightarrow a+1 = d+e \\\\\n& a+2+9+c = 9+c+e+f \\Rightarrow a+2 = e+f\n\\end{aligned}\n$$\n\nSubtraindo a $2^{\\mathrm{a}}$ igualdade da $3^{\\mathrm{a}}$, obtemos $f = 1 + d$. A nossa tabela então ficará assim:\n\n| 1 | $a$ | 2 |\n| :---: | :---: | :---: |\n| $b$ | 9 | $b-1$ |\n| $d$ | $e$ | $d+1$ |\n\nComo $a+1 = d+e$ e $\\{a, b, d, e\\} \\subset \\{3,4,5,6,7,8\\}$, temos os seguintes casos a considerar:\n\n| $a$ | $a+1 = d+e$ | Solução |   |   |\n| :---: | :---: | :---: | :---: | :---: |\n| 3 | 4 | não possui |   |   |\n| 4 | 5 | não possui |   |   |\n| 5 | 6 | não possui |   |   |\n|   | 7 | 1 | 6 | 2 |\n| 6 |   | 8 | 9 | 7 |\n|   |   |   | 3 | 5 |\n| 7 | 8 | não possui |   |   |\n| 8 | 9 | 1 | 8 | 2 |\n|   |   | 5 | 9 | 4 |\n|   |   |   | 3 | 7 |",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25998,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDiferença de quadrados - Se $(x+y)^2-(x-y)^2=20$, então $x y$ é igual a:\n(a) 0\n(b) 1\n(c) 2\n(d) 5\n(e) 10",
    "options": [],
    "answer": "d",
    "solution": "Solution:\n\nComo $(x+y)^2 = x^2 + 2 x y + y^2$ e $(x-y)^2 = x^2 - 2 x y + y^2$, temos:\n$$\n(x+y)^2 - (x-y)^2 = x^2 + 2 x y + y^2 - x^2 + 2 x y - y^2 = 4 x y = 20\n$$\nsegue-se que $x y = 5$. A opção correta é (d).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 25999,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDetermine all integer numbers $x$ and $y$ such that:\n$$\n\\frac{1}{x} + \\frac{1}{y} = \\frac{1}{19}\n$$",
    "options": [],
    "answer": "(38, 38), (380, 20), (20, 380), (-342, 18), (18, -342)",
    "solution": "Solution:\n\nA equação é equivalente a $x y = 19(x + y)$. Uma vez que estamos procurando soluções inteiras e $19$ é um número primo, esta igualdade implica que $x$ ou $y$ devem ser divisíveis por $19$. Como a equação é simétrica em relação às variáveis $x$ e $y$, podemos supor que $x$ é divisível por $19$. Isto é, $x = 19k$ para algum valor inteiro de $k$. Nessa condição, temos:\n$$\nx y = 19(x + y) \\quad \\Leftrightarrow \\quad k y = 19k + y\n$$\nDesta igualdade concluímos que $19k + y$ é divisível por $k$. Uma vez que $19k$ já é divisível por $k$, concluímos que $y$ é divisível por $k$ (prove isso). Isto é, $y = k m$ para algum valor inteiro de $m$. Assim,\n$$\nk y = 19k + y \\Rightarrow k^2 m = 19k + k m \\Rightarrow k m - m = 19 \\Rightarrow m(k - 1) = 19\n$$\nVisto que $m$ e $k$ são números inteiros e $19$ é um número primo, existem somente 4 possibilidades para a igualdade $m(k - 1) = 19$:\n- $m = 19$ e $k - 1 = 1$. Isto implica que $x = 38$ e $y = 38$;\n- $m = -19$ e $k - 1 = -1$. Isto implica que $x = 0$ e $y = 0$, que não é possível, pois na equação original, $x \\neq 0$ e $y \\neq 0$;\n- $m = 1$ e $k - 1 = 19$. Isto implica que $x = 380$ e $y = 20$;\n- $m = -1$ e $k - 1 = -19$. Isto implica que $x = -342$ e $y = 18$.\nDeste modo, obtemos as seguintes possibilidades para o par de números inteiros $(x, y)$ que são soluções da equação dada:\n$$\n(38, 38);\\ (380, 20);\\ (20, 380);\\ (-342, 18);\\ (18, -342)\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26000,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMultiplicação com letras - Na operação abaixo, as letras $a, b$ e $c$ são algarismos distintos e diferentes de 1.\n$$\n\\begin{array}{r}\na b b \\\\\\times \\quad c \\\\\\hline b c b 1\n\\end{array}\n$$\nDetermine os valores de $a, b$ e $c$.",
    "options": [],
    "answer": "a=5, b=3, c=7",
    "solution": "Solution:\n\nSe o produto de $b$ por $c$ termina em $1$, então $b \\times c$ pode ser $21$ ou $81$, segue que $b \\times c = 3 \\times 7$ ou $9 \\times 9$. A única possibilidade de escrever o produto de dois números distintos menores que $10$ é $21 = 3 \\times 7$. Assim temos dois possíveis casos:\n\n1o caso: $b = 3$ e $c = 7$ :\n$$\n\\begin{array}{r}\na 3 3 \\\\\\times \\quad 7 \\\\\\hline 3 7 3 1\n\\end{array}\n$$\nNeste caso $\\frac{3731}{7} = 533$ e, consequentemente, $a = 5$.\n\n2o caso: $b = 7$ e $c = 3$ :\nNesta caso $\\frac{7371}{3} = 2457$. Logo, este caso não ocorre.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26001,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEquação de duas variáveis - Determine todos os pares de inteiros $(x, y)$ tais que $9 x y - x^{2} - 8 y^{2} = 2005$.",
    "options": [],
    "answer": "(63, 58), (459, 58), (-63, -58), (-459, -58)",
    "solution": "Solution:\n\nTemos:\n$$\n\\begin{aligned}\n9 x y - x^{2} - 8 y^{2} = 2005 &\\Leftrightarrow x y - x^{2} + 8 x y - 8 y^{2} = 2005 \\\\\n&\\Leftrightarrow x(y - x) + 8 y(x - y) = 2005 \\\\\n&\\Leftrightarrow (x - y)(8 y - x) = 2005\\ (*)\n\\end{aligned}\n$$\nObservemos que a fatoração em primos de $2005$ é $5 \\cdot 401$.\nAlém disso, a soma dos fatores $x - y$ e $8 y - x$ é $7 y$, que é múltiplo de $7$. Devemos então escrever $2005$ como produto de dois fatores, cuja soma é um múltiplo de $7$. Para isso, os fatores devem ser $\\pm 5$ e $\\pm 401$. A soma dos fatores é $\\pm 406$.\nAssim, por $(*)$ temos:\n$$\n\\left\\{\n\\begin{array}{lll}\nx - y = 5 & \\text{e} & 8y - x = 401 \\\\\nx - y = 401 & \\text{e} & 8y - x = 5 \\\\\n\\text{ou} & & \\\\\nx - y = -5 & \\text{e} & 8y - x = -401 \\\\\n\\text{ou} & & \\\\\nx - y = -401 & \\text{e} & 8y - x = -5\n\\end{array}\n\\right.\n\\Leftrightarrow\n\\left\\{\n\\begin{array}{ll}\nx = 63 & \\text{e} \\ y = 58 \\\\\nx = 459 & \\text{e} \\ y = 58 \\\\\nx = -63 & \\text{e} \\ y = -58 \\\\\nx = -459 & \\text{e} \\ y = -58\n\\end{array}\n\\right.\n$$\nAs soluções são, portanto, $(63, 58)$, $(459, 58)$, $(-63, -58)$ e $(-459, -58)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26002,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFração e porcentagem - Se na fração $\\frac{x}{y}$ diminuirmos o numerador de $40\\%$ e o denominador $y$ de $60\\%$, então a fração $\\frac{x}{y}$ :\n\n(A) diminui $20\\%$\n(B) aumenta $20\\%$\n(C) diminui $50\\%$\n(D) aumenta $50\\%$",
    "options": [],
    "answer": "D",
    "solution": "Solution:\n\nA opção correta é (D).\nSe um número $x$ é diminuído de $40\\%$, ele passa a valer $60\\%$ de $x$, ou seja: $0,6x$. Do mesmo modo, quando um número $y$ é diminuído de $60\\%$, ele passa a valer $0,4y$. Portanto, a fração $\\frac{x}{y}$ passa a ter o valor $\\frac{0,6x}{0,4y} = \\frac{6}{4} \\frac{x}{y} = 1,5 \\frac{x}{y}$. Isto significa que a fração $\\frac{x}{y}$ aumentou $50\\%$ do seu valor.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26003,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMenor inteiro - Sejam $p$ e $q$ inteiros positivos tais que $\\frac{5}{8}<\\frac{p}{q}<\\frac{7}{8}$. Qual é o menor valor de $p$ para que $p+q=2005$ ?",
    "options": [],
    "answer": "772",
    "solution": "Solution:\n\n$$\n\\frac{5}{8}<\\frac{p}{2005-p}<\\frac{7}{8}\n$$\ndo qual segue que\n$$\n5(2005-p)<8p \\quad \\text{e} \\quad 8p<7(2005-p)\n$$\nLogo,\n$$\n\\frac{5 \\times 2005}{13}<p<\\frac{7 \\times 2005}{15} \\Rightarrow 771,15<p<935,66\n$$\nLogo 772 é o menor valor de $p$ que satisfaz as condições do problema.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26004,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO percurso de um atleta - Um atleta resolveu fazer uma corrida de $15~\\mathrm{km}$. Começou correndo $5~\\mathrm{km}$ na direção Sul, depois virou para direção Leste, correndo mais $5~\\mathrm{km}$ e, novamente, virou para a direção Norte, correndo os $5~\\mathrm{km}$ restantes. Após esse percurso, constatou, para seu espanto, que estava no ponto de onde havia partido.\nDescubra dois possíveis pontos sobre o Globo Terrestre de onde esse atleta possa ter iniciado sua corrida.",
    "options": [],
    "answer": "The North Pole; and any point on the parallel that is five kilometers north of a parallel whose circumference is five kilometers near the South Pole, so that the five-kilometer eastward leg loops around back to the same point.",
    "solution": "Solution:\n\nO Polo Norte da Terra é o ponto mais fácil de ser identificado como solução: Saindo o atleta do Polo Norte, correndo $5~\\mathrm{km}$ para o sul, depois $5~\\mathrm{km}$ para o leste e finalmente $5~\\mathrm{km}$ para o norte, ele volta novamente para o Polo Norte.\n\n![](attached_image_1.png)\n\nVamos determinar um outro ponto sobre a Terra que satisfaz as hipóteses do problema. Consideremos um paralelo (linha paralela ao Equador) de comprimento $5~\\mathrm{km}$. Existem dois deles: um próximo ao Polo Norte e outro próximo ao Polo Sul. Vamos denotar por $C_{1}$ o que está mais próximo do Polo Sul. Denotemos por $C_{2}$ o paralelo que está $5~\\mathrm{km}$ de distância de $C_{1}$, medida ao longo de um meridiano. Afirmamos que qualquer ponto $A$ sobre o paralelo $C_{2}$ satisfaz as hipóteses do problema. De fato, saindo de $A$ e caminhando $5~\\mathrm{km}$ para o sul, chega-se a um ponto $B$ do paralelo $C_{1}$. Como $C_{1}$ tem comprimento $5~\\mathrm{km}$, saindo de $B$ e caminhando $5~\\mathrm{km}$ para leste retorna-se novamente para $B$.\n\n![](attached_image_2.png)\n\nFinalmente, saindo de $B$ e caminhando $5~\\mathrm{km}$ para o norte, retorna-se novamente para o ponto de partida $A$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26005,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUma urna tem 9 bolas, numeradas com os números de 1 a 9. José e Maria retiram simultaneamente uma bola da urna. Com as bolas retiradas eles formam um número de 2 algarismos, sendo que o número que está escrito na bola de José é o algarismo das dezenas e o número que está escrito na bola de Maria é o algarismo das unidades. Qual a probabilidade deste número ser par?",
    "options": [],
    "answer": "4/9",
    "solution": "Solution:\n\nSejam $a$ e $b$ os números escritos nas bolas retiradas por José e Maria, respectivamente. Existem então 9 possibilidades para $a$ e 8 possibilidades para $b$. Deste modo, existem $9 \\times 8 = 72$ possibilidades para o número $ab$.\n\nPor outro lado, para contar quantos destes números são pares, precisamos analisar separadamente dois casos:\n- os números $a$ e $b$ são pares;\n- o número $a$ é ímpar e o número $b$ é par.\n\nNo primeiro caso, em que $a$ e $b$ são pares, existem 4 possibilidades para $a$ e 3 possibilidades para $b$. Deste modo, existem $4 \\times 3 = 12$ possibilidades.\n\nNo segundo caso, em que $a$ é ímpar e $b$ é par, existem 5 possibilidades para $a$ e 4 possibilidades para $b$. Assim, existem $5 \\times 4 = 20$ possibilidades.\n\nPortanto, a probabilidade do número $ab$ ser par é $\\frac{12+20}{72} = \\frac{32}{72} = \\frac{4}{9}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26006,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNo quadrilátero $ABCD$, tem-se: $AB=5$, $BC=17$, $CD=5$, $DA=9$, e a medida do segmento $DB$ é um inteiro. Determine $DB$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "13",
    "solution": "Solution:\n\nLembre que, num triângulo, qualquer lado é maior que a diferença e menor que a soma dos outros dois. Do triângulo $ADB$ temos $AD-AB < BD < AD+AB$, e do triângulo $CBD$ segue que $BC-CD < BD < BC+CD$. Substituindo os valores conhecidos obtemos:\n$$\n9-5 < BD < 5+9 \\text{ e } 17-5 < BD < 17+5\n$$\nou seja,\n$$\n4 < BD < 14 \\text{ e } 12 < BD < 22\n$$\nDas duas desigualdades concluímos que:\n$$\n12 < BD < 14\n$$\nComo $BD$ é inteiro, só podemos ter $BD=13$.\n\n![](attached_image_2.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26007,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA seguinte figura mostra um cubo.\n![](attached_image_1.png)\nCalcule o número de triângulos equiláteros que podem ser formados de modo que seus três vértices sejam vértices do cubo.",
    "options": [],
    "answer": "8",
    "solution": "Solution:\n\nSobre o cubo existem somente 3 distâncias possíveis entre os vértices:\n$$\nAB = \\ell, \\quad AR = \\ell \\sqrt{2} \\quad \\text{e} \\quad PD = \\ell \\sqrt{3}\n$$\nUm triângulo equilátero que usa os vértices do cubo devia então ter alguma dessas distâncias como a medida do seu lado.\n![](attached_image_2.png)\nVejamos se existem triângulos equiláteros de lado $\\ell$. Observe que se a distância entre dois vértices do cubo é $\\ell$, necessariamente eles estão unidos por uma aresta do cubo. Mas duas arestas do cubo são sempre perpendiculares. Então, não existem triângulos equiláteros de lado $\\ell$.\n\nVejamos agora se existem triângulos equiláteros de lado $\\ell \\sqrt{3}$. Os únicos pares de vértices à distância $\\ell \\sqrt{3}$ são $\\{A, Q\\}, \\{B, R\\}, \\{S, C\\}$ e $\\{P, D\\}$. Como não existem dois pares com algum vértice em comum, concluímos que não é possível ter um triângulo equilátero de lado $\\ell \\sqrt{3}$.\n\nContemos finalmente os triângulos equiláteros de lado $\\ell \\sqrt{2}$. Primeiro aqueles que têm $A$ como vértice. Os vértices à distância $\\ell \\sqrt{2}$ de $A$ são $P, R$ e $C$. Vemos que escolhendo qualquer par em $\\{P, R, C\\}$ obtemos um triângulo equilátero usando esses dois vértices junto com $A$. Contamos assim 3 triângulos equiláteros que usam o vértice $A$. Agora, se multiplicamos 8 vértices vezes 3 triângulos por cada vértice, consideraríamos todos os triângulos de lado $\\ell \\sqrt{2}$ e cada um deles estaria sendo contado exatamente 3 vezes (uma vez por cada vértice do triângulo contado). Concluímos, assim, que são\n$$\n\\frac{8 \\times 3}{3} = 8\n$$\nos triângulos de lado $\\ell \\sqrt{2}$. A resposta é então: 8 triângulos equiláteros.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26008,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA soma dos primos de 1 até 1000 é no máximo quanto?\n\nHá muitos anos, um professor que não queria dar aula, ordenou que seus alunos calculassem a soma dos números de 1 até 100. Um aluno muito esperto, chamado Gauss, descobriu um jeito muito simples de realizar a tarefa descobrindo a fórmula:\n$$\n1+2+3+\\ldots+n=\\frac{n(n+1)}{2}\n$$\nComo esta história já tem muito tempo, hoje os desafios dados aos alunos pedem tarefas mais elaboradas.\n\na. Verifique que todo número primo maior que 3 deixa resto 1 ou 5 na divisão por 6.\n\nb. Verifique que a soma dos números primos que são maiores que 1 e menores que 1000 é menor que 166338.\n\nc. Na estimativa acima, para ter menos complicações técnicas, não eliminamos alguns números que certamente não são primos. Elimine alguns desses números e verifique que a soma dos primos que são maiores que 1 e menores que 1000 é menor que 166000.",
    "options": [],
    "answer": "< 166000",
    "solution": "Solution:\n\na. Seja $p$ um primo maior que 3. Este primo, como qualquer número inteiro positivo, só pode deixar restos de 0 a 5 na divisão por 6. Veja que se ele deixasse os restos $0, 2, 3$ ou $4$, poderíamos fatorá-lo:\n$$\n\\begin{aligned}\n& p=6q+0=2 \\cdot 3q \\\\\n& p=6q+2=2 \\cdot (3q+1) \\\\\n& p=6q+3=3 \\cdot (2q+1) \\\\\n& p=6q+4=2 \\cdot (3q+2)\n\\end{aligned}\n$$\nComo $p$ é maior que 3, em cada um dos casos acima ele seria um número composto. Então os únicos restos possíveis de $p$ por 6 são 1 e 5.\n\nb. Para fazer uma estimativa superior, podemos somar 2, 3 e todos os números que deixam restos 1 ou 5 na divisão por 6. Usaremos a fórmula de soma dada no enunciado e, por simplicidade, representaremos um número que deixa resto 5 por um múltiplo de 6 menos 1:\n$$\n\\begin{aligned}\n2+3+(6 \\cdot 1-1)+(6 \\cdot 1+1)+\\ldots+(6 \\cdot 166+1) & =5+12 \\cdot (1+2+\\ldots+166) \\\\\n& =5+12 \\cdot \\frac{166 \\cdot 167}{2} \\\\\n& =5+166332 \\\\\n& =166337\n\\end{aligned}\n$$\nEntão a soma dos números primos menores que 1000 é menor que 166338.\n\nc. Entre os números considerados no item anterior, há vários que não são primos. Consideremos números que são múltiplos de 5 maiores que 5 dentre os que deixam resto 1 ou 5 por 6. Basta então subtrair alguns destes múltiplos de 5 menores que 100 para melhorarmos a estimativa:\n$$\n\\begin{aligned}\n25+35+55+65+85+95 & =60+120+180 \\\\\n& =360\n\\end{aligned}\n$$\nE, com isto, podemos concluir que a soma dos números primos menores que 1000 é menor que $166338-360<166000$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26009,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nQual o menor valor da fração\n$$\n\\frac{x^{4}+x^{2}+5}{\\left(x^{2}+1\\right)^{2}} ?\n$$",
    "options": [],
    "answer": "19/20",
    "solution": "Solution:\nTemos\n$$\n\\begin{aligned}\ny & =\\frac{x^{4}+x^{2}+5}{\\left(x^{2}+1\\right)^{2}} \\\\\n& =\\frac{\\left(x^{2}+1\\right)^{2}-\\left(x^{2}+1\\right)+5}{\\left(x^{2}+1\\right)^{2}} \\\\\n& =1-\\frac{1}{1+x^{2}}+\\frac{5}{\\left(1+x^{2}\\right)^{2}}\n\\end{aligned}\n$$\nSe $v=\\frac{1}{1+x^{2}}$, então o problema se resume a encontrarmos o mínimo de $y=1-v+5 v^{2}$. Como o gráfico, em função de $u$, da função $y=1-v+5 v^{2}$ é uma parábola, o mínimo será atingido quando $v=-\\frac{b}{2 a}=\\frac{1}{10}$. Outra maneira de ver isto, é fazer o completamento de quadrados:\n$$\n\\begin{aligned}\n1-v+5 v^{2} & =5\\left(v-\\frac{1}{10}\\right)^{2}+\\frac{19}{20} \\\\\n& \\geq \\frac{19}{20}\n\\end{aligned}\n$$\npois todo quadrado perfeito é não negativo. A igualdade ocorre apenas se $v=\\frac{1}{10}$ e o valor mínimo é, portanto, $\\frac{19}{20}$. Note que $\\frac{1}{1+x^{2}}=v=\\frac{1}{10}$ se, e somente se, $1+x^{2}=10$, ou seja, $x= \\pm 3$. A resposta é $\\frac{19}{20}$, que ocorre quando $x= \\pm 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26010,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA seguinte figura mostra um dodecágono regular.\n![](attached_image_1.png)\nResponda às seguintes perguntas:\na) Quantos triângulos equiláteros podem ser formados de modo que seus três vértices sejam vértices do dodecágono?\nb) Quantos triângulos escalenos podem ser formados de modo que seus três vértices sejam vértices do dodecágono?",
    "options": [],
    "answer": "a) 4; b) 168",
    "solution": "Solution:\nPodemos inscrever o dodecágono regular em um círculo como mostra a seguinte figura:\n![](attached_image_2.png)\nEntão os arcos que correspondem a cada um dos lados do dodecágono devem medir $\\left(360 / 12\\right)^{\\circ}=30^{\\circ}$.\n\na) Os triângulos equiláteros que podemos formar estarão também inscritos no círculo. Sabemos que a medida do arco que corresponde a um lado de um triângulo equilátero inscrito em uma circunferência deve ser $\\left(360 / 3\\right)^{\\circ}=120^{\\circ}$.\n![](attached_image_3.png)\nSe fixamos um vértice, digamos $A$ no gráfico, existirá então um único triângulo equilátero usando o vértice $A$.\n![](attached_image_4.png)\nPodemos contar assim 12 triângulos (um para cada vértice). Neste caso, cada triângulo seria contado 3 vezes. Concluímos que então existem $12 / 3=4$ triângulos equiláteros inscritos no dodecágono.\n\nb) Para contar o número de triângulos escalenos, podemos contar o número total de triângulos e subtrair o número de triângulos isósceles e equiláteros.\nPrimeiro, contaremos os triângulos isósceles que não são equiláteros. Fixemos um vértice, digamos, $A$. Existem 4 triângulos isósceles não equiláteros inscritos no dodecágono cujo vértice $A$ corresponde ao ângulo desigual.\n![](attached_image_5.png)\nPodemos contar, assim, 4 triângulos isósceles por cada vértice do dodecágono, isto é, $4 \\times 12$ triângulos isósceles não equiláteros no total.\nContemos agora o número total de triângulos. Para determinar um triângulo, devemos escolher três vértices no conjunto de 12 vértices do dodecágono. Para escolher o primeiro vértice do triângulo, temos 12 alternativas, para o segundo, 11 e para o terceiro, restam 10. Mas no produto $12 \\times 11 \\times 10$ estamos contando várias vezes um mesmo triângulo. Observe que há seis formas possíveis para ordenar os vértices de um triângulo $A B C$, as quais estão listadas abaixo:\n$$\nA B C, \\quad A C B, \\quad B C A, \\quad B A C, \\quad C A B \\quad \\text{e} \\quad C B A\n$$\nPortanto, no produto $12 \\times 11 \\times 10$, cada triângulo está sendo contado 6 vezes. Concluímos que há, no total, $\\frac{12 \\times 11 \\times 10}{6}=220$ triângulos inscritos no dodecágono.\nDaí, o número de triângulos escalenos será igual a\n$$\n220-4 \\times 12-4=168\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26011,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nNuma misteriosa ilha havia 13 camaleões cinza, 15 camaleões marrons e 17 camaleões vermelhos. Quando dois camaleões de cores diferentes se encontram, os dois tomam a terceira cor. Por exemplo, se um cinza se encontra com um vermelho, então os dois ficam marrons. Por causa de uma tempestade, ocorreram 2 encontros cinza-vermelho, 3 encontros marrom-vermelho e 1 encontro cinza-vermelho, quantos camaleões de cada cor ficaram na ilha?",
    "options": [],
    "answer": "16 gray, 18 brown, 11 red",
    "solution": "Solution:\n16 cinzas, 18 marrons e 11 vermelhos",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26012,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMaria está planejando participar do Triatlon-Brasil que consta de $800~\\mathrm{m}$ de nado, seguido de $20~\\mathrm{km}$ de bicicleta e, finalmente, $4~\\mathrm{km}$ de corrida. Maria corre a uma velocidade constante que é o triplo da velocidade com que nada e pedala 2,5 vezes mais rápido do que corre. Para terminar a prova em, no máximo, 1 hora e 20 minutos, qual deve ser sua velocidade mínima em cada uma das três modalidades?",
    "options": [],
    "answer": "Swimming: 60 m/min; Running: 180 m/min; Cycling: 450 m/min",
    "solution": "Solution:\n\nSeja $x$ a velocidade com que Maria nada, em metros por minuto. Então o tempo que Maria gasta nadando $800~\\mathrm{m}$ é dado por $800 / x$ minutos. Sabemos que sua velocidade na corrida é $3x$ e, na bicicleta, $2,5 \\times 3x = 7,5x$ metros por minuto. Assim, o tempo total que Maria gasta nas três etapas é\n$$\n\\underbrace{\\frac{800}{x}}_{\\text{nadando}} + \\underbrace{\\frac{20000}{7,5x}}_{\\text{pedalando}} + \\underbrace{\\frac{4000}{3x}}_{\\text{correndo}} = \\frac{800 \\times 7,5 + 20000 + 4000 \\times 2,5}{7,5x} = \\frac{4800}{x}\n$$\nminutos. Logo, para que ela vença as três etapas em, no máximo, uma hora e vinte minutos, ou seja, em $80~\\mathrm{min}$, devemos ter, no mínimo, $80 = 4800 / x$, ou seja, $x = 4800 / 80 = 60$ metros por minuto. Segue que $3x = 180$ e $7,5x = 450$ metros por minuto. Assim, para que Maria termine a prova em, no máximo, 1 hora e 20 minutos, ela deve desenvolver as seguintes velocidades mínimas:\n- nadar a uma velocidade mínima de $60~\\mathrm{m}/\\mathrm{min}$,\n- pedalar a uma velocidade mínima de $450~\\mathrm{m}/\\mathrm{min}$ e\n- correr a uma velocidade mínima de $180~\\mathrm{m}/\\mathrm{min}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26013,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nJúlia está treinando para olimpíadas de matemática. Um dia ela decide dividir $2014$ por cada um dos divisores inteiros positivos de $2015$. Para cada divisão, ela escreve o quociente no seu caderno e o resto em uma lousa.\n\na. Vamos ajudar Júlia. Escreva os oito divisores inteiros positivos de $2015$.\n\nb. Para cada um dos divisores, faça a divisão e escreva uma lista com os quocientes e outra com os restos obtidos.\n\nc. Ao terminar, Júlia percebeu uma grande \"coincidência\": os números escritos no caderno eram os mesmos que estavam no quadro, apenas escritos em uma ordem diferente. Seria uma coincidência? Mostre que, para qualquer número $n$ que Júlia escolher, se ela calcular o quociente e o resto da divisão de $n-1$ por cada um dos divisores de $n$ os números no caderno e na lousa serão exatamente os mesmos, estando apenas, possivelmente, escritos em uma ordem diferente.",
    "options": [],
    "answer": "a) Divisors of 2015: 1, 5, 13, 31, 65, 155, 403, 2015.\nb) Quotients when dividing 2014 by these divisors: 2014, 402, 154, 64, 30, 12, 4, 0. Remainders: 0, 4, 12, 30, 64, 154, 402, 2014.\nc) In general, for any n, if one divides n−1 by each positive divisor of n, the multiset of quotients equals the multiset of remainders; both consist exactly of each positive divisor of n decreased by one (accounting for multiplicity, with pairs of complementary divisors yielding swapped quotient and remainder, and a single instance when n is a perfect square).",
    "solution": "Solution:\n\na. A fatoração de $2015$ em primos é $5 \\cdot 13 \\cdot 31$. Então seus oito divisores inteiros positivos são:\n$$\n\\{1, 5, 13, 31, 65, 155, 403, 2015\\}\n$$\n\nb. Escrevendo as divisões na forma $n = m \\cdot q + r$, onde $q$ é o quociente e $r$ é o resto, teremos\n$$\n\\begin{aligned}\n& 2014 = 1 \\cdot 2014 + 0 \\\\\n& 2014 = 5 \\cdot 402 + 4 \\\\\n& 2014 = 13 \\cdot 154 + 12 \\\\\n& 2014 = 31 \\cdot 64 + 30 \\\\\n& 2014 = 65 \\cdot 30 + 64 \\\\\n& 2014 = 155 \\cdot 12 + 154 \\\\\n& 2014 = 403 \\cdot 4 + 402 \\\\\n& 2014 = 2015 \\cdot 0 + 2014\n\\end{aligned}\n$$\n\nc. Além dos números nas duas listas serem os mesmos, note que eles são exatamente os antecessores dos divisores de $2015$. De fato, seja $x$ um divisor de $n$, então existe um divisor $y$ tal que $n = x \\cdot y$. Dividindo $n-1$ por $x$ e por $y$, temos:\n$$\n\\begin{aligned}\nn & = x \\cdot y \\\\\nn-1 & = x \\cdot (y-1) + (x-1) \\\\\nn-1 & = y \\cdot (x-1) + (y-1)\n\\end{aligned}\n$$\nComo o resto na divisão de um número por outro é único e sabemos que $0 \\leq x-1 < x$ e $0 \\leq y-1 < y$, então os restos destas duas divisões são exatamente $x-1$ e $y-1$. O que define também, de forma única, os quocientes: $y-1$ e $x-1$. Estes números devem ser escritos na lousa e no caderno, mas na ordem trocada. Vale lembrar que se $x = y$, no caso em que $n$ é um quadrado perfeito, então em vez de duas divisões existirá apenas uma e o mesmo número será escrito na lousa e no caderno. Deste modo, as listas na lousa e no caderno possuem os mesmos números, que são os antecessores dos divisores positivos de $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26014,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nBruno tem uma lista com todos os números naturais de 10 dígitos que se podem formar utilizando apenas os dígitos 1, 2, 3 e 4 e além disto que possuem igual quantidade de algarismos 1 e 2, por exemplo, 3333333333, 1111342222 etc. Bernardo tem a lista de todos os números naturais de 20 dígitos formados por 10 dígitos 1 e 10 dígitos 2. Demonstre que a lista de Bruno tem a mesma quantidade de dígitos da lista de Bernardo.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nAbordemos inicialmente uma versão simplificada do problema. Suponha que Bruno quisesse listar todos os números de dois dígitos utilizando os mesmos quatro dígitos e também possuindo iguais quantidades de algarismos $1$ e $2$. Ele obteria a seguinte lista:\n$12, 21, 33, 44, 34, 43$.\n\nSuponha agora que Bernardo quisesse listar todos os números naturais de 4 dígitos formados por dois algarismos $1$ e dois algarismos $2$. Ele obteria a seguinte lista:\n$1122, 2211, 1212, 2121, 1221, 2112$.\n\nA comparação entre as duas listas nos sugere que pode existir alguma correspondência direta entre os elementos de cada lista. Considere a seguinte tabela de conversão:\n$$\n\\begin{array}{lll}\n1 & \\rightarrow & 11 \\\\\n2 & \\rightarrow & 22 \\\\\n3 & \\rightarrow & 12 \\\\\n4 & \\rightarrow & 21\n\\end{array}\n$$\nVeja que a troca de cada dígito de um número na primeira lista, seguindo a tabela de conversão, produz exatamente um número da segunda lista. Precisamos verificar que isso realmente ocorre com as listas originais de Bruno e Bernardo. Considere um número qualquer de 10 dígitos na lista de Bruno:\n$$\na_{1} a_{2} \\ldots a_{10}\n$$\nSuponha que o número acima possui $k$ dígitos iguais a $1$, $l$ dígitos iguais a $3$ e $m$ dígitos iguais a $4$. Como as quantidades de dígitos $1$ e $2$ é a mesma, temos $2k + l + m = 10$. Se trocarmos cada dígito $3$ pela sequência $12$, cada dígito $4$ por $21$, cada dígito $1$ por $11$ e cada dígito $2$ por $22$ teremos um número com o dobro de dígitos, ou seja, com $20$ dígitos e, além disto, com $2k + l + m = 10$ dígitos iguais a $1$ e também $2k + l + m = 10$ dígitos iguais a $2$.\n\nReciprocamente, cada número com $20$ dígitos possuindo a mesma quantidade de dígitos $1$ e $2$ pode ser convertido a um único número da lista de Bernardo simplesmente agrupando-se seus dígitos de dois em dois e convertendo os $10$ pares de dígitos consecutivos assim formado no sentido inverso da tabela anterior. Esta correspondência biunívoca entre as duas listas nos permite concluir que elas possuem a mesma quantidade de elementos.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26015,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nConsidere o seguinte hexágono regular $ABCDEF$, cujo lado mede $4$, e onde os pontos $P$ e $Q$ são os pontos médios dos lados $\\overline{BC}$ e $\\overline{DE}$, respectivamente.\n![](attached_image_1.png)\nCalcule o perímetro do hexágono $ABPQEF$.",
    "options": [],
    "answer": "22",
    "solution": "Solution:\nProlongamos os lados $\\overline{BC}$ e $\\overline{ED}$ do modo indicado na figura seguinte:\n![](attached_image_2.png)\nSendo $ABCDEF$ um hexágono regular, cada ângulo interno mede $120^\\circ$. Portanto, $\\measuredangle RCD = 60^\\circ$, $\\measuredangle RDC = 60^\\circ$, e concluímos que o triângulo $DRC$ é equilátero de lado $4$.\n![](attached_image_3.png)\nComo $|PR| = |QR| = 6$ e $\\measuredangle PRD = 60^\\circ$, temos que o triângulo $PQR$ é equilátero de lado $6$. Portanto, $|PQ| = 6$.\n![](attached_image_4.png)\nAgora é fácil calcular o perímetro do hexágono $ABPQEF$: $4 + 2 + 6 + 2 + 4 + 4 = 22$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26016,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nConsidere 2018 pontos em uma circunferência de raio 1. Verifique que existe um ponto $P$ da circunferência para o qual a soma das distâncias de $P$ aos 2018 pontos é pelo menos 2018.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nEscolha dois pontos $A$ e $B$ diametralmente opostos. Dado qualquer ponto $Q$ no círculo, temos $d(Q, A) + d(Q, B) > 2$, em que $d(X, Y)$ denota a distância entre $X$ e $Y$.\n\nConsidere as somas $S_{A}$ e $S_{B}$ das distâncias de todos os 2018 pontos até $A$ e $B$, respectivamente. Como $S_{A} + S_{B} > 2 \\cdot 2018$, pelo menos uma dessas somas é maior que 2018.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26017,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNum deserto há cobras, ratos e escorpiões. Cada manhã, cada cobra mata um rato. Cada meio-dia, cada escorpião mata uma cobra. Cada noite, cada rato mata um escorpião. Ao final de uma semana, à noite, só restava um rato. Quantos ratos havia na manhã no início da semana?",
    "options": [],
    "answer": "1873",
    "solution": "Solution:\n\n1873",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26018,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nUm número inteiro positivo é chamado \"equilibrado\" se ele tem quatro algarismos, e um desses algarismos é igual à média dos outros três. Por exemplo: o número $2631$ é equilibrado porque $3$ é a média de $2, 6$ e $1$; $4444$ também é equilibrado porque $4$ é a média de $4, 4$ e $4$.\n\na) Encontre os três menores números equilibrados.\n\nb) Quantos são os números equilibrados menores que $2014$?",
    "options": [],
    "answer": "a) 1003, 1012, 1021; b) 90",
    "solution": "Solution:\nUm número de quatro algarismos é equilibrado quando um de seus algarismos, digamos $a$, é a média dos outros três algarismos, digamos $b, c$ e $d$, ou seja, quando $(b+c+d) / 3 = a$ ou, equivalentemente, quando $(a+b+c+d) / 4 = a$. Vemos assim que um número $N$ de quatro algarismos é equilibrado quando ele satisfaz as duas seguintes condições:\n\n(i) A soma de seus algarismos, digamos $S$, é divisível por $4$;\n\n(ii) $S / 4$ é um dos algarismos de $N$.\n\na) Dos números da forma $\\overline{100x}$, os únicos que satisfazem a condição (i) são $1003$ e $1007$, e desses dois números, o único que satisfaz (ii) é $1003$. Dos números da forma $\\overline{101x}$, os únicos que satisfazem (i) são $1012$ e $1016$, e desses dois, o único que satisfaz (ii) é $1012$. Dos números da forma $\\overline{102x}$, os únicos que satisfazem (i) são $1021$, $1025$ e $1029$, e desses três, os que satisfazem (ii) são $1021$ e $1025$. Em conclusão, os três menores números equilibrados são\n$$\n1003, \\quad 1012 \\text{ e } 1021\n$$\n\nb) Analisemos primeiro os números equilibrados da forma $\\overline{1xyz}$. Pelas condições (i) e (ii), devemos ter $1+x+y+z$ divisível por $4$ e $(1+x+y+z) / 4$ deve coincidir com um dos algarismos $1, x, y$ ou $z$. Suponha que os algarismos $x, y$ e $z$ satisfazem essas condições.\n\n- Se $x, y$ e $z$ são distintos, podemos formar os $6$ números equilibrados $\\overline{1xyz}$, $\\overline{1xzy}$, $\\overline{1yxz}$, $\\overline{1yzx}$, $\\overline{1zxy}$ e $\\overline{1zyx}$.\n- Se $x = y$ e $z$ são distintos, podemos formar os $3$ números equilibrados $\\overline{1xxz}$, $\\overline{1xzx}$ e $\\overline{1zxx}$.\n- Se $x = y = z$, podemos formar somente o número equilibrado $\\overline{1xxx}$.\n\nNote agora que $1 = 1+0+0+0 \\leq 1+x+y+z \\leq 1+9+9+9 = 28$. Então a soma $1+x+y+z$ somente pode tomar os valores $4, 8, 12, 16, 20, 24$ e $28$. Separaremos em casos conforme os valores que pode tomar a soma $x+y+z$.\n\n- Se $x+y+z = 27$ : Nesse caso, devemos ter $x = y = z = 9$, mas o número $1999$ não satisfaz (ii), e então não é equilibrado. Assim, não temos números equilibrados.\n- Se $x+y+z = 23$ : Nesse caso, um dos algarismos deve ser igual a $(1+x+y+z) / 4 = 6$. Então o conjunto $\\{x, y, z\\}$ tem que ser $\\{6, 9, 8\\}$, e como os algarismos $6, 9$ e $8$ são distintos, temos $6$ números equilibrados.\n- Se $x+y+z = 19$ : Um dos algarismos deve ser igual a $(1+x+y+z) / 4 = 5$. Então o conjunto $\\{x, y, z\\}$ tem que ser $\\{5, 9, 5\\}, \\{5, 8, 6\\}$ ou $\\{5, 7, 7\\}$. Se o conjunto $\\{x, y, z\\}$ for $\\{5, 9, 5\\}$, temos $3$ números equilibrados, se for $\\{5, 8, 6\\}$ temos $6$ números equilibrados, e se for $\\{5, 7, 7\\}$, temos $3$ números equilibrados. Assim, temos no total $3+6+3=12$ números equilibrados.\n- Se $x+y+z = 15$ : Um dos algarismos deve ser igual a $(1+x+y+z) / 4 = 4$. Então o conjunto $\\{x, y, z\\}$ tem que ser $\\{4, 9, 2\\}, \\{4, 8, 3\\}, \\{4, 7, 4\\}$ ou $\\{4, 6, 5\\}$. Temos, então, $6+6+3+6=21$ números equilibrados.\n- Se $x+y+z = 11$ : Um dos algarismos deve ser igual a $(1+x+y+z) / 4 = 3$. Portanto, o conjunto $\\{x, y, z\\}$ tem que ser $\\{3, 8, 0\\}, \\{3, 7, 1\\}, \\{3, 6, 2\\}, \\{3, 5, 3\\}$ ou $\\{3, 4, 4\\}$. Temos, então, $6+6+6+3+3=24$ números equilibrados.\n- Se $x+y+z = 7$ : Um dos algarismos deve ser igual a $(1+x+y+z) / 4 = 2$. Então o conjunto $\\{x, y, z\\}$ tem que ser $\\{2, 5, 0\\}, \\{2, 4, 1\\}$ ou $\\{2, 3, 2\\}$. Temos então $6+6+3=15$ números equilibrados.\n- Se $x+y+z = 3$ : Um dos algarismos deve ser igual a $(1+x+y+z) / 4 = 1$. Então o conjunto $\\{x, y, z\\}$ tem que ser $\\{1, 2, 0\\}, \\{1, 1, 1\\}$, ou $\\{3, 0, 0\\}$. Temos, então, $6+1+3=10$ números equilibrados.\n\nFinalmente, é simples ver que dos números $2000, 2001, \\ldots, 2013$, os únicos que satisfazem as condições (i) e (ii) são $2006$ e $2011$. Então, a quantidade de números equilibrados menores que $2014$ é\n$$\n6+12+21+24+15+10+2=90\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26019,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOs elementos do conjunto $\\{1,2,3,4,5,6,7,8,9,10,11\\}$ podem ser separados nos conjuntos $\\{3,9,10,11\\}$ e $\\{1,2,4,5,6,7,8\\}$ de modo que cada um deles possua soma dos elementos igual a 33.\n\na. Exiba um modo de separar os elementos do conjunto $\\{1,2,3,4,5,6,7,8\\}$ em três conjuntos tais que as somas dos elementos de cada conjunto seja a mesma.\n\nb. Explique por que não é possível separar os números do conjunto $\\{1,2,3,4,5,6,7,8,9,10\\}$ em dois conjuntos de mesma soma.\n\nc. Para cada inteiro positivo $n \\geq 2$, determine o menor inteiro positivo $N$ tal que o conjunto $\\{1,2, \\ldots, N\\}$ pode ser separado em exatamente $n$ conjuntos de mesma soma.",
    "options": [],
    "answer": "N = 2n − 1",
    "solution": "Solution:\n\na. A soma dos números de $1$ até $8$ é igual a $36$, então cada um dos três conjuntos deve ter soma $12$. Basta separar nos conjuntos: $\\{4,8\\}$, $\\{5,7\\}$ e $\\{1,2,3,6\\}$.\n\nb. A soma dos números de $1$ até $10$ é igual a $55$. Para separar em dois conjuntos de mesma soma, cada um teria elementos inteiros positivos e soma $\\frac{55}{2}$. Como $\\frac{55}{2}$ não é inteiro, não é possível fazer tal separação.\n\nc. Se formarmos $n$ conjuntos de mesma soma, no máximo um deles pode ter um elemento e os demais $n-1$ conjuntos teriam pelo menos dois elementos. Isto nos permite concluir que $N \\geq 1+2 \\cdot(n-1)=2n-1$. Para verificar que $N=2n-1$ é o menor, temos que separar o conjunto $\\{1,2, \\ldots, 2n-1\\}$ em $n$ conjuntos de mesma soma. Considere os conjuntos $\\{2n-1\\}$, $\\{2n-2,1\\}$, $\\{2n-3,2\\}$, $\\ldots$, $\\{n+1, n-2\\}$ e $\\{n, n-1\\}$. São exatamente $n$ conjuntos e cada um possui soma dos elementos igual a $2n-1$. Assim, concluímos que $N=2n-1$ é o menor inteiro positivo que satisfaz às condições do problema.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26020,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm semicírculo de diâmetro $E F$, situado no lado $B C$ do triângulo $A B C$, é tangente aos lados $A B$ e $A C$ em $Q$ e $P$, respectivamente. As retas $E P$ e $F Q$ se encontram em $H$. Mostre que $A H$ é a altura do triângulo.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSejam $K$ o pé da perpendicular de $H$ ao segmento $B C$ e $O$ o centro do semicírculo. Suponha sem perda de generalidade que $K$ está no segmento $O C$.\n\n![](attached_image_2.png)\n\nComo $E F$ é um diâmetro, segue que $\\angle E Q F = \\angle H K E = 90^\\circ$ e consequentemente $E Q H K$ é um quadrilátero inscritível em um círculo de diâmetro $E H$. Daí segue que\n$$\n\\angle H K Q = \\angle Q E H = \\angle Q E P = \\frac{\\angle Q O P}{2}\n$$\nAnalisando os triângulos $A Q O$ e $A O P$, temos\n$$\nQ A = A P, \\quad Q O = O P \\text{ e } A O = A O\n$$\nPortanto, pelo caso de congruência $L L L$, os triângulos $A Q O$ e $A P O$ são congruentes. Assim $\\frac{\\angle Q O P}{2} = \\angle Q O A$ e\n$$\n\\angle Q K O = 90^\\circ - \\angle H K Q = 90^\\circ - \\frac{\\angle Q O P}{2} = 90^\\circ - \\angle Q O A = \\angle Q A O\n$$\npois $\\angle O Q A = 90^\\circ$. Consequentemente, $Q A K O$ é um quadrilátero inscritível. Lembrando que $\\angle O Q A = 90^\\circ$, o diâmetro de tal círculo é $A O$. Daí, $\\angle A K O = 90^\\circ$ e tanto $A K$ quanto $H K$ são perpendiculares a $B C$. Portanto, $A, H$ e $K$ são colineares e, finalmente, $A H$ é altura do triângulo.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26021,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\na) Qual o valor de\n$$\n\\frac{1}{1+\\sqrt{2}}+\\frac{1}{\\sqrt{2}+\\sqrt{3}}+\\frac{1}{\\sqrt{3}+\\sqrt{4}}+\\frac{1}{\\sqrt{4}+\\sqrt{5}}+\\frac{1}{\\sqrt{5}+\\sqrt{6}}+\\cdots+\\frac{1}{\\sqrt{99}+\\sqrt{100}} ?\n$$\nb) Se $x=\\sqrt{1+2 \\cdot \\sqrt{1+3 \\cdot \\sqrt{1+4 \\cdot \\sqrt{1+5 \\cdot \\sqrt{\\ldots}}}}}$ é um número real, qual o seu valor?",
    "options": [],
    "answer": "a) 9; b) 3",
    "solution": "Solution:\na) Perceba que\n$$\n\\begin{aligned}\n\\frac{1}{\\sqrt{n}+\\sqrt{n+1}} \\cdot \\frac{\\sqrt{n}-\\sqrt{n+1}}{\\sqrt{n}-\\sqrt{n+1}} & =\\frac{\\sqrt{n}-\\sqrt{n+1}}{(\\sqrt{n})^{2}-(\\sqrt{n+1})^{2}} \\\\\n& =\\frac{\\sqrt{n}-\\sqrt{n+1}}{n-(n-1)} \\\\\n& =-\\sqrt{n}+\\sqrt{n+1}\n\\end{aligned}\n$$\nPortanto,\n$$\n\\begin{aligned}\n& \\frac{1}{1+\\sqrt{2}}+\\frac{1}{\\sqrt{2}+\\sqrt{3}}+\\frac{1}{\\sqrt{3}+\\sqrt{4}}+\\frac{1}{\\sqrt{4}+\\sqrt{5}}+\\frac{1}{\\sqrt{5}+\\sqrt{6}}+\\cdots+\\frac{1}{\\sqrt{99}+\\sqrt{100}}= \\\\\n& -1+\\sqrt{2}-\\sqrt{2}+\\sqrt{3}-\\sqrt{3}+\\sqrt{4}-\\sqrt{4}+\\sqrt{5}-\\sqrt{5}+\\sqrt{6}-\\ldots-\\sqrt{99}+\\sqrt{100}= \\\\\n& \\sqrt{100}-1=9\n\\end{aligned}\n$$\n\nb) Observe a seguinte sequência de quadrados de inteiros positivos\n$$\n\\begin{array}{lll}\n3^{2}=1+8=1+2 \\cdot 4 & & \\\\\n4^{2}=1+15=1+3 \\cdot 5 & \\Rightarrow & 4=\\sqrt{1+3 \\cdot 5} \\\\\n5^{2}=1+24=1+4 \\cdot 6 & \\Rightarrow & 5=\\sqrt{1+4 \\cdot 6} \\\\\n6^{2}=1+35=1+5 \\cdot 7 & \\Rightarrow & 6=\\sqrt{1+5 \\cdot 7} \\\\\n\\vdots & \\vdots & \\vdots \\\\\nn^{2}=1+n^{2}-1=1+(n-1) \\cdot(n+1) & \\Rightarrow & n=\\sqrt{1+(n-1) \\cdot(n+1)}\n\\end{array}\n$$\nPodemos então iterar cada uma das equações anteriores de modo a obtermos\n$$\n\\begin{aligned}\n& 3^{2}=1+2 \\cdot 4 \\\\\n& 3^{2}=1+2 \\cdot \\sqrt{1+3 \\cdot 5} \\\\\n& 3^{2}=1+2 \\cdot \\sqrt{1+3 \\cdot \\sqrt{1+4 \\cdot 6}} \\\\\n& 3^{2}=1+2 \\cdot \\sqrt{1+3 \\cdot \\sqrt{1+4 \\cdot \\sqrt{1+5 \\cdot 7}}}\n\\end{aligned}\n$$\nPortanto,\n$$\n\\begin{aligned}\n3^{2} & =1+2 \\cdot \\sqrt{1+3 \\cdot \\sqrt{1+4 \\cdot \\sqrt{1+5 \\cdot \\sqrt{\\cdots}}}} \\\\\n3 & =\\sqrt{1+2 \\cdot \\sqrt{1+3 \\cdot \\sqrt{1+4 \\cdot \\sqrt{1+5 \\cdot \\sqrt{\\cdots \\cdots}}}}}\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26022,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSejam $a$ e $b$ números inteiros positivos tais que $a > b$. O professor Fernando disse ao aluno Raul que se ele calculasse o número $A = a^{2} + 4b + 1$, o resultado seria um quadrado perfeito. Raul, por engano, trocou os números $a$ e $b$ e calculou o número $B = b^{2} + 4a + 1$ que, por acaso, também é um quadrado perfeito.\n\na) Mostre que $A = (a+1)^{2}$.\n\nb) Encontre os números $a$, $b$, $A$ e $B$.",
    "options": [],
    "answer": "A = (a+1)^2; a = 8, b = 4, A = 81, B = 49",
    "solution": "Solution:\n\na) Note que\n$$\na^{2} < A = a^{2} + 4b + 1 < a^{2} + 4a + 1 < (a+2)^{2}\n$$\nAssim, $A$ é um quadrado perfeito entre os quadrados perfeitos $a^{2}$ e $(a+2)^{2}$, e então $A = (a+1)^{2}$.\n\nb) Pelo item anterior,\n$$\na^{2} + 4b + 1 = A = (a+1)^{2} = a^{2} + 2a + 1\n$$\ne então $a = 2b$. Substituindo em $B = b^{2} + 4a + 1$, obtemos $B = b^{2} + 8b + 1$. Note agora que\n$$\n(b+1)^{2} = b^{2} + 2b + 1 < b^{2} + 8b + 1 = B < b^{2} + 8b + 16 = (b+4)^{2}\n$$\nIsto é, $B$ é um quadrado perfeito entre os quadrados perfeitos $(b+1)^{2}$ e $(b+4)^{2}$. Temos assim dois possíveis casos: $B = (b+2)^{2}$ ou $B = (b+3)^{2}$.\n\nCaso I: $B = (b+2)^{2}$. Nesse caso, teríamos\n$$\nb^{2} + 4b + 4 = (b+2)^{2} = B = b^{2} + 8b + 1\n$$\ne, portanto, $b = 3/4$ não seria um número inteiro.\n\nCaso II: $B = (b+3)^{2}$. Nesse caso, teríamos\n$$\nb^{2} + 6b + 9 = (b+3)^{2} = B = b^{2} + 8b + 1\n$$\ne daí $b = 4$ e $a = 8$. Finalmente,\n$$\nA = 8^{2} + 4(4) + 1 = 9^{2} \\quad \\text{e}\\quad B = 4^{2} + 4(8) + 1 = 7^{2}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26023,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDetermine all distinct positive integers $x$ and $y$ such that\n$$\n\\frac{1}{x} + \\frac{1}{y} = \\frac{2}{7}\n$$",
    "options": [],
    "answer": "(x, y) = (4, 28) or (28, 4)",
    "solution": "Solution:\n\n$$\n2 x y = 7(x + y), \\quad \\text{com} \\quad x > 0, y > 0 \\text{ e } x \\neq y\n$$\nComo $2$ e $7$ são números primos, segue que $7$ divide $x$ ou $y$. Como a equação é simétrica em $x$ e $y$, podemos supor que $7$ divide $x$. Então, $x = 7k$, para algum $k > 0$ inteiro e decorre que $2 \\times 7k y = 7(7k + y)$, ou seja, simplificando, $2k y = 7k + y$ ou, ainda,\n$$\n(2k - 1) y = 7k = x\n$$\nSe $7$ dividisse $y$, teríamos $y = 7m$, para algum $m > 0$ inteiro. Nesse caso, teríamos $49 \\times 2k m = 2 x y = 7(x + y) = 49(k + m)$, acarretando $2k m = k + m$. Mas, então\n$$\n2 = \\frac{k + m}{k m} = \\frac{1}{k} + \\frac{1}{m} \\leq 1 + 1 = 2\n$$\no que significa que $k = m = 1$ e, portanto, $x = 7 = y$. Como queremos $x \\neq y$, concluímos que $7$ não divide $y$, de modo que $7$ divide $2k - 1$. Tomando $k = 4$, resulta $x = 28$ e\n$$\ny = \\frac{7k}{2k - 1} = \\frac{28}{7} = 4\n$$\nfornecendo a solução\n$$\n\\frac{1}{28} + \\frac{1}{4} = \\frac{2}{7}\n$$\n\n**Observação:** A solução obtida é única. De fato, como $2k - 1$ é, sempre, ímpar e $7$ divide $2k - 1$, o múltiplo de $7$ que é igual a $2k - 1$ deve ser ímpar. Assim, existe algum inteiro $n > 0$ tal que\n$$\n7(2n - 1) = 2k - 1\n$$\nIsso acarreta que $k = 7n - 3$ e, portanto,\n$$\ny = \\frac{7k}{2k - 1} = \\frac{7(7n - 3)}{7(2n - 1)} = \\frac{7n - 3}{2n - 1} = \\frac{3(2n - 1) + n}{2n - 1} = 3 + \\frac{n}{2n - 1}\n$$\nComo $y$ deve ser inteiro, concluímos que $2n - 1$ divide $n$, de modo que $2n - 1 \\leq n$. No entanto, $n \\geq 1$ e, portanto, $2n - 1 \\geq n$. A única possibilidade é $2n - 1 = n$ e, portanto, $n = 1$. Segue que $k = 4$ e $x = 28$ dão a única solução.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26024,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTabuleiro de xadrez - De quantas maneiras podemos colocar dois bispos de mesma cor num tabuleiro de xadrez em filas, colunas e casas de cores distintas?",
    "options": [],
    "answer": "768",
    "solution": "Solution:\n\nUm tabuleiro de xadrez é um quadrado reticulado de 64 quadradinhos, denominados casas, sendo 32 casas pretas e 32 brancas, posicionados alternadamente. Uma das peças do xadrez recebe o nome de bispo, havendo um par deles para cada jogador. Um dos dois bispos de um jogador só se movimenta pelas casas pretas e o outro só pelas brancas.\n\nInicialmente, é possível colocar um dos dois bispos em qualquer uma das 64 casas. Se o bispo estiver numa casa branca, então na fila em que ele está, bem como na coluna, temos quatro casas pretas que não podem ser ocupadas pelo segundo bispo, num total de oito casas. Assim, o segundo bispo pode ser colocado em qualquer uma das $32-8=24$ casas pretas restantes.\n\nNOTA: Aqui estamos entendendo que alternando a posição desses dois bispos não muda a configuração no tabuleiro de xadrez. Mais precisamente, os bispos têm a mesma cor, isto é, pertencem a um mesmo jogador.\n\nConcluímos, então, que se um dos bispos ocupar uma das 32 casas brancas, então o outro terá 24 casas pretas à disposição. Portanto, o número dessas configurações distintas que podem ser obtidas é $32 \\times 24=768$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26025,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCortando papéis - No início de uma brincadeira, André tinha sete pedaços de papel. Na primeira rodada da brincadeira, ele pegou alguns destes pedaços e cortou cada um deles em sete pedaços, que foram misturados aos pedaços de papel que não foram cortados nessa rodada. Na segunda rodada, ele novamente pegou alguns pedaços e cortou cada um deles em sete pedaços, que foram misturados aos demais papéis. Continuando dessa maneira, ao final de alguma rodada, André poderá ter exatamente $2009$ pedaços de papel?",
    "options": [],
    "answer": "No",
    "solution": "Solution:\n\nSe na primeira rodada André pega $n_{1}$ pedaços de papel para cortar cada um deles em sete pedaços, ao final dessa rodada ele ficará com $7-n_{1}$ pedaços sem cortar, mais $7 n_{1}$ pedaços cortados, totalizando $(7-n_{1})+7 n_{1}=7+6 n_{1}$ pedaços de papel. Analogamente, se na segunda rodada André pega $n_{2}$ pedaços de papel para cortar, ao final dessa rodada ele ficará com $7+6 n_{1}-n_{2}$ pedaços que não foram cortados nessa rodada, mais $7 n_{2}$ pedaços de papel provenientes dos cortes que ele fez nessa rodada. Assim, ao final da segunda rodada, André ficará com\n$$\n(7+6 n_{1}-n_{2})+7 n_{2}=7+6(n_{1}+n_{2})\n$$\nContinuando assim, conclui-se que, ao final de $k$ rodadas, André fica com\n$$\n7+6(n_{1}+n_{2}+\\cdots+n_{k})\n$$\npedaços de papel. Então, para ele ficar com $2009$ pedaços de papel ao final de alguma rodada, deveríamos ter essa última expressão igual a $2009$ ou, equivalentemente, subtraindo $7$ de cada lado, $6(n_{1}+n_{2}+\\cdots+n_{k})=2002$.\nNo entanto, $2002$ não é um múltiplo de $6$, de modo que essa equação não admite solução. Isso significa que André nunca poderá ficar com $2009$ pedaços ao final de alguma rodada de sua brincadeira.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26026,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm trapézio especial - A base $AD$ de um trapézio $ABCD$ mede $30~\\mathrm{cm}$. Suponhamos que exista um ponto $E$ na base $AD$ tal que os triângulos $\\triangle ABE$, $\\triangle BCE$ e $\\triangle CDE$ tenham perímetros iguais. Determine o comprimento de $BC$.",
    "options": [],
    "answer": "15 cm",
    "solution": "Solution:\n\nQueremos provar que $AE$ é igual a $BC$. Para isso, suponhamos que $AE$ seja maior do que $BC$ e escolhamos o ponto $A'$ sobre $AE$ tal que $EA' = BC$. Por construção, $EA'$ e $BC$ são paralelos, de modo que $A'BCE$ é um paralelogramo e, em particular,\n$$\nA'B = CE\n$$\nPela desigualdade triangular, temos\n$$\nA'A + AB > A'B\n$$\nLogo,\n![](attached_image_1.png)\n$$\n\\begin{aligned}\nEA + AB + BE & = EA' + A'A + AB + BE \\\\\n& > EA' + A'B + BE = BC + CE + EB\n\\end{aligned}\n$$\nDisso decorre que o perímetro do triângulo $\\triangle ABE$ é maior do que o perímetro do triângulo $\\triangle BCE$, contrário aos dados do problema.\n\nPor meio dessa contradição, estabelecemos que, diante das hipóteses do problema, $AE$ não pode ser maior do que $BC$. Por um processo totalmente análogo, também podemos estabelecer que, reciprocamente, $BC$ não pode ser maior do que $AE$, com o que concluímos que $BC = AE$. O mesmo raciocínio pode ser utilizado para mostrar que $BC = ED$. Assim,\n$$\nBC = \\frac{1}{2}(AE + ED) = 15~\\mathrm{cm}\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26027,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAlguns alunos do sétimo e oitavo ano de uma escola participam de um torneio de pingue-pongue, onde cada aluno joga contra todos os outros exatamente uma vez recebendo 1 por vitória e 0 ponto por derrota. Existem dez vezes mais alunos do oitavo ano do que do sétimo ano. A pontuação total dos alunos do oitavo ano é 4.5 vezes a pontuação total dos alunos do sétimo ano.\n\na) Verifique que se no torneio existem $k$ alunos, então o número de jogos é $\\frac{k(k-1)}{2}$.\n\nb) Qual é a soma das pontuações obtidas por todos os alunos do sétimo ano?",
    "options": [],
    "answer": "10",
    "solution": "Solution:\n\na) Cada um dos $k$ alunos irá jogar $k-1$ vezes. Somando-se a quantidade de jogos de cada um, obtemos $k(k-1)$. Entretanto, teremos contado cada jogo duas vezes, uma para cada um dos participantes da partida. Logo, o número de jogos é $\\frac{k(k-1)}{2}$.\n\nb) Seja $n$ o número de alunos do sétimo ano e suponha que este conjunto de alunos obteve $m$ pontos. Então o número de alunos do oitavo ano é $10 n$ e a pontuação obtida por eles é $4.5 m$. Como o número total de participantes do torneio é $n+10 n=11 n$ e cada partida vale exatamente 1 ponto, o número de pontos disputados é $\\frac{11 n(11 n-1)}{2}$. Por outro lado, sabemos que $m+4.5 m$ correspondem a todos os pontos disputados, daí\n$$\n\\begin{aligned}\n\\frac{11 n(11 n-1)}{2} & =5.5 m \\\\\nn(11 n-1) & =m\n\\end{aligned}\n$$\nComo cada participante disputou $11 n-1$ jogos e cada um deles vale 1 ponto, a soma dos pontos obtidos pelos alunos do sétimo ano é no máximo $n(11 n-1)$. Como este número coincide com a pontuação obtida por tal conjunto de participantes, podemos concluir que não existem dois alunos do sétimo ano que jogaram entre si, pois caso contrário devemos diminuir de $n(11 n-1)$ a contagem dupla de tal partida. Isto só é possível se existir apenas um aluno do sétimo ano. Portanto, $n=1$ e $m=1 \\cdot 10=10$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26028,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\na) Determine a soma das raízes reais da equação\n$$\nx^{2}+18 x+30=2 \\cdot \\sqrt{x^{2}+18 x+45}\n$$\nb) Resolva a equação $\\sqrt{5-\\sqrt{5-x}}=x$, com $0<x<5$.",
    "options": [],
    "answer": "a) -18; b) {(-1 + sqrt(21))/2}",
    "solution": "Solution:\na) Seja $x^{2}+18 x+30=y$. Temos, então\n$$\n\\begin{aligned}\ny & =2 \\cdot \\sqrt{y+15} \\\\\ny^{2} & =(2 \\cdot \\sqrt{y+15})^{2} \\\\\ny^{2} & =4 \\cdot(y+15) \\\\\ny^{2}-4 y-60 & =0\n\\end{aligned}\n$$\nAs raízes da última equação são $y=2 \\pm 8$, isto é, $y_{1}=10$ e $y_{2}=-6$. Como estamos buscando raízes reais, $y$ deve ser não negativo. Assim, devemos descartar $y_{2}=-6$ e obtemos $x^{2}+18 x+30=10$. Como o discriminante dessa equação é positivo, ambas as raízes são reais e ambas satisfazem a equação, pois $y_{1}+15 \\geq 0$. Assim, a soma das raízes é $-18$.\n\nb) Seja $x$ uma solução da equação. Então:\n$$\n\\begin{aligned}\n\\sqrt{5-\\sqrt{5-x}} & =x \\\\\n(\\sqrt{5-\\sqrt{5-x}})^{2} & =x^{2} \\\\\n5-\\sqrt{5-x} & =x^{2} \\\\\n5-x^{2} & =\\sqrt{5-x} \\\\\n\\left(5-x^{2}\\right)^{2} & =(\\sqrt{5-x})^{2} \\\\\n\\left(5-x^{2}\\right)^{2} & =(5-x)\n\\end{aligned}\n$$\nDesenvolvendo os produtos notáveis anteriores, tem-se:\n$$\n\\begin{aligned}\n\\left(5-x^{2}\\right)^{2} & =(5-x) \\\\\n5^{2}-2 \\cdot 5 \\cdot x^{2}+x^{4} & =5-x \\\\\n5^{2}-\\left(2 x^{2}+1\\right) \\cdot 5+\\left(x^{4}+x\\right) & =0\n\\end{aligned}\n$$\nFixado $x$, o número $5$ é raiz da equação do segundo grau:\n$$\nz^{2}-\\left(2 x^{2}+1\\right) \\cdot z+\\left(x^{4}+x\\right)=0\n$$\nO discriminante da equação anterior é:\n$$\n\\begin{aligned}\n\\Delta & =\\left[-\\left(2 x^{2}+1\\right)\\right]^{2}-4 \\cdot 1 \\cdot\\left(x^{4}+x\\right) \\\\\n& =4 x^{4}+4 x^{2}+1-4 x^{4}-4 x \\\\\n& =(2 x-1)^{2}\n\\end{aligned}\n$$\nComo $x>0$, $\\sqrt{\\Delta}= \\pm(2 x-1)$. Como $5$ é uma das raízes da equação, tem-se:\n$$\n5=\\frac{2 x^{2}+1 \\pm(2 x-1)}{2}\n$$\nTemos dois casos a considerar:\n$$\n\\begin{aligned}\n\\frac{2 x^{2}+1+(2 x-1)}{2} & =5 \\\\\nx^{2}+x-5 & =0 \\\\\nx & =\\frac{-1 \\pm \\sqrt{21}}{2}\n\\end{aligned}\n$$\nou\n$$\n\\begin{aligned}\n\\frac{2 x^{2}+1-(2 x-1)}{2} & =5 \\\\\nx^{2}-x-4 & =0 \\\\\nx & =\\frac{1 \\pm \\sqrt{17}}{2}\n\\end{aligned}\n$$\nComo $0<x<5$, precisa-se eliminar as duas raízes negativas. Além disso, é fácil verificar que as outras duas satisfazem a equação original. Portanto,\n$$\nS=\\left\\{\\frac{-1+\\sqrt{21}}{2}, \\frac{1+\\sqrt{17}}{2}\\right\\}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26029,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nJoão possui botões de três cores: azul, branco e cinza. Representaremos um botão azul por $A$, um botão branco por $B$ e um botão cinza por $C$. Ele coloca cinco botões nos vértices de um pentágono regular no sentido horário: $A, C, B, C$ e $B$, como na primeira figura a seguir. Ele sabe que pode fazer algumas trocas de botões obedecendo uma certa regra. Ao escolher um botão, ele pode trocá-lo por um de outra cor desde que não fiquem dois botões com uma mesma cor em vértices vizinhos. Para não ficar trocando botões sem objetivo, ele decide adotar como meta deixar os botões com as cores $A, B, C, B$ e $C$ no sentido horário como indicado na segunda figura a seguir.\n![](attached_image_1.png)\n\nApós fazer várias trocas, João não conseguiu chegar na situação final desejada. Decidiu então pedir ajuda ao seu professor Piraldo. O professor afirma que não é possível partir da configuração inicial e chegar na configuração final desejada. Para provar isto, o professor Piraldo pegou algumas cartas com os números $1,-1$ e 0 . Então ele disse para João colocar sobre cada lado do pentágono uma carta com o seguinte padrão:\ni) Olhando no sentido horário, se os vértices tiverem $B$ e $A$ ou tiverem $A$ e $C$ coloque uma carta com o número 1 .\nii) Ainda olhando no sentido horário, se os vértices tiverem $A$ e $B$ ou tiverem $C$ e $A$ coloque uma carta com o número -1 .\niii) Para os demais casos coloque a carta 0 .\nSempre que fizer uma troca, mude as cartas usando estas instruções. Após a explicação, o professor Piraldo disse que analisando o comportamento das cartas João agora poderia concluir porque é impossível atingir o objetivo inicial.\n\na) Fazendo todas as trocas possíveis de botões, de $A$ para $B$, de $B$ para $C$ ou de $C$ para $A$ e as trocas reversas, o que acontece com a soma das cartas colocadas sobre os lados segundo as instruções do professor Piraldo?\n\nb) Qual a soma das cartas na situação inicial e na situação final desejada? Conclua que não é possível partir da configuração inicial e chegar na situação final desejada.",
    "options": [],
    "answer": "a) The total sum of the edge cards remains constant under any allowed move.\nb) Initial configuration sum = 2; target configuration sum = −2. Since the invariant sums differ, it is impossible to reach the target from the initial configuration.",
    "solution": "Solution:\n\na) Como são apenas três cores, há apenas 3 trocas de botões possíveis. Além disto, os dois botões nos vértices vizinhos do vértice onde a troca é feita, devem ter botões de cores diferentes tanto da cor anterior quanto da cor após a troca. Como são apenas três cores, os dois vizinhos necessariamente possuem botões da mesma cor. Considerando o vértice em que o botão será trocado junto com seu anterior e seu sucessor no sentido horário, temos apenas três tipos de trocas possíveis de botões:\n$$\n\\begin{aligned}\n& C A C \\leftrightarrow C B C \\\\\n& A B A \\leftrightarrow A C A \\\\\n& B C B \\leftrightarrow B A B\n\\end{aligned}\n$$\nConsiderando agora a soma dos valores das duas cartas que estão entre o primeiro e o segundo e entre o segundo e terceiro vértice, teremos as seguintes possíveis alterações:\n$$\n\\begin{aligned}\n(-1)+1 & =0+0 \\\\\n(-1)+1 & =1+(-1) \\\\\n0+0 & =1+(-1)\n\\end{aligned}\n$$\nrespectivamente. Deste modo, concluímos que a soma dos números nas cartas, seguindo as instruções do professor Piraldo, não varia.\n\nb) Na figura a seguir, temos as cartas colocadas na situação inicial e na situação final.\n![](attached_image_2.png)\nComo a soma das cartas é diferente e já vimos no item anterior que as trocas não podem alterar esta soma, então não é possível chegar à situação final desejada partindo da situação inicial.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26030,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nConsideremos o conjunto $A=\\{1,2,3,4, \\ldots, n\\}$. Um subconjunto de $A$ é chamado hierárquico se satisfaz as seguintes duas propriedades:\n- O subconjunto deve ter mais de um número.\n- Há um número no subconjunto que coincide com a soma dos outros números do subconjunto.\nDeseja-se dividir o conjunto $A$ em subconjuntos hierárquicos.\na) Para $n=13$, mostre que não é possível fazer a divisão.\nb) Para $n=12$, mostre que tal divisão é possível.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na) Vamos supor que foi possível dividir o conjunto em $\\ell$ grupos. Em cada grupo, o maior coincide com a soma dos outros números do grupo. Então, a soma de todos os números do grupo seria duas vezes o maior número do grupo. Provamos assim que a soma dos números dentro de cada grupo é sempre um número par. Teríamos então que a soma de todos os números no conjunto $A$ pode ser escrita como a soma de números pares e, portanto, devia ser também par. Mas se $n=13$, a soma total é $13 \\times 14 / 2=13 \\times 7$, que não é um número par. Isso mostra que, para $n=13$, tal divisão não pode ser feita.\n\nb) Observe que não pode haver um grupo com dois números, porque pela segunda condição, esses números teriam que ser iguais e isso não é possível. Como há 12 números, isso implica que há, no máximo, 4 grupos. Além disso, depois da discussão feita no item a), sabemos que a soma dos números maiores em cada grupo deve ser a metade da soma de todos os números em $A$, isto é, a metade de $12 \\times 13 / 2=78$. Para que então a soma dos números maiores de cada grupo consiga ser $(78 / 2)=39$, é necessário ter ao menos 4 grupos, porque todos os números de $A$ são menores ou iguais a 12. Concluímos assim que devem existir exatamente 4 grupos, e que cada grupo deve ter exatamente 3 elementos. Um modo de conseguir a divisão é\n$$\nA=\\{12,9,3\\} \\cup\\{11,7,4\\} \\cup\\{10,8,2\\} \\cup\\{6,5,1\\}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26031,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nNo seguinte papel, foi desenhado um quadriculado de $4 \\times 6$ e depois traçada a diagonal de $A$ a $B$.\n![](attached_image_1.png)\nObserve que a diagonal $A B$ intersecta o quadriculado em 9 pontos:\n![](attached_image_2.png)\nSe o quadriculado fosse de tamanho $12 \\times 17$, em quantos pontos a diagonal $A B$ intersectaria o quadriculado?",
    "options": [],
    "answer": "29",
    "solution": "Solution:\nVamos enumerar as linhas verticais e as linhas horizontais do quadriculado do seguinte modo:\n![](attached_image_3.png)\nA diagonal $A B$ intersecta cada linha vertical em exatamente um ponto. Com isso, contaríamos 18 pontos de interseção. Também $A B$ intersecta cada linha horizontal em exatamente um ponto. Com isso, contaríamos 13 pontos de interseção. Sabemos que os pontos $A$ e $B$ são pontos de interseção em comum. Portanto, na soma $18+13$, estamos contando cada interseção duas vezes. Se não existissem outros pontos de interseção em comum a resposta seria\n$$\n18+13-2=29\n$$\nSuponhamos que existisse um outro ponto de interseção $P$, como mostrado no gráfico.\n![](attached_image_4.png)\nDigamos que o lado de cada quadradinho meça $1 u$. Então é claro que $|A C|=17$ $u$ e $|B C|=12 u$. Os triângulos $A B C$ e $P B Q$ são semelhantes porque possuem os mesmos ângulos. Portanto, $\\frac{|B Q|}{|Q P|}=\\frac{|B C|}{|C A|}$ ou, equivalentemente,\n$$\nB Q=\\frac{12 \\times Q P}{17}\n$$\nSe $P$ pertencesse a uma linha vertical e a uma linha horizontal, então tanto $Q P$ quanto $B Q$ deveriam ser números inteiros (quando escritos em termos de $u$ ). Para que $12 \\times|Q P| / 17$ seja inteiro, necessariamente 17 deve dividir $|Q P|$. Mas se $0<|Q P|<17$, isso é impossível. Concluímos assim que tal $P$ não existe. Portanto, a resposta final são 29 pontos de interseção de $A B$ com o quadriculado.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26032,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUma sequência especial - Na sequência $1, 3, 2, \\ldots$ cada termo depois dos dois primeiros é igual ao termo precedente, subtraído do termo que o precede, ou seja, se $n > 2$, então $a_{n} = a_{n-1} - a_{n-2}$. Qual é a soma dos cem primeiros termos dessa sequência?",
    "options": [],
    "answer": "5",
    "solution": "Solution:\n\nInicialmente escrevemos os primeiros termos dessa sequência, como segue.\n$$\n1, 3, 2, -1, -3, -2, 1, 3, 2, \\ldots\n$$\nO sétimo e o oitavo termos são, respectivamente, iguais ao primeiro e ao segundo. Isso significa que a sequência se repete de seis em seis termos. A soma dos seis primeiros termos é $1 + 3 + 2 - 1 - 3 - 2 = 0$ e, portanto, a soma dos 96 primeiros termos também é $0$. Assim, a soma dos 100 primeiros termos dessa sequência é igual à soma dos quatro últimos termos, ou seja, $1 + 3 + 2 - 1 = 5$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26033,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA seguinte figura mostra um quadrado $ABCD$, e dois pontos $P$ e $Q$ sobre os lados $\\overline{BC}$ e $\\overline{DA}$, respectivamente.\n![](attached_image_1.png)\nDobramos agora o quadrado ao longo do segmento $\\overline{PQ}$, levando o vértice $B$ até o ponto médio do segmento $\\overline{CD}$.\n![](attached_image_2.png)\nSabe-se que o lado do quadrado mede 24.\n\na) Calcule o comprimento do segmento $\\overline{PC}$.\n\nb) Calcule o comprimento do segmento $\\overline{AQ}$.\n\nc) Calcule o comprimento do segmento $\\overline{PQ}$.",
    "options": [],
    "answer": "PC = 9, AQ = 3, PQ = 12 sqrt(5)",
    "solution": "Solution:\na) Chamemos $x$ o comprimento do segmento $\\overline{PC}$. Como o lado do quadrado mede 24, então $|PB|=24-x$ e também $|PB'|=24-x$.\n![](attached_image_3.png)\nUsando o Teorema de Pitágoras no triângulo $PCB'$, obtemos a relação\n$$\nx^2 + 12^2 = (24-x)^2\n$$\nAssim, mostramos que $|PC|=9$. Observe também que $|PB'|=15$ (isso será usado nos próximos itens).\n\nb) Chamemos $\\alpha^\\circ$ a medida do ângulo $PB'C$. Então $\\measuredangle RB'D=(90-\\alpha)^\\circ$ e, portanto, $\\measuredangle DRB'=\\alpha^\\circ$.\n![](attached_image_4.png)\nIsso mostra que os triângulos $PCB'$ e $B'DR$ são semelhantes. Usando as relações de semelhança, vemos que $|RD|=12 \\times 12 / 9 = 16$ e então $|RB'|=20$ (por Pitágoras).\n![](attached_image_5.png)\nDado que $|A'B'|=|AB|=24$, obtemos que $|A'R|=4$. Além disso, podemos ver que $\\measuredangle QRA'=\\measuredangle DRB'=\\alpha^\\circ$ e, portanto, são também semelhantes os triângulos $RDB'$ e $RA'Q$. Usando a relação de semelhança, obtemos que $|QA'|=4 \\times 12 / 16 = 3$.\n![](attached_image_6.png)\nFinalmente, $|AQ|=|A'Q|=3$.\n\nc) Para calcular $PQ$, é suficiente observar o quadrilátero $A'QPB'$. Traçamos o segmento $\\overline{QS}$ perpendicular ao segmento $\\overline{PB'}$ de modo que $|QA'|=|SB'|=3$.\n![](attached_image_7.png)\nUsando Pitágoras no triângulo $PSQ$ temos\n$$\n|PQ|^2 = 24^2 + 12^2\n$$\nDeste modo, $|PQ|=12\\sqrt{5}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26034,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nResolva a equação\n$$\n\\sqrt{x+3-4 \\sqrt{x-1}}+\\sqrt{x+8-6 \\sqrt{x-1}}=1\n$$",
    "options": [],
    "answer": "[5, 10]",
    "solution": "Solution:\n\nTemos\n$$\n\\begin{aligned}\n\\sqrt{x+3-4 \\sqrt{x-1}}+\\sqrt{x+8-6 \\sqrt{x-1}} &= \\sqrt{(\\sqrt{x-1}-2)^{2}}+\\sqrt{(\\sqrt{x-1}-3)^{2}} \\\\\n&=|\\sqrt{x-1}-2|+|\\sqrt{x-1}-3|\n\\end{aligned}\n$$\nComo a distância entre 2 e 3 é 1, só podemos ter $|\\sqrt{x-1}-2|+|\\sqrt{x-1}-3|=1$ quando $\\sqrt{x-1}$ se situa entre estes dois números, ou seja,\n$$\n2 \\leq \\sqrt{x-1} \\leq 3\n$$\nIsto é equivalente a $4 \\leq x-1 \\leq 9$. Portanto, o conjunto solução é $\\{x \\in \\mathbb{R} \\mid 5 \\leq x \\leq 10\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26035,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm poliminó é uma sequência de quadradinhos $1 \\times 1$ justapostos compartilhando lados em comum com seus vizinhos e formando uma única peça. Os poliminós de dois quadradinhos são conhecidos como dominós e os poliminós com quatro quadradinhos são conhecidos como tetraminós, as pecinhas do famoso jogo Tetris. A figura a seguir mostra um quadrado $3 \\times 3$ com números de 1 até 9 escritos em cada um de seus quadradinhos $1 \\times 1$.\n\n| 1 | 2 | 3 |\n| :--- | :--- | :--- |\n| 4 | 5 | 6 |\n| 7 | 8 | 9 |\n\nSabendo que $1+2+\\ldots+9=45$, podemos tentar dividir o quadrado em 3 poliminós com mesma soma, cada um com soma $\\frac{45}{3}=15$. A figura a seguir, mostra uma maneira de fazer isto.\n\n| 1 | 2 | 3 |\n| :---: | :---: | :---: |\n| 4 | 5 | 6 |\n| 7 | 8 | 9 |\n\na. Mostre que não é possível dividir o quadrado $3 \\times 3$ em uma quantidade maior que três de poliminós de mesma soma.\n\nb. Considere o quadrado $4 \\times 4$ com os números de 1 até 16, escritos em ordem crescente como indicado na figura abaixo\n\n| 1 | 2 | 3 | 4 |\n| :---: | :---: | :---: | :---: |\n| 5 | 6 | 7 | 8 |\n| 9 | 10 | 11 | 12 |\n| 13 | 14 | 15 | 16 |\n\nMostre como dividir este quadrado em dois poliminós de modo que a soma dos números em cada um deles seja a mesma.\n\nc. Considere o quadrado $5 \\times 5$ com os números de 1 até 25 escritos em ordem crescente seguindo o padrão da figura anterior. Mostre que não é possível dividir este quadrado em dois ou mais poliminós com a mesma soma dos números em cada um deles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na. Para dividirmos o quadrado em $x$ poliminós de mesma soma, então $x$ deve obrigatoriamente ser um divisor da soma total dos números. No caso do quadrado $3 \\times 3$, como a soma é 45, se $x>5$, então cada poliminó teria soma menor que $\\frac{45}{5}=9$. Entretanto, o poliminó que tivesse o quadradinho 9 não poderia realizar esta soma. No caso $x=5$, devemos ter 5 pedaços com soma 9 em cada. Com isto, o 9 teria que ser deixado sozinho. Porém, o número 8 deixado sozinho não produziria a soma 9 e se o adicionarmos com qualquer um de seus vizinhos, formaremos um poliminó com soma maior que 9. Como 4 não divide 45, concluímos que não é possível dividir em mais que 3 pedaços de mesma soma.\n\nb. Como a soma dos números de 1 até 16 é igual a 136, então cada um dos dois pedaços deve somar 68. Somando os números da última linha teremos $13+14+15+16=58$. As 10 unidades que faltam podem ser incluídas com um único quadradinho, pois os quadrados com os números 10 e 14 possuem um lado em comum. Veja a divisão construída na figura a seguir.\n\n| 1 | 2 | 3 | 4 |\n| :---: | :---: | :---: | :---: |\n| 5 | 6 | 7 | 8 |\n| 9 | 10 | 11 | 12 |\n| 13 | 14 | 15 | 16 |\n\nc. Novamente o primeiro passo é calcular a soma dos números no quadrado, que no caso $5 \\times 5$, resulta em 325. Lembrando que para dividir o quadrado em $x$ poliminós de mesma soma é necessário que $x$ seja um divisor de 325. Se $x>13$, então cada parte deverá ter soma menor que $\\frac{325}{13}=25$, mas o poliminó com o quadrado 25 já ultrapassa este valor e impossibilita tal divisão. Entre os divisores de 325, os únicos entre 2 e 13 são 5 e 13. Se $x=13$, então 25 ficaria sozinho em um poliminó e, em seguida, o 24 não poderia formar poliminó com soma 25 em virtude de seus vizinhos. Se $x=5$, então a soma de cada poliminó deverá ser 65. Novamente, olhamos para o 25. Se ele estiver no mesmo poliminó que 24 a soma já será 49 e qualquer quadrado adicional faria a soma passar de 65. Então, considere o 25 junto com o 20. Se neste mesmo poliminó tivermos 15 ou 19, a soma passará de 59 e qualquer quadradinho que adicionarmos fará a soma passar de 65. Concluímos que não é possível dividir este quadrado $5 \\times 5$ em dois ou mais poliminós de mesma soma.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26036,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA seguinte figura mostra um triângulo $ABC$ que foi dividido em 3 triângulos isósceles.\n![](attached_image_1.png)\n\na) Mostre que todo triângulo retângulo pode ser dividido em 2 triângulos isósceles.\n\nb) Mostre que qualquer triângulo pode ser dividido em 4 triângulos isósceles.\n\nc) Mostre que qualquer triângulo pode ser dividido em 5 triângulos isósceles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na) Consideremos o triângulo retângulo com ângulos agudos $\\alpha^\\circ$ e $\\beta^\\circ$, como mostra a seguinte figura:\n![](attached_image_2.png)\n\nTraçamos um segmento $BM$, com $M$ sobre o lado $AC$, de modo que $\\measuredangle MBC = \\beta^\\circ$.\n![](attached_image_3.png)\n\nTemos então que $\\measuredangle MBA = 90^\\circ - \\beta^\\circ = \\alpha^\\circ$, e conseguimos assim os dois triângulos isósceles $AMB$ e $BMC$.\n\nb) Primeiro devemos observar que, em qualquer triângulo, existem sempre dois ângulos internos que são estritamente menores que $90^\\circ$. Se não fosse assim, existiriam dois ângulos internos maiores ou iguais a $90^\\circ$, e isso é impossível, dado que a soma dos ângulos internos deve ser $180^\\circ$. Na figura seguinte, estamos supondo que $\\measuredangle QPR$ e $\\measuredangle QRP$ são ambos menores que $90^\\circ$:\n![](attached_image_4.png)\n\nSe traçarmos a altura partindo de $Q$, a base da altura se encontrará no lado $\\overline{PR}$, como mostra a figura:\n![](attached_image_5.png)\n\nPelo item a), os triângulos retângulos $PHQ$ e $RHQ$ podem ser divididos em dois triângulos isósceles cada um. Feito isso, conseguiremos dividir o triângulo $PQR$ em 4 triângulos isósceles.\n\nc) Sejam $\\alpha^\\circ$ e $\\beta^\\circ$ as medidas do menor e do maior ângulo interno do triângulo, respectivamente. Vamos supor primeiro que $\\alpha < \\beta$ como na figura a seguir:\n![](attached_image_6.png)\n\nPodemos então traçar um segmento $BD$, como mostra a próxima figura, de modo que o triângulo $BDC$ seja isósceles.\n![](attached_image_7.png)\n\nAgora, pelo item b), sabemos que o triângulo $ABD$ pode ser dividido em 4 triângulos isósceles. Fazendo assim, teremos dividido o triângulo $ABC$ em 5 triângulos isósceles.\n\nDevemos considerar agora o caso em que $\\alpha = \\beta$. Isso acontece somente se o triângulo for equilátero. Podemos primeiro dividir o triângulo como mostra a seguinte figura, onde $I$ é o incentro do triângulo.\n![](attached_image_8.png)\n\nAssim, temos três triângulos isósceles. Finalmente, traçamos $IP$ e $IQ$ de modo que $\\measuredangle AIP = \\measuredangle CIQ = 30^\\circ$.\n![](attached_image_9.png)\n\nOs triângulos $AIP$ e $CIQ$ são claramente isósceles, enquanto o triângulo $IPQ$ é equilátero. Dividimos assim, o triângulo $ABC$ em 5 triângulos isósceles.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26037,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEm um torneio, quaisquer dois jogadores jogam entre si. Cada jogador obtém um ponto por vitória, $1/2$ por empate e $0$ ponto por derrota. Seja $S$ o conjunto das $10$ menores pontuações. Sabemos que cada jogador obteve metade da sua pontuação jogando contra jogadores de $S$.\n\na) Qual a soma das pontuações dos jogadores de $S$?\n\nb) Determine quantos participantes tem o torneio.\n\nObservação: Cada jogador joga apenas uma vez com cada adversário.",
    "options": [],
    "answer": "a) 90; b) 25",
    "solution": "Solution:\n\na) Os jogadores de $S$, em partidas disputadas apenas entre si, obtiveram $\\frac{10(10-1)}{2}=45$ pontos. Como os que eles obtiveram jogando entre si correspondem a metade dos pontos que cada um obteve no torneio, podemos concluir que a soma dos pontos dos jogadores de $S$ é $45+45=90$.\n\nb) Sejam $n$ o número de jogadores do torneio e $T$ o conjunto dos outros $n-10$ jogadores. Jogando apenas entre si, os jogadores de $T$ obtiveram $\\frac{(n-10)(n-11)}{2}$ pontos. Como eles obtiveram metades de seus pontos jogando com jogadores de $S$, podemos concluir que eles obtiveram a outra metade jogando entre si. Consequentemente, a soma das pontuações de todos os jogadores de $T$ é $(n-10)(n-11)$. Como a soma das pontuações de todos os jogadores corresponde ao total de jogos, temos\n$$\n\\begin{aligned}\n90+(n-10)(n-11) & =\\frac{n(n-1)}{2} \\\\\nn^{2}-41 n+400 & =0\n\\end{aligned}\n$$\nAs raízes desta equação do segundo grau são $16$ e $25$. Se $n=16$, os $16-10=6$ melhores jogadores que obtiveram, em conjunto, $(16-10)(16-11)=30$ pontos. Como cada jogador entre os $6$ melhores fez pelo menos tantos pontos quanto os $10$ piores, a média de pontos em $T$ é pelo menos a média de pontos de $S$, ou seja,\n$$\n5=\\frac{30}{6} \\geq \\frac{90}{10}=9\n$$\nIsto é um absurdo. Logo, só nos resta $n=25$. Resta apenas mostrar que é possível cumprirmos as exigências do torneio com $25$ jogadores.\n\nConsidere a divisão das $25$ pessoas em três grupos:\n$$\n\\begin{aligned}\n& G_{1}=\\{J_{1}, J_{2}, J_{3}, J_{4}, J_{5}, J_{6}\\} \\\\\n& G_{2}=\\{J_{7}, J_{8}, J_{9}, J_{10}, J_{11}, J_{12}, J_{13}, J_{14}, J_{15}\\} \\\\\n& G_{3}=\\{J_{16}, J_{17}, J_{18}, J_{19}, J_{20}, J_{21}, J_{22}, J_{23}, J_{24}, J_{25}\\}\n\\end{aligned}\n$$\nTodos os jogos entre elementos de $G_{2} \\cup G_{3}$ terminarão empatados. Vamos agora definir o resultado entre as partidas dos elementos de $G_{1}$ contra os outros competidores:\n\ni) Os elementos de $G_{1}$ vão ganhar todas as partidas contra elementos de $G_{3}$.\n\nii) Todos os jogos entre dois elementos de $G_{1}$ terminarão empatados.\n\niii) Cada elemento de $G_{1}$ irá vencer exatamente seis elementos de $G_{2}$ e empatará com os outros $3$ restantes. Para isto, representando uma vitória por meio de uma seta $(\\rightarrow)$, estabeleceremos:\n$$\n\\begin{aligned}\nJ_{1} & \\rightarrow J_{10}, J_{11}, J_{12}, J_{13}, J_{14}, J_{15} \\\\\nJ_{2} & \\rightarrow J_{10}, J_{11}, J_{12}, J_{13}, J_{14}, J_{15} \\\\\nJ_{3} & \\rightarrow J_{7}, J_{8}, J_{9}, J_{13}, J_{14}, J_{15} \\\\\nJ_{4} & \\rightarrow J_{7}, J_{8}, J_{9}, J_{13}, J_{14}, J_{15} \\\\\nJ_{5} & \\rightarrow J_{7}, J_{8}, J_{9}, J_{10}, J_{11}, J_{12} \\\\\nJ_{6} & \\rightarrow J_{7}, J_{8}, J_{9}, J_{10}, J_{11}, J_{12}\n\\end{aligned}\n$$\nCom esta distribuição, os jogadores de $G_{1}$, $G_{2}$ e $G_{3}$, obterão $20$, $10$ e $9$ pontos, respectivamente. Além disto, cada elemento destes três grupos terá obtido metade de seus pontos contra os elementos de $G_{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26038,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDetermine o termo mínimo da sequência\n$$\n\\sqrt{\\frac{7}{6}}+\\sqrt{\\frac{96}{7}}, \\sqrt{\\frac{8}{6}}+\\sqrt{\\frac{96}{8}}, \\sqrt{\\frac{9}{6}}+\\sqrt{\\frac{96}{9}}, \\ldots, \\sqrt{\\frac{95}{6}}+\\sqrt{\\frac{96}{95}}\n$$",
    "options": [],
    "answer": "4",
    "solution": "Solution:\nLembre que $(x-y)^2 \\geq 0$ para todos os reais $x$ e $y$. Assim, podemos reescrever a desigualdade como $\\frac{x^2+y^2}{2} \\geq x y$, e substituindo $x=\\sqrt{a}$ e $y=\\sqrt{b}$, com $a$ e $b$ reais não negativos, temos $\\frac{a+b}{2} \\geq \\sqrt{a b}$. Agora, observe que todos os termos são do tipo $\\sqrt{\\frac{n}{6}}+\\sqrt{16 \\cdot 6/n}=\\sqrt{\\frac{n}{6}}+4 \\sqrt{\\frac{6}{n}}$, com $n$ inteiro e maior do que ou igual a 7. Tomando $a=\\sqrt{\\frac{n}{6}}$ e $b=4 \\sqrt{\\frac{6}{n}}$, podemos fazer\n$$\n\\begin{aligned}\n\\frac{\\sqrt{\\frac{n}{6}}+4 \\sqrt{\\frac{6}{n}}}{2} & \\geq \\sqrt{\\sqrt{\\frac{n}{6}} \\cdot 4 \\sqrt{\\frac{6}{n}}} \\\\\n\\sqrt{\\frac{n}{6}}+4 \\sqrt{\\frac{6}{n}} & \\geq 2 \\cdot 2 \\cdot \\sqrt{\\sqrt{\\frac{n}{6} \\cdot \\frac{6}{n}}} \\\\\n& =4\n\\end{aligned}\n$$\nO valor mínimo de cada termo é 4 e ocorre quando $\\sqrt{\\frac{n}{6}}=4 \\sqrt{\\frac{6}{n}}$, ou seja, $n=24$. O termo procurado é $\\sqrt{\\frac{24}{6}}+\\sqrt{\\frac{96}{24}}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26039,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUma pilha de pedras está sobre uma mesa, Pedrinho joga o seguinte jogo: a cada momento, ele pode escolher uma pilha com pelo menos 3 pedras, retirar uma dessas pedras e dividir a pilha em duas pilhas não vazias. Por exemplo, se ele tem uma pilha com 15 pedras, ele pode dividir essa pilha em duas pilhas de 9 e 5 (ele tira uma pedra, ficando com 14 pedras na pilha e depois a divide). Ele pode continuar com o processo. Por exemplo, Pedrinho pode dividir a pilha com 9 pedras em duas, uma de 3 e uma de 5, ficando no final com três pilhas, uma de 3, e duas de 5.\n\na) Se no início há uma única pilha com 19 pedras sobre a mesa. Pedrinho consegue, depois de alguns movimentos, que todas as pilhas restantes tenham exatamente 3 pedras?\n\nb) E se houver uma única pilha com 1001 pedras?",
    "options": [],
    "answer": "a) Yes. b) No.",
    "solution": "Solution:\n\na) Sim, Pedrinho consegue! Basta fazer o seguinte processo: primeiro dividimos a pilha original em duas, uma com 3 e outra com 15. Depois, dividimos a pilha com 15 em uma com 3, e outra com 11, ficando com duas pilhas de 3, e uma de 11. Continuando o processo, dividimos a de 11 em uma de 3 e uma de 7. Por fim, dividimos a pilha de 7 em duas pilhas de 3, terminando com 5 pilhas de 3 pedras.\n\nb) Vamos provar que Pedrinho não consegue. Suponha que ele tenha feito $n$ divisões de pilhas. Como em cada divisão ele começa retirando uma pedra, Pedrinho retirará $n$ pedras no total. Além disso, a cada divisão é acrescentada uma pilha (uma pilha é dividida em duas), temos ao final $n+1$ pilhas. Então, se cada uma destas pilhas tem 3 pedras, Pedrinho, ao final, vai ter no total $3(n+1)$ pedras. Logo, concluímos que $3(n+1)=1001-n$, ou seja, o número de pedras ao final é igual ao número inicial de pedras, menos o número de pedras que foram retiradas. Portanto,\n$$\n\\begin{aligned}\n3n+3 &= 1001-n \\\\\n4n &= 998\n\\end{aligned}\n$$\nComo 998 não é múltiplo de 4, não existe este valor $n$. Logo, Pedrinho não consegue dividir a pilha original em várias pilhas de três pedras.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26040,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSe $n$ e $k$ são inteiros positivos, então\n$$\n(n+1) \\cdot(n+2) \\cdot \\ldots \\cdot(n+k)<(n+k)^{k}\n$$\nUse isto para determinar qual dos dois números a seguir é maior que o outro:\n$(100!)$ ! e $99!^{100!} \\cdot 100!^{99!}$.",
    "options": [],
    "answer": "99!^{100!} · 100!^{99!} is larger",
    "solution": "Solution:\nSe $a$ é um inteiro arbitrário, usando a dica do enunciado e multiplicando as 100 desigualdades a seguir\n$$\n\\begin{aligned}\n1 \\cdot 2 \\cdot 3 \\cdot \\ldots \\cdot a & <a^{a} \\\\\n(a+1)(a+2)(a+3) \\cdot \\ldots \\cdot(2 a) & <(2 a)^{a} \\\\\n(2 a+1)(2 a+2)(2 a+3) \\cdot \\ldots \\cdot(3 a) & <(3 a)^{a} \\\\\n& \\ldots \\\\\n(99 a+1)(99 a+2)(99 a+3) \\cdot \\ldots \\cdot(100 a) & <(100 a)^{a}\n\\end{aligned}\n$$\ntemos:\n$$\n(100 a)!<\\left(100!\\cdot a^{100}\\right)^{a}\n$$\nBasta agora substituirmos $a=99!$ para obtermos\n$$\n\\begin{aligned}\n(100!)! & <\\left(100!\\cdot(99!)^{100}\\right)^{99!} \\\\\n& =(100!)^{99!} \\cdot(99!)^{100!}\n\\end{aligned}",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26041,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nOs números $10$, $11$ e $12$ podem pertencer a uma mesma progressão geométrica?",
    "options": [],
    "answer": "No",
    "solution": "Solution:\nSuponha que $10$, $11$ e $12$ sejam termos de uma mesma progressão geométrica:\n$$\n\\begin{aligned}\n10 & = a r^{p-1} \\\\\n11 & = a r^{q-1} \\\\\n12 & = a r^{k-1}\n\\end{aligned}\n$$\nDaí,\n$$\n\\begin{aligned}\n& 11 / 10 = r^{q-p} \\\\\n& 12 / 11 = r^{k-q}\n\\end{aligned}\n$$\nAlém disso, temos\n$$\n\\begin{aligned}\n& \\left(\\frac{11}{10}\\right)^{k-q} = \\left(r^{q-p}\\right)^{k-q} \\\\\n& \\begin{aligned}\n\\left(\\frac{12}{11}\\right)^{q-p} & = \\left(r^{(q-p)(k-q)}\\right. \\\\\n&\n\\end{aligned} \\\\\n& = r^{(k-q)(q-p)}\n\\end{aligned}\n$$\nAssim,\n$$\n\\begin{aligned}\n\\left(\\frac{11}{10}\\right)^{k-q} & = \\left(\\frac{12}{11}\\right)^{q-p} \\\\\n(11)^{k-q+q-p} & = (10)^{k-q}(12)^{q-p} \\\\\n(11)^{k-p} & = 5^{k-q} \\cdot 2^{k-q} \\cdot 2^{2(q-p)} \\cdot 3^{q-p} \\\\\n& = 5^{k-q} \\cdot 2^{k+q-2p} \\cdot 3^{q-p}\n\\end{aligned}\n$$\nPelo Teorema Fundamental da Aritmética, temos\n$$\n\\left\\{\n\\begin{array}{l}\nk-p=0 \\\\\nk-q=0 \\\\\nk+q-2p=0 \\\\\nq-p=0\n\\end{array}\n\\right.\n$$\nResolvendo o sistema associado, temos $k=p=q$. Isso produz um absurdo, pois $p$, $q$ e $k$ são distintos. Portanto, $10$, $11$ e $12$ não podem pertencer a uma mesma progressão geométrica.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26042,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPedrinho faz uma lista de todos os números de 5 algarismos distintos que se formam com os dígitos $1,2,3,4,5$. Nesta lista os números estão ordenados de forma crescente.\n\na) Qual o número que ocupa a posição 10 da lista?\n\nb) Qual o número que ocupa a posição 85 da lista?",
    "options": [],
    "answer": "a) 13452; b) 43125",
    "solution": "Solution:\n\na) Começamos escrevendo os primeiros números da lista:\n12345, 12354, 12435, 12453, 12534, 12543, 13245, 13254, 13425, 13452.\nLogo, o décimo número é 13452.\n\nb) Para encontrar o número que ocupa a posição 85, percebemos que sempre que um número de 5 dígitos começa com o dígito 1, este número é menor do que um número de 5 dígitos que começa com o número 2. Por sua vez, este número é menor do que um que começa com 3, e assim por diante. Então, contaremos quantos destes números começam com 1, 2 etc.\n\nSe fixarmos o primeiro dígito, por exemplo 1, vamos ter 4 escolhas para o segundo dígito, a saber, cada um dos números $2,3,4$ e $5$. Para o terceiro dígito vamos ter só 3 escolhas, não podemos escolher um dígito já usado antes, e assim sucessivamente. Teremos 2 escolhas para o quarto dígito e uma única escolha para o último dígito. Logo, temos $4 \\times 3 \\times 2 \\times 1 = 24$ números de 5 dígitos que começam com 1 na lista de Pedrinho e, claramente, estes têm que ser os primeiros 24 números, pois começam com 1.\n\nAnalogamente, também temos 24 números que começam com 2 (é exatamente a mesma conta, trocando 1 por 2), que têm que ser os 24 seguintes números da lista. E o mesmo vale para 3, 4 e 5. Concluímos então que os primeiros 24 números começam com 1, depois temos 24 que começam com 2, depois 24 que começam com 3. Logo, o número que fica na posição 73 é o primeiro número que começa com 4, que é o número 41235, enquanto que o número na posição 96 é o último número que começa com 4, ou seja, o número 45321.\n\nConcluímos então que o número na posição 85 é um número que começa com 4. O que faremos agora é repetir o processo acima para o segundo dígito. Usando o mesmo argumento acima, temos $3 \\times 2 \\times 1 = 6$ números que começam com 41, e estes são claramente menores que os que começam com 42 (que também são 6). Como $72 + 6 + 6 = 84$, temos que o número que ocupa a posição 85 é o primeiro número que começa com 43, ou seja, é o número 43125.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26043,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSeja $a_{n}$ o número de maneiras de preencher um tabuleiro $n \\times n$ com os algarismos 0 e 1, de modo que a soma em cada linha e em cada coluna seja a mesma. Por exemplo, os tabuleiros $2 \\times 2$ que satisfazem essa regra são:\n\n| 0 | 0 |\n| :--- | :--- |\n| 0 | 0 |\n| 1 | 0 |\n| :--- | :--- |\n| 0 | 1 |\n| 0 | 1 |\n| :--- | :--- |\n| 1 | 0 |\n| 1 | 1 |\n| :--- | :--- |\n| 1 | 1 |\n\nLogo, $a_{2}=4$. Calcule os valores de $a_{3}$ e $a_{4}$.",
    "options": [],
    "answer": "a3 = 14, a4 = 140",
    "solution": "Solution:\nPara os tabuleiros $3 \\times 3$, dividiremos a solução em casos de acordo com a quantidade de números 1 por linha:\n\n- Caso 1: não há números 1. Neste caso, só temos um tabuleiro:\n\n| 0 | 0 | 0 |\n| :--- | :--- | :--- |\n| 0 | 0 | 0 |\n| 0 | 0 | 0 |\n\n- Caso 2: temos um número 1 por linha. Neste caso, temos que escolher um número 1 em cada linha. Para a primeira linha, temos 3 escolhas (cada uma das casas desta linha). Sem perda de generalidade, podemos supor que escolhemos a primeira casa. Note que na coluna que este 1 foi escolhido só podemos ter zeros, pois a soma em cada coluna também tem que ser 1.\n\n| 1 | 0 | 0 |\n| :--- | :--- | :--- |\n| 0 | $\\star$ | $\\star$ |\n| 0 | $\\star$ | $\\star$ |\n\nAgora, para a segunda linha só vamos ter duas escolhas possíveis (não podemos colocar dois números 1 na mesma coluna), enquanto que para a última linha só teremos uma. Então, neste caso, teremos $3 \\times 2 \\times 1=6$ tabuleiros.\n\n- Caso 3: agora, vemos que a quantidade de tabuleiros em que cada linha tem dois números 1 é igual à quantidade de tabuleiros com um número 1 em cada linha. Isso é verdade porque, dado um tabuleiro com dois números 1 em cada linha, basta trocarmos os zeros por uns e os números uns por zeros, e obteremos um tabuleiro com apenas um número 1 em cada linha.\n\n- Caso 4: o mesmo vale para a quantidade de tabuleiros só com uns, que é igual à quantidade de tabuleiros só com zeros. Portanto\n\n$$\na_{3}=1+6+6+1=14\n$$\n\nPara calcular $a_{4}$, faremos uma contagem semelhante ao caso do $a_{3}$. Se não houver números 1, só teremos um tabuleiro formado por zeros. O mesmo vale se não tivermos nenhum zero no tabuleiro. Se tivermos só um número 1 em cada linha, basta fazer uma conta análoga ao caso do $a_{3}$. Logo, temos 4 escolhas para a primeira linha, 3 escolhas para a segunda linha (não podemos ter dois números 1 na mesma coluna), 2 escolhas para a terceira linha e 1 escolha para a última linha. Daí, temos $4 \\times 3 \\times 2 \\times 1=24$ tabuleiros com um número 1 em cada linha. A mesma conta funciona para o caso com um zero em cada linha.\n\nAgora só falta o caso em que temos dois números 1 em cada linha. Temos então que colocar dois números na primeira linha. Podemos fazer isso de\n$$\n\\frac{4 \\times 3}{2}=6\n$$\nmaneiras. Sem perda de generalidade, vamos supor que a primeira linha seja da seguinte forma\n\n| 1 | 1 | 0 | 0 |\n| :--- | :--- | :--- | :--- | :--- |\n|  |  |  |  |\n|  |  |  |  |\n|  |  |  |  |\n\nNa primeira coluna deve haver mais um 1, o que pode ser feito de três maneiras:\n\n| 1 | 1 | 0 | 0 |\n| :--- | :--- | :--- | :--- |\n| $\\star$ |  |  |  |\n| $\\star$ |  |  |  |\n| $\\star$ |  |  |  |\n\nDe novo, sem perda de generalidade, podemos supor que o tabuleiro seja da forma\n\n| 1 | 1 | 0 | 0 |\n| :--- | :--- | :--- | :--- |\n| 1 |  |  |  |\n| 0 |  |  |  |\n| 0 |  |  |  |\n\nAgora, dividimos em dois casos:\n\n| 1 | 1 | 0 | 0 |\n| :--- | :--- | :--- | :--- |\n| 1 | 1 |  |  |\n| 0 |  |  |  |\n| 0 |  |  |  |\n\n| 1 | 1 | 0 | 0 |\n| :--- | :--- | :--- | :--- |\n| 1 | 0 |  |  |\n| 0 |  |  |  |\n| 0 |  |  |  |\n\nNo primeiro caso acima, é fácil ver que só podemos completar de um jeito, pois já temos dois números 1 na segunda coluna e na segunda linha, então temos que completá-las com zeros:\n\n| 1 | 1 | 0 | 0 |\n| :--- | :--- | :--- | :--- |\n| 1 | 1 | 0 | 0 |\n| 0 | 0 | 1 | 1 |\n| 0 | 0 | 1 | 1 |\n\nLogo, para este caso nós temos $6 \\times 3$ maneiras de preencher o tabuleiro, lembrando que 6 é o número de maneiras de se colocarem dois números 1 na primeira linha e 3 é a quantidade de escolhas para a posição do outro número 1 na primeira coluna.\n\nJá no segundo caso, ficamos com um tabuleiro do tipo:\n\n| 1 | 1 | 0 | 0 |\n| :--- | :--- | :--- | :--- |\n| 1 | 0 | $\\bullet$ | $\\bullet$ |\n| 0 | $\\star$ |  |  |\n| 0 | $\\star$ |  |  |\n\nTemos 2 maneiras de escolher onde fica o 1 na segunda linha (substituindo um dos $\\bullet$ acima) e 2 maneiras de escolher onde fica o 1 na segunda coluna (substituindo uma das $\\star$ acima). Sem perda de generalidade, vamos assumir que ficamos com um tabuleiro do tipo:\n\n| 1 | 1 | 0 | 0 |\n| :--- | :--- | :--- | :--- |\n| 1 | 0 | 1 | 0 |\n| 0 | 1 |  |  |\n| 0 | 0 | $\\star$ | $\\star$ |\n\nAgora, o tabuleiro está determinado. De fato, as duas estrelinhas acima têm que ser números 1, pois temos que ter dois números 1 na última linha. Terminar a partir daí é fácil:\n\n| 1 | 1 | 0 | 0 |\n| :--- | :--- | :--- | :--- |\n| 1 | 0 | 1 | 0 |\n| 0 | 1 | 0 | 1 |\n| 0 | 0 | 1 | 1 |\n\nEntão, neste caso temos 6 escolhas para a primeira linha, depois 3 para a primeira coluna. Em seguida, mais 2 escolhas para a segunda linha e 2 para a segunda coluna. Ficando com $6 \\times 3 \\times 2 \\times 2=72$ escolhas. Então, no total, ficamos com\n$$\na_{4}=1+24+(72+18)+24+1=140\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26044,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPirajuba possui 10 cidades, chamadas $H_{1}, H_{2}, \\ldots, H_{10}$, e algumas delas são ligadas por estradas de mão dupla. Sabe-se que é possível chegar de $H_{1}$ a $H_{10}$. Mostre que uma das situações abaixo ocorre:\n(i) Existe um caminho ligando $H_{1}$ a $H_{10}$ utilizando no máximo 3 estradas.\n(ii) Existem 2 cidades $H_{i}$ e $H_{j}, 2 \\leq i<j \\leq 9$, tais que todo caminho ligando $H_{1}$ a $H_{10}$ passa por $H_{i}$ ou $H_{j}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSe $H_{1}$ ou $H_{10}$ estão ligadas a no máximo 2 cidades, a condição (i) é claramente satisfeita, pois todo caminho ligando $H_{1}$ a $H_{10}$ deve passar por elas.\n\nSuponha então que $H_{1}$ está ligada a $A_{1}, A_{2}$ e $A_{3}$ e que $H_{10}$ está ligada a $B_{1}, B_{2}$ e $B_{3}$ (eventualmente podem existir mais cidades ligadas a $H_{1}$ e $H_{10}$).\n\nSe $\\{H_{1}, A_{1}, A_{2}, A_{3}\\} \\cap \\{H_{10}, B_{1}, B_{2}, B_{3}\\} \\neq \\varnothing$, então existe um caminho ligando $H_{1}$ a $H_{10}$ usando no máximo 2 estradas.\n\nSe isso não acontece e existe alguma estrada conectando uma cidade do conjunto $\\{A_{1}, A_{2}, A_{3}\\}$ a uma cidade de $\\{B_{1}, B_{2}, B_{3}\\}$, então teremos um caminho entre $H_{1}$ e $H_{10}$ utilizando 3 estradas e (i) é verificado.\n\nFinalmente, resta analisarmos a situação em que $\\{H_{1}, A_{1}, A_{2}, A_{3}\\} \\cap \\{H_{10}, B_{1}, B_{2}, B_{3}\\} = \\varnothing$ e que não existe uma estrada entre $A_{i}$ e $B_{j}$, para $i, j \\in \\{1,2,3\\}$.\n\nNesse caso, como existem apenas 10 cidades e é possível ir de uma delas a qualquer outra, um caminho que una $H_{1}$ a $H_{10}$ deve passar por uma das duas cidades que não está no conjunto\n$$\n\\{H_{1}, H_{10}, A_{1}, A_{2}, A_{3}, B_{1}, B_{2}, B_{3}\\}\n$$\nOu seja, (ii) é verificada.\n\nEm qualquer um dos casos analisados anteriormente, (i) ou (ii) é verificado.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26045,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSejam $ABCD$ e $EFGH$ quadrados de lados $33$ e $12$, com $EF$ sobre o lado $DC$ (como mostrado na figura abaixo). Seja $X$ o ponto de interseção dos segmentos $HB$ e $DC$. Suponha que $\\overline{DE} = 18$.\n![](attached_image_1.png)\n\na) Calcule o comprimento do segmento $\\overline{EX}$.\n\nb) Prove que os pontos $A, X$ e $G$ são colineares.",
    "options": [],
    "answer": "EX = 4; points A, X, and G are collinear.",
    "solution": "Solution:\na) Denote $\\overline{EX} = x$. Temos que $|\\overline{CX}| = 33 - 18 - x = 15 - x$.\n![](attached_image_2.png)\nAgora note que os triângulos $EXH$ e $CXB$ são semelhantes, logo:\n$$\n\\frac{|\\overline{EH}|}{|\\overline{CB}|} = \\frac{|\\overline{EX}|}{|\\overline{CX}|} \\Rightarrow \\frac{12}{33} = \\frac{x}{15-x}\n$$\nAgora encontramos $x$:\n$$\n\\begin{aligned}\n12(15-x) & = 33x \\\\\n4(15-x) & = 11x \\\\\n60 - 4x & = 11x \\\\\n15x & = 60 \\\\\nx & = 4\n\\end{aligned}\n$$\nPortanto, $|\\overline{EX}| = 4$.\n\nb) Seja $Y$ a interseção da reta $AG$ com o segmento $DC$. Para provar que $A, X$ e $G$ são colineares, basta mostrar que $Y = X$. E para isso, vamos calcular $|\\overline{EY}|$, e depois ver que isso é igual a $|\\overline{EX}|$. Denote $|\\overline{EY}| = y$. E, portanto, $|\\overline{FY}| = 12 - y$.\n![](attached_image_3.png)\nAnalogamente ao caso anterior, vemos que os triângulos $FYG$ e $DYA$ são semelhantes. Portanto:\n$$\n\\frac{|\\overline{FG}|}{|\\overline{DA}|} = \\frac{|\\overline{FY}|}{|\\overline{DY}|} \\quad \\Rightarrow \\frac{12}{33} = \\frac{12-y}{18+y}\n$$\nAgora encontramos $y$:\n$$\n\\begin{array}{ccc}\n12(18+y) & = & 33(12-y) \\\\\n4(18+y) & = & 11(12-y) \\\\\n72+4y & = & 132-11y \\\\\n15y & = & 60 \\\\\ny & = & 4.\n\\end{array}\n$$\nPortanto, $|\\overline{EY}| = |\\overline{EX}|$. Logo, $X = Y$, e concluímos que os pontos $A, X$ e $G$ são colineares.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26046,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSeja $ABC$ um triângulo inscrito na circunferência abaixo. Sejam também $I$ o incentro do triângulo $ABC$ e $D$ o ponto onde a reta $AI$ corta a circunferência. Mostre que $|\\overline{DB}| = |\\overline{DC}| = |\\overline{DI}|$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nComo $I$ é o incentro do triângulo $ABC$, então os segmentos $AI$, $BI$ e $CI$ são bissetrizes dos ângulos $\\hat{A}$, $\\hat{B}$ e $\\hat{C}$. Sejam então $\\alpha = \\angle BAI = \\angle CAI$ e $\\beta = \\angle ABI = \\angle CBI$.\n\n![](attached_image_2.png)\n\nComo o ângulo $\\angle BID$ é ângulo externo do triângulo $ABI$, temos que\n$$\n\\angle BID = \\angle BAI + \\angle ABI = \\alpha + \\beta\n$$\nPor outro lado, o ângulo $\\angle DBC$ está \"olhando para o arco $\\overparen{DC}$\", logo é igual ao ângulo $\\angle CAI = \\alpha$. Portanto, vale que\n$$\n\\angle IBD = \\angle IBC + \\angle CBD = \\beta + \\alpha\n$$\nAssim, segue que $\\angle IBD = \\angle BID$. Portanto, o triângulo $IBD$ é isósceles, o que implica que $|\\overline{DB}| = |\\overline{DI}|$. Analogamente, se prova que $|\\overline{DC}| = |\\overline{DI}|$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26047,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSão dadas 2017 retas separadas em três conjuntos de modo que retas em um mesmo conjunto são paralelas entre si. Qual é o maior número possível de triângulos que podemos formar com vértices nestas retas?",
    "options": [],
    "answer": "673 * 672^2",
    "solution": "Solution:\n\nSejam $a \\geq b \\geq c$ as quantidades de retas nos três conjuntos. Então $a+b+c=2017$ e o número de triângulos que podem ser formados é $a \\cdot b \\cdot c$, pois retas em um mesmo conjunto não se intersectam. Assim, queremos maximizar o produto anterior. Se $a>c+1$, podemos diminuir uma reta do conjunto com $a$ e aumentar uma reta no conjunto com $c$, mantendo assim a quantidade total de retas, mas aumentando o produto:\n$$\n\\begin{aligned}\na \\cdot b \\cdot c & < b \\cdot (a \\cdot c + a - c - 1) \\\\\n& = b \\cdot (a-1) \\cdot (c+1)\n\\end{aligned}\n$$\nPortanto, para ocorrer o produto máximo, $a=c$ ou $a=c+1$. Se tivermos $a=c$, teremos também $b=c$ e $3c=a+b+c=2017$. Como 2017 não é múltiplo de 3, isto não ocorre. Logo, $a=c+1$ e $b=c+1$ ou $b=c$. Assim, ou $a+b+c$ é $3c+1$ ou $3c+2$. O resto de 2017 por 3 é $1$ e isto nos permite concluir que $a+b+c=3c+1$. Logo, $c=672$ e $a \\cdot b \\cdot c=673 \\cdot 672^2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26048,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPedrinho escreveu dois números inteiros e positivos num pedaço de papel e mostrou para Joãozinho. Depois disso, Pedrinho calculou o dobro do produto destes dois números. Joãozinho somou 21 com o dobro do primeiro número e depois o resultado com o segundo número. Para surpresa dos dois, o resultado foi o mesmo. Quais são os possíveis números que Pedrinho escreveu no pedaço de papel?",
    "options": [],
    "answer": "(1, 23) and (6, 3)",
    "solution": "Solution:\n\nSejam $x$ e $y$ os números escritos por Pedrinho. Então o número que Pedrinho calculou foi $2 x y$, enquanto que Joãozinho calculou $21+2 x+y$. Como eles encontraram o mesmo resultado, vale que:\n$$\n2 x y=2 x+y+21\n$$\nVamos agora manipular a equação acima:\n$$\n\\begin{aligned}\n2 x y-2 x & =y+21 \\\\\n2 x(y-1) & =y+21\n\\end{aligned}\n$$\nNo segundo passo, colocamos $2 x$ em evidência. A ideia agora é fazer aparecer um $y-1$ no lado direito. Para isso basta somar e diminuir 1.\n$$\n\\begin{aligned}\n2 x(y-1) & =y-1+1+21 \\\\\n2 x(y-1) & =(y-1)+22 \\\\\n2 x(y-1)-(y-1) & =22 \\\\\n(2 x-1)(y-1) & =22\n\\end{aligned}\n$$\nNo último passo, colocamos o $(y-1)$ em evidência. Usando que $x$ e $y$ são inteiros positivos, temos que $2 x-1$ e $y-1$ são inteiros que não podem ser negativos. Portanto, temos dois números inteiros que, multiplicados, dão 22. Logo, temos que\n$$\n\\begin{aligned}\n& 2 x-1=1 \\quad \\text{e} \\quad y-1=22 \\quad \\text{ou} \\\\\n& 2 x-1=2 \\quad \\text{e} \\quad y-1=11 \\quad \\text{ou} \\\\\n& 2 x-1=11 \\quad \\text{e} \\quad y-1=2 \\quad \\text{ou} \\\\\n& 2 x-1=22 \\quad \\text{e} \\quad y-1=1\n\\end{aligned}\n$$\nMas como $2 x-1$ é sempre um número ímpar, o segundo e quarto casos acima não podem acontecer. Portanto, os números escritos por Pedrinho são 1 e 23 ou 6 e 3 .",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26049,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa cidade de Oropis existe um lago em forma de triângulo com cada um dos três lados sendo parte do perímetro de um terreno em forma de quadrado com áreas $370~\\mathrm{m}^2$, $116~\\mathrm{m}^2$ e $74~\\mathrm{m}^2$, como na primeira figura a seguir. O prefeito de Oropis, Arnaldo, deseja calcular a área do lago, mas não sabe como. O assistente do prefeito, Bernaldo, tem uma ideia. Ele diz que basta encontrar valores de $a, b, c$ e $d$ que satisfaçam as condições geométricas da segunda figura em que a área sombreada é congruente ao lago. Ele afirma que depois disto a tarefa se tornará muito mais simples.\n\n![](attached_image_1.png)\n\na) Determine valores inteiros de $a, b, c$ e $d$ que satisfaçam as condições geométricas da figura de Bernaldo.\n\nb) Determine a área do lago.",
    "options": [],
    "answer": "a=5, b=4, c=7, d=10; area of the lake = 11",
    "solution": "Solution:\n\n(a) Veja que os lados do lago ao quadrado resultam em $74$, $116$ e $370$. As condições que devem ser atendidas são as equações oriundas de três aplicações do Teorema de Pitágoras:\n$$\n\\begin{aligned}\na^2 + c^2 & = 74 \\\\\nb^2 + d^2 & = 116 \\\\\n(a+b)^2 + (c+d)^2 & = 370\n\\end{aligned}\n$$\nPodemos limitar os testes da primeira equação para números menores que $9$, pois $9^2 > 74$. Após uma análise de casos, obtemos $a=5$ e $c=7$ ou $a=7$ e $c=5$ como únicas possibilidades. Na segunda equação, também após uma análise de casos, temos $b=4$ e $d=10$ ou $b=10$ e $d=4$ e, por conseguinte, $a+b=5+4=9$ e $c+d=7+10=17$, pois $(5+10)^2 + (7+4)^2 \\neq 370$. Logo, $a=5$, $b=4$, $c=7$ e $d=10$.\n\n(b) Deste modo, a área do lago é a área do triângulo maior subtraído de dois triângulos e um retângulo. Ou seja, a área do lago, em $m^2$ é\n$$\n\\begin{aligned}\n\\frac{17 \\cdot 9}{2} - \\frac{7 \\cdot 5}{2} - \\frac{10 \\cdot 4}{2} - 7 \\cdot 4 & = \\frac{118}{2} - 20 - 28 \\\\\n& = 11\n\\end{aligned}\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26050,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPedrinho escolheu 8 números distintos entre 1 e 11 e os escreveu numa determinada ordem. Joãozinho, vendo os números que Pedrinho escreveu, notou o seguinte fato curioso: se fizermos a média dos $n$ primeiros números escritos por Pedrinho, $n=1, \\ldots, 8$, teremos como resultado sempre um número inteiro. Ou seja, se fizermos a média dos dois primeiros números, dos três primeiros, dos quatro primeiros números, e assim por diante, todas essas médias serão inteiras.\n\nQuais são as possíveis sequências de números que Pedrinho escreveu? (Dica: primeiro descubra quais são as possíveis somas para os 8 números, e depois tente descobrir de trás pra frente os números escolhidos.)",
    "options": [],
    "answer": "1,3,2,6,8,4,11,5; 3,1,2,6,8,4,11,5; 2,6,1,3,8,4,11,5; 6,2,1,3,8,4,11,5; 9,11,10,6,4,8,1,7; 11,9,10,6,4,8,1,7; 10,6,11,9,4,8,1,7; 6,10,11,9,4,8,1,7",
    "solution": "Solution:\n\nPrimeiro notamos que a média dos 8 números é simplesmente a soma deles dividida por 8. Como a média é um número inteiro, a soma tem que ser múltipla de 8. Agora, como estamos somando 8 números distintos entre 1 e 11, esta soma tem que ser no mínimo $1+2+\\ldots+8=36$ e no máximo $4+5+\\ldots+11=60$. E como tem que ser múltipla de 8, as únicas somas possíveis são $40,48,56$.\n\nVamos agora escrever a sequência como\n$$\na_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, a_{7}, a_{8}\n$$\ne dividir em casos.\n\nCaso 1) A soma é igual a 40:\n$$\na_{1}+a_{2}+\\ldots+a_{8}=40\n$$\nVamos tentar reconstruir a sequência a partir dos últimos números. Sabemos que a soma dos 7 primeiros números tem que ser múltipla de 7. Por outro lado, a soma dos 7 primeiros números é igual a 40 menos o último número, $a_{8}$ (lembre-se que 40 é a soma de todos os números). Ou seja,\n$$\na_{1}+a_{2}+\\ldots+a_{7}=40-a_{8}\n$$\nAgora, $40-a_{8}$ tem que ser múltiplo de 7, e como $1 \\leq a_{8} \\leq 11$, vale que $29 \\leq 40-a_{8} \\leq 39$. Mas o único múltiplo de 7 entre 29 e 39 é 35 e, portanto, $a_{8}=5$ e\n$$\na_{1}+a_{2}+\\ldots+a_{7}=35\n$$\nAgora repetimos o processo. A soma dos 6 primeiros números é $35-a_{7}$, e tem que ser múltipla de 6. Como antes, temos que $24 \\leq 35-a_{7} \\leq 34$, e os múltiplos de 6 entre 24 e 34 são 24 e 30. Logo $a_{7}=5$ ou $a_{7}=11$. Mas como $a_{7}$ é diferente de $a_{8}$, temos que a única possibilidade é $a_{7}=11$. Portanto,\n$$\na_{1}+a_{2}+\\ldots+a_{6}=24\n$$\nContinuando, a soma dos 5 primeiros números é $24-a_{6}$, e tem que ser múltipla de 5. Como $1 \\leq a_{6} \\leq 10$ ($a_{6}$ não pode ser 11, pois $a_{7}=11$), temos que $14 \\leq 24-a_{6} \\leq 23$, e os múltiplos de 5 entre 14 e 23 são 15 e 20. Logo, $a_{6}=4$ ou $a_{6}=9$.\n\nCaso 1.1) Caso $a_{6}=4$, ficamos com\n$$\na_{1}+a_{2}+a_{3}+a_{4}+a_{5}=20\n$$\nEntão, a soma dos quatro primeiros números é $20-a_{5}$, e vale que $10 \\leq 20-a_{5} \\leq 19$. Como os únicos múltiplos de 4 entre 10 e 19 são 12 e 16, temos que $a_{5}=4$ ou $a_{5}=8$. Mas $a_{6}$ já é igual a 4, então $a_{5}=8$, e\n$$\na_{1}+a_{2}+a_{3}+a_{4}=12\n$$\nPortanto,\n$$\na_{1}+a_{2}+a_{3}=12-a_{4}\n$$\nAgora notamos que $a_{1}+a_{2}+a_{3}$ é no mínimo $1+2+3=6$. Logo, $a_{4}=6$ ou $a_{4}=3$ (lembrando que $a_{1}+a_{2}+a_{3}$ tem que ser múltiplo de 3).\n\nSe $a_{4}=6$, então $a_{1}+a_{2}+a_{3}=6$. Logo, $a_{3}=2$, pois $6-a_{3}$ tem que ser par e $a_{3}$ não pode ser igual a $a_{6}=4$. E sobram 1 e 3 para os lugares de $a_{1}$ e $a_{2}$. Daí, temos as possíveis sequências:\n$$\n1,3,2,6,8,4,11,5 \\quad \\text{e} \\quad 3,1,2,6,8,4,11,5\n$$\nSe $a_{4}=3$, então $a_{1}+a_{2}+a_{3}=9$. Como $9-a_{3}$ tem que ser par, concluímos que $a_{3}$ tem que ser ímpar e diferente de todos os ímpares já usados na sequência (que são 3,5 e 11). Logo, $a_{3}=1$ ou $a_{3}=7$, mas se $a_{3}=7$, então $a_{1}+a_{2}=2$, o que não dá pra acontecer. Assim, $a_{3}=1$ e $a_{1}+a_{2}=8$. Agora o único par de números distintos que soma 8 que podemos ter é 2 e 6, pois 1 e 3 já foram usados. Logo, temos as possíveis sequências\n$$\n2,6,1,3,8,4,11,5 \\quad \\text{e} \\quad 6,2,1,3,8,4,11,5\n$$\n\nCaso 1.2) Se $a_{6}=9$, então\n$$\na_{1}+a_{2}+a_{3}+a_{4}+a_{5}=15\n$$\nPor outro lado, o número $5=a_{8}$ já foi usado. Então, $a_{1}+a_{2}+a_{3}+a_{4}+a_{5}$ é no mínimo $1+2+3+4+6=16$, logo $a_{6}$ não pode ser 9.\n\nCaso 2) A soma é igual a 48.\nComo no caso 1, vemos que a soma dos sete primeiros números é $48-a_{8}$. Como $37 \\leq 48-a_{8} \\leq 47$, temos que $a_{8}=6$ (o único múltiplo de 7 entre 37 e 47 é o 42). Logo,\n$$\na_{1}+a_{2}+\\ldots+a_{7}=42\n$$\nContinuando, a soma dos 6 primeiros números é $42-a_{7}$, e temos que $31 \\leq 42-a_{7} \\leq 41$. O único múltiplo de 6 entre 31 e 41 é o 36, logo $a_{7}=6$, o que não pode acontecer, por $a_{8}=6$. Então não existe nenhuma sequência com soma 48.\n\nCaso 3) A soma é igual a 56.\nComo nos casos anteriores, a soma dos sete primeiros números é $56-a_{8}$. Como $45 \\leq 56-a_{8} \\leq 55$, temos que $a_{8}=7$ (o único múltiplo de 7 entre 45 e 55 é o 49). Logo,\n$$\na_{1}+a_{2}+\\ldots+a_{7}=49\n$$\nContinuando, a soma dos seis primeiros números é $49-a_{7}$. Daí, temos que $38 \\leq 49-a_{7} \\leq 48$, e os únicos múltiplos de 6 entre 38 e 48 são 42 e 48. Como $a_{7}$ não pode ser igual a $a_{8}=7$, temos que $a_{7}=1$ e\n$$\na_{1}+a_{2}+\\ldots+a_{6}=48\n$$\nPortanto, a soma dos cinco primeiros números é $48-a_{6}$. Logo, $a_{6}=3$ ou $a_{6}=8$ (os únicos múltiplos de 5 entre 37 e 46 são 40 e 45).\n\nCaso 3.1) Caso $a_{6}=3$, então\n$$\na_{1}+a_{2}+a_{3}+a_{4}+a_{5}=45\n$$\nMas como $7=a_{8}$, a maior soma possível destes 5 números é $6+8+9+10+11=44$, logo $a_{6}$ não pode ser 3.\n\nCaso 3.2) Caso $a_{6}=8$, então\n$$\na_{1}+a_{2}+a_{3}+a_{4}+a_{5}=40\n$$\nContinuando com o processo anterior, vemos que a soma dos quatro primeiros números é $40-a_{5}$ e os únicos múltiplos de 4 entre 29 e 38 são 32, 36. Como $a_{5}$ não pode ser igual $8=a_{6}$, então $a_{5}=4$. Portanto,\n$$\na_{1}+a_{2}+a_{3}+a_{4}=36\n$$\nLogo, $36-a_{4}$ é múltiplo de 3 e, portanto, $a_{4}$ pode ser igual a 3,6 e 9. Mas como a soma $a_{1}+a_{2}+a_{3}$ é no máximo $9+10+11=30$, temos que $a_{4}$ não pode ser 3. Logo se $a_{4}=6$, então $a_{1}+a_{2}+a_{3}=30$. Portanto, $a_{1}, a_{2}$ e $a_{3}$ têm que ser os números 9, 10 e 11 em alguma ordem. Como $a_{1}+a_{2}=30-a_{3}$, $a_{3}$ tem que ser par, temos que $a_{3}=10$. Logo, temos as sequências\n$$\n9,11,10,6,4,8,1,7 \\quad \\text{e} \\quad 11,9,10,6,4,8,1,7\n$$\nAgora, se $a_{4}=9$, então $a_{1}+a_{2}+a_{3}=27$, e como 7,8 e 9 já foram usados, $a_{1}+a_{2}+a_{3}$ é no máximo $6+10+11=27$. Portanto, $a_{1}, a_{2}$ e $a_{3}$ têm que ser os números $6,10$ e 11 em alguma ordem. Como $30-a_{3}$ tem que ser par, então $a_{3}=11$. Logo, temos as sequências\n$$\n10,6,11,9,4,8,1,7 \\quad \\text{e} \\quad 6,10,11,9,4,8,1,7\n$$\n\nConcluindo, todas as sequências possíveis escritas por Pedrinho são\n$$\n\\begin{array}{rll}\n1,3,2,6,8,4,11,5 ; & 3,1,2,6,8,4,11,5 ; & 2,6,1,3,8,4,11,5 \\\\\n6,2,1,3,8,4,11,5 ; & 9,11,10,6,4,8,1,7 ; & 11,9,10,6,4,8,1,7 \\\\\n10,6,11,9,4,8,1,7 ; & 6,10,11,9,4,8,1,7\n\\end{array}\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26051,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSeja $n$ um inteiro positivo.\na) Um quadrado de lado $n$ é dividido em $n^{2}$ quadradinhos de lados unitários por retas paralelas aos seus lados. Determine o número de retângulos cujos vértices são vértices de quadradinhos e que possuem lados paralelos aos lados do quadrado original.\nb) Três quadrados de lado $n$ são arranjados como na figura a seguir e cada um deles é dividido em $n^{2}$ quadrados de lados unitários por retas paralelas aos seus lados. Determine o número de retângulos cujos vértices são vértices de quadradinhos e que possuem lados paralelos aos lados dos três quadrados originais.\n![](attached_image_1.png)",
    "options": [],
    "answer": "a) ((n(n+1))/2)^2. b) n^2(2n+1)^2 − n^4 − n^3(n+1) − (n(n+1)/2)^2.",
    "solution": "Solution:\n\na) Os vértices dos retângulos são unicamente determinados pelas interseções de duas das $n+1$ retas verticais e paralelas com duas das $n+1$ retas horizontais paralelas aos lados do quadrado que delimitam os quadradinhos unitários. Podemos escolher a primeira reta vertical de $n+1$ maneiras e a segunda de $n+1-1=n$ maneiras. Além disto, escolher uma reta $r$ e depois uma reta $s$ é o mesmo que escolher $s$ e depois $r$. Assim, o total de escolhas de pares de retas verticais é $\\frac{n \\cdot (n+1)}{2}$. De forma semelhante, o total de escolhas de retas horizontais é $\\frac{n(n+1)}{2}$. Portanto, o número procurado de retângulos é\n$$\n\\left(\\frac{n \\cdot (n+1)}{2}\\right)^{2}\n$$\n\nb) Completemos a figura dada de modo a obtermos um quadrado de lado $2n$ subdividido em $(2n)^{2}$ quadradinhos de lados unitários. Já sabemos que em tal configuração existem\n$$\n\\left(\\frac{2n(2n+1)}{2}\\right)^{2} = n^{2} \\cdot (2n+1)^{2}\n$$\nretângulos com vértices nos quadradinhos unitários e lados paralelos aos lados do quadrado. Queremos remover desta contagem, retângulos que tenham pelo menos um vértice na nova região inserida na figura dada. Estes retângulos indesejados podem possuir um, dois ou quatro vértices dentro do quadrado de lado $n$ acrescentado à figura. Vejamos a quantidade de tais retângulos em cada caso:\n\ni) Para contar os retângulos com quatro vértices no novo quadrado, basta apelar novamente para o item a) e obter\n$$\n\\left(\\frac{n \\cdot (n+1)}{2}\\right)^{2}\n$$\nretângulos.\n\nii) Para contar os retângulos com dois vértices no novo quadrado, note que eles ou estarão em uma reta vertical ou horizontal. Suponha inicialmente que os dois vértices na região nova estão numa reta vertical. Podemos escolhê-la de $n$ formas, correspondendo exatamente às $n$ últimas retas verticais da direita. A outra reta vertical que contém os outros vértices fora da nova região, pode ser escolhida de $n$ formas, correspondendo exatamente às $n$ primeiras retas verticais da esquerda. Assim, temos $n \\cdot n = n^{2}$ escolhas para as retas verticais. Para escolhermos as retas horizontais, como estamos supondo que existem exatamente dois vértices na região nova, eles devem ser escolhidos entre as $n+1$ primeiras retas horizontais de cima. Como a ordem de escolha delas não importa, temos $\\frac{n \\cdot (n+1)}{2}$ possíveis escolhas de retas horizontais. A outra possibilidade, em que os dois vértices estão numa reta horizontal, é análoga e basta apenas dobrar a contagem que fizemos anteriormente para incluí-los. Consequentemente, o total de retângulos indesejados neste caso é\n$$\n2 \\cdot n \\cdot n \\cdot \\frac{n \\cdot (n+1)}{2} = n^{3}(n+1)\n$$\n\niii) Finalmente, para contar os retângulos com apenas um vértice na nova região, note que ele é a interseção de uma das $n$ últimas retas verticais com uma das $n$ primeiras retas horizontais. Portanto, podemos escolher estas retas de $n \\cdot n = n^{2}$ maneiras. Para escolher as outras retas que determinarão os outros vértices do retângulo, devemos escolher uma reta vertical dentre as $n$ primeiras e uma reta horizontal dentre as $n$ últimas. Portanto, o total de retângulos neste caso é\n$$\nn \\cdot n \\cdot n^{2} = n^{4}\n$$\n\nPara finalizar a contagem, basta eliminarmos os retângulos indesejados que acabamos de calcular e assim obtemos como resposta o número:\n$$\nn^{2} \\cdot (2n+1)^{2} - n^{4} - n^{3}(n+1) - \\left(\\frac{n(n+1)}{2}\\right)^{2}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26052,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEm um certo país há 21 cidades e o governo pretende construir $n$ estradas (todas de mão dupla), sendo que cada estrada liga exatamente 2 das cidades do país. Qual o menor valor de $n$ para que, independente de como as estradas sejam construídas, seja possível viajar entre quaisquer 2 cidades (passando, possivelmente, por cidades intermediárias)?",
    "options": [],
    "answer": "191",
    "solution": "Solution:\n\nDiremos que um grupo de cidades é conexo se é possível viajar de uma delas para qualquer outra por meio de estradas. Naturalmente grupos de cidades conexas distintos não podem possuir cidades em comum. Se o governo construir todas as estradas de um grupo de 20 cidades e deixar uma cidade isolada sem estradas, a propriedade de todas as cidades pertencerem a um mesmo grupo conexo não será estabelecida. Portanto, $n > \\binom{20}{2} = 190$. Mostraremos agora que se o governo construir 191 estradas todas as cidades farão parte de um mesmo grupo conexo. De fato, suponha que exista mais de um grupo conexo e seja $k$, com $1 \\leq k \\leq 20$ o número de vértices de um deles, que chamaremos de $A$. Recordamos que se ligarmos todas as estradas possíveis em um grupo de $n$ cidades, obteremos $\\frac{n(n-1)}{2}$ estradas. Como não devem existir estradas entre o grupo $A$ e os demais grupos que contém as outras $21-k$ cidades, o total de estradas não é maior que\n$$\n\\begin{aligned}\n\\frac{k(k-1)}{2} + \\frac{(21-k)(20-k)}{2} & = \\\\\nk^{2} - 21k + 210 & = \\\\\n(k-21/2)^{2} + 399/4 & \\leq \\\\\n(19/2)^{2} + 399/4 & = 190\n\\end{aligned}\n$$\nIsso é um absurdo, pois o governo construiu 191 estradas. Assim, o menor valor de $n$ é 191.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26053,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA figure shows a grid formed by squares of $1~\\mathrm{cm}$ side. What is the ratio between the shaded area and the non-shaded area?\n(a) $\\frac{1}{4}$\n(b) $\\frac{1}{5}$\n(c) $\\frac{1}{6}$\n(d) $\\frac{2}{5}$\n(e) $\\frac{2}{7}$\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "(a)",
    "solution": "Solution:\nThe correct option is (a).\nThe grid is a square with side equal to $5~\\mathrm{cm}$, so its area is $25~\\mathrm{cm}^2$. The shaded part of the grid is formed by four triangles, two of which have base $1~\\mathrm{cm}$ and height $2~\\mathrm{cm}$, and the other two have base $1~\\mathrm{cm}$ and height $3~\\mathrm{cm}$. Therefore, the shaded area is $2 \\times \\frac{1}{2}(1 \\times 2) + 2 \\times \\frac{1}{2}(1 \\times 3) = 5~\\mathrm{cm}^2$ and the non-shaded area is $25 - 5 = 20~\\mathrm{cm}^2$. Thus, the required ratio is $5 / 20 = 1 / 4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26054,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm grupo de rapazes e moças saiu para comer pizza em dois dias consecutivos. No restaurante em que foram, as pizzas são cortadas em doze pedaços iguais. Maria observou que no primeiro dia cada rapaz comeu 7 pedaços, e cada moça 3 pedaços. Já no segundo dia, cada rapaz comeu 6 pedaços e cada moça 2 pedaços. Curiosamente, em ambos os dias eles pediram quatro pizzas que foram totalmente consumidas e depois pediram mais uma, da qual sobraram alguns pedaços (ou seja, foi comido pelo menos um pedaço e sobrou pelo menos um pedaço). Quantos rapazes e moças foram à pizzaria?",
    "options": [],
    "answer": "8 boys and 1 girl",
    "solution": "Solution:\n\nSejam $x$ e $y$ o número de rapazes e moças, respectivamente. Sabemos que o número total de pedaços consumidos foi no mínimo $49$ (4 pizzas e um pedaço da última pizza) e no máximo $59$ (4 pizzas mais 11 pedaços, lembre que sobrou pelo menos um pedaço da última pizza). Por outro lado,\n$$\n\\begin{aligned}\n& 7x + 3y \\leq 59 \\\\\n& 6x + 2y \\geq 49\n\\end{aligned}\n$$\nMultiplicando a última equação por $-1$, temos que trocar a desigualdade, ficando com\n$$\n\\begin{aligned}\n7x + 3y &\\leq 59 \\\\\n-6x - 2y &\\leq -49\n\\end{aligned}\n$$\nE somando as duas desigualdades, chegamos a\n$$\nx + y \\leq 10\n$$\nSubstituindo na segunda desigualdade, ficamos com\n$$\n4x + 10 + 10 \\geq 4x + (x + y) + (x + y) = 6x + 2y \\geq 49\n$$\nLogo, $4x \\geq 29$, e portanto $x \\geq 8$. Por outro lado, da primeira equação temos que\n$$\n7x \\leq 7x + 3y \\leq 59\n$$\nDaí, $7x \\leq 59$, que implica $x \\leq 8$. Portanto, $x = 8$ e, substituindo, ficamos com\n$$\n3y \\leq 3 \\quad \\text{e} \\quad 2y \\geq 1\n$$\nIsso nos dá, $y = 1$. Portanto, foram $8$ rapazes e $1$ moça à pizzaria.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26055,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEm um torneio de xadrez, cada um dos participantes jogou exatamente uma vez com cada um dos demais e não houve empates. Mostre que existe um jogador $P$ tal que, para qualquer outro jogador $Q$, distinto de $P$, uma das situações a seguir ocorre:\n\ni) $Q$ perdeu de $P$;\nii) $Q$ perdeu de alguém que perdeu de $P$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSeja $P$ o jogador que mais venceu partidas no torneio. Digamos que $P$ tenha vencido os jogadores do conjunto $S=\\{P_{1}, P_{2}, \\ldots, P_{k}\\}$. Considere um jogador qualquer $Q$ diferente de $P$. Se $Q$ perdeu para $P$, ele satisfaz o item $i$).\n\nSe além de vencer $P$, $Q$ também ganhou de todos os elementos de $S$, então ele terá mais vitórias que $P$. Esse absurdo mostra que se $Q$ tiver ganho de $P$, então ele perdeu para alguém de $S$ e assim $ii$) é verdadeira. Em qualquer caso, $i$) ou $ii$) é satisfeita.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26056,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nConsidere cinco números reais positivos ordenados por $0<a \\leq b \\leq c \\leq d \\leq e$. Sabe-se que sempre que tiramos um destes números, podemos separar os outros quatro em dois grupos tais que a soma dos números de um grupo é igual à soma dos números do outro grupo. Se uma sequência $(a, b, c, d, e)$ satisfaz esta condição, dizemos que ela é quase-equilibrada. Existem sequências que atendem a uma condição mais restrita: se retirarmos um número podemos separar os quatro números restantes em dois grupos com a mesma quantidade de números tais que a soma dos números de um grupo é igual à soma dos números do outro grupo. Se uma sequência de números reais positivos $(a, b, c, d, e)$ satisfaz esta condição mais restrita, dizemos que essa sequência é equilibrada.\n\na) Mostre um exemplo de cinco números reais positivos ordenados por $0<a \\leq b \\leq c \\leq d \\leq e$, com $a<e$, que formam uma sequência quase-equilibrada. Veja que podemos fazer alguns deles iguais se isto for conveniente.\n\nb) Se uma sequência equilibrada possui três termos iguais, mostre que os cinco números são obrigatoriamente iguais.\n\nc) Considere uma sequência equilibrada $(a, b, c, d, e)$ com $0<a \\leq b \\leq c \\leq d \\leq e$. Sabe-se ainda que $(e+c)=(b+d)$ e que $(e+a)=(c+d)$. Prove que os cinco números são iguais.",
    "options": [],
    "answer": "a) An example is 1, 1, 1, 3, 3.\nb) If three terms are equal in a balanced sequence, then all five terms are equal.\nc) Under the stated conditions, all five numbers are equal.",
    "solution": "Solution:\n\n(a) Considere os números $0<1 \\leq 1 \\leq 1 \\leq 3 \\leq 3$. Veja que se tirarmos um dos números iguais a $1$, podemos separar os restantes em dois grupos iguais a $(1,3)$. Se tirarmos um $3$, então podemos separar os restantes nos grupos $(1,1,1)$ e $(3)$.\n\n(b) Suponha que temos três números iguais a $x$ e dois restantes $y$ e $z$. Se retirarmos o $z$, teremos $x+x=x+y$, ou seja, $x=y$. Se separarmos o $y$, teremos $x+x=x+z$, ou seja, $x=z$. Concluímos assim que os cinco números são obrigatoriamente iguais a $x$.\n\n(c) Veja que $e \\geq d$ e $c \\geq b$, logo:\n$$\nb+d=e+c \\geq d+b\n$$\nAssim, segue que $e=d$ e $b=c$, pois caso contrário chegaríamos a uma contradição. Como $e=d$ e $e+a=c+d$, temos $a=c$. Deste modo, $a=b=c$ e, usando o item anterior, podemos concluir que os cinco números são iguais.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26057,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPedrinho está brincando de fazer arranjos com palitos. Ele dispõe seus palitos formando triângulos equiláteros, como mostra a figura abaixo:\n![](attached_image_1.png)\nPedrinho quer pintar cada palito de seu arranjo de tal forma que cada triângulo tenha seus lados pintados de exatamente duas cores diferentes. Para isso, ele dispõe de tintas vermelha, azul e preta. De quantos modos ele pode pintar o arranjo?",
    "options": [],
    "answer": "46656",
    "solution": "Solution:\n\nComeçamos pintando os palitos de dentro do hexágono. Cada um desses palitos pode ser pintado com 3 cores distintas. Temos então $3^{6}$ maneiras de pintar estes palitos interiores.\n\nNotamos agora que cada um dos seis triângulos formados podem ser completados de duas maneiras diferentes, independentemente das cores dos dois palitos pintados. Se pintamos os dois palitos que formam este triângulo de uma mesma cor, digamos azul, então o palito que sobra pode ser pintado de vermelho ou preto. Por outro lado, se pintamos os dois palitos de cores diferentes, digamos azul e vermelho, então o palito que sobra só pode ser pintado de azul ou vermelho. Logo, o número total de maneiras de pintar o arranjo é igual a $3^{6} \\times 2^{6} = 46656$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26058,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSejam $O$ e $H$ o circuncentro e o ortocentro do triângulo $\\triangle ABC$, respectivamente. Sejam $O_1$, $O_2$ tais que $AC$ é mediatriz de $OO_1$ e $BC$ é mediatriz de $OO_2$, respectivamente.\n\na) Verifique que $\\angle BAH = \\angle OAC$.\n\nb) Se $M$ é o ponto médio de $BC$, mostre que $OM = \\frac{AH}{2}$.\n\nc) Encontre o ângulo $\\angle O_1 H O_2$ sabendo que $\\angle BAC = 60^\\circ$ e $\\angle ABC = 80^\\circ$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "80°",
    "solution": "Solution:\n\na) Como o triângulo $ABF$ é retângulo, temos $\\angle BAH = 90^\\circ - \\angle ABC$. No triângulo isósceles $AOC$, o ângulo da base $\\angle OAC$ mede $\\frac{180^\\circ - \\angle AOC}{2} = 90^\\circ - \\angle ABC$. Portanto, $\\angle BAH = \\angle OAC$.\n\nb) Sejam $D$ e $G$ as interseções, diferentes de $A$, de $AO$ e $AF$, respectivamente, com o circuncírculo do triângulo $ABC$. Como $\\angle BAG = \\angle DAC$ e\n$$\n\\begin{aligned}\n\\angle BAD & = \\angle BAG + \\angle GAD \\\\\n& = \\angle GAD + \\angle DAC \\\\\n& = \\angle GAC\n\\end{aligned}\n$$\nsegue que $BG = DC$ e $BD = GC$. Além disto, $\\angle GBC = \\angle GAC = \\angle BAD = \\angle BCD$ e, pelo caso de congruência $LAL$, segue que $\\triangle BGC \\equiv \\triangle CDB$. Consequentemente, $BG = CD$. Temos também $\\angle EBC = 90^\\circ - \\angle BCA = \\angle GAC = \\angle GBC$. Assim, pelo caso de congruência $ALA$, dado que $\\angle HFB = \\angle BFG$ e $\\angle HBF = \\angle GBF$, temos $\\triangle BHF \\equiv \\triangle BGF$ e, como consequência, $BH = BG$. Observando os triângulos $BHM$ e $MCD$, temos\n$$\n\\begin{aligned}\nBH & = CD \\\\\nBM & = MC \\\\\n\\angle HBM = 90^\\circ - \\angle BCA & = \\angle ACD - \\angle BCA = \\angle MCD\n\\end{aligned}\n$$\nNovamente pelo caso de congruência $LAL$, podemos concluir que os triângulos $BHM$ e $CMD$ são congruentes. Isto implica que:\n$$\n\\begin{aligned}\n\\angle HMB & = \\angle CMD \\\\\nHM & = MD\n\\end{aligned}\n$$\nAssim, $H$, $M$ e $D$ são colineares e $OM$ é base média do triângulo $AHD$. Portanto, $OM = \\frac{AH}{2}$.\n\nc) Como $BC$ é mediatriz de $OO_2$, segue que $OO_2 = 2 \\cdot OM = AH$. Além disto, como $OM$ e $AH$ são perpendiculares a $BC$, segue que $AH$ e $OO_2$ são paralelos e congruentes. Consequentemente $AHO_2O$ é um paralelogramo e $AO \\parallel HO_2$. De modo semelhante, $HOO_1OB$ é um paralelogramo e $HO_1 \\parallel BO$. Assim,\n$$\n\\begin{aligned}\n\\angle O_1 H O_2 & = \\angle AOB \\\\\n& = 2 \\cdot \\angle ACB \\\\\n& = 2 \\cdot (180^\\circ - \\angle BAC - \\angle ABC) \\\\\n& = 80^\\circ\n\\end{aligned}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26059,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\na) Num triângulo de lados $a$, $b$ e $c$, vale sempre que a soma de dois lados é maior do que o terceiro lado. Por exemplo, no triângulo a seguir, de lados $a$, $b$ e $c$,\n![](attached_image_1.png)\nvale a desigualdade $a < b + c$. Além disso, valem outras duas desigualdades. Quais são?\n\nb) Na figura abaixo pode-se observar um retângulo cujo lado menor mede 3 e cujo lado maior mede 8.\n![](attached_image_2.png)\nSuponha que $a \\neq b$. Mostre que $a + b > 10$. Sugestão: copie um retângulo igual ao desenhado em cima dele!\n\nc) Em cada lado de um quadrado é escolhido um ponto. Em seguida, estes pontos são ligados formando um quadrilátero, conforme mostrado na figura abaixo:\n![](attached_image_3.png)\nMostre que o perímetro deste quadrilátero (soma dos comprimentos dos lados) é maior ou igual a duas vezes o comprimento da diagonal do quadrado.\nSugestão: desenhe vários quadrados iguais ao quadrado dado!",
    "options": [],
    "answer": "b < a + c and c < a + b; a + b > 10; and the quadrilateral’s perimeter is at least twice the square’s diagonal.",
    "solution": "Solution:\na) As outras desigualdades são $b < a + c$ e $c < a + b$.\n\nb) Conforme a sugestão, desenhamos um retângulo idêntico em cima do retângulo original, com o segmento de comprimento $b$ refletido:\n![](attached_image_4.png)\nEm seguida, traçamos o segmento tracejado abaixo:\n![](attached_image_5.png)\nAplicando o Teorema de Pitágoras à diagonal, temos que $d^2 = 6^2 + 8^2$. Portanto, $d^2 = 100$, e daí $d = 10$. Aplicando a desigualdade triangular ao triângulo cujos lados são $a$, $b$ e a diagonal, obtemos $a + b > 10$.\n\nc) Vamos desenhar quatro quadrados da seguinte maneira:\n![](attached_image_6.png)\nEm seguida, vamos copiar os segmentos de comprimentos $a$, $b$, $c$ e $d$ da seguinte maneira:\n![](attached_image_7.png)\nObserve que os segmentos de comprimento $b$ e $d$ foram refletidos pelo eixo vertical. Como a menor distância entre dois pontos é dado pelo comprimento do segmento de reta que liga os dois pontos, temos que $a + b + c + d$ é maior ou igual a duas vezes o comprimento da diagonal do quadrado.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26060,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDiogo recortou uma cruz de cartolina como mostrado abaixo. Nesta cruz, todos os lados têm comprimento igual a $1~\\mathrm{cm}$, e todos os ângulos são retos. Fernanda desafiou Diogo a fazer dois cortes em linha reta nesta cruz, de modo a formar quatro peças que possam ser reencaixadas de modo a formar um quadrado.\n\n![](attached_image_1.png)\n\na) Qual será a área do quadrado obtido?\n\nb) Qual será o lado do quadrado obtido?\n\nc) Mostre como Diogo pode fazer estes dois cortes em linha reta de modo a poder formar o quadrado com as quatro peças obtidas pelo corte!",
    "options": [],
    "answer": "Area: 5 square centimeters; side length: square root of 5 centimeters.",
    "solution": "Solution:\n\na) A área do quadrado obtido deve ser a mesma da cruz de cartolina. Como a cruz de cartolina pode ser dividida em cinco quadrados de lados iguais a um, sua área será igual a $5~\\mathrm{cm}^2$.\n\nb) Para que a área do quadrado obtido seja igual a $5~\\mathrm{cm}^2$, é necessário que seu lado seja igual a $\\sqrt{5}~\\mathrm{cm}$.\n\nc) Os dois cortes que devem ser feitos são:\n\n![](attached_image_2.png)",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26061,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm barco motorizado solta uma boia em um rio de margens retilíneas e paralelas às 10:00 e começa a navegar, na direção determinada pelo rio, contra a correnteza até às 10:15. Depois disto, ele retorna, também na direção determinada pelo rio. Em que instante o barco encontrará novamente a boia?",
    "options": [],
    "answer": "10:30",
    "solution": "Solution:\n\nComo tanto o barco quanto a boia vão estar sujeitos aos mesmos efeitos da correnteza do rio, para efeitos práticos, podemos considerar apenas a velocidade relativa do barco em relação à boia e supor que a correnteza é nula. Neste caso, se o barco levou 15 minutos para ir em correnteza parada, ele também levará 15 para voltar e assim ele encontrará a boia às 10:30.\n\nPodemos também resolver o problema analisando o sistema de equações produzido pelas informações do enunciado. Sejam $c$ a velocidade da correnteza e $v$ a velocidade do barco, ambos medidos em quilômetros por hora. Seja ainda $x$ o tempo que leva para o barco encontrar a boia desde o momento em que ele a solta no rio. Como o barco navega contra a correnteza no primeiro um quarto de hora, a distância que ele percorre é $\\frac{v}{4}-\\frac{c}{4}$, pois o motor do barco produz o deslocamento de $\\frac{v}{4}$ e a correnteza do rio o faz retroceder $\\frac{c}{4}$. No movimento de volta, que dura $x-\\frac{1}{4}$ horas, o barco percorre, agora no sentido da correnteza, $\\left(x-\\frac{1}{4}\\right) v+\\left(x-\\frac{1}{4}\\right) c$, onde a primeira parcela é a contribuição do motor do barco no deslocamento e a segunda a da correnteza. Durante o movimento de ida e volta do barco, a boia foi deslocada pela correnteza por uma distância de $x c$ quilômetros, portanto, este valor corresponde à diferença entre as distâncias de ida e volta:\n$$\n\\begin{aligned}\n\\frac{v}{4}-\\frac{c}{4}+x c & =\\left(x-\\frac{1}{4}\\right) v+\\left(x-\\frac{1}{4}\\right) c \\\\\nx v & =\\frac{v}{2} \\\\\nx & =\\frac{1}{2}\n\\end{aligned}\n$$\nAssim, o barco gastou 1/2 de uma hora, ou seja, 30 minutos, para encontrar a boia.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26062,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSeja $S(n)$ a soma dos dígitos de um inteiro $n$. Por exemplo, $S(327)=3+2+7=12$. Encontre o valor de\n$$\nA=S(1)-S(2)+S(3)-S(4)+\\ldots-S(2016)+S(2017)\n$$",
    "options": [],
    "answer": "1009",
    "solution": "Solution:\nSe $m$ é par, o número $m+1$ possui os mesmos dígitos que $m$ com exceção do dígito das unidades, que é uma unidade maior. Portanto, $S(m+1)-S(m)=1$. Isso nos permite agrupar os termos da sequência em pares com diferença igual a 1 :\n$$\n\\begin{array}{r}\nS(1)-S(2)+S(3)-S(4)+\\ldots-S(2016)+S(2017)= \\\\\nS(1)+(S(3)-S(2))+(S(5)-S(4))+\\ldots+(S(2017)-S(2016))= \\\\\n1+1+1+\\ldots+1= \\\\\n1009\n\\end{array}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26063,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNo ponto de ônibus - Um grupo de meninos e meninas aguarda em um ponto pelo ônibus. No primeiro ônibus que passa embarcam somente 15 meninas, e ficam 2 meninos para cada menina no ponto de ônibus. No segundo ônibus que passa, embarcam somente 45 meninos, e ficam 5 meninas para cada menino no ponto de ônibus. Determine o número de meninos e meninas que estavam no ponto antes da parada do primeiro ônibus.",
    "options": [],
    "answer": "50 boys and 40 girls",
    "solution": "Solution:\n\nVamos representar por $M$ o número de meninas e por $H$ o número de meninos que estavam no ponto antes da parada do primeiro ônibus. Depois do embarque das 15 meninas no primeiro ônibus, ficaram no ponto $M-15$ meninas e $H$ meninos. Uma vez que, neste momento, ficam no ponto 2 meninos para cada menina, temos: $H=2(M-15)$.\n\nNo segundo ônibus embarcam 45 meninos, e ficaram no ponto $M-15$ meninas e $H-45$ meninos. Como, neste momento, ficaram no ponto 5 meninas para cada menino, temos: $M-15=5(H-45)$.\n\nDeste modo, obtemos o sistema linear\n$$\n\\left\\{\\begin{array}{l}\nH=2(M-15) \\\\\nM-15=5(H-45)\n\\end{array}\\right.\n$$\nSubstituindo a primeira equação na segunda obtemos: $M-15=5(2M-30-45)$. Logo:\n$$\n375-15=10M-M \\Rightarrow M=40\n$$\ne $H=2(40-15)=50$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26064,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMistério das bolas - Henrique têm duas urnas. A primeira urna contém somente bolas pretas e a segunda somente bolas brancas. Henrique retirou um número de bolas da primeira urna e as colocou na segunda. Em seguida, retirou o mesmo número de bolas da segunda urna e as colocou na primeira. Depois disso o número de bolas brancas na primeira urna é maior, menor ou igual ao número de bolas pretas na segunda urna?",
    "options": [],
    "answer": "equal",
    "solution": "Solution:\n\nSeja $m$ o número de bolas pretas na primeira urna e $n$ o de bolas brancas na segunda urna. Inicialmente, Henrique retirou $k$ bolas pretas da primeira urna e as colocou na segunda urna. Nesse ponto a situação é a seguinte:\n- na 1a urna temos:\n$$\n\\underbrace{m-k}_{\\text{pretas}}\n$$\n- na $2^{\\mathrm{a}}$ urna temos:\n$$\n\\underbrace{n}_{\\text{brancas}}+\\underbrace{k}_{\\text{pretas}}\n$$\nDepois, ele retirou $k$ bolas da segunda urna e as colocou na primeira urna. Agora esse grupo de $k$ bolas pode ter bolas brancas e pretas. Assim chamemos de $p$ o número de bolas pretas e de $b$ o de bolas brancas retiradas da $2^{\\mathrm{a}}$ urna, e logo $k=b+p$. Temos então:\n- na 1a urna temos: $\\underbrace{m-k}_{\\text{pretas}}+\\underbrace{p}_{\\text{pretas}}+\\underbrace{b}_{\\text{brancas}}=\\underbrace{m-k+p}_{\\text{pretas}}+\\underbrace{b}_{\\text{brancas}}$\n- na 2a urna temos:\n$$\n\\underbrace{n}_{\\text{brancas}}+\\underbrace{k}_{\\text{pretas}}-\\underbrace{b}_{\\text{brancas}}-\\underbrace{p}_{\\text{pretas}}=\\underbrace{n-b}_{\\text{brancas}}+\\underbrace{k-p}_{\\text{pretas}}\n$$\nAssim, ele ficou com $b$ bolas brancas na primeira urna e $k-p$ bolas pretas na segunda urna. Mas, $k=p+b$, ou seja, $b=k-p$. Logo, o número de bolas brancas na primeira urna é igual ao número de bolas pretas na segunda urna.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26065,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO preço da gasolina - Em 1972 encher o tanque de gasolina de um carro pequeno custava $R\\$ 29,90$, e em 1992, custava $\\$ 149,70$ para encher o mesmo tanque. Qual dos valores abaixo melhor aproxima o percentual de aumento no preço da gasolina nesse período de 20 anos?\n(a) $20\\%$\n(b) $125\\%$\n(d) $300\\%$\n(d) $400\\%$\n(e) $500\\%$",
    "options": [],
    "answer": "(d)",
    "solution": "Solution:\n\nO aumento do valor foi\n$$\n149,70 - 29,90 = 119,80 \\text{ reais }\n$$\nque corresponde a:\n$$\n\\frac{119,80}{29,90} \\times 100\\% = 400,66\\%\n$$\nA opção correta é (d).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26066,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm queijo triangular - Osvaldo comprou um queijo em forma de um triângulo equilátero. Ele quer dividir o queijo igualmente entre ele e seus quatro primos. Faça um desenho indicando como ele deve fazer essa divisão.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nUm queijo triangular - Para dividir o queijo em 5 partes iguais, é suficiente dividi-lo em $5k$ partes iguais e dar $k$ partes a cada um. Uma forma de fazer essa partição, é mostrada na figura, onde o queijo foi partido em $25 = 5 \\times 5$ triângulos.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26067,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAumento de preço - Se o preço de um produto subiu de $R\\$ 5,00$ para $R\\$ 5,55$, qual foi a taxa percentual de aumento?",
    "options": [],
    "answer": "11%",
    "solution": "Solution:\n\nO aumento em reais foi $5,55 - 5 = 0,55$; então o percentual de aumento foi\n$$\n\\frac{0,55}{5} = \\frac{0,55 \\times 20}{5 \\times 20} = \\frac{11}{100} = 11\\%.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26068,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTortas da vovó - Sofia foi levar uns docinhos para sua avó; são 7 docinhos de amora, 6 de côco e 3 de chocolate. Durante o caminho, a gulosa Sofia come 2 docinhos. Qual das situações abaixo é possível?\n\n(A) Vovó não recebeu docinhos de chocolate.\n(B) Vovó recebeu menos docinhos de côco do que de chocolate.\n(C) Vovó recebeu o mesmo número de docinhos de cada uma das 3 variedades.\n(D) Existem 2 variedades de docinhos das quais vovó recebeu o mesmo número.\n(E) O número de docinhos de amora que vovó recebeu é maior que o dos outros 2 somados.",
    "options": [],
    "answer": "D",
    "solution": "Solution:\n\nVamos examinar cada uma das situações propostas. Lembre que no final vovó recebeu $7+6+3-2=14$ docinhos.\n\n(A) Impossível porque ela recebeu no mínimo $3-2=1$ docinho de chocolate.\n\n(B) Impossível porque ela recebeu no mínimo $6-2=4$ docinhos de côco.\n\n(C) Impossível porque $7-2=5>3$.\n\n(D) Possível porque Sofia pode ter comido 1 docinho de amora e 1 de chocolate, restando para vovó: 6 de amora, 6 de côco e 2 de chocolate.\n\n(E) Impossível porque 7 não é maior do que $6+2-3$.\n\nLogo, a única situação possível é (D).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26069,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nBolas brancas e pretas - Uma caixa tem exatamente $100$ bolas pretas e $100$ bolas brancas. Repetidamente, $3$ bolas são retiradas da caixa e substituídas por outras bolas que estão em um saco da seguinte maneira:\n\n| BOLINHAS REMOVIDAS | SUBSTITUÍDAS POR           |\n|--------------------|----------------------------|\n| $3$ pretas         | $1$ preta                  |\n| $2$ pretas e $1$ branca | $1$ preta e $1$ branca |\n| $1$ preta e $2$ brancas | $2$ brancas           |\n| $3$ brancas        | $1$ preta e $1$ branca     |\n\nQual pode ser o conteúdo da caixa depois de seguidas aplicações desse procedimento?\n\n(a) $2$ pretas\n(b) $2$ brancas\n(c) $1$ preta\n(d) $1$ preta e $1$ branca\n(e) $1$ branca.",
    "options": [],
    "answer": "b",
    "solution": "Solution:\n\nInicialmente observe que depois de cada substituição o número de bolas brancas ou permanece o mesmo ou decresce de $2$. Logo o número de bolas brancas permanece par. Por outro lado, cada grupo de bolas removidas que contém pelo menos $1$ bola branca é substituído por outro que também contém $1$ bola branca, o número de bolas brancas nunca é zero. Agora observe que a opção (b) é a única incluindo pelo menos $2$ bolas brancas, logo ela é a opção correta. Um modo de obter esse resultado é remover $3$ bolas brancas $49$ vezes até obter $149$ pretas e $2$ brancas, e depois, remover $1$ preta e $2$ brancas $149$ vezes.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26070,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAmigos do século $XX$ - Dois amigos nasceram no século XX, com uma semana de intervalo e no mesmo mês e ano. Escrevendo da esquerda para a direita a data na forma o (ou os) algarismo(s) do dia, (ou os) algarismo(s) do mês, e os dois últimos algarismos do ano, obtemos dois números sendo um o sêxtuplo do outro. Não colocamos 0 na frente dos 9 primeiros meses. Qual é a data de nascimento do amigo mais velho?",
    "options": [],
    "answer": "1 April 1900",
    "solution": "Solution:\n\nOs dois amigos nasceram no mesmo mês e no mesmo ano, com uma diferença de 7 dias, assim um nasceu no dia $d/m/a$ e o outro no dia $(d+7)/m/a$. Com estas datas formamos os números $(d)(m)(a)$ e $(d+7)(m)(a)$. Sabemos que:\n$$\n(d+7)(m)(a) = (d)(m)(a) + 7 \\times 10^{k}\n$$\nAssim,\n$$\n(d+7)(m)(a) = 6 \\times (d)(m)(a), \\quad \\Longrightarrow \\quad 7 \\times 10^{k} = 5(d)(m)(a)\n$$\nLogo, $k=3$ se o mês tem 1 algarismo e $k=4$ se o mês tem 2 algarismos. No primeiro caso, quando $k=3$, temos que $\\frac{7000}{5} = 1400$, isto é, 1 de abril de 1900. Logo, seu amigo nasceu em 8 de abril de 1900. No segundo caso, quando $k=4$, $\\frac{70000}{5} = 14000$ não é uma data válida.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26071,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSeja $\\triangle ABC$ um triângulo tal que a altura relativa ao lado $BC$ não é menor do que o lado $BC$ e a altura relativa ao lado $AB$ não é menor do que o lado $AB$. Determine as medidas dos ângulos deste triângulo.",
    "options": [],
    "answer": "45°, 45°, 90°",
    "solution": "Solution:\n\nSejam $h_{a}$ e $h_{c}$ as alturas relativas aos lados $BC = a$ e $AB = c$, respectivamente. Por hipótese temos que $h_{a} \\geq a$ e $h_{c} \\geq c$. Como $h_{a}$ e $h_{c}$ são os comprimentos das alturas, então $h_{a} \\leq c$ e $h_{c} \\leq a$.\n\nUm dos lados considerados é maior ou igual ao outro: digamos $a \\geq c$. Das desigualdades acima temos:\n$$\nh_{a} \\geq a \\geq c \\geq h_{a} \\quad \\Rightarrow \\quad a = c = h_{a}\n$$\nDaí, segue que $AB$ é perpendicular a $BC$. Logo, o triângulo é retângulo isósceles e portanto, os ângulos medem $45^{\\circ}, 45^{\\circ}$ e $90^{\\circ}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26072,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDesenhe duas circunferências de mesmo centro, uma de raio $1~\\mathrm{cm}$ e a outra de raio $3~\\mathrm{cm}$. Na região exterior à circunferência de raio $1~\\mathrm{cm}$ e interior à de raio $3~\\mathrm{cm}$, desenhe circunferências que sejam simultaneamente tangentes às duas circunferências, como mostrado na figura a seguir.\n\na) Qual deve ser o raio dessas circunferências?\n\nb) Qual o número máximo dessas circunferências, caso elas não se sobreponham?\n\n![](attached_image_1.png)\n",
    "options": [],
    "answer": "radius 1 cm; maximum 6",
    "solution": "Solution:\n\n(a) Como as circunferências de raios $1~\\mathrm{cm}$ e $3~\\mathrm{cm}$ são concêntricas, as outras circunferências mostradas na figura devem ter raio igual a $1~\\mathrm{cm}$.\n\n(b) Os centros das 3 circunferências de raio $1~\\mathrm{cm}$ mostradas na figura formam um triângulo equilátero de lado $2~\\mathrm{cm}$. Logo, seus ângulos internos medem $60^\\circ$.\n\n![](attached_image_2.png)\n\nComo $\\frac{360}{60} = 6$, concluímos que até 6 circunferências podem ser dispostas nas condições exigidas.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26073,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nÁrea de triângulo - Se $AC = 1{,}5\\ \\mathrm{cm}$ e $AD = 4\\ \\mathrm{cm}$, qual é a relação entre as áreas dos triângulos $\\triangle ABC$ e $\\triangle DBC$?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "3/5",
    "solution": "Solution:\n\nOs triângulos $\\triangle ABC$ e $\\triangle DBC$ têm bases $AC$ e $CD$ respectivamente, e a mesma altura $h$ em relação a essas bases.\n\n![](attached_image_2.png)\n\nAssim temos:\n$$\n\\text{área } \\triangle ABC = \\frac{AC \\times h}{2} \\quad \\text{e área } \\triangle DBC = \\frac{CD \\times h}{2}.\n$$\nLogo, a relação entre as áreas é dada por:\n$$\n\\frac{\\text{área } \\triangle ABC}{\\text{área } \\triangle DBC} = \\frac{\\frac{AC \\times h}{2}}{\\frac{CD \\times h}{2}} = \\frac{AC}{CD} = \\frac{1{,}5}{4-1{,}5} = \\frac{15}{25} = \\frac{3}{5}\n$$\n\n**LEMBRE-SE:** A área de um triângulo é a metade do produto de um dos seus lados pela altura $h$ relativa a este lado, como exemplificado nas duas figuras a seguir.\n\n![](attached_image_3.png)\n\nÁrea do $\\triangle BCD = \\frac{CD \\times h}{2}$\n\n![](attached_image_4.png)\n\nÁrea do $\\triangle ABC = \\frac{AC \\times h}{2}$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26074,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOs lados de um triângulo têm comprimentos: $a$, $a+2$ e $a+5$, onde $a>0$. Determine os possíveis valores de $a$.",
    "options": [],
    "answer": "a > 3",
    "solution": "Solution:\n\nComo a soma dos comprimentos dos lados menores deve ser maior que o comprimento do lado maior, então temos que $a + (a+2) > a+5$, assim $a > 3$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26075,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm caminho retangular - Janete passeia por um caminho de forma retangular $ABCD$ com largura $AB = 1992~\\mathrm{m}$. Ela gasta 24 minutos para percorrer a largura $AB$. Depois, com a mesma velocidade, ela percorre o comprimento $BC$ e a diagonal $CA$ em 2 horas e 46 minutos. Qual é o comprimento $BC$?",
    "options": [],
    "answer": "6745",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26076,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAmélia, Bruno, Constância e Denise são 4 amigos que moram em Estados diferentes e se encontram sentados numa mesa quadrada, cada um ocupa um lado da mesa.\n- À direita de Amélia está quem mora no Amazonas;\n- Em frente à Constância está a pessoa que mora em São Paulo;\n- Bruno e Denise estão um ao lado do outro;\n- Uma mulher está à esquerda da pessoa que mora no Ceará.\n- Um dos quatro mora na Bahia. Quem?",
    "options": [],
    "answer": "Bruno ou Amélia",
    "solution": "Solution:\n\nBruno ou Amélia (O desafio tem duas soluções).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26077,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA sequência $\\{a, b, c\\}$ - A lei de formação da sequência $10, a, 30, b, c, \\ldots$ é: cada termo, começando com o $30$, é o dobro da soma dos dois termos imediatamente anteriores. Qual o valor de $c$?",
    "options": [],
    "answer": "200",
    "solution": "Solution:\n\nSabemos que $30 = 2(10 + a)$, logo $a = 5$.\n\n$$\nb = 2(30 + a) = 2(30 + 5) = 70\n$$\n\ne\n$$\nc = 2(b + 30) = 2(70 + 30) = 200\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26078,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nRosa, Margarida e Dália são três constelações em forma de buquês de flores. Sabemos que:\n\na. O número de estrelas de Dália, que é a menor das três, é o quadrado de um quadrado;\nb. O número de estrelas de Rosa é também o quadrado de um quadrado;\nc. Margarida tem 28561 estrelas;\nd. Dália e Rosa têm juntas o mesmo número de estrelas do que Margarida.\nQuantas estrelas possuem Dália e Rosa cada uma?",
    "options": [],
    "answer": "Dália: 4225; Rosa: 24336",
    "solution": "Solution:\n\n$\\{D\\} = 4225 = 25 \\times 169$\ne $\\{R\\} = 144 \\times 169 = 24336$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26079,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO valor de $\\frac{\\sqrt[3]{-0,001} \\times \\sqrt{400}}{\\sqrt{0,25}}-\\frac{\\sqrt{0,036}-\\sqrt{0,4}}{\\sqrt{0,4}}$ é:\n\n(a) $-3,3$\n(b) $-4,7$\n(c) $-4,9$\n(d) $-3,8$\n(e) $-7,5$",
    "options": [],
    "answer": "a",
    "solution": "Solution:\n\nTemos:\n$$\n\\frac{-0,1 \\times 20}{0,5}-\\frac{\\sqrt{0,4}(\\sqrt{0,09}-1)}{\\sqrt{0,4}}=-\\frac{20}{5}-(0,3-1)=-4-0,3+1=-3,3.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26080,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm número sortudo é aquele cuja soma de seus algarismos é divisível por $7$. Por exemplo, $7$, $25$ e $849$ são números sortudos. O menor par de números sortudos é $7$ e $16$.\n\na. Encontre oito números consecutivos, dos quais dois são números sortudos.\n\nb. Encontre $12$ números consecutivos, tal que nenhum seja sortudo.\n\nc. Mostre que qualquer sequência de $13$ números consecutivos contém pelo menos um número sortudo.",
    "options": [],
    "answer": "a) 52–59 (lucky: 52 and 59); another example: 55–62 (lucky: 59 and 61).\nb) 994–1005 or 7994–8005 (none lucky).\nc) Any sequence of 13 consecutive integers contains at least one lucky number.",
    "solution": "Solution:\n\na. A sequência de oito números consecutivos de $52$ a $59$ tem exatamente dois números sortudos: $52$ e $59$. Outro exemplo é qualquer sequência de $8$ números que contenha $59$ e $61$, por exemplo: $55, 56, 57, 58, 59, 60, 61, 62$.\n\nb. Dois exemplos: $994, \\ldots, 1005$ e $7994, \\ldots, 8005$. Existem mais exemplos, encontre alguns.\n\nc. Chamemos de década qualquer sequência de $10$ números consecutivos cujo primeiro termo é múltiplo de $10$:\n$$\n\\begin{aligned}\n& 10, 11, 12, 13, 14, 15, 16, 17, 18, 19 \\\\\n& 140, 141, 142, 143, 144, 145, 146, 147, 148, 149\n\\end{aligned}\n$$\nNote que qualquer sequência de $7$ números consecutivos numa década contém pelo menos um número sortudo porque a soma de seus algarismos é uma sequência de $7$ números consecutivos, um dos quais tem de ser divisível por $7$. Finalmente, qualquer sequência de $13$ números consecutivos contém pelo menos $7$ números consecutivos de uma década, que sempre contém um número sortudo. Examine alguns exemplos para melhor entender essa justificativa.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26081,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoma de potências de 2 - Determine um valor de $n$ para o qual o número $2^{8}+2^{11}+2^{n}$ seja um quadrado perfeito.",
    "options": [],
    "answer": "12",
    "solution": "Solution:\n\nObserve que\n$$\n2^{8}+2^{11}+2^{n} = \\left(2^{4}\\right)^{2} + 2 \\times 2^{4} \\times 2^{6} + \\left(2^{\\frac{n}{2}}\\right)^{2}\n$$\nLogo, se $n=12$, temos\n$$\n2^{8}+2^{11}+2^{12} = \\left(2^{4}+2^{6}\\right)^{2}\n$$\nLogo $n=12$ é uma solução.\n\nSolução Geral: Se $2^{8}+2^{11}+2^{n}=k^{2}$, então:\n$$\n\\begin{aligned}\n2^{8}+2^{3} \\times 2^{8}+2^{n} & = k^{2} \\\\\n9 \\times 2^{8}+2^{n} & = k^{2} \\\\\n2^{n} & = k^{2} - \\left(3 \\times 2^{4}\\right)^{2} \\\\\n2^{n} & = \\left(k-3 \\times 2^{4}\\right)\\left(k+3 \\times 2^{4}\\right)\n\\end{aligned}\n$$\nLogo, $\\left(k-3 \\times 2^{4}\\right)$ e $\\left(k+3 \\times 2^{4}\\right)$ são potências de 2, ou seja:\n$$\nk+3 \\times 2^{4}=2^{a}, \\quad k-3 \\times 2^{4}=2^{b} \\text{ e } a+b=n\n$$\nTemos:\n$$\n2^{a}-2^{b}=\\left(k+3 \\times 2^{4}\\right)-\\left(k-3 \\times 2^{4}\\right)=3 \\times 2^{5}=96\n$$\nExaminemos a lista das potências de 2:\n$$\n1, 4, 8, 16, 32, 64, 128, 256, \\cdots\n$$\nConstatamos que $128-32=96$. Logo, $a=7$, $b=5$ e $n=12$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26082,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm quadrilátero - O quadrilátero $ABCD$ da figura é um paralelogramo?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nPara que $ABCD$ seja um paralelogramo, seus lados devem ser dois a dois paralelos, isto é: $AB \\parallel CD$ e $AD \\parallel BC$.\n\nComo\n$$\n\\widehat{DAB} + \\widehat{ABC} = 180^\\circ\n$$\nentão as retas $AD$ e $BC$ são paralelas. Além disso, temos dois ângulos alternos internos de $45^\\circ$ entre as retas $AB$ e $DC$, segue-se que elas são paralelas. Logo $ABCD$ é um paralelogramo.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26083,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCanelonis de ricota - Todo domingo, Pedro prepara canelonis para o almoço. Primeiro ele corta retângulos de massa de $16~\\mathrm{cm}$ por $12~\\mathrm{cm}$ e depois cola os dois lados mais longos, superpondo uma faixa de $2~\\mathrm{cm}$.\n\n![](attached_image_1.png)\n\nDessa forma ele obtém cilindros que ele recheia com ricota, ele já sabe que sempre gasta $500~\\mathrm{g}$ de ricota. Num belo domingo, com o mesmo número de retângulos de massa de $16~\\mathrm{cm}$ por $12~\\mathrm{cm}$, ele decide produzir os cilindros colando os lados menores, sempre superpondo uma faixa de $2~\\mathrm{cm}$. Nessa situação, ele vai gastar mais ou menos ricota que antes? Quanto?\n",
    "options": [],
    "answer": "735 g (more than before)",
    "solution": "Solution:\n\nColando os retângulos de massa ao longo do maior lado, Pedro obtém um cilindro de base circular com $10~\\mathrm{cm}$ de comprimento e $16~\\mathrm{cm}$ de altura. O volume então que ele recheia com ricota é o volume desse cilindro:\n$$\nV=\\text{área da base} \\times \\text{altura}\n$$\nA área da base é dada por $\\pi \\times r^{2}$, onde $r$ é o raio da base. Vamos então calcular o raio sabendo que o comprimento da base é $10~\\mathrm{cm}$; temos:\n$$\n2 \\pi r=10 \\Rightarrow r=\\frac{5}{\\pi}\n$$\nLogo, o volume de ricota para cada caneloni é\n$$\nV=\\pi \\times \\frac{5^{2}}{\\pi^{2}} \\times 16=\\frac{16 \\times 25}{\\pi}=\\frac{400}{\\pi}~\\mathrm{cm}^{3}\n$$\nAgora, colando os retângulos de massa ao longo do menor lado, Pedro obtém um cilindro de base circular com $14~\\mathrm{cm}$ de comprimento e $12~\\mathrm{cm}$ de altura. O raio da base é $r' = \\frac{14}{2 \\pi} = \\frac{7}{\\pi}$, logo o volume de ricota para cada caneloni será:\n$$\nV' = \\pi \\times \\frac{7^{2}}{\\pi^{2}} \\times 12 = \\frac{588}{\\pi}~\\mathrm{cm}^{3}\n$$\n![](attached_image_2.png)\n\nFinalmente, para calcular o novo gasto com ricota, usamos a seguinte Regra de Três direta:\n$$\n\\begin{array}{ccc}\n\\text{Volume}~(\\mathrm{cm}^{3}) & & \\text{Ricota}~(\\mathrm{g}) \\\\\n\\frac{400}{\\pi} & \\longrightarrow & 500 \\\\\n\\frac{588}{\\pi} & \\longrightarrow & x\n\\end{array}\n$$\nSegue que\n$$\nx=\\frac{500 \\times 588}{400}=735~\\mathrm{g}.\n$$\n\n![](attached_image_2.png)",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26084,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nMostre que se o produto $N = (n + 6m)(2n + 5m)(3n + 4m)$ é múltiplo de $7$, com $m$ e $n$ números naturais, então $N$ é múltiplo de $7^{3} = 343$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nInicialmente, observemos que:\n$$\n\\begin{aligned}\nN & = (n + 6m)(2n + 5m)(3n + 4m) \\\\\n  & = (n + 7m - m)(2n + 7m - 2m)(3n + 7m - 3m) \\\\\n  & = (n - m + 7m)[2(n - m) + 7m][3(n - m) + 7m] \\\\\n  & = (k + 7m)(2k + 7m)(3k + 7m)\n\\end{aligned}\n$$\nonde $k = n - m$.\nAfirmamos que se $N$ é múltiplo de $7$, então $k$ é múltiplo de $7$. De fato, como $7$ é primo e divide $N$, então um dos fatores $k + 7m$, $2k + 7m$ ou $3k + 7m$ é múltiplo de $7$. Temos:\n\n(i) Se $k + 7m$ é múltiplo de $7$, então $\\frac{k + 7m}{7} = \\frac{k}{7} + m$ é inteiro, logo $k$ é múltiplo de $7$. Segue que $2k$ e $3k$ também são múltiplos de $7$ e portanto os três fatores $k + 7m$, $2k + 7m$ e $3k + 7m$ são múltiplos de $7$. Concluímos que $N$ é múltiplo de $7^{3}$.\n\n(ii) Se $2k + 7m$ é múltiplo de $7$, então $\\frac{2k + 7m}{7} = \\frac{2k}{7} + m$ é inteiro, logo $2k$ é múltiplo de $7$. Como $2$ e $7$ são primos entre si, segue que $k$ é múltiplo de $7$, o que leva ao caso anterior.\n\n(iii) Se $3k + 7m$ é múltiplo de $7$, analogamente concluímos que $k$ é múltiplo de $7$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26085,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO triângulo de Reuleaux é um disco formado a partir de um triângulo equilátero, agregando arcos de circunferência com centros nos vértices do triângulo e raios iguais ao lado do triângulo.\n\n![](attached_image_1.png)\n\nQual é a área de um triângulo de Reuleaux, se o triângulo equilátero tem lado de medida $1~\\mathrm{cm}$ ?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "pi/2 - sqrt(3)/2 cm^2",
    "solution": "Solution:\n\nO triângulo de Reuleaux é formado por 4 regiões: um triângulo equilátero e três calotas. Cada calota é um sexto de um círculo de raio $1$ do qual foi retirado um triângulo equilátero de lado $1$.\n\nPelo Teorema de Pitágoras, a altura do triângulo equilátero é:\n$$\nh = \\sqrt{1 - \\left(\\frac{1}{2}\\right)^2} = \\frac{\\sqrt{3}}{2}\n$$\nlogo a área do triângulo vale:\n$$\n\\frac{1 \\times \\frac{\\sqrt{3}}{2}}{2} = \\frac{\\sqrt{3}}{4}~\\mathrm{cm}^2\n$$\nA área de um setor circular é um sexto da área do círculo, ou seja, igual a $\\frac{\\pi}{6}$. Logo, a área da calota é a diferença:\n$$\n\\frac{\\pi}{6} - \\frac{\\sqrt{3}}{4}\n$$\nPortanto, a área do triângulo de Reuleaux é\n$$\n3\\left(\\frac{\\pi}{6} - \\frac{\\sqrt{3}}{4}\\right) + \\frac{\\sqrt{3}}{4} = \\frac{\\pi}{2} - \\frac{\\sqrt{3}}{2}~\\mathrm{cm}^2\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26086,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nExiste um número de 8 algarismos da forma\n$$\n9999 * * * *\n$$\nque é um quadrado perfeito?",
    "options": [],
    "answer": "No",
    "solution": "Solution:\nSeja $x$ um número de oito algarismos da forma\n$$\nx = 9999 * * * *\n$$\nComo o menor desses números é $99990000$ e o maior é $99999999$, temos que:\n$$\n99990000 \\leq x \\leq 99999999\n$$\nObservemos que $10^{8} = 100000000 = 99999999 + 1$. Então $99990000 \\leq x < 10^{8}$. Como $10^{8} = (10^{4})^{2} = 10000^{2}$, temos que $99990000 \\leq x < 10000^{2}$. Agora, o maior quadrado perfeito menor que $10000^{2}$ é igual a\n$$\n9999^{2} = (10000 - 1)^{2} = 10000^{2} - 20000 + 1 = 100000000 - 20000 + 1 = 99980001.\n$$\nComo $99980001 < 99990000$ concluímos que $9999^{2} < x < 10000^{2}$. Isto mostra que $x$ está compreendido entre dois quadrados perfeitos consecutivos. Portanto, $x$ não pode ser um quadrado perfeito.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26087,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSeja $ABCD$ um trapézio retângulo de bases $AB$ e $CD$, com ângulos retos em $A$ e $D$. Dado que a diagonal menor $BD$ é perpendicular ao lado $BC$, determine o menor valor possível para a razão $\\frac{CD}{AD}$.",
    "options": [],
    "answer": "2",
    "solution": "Solution:\n\nSeja $A\\widehat{B}D = B\\widehat{D}C = \\alpha$. Então temos que $DC = \\frac{BD}{\\cos \\alpha}$ e $AD = BD \\sen \\alpha$, donde\n\n$$\n\\frac{DC}{AD} = \\frac{\\frac{BD}{\\cos \\alpha}}{BD \\sen \\alpha} = \\frac{1}{\\sen \\alpha \\cos \\alpha} = \\frac{2}{\\sen 2\\alpha} \\geq 2.\n$$\n\nA igualdade ocorre quando $\\sen 2\\alpha = 1$, ou seja, quando $\\alpha = 45^\\circ$.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26088,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDois triângulos retângulos congruentes possuem catetos de medidas $4~\\mathrm{cm}$ e $7~\\mathrm{cm}$. Na figura abaixo, à esquerda, os triângulos foram desenhados de modo a coincidirem os catetos de $7~\\mathrm{cm}$. Assim, $AB=7~\\mathrm{cm}$ e $AD=BC=4~\\mathrm{cm}$. Já na figura à direita, eles foram desenhados de modo a coincidirem as hipotenusas donde, $AD=BC=4~\\mathrm{cm}$ e $AC=BD=7~\\mathrm{cm}$.\n![](attached_image_1.png)\nCalcule as áreas sombreadas nas duas figuras.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "First figure: 21 cm^2; Second figure: 131/7 cm^2",
    "solution": "Solution:\n\nOs pontos $A, B, C$ e $D$ formam o retângulo $ABCD$.\n![](attached_image_2.png)\nComo as diagonais de um retângulo o dividem em quatro triângulos de mesma área, a área sombreada é igual a três quartos da área do retângulo $ABCD$. Portanto, a área sombreada é igual a $\\frac{3}{4}(7 \\times 4)=21~\\mathrm{cm}^2$.\n\nVejamos agora o caso da outra figura. Sejam $E$ o ponto de interseção dos segmentos $AC$ e $BD$, $x=DE=CE$ e $y=AE=BE$.\n![](attached_image_3.png)\nA área sombreada é a soma das áreas dos triângulos $ADE$ e $ABC$, ou seja:\n$$\n\\frac{4 \\times x}{2}+\\frac{4 \\times 7}{2}=2x+14\n$$\nLogo, basta calcularmos $x$. Temos que $x+y=7$ e, pelo Teorema de Pitágoras aplicado ao triângulo $AED$, $y^2=x^2+4^2$. Substituindo $y=7-x$ nessa última equação obtemos:\n$$\n(7-x)^2=x^2+16 \\Rightarrow 49-14x+x^2=x^2+16 \\Rightarrow x=\\frac{49-16}{14}=\\frac{33}{14}\n$$\nFinalmente, a área sombreada é:\n$$\n2 \\times \\frac{33}{14}+14=\\frac{33}{7}+14=4\\frac{5}{7}+14=18\\frac{5}{7}.\n$$\n\n![](attached_image_2.png)\n\n![](attached_image_3.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26089,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm número perfeito - Um número natural $n$ é dito perfeito se a soma de todos os seus divisores próprios, isto é, diferentes de $n$, é igual a $n$. Por exemplo, $6$ e $28$ são perfeitos, pois: $6=1+2+3$ e $28=1+2+4+7+14$. Sabendo que $2^{31}-1$ é um número primo, mostre que $2^{30}\\left(2^{31}-1\\right)$ é um número perfeito.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSe $2^{31}-1$ é um número primo, seu único divisor próprio é o número $1$. Então os divisores próprios de $2^{30}\\left(2^{31}-1\\right)$ são:\n$$\n1, 2, 2^{2}, 2^{3}, \\ldots, 2^{29}, 2^{30}, \\left(2^{31}-1\\right), 2\\left(2^{31}-1\\right), 2^{2}\\left(2^{31}-1\\right), \\ldots, 2^{29}\\left(2^{31}-1\\right)\n$$\nA soma $S$ desses divisores é:\n$$\nS=\\left[1+2+2^{2}+2^{3}+\\cdots+2^{29}+2^{30}\\right]+\\left(2^{31}-1\\right)\\left[1+2+2^{2}+2^{3}+\\cdots+2^{29}\\right]\n$$\nEm cada um dos dois colchetes aparece a soma $S_{n}$ de uma progressão geométrica de primeiro termo igual a $1$ e razão $2$.\nO primeiro colchete, $S_{31}$, contém $31$ termos e o segundo, $S_{30}$, contém $30$ termos. Usando a fórmula da soma dos termos de uma progressão geométrica, temos:\n$$\nS_{31}=\\frac{2^{31}-1}{2-1}=2^{31}-1 \\quad \\text{e} \\quad S_{30}=\\frac{2^{30}-1}{2-1}=2^{30}-1\n$$\nEntão a soma dos divisores próprios de $2^{30}\\left(2^{31}-1\\right)$ é:\n$$\nS=\\left(2^{31}-1\\right)+\\left(2^{31}-1\\right)\\left[2^{30}-1\\right]=\\left(2^{31}-1\\right)\\left(1+2^{30}-1\\right)=2^{30}\\left(2^{31}-1\\right)\n$$\nLogo, essa soma é igual a $2^{30}\\left(2^{31}-1\\right)$, como queríamos provar.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26090,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa figura ao lado, o triângulo $\\triangle ABC$ é retângulo e os semicírculos desenhados têm diâmetros $AB$, $BC$ e $AC$.\nMostre que a área sombreada é igual à área do triângulo $\\triangle ABC$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSejam $T$ a área do triângulo $\\triangle ABC$, $a$ e $c$ as áreas sombreadas na figura dada e $b$ e $d$ as áreas compreendidas entre os catetos do triângulo e o semicírculo de diâmetro $AB$.\n\nA área $a+b$ é a área do semicírculo de diâmetro $AB$:\n$$\na+b=\\frac{1}{2} \\pi\\left(\\frac{AB}{2}\\right)^{2}=\\frac{\\pi AB^{2}}{8}\n$$\nA área $c+d$ é a área do semicírculo de diâmetro $BC$:\n$$\nc+d=\\frac{1}{2} \\pi\\left(\\frac{BC}{2}\\right)^{2}=\\frac{\\pi BC^{2}}{8}\n$$\n![](attached_image_2.png)\nA área $b+d+T$ é a área do semicírculo de diâmetro $AC$:\n$$\nb+d+T=\\frac{1}{2} \\pi\\left(\\frac{AC}{2}\\right)^{2}=\\frac{\\pi AC^{2}}{8}\n$$\nPortanto,\n$$\n(a+b)+(c+d)=\\frac{\\pi AB^{2}}{8}+\\frac{\\pi BC^{2}}{8}\n$$\nComo $b+d=\\frac{\\pi AC^{2}}{8}-T$ temos\n$$\n(a+c)+\\left(\\frac{\\pi AC^{2}}{8}-T\\right)=\\frac{\\pi AB^{2}}{8}+\\frac{\\pi BC^{2}}{8}\n$$\nou equivalentemente,\n$$\n(a+c)+\\frac{\\pi}{8} AC^{2}=\\frac{\\pi}{8}\\left(AB^{2}+BC^{2}\\right)+T\n$$\nUma vez que $AC^{2}=AB^{2}+BC^{2}$, pelo Teorema de Pitágoras, podemos simplificar a igualdade acima e obter $a+c=T$. Esta igualdade implica que a soma das áreas sombreadas é igual à área do triângulo retângulo $\\triangle ABC$.\n\n![](attached_image_2.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26091,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nExistem 1999 cidades e 4000 estradas em um certo país (cada estrada conecta 2 cidades). Prove que existe um caminho fechado passando através de não mais que 20 cidades.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nInicialmente destrua as cidades, juntamente com suas estradas incidentes, se delas partem menos que 3 estradas. Como $2 \\cdot 1999 < 4000$, eventualmente irão sobrar estradas e cidades após esse processo. Assim, podemos assumir que de todas as cidades restantes partem pelo menos 3 estradas. Escolha uma cidade qualquer, que será chamada de patamar $T_{0}$. A seguir, descreveremos uma construção ordenada de conjuntos de cidades. As cidades que podem ser acessadas a partir de $T_{0}$ serão denotadas por $T_{1}$. As que podem ser acessadas a partir de $T_{1}$ e ainda não estão em $\\{T_{0}, T_{1}\\}$ serão chamadas de $T_{2}$. Continuando esse processo, podemos definir $T_{3}, T_{4}, \\ldots$.\n\nSe não existir um ciclo com não mais que 20 cidades, como cada cidade possui grau pelo menos 3, então para cada um dos primeiros patamares $T_{i}$ possui pelo menos o dobro de cidades de $T_{i-1}$.\n\n![](attached_image_1.png)\n\nSe existirem mais de 10 patamares, o número de cidades será de pelo menos\n$$\n\\begin{array}{r}\n|T_{0}| + |T_{1}| + |T_{2}| + \\ldots + |T_{10}| \\geq \\\\\n1 + 2 + 4 + \\ldots + 2^{10} = \\\\\n2047 > \\\\\n1999\n\\end{array}\n$$\nEsse absurdo mostra que existem no máximo 10 patamares. Isso significa que, para algum $i \\leq 9$, as cidades que podem ser acessadas a partir de $T_{i}$ estão em $\\{T_{0}, T_{1}, \\ldots, T_{i-1}\\}$ e isso gera um ciclo com não mais que 20 cidades.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 26092,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSeja $ABCD$ um paralelogramo, e $ABF$ e $ADE$ triângulos equiláteros construídos exteriormente ao paralelogramo. Prove que $FCE$ também é equilátero.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSeja $\\alpha = \\angle ABC$. Como $ABCD$ é um paralelogramo, ele tem ângulos opostos iguais, logo $\\angle ADC = \\alpha$, e ângulos adjacentes suplementares, logo $\\angle BAD = 180^\\circ - \\angle ABC = 180^\\circ - \\alpha$. Portanto conseguimos calcular o ângulo $\\angle FAE$, usando o fato de que todo triângulo equilátero tem ângulos de $60^\\circ$. Logo,\n$$\n\\begin{aligned}\n\\angle FAE + \\angle EAD + \\angle DAB + \\angle FAB & = 360^\\circ \\\\\n\\angle FAE + 60^\\circ + 180^\\circ - \\alpha + 60^\\circ & = 360^\\circ \\\\\n\\angle FAE & = 60^\\circ + \\alpha\n\\end{aligned}\n$$\n![](attached_image_2.png)\nPortanto, os ângulos $\\angle FAE$, $\\angle FBC$ e $\\angle CDE$ são iguais a $60^\\circ + \\alpha$.\n\nMais ainda, todo paralelogramo tem lados opostos iguais, logo $|\\overline{AB}| = |\\overline{DC}|$, e como o triângulo $ABF$ é equilátero, vale que\n$$\n|\\overline{DC}| = |\\overline{AB}| = |\\overline{AF}| = |\\overline{BF}|\n$$\nAnalogamente, temos que\n$$\n|\\overline{BC}| = |\\overline{AD}| = |\\overline{AE}| = |\\overline{DE}|\n$$\nLogo, concluímos que os triângulos $FAE$, $FBC$ e $CDE$ são congruentes pelo caso lado-ângulo-lado. Portanto, vale que\n$$\n|\\overline{EF}| = |\\overline{CF}| = |\\overline{EC}|\n$$\nLogo, o triângulo $FCE$ é equilátero.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26093,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPedrinho tem várias peças de madeira na forma de um triângulo retângulo de catetos $1~\\mathrm{cm}$ e $2~\\mathrm{cm}$. Com 4 dessas peças, ele consegue montar um quadrado de lado $2~\\mathrm{cm}$, como na figura abaixo.\n\n![](attached_image_1.png)\n\nBrincando com mais peças, ele conseguiu montar um quadrado usando exatamente 20 peças. Monte você também um quadrado usando 20 peças. (Dica: calcule a área dos triângulos, e com isso calcule o lado do quadrado. Depois compare com a hipotenusa das peças.)",
    "options": [],
    "answer": "2√5 cm",
    "solution": "Solution:\n\nCada peça tem área $\\frac{1 \\times 2}{2}~\\mathrm{cm}^2 = 1~\\mathrm{cm}^2$. Logo, se usarmos 20 das peças, teremos um quadrado de área $20~\\mathrm{cm}^2$. Portanto, seu lado mede $\\sqrt{20} = 2\\sqrt{5}~\\mathrm{cm}$. Por outro lado, pelo Teorema de Pitágoras, vale que a hipotenusa de cada triângulo retângulo mede $\\sqrt{1^2 + 2^2} = \\sqrt{5}~\\mathrm{cm}$. Daqui, vemos que para construir esse quadrado, vamos ter que usar as hipotenusas para formar os lados do quadrado. Uma possível solução é a seguinte:\n\n![](attached_image_2.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26094,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nNum triângulo $ABC$, tomamos pontos $X, Y$ sobre os lados $AB, BC$, respectivamente. Se $AY$ e $CX$ se intersectam em $Z$ e\n$$\nAY = YC \\text{ e } AB = ZC\n$$\nverifique que\n$$\n\\angle CZY = \\angle ABY\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n![](attached_image_1.png)\nSejam $l$ a mediatriz do segmento $AC$ e $B'$ o simétrico de $B$ com respeito a $l$. Então, como $A$ é o simétrico de $C$, os triângulos $ABY$ e $CB'Y$ são congruentes, donde\n$$\n\\begin{aligned}\n\\angle CB'Y & = \\angle ABY \\\\\nCB' & = AB\n\\end{aligned}\n$$\nAlém disso, os pontos $A, Y, B'$ são colineares (pois suas reflexões $C, Y, B$ são). A última igualdade, juntamente com a segunda relação do enunciado, dá que\n$$\nCB' = AB = CZ\n$$\nComo os ângulos da base do triângulo isósceles $\\triangle ZB'C$ são iguais, segue que\n$$\n\\angle CZY = \\angle CB'Y = \\angle ABY\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26095,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDefina $f(n, k)$ como o número de maneiras de distribuir $k$ chocolates para $n$ crianças em que cada criança recebe 0, 1 ou 2 chocolates. Por exemplo, $f(3,4)=6$, $f(3,6)=1$ e $f(3,7)=0$.\n\na) Exiba todas as 6 maneiras de distribuir 4 chocolates para 3 crianças com cada uma ganhando no máximo dois chocolates.\n\nb) Considerando 2015 crianças, verifique que $f(2015, k)=0$ para todo $k$ maior ou igual a um valor apropriado.\n\nc) Mostre que a equação\n$$\nf(2016, k)=f(2015, k)+f(2015, k-1)+f(2015, k-2)\n$$\né verdadeira para todo $k$ inteiro positivo maior ou igual a 2.\n\nd) Calcule o valor da expressão\n$$\nf(2016,1)+f(2016,4)+f(2016,7)+\\ldots+f(2016,4027)+f(2016,4030)\n$$",
    "options": [],
    "answer": "3^2015",
    "solution": "Solution:\n\n(a) Vamos representar cada distribuição por uma tripla ordenada de números $(a, b, c)$ em que cada número representa a quantia de chocolates que cada criança receberá. As seis possibilidades são:\n$$\n(2,2,0);\\ (2,0,2);\\ (0,2,2);\\ (2,1,1);\\ (1,2,1);\\ (1,1,2)\n$$\n\n(b) Se $k \\geq 2 \\cdot 2015 + 1 = 4031$, então pelo Princípio da Casa dos Pombos, se forem distribuídos $k$ chocolates para 2015 crianças, pelo menos uma delas ganhará mais que 2 chocolates. Em outras palavras, é impossível que cada uma ganhe no máximo dois chocolates. Então, para $k \\geq 4031$, temos $f(2015, k) = 0$.\n\n(c) Vamos considerar as possibilidades de chocolates para a primeira criança. Se ela ganhar 0, então restam $k$ chocolates para as outras 2015. Se ela ganhar 1, restam $k-1$ para as outras. E se ela ganhar 2, então restam $k-2$ para as demais. Esta contagem em três casos corresponde à seguinte equação:\n$$\nf(2016, k) = f(2015, k) + f(2015, k-1) + f(2015, k-2)\n$$\n\n(d) Chamaremos esta soma de $S$. Usando o item anterior, temos\n$$\n\\begin{aligned}\nf(2016,1) &= f(2015,1) + f(2015,0) \\\\\nf(2016,4) &= f(2015,4) + f(2015,3) + f(2015,2) \\\\\nf(2016,7) &= f(2015,7) + f(2015,6) + f(2015,5) \\\\\n& \\cdots \\\\\nf(2016,4027) &= f(2015,4027) + f(2015,4026) + f(2015,4025) \\\\\nf(2016,4030) &= f(2015,4030) + f(2015,4029) + f(2015,4028)\n\\end{aligned}\n$$\nSomando tudo, teremos\n$$\nS = f(2015,0) + f(2015,1) + \\ldots + f(2015,4029) + f(2015,4030)\n$$\nVeja que a soma representa todas as maneiras de distribuirmos 0, 1 ou 2 chocolates para cada criança de um grupo de 2015 crianças. Portanto, usando o princípio multiplicativo, esta soma é igual a $3^{2015}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26096,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAlguns alunos de uma escola foram divididos em equipes satisfazendo as seguintes condições:\ni) Quaisquer 2 equipes diferentes possuem exatamente 2 membros em comum.\nii) Toda equipe possui exatamente 4 elementos.\niii) Para quaisquer 2 alunos, existe uma equipe da qual ambos não fazem parte.\na) Explique por que um par qualquer de estudantes pode participar de no máximo 3 equipes.\nb) Qual o número máximo de equipes?",
    "options": [],
    "answer": "7",
    "solution": "Solution:\n\na) Suponha que existem estudantes $A$ e $B$ participando de 4 equipes chamadas de $E_{1}, E_{2}$, $E_{3}$ e $E_{4}$ :\n$$\n\\begin{aligned}\nE_{1} &= \\{A, B, C, D\\} \\\\\nE_{2} &= \\{A, B, E, F\\} \\\\\nE_{3} &= \\{A, B, G, H\\} \\\\\nE_{4} &= \\{A, B, I, J\\}\n\\end{aligned}\n$$\nPela condição (iii), existe uma equipe $E_{5}$ que não possui simultaneamente os estudantes $A$ e $B$. Pela condição (i), a equipe $E_{5}$ deve possuir exatamente dois membros em comum com $E_{1}, E_{2}, E_{3}$ e $E_{4}$. Se $E_{5}$ não possuir $A$ e nem $B$, então deverá possuir os elementos dos seguintes 4 conjuntos disjuntos $\\{C, D\\},\\{E, F\\},\\{G, H\\}$ e $\\{I, J\\}$. Isso obriga $E_{5}$ a ter mais de 4 elementos. Digamos que $E_{5}$ não possua o estudante $A$, mas possua o estudante $B$ (como $A$ e $B$ desempenham o mesmo papel nessa análise, o estudo que faremos para $A$ é o mesmo para $B$ ). Assim, $E_{5}$ deverá possuir pelo menos um estudante de cada um dos mesmos 4 conjuntos anteriores e novamente essa equipe será forçada a ter mais que 4 elementos. Esse absurdo mostra que um par qualquer de estudantes não pode participar de 4 equipes.\n\nb) Considere um par de estudantes ($A, B$) que participa de mais de uma equipe, cuja existência é garantida pelo item (i), e estudemos as seguintes possibilidades a partir das informações do item anterior.\n\n1) O par ($A, B$) participa de exatamente três outras equipes que são chamadas de $E_{1}$, $E_{2}, E_{3}$ :\n$$\n\\begin{aligned}\n& E_{1}=\\{A, B, C, D\\} \\\\\n& E_{2}=\\{A, B, E, F\\} \\\\\n& E_{3}=\\{A, B, G, H\\}\n\\end{aligned}\n$$\nIremos mostrar que nesse caso não podem existir mais que 4 outras equipes. Repetindo o argumento do item anterior, se uma equipe não possuir $A$ e nem $B$, ela possuirá todos os elementos do conjunto $\\{C, D, E, F, G, H\\}$ e isso produzirá um absurdo com a condição (ii). Logo, todas as demais equipes possuem $A$ ou $B$ e, pela condição (i), qualquer outra equipe possui exatamente um elemento de cada um dos três conjuntos: $\\{C, D\\},\\{E, F\\}$ e $\\{G, H\\}$. Analisando as possíveis combinações dos 3 elementos restantes, em princípio, teríamos no máximo $2 \\cdot 2 \\cdot 2=8$ escolhas associadas aos conjuntos anteriores. Entretanto, existem alguns pares de escolhas que não podem estar simultaneamente presentes para que a condição (i) seja satisfeita:\n$$\n\\begin{array}{ccc}\n\\{C, E, G\\} & \\text{ e } & \\{D, F, H\\} \\\\\n\\{C, E, H\\} & \\text{ e } & \\{D, F, G\\} \\\\\n\\{C, F, G\\} & \\text{ e } & \\{D, E, H\\} \\\\\n\\{C, F, H\\} & \\text{ e } & \\{D, E, G\\}\n\\end{array}\n$$\nAssim, podem existir no máximo mais 4 equipes além de $E_{1}, E_{2}$ e $E_{3}$.\n\n2) O par $(A, B)$ participa de exatamente duas outras equipes que são chamadas de $E_{1}$, $E_{2}$ :\n$$\n\\begin{aligned}\n& E_{1}=\\{A, B, C, D\\} \\\\\n& E_{2}=\\{A, B, E, F\\}\n\\end{aligned}\n$$\nEm virtude da condição (i), a única equipe que não pode possuir simultaneamente $A$ e $B$ é a $\\{C, D, E, F\\}$. Considerando uma equipe distinta dela, novamente pelo item (i), um elemento de cada um dos conjuntos $\\{A, B\\},\\{C, D\\}$ e $\\{E, F\\}$ deve estar presente. Isso novamente gera $2 \\cdot 2 \\cdot 2=8$ escolhas que podem ser divididas nos seguintes pares:\n$$\n\\begin{array}{lcc}\n\\{A, C, E\\} & \\text{ e } & \\{B, D, F\\} \\\\\n\\{A, C, F\\} & \\text{ e } & \\{B, D, E\\} \\\\\n\\{A, D, E\\} & \\text{ e } & \\{B, C, F\\} \\\\\n\\{A, D, F\\} & \\text{ e } & \\{B, C, E\\}\n\\end{array}\n$$\nDe cada par, apenas uma escolha pode estar presente nas equipes de estudantes e isso nos mostra que existem no máximo mais 4 equipes que contém $A$ ou $B$. Considerando as duas equipes que contém $\\{A, B\\}$, a eventual equipe $\\{C, D, E, F\\}$ e as possíveis 4 outras equipes, temos um total de no máximo $2+1+4=7$ equipes.\n\nEm ambos os casos, o máximo de equipes é 7. Para mostrar que é possível satisfazermos as três condições com esse número, basta considerar o seguinte exemplo:\n$$\n\\{A, B, C, H\\},\\{A, D, F, H\\},\\{A, E, G, H\\},\\{B, D, G, H\\},\\{B, E, F, H\\},\\{C, D, E, H\\},\\{C, F, G, H\\}\n$$\nem que cada letra, como anteriormente, representa um certo estudante.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26097,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n(a) Um quadrilátero com lados opostos iguais é um paralelogramo. As figuras a seguir mostram dois paralelogramos com os mesmos lados, sendo o segundo um retângulo.\nDetermine a maior área possível de um paralelogramo de lados $b$ e $h$.\n\n![](attached_image_1.png)\n\n(b) Considere as duas figuras a seguir, onde $a, b, x$ e $y$ são números reais positivos. Mostre que a segunda figura possui maior área.\n\n![](attached_image_2.png)\n\n(c) Usando o resultado do item anterior, prove a desigualdade de Cauchy-Schwarz para dois termos, descrita pela desigualdade a seguir:\n$$\na x+b y \\leq \\sqrt{a^{2}+b^{2}} \\cdot \\sqrt{x^{2}+y^{2}}\n$$",
    "options": [],
    "answer": "b*h",
    "solution": "Solution:\n\n(a) Considerando todos os paralelogramos com lados $b$ e $h$, observe que o retângulo é o que possui maior altura em relação à base fixa $b$. Na figura a seguir, pela Desigualdade Triangular, temos $h' < h$.\n\n![](attached_image_3.png)\n\nComo cada triângulo representa metade do paralelogramo, então $h' < h$ implica\n$$\n\\frac{b \\cdot h'}{2} < \\frac{b \\cdot h}{2}\n$$\nlogo o retângulo possui maior área.\n\n(b) O resultado é consequência do item anterior. A diferença entre as duas figuras é dada pelos quadriláteros. Na segunda figura, temos um retângulo com mesmos lados que o paralelogramo da primeira figura, então a área da segunda figura é maior.\n\n(c) Comparando as áreas das duas figuras do item anterior, temos:\n$$\n\\begin{aligned}\n(a+y)(b+x) & \\leq 2 \\cdot \\frac{a b}{2} + 2 \\cdot \\frac{x y}{2} + \\sqrt{a^{2}+b^{2}} \\cdot \\sqrt{x^{2}+y^{2}} \\\\\na b + a x + b y + x y & \\leq a b + x y + \\sqrt{a^{2}+b^{2}} \\cdot \\sqrt{x^{2}+y^{2}} \\\\\na x + b y & \\leq \\sqrt{a^{2}+b^{2}} \\cdot \\sqrt{x^{2}+y^{2}}\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26098,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA figura a seguir mostra um triângulo $ABC$ com lados $AB=13~\\mathrm{cm}$, $BC=14~\\mathrm{cm}$ e $CA=15~\\mathrm{cm}$. A circunferência de centro $I$ tangencia os lados $AB$, $BC$ e $CA$ nos pontos $F$, $D$ e $E$, respectivamente. Os pontos $R$, $S$, $T$, $U$, $V$ e $X$ são marcados nos prolongamentos dos lados do triângulo $ABC$ tal que $AR=AS=14~\\mathrm{cm}$, $BT=BU=15~\\mathrm{cm}$ e $CV=CX=13~\\mathrm{cm}$.\n![](attached_image_1.png)\n(a) Prove que $AE=AF$.\n(b) Determine a medida do segmento $BF$.\n(c) Prove que o hexágono $RSTUVX$ é inscritível, ou seja, mostre que existe uma circunferência passando pelos seus seis vértices.",
    "options": [],
    "answer": "6",
    "solution": "Solution:\n(a) O Teorema do Bico diz que as distâncias de um ponto exterior a uma circunferência aos pontos onde suas tangentes tocam a circunferência são iguais. Aplicando este resultado aos pontos $A$, $B$ e $C$ na circunferência que tangencia os três lados do triângulo $ABC$, temos\n$$\n\\begin{aligned}\n& AE=AF=x \\\\\n& BF=BD=y \\\\\n& CD=CE=z\n\\end{aligned}\n$$\n\n(b)\n![](attached_image_2.png)\nEm virtude do item anterior, temos as seguintes equações:\n$$\n\\begin{aligned}\n& x+y=13 \\\\\n& y+z=14 \\\\\n& z+x=15\n\\end{aligned}\n$$\nResolvendo o sistema, temos $x=7$, $y=6$ e $z=8$. Portanto, $BF=6$.\n\n(c) Seja $r$ o raio da circunferência que tangencia os três lados do triângulo. Usando o Teorema de Pitágoras, podemos calcular os quadrados das distâncias do ponto $I$ até os pontos $R$, $T$ e $V$ :\n$$\n\\begin{aligned}\nIR^{2} & =r^{2}+(14+x)^{2} \\\\\n& =r^{2}+21^{2} \\\\\nIT^{2} & =r^{2}+(15+y)^{2} \\\\\n& =r^{2}+21^{2} \\\\\nIV^{2} & =r^{2}+(13+z)^{2} \\\\\n& =r^{2}+21^{2}\n\\end{aligned}\n$$\nO mesmo pode ser feito para os pontos $S$, $U$ e $X$. Logo,\n$$\nIR=IS=IT=IU=IV=IX=\\sqrt{r^{2}+21^{2}}\n$$\nConcluímos que o hexágono $RSTUVX$ é inscritível, pois os seis pontos estão sobre uma circunferência de centro $I$ e raio $\\sqrt{r^{2}+21^{2}}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26099,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUsando a fatoração da diferença de quadrados, $x^{2}-y^{2}=(x-y)(x+y)$, podemos escrever\n$$\nx^{2^{n}}-y^{2^{n}}=\\left(x^{2^{n-1}}+y^{2^{n-1}}\\right)\\left(x^{2^{n-1}}-y^{2^{n-1}}\\right) \\text{.}\n$$\na) Explique por que $3^{2^{2018}}-2^{2^{2018}}$ pode ser escrito como produto de 2018 inteiros maiores que 1 e distintos.\nb) Verifique que $3^{2^{n}}-1=\\left(3^{2^{n-1}}+1\\right)\\left(3^{2^{n-2}}+1\\right)\\left(3^{2^{n-3}}+1\\right) \\ldots\\left(3^{2}+1\\right)\\left(3^{1}+1\\right)\\left(3^{1}-1\\right)$.\nc) Usando o item anterior, verifique que $2^{n+1}$ é um divisor de $3^{2^{n}}-1$.\nd) Conhecendo a fatoração\n$$\nx^{m}-y^{m}=(x-y)\\left(x^{m-1}+x^{m-2} y+x^{m-3} y^{2}+\\ldots+x y^{m-2}+y^{m-2}\\right)\n$$\nencontre um inteiro positivo $n$ com mais de 2018 divisores positivos tal que $3^{n-1}-2^{n-1}$ é múltiplo de $n$.",
    "options": [],
    "answer": "3^{2^{2017}} - 2^{2^{2017}}",
    "solution": "Solution:\n\na) Usando a fatoração da diferença de quadrados, temos:\n$$\n\\begin{aligned}\nx^{2^{n}}-y^{2^{n}} & =\\left(x^{2^{n-1}}+y^{2^{n-1}}\\right)\\left(x^{2^{n-1}}-y^{2^{n-1}}\\right) \\\\\nx^{2^{n-1}}-y^{2^{n-1}} & =\\left(x^{2^{n-2}}+y^{2^{n-2}}\\right)\\left(x^{2^{n-2}}-y^{2^{n-2}}\\right) \\\\\nx^{2^{n-2}}-y^{2^{n-2}} & =\\left(x^{2^{n-3}}+y^{2^{n-3}}\\right)\\left(x^{2^{n-3}}-y^{2^{n-3}}\\right) \\\\\n& \\cdots \\\\\nx^{2}-y^{2} & =(x+y)(x-y)\n\\end{aligned}\n$$\nMultiplicando todas as equações e cancelando os fatores repetidos, obtemos\n$$\nx^{2^{n}}-y^{2^{n}}=\\left(x^{2^{n-1}}+y^{2^{n-1}}\\right)\\left(x^{2^{n-2}}+y^{2^{n-2}}\\right) \\ldots\\left(x^{2}+y^{2}\\right)(x+y)(x-y)\n$$\nAssim, em geral, substituindo $x=3$ e $y=2,3^{2^{n}}-2^{2^{n}}$ é o produto de $n$ fatores da forma $2^{2^{i}}+3^{2^{i}}$ com $i$ pertencente ao conjunto $\\{0,1,2, \\ldots, n-1\\}$.\n\nb) Na fatoração obtida no item anterior, basta fazer $x=3$ e $y=1$.\n\nc) Como cada termo da forma $3^{2^{i}}+1$ possui o fator 2 por ser par, podemos concluir que o produto de todos os números da forma $3^{2^{i}}+1$, com $i \\in\\{1,2, \\ldots, n-1\\}$ é múltiplo de $2^{n-1}$. Além disso, o fator $3^{2^{0}}+1=2^{2}$ possui dois fatores 2 . Assim, usando a expressão encontrada no item anterior, podemos concluir que $2^{n-1} \\cdot 2^{2}=2^{n+1}$ é um divisor de $3^{2^{n}}+1$.\n\nd) Seja $n=3^{2^{2017}}-2^{2^{2017}}$. Pelo primeiro item, todos os 2019 números da lista\n$$\n1,2^{2^{0}}+3^{2^{0}}, 2^{2^{1}}+3^{2^{1}}, \\ldots, 3^{2^{2017}}+2^{2^{2017}}\n$$\nsão divisores de $n$. A partir da fatoração dada, podemos concluir que se $x$ e $y$ são inteiros, então $x-y$ é um divisor de $x^{m}-y^{m}$. Fazendo $x=2^{2^{2017}}$ e $y=3^{2^{2017}}$, obtemos que $n=3^{2^{2017}}-2^{2^{2017}}$ é um divisor de $3^{2^{2017} m}-2^{2^{2017} m}$, para qualquer inteiro positivo $m$. Note agora que\n$$\nn-1=\\left(3^{2^{2017}}-1\\right)-2^{2^{2017}}\n$$\nPelo terceiro item, $3^{2^{2017}}-1$ é um múltiplo de $2^{2017}$. Como $2017<2^{2017}$, também temos que $2^{2017}$ é um divisor de $2^{2^{2017}}$. Assim, $\\left(3^{2^{2017}}-1\\right)-2^{2^{2017}}$ é um número da forma $2^{2017} m$. Daí, $n$ divide $3^{n-1}-2^{n-1}$ e, além disso, possui pelo menos de 2019 divisores positivos.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26100,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSeja $Q$ um ponto na circunferência de diâmetro $AB$, sendo $Q$ diferente de $A$ e $B$. Seja $QH$ a reta perpendicular a $AB$ que passa por $Q$, sendo $H$ pertencente a $AB$. Os pontos de interseção da circunferência de diâmetro $AB$ e a circunferência de centro $Q$ e raio $QH$ são $C$ e $D$. Prove que $CD$ passa pelo ponto médio de $QH$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nConsideremos as duas figuras a seguir. Sejam $\\angle CQH = \\alpha$, $\\angle QCD = \\beta$ e $E$ a interseção de $QH$ com a circunferência de diâmetro $AB$. Delas, podemos concluir que:\n\ni) Temos $\\angle CDH = \\frac{\\alpha}{2}$, pois na circunferência de centro em $Q$ o seu ângulo central $\\angle CQH$ é correspondente ao $\\overparen{CH} = \\alpha$.\n\n![](attached_image_1.png)\n\nii) $\\angle CDE = \\alpha$, pois ele está inscrito ao arco $\\overparen{CE}$, assim como o ângulo $\\angle CQE$, considerando a circunferência de diâmetro $AB$. Com isso, temos que:\n$$\n\\angle HDE = \\alpha - \\frac{\\alpha}{2} = \\frac{\\alpha}{2}\n$$\n\niii) $\\angle QED = \\beta$, pois está inscrito ao arco $\\overparen{QD}$, assim como o ângulo $\\angle QCD = \\beta$.\n\n![](attached_image_2.png)\n\niv) $QH = HE$, pois $QE$ é perpendicular ao diâmetro $AB$.\n\nPor fim,\n- note que $\\triangle QMC \\sim \\triangle DME$. Daí, podemos concluir que:\n$$\n\\begin{aligned}\n\\frac{QM}{DM} &= \\frac{QC}{DE} \\\\\nQM &= QC \\cdot \\frac{DM}{DE}\n\\end{aligned}\n$$\n- Aplicando o Teorema da Bissetriz Interna no $\\triangle MDE$, temos que:\n$$\n\\begin{aligned}\n\\frac{DM}{HM} &= \\frac{DE}{HE} \\\\\nHM &= HE \\cdot \\frac{DM}{DE}\n\\end{aligned}\n$$\nPelas duas conclusões anteriores e sabendo que\n$$\nQC = HE = QH = \\text{raio da circunferência de centro } Q\n$$\npodemos concluir que $QM = HM$. Portanto, $CD$ passa pelo ponto médio de $QH$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26101,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSeja $ABC$ um triângulo acutângulo com $AB \\neq AC$. Considere todos os retângulos com dois vértices sobre o lado $BC$, um sobre o lado $AB$ e um sobre o lado $AC$. Chamaremos de centro do retângulo o ponto de encontro das diagonais. Na figura a seguir, o centro do retângulo $DEFG$ é o ponto $P$.\n\n![](attached_image_1.png)\n\nDeseja-se determinar qual o lugar geométrico determinado pelos centros destes retângulos, ou seja, se fossem marcados todos os centros possíveis, qual figura seria formada pela união destes pontos. Para resolver este problema, usaremos geometria analítica. Seja $O$ o pé da altura por $A$ ao lado $BC$. Considere o sistema de coordenadas com origem em $O$, eixo $x$ sobre a reta $BC$ e eixo $y$ sobre a reta $OA$. Então podemos marcar os pontos $O=(0,0), A=(0, a)$, $B=(b, 0)$ e $C=(c, 0)$. Vale ressaltar que de $AB \\neq AC$ podemos afirmar que $b \\neq -c$, ou seja, $b+c \\neq 0$.\n\na) Determine as coordenadas dos pontos médios $M$ e $N$ dos segmentos $OA$ e $BC$, respectivamente. Em seguida, determine a equação da reta $MN$.\n\nb) Se considerarmos as coordenadas do ponto $D=(z, w)$, quais as coordenadas dos pontos $E, F$ e $G$?\n\nc) O centro do retângulo é o encontro das diagonais e, como todo retângulo é um paralelogramo, ele coincide com o ponto médio da diagonal $DF$. Determine as coordenadas deste ponto e conclua que ele está sobre a reta $MN$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_2.png)\n\nComo as coordenadas do ponto médio de um segmento são as médias aritméticas das coordenadas de seus extremos, obtemos:\n$$\n\\begin{aligned}\nM & =\\left(0, \\frac{a}{2}\\right) \\\\\nN & =\\left(\\frac{b+c}{2}, 0\\right)\n\\end{aligned}\n$$\nPara determinar a equação da reta $MN$, devemos determinar os números reais $m$ e $n$ tais que as coordenadas dos pontos $M$ e $N$ satisfazem $y=m \\cdot x+n$. Para isto, basta resolver o sistema a seguir:\n$$\n\\left\\{\\begin{array}{l}\n\\frac{a}{2}=m \\cdot 0+n \\\\\n0=m \\cdot \\frac{b+c}{2}+n\n\\end{array}\\right.\n$$\ncuja solução é $(n, m)=\\left(a / 2,-\\frac{a}{b+c}\\right)$.\nPortanto, a equação da reta $MN$ é\n$$\ny=-\\frac{a}{b+c} \\cdot x+\\frac{a}{2}\n$$\n\nb)\nComo os pontos $E$ e $F$ estão sobre o eixo $x$, suas ordenadas são nulas. Sabendo que os segmentos $DE$ e $GF$ são perpendiculares ao eixo $x$, temos $x_{E}=x_{D}=z$ e $x_{G}=x_{F}$. Como o segmento $DG$ é perpendicular ao eixo $y$, temos $y_{G}=y_{D}=w$. Para determinar $x_{G}$, basta estudar as equações das retas $AC$ e $GF$. A equação da reta $AC$ é $y=-\\frac{a}{c} \\cdot x+a$. O ponto $G$ está sobre esta reta e possui $y_{G}=w$, então\n$$\nw=-\\frac{a}{c} \\cdot x_{G}+a \\Rightarrow x_{G}=\\frac{c}{a} \\cdot(a-w)\n$$\nAs coordenadas dos pontos $E, F$ e $G$ são:\n$$\n\\begin{aligned}\nE & =(z, 0) \\\\\nF & =\\left(\\frac{c}{a} \\cdot(a-w), 0\\right) \\\\\nG & =\\left(\\frac{c}{a} \\cdot(a-w), w\\right)\n\\end{aligned}\n$$\n\nc)\nO ponto médio de $DF$ possui coordenadas\n$$\nP=\\left(\\frac{1}{2}\\left(z+\\frac{c}{a} \\cdot(a-w)\\right), \\frac{1}{2}(w+0)\\right)=\\left(\\frac{z a+c a-w c}{2 a}, \\frac{w}{2}\\right)\n$$\nComo o ponto $D$ está sobre a reta $AB$, temos, por semelhança de triângulos, que\n$$\n\\begin{aligned}\n\\frac{b}{a} & =\\frac{b-z}{w} \\\\\nw b & =a(b-z)\n\\end{aligned}\n$$\nAgora, devemos provar que as coordenadas do ponto $P$ satisfazem a equação da reta $MN$ encontrada no primeiro item deste problema. Ou seja,\n$$\n\\begin{aligned}\nm \\cdot x_{P}+n & =-\\frac{a}{b+c} \\cdot \\frac{z a+c a-w c}{2 a}+\\frac{a}{2} \\\\\n& =\\frac{-z a-c a+w c}{2(b+c)}+\\frac{a}{2} \\\\\n& =\\frac{-z a-c a+w c+a b+a c}{2(b+c)} \\\\\n& =\\frac{w c+a(b-z)}{2(b+c)} \\\\\n& =\\frac{w c+w b}{2(b+c)} \\\\\n& =\\frac{w}{2} \\\\\n& =y_{P}\n\\end{aligned}\n$$\nConcluímos assim que o ponto $P$ está sobre a reta $MN$. É imediato verificar que todo ponto do segmento $MN$ pode ser o centro de algum retângulo. Portanto, o lugar geométrico dos centros dos retângulos é o segmento $MN$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26102,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nArtur e Dinah vão disputar o jogo do par ou ímpar maluco. Dinah escolhe \"par\" e Artur escolhe \"ímpar\". Em seguida, cada um escreve um número inteiro positivo em uma folha de papel sem que o outro a veja. Emílio recolhe as duas folhas, multiplica os números e declara Dinah vencedora se o resultado for par e Artur vencedor se for ímpar.\n\na) Como deve fazer Dinah para que ela sempre ganhe o jogo?\n\nEmílio sugere uma modificação na disputa. Primeiramente ele pede que Artur e Dinah escrevam apenas números que não sejam divisíveis por três. Ele recolhe as folhas, multiplica os dois números, divide o resultado por três e declara Dinah vencedora se o resto da divisão for igual a 1 e Artur vencedor se esse resto for igual a 2.\n\nb) Mostre que Dinah não pode mais ter uma estratégia vencedora.\n\nc) Mostre que Artur e Dinah têm a mesma probabilidade de ganhar.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na) Temos que o produto de número par por um número ímpar é sempre par! Portanto, se Dinah pedir par e escrever no papel um número par, ela certamente ganhará.\n\nb) Dinah escolhe um número natural, digamos $3q_1 + r_1$, onde $q_1$ e $r_1$ são naturais e $r_1$ é o resto na divisão desse número por três. Ou seja, $r_1$ pode ser 1 ou 2 (não pode ser zero pela regra do jogo). O mesmo para Artur, ou seja, ele escolhe um número natural da forma $3q_2 + r_2$, onde $r_2$ pode ser 1 ou 2. O produto desses dois números será:\n$$\n(3q_1 + r_1)(3q_2 + r_2) = 3(3q_1q_2 + q_1r_2 + q_2r_1) + r_1r_2\n$$\nOu seja, quando dividimos o resultado por três, o resto na divisão por 3 será igual ao resto que obtemos quando dividimos $r_1 r_2$ por 3. Temos os casos:\n\n| $r_1$ | $r_2$ | Resto na divisão por 3 do resultado |\n| :---: | :---: | :---: |\n| 1 | 1 | 1 |\n| 1 | 2 | 2 |\n| 2 | 1 | 2 |\n| 2 | 2 | 1 |\n\nLogo, Dinah não tem estratégia vencedora. Para qualquer número que Dinah escolha, Artur tem uma opção que o torna vencedor.\n\nc) Como são quatro casos (veja a tabela no item anterior) e Dinah ganha em dois casos e Artur ganha em dois casos, concluímos que os dois têm a mesma probabilidade de ganhar.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26103,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO número $1089$ tem uma propriedade interessante. Quando fazemos a multiplicação deste número por $9$, como é mostrado a seguir,\n\n| 1 | 0 | 8 | 9 |\n| :--- | :--- | :--- | :--- |\n|   |   | $\\times$ | 9 |\n| 9 | 8 | 0 | 1 |\n\nobtemos o número $9801$ que é o número $1089$ com os seus algarismos escritos da esquerda para direita!\n\na) Encontre um número de cinco algarismos $A B C D E$ tal que sua multiplicação por $9$ seja igual ao número que tem os dígitos de $A B C D E$ escritos da direita para a esquerda, ou seja,\n\n| $A$ | $B$ | $C$ | $D$ | $E$ |\n| :---: | :---: | :---: | :---: | :---: |\n|   |   |   | $\\times$ | 9 |\n| $E$ | $D$ | $C$ | $B$ | $A$ |\n\nb) Encontre todos os números de sete algarismos cuja multiplicação por $9$, como anteriormente, inverte a posição de seus algarismos.",
    "options": [],
    "answer": "a) 10989; b) 1099989",
    "solution": "Solution:\n\na) Estamos procurando um número de cinco algarismos $A B C D E$ tal que\n$$\nA B C D E \\times 9 = E D C B A\n$$\nSomando $A B C D E$ dos dois lados dessa igualdade, obtemos que $A B C D E \\times 10 = E D C B A + A B C D E$. Mas o número $A B C D E \\times 10$ pode ser representado ainda pelos algarismos $A B C D E 0$. Sendo assim, obtivemos que $E D C B A + A B C D E = A B C D E 0$. Para visualizar melhor, mostramos essa igualdade da seguinte forma:\n$$\n\\begin{array}{rccccc} \n& E & D & C & B & A \\\\\n+ & A & B & C & D & E \\\\\n\\hline A & B & C & D & E & 0\n\\end{array}\n$$\nQuando somamos dois números contendo cinco algarismos e encontramos como resultado um número contendo seis algarismos, esse último número tem o seu primeiro algarismo igual a $1$. Assim, temos que $A=1$ :\n$$\n\\begin{array}{rccccc} \n& E & D & C & B & 1 \\\\\n+ & 1 & B & C & D & E \\\\\n\\hline 1 & B & C & D & E & 0\n\\end{array}\n$$\nComo $A=1$ e o último algarismo do resultado da soma é igual a $0$, necessariamente devemos ter $E=9$ :\n$$\n\\begin{array}{rrrrrr} \n& & & & 1 \\\\\n& 9 & D & C & B \\\\\n+ & 1 & B & C & D & 9 \\\\\n\\hline 1 & B & C & D & 9 & 0\n\\end{array}\n$$\no que implica que:\n$$\n\\begin{array}{rrrr} \n& & & 1 \\\\\n& D & C & B \\\\\n+ & B & C & D \\\\\n\\hline B & C & D & 9\n\\end{array}\n$$\nAgora, temos duas possibilidades para o valor de $B$. Devemos ter então $B=1$ ou $B=0$.\n\nNo caso em que $B=1$, temos que:\n$$\n\\begin{aligned}\n& 1 \\\\\n& \\begin{array}{lll}\nD & C & 1\n\\end{array} \\\\\n& \\begin{array}{cccc}\n+ & 1 & C & D \\\\\n\\hline 1 & C & D & 9\n\\end{array}\n\\end{aligned}\n$$\nAssim, devemos ter que $D=7$ e então:\nou de maneira mais simples:\n$$\n\\begin{array}{r}\n7 C \\\\\n+\\quad 1 C \\\\\n\\hline 1 C 7\n\\end{array}\n$$\no que é impossível, já que a soma de dois números que terminam com o mesmo algarismo deve resultar em um número par.\n\nVamos nos concentrar no caso $B=0$, do qual obtemos então:\n\n| $D$ | $C$ |\n| :---: | :---: |\n| + | $C$ |\n\nAssim, temos que $D=8$ :\n$$\n\\begin{array}{r} \\\\\n8 C 0 \\\\\n8 \\quad \\\\\n+\\quad C 8 \\\\\n\\hline C 89\n\\end{array}\n$$\no que nos dá $C=9$. Concluímos, portanto, que $A B C D E = 10989$.\n\nb) A solução desse item é semelhante àquela do item anterior. Estamos procurando um número de sete algarismos $A B C D E F G$ tal que $A B C D E F G \\times 9 = G F E D C B A$. Somando $A B C D E F G$ dos dois lados dessa igualdade, obtemos que:\n\n|   | $G$ | $F$ | $E$ | $D$ | $C$ | $B$ | $A$ |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| + | $A$ | $B$ | $C$ | $D$ | $E$ | $F$ | $G$ |\n| $A$ | $B$ | $C$ | $D$ | $E$ | $F$ | $G$ | 0 |\n\nLogo, concluímos que $A=1$ e que $G=9$, obtendo assim:\n\n|   |   |   |   |   | 1 |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n|   | $F$ | $E$ | $D$ | $C$ | $B$ |\n| + | $B$ | $C$ | $D$ | $E$ | $F$ |\n| $B$ | $C$ | $D$ | $E$ | $F$ | 9 |\n\nDaí, concluímos que $B=1$ ou que $B=0$. Se fosse correto que $B=1$, então, para que o último algarismo do resultado fosse igual a $9$, deveríamos ter que $F=7$. Logo, a última soma mostrada tem um número começando com o algarismo $7$ e outro com o algarismo $1$ e não poderia resultar em um número contendo seis algarismos. Assim, devemos abandonar a possibilidade de que $B=1$ e nos concentrar no caso $B=0$. Nesse caso, temos que $F=8$ e então a soma fica:\n$$\n\\begin{array}{r}\n8 E D C \\\\\n+\\quad C D E \\\\\n\\hline C D E\n\\end{array}\n$$\nTemos agora duas possibilidades: $C=8$ ou $C=9$. Se fosse correto $C=8$ então teríamos que $E=0$. Nesse caso, a soma acima ficaria:\n\n| $80 D 8$ |\n| ---: |\n| $+\\quad 8 D 0$ |\n| $8 D 08$ |\n\no que é impossível, já que para que $D+D$ termine com o algarismo $0$ deveríamos ter que $D=5$ e assim:\n\n| 8058 |\n| ---: |\n| $+\\quad 850$ |\n| 8508 |\n\no que é, claramente, errado! Logo devemos abandonar a possibilidade de que seja $C=8$ e nos concentrar no caso $C=9$. Nesse caso, temos que $E=9$, logo:\n![](attached_image_1.png)\nque só pode ser satisfeita se $D=9$.\nAssim, a única possibilidade é $A B C D E F G = 1099989$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26104,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEm um torneio com 10 times, cada um deles se enfrenta uma única vez. Além disso, não ocorrem empates e cada um possui $50\\%$ de chance de ganhar qualquer partida. Qual a probabilidade de, após contabilizadas as pontuações dos $\\frac{10 \\cdot 9}{2}=45$ jogos, não existirem dois jogadores com o mesmo número de vitórias?",
    "options": [],
    "answer": "10!/2^45",
    "solution": "Solution:\n\nComo cada uma das 45 partidas possui dois resultados possíveis, existem $2^{45}$ maneiras distintas de distribuir as vitórias do torneio. Cada time pode vencer de 0 até 9 partidas. Como são 10 times, para que todos os números de vitórias sejam distintos, cada inteiro do conjunto $\\{0,1,2, \\ldots, 9\\}$ deve ser associado como número de vitórias de um único jogador. Isso pode ser feito de $10!$ formas. Resta saber se essa distribuição de números de vitórias é realizável neste torneio. Se $j_{i}$ é o jogador que deverá vencer apenas $i$ partidas, com $i \\in \\{0,1,2, \\ldots, 9\\}$, então faça $j_{i}$ ganhar de todos os jogadores $j_{0}, j_{1}, \\ldots, j_{i-1}$ e perder de todos os jogadores $j_{i+1}, j_{i+2}, \\ldots, j_{9}$. Isso garantirá a distribuição planejada. Portanto, a probabilidade é $\\dfrac{10!}{2^{45}}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26105,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDuas tangentes são desenhadas de um ponto $A$ a um círculo de centro $O$, tocando-o em $B$ e $C$. Seja $H$ o ortocentro do triângulo $ABC$, sabendo que $\\angle BAC = 40^\\circ$, encontre o valor do ângulo $\\angle HCO$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "40°",
    "solution": "Solution:\n\nComo $AC$ é tangente ao círculo, temos $\\angle ACO = 90^\\circ$. Assim\n$$\n\\begin{aligned}\n\\angle HCO & = 90^\\circ - \\angle ACF \\\\\n& = \\angle CAF \\\\\n& = 40^\\circ\n\\end{aligned}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26106,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\na) Observe as somas:\n![](attached_image_1.png)\nVerifique que vale:\n$$\n\\begin{gathered}\n\\frac{(k+1) \\cdot (k+2) \\cdot (k+3) \\cdot \\ldots \\cdot (k+901)}{901} + (k+2) \\cdot (k+3) \\cdot (k+4) \\cdot \\ldots \\cdot (k+901) = \\\\\n\\frac{(k+2) \\cdot (k+3) \\cdot \\ldots \\cdot (k+901) \\cdot (k+902)}{901}\n\\end{gathered}\n$$\n\nb) Seja $N$ a soma dos números:\n$$\n\\begin{array}{rrrrrrrrrr} \n& 1 & \\times & 2 & \\times & 3 & \\times & \\ldots & \\times & 900 \\\\\n+ & 2 & \\times & 3 & \\times & 4 & \\times & \\ldots & \\times & 901 \\\\\n+ & \\ldots & & & & & & & & \\\\\n+ & 1116 & \\times & 1117 & \\times & 1118 & \\times & \\ldots & \\times & 2015\n\\end{array}\n$$\nMostre que $901 \\cdot N$ é divisível por todo elemento do conjunto $\\{1116,1117, \\ldots, 2016\\}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\na)\n$$\n\\begin{array}{r}\n\\frac{(k+1) \\cdot (k+2) \\cdot (k+3) \\cdot \\ldots \\cdot (k+901)}{901} + (k+2) \\cdot (k+3) \\cdot (k+4) \\cdot \\ldots \\cdot (k+901) \\\\\n= \\frac{(k+2) \\cdot (k+3) \\cdot \\ldots \\cdot (k+901) \\cdot [(k+1)+901]}{901} \\\\\n= \\frac{(k+2) \\cdot (k+3) \\cdot \\ldots \\cdot (k+901) \\cdot (k+902)}{901}\n\\end{array}\n$$\n\nb) Usando o item anterior 1116 vezes para $k=0,1,2, \\ldots, 1115$, podemos reduzir sucessivamente a quantidade de parcelas que constituem $N$ sempre operando com os dois primeiros termos da soma. Obtemos assim:\n$$\n\\begin{aligned}\nN &= \\frac{1116 \\cdot 1117 \\cdot 1118 \\cdot \\ldots \\cdot 2016}{901} \\\\\n901 \\cdot N &= 1116 \\cdot 1117 \\cdot 1118 \\cdot \\ldots \\cdot 2016\n\\end{aligned}\n$$\nPortanto, cada um dos números do conjunto $\\{1116,1117, \\ldots, 2016\\}$ é um divisor de $901 \\cdot N$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26107,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO triângulo de latas - Um menino tentou alinhar 480 latas em forma de um triângulo com uma lata na $1^{a}$ linha, 2 latas na $2^{a}$ e assim por diante. No fim sobraram 15 latas. Quantas linhas tem esse triângulo?",
    "options": [],
    "answer": "30",
    "solution": "Solution:\n\nSuponhamos que o triângulo está composto por $n$ linhas, logo foram usadas $1+2+3+\\cdots+n$ latas, assim\n$$\n480-15=1+2+\\cdots+n=\\frac{n(n+1)}{2} \\Longrightarrow n^{2}+n-930=0\n$$\nResolvendo a equação $n^{2}+n-930=0$, obtemos:\n$$\nn=\\frac{-1 \\pm \\sqrt{1+4 \\times 930}}{2}=\\frac{-1 \\pm 61}{2}\n$$\nAssim, $n=30$ que é única solução positiva desta equação. Logo o triângulo tem 30 linhas.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26108,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nJorge ganhou 15 roseiras para seu jardim, com a condição de plantá-las em 6 filas de 5 roseiras cada uma. Isso é possível? Em caso afirmativo faça um desenho indicando para Jorge como plantar as roseiras.",
    "options": [],
    "answer": "Yes. Take six straight lines with no two parallel and no three concurrent; plant at the fifteen pairwise intersection points so that each of the six lines forms a row containing five plants.",
    "solution": "Solution:\n\nÉ possível plantar as roseiras em 6 filas de 5 roseiras cada uma, conforme mostra o desenho a seguir.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26109,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nConstruindo quadrados perfeitos - Observe as seguintes igualdades:\n![](attached_image_1.png)\nSerá que isso é sempre verdadeiro? Isto é: o produto de quatro números inteiros consecutivos, mais $1$, é sempre um quadrado perfeito?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSim, será sempre um quadrado perfeito. De fato, se $n-1$, $n$, $n+1$ e $n+2$ são quatro inteiros consecutivos, então seu produto mais $1$ é dado por:\n$$\n\\begin{aligned}\n(n-1) n(n+1)(n+2)+1 & = n\\left(n^{2}-1\\right)(n+2)+1 \\\\\n& = n\\left(n^{3}+2 n^{2}-n-2\\right)+1 \\\\\n& = n^{4}+2 n^{3}-n^{2}-2 n+1 \\\\\n& = n^{4}+2 n^{3}+(n^{2}-2 n^{2})-2 n+1 \\\\\n& = \\left(n^{4}+2 n^{3}+n^{2}\\right)-2 n^{2}-2 n+1 \\\\\n& = \\left(n^{2}+n\\right)^{2}-2\\left(n^{2}+n\\right)+1 \\\\\n& = \\left[\\left(n^{2}+n\\right)-1\\right]^{2}\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26110,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFrações egípcias - Encontre números inteiros positivos $a$ e $b$, com $a > b$, tais que:\n$$\n\\frac{2}{7} = \\frac{1}{a} + \\frac{1}{b}\n$$",
    "options": [],
    "answer": "a = 28, b = 4",
    "solution": "Solution:\n\nA equação é equivalente a $2ab = 7(a + b)$. Como $2$ e $7$ são números primos entre si, segue que $ab$ é múltiplo de $7$ e que $a + b$ é múltiplo de $2$. Mas, para $ab$ ser múltiplo de $7$, a única possibilidade é $a$ ser múltiplo de $7$ ou $b$ ser múltiplo de $7$. Suponhamos primeiramente que $a$ é múltiplo positivo de $7$, ou seja, $a = 7k$ para algum inteiro positivo $k$. Daí obtemos:\n$$\n2ab = 7(a + b) \\Rightarrow 2k b = 7k + b \\Rightarrow (2k - 1) b = 7k\n$$\nEsta última equação implica que $b$ ou $2k - 1$ são múltiplos de $7$. Se $b$ é múltiplo de $7$, $b = 7m$, e assim\n$$\n\\frac{2}{7} = \\frac{1}{a} + \\frac{1}{b} \\Rightarrow \\frac{2}{7} = \\frac{1}{7k} + \\frac{1}{7m} \\Rightarrow \\frac{1}{k} + \\frac{1}{m} = 2\n$$\nMas, $\\frac{1}{k} \\leq 1$ e $\\frac{1}{m} \\leq 1$. Assim $\\frac{1}{k} + \\frac{1}{m} \\leq 2$. Portanto, a equação $\\frac{1}{k} + \\frac{1}{m} = 2$ possui única solução inteira positiva, a saber, $k = 1 = m$. Entretanto, esta solução não nos interessa, pois neste caso $a = b$.\n\nPassemos então ao caso em que $2k - 1$ é múltiplo de $7$.\n\nUma possibilidade para $2k - 1$ ser múltiplo de $7$ ocorre quando $k = 4$. Neste caso, temos que\n$$\n\\begin{gathered}\nk = 4 \\Rightarrow a = 7k = 28 \\\\\n(2k - 1) b = 7k \\Rightarrow 7b = 28 \\Rightarrow b = 4\n\\end{gathered}\n$$\nObtemos então\n$$\n\\frac{2}{7} = \\frac{1}{28} + \\frac{1}{4}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26111,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFamília Sétimo - O Sr. e Sra. Sétimo têm 7 filhos, todos nascidos em 1º de abril, na verdade em seis 1º de abril consecutivos. Este ano, para seus aniversários, a Sra. Sétimo fez um bolo com velinhas para cada um - o número de velas igual ao número de anos de cada um. João Sétimo, o filho que mais gosta de Matemática, reparou que nesse ano o número total de velinhas é o dobro do que havia 2 anos atrás e que há 2 bolos a mais. Quantas velinhas serão acesas esse ano?",
    "options": [],
    "answer": "26",
    "solution": "Solution:\n\nOs nascimentos ocorreram em seis 1º de abril, logo existem irmãos gêmeos. Como nesse ano temos 2 bolos a mais que há 2 anos atrás, então há 2 anos atrás o mais jovem ainda não tinha nascido, o penúltimo filho tinha acabado de nascer, e os gêmeos já tinham nascido. Atualmente o mais jovem tem 1 ano e os gêmeos têm $x$ anos com $x \\geq 3$. Temos:\n$$\n\\underbrace{1+2+3+4+5+6+x}_{\\text{número de velas nesse ano}} = 2 \\times \\underbrace{(1+2+3+4+x-2)}_{\\text{número de velas 2 anos atrás}} \\Rightarrow x=5\n$$\nLogo serão acesas $1+2+3+4+2 \\times 5+6=26$ velinhas.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26112,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNum triângulo $\\triangle ABC$, o ponto $F$ está sobre o lado $AC$ e $FC = 2 AF$. Se $G$ é o ponto médio do segmento $BF$ e $E$ o ponto de interseção da reta passando por $A$ e $G$ com o segmento $BC$, calcule a razão $\\frac{EC}{EB}$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "3",
    "solution": "Solution:\n\nTemos que $\\frac{FC}{AF} = 2$. Agora, trace o segmento $FH$, paralelo ao segmento $AE$ onde $H$ está sobre o segmento $BC$, como na figura a seguir.\n\nOs triângulos $\\triangle AEC$ e $\\triangle FHC$ são semelhantes pois têm lados paralelos. Isto implica que $CH = 2 EH$.\n\nPor outro lado, os triângulos $\\triangle BFH$ e $\\triangle BGE$ também são semelhantes, pois têm lados paralelos. Dessa semelhança e do fato que $G$ é ponto médio do segmento $BF$ concluímos que $E$ é ponto médio do segmento $BH$.\n\nAssim, $BE = EH$ e, portanto, $EC = EH + CH = EH + 2 EH = 3 EH = 3 EB$. Consequentemente, $\\frac{EC}{EB} = 3$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26113,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nReverso de um número - O reverso de um número inteiro de dois algarismos é o número que se obtém invertendo a ordem de seus algarismos. Por exemplo, 34 é o reverso de 43. Quantos números existem que somados ao seu reverso dão um quadrado perfeito?",
    "options": [],
    "answer": "8",
    "solution": "Solution:\n\nDenotemos por $ab$ e $ba$ o número e seu reverso. Temos que\n$$\nab + ba = 10a + b + 10b + a = 11(a + b)\n$$\nPor outro lado, $a \\leq 9$ e $b \\leq 9$, logo, $a + b \\leq 18$. Como $11$ é um número primo e $a + b \\leq 18$, para que $11(a + b)$ seja um quadrado perfeito, só podemos ter $a + b = 11$.\n\nAssim, temos 8 números satisfazendo a condição do problema: $29$, $38$, $47$, $56$, $65$, $74$, $83$ e $92$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26114,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nNa tabela ao lado, com 6 colunas e diversas linhas, estão escritos os números $1,2,3,4, \\ldots$ Qual é a posição do número $1000$?\n\n| 1 | 2 | 3 | 4 | 5 | 6 |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| 7 | 8 | 9 | 10 | 11 | 12 |\n| 13 | 14 | $\\cdots$ |  |  |  |\n|  |  |  |  |  |  |\n|  |  |  |  |  |  |\n| $\\vdots$ | $\\vdots$ | $\\vdots$ | $\\vdots$ | $\\vdots$ | $\\vdots$ |",
    "options": [],
    "answer": "row 167, column 4",
    "solution": "Solution:\nComo a tabela tem 6 colunas, em cada linha escrevemos 6 números consecutivos. Dividindo $1000$ por $6$ obtemos\n$$\n1000 = 6 \\times 166 + 4\n$$\n\n| $1^{\\text{a}}$ linha | 1 | 2 | 3 | 4 | 5 | 6 |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| $2^{\\text{a}}$ linha | 7 | 8 | 9 | 10 | 11 | 12 |\n| $3^{\\text{a}}$ linha | 13 | 14 | 15 | 16 | 17 | 18 |\n| $\\vdots$ | $\\vdots$ | $\\vdots$ | $\\vdots$ | $\\vdots$ | $\\vdots$ | $\\vdots$ |\n| $167^{\\text{a}}$ linha | 997 | 998 | 999 | 1000 |  |  |\n| $\\vdots$ | $\\vdots$ | $\\vdots$ | $\\vdots$ | $\\vdots$ | $\\vdots$ | $\\vdots$ |\n\nDeste modo, para escrever o número $1000$ na tabela serão necessárias $166$ linhas completas (terminando no número $6 \\times 166 = 996$) e mais uma linha com os $4$ números: $997$, $998$, $999$ e $1000$. Logo, $1000$ está escrito na $167^{\\text{a}}$ linha e na $4^{\\text{a}}$ coluna.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26115,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA superfície do globo terrestre consiste de água (70\\%) e de terra (30\\%). Dois quintos da terra são desertos ou cobertos por gelo e, um terço é pastagem, floresta ou montanha; o resto é cultivado. Que percentual da superfície total do globo terrestre é cultivada?",
    "options": [],
    "answer": "8%",
    "solution": "Solution:\nA fração da terra que é cultivada é\n$$\n1-\\frac{2}{5}-\\frac{1}{3}=\\frac{15-6-5}{15}=\\frac{4}{15}\n$$\nComo a terra é $\\frac{3}{10}$ do globo, temos que área cultivada é $\\frac{4}{15} \\times \\frac{3}{10}=\\frac{2}{25}$ do globo, isto é o $\\frac{2}{25} \\times 100 \\%=8 \\%$ do globo terrestre.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26116,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNuma divisão, aumentando o dividendo de $1989$ e o divisor de $13$, o quociente e o resto não se alteram. Qual é o quociente?",
    "options": [],
    "answer": "153",
    "solution": "Solution:\n\n$153$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26117,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoma dos terminados em 9 - A soma $S_{n}=9+19+29+39+\\cdots+a_{n}$ denota a soma dos primeiros $n$ números naturais terminados em $9$. Qual é o menor valor de $n$ para que $S_{n}$ seja maior do que $10^{5}$?",
    "options": [],
    "answer": "142",
    "solution": "Solution:\n\nNote que $S_{n}$ é a soma dos $n$ primeiros termos de uma progressão aritmética cujo primeiro termo é $a_{1}=9$ e a razão é $r=10$. Substituindo esses dados na fórmula $a_{n}=a_{1}+(n-1) r$ obtemos $a_{n}=9+10(n-1)$. Por outro lado, note que:\n\n$$\n\\begin{aligned}\n& 9=9+0 \\cdot 10 \\\\\n& 19=9+1 \\cdot 10 \\\\\n& 29=9+2 \\cdot 10 \\\\\n& 39=9+3 \\cdot 10 \\\\\n& \\ldots \\\\\n& a_{n}=9+(n-1) \\cdot 10\n\\end{aligned}\n$$\n\nEntão temos:\n\n$$\n\\begin{aligned}\nS_{n} & =9+19+29+\\cdots+a_{n} \\\\\n& =(9+0)+(9+10)+(9+2 \\cdot 10)+\\cdots+[9+(n-1) \\cdot 10] \\\\\n& =9 n+[1 \\cdot 10+2 \\cdot 10+3 \\cdot 10+\\cdots+(n-1) \\cdot 10] \\\\\n& =9 n+[1+2+3+\\cdots+(n-1)] \\cdot 10 \\\\\n& =9 n+\\frac{n(n-1)}{2} \\cdot 10 \\\\\n& =9 n+5 \\cdot n(n-1) \\\\\n& =5 n^{2}+4 n\n\\end{aligned}\n$$\n\nComo queremos que $S_{n} \\geq 10^{5}$ então precisamos encontrar o menor valor inteiro positivo $n$ tal que $5 n^{2}+4 n \\geq 10^{5}$, ou equivalentemente, $5 n^{2}+4 n-10^{5} \\geq 0$. Resolvendo a equação $5 x^{2}+4 x-10^{5}=0$ temos que\n\n$$\nx=\\frac{-4 \\pm \\sqrt{16+20 \\cdot 10^{5}}}{10}\n$$\n\nA raiz positiva é $x_{1}=\\frac{-4+\\sqrt{16+20 \\cdot 10^{5}}}{10} \\simeq 141,02$.\n\nMas, $5 x^{2}+4 x-10^{5}$ é positiva fora das raízes. Como estamos procurando o menor número positivo e inteiro tal que $5 x^{2}+4 x-10^{5} \\geq 0$, então $n=142$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26118,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO preço do feijão - A tabela e o gráfico, dados a seguir, mostram a evolução do preço médio de três tipos de feijão, $A$, $B$ e $C$, na bolsa de alimentos durante os primeiros quatro meses de certo ano: Desses 3 tipos, os que apresentam, respectivamente, o maior e o menor crescimento percentual no preço nesse período são:\n\n(a) $A$ e $B$\n(b) $A$ e $C$\n(c) $B$ e $C$\n(d) $C$ e $A$\n(e) $C$ e $B$\n\n![](attached_image_1.png)\n\n|   | jan   | fev   | mar   | abr    |\n|:-:|:-----:|:-----:|:-----:|:------:|\n| A | 65,67 | 83,33 | 96,67 | 103,33 |\n| B | 73,30 | 80,50 | 99,55 | 109,50 |\n| C | 64,50 | 71,57 | 89,55 | 100,00 |",
    "options": [],
    "answer": "a",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26119,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO perímetro de um retângulo $ABCD$ é $20~\\mathrm{m}$. O menor comprimento, em metros, que a diagonal $AC$ pode ter é:\n\n(a) $0$\n(b) $\\sqrt{50}$\n(c) $10$\n(d) $\\sqrt{200}$\n(e) $20 \\sqrt{5}$",
    "options": [],
    "answer": "b",
    "solution": "Solution:\n\nDenotemos por $a$ e $b$ os comprimentos dos lados do retângulo, assim $2a + 2b = 20$, logo $a + b = 10$. Por outro lado, o quadrado do comprimento da diagonal pode ser calculado usando o teorema de Pitágoras, assim $d^{2} = a^{2} + b^{2}$. Como\n\n$$\n\\begin{aligned}\n2d^{2} &= 2a^{2} + 2b^{2} = (a^{2} + 2ab + b^{2}) + (a^{2} - 2ab + b^{2}) \\\\\n&= (a + b)^{2} + (a - b)^{2} \\\\\n&= 100 + (a - b)^{2}\n\\end{aligned}\n$$\n\ntemos que o comprimento da diagonal é mínimo quando $a = b$, e neste caso $2d^{2} = 100$ e $d = \\sqrt{50}$. A opção correta é (b).",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26120,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$O$ cubo - Alice tem uma folha de cartolina de $60~\\mathrm{cm}$ por $25~\\mathrm{cm}$. Ela quer cortar a folha para montar um cubo. Qual o cubo de maior volume que ela pode construir?",
    "options": [],
    "answer": "Edge length 15 cm; volume 3375 cm^3",
    "solution": "Solution:\n\nPodemos cortar um retângulo de $60~\\mathrm{cm}$ por $2{,}5~\\mathrm{cm}$ e os pedaços marcados com $\\circledast$ de dimensões $15~\\mathrm{cm}$ por $7{,}5~\\mathrm{cm}$. Assim na figura a linha pontilhada indica dobradura e a linha continua indica corte e com os pedaços de cartolina marcados com $\\circledast$ formamos a tampa.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 26121,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSexta-feira 13 - Qual o número máximo de sexta-feiras 13 que podem ocorrer num ano não bissexto? Neste caso, qual é o $10^{\\circ}$ dia do ano?",
    "options": [],
    "answer": "3; Saturday",
    "solution": "Solution:\n\nDado que os dias da semana se repetem a cada 7 dias, então a diferença entre os dias da semana é dada pelo resto ao dividir por 7 o número de dias transcorridos.\n\nNa tabela seguinte temos:\n- na primeira linha o número de dias entre o dia 13 de um mês e o dia 13 do mês seguinte;\n- na segunda linha o resto quando dividimos esse número por 7;\n- na terceira linha o resto quando dividimos por 7 o número de dias entre o 13 de janeiro e o 13 do mês correspondente, ou seja, é obtida somando os resultados obtidos na linha anterior desde janeiro até o mês correspondente e depois calculando o resto ao dividir por 7.\n\n| J-F | F-M | M-A | A-M | M-J | J-J | J-A | A-S | S-O | O-N | N-D |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| 31 | 28 | 31 | 30 | 31 | 30 | 31 | 31 | 30 | 31 | 30 |\n| 3 | 0 | 3 | 2 | 3 | 2 | 3 | 3 | 2 | 3 | 2 |\n| 3 | 3 | 6 | 1 | 4 | 6 | 2 | 5 | 0 | 3 | 5 |\n\nOs valores iguais na última linha, significam que nestes meses o dia 13 caiu no mesmo dia da semana. Em particular esta última linha nos diz que 13 de fevereiro, 13 de março e 13 de novembro correspondem ao mesmo dia da semana. Logo, temos no máximo três sexta-feiras treze.\n\nNesse caso temos que 13 de janeiro ocorreu 3 dias antes de sexta-feira, isto é, terça-feira e o dia 10 de janeiro aconteceu 3 dias antes, isto é, no sábado.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26122,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOperação em uma fração - Que número se deve somar aos dois termos de uma fração para se obter o inverso dessa mesma fração?",
    "options": [],
    "answer": "c = -(a + b), with the special case a = b (fraction equal to 1) where any c works",
    "solution": "Solution:\n\nSeja $\\frac{a}{b}$ a fração procurada e seja $c$ um número tal que $\\frac{a+c}{b+c}=\\frac{b}{a}$. Esta igualdade é equivalente a $(a+c) a=(b+c) b$. Assim temos:\n$$\n(a+c) a=(b+c) b \\Longrightarrow a^{2}+a c-b^{2}-b c=0 \\Longrightarrow\\left(a^{2}-b^{2}\\right)+c(a-b)=0\n$$\nDonde\n$$\n0=\\left(a^{2}-b^{2}\\right)+c(a-b)=(a-b)(a+b)+c(a-b)=(a-b)(a+b+c)\n$$\nPortanto $(a-b)(a+b+c)=0$. Temos dois casos:\n$\\left.1^{o}\\right) a-b=0 \\Longrightarrow a=b$. Nesse caso a fração é igual a $1=\\frac{a}{a}$ e podemos somar qualquer número.\n$\\left.2^{o}\\right) a+b+c=0 \\Longrightarrow c=-(a+b)$. Nesse caso temos que somar $-a-b$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26123,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm quadriculado - O retângulo quadriculado na figura é feito de 31 segmentos de $0{,}5~\\mathrm{cm}$, e compreende 12 quadrados. Rosa desenhou numa folha retangular de $21~\\mathrm{cm}$ por $29{,}7~\\mathrm{cm}$ quadriculada com quadrados de lado $0{,}5~\\mathrm{cm}$, um grande retângulo quadriculado feito com 1997 segmentos. Quantos quadrados tem esse retângulo?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "798, 936, or 966",
    "solution": "Solution:\n\nSejam $m$ e $n$ respectivamente, o número de segmentos de $0{,}5~\\mathrm{cm}$ sobre dois lados consecutivos do retângulo. Sabemos que o número total de segmentos de $0{,}5~\\mathrm{cm}$ na divisão do retângulo em $m \\times n$ quadrados de lado $0{,}5~\\mathrm{cm}$ é: $m(n+1)+n(m+1)$ (prove isso). Assim,\n$$\nm(n+1)+n(m+1)=1997 \\Rightarrow n=\\frac{1997-m}{2m+1}\n$$\nAlém disso, um dos lados considerados é menor ou igual ao outro, digamos: $m \\leq n$. Nesse caso podemos concluir que $m \\leq 31$, pois\n$$\nn \\geq m \\Rightarrow n(m+1)+m(n+1) \\geq 2m(m+1)\n$$\nLogo $1997 \\geq 2m(m+1)$ e como $1998>1997$ segue que\n$$\n1998>2m(m+1) \\Rightarrow 999>m(m+1)\n$$\nDaí concluímos que $m<32$.\n\nPor outro lado temos que:\n$$\nn=\\frac{1997-m}{2m+1} \\Rightarrow 2n=\\frac{3994-2m}{2m+1}=\\frac{3995-(2m+1)}{2m+1} \\Rightarrow 2n=\\frac{3995}{2m+1}-1\n$$\nAssim, a questão se resume agora em pesquisar os divisores de $3995=5 \\times 17 \\times 47$. Os únicos valores de $m$ que atendem a condição $1 \\leq m \\leq 31$ são $m=2, m=8$ e $m=23$, que correspondem, respectivamente, aos divisores 5, 17 e 47. Para esses valores de $m$ temos $n=399, n=117$ e $n=42$ respectivamente. Os outros divisores darão configurações equivalentes (trocando $m$ por $n$).\n\nPortanto, Rosa pode ter construído 3 configurações diferentes com os 1997 segmentos. A primeira com $2 \\times 399$ quadrados, a segunda com $8 \\times 117$ quadrados e a terceira com $23 \\times 42$ quadrados.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26124,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPara a festa de aniversário da escola, Ana, Pedro, Miriam e Fábio levaram juntos 90 docinhos. A professora deles observou que:\n- se Ana tivesse levado 2 docinhos a mais;\n- se Pedro tivesse levado 2 docinhos a menos;\n- se Miriam tivesse levado o dobro;\n- se Fábio tivesse levado a metade;\nos 4 amigos teriam levado todos o mesmo número de docinhos. Quantos docinhos levou cada um dos amigos?",
    "options": [],
    "answer": "Ana 18, Pedro 22, Miriam 10, Fábio 40",
    "solution": "Solution:\n\nRepresentando o número de docinhos que cada um dos 4 amigos levou pela inicial de seu nome temos:\n$$\n\\left\\{\\begin{array}{l}\nA+P+M+F=90 \\\\\nA+2=P-2=2M=\\frac{F}{2}\n\\end{array}\\right.\n$$\nSegue da segunda equação que:\n$$\nP=A+4 \\quad ; \\quad M=\\frac{A+2}{2} \\quad ; \\quad F=2(A+2)\n$$\nSubstituindo esses valores na primeira equação obtemos:\n$$\nA+A+4+\\frac{A+2}{2}+2(A+2)=90 \\Rightarrow 9A=180-18 \\Rightarrow A=18\n$$\nLogo:\n$$\nP=18+4=22 ; \\quad M=\\frac{18+2}{2}=10 \\quad \\text{e} \\quad F=2(18+2)=40\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26125,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrês circunferências de raios $1~\\mathrm{cm}$, $2~\\mathrm{cm}$ e $3~\\mathrm{cm}$ são duas a duas tangentes exteriormente, como na figura ao lado.\nDetermine o raio da circunferência tangente exteriormente às três circunferências.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "6 cm",
    "solution": "Solution:\n\nLigando os centros das três circunferências obtemos o triângulo $\\triangle ABC$ de lados $AB=3~\\mathrm{cm}$, $AC=4~\\mathrm{cm}$ e $BC=5~\\mathrm{cm}$. Como $3^{2}+4^{2}=5^{2}$, esse triângulo é retângulo, com hipotenusa $BC$.\n\n![](attached_image_2.png)\n\nConstrua o retângulo $ABDC$, fazendo uma cópia $\\triangle BCD$, congruente ao triângulo $\\triangle ABC$ e com lado comum $BC$.\n\n![](attached_image_3.png)\n\nUma vez que $DC=AB=3$ e que a circunferência de centro $C$ também tem raio $3~\\mathrm{cm}$, vemos que o ponto $D$ está sobre essa circunferência.\n\nLigando o ponto $D$ a cada um dos vértices do triângulo $\\triangle ABC$ e prolongando esses segmentos até interceptarem as circunferências, obtemos os pontos $P_{1}, P_{2}$ e $P_{3}$.\n\n![](attached_image_4.png)\n\nTemos que:\n- $DP_{2}=DB+BP_{2}=CA+BP_{2}=4+2=6$.\n- $DP_{1}=DA+AP_{1}=5+1=6$.\n- $DP_{3}=DC+CP_{3}=3+3=6$.\n\nDeste modo $DP_{1}=DP_{2}=DP_{3}=6$. Assim se considerarmos a circunferência de centro $D$ e raio $6~\\mathrm{cm}$ vemos que esta circunferência passa por $P_{1}, P_{2}$ e $P_{3}$. Além disso, como os pontos $\\{D, A, P_{1}\\}$, $\\{D, B, P_{2}\\}$ e $\\{D, C, P_{3}\\}$ estão alinhados, segue que a circunferência de centro $D$ e raio $6~\\mathrm{cm}$ é tangente às três circunferências dadas de centros $A, B$ e $C$.\n\n![](attached_image_5.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26126,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCálculo de segmentos - As medidas do retângulo $ABCD$ são $1200~\\mathrm{m}$ por $150~\\mathrm{m}$. Além disso, $P$ está no prolongamento do lado $BC$ e dista $350~\\mathrm{m}$ de $C$. Determine $AP$, $PQ$, $PD$, $CQ$ e $DP$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "AP = 1300 m, PQ = 910 m, CQ = 840 m, DP = 1250 m, PD = 1250 m",
    "solution": "Solution:\n\nO triângulo $\\triangle ABP$ é retângulo com catetos $AB = 1200$ e $BP = 150 + 350 = 500$. Pelo Teorema de Pitágoras, temos:\n$$\nAP^2 = 1200^2 + 500^2 = (144 + 25) \\times 10^4 = 169 \\times 10^4 = (13 \\times 10^2)^2\n$$\nLogo, $AP = 13 \\times 10^2 = 1300~\\mathrm{m}$.\n\nAnalogamente, considerando o triângulo retângulo $\\triangle PCD$, temos:\n$$\nDP^2 = 350^2 + 1200^2 = (7^2 + 12^2 \\times 2^2)(5^2 \\times 10^2) = 25^2 \\times 50^2 \\Longrightarrow DP = 1250~\\mathrm{m}\n$$\n\nOs triângulos $\\triangle PCQ$ e $\\triangle PBA$ são retângulos com um ângulo em comum, logo são semelhantes; segue que:\n$$\n\\frac{PQ}{PA} = \\frac{PC}{PB} = \\frac{CQ}{AB}\n$$\nSubstituindo os valores conhecidos temos:\n$$\n\\frac{PQ}{1300} = \\frac{350}{500} = \\frac{CQ}{1200}\n$$\nLogo,\n$$\nPQ = \\frac{350 \\times 1300}{500} = 910~\\mathrm{m}\n$$\ne\n$$\nCQ = \\frac{350 \\times 1200}{500} = 840~\\mathrm{m}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26127,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nQual o menor número inteiro positivo $n$ tal que\n$$\n\\sqrt{n} - \\sqrt{n-1} < 0,01 ?\n$$",
    "options": [],
    "answer": "2501",
    "solution": "Solution:\n\nA inequação $\\sqrt{n} - \\sqrt{n-1} < 0,01$ é equivalente a $\\sqrt{n} < 0,01 + \\sqrt{n-1}$. Como os dois lados desta inequação são números positivos, podemos elevar esses dois membros ao quadrado para obter a inequação equivalente:\n$$\n(\\sqrt{n})^{2} < (0,01 + \\sqrt{n-1})^{2} \\Leftrightarrow n < 0,01^{2} + 0,02 \\sqrt{n-1} + n - 1\n$$\nDaí obtemos\n$$\n\\sqrt{n-1} > \\frac{1 - 0,01^{2}}{0,02} = \\frac{1 - \\frac{1}{100^{2}}}{\\frac{2}{100}} = \\frac{100^{2} - 1}{200}\n$$\nElevando novamente ao quadrado os dois lados (não negativos) desta inequação, obtemos:\n$$\nn-1 > \\frac{\\left(100^{2} - 1\\right)^{2}}{200^{2}} = \\frac{100^{4} - 2 \\times 100^{2} + 1}{4 \\times 100^{2}} = \\frac{100^{2}}{4} - \\frac{1}{2} + \\frac{1}{4 \\times 100^{2}} \n$$\nou seja,\n$n-1 > 2500 - \\frac{1}{2} + \\frac{1}{40000} \\Leftrightarrow n > 2500 - \\frac{1}{2} + \\frac{1}{40000} + 1 \\Leftrightarrow n > 2500 + \\frac{1}{2} + \\frac{1}{40000}$.\nUma vez que $\\frac{1}{2} + \\frac{1}{40000} < 1$, temos que o menor número inteiro maior que $2500 + \\frac{1}{2} + \\frac{1}{40000}$ é $2501$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26128,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDois motoristas viajam da cidade $A$ até a cidade $B$ e, imediatamente, regressam à cidade $A$. O primeiro motorista viaja com velocidade constante de $80~\\mathrm{km}/\\mathrm{h}$, tanto na ida quanto na volta. O segundo motorista viaja até a cidade $B$ com velocidade constante de $90~\\mathrm{km}/\\mathrm{h}$ e retorna com velocidade constante de $70~\\mathrm{km}/\\mathrm{h}$. Qual desses motoristas gasta menos tempo no percurso de ida e volta?",
    "options": [],
    "answer": "The driver who maintains a constant speed of 80 km/h both ways takes less total time.",
    "solution": "Solution:\n\nSeja $d$ a distância entre as cidades $A$ e $B$, e lembre que tempo $=$ distância $/$ velocidade.\n\n- O primeiro motorista viaja a distância de $2d$ com velocidade constante igual a $80~\\mathrm{km}/\\mathrm{h}$. Logo, o tempo total gasto por ele é:\n$$\nt = \\frac{2d}{80} = \\frac{d}{40}\n$$\n\n- O segundo motorista percorre a distância $d$, na ida, com velocidade igual a $90~\\mathrm{km}/\\mathrm{h}$ e, na volta, a mesma distância com velocidade de $70~\\mathrm{km}/\\mathrm{h}$. Logo o tempo gasto na ida e volta é:\n$$\nt' = \\frac{d}{70} + \\frac{d}{90} = \\frac{16d}{630} = \\frac{8d}{315}\n$$\n\nComo\n$$\n\\frac{d}{40} = \\frac{8d}{320} < \\frac{8d}{315}\n$$\n\nconclui-se que o motorista que viaja com velocidade constante de $80~\\mathrm{km}/\\mathrm{h}$ é o que gasta menos tempo no percurso de ida e volta.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26129,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nInterseção entre circunferências - Com centros nos vértices do triângulo equilátero $\\triangle ABC$ de lado $a$, foram desenhadas três circunferências de raio $r$.\nSe $r < a$ e $2r > a$, estas três circunferências são duas a duas concorrentes nos pontos $X, Y$ e $Z$, exteriores ao triângulo $\\triangle ABC$. Mostre que $\\triangle XYZ$ é um triângulo equilátero e calcule o comprimento do seu lado em termos de $a$ e $r$.\n\n![](attached_image_1.png)\n",
    "options": [],
    "answer": "a/2 + sqrt(3)*sqrt(r^2 - a^2/4)",
    "solution": "Solution:\n\nSeja $G$ o baricentro (encontro das medianas) do triângulo $ABC$. Como a figura é invariante por rotações de $60^\\circ$ ao redor do ponto $G$, temos que o triângulo $\\triangle XYZ$ é equilátero, e que $G$ também é o seu baricentro.\n\nVamos calcular o comprimento $L$ do seu lado.\n\nSeja $CM$ a altura do triângulo $\\triangle ABC$ em relação à base $AB$. Uma vez que a altura de um triângulo equilátero de lado $a$ tem medida $\\frac{a \\sqrt{3}}{2}$, e que o baricentro divide a altura em dois segmentos, um com o dobro do comprimento do outro, temos que:\n$$\nCM = \\frac{a \\sqrt{3}}{2}, \\quad GM = \\frac{1}{3} CM = \\frac{a \\sqrt{3}}{6} \\quad \\text{e} \\quad CG = \\frac{2}{3} CM = \\frac{a \\sqrt{3}}{3}\n$$\n![](attached_image_2.png)\n\nComo $AZ = BZ = r$ vemos que o ponto $Z$ está na reta mediatriz do segmento $AB$. Entretanto, esta mediatriz é a reta suporte da altura $CM$ do triângulo $ABC$. Isto implica que os pontos $C, G, M$ e $Z$ estão alinhados, e que o triângulo $\\triangle MZB$ é retângulo.\n\n![](attached_image_3.png)\n\nPelo Teorema de Pitágoras, obtemos:\n$$\nMZ = \\sqrt{ZB^2 - MB^2} = \\sqrt{r^2 - \\frac{a^2}{4}}\n$$\n$$\nGZ = GM + MZ = \\frac{a \\sqrt{3}}{6} + \\sqrt{r^2 - \\frac{a^2}{4}}\n$$\nAgora vamos considerar a altura $NZ$ do triângulo $XYZ$ em relação à sua base $XY$. Como esta altura tem comprimento $\\frac{L \\sqrt{3}}{2}$, e como $GZ = \\frac{2}{3} NZ = \\frac{L \\sqrt{3}}{3}$, concluímos que\n$$\nGZ = \\frac{L \\sqrt{3}}{3} = \\frac{a \\sqrt{3}}{6} + \\sqrt{r^2 - \\frac{a^2}{4}}\n$$\nEsta última igualdade implica que\n$$\nL = \\frac{a}{2} + \\sqrt{3} \\sqrt{r^2 - \\frac{a^2}{4}}\n$$\n\n![](attached_image_2.png)\n\n![](attached_image_3.png)",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26130,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nJogos de futebol - Os doze alunos de uma turma de olimpíada saíam para jogar futebol todos os dias após a aula de matemática, formando dois times de 6 jogadores cada e jogando entre si. A cada dia eles formavam dois times diferentes dos times formados em dias anteriores. Ao final do ano, eles verificaram que cada 5 alunos haviam jogado juntos num mesmo time exatamente uma vez. Quantos times diferentes foram formados ao longo do ano?",
    "options": [],
    "answer": "132",
    "solution": "Solution:\n\nPara cada grupo de 5 alunos, existe um único time formado que os contém. Logo, contamos $C_{12}^{5} = \\frac{12 \\cdot 11 \\cdot 10 \\cdot 9 \\cdot 8}{5!} = 792$ times para cada 5 alunos escolhidos. Por outro lado, em cada time de 6 jogadores, temos $C_{6}^{5} = 6$ modos de escolhermos cinco jogadores, ou seja, existem 6 grupos de 5 jogadores que eram mesmo time na nossa primeira contagem. Logo, o total de times formados é igual a $\\frac{792}{6} = 132$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26131,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLados de um paralelepípedo - Se $x$ e $y$ são números inteiros positivos tais que $x y z=240$, $x y+z=46$ e $x+y z=64$, qual é o valor de $x+y+z$?\n\n(a) 19\n(b) 20\n(c) 21\n(d) 24\n(e) 36",
    "options": [],
    "answer": "b",
    "solution": "Solution:\n\nSolução 1: De $x y z=240$, segue que $x y=\\frac{240}{z}$. Substituindo em $x y+z=46$, obtemos $\\frac{240}{z}+z=46$, ou seja, $z^{2}-46z+240=0$. As raízes dessa equação são números cuja soma é 46 e cujo produto é 240, e é fácil verificar que essas raízes são 6 e 40. Logo, $z=6$ ou $z=40$. De maneira completamente análoga, a substituição de $y z=\\frac{240}{x}$ em $x+y z=64$ nos leva a $x=4$ ou $x=60$.\n\nAgora, de $x y z=240$, segue que $y=\\frac{240}{x z}$. Como $y$ é um número inteiro, então $x z$ é um divisor de 240. De $x=4$ ou $x=60$ e $z=6$ ou $z=40$ segue que as possibilidades para $x z$ são\n$$\n\\underbrace{4}_{x} \\times \\underbrace{6}_{z}=24, \\underbrace{4}_{x} \\times \\underbrace{40}_{z}=160, \\underbrace{60}_{x} \\times \\underbrace{6}_{z}=360, \\underbrace{60}_{x} \\times \\underbrace{40}_{z}=2400\n$$\nVemos que só podemos ter $x=4$ e $z=6$, pois em qualquer outro caso o produto $x z$ não é um divisor de 240. Segue que $y=\\frac{240}{x z}=\\frac{240}{4 \\times 6}=10$, donde\n$$\nx+y+z=4+10+6=20\n$$\n\n\nSolução 2: Somando $x y+z=46$ e $x+y z=64$, obtemos\n$$\n(x+z)(y+1)=(x+z) y+(x+z)=x y+z+x+y z=46+64=110\n$$\ne vemos que $y+1$ é um divisor de 110. Logo, temos as possibilidades\n$$\ny+1=1,2,5,10,11,22,55 \\text{ e } 110\n$$\nou seja, $y=0,1,4,9,10,21,54$ e 109. Por outro lado, $y$ é um divisor de 240, porque $x y z=240$ e, além disso, $y$ é positivo, que nos deixa com as únicas possibilidades $y=1,4$ e 10. Examinemos cada caso de $y$.\n- Se $y=1$, então $110=(x+z)(y+1)=(x+z) \\times 2$, portanto, $x+z=55$. Como também $46=x y+z=x+z$, esse caso $y=1$ não é possível.\n- Se $y=4$, então $110=(x+z)(y+1)=(x+z) \\times 5$, portanto, $x+z=22$. Mas $240=x y z=4 x z$, portanto, $x z=60$. Podemos verificar (por exemplo, com uma lista de divisores de 60 ou, então, resolvendo a equação $w^{2}-22w+60=0$) que não há valores inteiros positivos de $x$ e $z$ que verifiquem essas duas condições $x+z=22$ e $x z=60$. Logo, esse caso $y=4$ também não é possível.\n- Se $y=10$, então $110=(x+z)(y+1)=(x+z) \\times 10$, portanto, $x+z=11$. Mas $240=x y z=10 x z$, portanto, $x z=24$. Podemos verificar (por exemplo, com uma lista de divisores de 24 ou, então, resolvendo a equação $w^{2}-11w+24=0$) que os únicos valores inteiros positivos de $x$ e $z$ que verifiquem essas duas condições $x+z=11$ e $x z=24$ são $x=4$ e $z=6$.\nAssim, a única possibilidade é $x=4, y=10$ e $z=6$, com o que $x+y+z=20$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26132,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO salta-ficha - Temos dez fichas numeradas colocadas em linha reta, como na figura dada.\n## (1) (2) (3) (4) (5) (6) (7) (8) (9) (10)\nQueremos arrumá-las em cinco pilhas, com duas fichas em cada pilha. A regra para isso é que só podemos movimentar uma ficha fazendo-a saltar sobre uma ou mais fichas, ou sobre uma única pilha já formada. Um exemplo de três movimentos é o seguinte.\n- A ficha 7 pode saltar sobre a ficha 8 e formar uma pilha com a 9 ;\n- a ficha 5 pode saltar sobre as fichas 6 e 7 e formar uma pilha com a 8 e\n- a ficha 6 pode saltar sobre as fichas 5,4 e 3 formar uma pilha com a 2 .\n![](attached_image_1.png)\nComo formar cinco pilhas de duas fichas com apenas cinco movimentos?",
    "options": [],
    "answer": "A valid sequence of five moves is:\n1) Move 7 over 8 and 9 to stack with 10.\n2) Move 4 over 5 and 6 to stack with 8.\n3) Move 6 over 3 and 5 to stack with 2.\n4) Move 5 over the stack (4,8) to stack with 9.\n5) Move 1 over the stack (6,2) to stack with 3.",
    "solution": "Solution:\n\n(a) A ficha 7 salta sobre as fichas 8 e 9 formando uma pilha com a ficha 10;\n\n(b) a ficha 4 salta sobre as fichas 5 e 6 formando uma pilha com a ficha 8 ;\n\n(c) a ficha 6 salta sobre as fichas 3 e 5 formando uma pilha com a ficha 2 ;\n\n(d) a ficha 5 salta sobre a pilha $(4,8)$ formando uma pilha com a ficha 9 e\n\n(e) a ficha 1 salta sobre a pilha $(6,2)$ formando uma pilha com a ficha 3.\n\nO resultado segue.\n\n![](attached_image_2.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26133,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nQual é o menor número, $5^{2002}$ ou $3^{2002}+4^{2002}$?",
    "options": [],
    "answer": "3^{2002}+4^{2002}",
    "solution": "Solution:\nComo $5^{2}=3^{2}+4^{2}$, temos $5^{2002}=(3^{2}+4^{2})^{1001}$. Sabemos que, para $a>0$ e $b>0$,\n$$\n(a+b)^{1001}>a^{1001}+b^{1001}\n$$\nAssim, $5^{2002}>3^{2002}+4^{2002}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26134,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDivisão de números grandes - Determine o valor de $123456123456 \\div 10000001$.",
    "options": [],
    "answer": "123456",
    "solution": "Solution:\n\nÉ claro que com números tão grandes, o objetivo da questão não é efetuar a divisão. Em vez disso, decompomos o número em partes convenientes.\n$$\n\\begin{aligned}\n123456123456 & = 123456000000 + 123456 = 123456 \\times 1000000 + 123456 \\\\\n& = 123456 \\times (1000000 + 1) = 123456 \\times 1000001\n\\end{aligned}\n$$\nLogo, $123456123456 \\div 1000001 = 123456$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26135,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDivisão de potências - Qual é o quociente de $50^{50}$ por $25^{25}$?\n\n(a) $25^{25}$\n(b) $10^{25}$\n(c) $100^{25}$\n(d) $2^{25}$\n(e) $2 \\times 25^{25}$",
    "options": [],
    "answer": "c",
    "solution": "Solution:\n\nSolução 1:\n$$\n\\frac{50^{50}}{25^{25}} = \\frac{\\left(2 \\times 5^{2}\\right)^{50}}{\\left(5^{2}\\right)^{25}} = \\frac{2^{50} \\times 5^{100}}{5^{50}} = 2^{50} \\times 5^{50} = \\left(2^{2} \\times 5^{2}\\right)^{25} = 100^{25}\n$$\n\n\nSolução 2:\n$$\n\\frac{50^{50}}{25^{25}} = \\frac{(2 \\times 25)^{50}}{25^{25}} = \\frac{2^{50} \\times 25^{50}}{25^{25}} = 2^{25} \\times 2^{25} \\times 25^{25} = 100^{25}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26136,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm baralho possui 32 cartas divididas em 4 tipos, cada um com 8 cartas. De quantas formas podemos escolher 6 cartas de modo que todos os quatro tipos de cartas estejam entre elas?",
    "options": [],
    "answer": "415744",
    "solution": "Solution:\n\nVamos dividir as escolhas apropriadas de cartas em dois grupos:\n\na) Grupo $S_{1}$: Dois tipos são representados por duas cartas e os outros dois tipos restantes por apenas uma carta cada.\n\nb) Grupo $S_{2}$: Um tipo é representado por três cartas e os outros três tipos restantes por apenas uma carta cada.\n\nPara contar a quantidade de cartas de $S_{1}$, podemos primeiramente escolher os dois tipos que serão representados por duas cartas de $\\binom{4}{2}$ maneiras. Em seguida, podemos escolher duas cartas dentre as 8 de cada um destes tipos de $\\binom{8}{2} \\cdot \\binom{8}{2}$. Finalmente, podemos escolher as cartas restantes nos grupos que restaram de $\\binom{8}{1} \\cdot \\binom{8}{1}$ maneiras. Portanto, a quantidade de elementos de $S_{1}$ é\n$$\n\\binom{4}{2} \\cdot \\binom{8}{2}^2 \\cdot 8^2\n$$\n\nPara contar a quantidade de cartas de $S_{2}$, podemos inicialmente escolher o tipo que será representado por 3 cartas de $\\binom{4}{1}$ maneiras. Em seguida, as três cartas deste tipo podem ser escolhidas de $\\binom{8}{3}$ maneiras. Finalmente, podemos escolher as cartas restantes dos três grupos que restaram de $\\binom{8}{1} \\cdot \\binom{8}{1} \\cdot \\binom{8}{1}$ maneiras. Portanto, a quantidade de elementos de $S_{2}$ é\n$$\n\\binom{4}{1} \\cdot \\binom{8}{3} \\cdot 8^3\n$$\n\nAssim, a quantidade de escolhas é\n$$\n\\binom{4}{2} \\cdot \\binom{8}{2}^2 \\cdot 8^2 + \\binom{4}{1} \\cdot \\binom{8}{3} \\cdot 8^3 = 415744\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26137,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDizemos que um número inteiro positivo de três dígitos é três estrelas se ele for o resultado do produto de três números primos distintos. Por exemplo, $286=2 \\cdot 11 \\cdot 13$ é um número três estrelas, mas $30=2 \\cdot 3 \\cdot 5$ e $275=5 \\cdot 5 \\cdot 13$ não são números três estrelas, pois o primeiro só possui dois dígitos e o segundo não é o produto de três primos distintos.\n\na) Qual o menor número três estrelas?\n\nb) Mostre que cada número três estrelas possui algum divisor em comum com $30$ maior que $1$.",
    "options": [],
    "answer": "102",
    "solution": "Solution:\n\n(a) Os dois primeiros números de três dígitos são $100=2 \\cdot 2 \\cdot 5 \\cdot 5$ e $101=101$ (que é primo). Ao testar $102$, temos $102=2 \\cdot 3 \\cdot 17$, que é o menor número três estrelas.\n\n(b) Basta mostrar que todo número três estrelas possui pelo menos um dos fatores primos do conjunto $\\{2,3,5\\}$. Se tomarmos um número três estrelas que não tenha pelo menos um destes fatores, ele será pelo menos $7 \\cdot 11 \\cdot 13=1001$ e, consequentemente, possuirá mais que três dígitos e não será um número três estrelas.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26138,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDiogo e Helen jogam o Jogo do Tira, que consiste no seguinte. Dado um quadriculado de quadrados $1 \\times 1$, cada jogador, em sua vez, tem o direito de escolher um quadrado e então retirar do quadriculado todos os quadrados abaixo dele, todos os quadrados à esquerda dele, e todos os outros que estejam abaixo e à esquerda dele. Por exemplo, dado o quadriculado abaixo,\n![](attached_image_1.png)\no jogador que tem a vez pode selecionar o quadrado abaixo marcado em cinza, deixando para seu adversário os quadrados mostrados.\n![](attached_image_2.png)\nPerde quem tira o último quadrado.\n\na) Dado o quadriculado abaixo, Helen começa jogando. Mostre uma estratégia para que ela ganhe a partida, independente da estratégia de Diogo.\n![](attached_image_3.png)\n\nb) Dado o quadriculado abaixo, Helen começa jogando. Mostre uma estratégia para que ela ganhe a partida, independente da estratégia de Diogo.\n![](attached_image_4.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na) Helen começa tirando o quadrado abaixo:\n![](attached_image_5.png)\nDiogo tem duas opções agora: se tira o quadrado abaixo, então Helen tira o quadrado seguinte e ganha a partida, pois Diogo terá que tirar o último quadrado.\n![](attached_image_6.png)\nE se Diogo tira o quadrado a seguir, então Helen tira o quadrado seguinte e também ganha.\n![](attached_image_7.png)\n\nb) Vamos descrever as configurações que fazem perder o jogador que as tem (em sua vez de jogar), o que também nos mostrará qual é a estratégia vencedora. Vejamos:\n![](attached_image_8.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26139,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nInteiros de 4 algarismos - Sabendo que $a$ é um número natural, e que $4 a^{2}$ e $\\frac{4}{3} \\times a^{3}$ são números naturais de 4 algarismos, determine $a$.",
    "options": [],
    "answer": "18",
    "solution": "Solution:\n\nTemos que $1000 \\leq 4 a^{2} < 10000$ e também $1000 \\leq \\frac{4}{3} \\times a^{3} < 10000$.\n\nDe $1000 \\leq 4 a^{2} < 10000$ segue que $250 \\leq a^{2} < 2500$. Sendo $a$ um número inteiro e $15^{2} = 225$, $16^{2} = 256$ e $50^{2} = 2500$, temos que $15 < a < 50$.\n\nDe $1000 \\leq \\frac{4}{3} a^{3} < 10000$ temos $750 \\leq a^{3} < 7500$. Mas, $10^{3} = 1000$, $9^{3} = 729$, $20^{3} = 8000$ e $19^{3} = 6859$. Assim, $9 < a < 20$.\n\nPortanto, temos $a = 16, 17, 18$ ou $19$.\n\nPor outro lado, como $\\frac{4}{3} \\times a^{3}$ é um número inteiro, concluímos que $a^{3}$ é múltiplo de $3$ e consequentemente, $a$ é múltiplo de $3$. Logo, $a = 18$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26140,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa Feira de Ciências de uma escola, observou-se que metade dos alunos do ensino fundamental e um quarto dos alunos do ensino médio presentes nesse evento compraram um adesivo cada.\n\n| FEIRA DE CIÊNCIAS |  |\n| :---: | :---: |\n| Preço dos Adesivos (unidade) |  |\n| $R\\$ 0,30$ | alunos do ensino fundamental |\n| $R\\$ 0,50$ | alunos do ensino médio |\n\nNotou-se também que o número de alunos do ensino médio presentes que não compraram adesivos foi o dobro do número de alunos do ensino fundamental que não compraram adesivos. Sabendo que arrecadou-se $R\\$ 38,00$ na venda de adesivos para os alunos desses dois níveis, quantos alunos de cada nível participaram da feira?",
    "options": [],
    "answer": "120 alunos do ensino fundamental e 160 alunos do ensino médio",
    "solution": "Solution:\n\nSejam $x$ e $y$ o número de alunos do ensino fundamental e do médio respectivamente, presentes na feira. Logo, o número daqueles que compraram um adesivo é:\n$$\n\\frac{x}{2} \\text{ do ensino fundamental e } \\quad \\frac{y}{4} \\text{ do ensino médio; }\n$$\ne os que não compraram foram\n$$\n\\frac{x}{2} \\text{ do ensino fundamental e } \\frac{3y}{4} \\text{ do ensino médio. }\n$$\nDentre os alunos que não compraram adesivos, os do ensino médio representam o dobro dos do ensino fundamental. Logo,\n$$\n\\frac{3y}{4} = 2 \\times \\frac{x}{2} \\Rightarrow \\frac{x}{2} = \\frac{3y}{8}\n$$\nSabendo que o total arrecadado foi de $R\\$ 38,00$, concluímos que:\n$$\n\\begin{aligned}\n38 &= 0,30 \\frac{x}{2} + 0,50 \\frac{y}{4} \\\\\n   &= 0,30 \\frac{3y}{8} + 0,50 \\frac{y}{4} \\\\\n   &\\Rightarrow 1,90y = 8 \\times 38 \\\\\n   &\\Rightarrow y = 160\n\\end{aligned}\n$$\nAgora, de $x = \\frac{3y}{4}$, segue que $x = 120$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26141,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nInflação - Márcia está numa loja comprando um gravador que ela queria há muito tempo. Quando o caixa registra o preço ela exclama: \"Não é possível, você registrou o número ao contrário, trocou a ordem de dois algarismos, lembro que na semana passada custava menos que 50 reais!\" Responde o caixa: Sinto muito, mas ontem todos os nossos artigos tiveram um aumento de $20\\%$. Qual é o novo preço do gravador?",
    "options": [],
    "answer": "54",
    "solution": "Solution:\n\nO preço antigo era menor que 50 reais e sofreu um acréscimo de $20\\%$. Logo, o novo preço ainda é um número de 2 algarismos. Vamos representá-lo por $ab$, onde $a$ é o algarismo das dezenas e $b$ é o algarismo das unidades. Logo, o novo preço é $ba$, e temos:\n$$\n10b + a = 1,2(10a + b)\n$$\n$$\n10b - 1,2b = 12a - a\n$$\nPortanto:\n$$\n8,8b = 11a \\quad \\Rightarrow \\quad a = \\frac{8}{10} \\times b = \\frac{4}{5} \\times b\n$$\nComo $a$ e $b$ são algarismos, só podemos ter $a = 4$ e $b = 5$. Logo, o novo preço é $R\\$ 54,00$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26142,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrês cilindros têm alturas e raios das bases iguais a $10~\\mathrm{cm} \\times 10~\\mathrm{cm}$, $10~\\mathrm{cm} \\times 5~\\mathrm{cm}$ e $20~\\mathrm{cm} \\times 5~\\mathrm{cm}$, e volumes $V_{1}, V_{2}$ e $V_{3}$, respectivamente.\n\n![](attached_image_1.png)\n\na) Escreva em ordem crescente os volumes $V_{1}, V_{2}$ e $V_{3}$ dos três cilindros.\n\nb) Dê as dimensões de um cilindro $V_{4}$ cujo volume esteja entre $V_{2}$ e $V_{3}$.\n\nc) Dê as dimensões de um cilindro $V_{5}$ cujo volume esteja entre $V_{1}$ e $V_{3}$.",
    "options": [],
    "answer": "(a) V2 < V3 < V1. (b) One example: radius 5 cm and height 15 cm. (c) One example: radius 8 cm and height 10 cm.",
    "solution": "Solution:\n\n(a) Dado que o volume de um cilindro de raio $R$ e altura $h$ é $\\pi R^{2} h$, temos que os volumes $V_{1}, V_{2}$ e $V_{3}$ são:\n\n$V_{1} = \\pi \\times 10^{2} \\times 10 = 1000\\pi$\n\n$V_{2} = \\pi \\times 5^{2} \\times 10 = 250\\pi$\n\n$V_{3} = \\pi \\times 5^{2} \\times 20 = 500\\pi$\n\nAssim, temos então que $V_{2} < V_{3} < V_{1}$.\n\n(b) Como os dois cilindros têm o mesmo raio, basta manter o raio do cilindro com $5~\\mathrm{cm}$ e a altura entre $10~\\mathrm{cm}$ e $20~\\mathrm{cm}$, por exemplo: $h = 15~\\mathrm{cm}$. Neste caso, o volume $V_{4}$ é:\n\n$V_{4} = \\pi \\times 5^{2} \\times 15 = 375\\pi~\\mathrm{cm}^{3}$.\n\n(c) Para construir um cilindro de volume $V_{5}$ entre $V_{3}$ e $V_{1}$, podemos diminuir o raio do cilindro para $8~\\mathrm{cm}$ e tomar como altura $10~\\mathrm{cm}$, a menor das duas alturas, obtendo um cilindro de volume:\n\n$V_{5} = \\pi \\times 8^{2} \\times 10 = 640\\pi~\\mathrm{cm}^{3}$.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 26143,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNo planeta Staurus, os anos têm $228$ dias ($12$ meses de $19$ dias). Cada semana tem $8$ dias: Zerum, Uni, Duodi, Trio, Quati, Quio, Seise e Sadi. Sybock nasceu num duodi que foi o primeiro dia do quarto mês. Que dia da semana ele festejará seu primeiro aniversário?",
    "options": [],
    "answer": "Seise",
    "solution": "Solution:\n\nSeise",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26144,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO número 119 tem a seguinte propriedade:\n- a divisão por 2 deixa resto 1 ;\n- a divisão por 3 deixa resto 2 ;\n- a divisão por 4 deixa resto 3 ;\n- a divisão por 5 deixa resto 4 ;\n- a divisão por 6 deixa resto 5 .\nQuantos inteiros positivos menores que 2007 satisfazem essa propriedade?",
    "options": [],
    "answer": "33",
    "solution": "Solution:\n\nInicialmente note que se $N$ dividido por $d$ deixa resto $r$, então somando a $N$ um múltiplo de $d$, o resto não se altera, isto é:\n$$\n\\frac{(N+\\text{ múltiplo de } d)}{d} \\text{ também deixa resto } r\n$$\nPor exemplo: $38$ dividido por $3$ deixa resto $2$, logo o resto da divisão de $(38+5 \\times 3)$ também é $2$.\n\nAssim, se somamos a $119$ um número que seja múltiplo simultaneamente de $2, 3, 4, 5$ e $6$, esse número deixa os mesmos restos que $119$ quando dividido por $2, 3, 4, 5$ e $6$. O menor múltiplo comum de $2, 3, 4, 5$ e $6$ é $60$, logo todo número da forma\n$$\n119 + (\\text{ múltiplo de } 60)\n$$\nsatisfaz as cinco condições do enunciado.\n\nDa divisão de $2007$ por $60$ temos:\n$$\n2007 = 33 \\times 60 + 27 = 32 \\times 60 + 87 = 31 \\times 60 + 147\n$$\nComo $119$ está entre $87$ e $147$, temos que os números\n$$\n59, 119, 179, \\ldots, 31 \\times 60 + 119\n$$\ncumprem a mesma propriedade que $119$. Logo, temos $33$ possíveis números.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26145,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUma fábrica produz blusas a um custo de $R\\$ 2{,}00$ por unidade além de uma parte fixa de $R\\$ 500{,}00$. Se cada unidade produzida é comercializada a $R\\$ 2{,}50$, a partir de quantas unidades produzidas a fábrica obtém lucro?\n(a) 250\n(b) 500\n(c) 1000\n(d) 1200\n(e) 1500",
    "options": [],
    "answer": "c",
    "solution": "Solution:\n\nDenotemos por $x$ o número de unidades produzidas. Assim o custo de produção é $500+2x$ reais. Pela venda o fabricante está recebendo $2{,}5x$. Assim, ele terá lucro quando\n$$\n2{,}5x > 500 + 2x\n$$\nisto é, $0{,}5x > 500$. Portanto $x > 1000$. Logo, a opção correta é (c).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26146,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nEntre $1$ e $2$ - Complete os numeradores com inteiros positivos para satisfazer as condições: $\\frac{a}{5}$ e $\\frac{b}{7}$ são menores do que $1$, e $1 < \\frac{a}{5} + \\frac{b}{7} < 2$.",
    "options": [],
    "answer": "All solutions: (a,b) = (1,6), (2,5), (2,6), (3,3), (3,4), (3,5), (3,6), (4,2), (4,3), (4,4), (4,5), (4,6).",
    "solution": "Solution:\nComo as duas frações são positivas e menores do que $1$, seus numeradores devem ser respectivamente menores que seus denominadores, logo devemos ter:\n$$\n0 < a < 5 \\text{ e } 0 < b < 7\n$$\nTemos $\\frac{a}{5} + \\frac{b}{7} = \\frac{7a + 5b}{35}$, portanto:\n$$\n1 < \\frac{7a + 5b}{35} < 2 \\Rightarrow 35 < 7a + 5b < 70\n$$\nVejamos as opções para que $a$ e $b$ sejam inteiros positivos e satisfaçam (1) e (2):\n- $a = 1 \\Rightarrow 7a + 5b = 7 + 5b$. Logo,\n$$\n35 < 7 + 5b < 70 \\Rightarrow 28 < 5b < 63 \\Rightarrow \\frac{28}{5} < b < \\frac{63}{5} \\Rightarrow 5,6 < b < 12,6\n$$\nComo $b$ é um número inteiro, concluímos que $b = 6, 7, 8, \\ldots, 12$. No entanto, só podemos escolher $b = 6$, pois $b < 7$.\n- $a = 2 \\Rightarrow 7a + 5b = 14 + 5b$. Logo,\n$$\n35 < 14 + 5b < 70 \\Rightarrow 21 < 5b < 56 \\Rightarrow \\frac{21}{5} < b < \\frac{56}{5} \\Rightarrow b = 5, 6, 7, 8, \\ldots, 11\n$$\nNesse caso, só podemos escolher $b = 5$ e $b = 6$, pois $b < 7$.\n- $a = 3 \\Rightarrow 7a + 5b = 21 + 5b$. Logo,\n$$\n35 < 21 + 5b < 70 \\Rightarrow 14 < 5b < 49 \\Rightarrow \\frac{14}{5} < b < \\frac{49}{5} \\Rightarrow b = 3, 4, 5, 6, \\ldots, 9\n$$\nAqui, podemos escolher $b = 3, 4, 5, 6$.\n- $a = 4 \\Rightarrow 7a + 5b = 28 + 5b$. Logo,\n$$\n35 < 28 + 5b < 70 \\Rightarrow 7 < 5b < 42 \\Rightarrow \\frac{7}{5} < b < \\frac{42}{5} \\Rightarrow b = 2, 3, 4, 5, 6, \\ldots, 8\n$$\nPodemos escolher $b = 2, 3, 4, 5, 6$.\nPara finalizar, exibimos as soluções na tabela abaixo:\n\n| $a$ | $b$ | $\\frac{a}{5} + \\frac{b}{7}$ |\n| :---: | :---: | :---: |\n| 1 | 6 | $\\frac{1}{5} + \\frac{6}{7} = \\frac{37}{35}$ |\n| 2 | 5 | $\\frac{2}{5} + \\frac{5}{7} = \\frac{39}{35}$ |\n|   | 6 | $\\frac{2}{5} + \\frac{6}{7} = \\frac{44}{35}$ |\n| 3 | 3 | $\\frac{3}{5} + \\frac{3}{7} = \\frac{36}{35}$ |\n|   | 4 | $\\frac{3}{5} + \\frac{4}{7} = \\frac{41}{35}$ |\n|   | 5 | $\\frac{3}{5} + \\frac{5}{7} = \\frac{46}{35}$ |\n|   | 6 | $\\frac{3}{5} + \\frac{6}{7} = \\frac{51}{35}$ |\n| 4 | 2 | $\\frac{4}{5} + \\frac{2}{7} = \\frac{38}{35}$ |\n|   | 3 | $\\frac{4}{5} + \\frac{3}{7} = \\frac{43}{35}$ |\n|   | 4 | $\\frac{4}{5} + \\frac{4}{7} = \\frac{48}{35}$ |\n|   | 5 | $\\frac{4}{5} + \\frac{5}{7} = \\frac{53}{35}$ |\n|   | 6 | $\\frac{4}{5} + \\frac{6}{7} = \\frac{58}{35}$ |",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26147,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa figura, $ABCD$ é um quadrado cuja área é $\\frac{7}{32}$ da área do triângulo $XYZ$. Qual é a razão entre $XA$ e $XY$?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "7/8 or 1/8",
    "solution": "Solution:\n\nSejam $l$ o comprimento do lado do quadrado, $h$ a altura do triângulo $\\triangle XAB$, $H$ a altura do triângulo $\\triangle XYZ$ e $b$ o comprimento do lado $YZ$.\nA área do quadrado é $l^2$ e a área do triângulo $\\triangle XYZ$ é $\\frac{bH}{2}$. Como os triângulos $XYZ$ e $ABC$ são semelhantes, temos\n$$\n\\frac{b}{l} = \\frac{H}{h} = \\frac{XY}{XA}\n$$\nPortanto $b = \\frac{Hl}{h} = \\frac{(h+l)l}{h}$.\n\n![](attached_image_2.png)\n\nAssim, a área do triângulo $\\triangle XYZ$ é: $\\frac{bH}{2} = \\frac{(h+l)^2 l}{2h}$ e a razão $\\frac{XA}{XY}$ é\n$$\n\\frac{XA}{XY} = \\frac{h}{H} = \\frac{h}{h+l} = \\frac{1}{1+\\frac{l}{h}}\n$$\nLogo, basta calcular $\\frac{l}{h}$.\nComo a razão entre a área do triângulo $\\triangle XYZ$ e a área do quadrado é $\\frac{32}{7}$, então\n$$\n\\frac{\\frac{(h+l)^2 l}{2h}}{l^2} = \\frac{32}{7} \\Longrightarrow (h+l)^2 = \\frac{64}{7}hl \\Longrightarrow l^2 - \\frac{50}{7}hl + h^2 = 0\n$$\nDividindo por $h^2$ obtemos a equação quadrática $\\left(\\frac{l}{h}\\right)^2 - \\frac{50}{7}\\left(\\frac{l}{h}\\right) + 1 = 0$, que tem como soluções\n$$\n\\frac{l}{h} = \\frac{\\frac{50}{7} \\pm \\sqrt{\\left(\\frac{50}{7}\\right)^2 - 4}}{2} = \\frac{25 \\pm \\sqrt{25^2 - 7^2}}{7} = \\frac{25 \\pm 24}{7}\n$$\nAssim, $\\frac{l}{h}$ tem dois possíveis valores $\\frac{1}{7}$ e $7$, e em cada um destes casos $\\frac{XA}{XY}$ é $\\frac{7}{8}$ e $\\frac{1}{8}$, respectivamente.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26148,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUma faixa retangular de cartolina tem $5~\\mathrm{cm}$ por $1~\\mathrm{cm}$. Corte a faixa com 4 cortes retilíneos de modo a poder montar um quadrado com as peças obtidas (não vale superposição das peças).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26149,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n5. Calculadora diferente - Uma fábrica produziu uma calculadora original que efetua duas operações:\n- a adição usual $+$\n- a operação $\\circledast$\n\nSabemos que para todo número natural $a$ tem-se:\n$$\n\\text{(i)}\\ a \\circledast a = a \\quad \\text{e (ii)}\\ a \\circledast 0 = 2a\n$$\ne, para quaisquer quatro naturais $a, b, c$ e $d$\n$$\n\\text{(iii)}\\ (a \\circledast b) + (c \\circledast d) = (a + c) \\circledast (b + d)\\text{.}\n$$\nQuais são os resultados das operações $(2+3) \\circledast (0+3)$ e $1024 \\circledast 48$ ?",
    "options": [],
    "answer": "7; 2000",
    "solution": "Solution:\n\nPara calcular $(2+3) \\circledast (0+3)$ utilizaremos a propriedade (iii), e temos:\n$$\n(2+3) \\circledast (0+3) = (2 \\circledast 0) + (3 \\circledast 3)\n$$\nAgora, por (ii) temos $2 \\circledast 0 = 2 \\times 2 = 4$, e por (i) temos $3 \\circledast 3 = 3$. Portanto,\n$$\n(2+3) \\circledast (0+3) = 4 + 3 = 7\n$$\nAgora, para calcular $1024 \\circledast 48$ vamos usar a mesma estratégia que acima. Para isso, note que $1024 = 976 + 48$ e $48 = 0 + 48$.\n$$\n\\begin{aligned}\n1024 \\circledast 48 &= (976 + 48) \\circledast (0 + 48) \\\\\n&= (976 \\circledast 0) + (48 \\circledast 48) \\\\\n&= 2 \\times 976 + 48 \\\\\n&= 1952 + 48 = 2000\n\\end{aligned}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26150,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCircunferência e triângulo retângulo - Inscreve-se uma circunferência num triângulo retângulo. $O$ ponto de tangência divide a hipotenusa em dois segmentos de comprimentos $6~\\mathrm{cm}$ e $7~\\mathrm{cm}$. Calcule a área do triângulo.",
    "options": [],
    "answer": "42",
    "solution": "Solution:\n\nSeja $r$ o raio da circunferência inscrita. Usando o teorema de Pitágoras temos que $(6+7)^2 = (r+6)^2 + (r+7)^2 = r^2 + 12r + 36 + r^2 + 14r + 49 = 2(r^2 + 13r) + 85$, assim temos que $r^2 + 13r = \\frac{169 - 85}{2} = 42$.\n\nPor outro lado, a área do triângulo é\n$$\n\\frac{(r+6)(r+7)}{2} = \\frac{r^2 + 13r + 42}{2} = \\frac{42 + 42}{2} = 42.\n$$\n\n![](attached_image_1.png)",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26151,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTriângulos com lados inteiros - Quantos triângulos existem cujos lados são números inteiros e o perímetro é $12$?\n(a) $1$\n(b) $3$\n(c) $5$\n(d) $7$\n(e) $9$",
    "options": [],
    "answer": "b",
    "solution": "Solution:\n\nPara que três números $a$, $b$, $c$ sejam os comprimentos dos lados do triângulo, cada um deles deve ser maior que a diferença e menor que a soma dos outros dois.\n\nSejam $a \\leq b \\leq c$ os comprimentos dos lados do triângulo. Assim, $c < a + b$.\n\nAgora, somando $c$ a ambos os membros temos que: $2c < a + b + c = 12$, ou seja, $2c < 12$, logo $c < 6$.\n\nAlém disso, como $3c \\geq a + b + c = 12$ temos que: $c \\geq 4$. Logo, $4 \\leq c < 6$.\n\nNo caso de $c = 5$, temos que $a + b = 7$. Os possíveis valores de $a$ e $b$ são: $a = 2$ e $b = 5$ ou $a = 3$ e $b = 4$.\n\nNo caso de $c = 4$, temos que $a + b = 8$, e portanto temos somente a solução $a = b = 4$.\n\nAssim temos $3$ possíveis triângulos. A opção correta é (b).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26152,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm quadrilátero especial - Dois lados consecutivos de um quadrilátero medem $10~\\mathrm{cm}$ e $15~\\mathrm{cm}$. Se cada diagonal divide o quadrilátero em duas regiões de mesma área, calcule o seu perímetro.",
    "options": [],
    "answer": "50 cm",
    "solution": "Solution:\n\nSe cada diagonal divide o quadrilátero em duas regiões de mesma área temos:\n$$\n\\text{Área}(\\triangle ABD) = \\text{Área}(\\triangle BCD) \\text{ e Área}(\\triangle ABC) = \\text{Área}(\\triangle ACD).\n$$\nMas,\n$$\n\\begin{aligned}\n\\text{Área}(\\triangle ABD) & = X + W \\\\\n\\text{Área}(\\triangle BCD) & = Y + Z \\\\\n\\text{Área}(\\triangle ABC) & = Z + W \\\\\n\\text{Área}(\\triangle CDA) & = X + Y\n\\end{aligned}\n$$\n\n![](attached_image_1.png)\n\nAssim,\n$$\nZ - X = \\text{Área}(\\triangle ABC) - \\text{Área}(\\triangle ABD) = \\text{Área}(\\triangle CDA) - \\text{Área}(\\triangle BCD) = X - Z\n$$\ne portanto, $Z = X$. Consequentemente, também temos $Y = W$.\n\nComo as áreas opostas são iguais, resulta da semelhança de triângulos que:\n$$\nAE \\times ED = BE \\times EC \\text{ e } AE \\times EB = CE \\times ED\n$$\nDividindo estas duas equações obtemos:\n$$\n\\frac{ED}{EB} = \\frac{EB}{ED} \\quad \\Rightarrow \\quad ED = EB\n$$\nAnalogamente podemos mostrar que $EA = EC$. Logo, as diagonais se cortam no ponto médio, e consequentemente o quadrilátero é um paralelogramo donde, o perímetro é igual a $2 \\times 10 + 2 \\times 15 = 50~\\mathrm{cm}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26153,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nO número\n$$\nA=(\\sqrt{6}+\\sqrt{2})(\\sqrt{3}-2) \\sqrt{\\sqrt{3}+2}\n$$\né igual a:\n(a) $-\\sqrt{3}$\n(b) $-\\sqrt{2}$\n(c) -2\n(d) 1\n(d) 2",
    "options": [],
    "answer": "c",
    "solution": "Solution:\nComo\n$$\n\\begin{aligned}\nA^{2} & =[(\\sqrt{6}+\\sqrt{2})(\\sqrt{3}-2) \\sqrt{\\sqrt{3}+2}]^{2} \\\\\n& =(\\sqrt{6}+\\sqrt{2})^{2}(\\sqrt{3}-2)^{2}(\\sqrt{\\sqrt{3}+2})^{2} \\\\\n& =(\\sqrt{6}+\\sqrt{2})^{2}(\\sqrt{3}-2)^{2}(\\sqrt{3}+2) \\\\\n& =(\\sqrt{6}+\\sqrt{2})^{2}(\\sqrt{3}-2)[(\\sqrt{3}-2)(\\sqrt{3}+2)] \\\\\n& =(6+2 \\sqrt{12}+2)(\\sqrt{3}-2)((\\sqrt{3})^{2}-2^{2}) \\\\\n& =(6+2 \\sqrt{12}+2)(\\sqrt{3}-2)(-1) \\\\\n& =(8+4 \\sqrt{3})(2-\\sqrt{3}) \\\\\n& =4(2+\\sqrt{3})(2-\\sqrt{3}) \\\\\n& =4\\left(2^{2}-(\\sqrt{3})^{2}\\right)=4 \\times 1=4\n\\end{aligned}\n$$\nAssim $A^{2}=4$ e logo, $A$ pode ser 2 ou -2. Como $\\sqrt{3}-2$ é negativo, temos que $A$ tem que ser negativo, e portanto $A=-2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26154,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAs idades numa classe - Numa classe na escola, todos os alunos têm a mesma idade, exceto sete que têm 1 ano a menos e dois que têm 2 anos a mais.\n\nA soma das idades de todos os alunos dessa classe é 330. Quantos alunos tem essa classe?",
    "options": [],
    "answer": "37",
    "solution": "Solution:\n\nDenotemos por $a$ a idade comum dos alunos e $n$ o número de alunos, assim temos 7 alunos com $a-1$ anos, 2 com $a+2$ anos e o resto, isto é, $n-9$ com $a$ anos. Assim a soma das idades é\n\n$$\n7(a-1) + 2(a+2) + (n-9)a = n a - 3 = 330\n$$\n\n$$\nn a = 333 = 9 \\times 37\n$$\n\nComo a classe tem mais do que 9 alunos, então $a = 9$ e $n = 37$, portanto a classe tem 37 alunos.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26155,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nInterseç̧ão de triângulos - Os $3$ triângulos da figura se cortam em $12$ pontos diferentes. Qual é o número máximo de pontos de intersecção de $3$ triângulos?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "18",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26156,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nQuais são os números? - Descubra quais números inteiros positivos $x$ e $y$ satisfazem a equação $x^{4}=y^{2}+71$.",
    "options": [],
    "answer": "x = 6, y = 35",
    "solution": "Solution:\n\nA equação pode ser escrita na forma $x^{4}-y^{2}=71$. Agora, fatorando $x^{4}-y^{2}$ temos:\n$$\n\\left(x^{2}-y\\right)\\left(x^{2}+y\\right)=71\\ (*)\n$$\nComo $x$ e $y$ são inteiros, então cada um dos fatores $\\left(x^{2}-y\\right)$ e $\\left(x^{2}+y\\right)$ também é um número inteiro. Logo em $\\left(*)\\right)$ escrevemos $71$ como o produto de $2$ números inteiros. Como $71$ é um número primo, ele só pode ser escrito como produto de inteiros na forma: $71=1 \\times 71$. Temos então dois casos a considerar: $\\left(x^{2}-y\\right)=1$ e $\\left(x^{2}+y\\right)=71$, ou $\\left(x^{2}-y\\right)=71$ e $\\left(x^{2}+y\\right)=1$.\n\nVamos estudar cada caso.\n\n1o caso:\n$$\n\\left\\{\n\\begin{array}{l}\nx^{2}-y=1 \\\\\nx^{2}+y=71\n\\end{array}\n\\right.\n$$\nSomando as duas equações obtemos: $2 x^{2}=72$, o que implica $x= \\pm 6$. Portanto, $y=( \\pm 6)^{2}-1=35$. Como $x, y$ são inteiros positivos, concluímos que a solução nesse primeiro caso é: $x=6$ e $y=35$.\n\n2o caso:\n$$\n\\left\\{\n\\begin{array}{l}\nx^{2}-y=71 \\\\\nx^{2}+y=1\n\\end{array}\n\\right.\n$$\nSe $x^{2}+y=1$, então $x=0$ e $y=1$ ou $x=1$ e $y=0$ já que $x, y$ são inteiros positivos. Por outro lado, é fácil verificar que tais valores não satisfazem a equação $x^{4}=y^{2}+71$.\n\nLogo, a solução para o problema é: $x=6$ e $y=35$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26157,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEncontre todos os pares de inteiros positivos $x, y$ tais que $x$ e $y$ são primos entre si, $x<y$ e $2000\\left(\\frac{x}{y}+\\frac{y}{x}\\right)$ é um inteiro ímpar.",
    "options": [],
    "answer": "(1,16), (1,80), (1,400), (1,2000), (5,16), (16,25), (16,125)",
    "solution": "Solution:\n\nSolução 1: Temos:\n$$\n2000\\left(\\frac{x}{y}+\\frac{y}{x}\\right)=16 \\times 125\\left(\\frac{x^{2}+y^{2}}{x y}\\right)\n$$\nComo $x$ e $y$ são primos entre si, concluímos que $x y$ e $x^{2}+y^{2}$ não têm fatores em comum (prove isso). Logo, $x$ e $y$ são divisores de $2000=16 \\times 125=2^{4} \\times 5^{3}$. Além disso, 16 tem que dividir $x y$ porque a expressão tem que ser ímpar. Portanto, $16=2^{4}$ divide $x$ ou $y$.\n\n1o caso: 16 divide $x$.\nSe $x>16$ então, $x$ é no mínimo $16 \\times 5=80$, já que $x$ divide 2000. Nesse caso, $x>y$ pois $x y$ divide 2000. Logo, $x=16$. Assim, como $x$ e $y$ são primos entre si, $x<y$ e $y$ divide 2000 concluímos que as únicas possibilidades são $y=25$ ou $125$.\n\n2o caso: 16 divide $y$.\nComo no item anterior fazemos:\nSe $y>16$ então as possibilidades seriam: $y=16 \\times 5$ e $x=1$; $y=16 \\times 25$ e $x=1$; $y=16 \\times 125$ e $x=1$.\nSe $y=16$ então as possibilidades para $x$ seria: $x=1$; $x=5$.\n\nLogo, os pares $(x, y)$ satisfazendo as condições do problema são:\n$$\n(16,25) ;(16,125) ;(5,16) ;(1,16) ;(1,80) ;(1,400) ;(1,2000)\n$$\nSolution:\n\nTemos $N=2000\\left(\\frac{x}{y}+\\frac{y}{x}\\right)=\\frac{16 \\times 125 \\times\\left(x^{2}+y^{2}\\right)}{x y}=\\frac{2^{4} \\times 5^{3}}{x y} \\times\\left(x^{2}+y^{2}\\right)$.\nComo $N$ é ímpar segue que $\\frac{2^{4} \\times 5^{3}}{x y}$ e $x^{2}+y^{2}$ são ímpares. As opções para isso são:\n$x y=2^{4}, 2^{4} \\times 5,2^{4} \\times 5^{2}, 2^{4} \\times 5^{3}$ e, $x$ e $y$ têm paridades distintas. Vamos determinar $x$ e $y$ para cada uma dessas opções:\n\n| $x y$ | $x$ | $y$ |\n| :---: | :---: | :---: |\n| $2^{4}$ | 1 | $2^{4}$ |\n| $2^{4} .5$ | 1 | $2^{4} .5$ |\n|  | $2^{4}$ | 5 |\n| $2^{4} .5^{2}$ | 1 | $2^{4} .5^{2}$ |\n|  | $2^{4}$ | $5^{2}$ |\n| $2^{4} .5^{3}$ | 1 | $2^{4} .5^{3}$ |\n|  | $2^{4}$ | $5^{3}$ |\n\nLogo, os pares $(x, y)$ satisfazendo as condições do problema são:\n$$\n(1,16) ;(1,80) ;(16,5) ;(1,400) ;(16,25) ;(1,2000) ;(16,125)\n$$\nPorém, como podemos trocar $x$ e $y$ em vista da simetria da expressão, temos ainda as soluções:\n$$\n(16,1) ;(80,1) ;(5,16) ;(400,1) ;(25,16) ;(2000,1) ;(125,16)\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26158,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA soma dos algarismos de um número - Denotemos por $s(n)$ a soma dos algarismos do número $n$. Por exemplo $s(2345)=2+3+4+5=14$. Observemos que:\n$40-s(40)=36=9 \\times 4 ; 500-s(500)=495=9 \\times 55 ; 2345-s(2345)=2331=9 \\times 259$.\n\na. O que podemos afirmar sobre o número $n-s(n)$ ?\n\nb. Usando o item anterior calcule $s\\left(s\\left(s\\left(2^{2009}\\right)\\right)\\right.$ ).\n\nSugestão: Mostre que o número procurado é menor do que 9.",
    "options": [],
    "answer": "n − s(n) is divisible by 9; s(s(s(2^2009))) = 5",
    "solution": "Solution:\n\n(a) Observe esses dois exemplos:\n$$\n\\underbrace{2000}_{2 \\cdot 10^3}-\\underbrace{s(2000)}_{2}=1998, \\underbrace{60000}_{6 \\cdot 10^4}-\\underbrace{s(60000)}_{6}=59994\n$$\nA partir deles é fácil entender que se $a$ é um algarismos entre 1 e 9 , então $s\\left(a \\cdot 10^{k}\\right)=a$.\nDaí temos:\n$$\na \\cdot 10^{k}-s\\left(a \\cdot 10^{k}\\right)=a \\cdot 10^{k}-a=a\\left(10^{k}-1\\right)=a \\times \\underbrace{9 \\cdots 9}_{k \\text{ noves }}=a \\times 9 \\times \\underbrace{1 \\cdots 1}_{k \\text{ uns }}\n$$\nComo todo número pode ser decomposto em unidades, dezenas, centenas etc, isto é, todo número pode ser escrito na forma:\n$$\nn=a_{0}+a_{1} \\cdot 10+a_{2} \\cdot 10^{2}+\\cdots+a_{k} \\cdot 10^{k}\n$$\ntemos que\n$$\nn-s(n)=a_{1} \\times 9+a_{2} \\times 99+\\cdots a_{k} \\times \\underbrace{9 \\cdots 9}_{k \\text{ noves }}\n$$\nLogo, a diferença $n-s(n)$ é sempre divisível por 9 .\n\n(b) Seguindo o mesmo raciocínio temos que: $s(n)-s(s(n))$ e $s(s(n))-s(s(s(n)))$ são divisíveis por 9 , logo $n-s\\left(s\\left(s(n)\\right.\\right.$ é divisível por 9 . Em particular $2^{2009}-s\\left(s\\left(s\\left(2^{2009}\\right)\\right)\\right)$ é divisível por 9 , ou equivalentemente, $2^{2009}$ e $s\\left(s\\left(s\\left(2^{2009}\\right)\\right)\\right)$ deixam o mesmo resto quando são divididos por 9 .\nComo $2^{6}-1=63$ é divisível por 9 então, $\\left(2^{6}\\right)^{334}-1=2^{2004}-1$ é divisível por $9 \\mathrm{e}$, portanto, $2^{2009}-2^{5}$ é divisível por 9 . Como $2^{5}=32$ deixa resto 5 quando dividido por 9 , temos que $2^{2009}$ deixa resto 5 quando dividido por 9 .\nPor outro lado\n$$\n2^{2009}<\\left(2^{9}\\right)^{224}<\\left(10^{3}\\right)^{224}=10^{672}\n$$\nAssim, $2^{2009}$ tem menos que 672 algarismos e, portanto,\n$$\n\\begin{aligned}\ns\\left(2^{2009}\\right) & <9 \\times 672=6048 \\\\\ns\\left(s\\left(2^{2009}\\right)\\right) & \\leq 5+9+9+9=32 \\\\\ns\\left(s\\left(s\\left(2^{2009}\\right)\\right)\\right) & \\leq 2+9=13\n\\end{aligned}\n$$\nComo o único número menor ou igual a 13 que deixa resto 5 quando dividido por 9 é o 5 temos que $s\\left(s\\left(s\\left(2^{2009}\\right)\\right)\\right)=5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26159,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCercando o Globo Terrestre - O raio do Globo Terrestre é aproximadamente $6670~\\mathrm{km}$. Suponhamos que um fio esteja ajustado exatamente sobre o Equador, que é um círculo de raio aproximadamente igual a $6670~\\mathrm{km}$.\n\n![](attached_image_1.png)\n\nEm seguida, suponhamos que o comprimento do fio seja aumentado em $1~\\mathrm{m}$, de modo que o fio e o Equador fiquem como círculos concêntricos ao redor da Terra. Um homem em pé, uma formiga ou um elefante são capazes de passar por baixo desse fio?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Gap height = 1/(2π) meters ≈ 0.16 m (about 16 cm); only the ant can pass under.",
    "solution": "Solution:\n\nCercando o Globo Terrestre - Como o raio da Terra é muito grande, e foi dado apenas um acréscimo de $1~\\mathrm{m}$ no comprimento do fio, parece que a folga entre o fio e o Equador é muito pequena. Mais ainda, se trocarmos o Globo Terrestre por Júpiter ou por uma bolinha de gude e realizarmos esta mesma experiência, parece que a altura da folga entre o fio aumentado e o equador da esfera também muda, sendo que quanto maior a esfera considerada, menor é a folga entre o fio e o equador da esfera.\n\nVejamos que esta ideia intuitiva é falsa e que a altura da folga, entre o fio e o Equador, é de aproximadamente $16~\\mathrm{cm}$, independentemente do raio da esfera em que a experiência é realizada.\n\nConsideremos um círculo de raio $R$. Seu comprimento é igual a $2\\pi R$. Vamos considerar também um círculo de mesmo centro, mas que tenha comprimento igual a $2\\pi R + 1$.\n\n![](attached_image_2.png)\n\nEste círculo tem raio igual a $R + h$, sendo $h$ a altura da folga entre os dois círculos. Como um círculo de raio $R + h$ tem comprimento $2\\pi(R + h)$, obtemos a igualdade $2\\pi(R + h) = 2\\pi R + 1$. Simplificando esta expressão obtemos $h = \\frac{1}{2\\pi} \\approx \\frac{1}{6.28} \\approx 0,16$. Portanto, para qualquer valor de $R$, a altura da folga é de aproximadamente $16~\\mathrm{cm}$.\n\nAssim, somente a formiga é capaz de passar por debaixo do fio.\n\n![](attached_image_2.png)",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26160,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nConjunto de Cantor - Desenhe um segmento de reta de comprimento $1$, e denote-o por $C_{1}$. Remova o terço central (sem remover os extremos). Denote por $C_{2}$ o que sobrou. Agora, remova o terço central (sem os extremos) de cada segmento de reta de $C_{2}$. Denote por $C_{3}$ o que sobrou. Podemos continuar esse processo, em cada estágio removendo o terço central de cada segmento em $C_{n}$ para formar $C_{n+1}$.\n\n![](attached_image_1.png)\n\na. Desenhe $C_{1}, C_{2}$ e $C_{3}$, indicando os números nos extremos dos segmentos.\n\nb. Quais dos seguintes pontos pertencem ao conjunto de Cantor? $\\frac{1}{3}, \\frac{4}{9}, \\frac{3}{81}, \\frac{4}{81}$.\n\nc. Quais são os comprimentos de $C_{3}, C_{4}$ e $C_{5}$? Você pode achar uma expressão para o comprimento de $C_{n}$?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "a) C1: [0, 1]. C2: [0, 1/3] and [2/3, 1]. C3: [0, 1/9], [2/9, 1/3], [2/3, 7/9], [8/9, 1].\nb) 1/3 ∈ Cantor set; 3/81 = 1/27 ∈ Cantor set; 4/9 ∉ Cantor set; 4/81 ∉ Cantor set.\nc) Lengths: |C3| = 4/9, |C4| = 8/27, |C5| = 16/81. In general, |Cn| = (2/3)^(n−1).",
    "solution": "Solution:\n\na. De acordo com a definição do Conjunto de Cantor temos os seguintes desenhos:\n\n![](attached_image_2.png)\n\nb. $1/3$ é um extremo de $C_{2}$, logo pertence ao conjunto de Cantor.\n\n$3/81 = 1/27$ e $1/27$ é um extremo de $C_{4}$, logo $3/81$ pertence ao conjunto de Cantor.\n\n$4/9$ está entre $1/3$ e $2/3$, logo está no terço central de $C_{1}$ e é removido de $C_{2}$, logo $4/9$ não pertence ao conjunto de Cantor.\n\n$4/81$ está entre $1/27$ e $2/27$, e portanto está no terço central de $C_{3}$ e é removido de $C_{4}$. Assim, $4/81$ não pertence ao conjunto de Cantor.\n\nc. Vamos tentar achar um padrão para os comprimentos dos segmentos. Por exemplo, $C_{1}$ tem comprimento $1$ e $C_{2}$ tem comprimento $2/3$. Será que isso já fornece um padrão, ou seja o numerador é obtido multiplicando por $2$ e o denominador por $3$, ou seja por $2/3$?\n\nAgora $C_{3}$ tem comprimento $4/9$, $C_{4}$ comprimento $8/27$ e $C_{5}$ comprimento $16/81$. Logo, o comprimento de $C_{n}$ é $\\left(\\frac{2}{3}\\right)^{n-1}$\n\nNote que os comprimentos de $C_{1}, C_{2}, C_{3}, \\ldots, C_{n}, \\ldots$, formam uma progressão geométrica de razão $q=2/3$ e primeiro termo $a_{1}=1$.\n\n$$\n1, \\frac{2}{3}, \\left(\\frac{2}{3}\\right)^{2}, \\left(\\frac{2}{3}\\right)^{3}, \\left(\\frac{2}{3}\\right)^{4}, \\ldots, \\left(\\frac{2}{3}\\right)^{n}, \\ldots\n$$\n\n![](attached_image_2.png)",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26161,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDetermine 4 números distintos $a_{1}, a_{2}, a_{3}$ e $a_{4}$ que sejam termos consecutivos de uma progressão aritmética e que os números $a_{1}, a_{3}$ e $a_{4}$ formem uma progressão geométrica.",
    "options": [],
    "answer": "All solutions are given by 2x, 3x/2, x, x/2 with x ≠ 0 (e.g., 2, 3/2, 1, 1/2).",
    "solution": "Solution:\n\nOs 4 termos de uma progressão aritmética de razão $r$ podem ser escritos como:\n$$\nx-2r,\\ x-r,\\ x,\\ x+r\n$$\nLogo, os 3 termos da progressão geométrica de razão $q$ serão\n$$\nx-2r,\\ x,\\ x+r\n$$\nonde\n$$\nx = (x-2r)q \\text{ e } x+r = xq\n$$\nDaí segue que:\n$$\nx = xq - 2rq \\Rightarrow x = x + r - 2rq \\Rightarrow q = \\frac{1}{2}\n$$\nObtemos que $x+r = \\frac{x}{2} \\Rightarrow r = -\\frac{x}{2}$. Logo a progressão aritmética é da forma:\n$$\n2x,\\ \\frac{3x}{2},\\ x,\\ \\frac{x}{2}.\n$$\nEscolhendo um valor para $x$, por exemplo $x=1$, obtemos 4 números formando uma progressão aritmética $2, \\frac{3}{2}, 1, \\frac{1}{2}$ de razão $-\\frac{1}{2}$ tais que $2, 1, \\frac{1}{2}$ formam uma progressão geométrica de razão $\\frac{1}{2}$. Note que esse problema tem uma solução para cada escolha de $x$, portanto tem uma infinidade de soluções.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26162,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nCada um dos números $x_{1}, x_{2}, \\ldots, x_{2004}$ pode ser igual a $\\sqrt{2}-1$ ou a $\\sqrt{2}+1$. Quantos valores inteiros distintos a soma\n$$\n\\sum_{k=1}^{1002} x_{2k-1} x_{2k} = x_{1} x_{2} + x_{3} x_{4} + x_{5} x_{6} + \\cdots + x_{2003} x_{2004}\n$$\npode assumir?",
    "options": [],
    "answer": "502",
    "solution": "Solution:\nTemos que os possíveis produtos $x_{2k-1} x_{2k}$ onde $k \\in \\{1,2, \\ldots, 1002\\}$ são $(\\sqrt{2}-1)(\\sqrt{2}-1) = 3-2\\sqrt{2}$, $(\\sqrt{2}+1)(\\sqrt{2}+1) = 3+2\\sqrt{2}$ e $(\\sqrt{2}-1)(\\sqrt{2}+1) = 1$.\n\nSuponha que $a$ produtos são iguais a $3-2\\sqrt{2}$, $b$ produtos são iguais a $3+2\\sqrt{2}$ e $1002-a-b$ produtos são iguais a $1$.\n\nA soma é igual a\n$$\na(3-2\\sqrt{2}) + b(3+2\\sqrt{2}) + (1002-a-b) \\cdot 1 = 1002 + 2a + 2b + 2(b-a)\\sqrt{2}\n$$\nAssim, para que a soma seja inteira, devemos ter $a = b$. Logo a soma é igual a $1002 + 4a$. Como $a$ varia de $0$ a $501$ (pois $a+b$ não pode ser maior que $1002$), a soma pode assumir $502$ valores inteiros distintos.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26163,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPara qual número natural $k$ a expressão $\\frac{k^{2}}{1,001^{k}}$ atinge seu maior valor?",
    "options": [],
    "answer": "k = 2001",
    "solution": "Solution:\n\nEstamos procurando o valor de $k$ para o qual é máximo o termo da sequência:\n$$\n\\frac{1^{2}}{1,001}, \\frac{2^{2}}{1,001^{2}}, \\frac{3^{2}}{1,001^{3}}, \\ldots, \\frac{k^{2}}{1,001^{k}}, \\ldots\n$$\nConsidere as seguintes inequações equivalentes:\n$$\n\\frac{(k+1)^{2}}{1,001^{k+1}}<\\frac{k^{2}}{1,001^{k}} \\Leftrightarrow \\frac{(k+1)^{2}}{1,001^{k+1}}-\\frac{1,001 k^{2}}{1,001^{k+1}}<0\n$$\nA segunda inequação tem denominadores iguais e positivos, logo ela é equivalente a\n$$\n(k+1)^{2}-1,001 k^{2}<0 \\Leftrightarrow k(k-2000)>1000 \\Leftrightarrow k>2000\n$$\nAssim, a sequência decresce estritamente para $k \\geq 2001$ e cresce estritamente para $k \\leq 2000$. Logo, o maior termo da sequência corresponde a $k=2001$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26164,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nArqueólogos encontraram um colar de ouro feito de placas no formato de pentágonos regulares. Cada uma destas placas está conectada a outras duas placas, como ilustra a figura.\n\n![](attached_image_1.png)\n\nFigura 51.1\n\nQuantas placas formam o colar?",
    "options": [],
    "answer": "10",
    "solution": "Solution:\n\nO ângulo interno de um pentágono regular mede $108^{\\circ}$. Assim, o ângulo interno do polígono determinado pelo colar mede $360^{\\circ} - 108^{\\circ} - 108^{\\circ} = 144^{\\circ}$. Devemos então encontrar $n$ tal que\n$$\n\\frac{180^{\\circ}(n-2)}{n} = 144^{\\circ}\n$$\nResolvendo esta equação, obtemos $n = 10$. Portanto, dez placas formam o colar.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 26165,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm pirata resolveu enterrar um tesouro em uma ilha. Para tal, ele caminhou da árvore $A$ para a rocha $R_{1}$, e depois a mesma distância e na mesma direção até o ponto $X$. Ele fez o mesmo em relação à entrada da caverna $C$ e em relação à rocha $R_{2}$, alcançando os pontos $Y$ e $Z$, respectivamente. Ele enterrou o tesouro em $T$, ponto médio de $AZ$.\n\n![](attached_image_1.png)\nFigura 54.1\n\nAo voltar à ilha para desenterrar o tesouro, o pirata encontrou as rochas e a caverna, mas não encontrou a árvore. Como o pirata pode descobrir o tesouro?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nA chave para o pirata encontrar o tesouro está no seguinte fato geométrico:\n\nAfirmação: Em todo quadrilátero, os pontos médios dos lados são vértices de um paralelogramo.\n\nIsto significa que a posição $T$ do tesouro independe da posição da árvore. No quadrilátero $A X Y Z$, $R_{1}$, $C$, $R_{2}$ e $T$ são os pontos médios dos lados. Portanto, $R_{1} C R_{2} T$ é um paralelogramo.\n\nO pirata pode começar de um ponto qualquer e repetir os procedimentos, ou pode determinar $T$ traçando uma reta paralela a $R_{1} C$ por $R_{2}$ e uma paralela a $C R_{2}$ por $R_{1}$. O ponto de interseção das paralelas é o ponto $T$, localização do tesouro.\n\n![](attached_image_2.png)\nFigura 54.2\n\nDemonstração da Afirmação: Seja $A B C D$ um quadrilátero convexo e $M$, $N$, $P$ e $Q$ os pontos médios dos lados $AB$, $BC$, $CD$ e $DA$, respectivamente. Vamos provar que $MNPQ$ é um paralelogramo.\n\nConsiderando o triângulo $A B C$, o segmento $M N$ é a base média relativa ao lado $AC$, sendo paralelo ao mesmo e medindo a metade de $A C$.\n\nAnalogamente, olhando para o triângulo $C D A$, o segmento $P Q$ é a base média relativa ao lado $A C$, e portanto é paralelo a $A C$ e mede a metade de $A C$.\n\nSegue que os segmentos $M N$ e $P Q$ são iguais e paralelos, mostrando que o quadrilátero $MNPQ$ é um paralelogramo.\n\n![](attached_image_3.png)\nFigura 54.3",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26166,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSeja $ABC$ um triângulo com $AB = 13$, $BC = 15$ e $AC = 9$. Seja $r$ a reta paralela a $BC$ traçada por $A$. A bissetriz do ângulo $\\angle BCA$ corta a reta $r$ em $E$ e a bissetriz do ângulo $\\angle CBA$ corta $r$ em $F$. Calcular a medida do segmento $EF$.",
    "options": [],
    "answer": "22",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nComo a reta $EF$ é paralela ao lado $BC$, os ângulos alternos internos gerados pela transversal $CF$ são iguais, isto é, $\\angle FCB = \\angle CFA$. Por outro lado, como $CF$ é bissetriz, temos $\\angle FCE = \\angle FCA$ e assim, $\\angle FCA = \\angle CFA$, donde o triângulo $CAF$ é isósceles de base $CF$. Portanto, $AF = AC = 9$.\n\nAnalogamente, concluímos que o triângulo $BAE$ é isósceles de base $BE$ e $AE = AB = 13$. Assim, $EF = EA + AF = 22$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26167,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNo interior de um triângulo $ABC$, toma-se um ponto $E$ tal que $AE = BE$ e $AB = EC$. Se $A\\hat{BE} = \\alpha = E\\hat{CA}$, $E\\hat{AC} = 2\\alpha$ e $E\\hat{BC} = 5\\alpha$, determine $\\alpha$.",
    "options": [],
    "answer": "15°",
    "solution": "Solution:\n\nSabendo que a soma dos ângulos internos de um triângulo é $180^{\\circ}$, obtemos\n\n$$\n\\begin{cases}\nA\\hat{E}B = 180^{\\circ} - (\\alpha + \\alpha) = 180^{\\circ} - 2\\alpha \\\\\nA\\hat{E}C = 180^{\\circ} - (\\alpha + 2\\alpha) = 180^{\\circ} - 3\\alpha\n\\end{cases}\n$$\n\nAssim, temos que\n\n$$\n\\begin{aligned}\nC\\hat{E}B &= 360^{\\circ} - (A\\hat{E}C + A\\hat{E}B) \\\\\n&= 360^{\\circ} - \\left[(180^{\\circ} - 3\\alpha) + (180^{\\circ} - 2\\alpha)\\right] \\\\\n&= 360^{\\circ} - (360^{\\circ} - 5\\alpha) \\\\\n&= 5\\alpha\n\\end{aligned}\n$$\n\nComo o ângulo $E\\hat{BC}$ também mede $5\\alpha$, segue que o triângulo $BEC$ é isósceles. Assim, $AB = CE = BC$, isto é, o triângulo $ABC$ também é isósceles.\n\nLogo, $BC\\hat{A} = B\\hat{A}C = 3\\alpha$ e $B\\hat{C}A + C\\hat{A}B + A\\hat{B}C = 180^{\\circ}$, isto é, $3\\alpha + 3\\alpha + 6\\alpha = 180^{\\circ}$, o que resulta em $\\alpha = 15^{\\circ}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26168,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSe dois lados de um triângulo medem $5$ e $7$ cm, então o terceiro lado não pode medir quantos centímetros?\n(a) $11$\n(b) $10$\n(c) $6$\n(d) $3$\n(e) $1$",
    "options": [],
    "answer": "(e)",
    "solution": "Solution:\n\nA opção correta é (e).\n\nLembre que, num triângulo, a soma de dois lados quaisquer deve ser maior que o terceiro lado. Como $1+5$ não é maior do que $7$, o terceiro lado não pode medir $1~\\mathrm{cm}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26169,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAsterisco $-\\mathrm{Se}$ $\\frac{*}{24}-\\frac{3}{8}-\\frac{2}{3}=\\frac{1}{6}$, quanto vale $*$ ?\n\n(a) 20\n(b) 21\n(c) 23\n(d) 25\n(e) 29",
    "options": [],
    "answer": "e",
    "solution": "Solution:\n\nA opção correta é (e).\n$$\n\\frac{1}{6}=\\frac{*}{24}-\\frac{3}{8}-\\frac{2}{3}=\\frac{*}{24}-\\left(\\frac{3}{8}+\\frac{2}{3}\\right)=\\frac{*}{24}-\\frac{25}{24}=\\frac{*-25}{24}\n$$\nLogo, $\\frac{*-25}{24}=\\frac{1}{6}=\\frac{4}{24}$, donde $*-25=4$, ou seja, $*=29$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26170,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA figura ilustra um polígono regular de 9 lados. A medida do lado do polígono é $a$, a medida da menor diagonal é $b$ e a medida da maior diagonal é $d$.\n![](attached_image_1.png)\nFigura 58.1\n\na) Determine a medida do ângulo $B\\hat{A}E$.\n\nb) Mostre que $d = a + b$.\n\n\nFigura 58.2",
    "options": [],
    "answer": "∠BAE = 60° and d = a + b",
    "solution": "Solution:\n\na) A medida do ângulo interno do eneágono regular (9 lados) é igual a $180^{\\circ} \\times 7 / 9 = 140^{\\circ}$.\n\nConsidere agora o pentágono $ABCDE$, como indicado na figura. A soma de seus ângulos internos é $180^{\\circ}(5-2) = 540^{\\circ}$. Sabemos que $\\angle ABC = \\angle BCD = \\angle CDE = 140^{\\circ}$ e, pela simetria da figura, sabemos que $\\angle EAB = \\angle AED = \\alpha$. Portanto,\n$$\n2\\alpha + 3 \\times 140^{\\circ} = 540^{\\circ}\n$$\ndonde $\\alpha = 60^{\\circ}$.\n\nb) Seja $X$ o ponto de interseção das retas $AB$ e $DE$. Como $\\angle XAE = 60^{\\circ}$, o triângulo $AXE$ é equilátero. O triângulo $BXD$ também é equilátero, pois a reta $AE$ é paralela à reta $BD$.\nAssim, temos $AX = AE$ e $BX = BD$. Como $AX = AB + BX$, temos $AE = AB + BD$, ou seja, $d = a + b$.\n\n![](attached_image_2.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26171,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nExpressões algébricas - $\\mathrm{O}$ que representam, geometricamente, na figura dada, as expressões\n$$\na^{2}+1,5 a\n$$\ne\n$$\n4 a+3\n$$\n![](attached_image_1.png)",
    "options": [],
    "answer": "a^2 + 1.5a is the total area of the figure; 4a + 3 is the perimeter of the figure.",
    "solution": "Solution:\n\nNote que a figura é um retângulo formado por um quadrado de lado $a$ e um retângulo de lados $1{,}5$ e $a$. Logo, $a^{2}$ é a área do quadrado e $1{,}5 a$ é a área do retângulo. Assim, $a^{2}+1{,}5 a$ representa a soma dessas duas áreas, ou seja, a área total da figura.\n\nJá $4 a+3=3 a+1{,}5+a+1{,}5$ é o perímetro da figura.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26172,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTodos os ângulos de um hexágono $A B C D E F$ são iguais. Mostre que $\\mathrm{AB}-\\mathrm{DE}=\\mathrm{EF}-\\mathrm{BC}=\\mathrm{CD}-\\mathrm{FA}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nProlonguemos os segmentos $AF$, $BC$ e $DE$ determinando os pontos de intersecção $X$, $Y$, $Z$, como mostrado na figura.\n\nComo a soma dos ângulos internos de um hexágono convexo é $180^{\\circ} \\times (6-2) = 720^{\\circ}$, cada ângulo interno deste hexágono mede $720^{\\circ} / 6 = 120^{\\circ}$. Assim, $X \\hat{A} B = 180^{\\circ} - 120^{\\circ} = 60^{\\circ}$ e do mesmo modo $X \\hat{B} A = 60^{\\circ}$. Segue que o ângulo $A \\hat{X} B$ mede $60^{\\circ}$ e de igual forma os ângulos em $Y$ e $Z$ medem $60^{\\circ}$.\n\nPortanto, os triângulos $XAB$, $YCD$, $ZFE$ e $XYZ$ são equiláteros. Em particular, $XY = XZ$. Mas\n\n$$\n\\begin{aligned}\n& X Y = X B + B C + C Y = A B + B C + C D \\\\\n& X Z = X A + A F + F Z = A B + A F + E F\n\\end{aligned}\n$$\n\nIgualando obtemos $B C + C D = A F + E F$, donde obtemos $E F - B C = C D - F A$. Pelo mesmo processo, de $X Y = Y Z$, obtemos $A B - D E = E F - B C$.\n\nAssim, $A B - D E = E F - B C = C D - F A$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26173,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA professora Jussara gosta de fazer cópias reduzidas. Ela começa desenhando o triângulo retângulo $A B C$ abaixo:\n![](attached_image_1.png)\nEm seguida, a professora Jussara traça um segmento $A P_{1}$ de forma que $A P_{1}$ seja perpendicular a $B C$, conforme mostra a figura abaixo.\n![](attached_image_2.png)\nJussara afirma então que o triângulo $A P_{1} C$ é semelhante ao triângulo $A B C$ (ou seja, tem ângulos correspondentes iguais).\na) Mostre que a professora Jussara está certa!\nb) Calcule o comprimento do segmento $A P_{1}$.\nc) Calcule a razão entre a área do triângulo $A B C$ e a área do triângulo $A P_{1} C$.\nd) A professora Jussara repete o processo, agora traçando um segmento $P_{1} P_{2}$ perpendicular ao lado $A C$, conforme a figura abaixo.\n![](attached_image_3.png)\nQual a razão entre as áreas dos triângulos $P_{1} P_{2} C$ e $A P_{1} C$ ?\ne) A professora Jussara repete o processo mais duas vezes, conforme mostra a figura abaixo.\n![](attached_image_4.png)\nQual a área do triângulo $P_{5} P_{6} C$ ?",
    "options": [],
    "answer": "a) The claim is correct: triangles APC and ABC are similar. b) AP1 = 12/5. c) Area(ABC) : Area(AP1C) = 25/16. d) Area(P1P2C) : Area(AP1C) = 16/25. e) Area(P5P6C) = 6 × (16/25)^6.",
    "solution": "Solution:\na) Usaremos aqui que a soma dos ângulos internos de um triângulo é igual a $180^{\\circ}$. Considere a seguinte figura:\n![](attached_image_5.png)\nUsando os ângulos internos do triângulo $A B C$, concluímos que $\\alpha+\\beta+90^{\\circ}=180^{\\circ}$. Logo temos que $\\alpha+\\beta=90^{\\circ}$. Um mesmo raciocínio nos fornece que $\\alpha+\\gamma=90^{\\circ}$. Assim, temos que $\\beta=\\gamma$. Logo, os triângulos $B A C$ e $A P_{1} C$ têm ângulos com mesmas medidas, logo são semelhantes.\n\nb) Vimos no item a) que os triângulos $B A C$ e $A P_{1} C$ são semelhantes. Assim, obtemos que a razão entre os catetos adjacentes aos ângulos correspondentes deve ser igual à razão entre as hipotenusas. Ou seja,\n$$\n\\frac{\\left|A P_{1}\\right|}{|A B|}=\\frac{|A C|}{|B C|}\n$$\nSegue daí que\n$$\n\\left|A P_{1}\\right|=\\frac{|A C| \\times|A B|}{|B C|}=\\frac{3 \\times 4}{5}=\\frac{12}{5}\n$$\nObservação: Note que\n$$\n\\frac{\\left|A P_{1}\\right|}{|A B|}=\\frac{|A C|}{|B C|}=\\frac{4}{5}\n$$\nou seja, a razão de semelhança entre esses dois triângulos é igual a 4/5.\n\nc) Como a razão de semelhança entre os triângulos é igual a $4 / 5$, temos que a base e a altura do triângulo $A P_{1} C$ medem $4 / 5$ da base e altura do triângulo $A B C$ respectivamente. Assim temos que:\n$$\n\\begin{aligned}\n\\text { Área }\\left(A P_{1} C\\right) & =\\frac{\\operatorname{base}\\left(A P_{1} C\\right) \\times \\text { altura }\\left(A P_{1} C\\right)}{2}=\\frac{4 / 5 \\times \\operatorname{base}(A B C) \\times 4 / 5 \\times \\text { altura }(A B C)}{2} \\\\\n& =\\frac{16}{25} \\frac{\\text { base }(A B C) \\times \\operatorname{altura}(A B C)}{2}=\\frac{16}{25} \\times \\operatorname{Área}(A B C)\n\\end{aligned}\n$$\nSegue daí que a razão entre as áreas dos triângulos $A P_{1} C$ e $A B C$ é igual a 16/25.\n\nd) Usando um argumento semelhante ao do item a), conclui-se que os triângulos $A P_{1} C$ e $P_{1} P_{2} C$ são semelhantes. Pela figura abaixo,\n![](attached_image_6.png)\nconclui-se que\n$$\n\\frac{\\left|P_{1} P_{2}\\right|}{\\left|A P_{1}\\right|}=\\frac{\\left|P_{1} C\\right|}{|A C|}\n$$\nComo a razão de semelhança entre os triângulos $A P_{1} C$ e $A B C$ é igual a 4/5, vale que\n$$\n\\frac{\\left|P_{1} P_{2}\\right|}{\\left|A P_{1}\\right|}=\\frac{4}{5}\n$$\nDaí, concluímos que a razão de semelhança entre os triângulos $P_{1} P_{2} C$ e $A P_{1} C$ também é igual a $\\frac{4}{5}$. Portanto, a proporção entre as suas áreas é também igual a $16 / 25$.\n\ne) Através da figura\n![](attached_image_7.png)\ne um procedimento análogo ao dos itens a) e d), vemos que os triângulos $P_{5} P_{6} C$, $P_{4} P_{5} C, P_{3} P_{4} C, P_{2} P_{3} C, P_{1} P_{2} C, A P_{1} C$ e $B A C$ são todos semelhantes, sendo a razão de semelhança entre dois desses triângulos consecutivos igual a 4/5. Logo, a razão entre as áreas de dois desses triângulos consecutivos é igual a 16/25. Daí,\n$$\n\\begin{aligned}\n\\text { Área }\\left(P_{5} P_{6} C\\right) & =\\frac{16}{25} \\times \\operatorname{Área}\\left(P_{4} P_{5} C\\right)=\\left(\\frac{16}{25}\\right)^{2} \\times \\operatorname{Área}\\left(P_{3} P_{4} C\\right)=\\cdots= \\\\\n& =\\left(\\frac{16}{25}\\right)^{4} \\times \\operatorname{Área}\\left(P_{1} P_{2} C\\right)=\\left(\\frac{16}{25}\\right)^{5} \\times \\text { Área }\\left(A P_{1} C\\right)= \\\\\n& =\\left(\\frac{16}{25}\\right)^{6} \\times \\operatorname{Área}(A B C)=\\left(\\frac{16}{25}\\right)^{6} \\times \\frac{3 \\times 4}{2}\n\\end{aligned}\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26174,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nAugusto tem um arame com $10~\\mathrm{m}$ de comprimento. Ele realiza um corte em um ponto do arame obtendo assim dois arames. Um com comprimento $x$ e outro com comprimento $10-x$ como mostra a figura abaixo:\n![](attached_image_1.png)\nAugusto usa os dois pedaços do arame para fazer dois quadrados.\n\na) Qual é o comprimento do lado de cada um dos quadrados? Qual é a área de cada um?\n\nb) Qual é o valor do comprimento de cada um dos dois pedaços do arame para que a soma das áreas dos quadrados obtidos seja mínima?\n\nc) Suponha que Augusto corte o arame em dez pedaços e use cada um deles para fazer um quadrado. Qual deve ser o tamanho de cada um dos pedaços para que a soma das áreas dos quadrados obtidos seja mínima?",
    "options": [],
    "answer": "a) Lados: x/4 e (10 − x)/4. Áreas: x^2/16 e (10 − x)^2/16.\nb) Corte no meio: comprimentos 5 e 5.\nc) Dez pedaços iguais: cada um com comprimento 1.",
    "solution": "Solution:\nNessa questão todos os comprimentos são dados em metros e as áreas em metros quadrados.\n\na) Um pedaço de corda tem comprimento $x$ e outro pedaço de corda tem comprimento $10-x$. Como um quadrado tem quatro lados de tamanhos iguais, um quadrado terá lado de comprimento igual a $\\frac{x}{4}$ e outro quadrado terá lado de comprimento igual a $\\frac{10-x}{4}$.\nA área de um quadrado de lado $\\ell$ é igual a $\\ell^{2}$. Portanto, um quadrado terá área igual a $\\left(\\frac{x}{4}\\right)^{2}=\\frac{x^{2}}{16}$ enquanto o outro quadrado terá área igual a $\\left(\\frac{10-x}{4}\\right)^{2}=\\frac{100-20x+x^{2}}{16}$.\n\nb) Seja $S(x)$ a soma das áreas dos dois quadrados. Pelo item anterior, temos que\n$$\nS(x)=\\frac{x^{2}}{16}+\\frac{100-20x+x^{2}}{16}=\\frac{100-20x+2x^{2}}{16}=\\frac{1}{8}x^{2}-\\frac{5}{4}x+\\frac{25}{4}\n$$\né uma função do segundo grau. O mínimo de uma função do tipo\n$$\nf(x)=ax^{2}+bx+c\n$$\ncom $a>0$ é atingido em $x=\\frac{-b}{2a}$. Assim, a área mínima será atingida se\n$$\nx=-\\frac{\\left(-\\frac{5}{4}\\right)}{2\\cdot\\frac{1}{8}}=5\n$$\nOu seja, se a corda for cortada exatamente no meio!\n\nc) Pelo item anterior, sabemos que para minimizar a soma das áreas é necessário cortar exatamente no meio. Bem, afirmamos que para minimizar a área com nove cortes (ou seja, criando dez quadrados) é necessário que os pedaços de corda sejam todos iguais. Para mostrar isso, vejamos o seguinte argumento: se dois dos dez pedaços de corda fossem diferentes, seria possível diminuir a área cortando os pedaços de corda de modo que esses dois fossem iguais (estamos usando o item anterior). Portanto, dois pedaços de corda quaisquer devem ser iguais. Logo, todos devem ser iguais!",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26175,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nNa figura a seguir, o círculo de centro $B$ é tangente ao círculo de centro $A$ em $X$. O círculo de centro $C$ é tangente ao círculo de centro $A$ em $Y$. Além disto, os círculos de centros $B$ e $C$ também são tangentes. Se $A B=6, A C=5$ e $B C=9$, quanto mede $A X$ ?\n![](attached_image_1.png)",
    "options": [],
    "answer": "10",
    "solution": "Solution:\nSejam $r_{a}, r_{b}$ e $r_{c}$ os raios dos círculos de centros $A, B$ e $C$, respectivamente. Se $Z$ é o ponto de tangência dos círculos de centros $B$ e $C$, os dados do problema nos permitem montar o seguinte sistema de equações:\n$$\n\\begin{aligned}\nA B & = A X - B X \\\\\n6 & = r_{a} - r_{b} \\\\\nA C & = A Y - C Y \\\\\n5 & = r_{a} - r_{c} \\\\\nB C & = B Z + Z C \\\\\n9 & = r_{b} + r_{c}\n\\end{aligned}\n$$\nEntão\n$$\n\\begin{aligned}\n6 + 5 & = (r_{a} - r_{b}) + (r_{a} - r_{c}) \\\\\n& = 2 r_{a} - (r_{b} + r_{c}) \\\\\n& = 2 r_{a} - 9 \\\\\nr_{a} & = 10\n\\end{aligned}\n$$\nPortanto, $A X = 10$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26176,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSe $m!$ termina com exatamente $n$ zeros, dizemos que $n$ é a cauda do fatorial $m!$. Observe os exemplos e responda:\n- $5! = 5 \\times 4 \\times 3 \\times 2 \\times 1 = 120$, termina em um zero, por isso, a cauda do fatorial $5!$ é $1$;\n- $10! = 3628800$, termina em dois zeros, logo a cauda do fatorial $10!$ é igual a $2$.\n\na) Quais são as caudas dos fatoriais de $20!$ e $25!$?\n\nb) Qual o dígito das dezenas de $7! + 8! + 9! + \\ldots + 2018!$?\n\nc) Algum $m!$ tem cauda do fatorial igual a $5$?\n\nd) Qual a cauda do fatorial de $2018!$ (dois mil e dezoito fatorial)?\n\ne) Quantos inteiros positivos menores que $2018$ não são cauda do fatorial?",
    "options": [],
    "answer": "a) 4 and 6; b) 4; c) no; d) 502; e) 121",
    "solution": "Solution:\n\na) Como $10 = 2 \\cdot 5$, basta contar as quantidades de fatores $2$ e $5$ nesses números. Observe que $20!$ é o mesmo que $20 \\cdot \\ldots \\cdot 15 \\cdot \\ldots \\cdot 10 \\cdot \\ldots \\cdot 5 \\cdot \\ldots \\cdot 1$, como são $4$ múltiplos de $5$ em sua representação fatorial e existem mais de $4$ fatores $2$ nesse produto, teremos $4$ zeros à sua direita. Expandindo $25!$, temos $5$ múltiplos de $5$, sendo um múltiplo de $5^{2}$. Como a quantidade de fatores $2$ é maior, logo ele termina em $6$ zeros. Como existem mais fatores $2$ do que $5$, a cauda do fatorial de $m!$ é determinada pelo expoente de $5$ na fatoração de $m!$.\n\nb) Como $10!$ é múltiplo de $100$, qualquer fatorial associado a um número maior que $10$ é também múltiplo de $100$. Portanto, os últimos dois dígitos da soma\n$$\n10! + 11! + 12! + \\ldots + 2018!\n$$\nsão $00$. Os últimos dois dígitos de $7! + 8! + 9! = 5040 + 40320 + 362880 = 408240$ são $40$. Logo o dígito das dezenas procurado é $4$.\n\nc) Pelos itens anteriores, com inclusão do produto por $25$, o expoente do $5$ na expansão do fatorial pula de $4$ para $6$. Os próximos fatoriais terão pelo menos $6$ zeros e assim a não existe cauda do fatorial de tamanho $5$.\n\nObservação: Perceba que a cada novo múltiplo de $5$, a cauda \"pula\" uma unidade, para múltiplos de $5^{2}$ teremos duas unidades, para os de $5^{3}$ serão três e assim por diante.\n\nd) Devemos analisar os múltiplos das potências de $5$ menores ou iguais a $2018$: $5$, $25$, $125$, $625$. São $403$ múltiplos de $5$, $80$ de $25$, $16$ de $125$ e $3$ de $625$. Portanto, a cauda será igual a\n$$\n403 + 80 + 16 + 3 = 502\n$$\n\ne) A cada pulo duplo associado a um múltiplo de $5^{2}$, perdemos um inteiro como possível cauda do fatorial. A cada pulo triplo associado a um múltiplo de $5^{3}$, perdemos dois possíveis inteiros como cauda do fatorial e assim sucessivamente. Analisando os múltiplos de $5$ menores ou iguais a $2018$, temos $80$ pulos duplos, $16$ triplos e $3$ quádruplos. Portanto, chegamos a\n$$\n80 \\times 1 + 16 \\times 2 + 3 \\times 3 = 121\n$$\ninteiros que não são cauda do fatorial.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26177,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPara comemorar seu aniversário, Ana vai preparar tortas de pera e tortas de maçã. No mercado, uma maçã pesa $300~\\mathrm{g}$ e uma pera $200~\\mathrm{g}$. A sacola de Ana aguenta um peso máximo de $7~\\mathrm{kg}$. Qual é o número máximo de frutas que ela pode comprar para poder fazer tortas das duas frutas?",
    "options": [],
    "answer": "34",
    "solution": "Solution:\n\nDenotemos por $m$ o número de maçãs e $p$ o número de peras que Ana comprou, assim o peso que ela leva na sacola é $300 m + 200 p$ gramas. Como a sacola aguenta no máximo $7000$ gramas, temos que\n$$\n300 m + 200 p \\leq 7000, \\text{ que é equivalente a } 3 m + 2 p \\leq 70\n$$\nComo as peras pesam menos, Ana tem que levar a máxima quantidade de peras, e portanto, a mínima quantidade de maçãs. Assim, se ela levar $1$ maçã, temos:\n$$\n2 p \\leq 70 - 3 = 67 \\Longrightarrow p \\leq 33,5\n$$\nLogo, levando $1$ maçã, ela pode levar $33$ peras. Então, o número máximo de frutas é $34$.\n\nNa tabela abaixo vemos que Ana pode também levar $2$ maçãs e $32$ peras.\n\n| $p$ | $m$ | $300 m + 200 p$ | $p + m$ |\n| :---: | :---: | :---: | :---: |\n| 34 | 0 | 6800 | 34 |\n| 33 | 1 | 6900 | 34 |\n| 32 | 2 | 7000 | 34 |",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26178,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoma de razão $\\frac{1}{2}$ - Se $S_{n}=\\frac{1}{2}+\\frac{1}{2^{2}}+\\cdots+\\frac{1}{2^{n}}$, qual é o menor número inteiro positivo $n$ tal que $S_{n}>0,99$ ?",
    "options": [],
    "answer": "7",
    "solution": "Solution:\n\nComo\n$$\nS_{n}=\\frac{1}{2}+\\frac{1}{4}+\\frac{1}{8}+\\cdots+\\frac{1}{2^{n}}\n$$\nsegue que\n$$\n\\frac{1}{2} S_{n}=\\frac{1}{2} \\times\\left(\\frac{1}{2}+\\frac{1}{4}+\\frac{1}{8}+\\cdots+\\frac{1}{2^{n}}\\right)=\\frac{1}{4}+\\frac{1}{8}+\\cdots+\\frac{1}{2^{n+1}}\n$$\nLogo,\n$$\n\\frac{1}{2} S_{n}=S_{n}-\\frac{1}{2} S_{n}=\\frac{1}{2}-\\frac{1}{2^{n+1}}\n$$\nAssim\n$$\nS_{n}=1-\\frac{1}{2^{n}}\n$$\nComo queremos $S_{n}>0,99$, isto é equivalente a encontrar o menor $n$ tal que\n$$\n1-\\frac{1}{2^{n}}>0,99\n$$\ne assim\n$$\n2^{n}>100\n$$\nLogo, devemos ter $n \\geq 7$ porque $128=2^{7}>100>2^{6}=64$.\n\nObservação: Outro modo de calcular $S_{n}$, é notar que é a soma de uma progressão geométrica com $a_{1}=1 / 2$ e razão $q=1 / 2$. Aplicando a fórmula, temos:\n$$\nS_{n}=\\frac{\\frac{1}{2}-\\frac{1}{2^{n}} \\frac{1}{2}}{1-\\frac{1}{2}}=1-\\frac{1}{2^{n}}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26179,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNuma família cada menino tem o mesmo número de irmãos que de irmãs, e cada menina tem o dobro de irmãos que de irmãs. Qual é a composição dessa família?",
    "options": [],
    "answer": "3 girls and 4 boys",
    "solution": "Solution:\n\n3 meninas e 4 meninos",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26180,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUma brincadeira - É feita uma brincadeira com quatro números inteiros da seguinte maneira: some três desses números, divida essa soma por $3$ e o resultado some com o quarto número. Existem quatro formas de fazer esta brincadeira, obtendo os seguintes resultados: $17$, $21$, $23$ e $29$. Qual é o maior dos quatro números?",
    "options": [],
    "answer": "21",
    "solution": "Solution:\n\nSejam $a$, $b$, $c$ e $d$ os números procurados. São dados os números\n$$\n\\frac{a+b+c}{3} + d, \\quad \\frac{a+b+d}{3} + c, \\quad \\frac{a+c+d}{3} + b, \\quad \\text{e} \\quad \\frac{b+c+d}{3} + a\n$$\nmas não sabemos a ordem deles. Como\n$$\n\\begin{aligned}\n\\frac{a+b+c}{3} + d + \\frac{a+b+d}{3} + c + \\frac{a+c+d}{3} + b + \\frac{b+c+d}{3} + a &= 2(a+b+c+d) \\\\\n&= 17 + 21 + 23 + 29\n\\end{aligned}\n$$\nLogo,\n$$\n2(a+b+c+d) = 90 \\Longrightarrow a+b+c+d = 45\n$$\nAgora, seja $d$ o maior dentre os números $a$, $b$, $c$ e $d$. Assim,\n$$\nd = 29 - \\frac{a+b+c}{3} = 29 - \\frac{45-d}{3} \\Longrightarrow d = 21\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26181,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUma urna tem 6 bolas numeradas de 1 a 6. Se duas bolas são extraídas, qual é a probabilidade da diferença entre os números dessas 2 bolas ser 1?",
    "options": [],
    "answer": "1/3",
    "solution": "Solution:\n\nObservemos que se extraímos a primeira bola com um número entre 2 e 5, então dentre as 5 bolas que ficam na urna temos duas possíveis bolas que cumprem a condição do problema, logo neste caso a probabilidade que a segunda bola cumpra a condição é $\\frac{2}{5}$ e a probabilidade que a primeira bola tenha um número entre 2 e 5 é $\\frac{4}{6}$. Por outro lado, se a primeira bola extraída é 1 ou 6, só temos uma bola na urna que cumpre a condição, logo neste caso a probabilidade para a escolha da segunda bola é $\\frac{1}{5}$ e a probabilidade da primeira bola ser 1 ou 6 é $\\frac{2}{6}$. Portanto, a probabilidade das bolas serem consecutivas é\n$$\n\\frac{4}{6} \\times \\frac{2}{5} + \\frac{2}{6} \\times \\frac{1}{5} = \\frac{1}{3}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26182,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm número de 3 algarismos e seu sêxtuplo são formados pelos mesmos algarismos. A soma dos algarismos desse número é 17 e a de seu sêxtuplo é 21. Qual é esse número? Existe mais do que um?",
    "options": [],
    "answer": "746; unique",
    "solution": "Solution:\n\n746 (solução única?)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26183,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFonte com 3 torneiras - Sílvia vai a uma fonte que tem três torneiras, encher os seus dez garrafões. Um dos garrafões demora um minuto para encher, outro dois minutos, outro três minutos e assim por diante. Como Sílvia deverá distribuir os garrafões pelas torneiras de modo a gastar o menor tempo possível? Qual é esse tempo?",
    "options": [],
    "answer": "Minimum time: 19 minutes. One optimal assignment is: Tap 1 fills the ten-minute and nine-minute containers; Tap 2 fills the eight-minute, seven-minute, and three-minute containers; Tap 3 fills the five-minute, four-minute, two-minute, and one-minute containers.",
    "solution": "Solution:\n\nSolução 1: Para simplificar, numeramos os garrafões de acordo com os respectivos tempos que gastam para ficar cheios. A ideia é utilizar o \"tempo que sobra\" de um garrafão para encher outro garrafão, enchendo simultaneamente outros. As figuras ilustram a solução.\n\n![](attached_image_1.png)\n\nNa figura I as 3 torneiras gastam 10 minutos para encher os garrafões 10, 9, 8, 1 e 2. Na figura II as 3 torneiras gastam 9 minutos para encher os garrafões $7, 6, 5, 2, 3$ e 4. Logo, o tempo total gasto é de 19 minutos.\n\n\nSolução 2: Se tivéssemos uma torneira só, o tempo gasto para encher os 10 garrafões é $1+2+\\cdots+9+10=55$ minutos. Como $55=18 \\times 3+1$, se temos 3 torneiras devemos gastar pelo menos 19 minutos. A seguinte tabela mostra a forma de fazer o trabalho em 19 minutos.\n\n| Torneira 1 | 10 | 9 |   |   |\n| :---:      |:---:|:---:|:---:|:---:|\n| Torneira 2 | 8  | 7  | 3 |   |\n| Torneira 3 | 5  | 4  | 2 | 1 |",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26184,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n6. Percentagem de mortalidade - Se $15\\%$ dos membros de uma população afetados por uma doença $8\\%$ morreram, a percentagem da mortalidade em relação à população inteira é:\n(a) $1,2\\%$\n(b) $1,8\\%$\n(c) $8\\%$\n(d) $12\\%$\n(e) $23\\%$",
    "options": [],
    "answer": "a",
    "solution": "Solution:\n\nA proporção de população que fica doente pela enfermidade é $\\frac{15}{100}$ e dos que ficam doentes, a proporção que morre é $\\frac{8}{100}$. Logo, a proporção de população que morre pela doença é $\\frac{15}{100} \\times \\frac{8}{100}$, que corresponde a\n\n$$\n\\frac{15 \\times 8}{100^{2}}=\\frac{120}{10000}=\\frac{12}{1000}=\\frac{1,2}{100}=1,2\\%\n$$\n\nA opção correta é (a).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26185,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDois quadrados - Na figura ao lado, a área do quadrado maior é $10~\\mathrm{cm}^2$ e do menor é $4~\\mathrm{cm}^2$. As diagonais do quadrado maior contêm as diagonais do quadrado menor. Quanto mede a área da região tracejada?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "21",
    "solution": "Solution:\n\nObservemos que a área do quadrado maior menos a área do quadrado menor é igual a 4 vezes a área procurada. Logo a área tracejada é\n$$\n\\frac{10^2-4^2}{4}=\\frac{100-16}{4}=25-4=21\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26186,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nQual dos seguintes triângulos não pode existir?\n\na) triângulo agudo isósceles\nb) triângulo retângulo isósceles\nc) triângulo retângulo obtusângulo\nd) triângulo retângulo escaleno\ne) triângulo escaleno obtusângulo",
    "options": [],
    "answer": "c",
    "solution": "Solution:\n\nA soma dos três ângulos internos de um triângulo é $180^\\circ$. Logo, se um deles mede $90^\\circ$, a soma dos outros dois é $90^\\circ$, e por isso não podem ser maiores do que $90^\\circ$. Portanto, não existem triângulos retângulos obtusângulos.\n\nOs seguintes exemplos de comprimentos de lados mostram que os outros casos podem ocorrer:\n\na) $2, 3, 3$;\nb) $1, 1, \\sqrt{2}$;\nd) $3, 4, 5$;\ne) $3, 4, 6$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26187,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMostre que é possível construir um pentágono com todos os lados de mesma medida e cujos ângulos internos meçam $60^{\\circ}$, $80^{\\circ}$, $100^{\\circ}$, $140^{\\circ}$ e $160^{\\circ}$, em alguma ordem.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSuponhamos que já construímos o pentágono $A B C D E$ e que o ângulo em $A$ mede $60^{\\circ}$. Traçando a reta $B E$, concluímos que o triângulo $A B E$ é equilátero, pois $A B = A E$ e $E \\hat{A} B = 60^{\\circ}$. Logo, $B E = A B$ e, portanto, $B C D E$ tem todos os seus lados com a mesma medida, isto é, $BCDE$ é um losango.\n\nEm particular, os ângulos opostos do losango são iguais. Isto implica que, no pentágono, o ângulo em $B$ é igual ao ângulo em $D$ mais $60^{\\circ}$ e o ângulo em $E$ é igual ao ângulo em $C$ mais $60^{\\circ}$.\n\nComo $160^{\\circ} = 100^{\\circ} + 80^{\\circ}$ e $140^{\\circ} = 80^{\\circ} + 60^{\\circ}$, concluímos que os ângulos em $C$ e $D$ devem assumir os valores $80^{\\circ}$ e $100^{\\circ}$, não necessariamente nessa ordem, enquanto $B$ e $E$ assumem os respectivos valores de $D$ e $C$, adicionados de $60^{\\circ}$.\n\nPortanto, para construir tal pentágono basta construir um triângulo equilátero $A B E$ e um losango $B C D E$ com ângulos de medidas $100^{\\circ}$ e $80^{\\circ}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26188,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDesigualdade racional - Quais são os valores de $x$ que satisfazem a desigualdade $\\frac{1}{x-2}<4$ ?\n\n(a) $x>\\frac{9}{4}$\n\n(c) $x<2$ ou $x>\\frac{9}{4}$\n\n(e) $x<2$\n\n(b) $2<x$ e $x<\\frac{9}{4}$\n\n(d) $x<-2$",
    "options": [],
    "answer": "(c)",
    "solution": "Solution:\n\nA opção correta é (c).\n\nTemos $\\frac{1}{x-2}<4 \\Longleftrightarrow \\frac{1}{x-2}-4<0 \\Longleftrightarrow \\frac{1-4(x-2)}{x-2}<0 \\Longleftrightarrow \\frac{9-4x}{x-2}<0$.\n\nPara que uma fração seja negativa, o numerador e o denominador devem ter sinais contrários.\n\n1o Caso: $9-4x>0$ e $x-2<0$. Devemos ter $x<(9/4)$ e $x<2$. Portanto, $x<2$, pois sendo menor do que 2, automaticamente $x$ será menor do que $9/4$. Concluímos que todo $x<2$ satisfaz a desigualdade.\n\n2o Caso: $9-4x<0$ e $x-2>0$. Devemos ter $x>(9/4)$ e $x>2$. Portanto, $x>(9/4)$, pois sendo maior do que $9/4$, automaticamente $x$ será maior do que 2. Concluímos que todo $x>(9/4)$ satisfaz a desigualdade.\n\nJuntando os dois casos, concluímos que $x$ satisfaz a desigualdade se, e só se, $x<2$ ou $x>(9/4)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26189,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nQuantos números inteiros e positivos satisfazem a dupla inequação $2000<\\sqrt{n(n-1)}<2005$?\n\n(a) 1\n(b) 2\n(c) 3\n(d) 4\n(e) 5",
    "options": [],
    "answer": "e",
    "solution": "Solution:\n\nA opção correta é (e).\nComo os números que aparecem são todos positivos, podemos elevá-los ao quadrado mantendo o sentido das desigualdades, obtendo\n$$\n2000 \\times 2000 = 2000^{2} < n(n-1) < 2005^{2} = 2005 \\times 2005\n$$\nObserve que $n$ e $n-1$ são inteiros consecutivos, portanto, as únicas opções são as seguintes.\n- $2000^{2} < 2000 \\times 1999 < 2005^{2}$\n- $2000^{2} < 2001 \\times 2000 < 2005^{2}$\n- $2000^{2} < 2002 \\times 2001 < 2005^{2}$\n- $2000^{2} < 2003 \\times 2002 < 2005^{2}$\n- $2000^{2} < 2004 \\times 2003 < 2005^{2}$\nLogo, temos cinco possibilidades para $n$, a saber, $2000$, $2001$, $2002$, $2003$ e $2004$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26190,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm anel simétrico com $m$ polígonos regulares de $n$ lados cada é formado de acordo com as regras:\ni) cada polígono no anel encontra dois outros;\nii) dois polígonos adjacentes têm apenas um lado em comum;\niii) o perímetro da região interna delimitada pelos polígonos consiste em exatamente dois lados de cada polígono.\nO exemplo na figura a seguir mostra um anel com $m=6$ e $n=9$. Para quantos valores diferentes de $n$ é possível construir esse anel?\n![](attached_image_1.png)",
    "options": [],
    "answer": "6",
    "solution": "Solution:\n\nSeja $\\alpha=\\frac{360^{\\circ}}{n}$ a medida de cada ângulo externo de um polígono regular $A B C D E \\ldots$. Suponha que o caminho $B C D$ faz parte do perímetro da região interna de algum anel. Se $O$ é o encontro dos prolongamentos de $A B$ e $D E$, por simetria, ele é o centro da região interna. Como são $m$ polígonos regulares, conclui-se que $\\angle B O D=\\frac{360^{\\circ}}{m}$. Agora, a soma dos ângulos internos de $B C D O$ fica\n$$\n\\begin{aligned}\n360^{\\circ} & =\\frac{360^{\\circ}}{m}+\\alpha+(180+\\alpha)+\\alpha \\\\\n360^{\\circ} & =\\frac{360^{\\circ}}{m}+\\frac{3 \\cdot 360^{\\circ}}{n}+180^{\\circ} \\\\\n1 & =\\frac{2}{m}+\\frac{6}{n}\n\\end{aligned}\n$$\nDaí, multiplicando a equação anterior por $m n$ e fatorando a expressão encontrada, temos\n$$\n\\begin{aligned}\nm n & =6 m+2 n \\\\\n(m-2)(n-6) & =12\n\\end{aligned}\n$$\n![](attached_image_2.png)\nPara o anel existir, precisamos de $m>2$. Além disso, $(m-2)$ e $(n-6)$ precisam dividir 12 e $n>0$, então ( $n-6) \\in\\{1,2,3,4,6,12\\}$, o que faz $n \\in\\{7,8,9,10,12,18\\}$, sendo 6 valores possíveis diferentes para $n$. Para verificar que todos eles são soluções, basta considerar os seguintes exemplos de anéis\n![](attached_image_3.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26191,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDado um quadrilátero convexo, se as quatro bissetrizes de seus ângulos formam um novo quadrilátero $H I J E$, calcule a soma dos ângulos opostos $\\angle H I J + \\angle J E H$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "180°",
    "solution": "Solution:\n\n![](attached_image_2.png)\nComo a soma dos ângulos internos de um quadrilátero é $360^\\circ$, temos:\n$$\n\\begin{aligned}\n\\alpha + \\beta & = 360^\\circ - \\angle I H E - \\angle I J E \\\\\n& = 360^\\circ - \\angle D H A - \\angle C J B \\\\\n& = 360^\\circ - \\left(180^\\circ - \\angle A D H - \\angle D A H\\right) - \\left(180^\\circ - \\angle J C B - \\angle J B C\\right) \\\\\n& = \\angle A D H + \\angle D A H + \\angle J C B + \\angle J B C \\\\\n& = \\frac{\\angle A D C}{2} + \\frac{\\angle D A B}{2} + \\frac{\\angle D C B}{2} + \\frac{\\angle C B A}{2} \\\\\n& = \\frac{360^\\circ}{2} \\\\\n& = 180^\\circ\n\\end{aligned}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26192,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\na) Uma calculadora do país de Cincolândia tem apenas os algarismos de 0 a 9 e dois botões $\\square$ e $\\triangle$. O botão $\\square$ eleva ao quadrado o número que está no visor da calculadora. O botão $\\triangle$ subtrai 5 do número que está no visor da calculadora. Mônica digita o número 7 e depois aperta $\\square$, em seguida, aperta o botão $\\triangle$. Qual o resultado mostrado pela calculadora?\n\nb) Mostre que se um número natural $x$ deixa resto 4 quando dividido por 5, então o número $x^{2}$ deixa resto 1 quando dividido por 5.\n\nc) Na calculadora de Cincolândia, é possível digitar o número 9 e depois chegar ao resultado 7 apertando os botões $\\square$ ou $\\triangle$ de maneira adequada?",
    "options": [],
    "answer": "a) 44; b) remainder 1 when divided by 5; c) no",
    "solution": "Solution:\na) Mônica começa digitando o número 7. Daí,\n$$\n7 \\xrightarrow{\\square} 7^{2}=49 \\xrightarrow{\\triangle} 49-5=44.\n$$\nLogo, o resultado final que aparece na calculadora é o número 44.\n\nb) Se um número natural $x$ deixa resto 4 quando dividido por 5, isso quer dizer que $x$ é da forma\n$$\nx=5q+4\n$$\nonde $q$ é um número natural. Elevando ao quadrado, obtemos\n$$\n\\begin{aligned}\nx^{2} & =(5q+4)^{2} \\\\\n& =25q^{2}+2 \\cdot 5q \\cdot 4+4^{2} \\\\\n& =5\\left(5q^{2}+8q\\right)+16 \\\\\n& =5\\left(5q^{2}+8q\\right)+15+1 \\\\\n& =5\\left(5q^{2}+8q+3\\right)+1\n\\end{aligned}\n$$\no que quer dizer que $x^{2}$ deixa resto 1 na divisão por 5.\n\nc) O número 9 deixa resto 4 na divisão por 5, pois $9=5 \\cdot 1+4$. O número 7 deixa resto 2 na divisão por 5, pois $7=5 \\cdot 1+2$.\nObserve que se um número deixa resto 1 na divisão por 5, o seu quadrado também deixa resto 1 na divisão por 5. De fato, seja $x$ um número que deixa resto 1 na divisão por 5. Daí, $x=5q+1$. Portanto,\n$$\n\\begin{aligned}\nx^{2} & =(5q+1)^{2} \\\\\n& =25q^{2}+2 \\cdot 5q \\cdot 1+1^{2} \\\\\n& =5\\left(5q^{2}+2q\\right)+1\n\\end{aligned}\n$$\no que mostra que $x^{2}$ também deixa resto 1 na divisão por 5.\nComeçamos com o número 9 na tela da calculadora. Se apertarmos a tecla $\\square$, o resultado deixará resto 1 na divisão por 5, pelo item anterior. Se apertarmos a tecla $\\triangle$, o resultado continuará deixando resto 4 na divisão por 5, pois subtrair 5 não muda o resto na divisão por 5. Se em algum momento o resto for 1, então continuará sendo 1, para sempre, pois nenhuma das duas operações $\\square$ ou $\\triangle$ alterará o resto na divisão por 5.\nAssim, começando com o 9, o resto na divisão por 5 será sempre 4 ou 1. Como 7 deixa resto 2 na divisão por 5, não é possível obtê-lo!",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26193,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOs pontos $M$, $N$ e $P$ são escolhidos sobre os lados $BC$, $CA$ e $AB$ do triângulo $ABC$ de modo que $BM = BP$ e $CM = CN$. A perpendicular baixada de $B$ à $MP$ e a perpendicular baixada de $C$ à $MN$ se intersectam em $I$. Prove que os ângulos $\\angle IPA$ e $\\angle INC$ são congruentes.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nComo o triângulo $MNC$ é isósceles, $CI$ além de altura é também a bissetriz relativa ao vértice $C$. Consequentemente, $\\angle ICN = \\angle ICM$. Pelo caso de congruência $LAL$, segue que $\\triangle ICM \\equiv \\triangle ICN$. Daí, $\\angle IMC = \\angle INC$. De forma semelhante, temos $\\angle BPI = \\angle BMI$. Assim,\n$$\n\\angle IPA = 180^\\circ - \\angle IPB = 180^\\circ - \\angle IMB = \\angle IMC = \\angle INC\n$$\n\n![](attached_image_2.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26194,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n(a) Se $r=\\sqrt{2}+\\sqrt{3}$, mostre que $\\sqrt{6}=\\frac{r^{2}-5}{2}$.\n\n(b) Se $s=\\sqrt{215}+\\sqrt{300}$, mostre que $s^{2}>1015$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n(a) Como\n$$\nr^{2}=(\\sqrt{2}+\\sqrt{3})^{2}=(\\sqrt{2})^{2}+2(\\sqrt{2})(\\sqrt{3})+(\\sqrt{3})^{2}=2+2 \\sqrt{6}+3=5+2 \\sqrt{6}\n$$\nportanto $r^{2}-5=2 \\sqrt{6} \\Longrightarrow \\sqrt{6}=\\frac{r^{2}-5}{2}$.\n\n(b) Pelo mesmo argumento temos que\n$$\n\\begin{aligned}\ns^{2} & =(\\sqrt{215}+\\sqrt{300})^{2}=215+2 \\sqrt{215 \\cdot 300}+300 \\\\\n& =515+10 \\sqrt{43 \\cdot 60}=515+10 \\sqrt{2580}> \\\\\n& >515+10 \\sqrt{2500}=515+500=1015\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26195,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUma loja estava vendendo um brinquedo por $R\\$ 13{,}00$ a unidade. Para conseguir vender todo o seu estoque que não era superior a 100 unidades, resolveu abaixar o preço de um número inteiro de reais. Com isso, conseguiu vender todo o estoque por $R\\$ 781{,}00$. Qual foi a redução do preço, por unidade?",
    "options": [],
    "answer": "2",
    "solution": "Solution:\n\nSe $x$ é o desconto em reais e $y$ é o número de peças, então\n$$\n(13-x) \\times y = 781 \\text{ e } y < 100\n$$\nAssim, $(13-x)$ e $y$ são divisores de $781$. Como $781 = 11 \\times 71$, a única solução é $y = 71$ e $13-x = 11$. Logo, a redução foi de $R\\$ 2,00$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26196,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDois tipos de vela têm o mesmo comprimento mas são feitas de material diferente; uma queima completamente em 3 horas e a outra em 4 horas, ambas queimam com velocidade uniforme. A que horas as velas devem ser acesas de modo que às 16 horas o comprimento de uma seja o dobro do da outra?\n\n(a) $1: 24$\n(b) $1: 28$\n(c) $1: 36$\n(d) $1: 40$\n(e) $1: 48$",
    "options": [],
    "answer": "c",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26197,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA seqüência $x y z$ - Na seqüência $\\frac{1}{2}, \\frac{5}{8}, \\frac{3}{4}, \\frac{7}{8}, x, y, z, \\ldots$ os valores de $x, y$ e $z$ são...",
    "options": [],
    "answer": "x = 1, y = 9/8, z = 5/4",
    "solution": "Solution:\n\nIgualando os denominadores, verificamos que a seqüência dada é a mesma que a seqüência\n$$\n\\frac{4}{8}, \\frac{5}{8}, \\frac{6}{8}, \\frac{7}{8}, x, y, z, \\ldots\n$$\nAssim, o denominador é $8$ e os numeradores são números consecutivos. Logo $x=\\frac{8}{8}=1$, $y=\\frac{9}{8}$ e $z=\\frac{10}{8}=\\frac{5}{4}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26198,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSete números inteiros positivos estão escritos em ordem crescente numa mesma linha. Coloque entre esses números cinco sinais de \"+\" e um só de \"$=$\" para obter uma igualdade.",
    "options": [],
    "answer": "9+10+11+12=13+14+15",
    "solution": "Solution:\n\nSolução 1 - Os 7 números podem ser escritos como\n\n$$\n\\underbrace{n-3, n-2, n-1}_{3 n-6}, n, \\underbrace{n+1, n+2, n+3}_{3 n+6}\n$$\n\nObservando que $3 n-6+12=3 n+6$, concluímos que $n=12$. Logo, os números são\n\n$$\n9+10+11+12=13+14+15\n$$\n\n\nSolução 2 - Seja $n+1, n+2, \\ldots, n+7$ os sete números consecutivos e suponhamos que\n\n$$\n(n+1)+\\cdots+(n+k)=(n+k+1)+\\cdots+(n+7)\n$$\n\nComo os números à esquerda são menores, então tem mais somandos à esquerda, assim $k \\geq 4$. Supondo $k=4$, a igualdade anterior é\n$$\n4 n+1+2+3+4=3 n+5+6+7\n$$\n\nLogo $n=8$. No caso $k=5$ temos que\n$$\n5 n+1+2+3+4+5=2 n+6+7\n$$\nque não gera solução inteira. De igual forma $k=6$ não gera solução inteira positiva. Portanto a única solução é\n\n$$\n9+10+11+12=13+14+15\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26199,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOs dois quadrados - As medidas em centímetros dos lados de cada um dos dois quadrados são números inteiros. Se o menor quadrado tivesse $2001~\\mathrm{cm}^2$ a mais de área, os dois quadrados seriam iguais. Quanto pode medir o lado do maior quadrado?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "1001, 335, 55, 49",
    "solution": "Solution:\n\nSe $a$ é a medida do lado do quadrado maior e $b$ a medida do lado do quadrado menor, então pelo enunciado temos\n$$\na^2 = b^2 + 2001\n$$\nLogo:\n$$\n2001 = a^2 - b^2 = (a+b)(a-b)\n$$\nComo $a$ e $b$ são números inteiros, temos que $a+b$ e $a-b$ são divisores de $2001$. Mas, $2001 = 3 \\times 23 \\times 29$, segue que\n$$\n(a+b)(a-b) = 2001 \\times 1 = 667 \\times 3 = 87 \\times 23 = 69 \\times 29\n$$\nConsequentemente, temos 4 possíveis formas de fatorar $2001$ em dois fatores:\n- se $a+b = 2001$ e $a-b = 1 \\Longrightarrow a = \\frac{(a+b)+(a-b)}{2} = 1001$;\n- se $a+b = 667$ e $a-b = 3 \\Longrightarrow a = \\frac{667+3}{2} = 335$;\n- se $a+b = 87$ e $a-b = 23 \\Longrightarrow a = \\frac{87+23}{2} = 55$;\n- se $a+b = 69$ e $a-b = 29 \\Longrightarrow a = \\frac{69+29}{2} = 49$.\n\nAssim as possibilidades para o lado maior são: $1001, 335, 55$ e $49$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26200,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOs doze pontos - Doze pontos estão marcados numa folha de papel quadriculada, conforme mostra a figura. Qual o número máximo de quadrados que podem ser formados unindo quatro desses pontos?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "11",
    "solution": "Solution:\n\nOs doze pontos - No total, temos 11 possíveis quadrados como mostrado a seguir.\n\n![](attached_image_2.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26201,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDistância na reta - Cinco pontos estão sobre uma mesma reta. Quando listamos as dez distâncias entre dois desses pontos, da menor para a maior, encontramos $2,4,5,7,8, k, 13,15,17,19$. Qual o valor de $k$ ?",
    "options": [],
    "answer": "12",
    "solution": "Solution:\n\nSolução 1: - Essa solução é um pouco difícil de escrever porque é feita na base de \"tentativa e erro\". Começamos desenhando uma reta numérica e colocando os pontos $0$ e $19$. Como a primeira distância é $2$, marcamos nossos primeiros três pontos:\n\n![](attached_image_1.png)\n\nComo temos que ter uma distância $7$, colocamos o ponto $7$ na reta. Isso nos dá distâncias que não são incompatíveis com o problema:\n\n![](attached_image_2.png)\n\nAs distâncias entre esses $4$ pontos são: $2, 7, 19, 5, 17$ e $12$. Finalmente, colocando o ponto $15$ na reta obtemos o seguinte:\n\n![](attached_image_3.png)\n\nCom esses pontos as distâncias são: $2,7,15,19,5,13,17,8,12,4$, que são compatíveis com os dados do problema. Logo, $k=12$.\n\nNote que temos também uma outra distribuição dos números, a saber:\n\n![](attached_image_4.png)\n\nNessa distribuição também obtemos $k=12$.\n\n\nSolução 2: Como a maior distância é $19$ podemos, supor que um ponto é o $0$ e outro é $19$.\nSe $a$ é um outro ponto, então na lista das distâncias temos os números: $a-0=a$ e $19-a$. De fato, na lista aparecem os pares $2$ e $17$, assim podemos supor que o número $2$ é outro ponto sobre a reta.\nDa mesma forma, como $4$ e $15$ estão na lista das distâncias, temos que $4$ ou $15$ é outro ponto na reta. Mas, $4$ não pode ser um dos pontos porque a distância $2$ não apareceu duas vezes. Logo, $15$ é outro ponto na reta.\nPor último o quinto ponto tem que estar a uma distância $5$ de um dos pontos e a $7$ de outro, logo o ponto que falta é o ponto $7$ e a distância desconhecida é $k=19-7=12$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26202,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEliane quer escolher o seu horário para a natação. Ela quer ir a duas aulas por semana, uma de manhã e a outra de tarde, não sendo no mesmo dia nem em dias seguidos. De manhã, há aulas de natação de segunda-feira a sábado, às $9 h$, às $10 h$ e às $11 h$ e de tarde, de segunda-feira a sexta-feira, às $17 h$ e às $18 h$. De quantas maneiras distintas pode Eliane escolher o seu horário?",
    "options": [],
    "answer": "96",
    "solution": "Solution:\n\nSe a aula da manhã é segunda ou sexta (em qualquer dos três horários), então o dia da aula de tarde pode ser escolhida de 3 formas diferentes (em qualquer dos dois horários), assim temos $2 \\times 3 \\times 3 \\times 2 = 36$ formas diferentes de escolher o horário.\n\nNo caso em que a aula de manhã seja sábado então o dia da aula da tarde pode ser qualquer dia de segunda a quinta, assim temos $3 \\times 4 \\times 2 = 24$ possíveis formas.\n\nPor último, se a aula da manhã é terça, quarta ou quinta, então a aula da tarde só pode ser escolhida de duas formas, assim temos $3 \\times 3 \\times 2 \\times 2 = 36$ formas.\n\nLogo a Eliana pode escolher seu horário de $36 + 24 + 36 = 96$ formas distintas.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26203,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoma das raízes de uma equação - Determine a soma das raízes distintas da equação $x^{2} + 3x + 2 = |x + 1|$.",
    "options": [],
    "answer": "-4",
    "solution": "Solution:\n\nTemos que considerar dois casos.\n\nCaso 1: $x \\geq -1$.\n\nNesse caso, $x^{2} + 3x + 2 = x + 1$, e $x^{2} + 2x + 1 = 0$ que só possui a solução $x = -1$.\n\nCaso 2: $x < -1$.\n\nNesse caso, $x^{2} + 3x + 2 = -x - 1$, logo $x^{2} + 4x + 3 = 0$ que tem, no intervalo, apenas a solução $x = -3$.\n\nAssim as únicas soluções distintas da equação são $-1$ e $-3$, cuja soma é $-4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26204,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO maior resultado - Qual o maior resultado que podemos encontrar quando dividimos um número de 2 algarismos pela soma de seus algarismos?",
    "options": [],
    "answer": "10",
    "solution": "Solution:\n\nO maior resultado - Estamos procurando o maior valor de $\\frac{10a+b}{a+b}$, onde $a$ e $b$ representam algarismos, pelo menos um diferente de $0$. Temos\n$$\n\\frac{10a+b}{a+b} = \\frac{10a+10b-9b}{a+b} = \\frac{10a+10b}{a+b} - \\frac{9b}{a+b} = 10 - \\frac{9b}{a+b} \\leq 10\n$$\nLogo, se conseguirmos encontrar $a$ e $b$ tais que $\\frac{10a+b}{a+b}=10$, teremos o maior resultado. Note que isso ocorre quando $b=0$, ou seja:\n$$\n\\frac{10}{1} = \\frac{20}{2} = \\frac{30}{3} = \\frac{40}{4} = \\frac{50}{5} = \\frac{60}{6} = \\frac{70}{7} = \\frac{80}{8} = \\frac{90}{9} = 10.\n$$\nLogo, a resposta é $10$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26205,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSinais e radicais - Quais dos números dados são negativos?\n\na. $10-3 \\sqrt{11}$\n\nb. $3 \\sqrt{11}-10$\n\nc. $18-5 \\sqrt{13}$\n\nd. $51-10 \\sqrt{26}$\n\ne. $10 \\sqrt{26}-51$",
    "options": [],
    "answer": "b, c, e",
    "solution": "Solution:\n\nTemos $3 \\sqrt{11}=\\sqrt{9 \\times 11}=\\sqrt{99}$. Como $100>99$, obtemos $10=\\sqrt{100}>\\sqrt{99}=3 \\sqrt{11}$, portanto, $10-3 \\sqrt{11}>0$ e $3 \\sqrt{11}-10<0$.\n\nAnalogamente, temos $10 \\sqrt{26}=\\sqrt{100 \\times 26}=\\sqrt{2600}$. Como $2601>2600$, obtemos\n$$\n51=\\sqrt{2601}>\\sqrt{2600}=10 \\sqrt{26}\n$$\nportanto, $51-10 \\sqrt{26}>0$ e $10 \\sqrt{26}-51<0$.\n\nFinalmente, $18^{2}=324<325=25 \\times 13$ garante que $18<5 \\sqrt{13}$, de modo que $18-5 \\sqrt{13}<0$.\n\nOs números negativos são (b) $3 \\sqrt{11}-10$, (c) $10 \\sqrt{26}-51$ e (e) $18-5 \\sqrt{13}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26206,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCinco bolas iguais estão se movendo na mesma direção ao longo de uma reta fixa, mantendo uma certa distância de uma para outra. Na mesma direção, mas no sentido oposto, outras cinco bolas se movem de encontro às primeiras. As velocidades de todas as bolas são iguais. Quando duas bolas colidem, voltam na mesma velocidade de antes, ao longo da mesma direção. Quantas colisões entre bolas vão ocorrer?",
    "options": [],
    "answer": "25",
    "solution": "Solution:\n\nUma solução clara para o problema seria fazer todo o percurso das bolas, mas adotaremos outra estratégia.\nImagine que quando há a colisão de duas bolas, ao invés de gerar a volta das mesmas, uma bola se transforma na outra, como se não houvesse a colisão. Chamaríamos a esse processo de transmutação.\nÉ claro que cada colisão do problema inicial corresponde a uma transmutação na nossa interpretação.\nMas o número de transmutações é bem mais fácil de calcular, porque as bolas não mudam de direção. As cinco bolas à esquerda encontrarão as cinco bolas à direita e o número procurado será então $5 \\times 5 = 25$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26207,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA estrada que liga dois vilarejos em uma montanha é formada somente por trechos de subida ou descida. Um ônibus sempre viaja a $15~\\mathrm{km}/\\mathrm{h}$ em trechos de subida e a $30~\\mathrm{km}/\\mathrm{h}$ em trechos de descida. Encontre a distância entre os vilarejos se o ônibus leva exatamente 4 horas para fazer a viagem completa de ida e volta.",
    "options": [],
    "answer": "40 km",
    "solution": "Solution:\n\nObserve que os trechos de subida no percurso de ida são exatamente os trechos de descida para a volta e vice-versa. Assim, em uma viagem de ida e volta a distância percorrida nas subidas é igual à distância percorrida nas descidas.\n\nChamemos de $d$ a distância entre os dois vilarejos. Como a distância total percorrida foi igual a $2d$, então o tempo gasto subindo foi $d/15$ horas e o tempo gasto descendo foi $d/30$ horas. Como o tempo total foi 4 horas, temos\n\n$$\n\\frac{d}{15} + \\frac{d}{30} = 4\n$$\n\nResolvendo a equação, encontramos $d = 40$, ou seja, a distância entre os vilarejos é igual a $40~\\mathrm{km}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26208,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOrdenando frações - Qual dos números fica entre $2/5$ e $3/4$?\n(a) $1/6$\n(b) $4/3$\n(c) $5/2$\n(d) $4/7$\n(e) $1/4$",
    "options": [],
    "answer": "d",
    "solution": "Solution:\n\nA opção correta é (d).\n\nLembre que a ordem entre frações constituídas de inteiros positivos é determinada pelo produto cruzado dos inteiros, ou seja, $\\frac{a}{b} < \\frac{c}{d}$ equivale à afirmação $a \\times d < b \\times c$. Desse modo, temos\n$$\n\\frac{1}{6} < \\frac{1}{4} < \\frac{2}{5} < \\frac{4}{7} < \\frac{3}{4} < 1 < \\frac{4}{3} < \\frac{5}{2}\n$$\nUm número $x$ que “fica entre” $2/5$ e $3/4$ é um número maior do que $2/5$ e menor do que $3/4$, ou seja,\n$$\n\\frac{2}{5} < x < \\frac{3}{4}\n$$\njá que $4 < 6$, $5 < 8$, $14 < 20$, $16 < 21$, $3 < 4$ (duas vezes) e $8 < 15$, respectivamente. Assim, $1/6$ e $1/4$ ficam à esquerda de $2/5$, $4/3$ e $5/2$ ficam à direita de $3/4$ e só $4/7$ fica entre $2/5$ e $3/4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26209,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nConsidere um grupo de 15 pessoas. É possível que cada uma delas conheça exatamente:\n(a) 4 pessoas do grupo?\n(b) 3 pessoas do grupo?\n(Admita que se $A$ conhece $B$ então $B$ conhece $A$.)",
    "options": [],
    "answer": "a: yes; b: no",
    "solution": "Solution:\n(a) É possível. Representamos as 15 pessoas por pontos, conforme o diagrama ao lado. Um arco entre dois pontos significa que as duas pessoas representadas se conhecem. Como cada ponto está ligado a dois pontos à esquerda e a dois pontos à direita, saem quatro arcos de cada ponto, o que significa que é possível que cada pessoa conheça exatamente 4 pessoas do grupo.\n\n(b) Não é possível! Vamos representar as pessoas por pontos. Ligamos dois pontos se as pessoas representadas se conhecem. Quantos arcos vamos precisar traçar para representar todas as amizades? Cada ponto é extremidade de 3 arcos, resultando num total de $15 \\times 3 = 45$ arcos que saem de todos os pontos. Porém, nesta contagem, cada arco foi contado duas vezes, nas duas extremidades. Portanto, o número de segmentos deve ser $45/2$, o que é um absurdo, pois este número não é inteiro.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26210,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA distância entre Coco da Selva e Quixajuba é $24~\\mathrm{km}$. Dois amigos precisam ir de Quixajuba a Coco da Selva e um terceiro amigo precisa ir de Coco da Selva a Quixajuba. Eles possuem uma bicicleta que inicialmente está em Quixajuba. Cada um deles pode ir caminhando a velocidade de $6~\\mathrm{km}/\\mathrm{h}$, ou de bicicleta a velocidade de $18~\\mathrm{km}/\\mathrm{h}$. Além disso, podem deixar a bicicleta em qualquer ponto do trajeto.\n\n![](attached_image_1.png)\n\nMostre como eles podem proceder para chegarem a seus destinos em no máximo $2~\\mathrm{h}~40~\\mathrm{min}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nChamaremos de $A$ e $B$ os amigos que estão em Quixajuba e $C$ o que está em Coco da Selva. Nossos personagens podem seguir a seguinte estratégia:\n- Na primeira hora, $A$ vai de bicicleta enquanto $B$ e $C$ irão caminhando. Depois dessa hora, $A$ e $C$ se encontram no quilômetro $18$ (medido desde Quixajuba) e $B$ está no quilômetro $6$.\n\n- $A$ continua caminhando e chegará a seu destino depois de uma hora. Enquanto isso, $C$ continua de bicicleta e $B$ fica parado esperando $C$ chegar. Como a distância entre $C$ e $B$ é de $12~\\mathrm{km}$, isso acontecerá depois de $12/18 = 2/3~\\mathrm{h}$, isto é, $40$ minutos.\n\n- Nesse ponto, $C$ passa a bicicleta para $B$ e cada um continua seu trajeto chegando a seus destinos em uma hora.\n\nAssim o tempo total empregado por $B$ e $C$ foi de $2~\\mathrm{h}~40~\\mathrm{min}$, enquanto $A$ gastou $2~\\mathrm{h}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26211,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nEm uma circunferência foram marcados $15$ pontos brancos e $1$ ponto preto. Consideremos todos os possíveis polígonos (convexos) com seus vértices nestes pontos.\nVamos separá-los em dois tipos:\n- Tipo 1: os que possuem somente vértices brancos.\n- Tipo 2: os que possuem o ponto preto como um dos vértices.\nExistem mais polígonos do tipo 1 ou do tipo 2? Quantos existem a mais?",
    "options": [],
    "answer": "Type 2 has 105 more polygons.",
    "solution": "Solution:\nObserve que para cada polígono do tipo $1$ podemos construir um polígono do tipo $2$ adicionando o ponto preto.\nPor outro lado, se temos um polígono do tipo $2$ e retirarmos o ponto preto, a única forma de não gerar um polígono é se sobrarem exatamente dois pontos brancos.\nPortanto, existem mais polígonos do tipo $2$ do que do tipo $1$.\nPara calcular a diferença, basta contar o número de pares de pontos brancos. Para isso, observe que cada ponto branco pode formar um par com cada um dos outros $14$ pontos brancos. Assim, como existem $15$ pontos brancos, teremos $15 \\times 14$ pares ordenados. Segue que temos $15 \\times 14 / 2 = 105$ pares de pontos.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26212,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA partir de hoje, o grande apostador Carlo Pietro decidiu frequentar cassinos diariamente. No primeiro dia, ele apostará em um jogo cuja probabilidade de ganhar é igual a $\\frac{1}{2}$. Nos segundo, terceiro e quarto dias, ele apostará em jogos diferentes cujas probabilidades de vitória são, respectivamente, iguais a $\\frac{1}{3}, \\frac{1}{4}, \\frac{1}{5}$ e assim por diante nos dias que se seguirem.\na) Qual é a probabilidade de que ele não tenha ganhado até o terceiro dia?\nb) Qual é a probabilidade de que ele não tenha ganhado até o quinto dia?\nc) Qual é a probabilidade de que ele não tenha ganhado até o $2013^{\\underline{0}}$ dia?",
    "options": [],
    "answer": "a) 1/4; b) 1/6; c) 1/2014",
    "solution": "Solution:\na) Para que Pietro não tenha ganho até o terceiro dia, é necessário que ele tenha perdido no primeiro, no segundo e no terceiro dia. A probabilidade de que Pietro ganhe no primeiro dia é $\\frac{1}{2}$. Logo, a probabilidade de que Pietro tenha perdido no primeiro dia é\n$$\n1-\\frac{1}{2}=\\frac{1}{2}\n$$\nA probabilidade de que Pietro ganhe no segundo dia é $\\frac{1}{3}$. Logo, a probabilidade de que Pietro tenha perdido no segundo dia é\n$$\n1-\\frac{1}{3}=\\frac{2}{3}\n$$\nA probabilidade de que Pietro ganhe no terceiro dia é $\\frac{1}{4}$. Logo, a probabilidade de que Pietro tenha perdido no terceiro dia é\n$$\n1-\\frac{1}{4}=\\frac{3}{4}\n$$\nPortanto, a probabilidade de que Pietro tenha perdido no primeiro, segundo e terceiro dia é igual ao produto dessas probabilidades, que nos dá\n$$\n\\frac{1}{2} \\cdot \\frac{2}{3} \\cdot \\frac{3}{4}=\\frac{1}{4}\n$$\n\nb) Repetimos o mesmo argumento de antes! Agora até o quinto dia, o que nos dá como probabilidade de Pietro não haver ganhado:\n$$\n\\frac{1}{2} \\cdot \\frac{2}{3} \\cdot \\frac{3}{4} \\cdot \\frac{4}{5} \\cdot \\frac{5}{6}=\\frac{1}{6}\n$$\n\nc) Pensando indutivamente, a probabilidade de que Pietro não tenha ganho até o $2013^{\\circ}$ dia é igual a\n$$\n\\frac{1}{2} \\cdot \\frac{2}{3} \\cdot \\frac{3}{4} \\cdot \\frac{4}{5} \\cdots \\frac{2011}{2012} \\cdot \\frac{2012}{2013} \\cdot \\frac{2013}{2014}=\\frac{1}{2014}\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26213,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUma linha de trem está dividida em 10 trechos pelas estações $A, B, C, D, E, F, G, H, I, J$ e $K$. A distância de $A$ até $K$ é igual a $56~\\mathrm{km}$. O trajeto de dois trechos consecutivos é sempre menor ou igual a $12~\\mathrm{km}$ e o trajeto de três trechos consecutivos sempre é maior ou igual a $17~\\mathrm{km}$. Determine as distâncias:\n\na) de $J$ até $K$;\nb) de $D$ até $H$;\nc) de $B$ a $G$.",
    "options": [],
    "answer": "a) 5 km; b) 22 km; c) 29 km",
    "solution": "Solution:\n\na) Podemos escrever $AK = 56$ e $AK = AD + DG + GJ + JK$. Como $AD, DG$ e $GJ \\geq 17$, então $JK \\leq 5$. Daí, para $HK \\geq 17$, devemos ter $HJ \\geq 12$. Mas, sabemos que $HJ \\leq 12$, assim $HJ = 12$. A partir de $HK \\geq 17$ e $HJ = 12$, concluímos $JK \\geq 5$ e a única possibilidade é $JK = 5$.\n\nb) Por simetria, teremos $AB = 5$ e $BD = 12$. Agora, seguimos com\n$$\n\\begin{aligned}\nDH & = AK - AB - BD - HJ - JK \\\\\n   & = 56 - 5 - 12 - 5 - 12 = 22\n\\end{aligned}\n$$\n\nc) Por fim, $GJ \\geq 17$, mas $HJ = 12$. Assim $GH \\geq 5$. A partir de $DG \\geq 17$ e $DH = DG + GH = 22$, obtemos $DG = 17$ e $GH = 5$. Portanto\n$$\n\\begin{aligned}\nBG & = BD + DG \\\\\n   & = 12 + 17 \\\\\n   & = 29\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26214,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nMostre que existe um múltiplo de 2017 que termina em 2016.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nConsidere a seguinte lista contendo 2018 números:\n$$\n\\begin{aligned}\na_{1} & =2016 \\\\\na_{2} & =20162016 \\\\\na_{3} & =201620162016 \\\\\\vdots & \\vdots \\\\\na_{2018} & =\\underbrace{20162016 \\ldots 20162016}_{2018 \\text{ vezes}} .\n\\end{aligned}\n$$\nComo existem apenas 2017 restos possíveis na divisão por 2017, dois dos números da lista anterior devem possuir o mesmo resto por $2017$ e, consequentemente, a diferença entre eles deve ser um múltiplo de 2017. Digamos que eles sejam $a_{i}$ e $a_{j}$ :\n$$\n\\begin{aligned}\na_{i}-a_{j} & =\\underbrace{20162016 \\ldots 20162016}_{i-j \\text{ vezes } 2016}\\underbrace{00 \\ldots 0}_{4 j \\text{ vezes}} \\\\\n& =\\underbrace{20162016 \\ldots 20162016}_{i-j \\text{ vezes } 2016} \\cdot 10^{4 j} .\n\\end{aligned}\n$$\nComo $\\operatorname{MDC}\\left(2017,10^{4 j}\\right)=1$, segue que $\\underbrace{20162016 \\ldots 20162016}_{i-j \\text{ vezes } 2016}$ deve ser divisível por 2017.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26215,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUma mesa circular tem 60 cadeiras em sua volta. Existem $N$ pessoas sentadas nessas cadeiras de tal modo que a próxima pessoa a se sentar vai ter que se sentar ao lado de alguém. Qual é o menor valor possível para $N$?",
    "options": [],
    "answer": "20",
    "solution": "Solution:\n\nSe a próxima pessoa a se sentar vai ter que se sentar ao lado de uma cadeira ocupada, isso significa que existem no máximo 2 cadeiras desocupadas consecutivas. Veja na figura: as cadeiras ocupadas estão representadas por quadradinhos brancos e as desocupadas por quadradinhos pretos. Podemos então pensar nas cadeiras\n\n![](attached_image_1.png)\n\nem grupos de 3 e a terceira está ocupada. Logo, o menor valor de $N$ é $60 \\div 3 = 20$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26216,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nJúlio faz multiplicações usando apenas os quadrados dos números. Ele tem que calcular o produto $85 \\times 135$. Para isso, ele desenha um retângulo de $85 \\mathrm{~mm}$ por $135 \\mathrm{~mm}$ e traça nesse retângulo o maior quadrado possível; faz o mesmo no quadrado restante e assim sucessivamente. Dessa maneira ele obtém oito quadrados. Desenhe a figura feita por Júlio e escreva $85 \\times 135$ como a soma de oito quadrados: $85 \\times 135=85^{2}+\\ldots$",
    "options": [],
    "answer": "85^2 + 50^2 + 35^2 + 15^2 + 15^2 + 5^2 + 5^2 + 5^2",
    "solution": "Solution:\n\nO maior quadrado no retângulo de $85 \\times 135$ é aquele de $85 \\times 85$. Sobra então um retângulo de $50 \\times 85$, onde o maior quadrado é de $50 \\times 50$. Continuando assim, obtemos:\n$$\n85 \\times 135=85^{2}+50^{2}+35^{2}+15^{2}+15^{2}+5^{2}+5^{2}+5^{2}\n$$\n\n|   |   |   | $5^{2}$ $5^{2}$ $5^{2}$ |   |\n| :--- | :--- | :--- | :--- | :--- |\n|   | $35^{2}$ | $15^{2}$ |   |   |\n|   |   |   | $15^{2}$ |   |\n|   |   |   |   |   |\n|   |   | $50^{2}$ |   |   |\n|   |   |   |   |   |\n|   |   |   |   |   |",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26217,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCara ou Coroa - Jerônimo joga no tabuleiro ao lado da seguinte maneira: Ele coloca uma peça na casa \"PARTIDA\" e ele move a peça da seguinte maneira: ele lança uma moeda, se der CARA ele avança duas casas, e se der COROA ele recua uma casa. Jerônimo lançou a moeda 20 vezes e conseguiu chegar na casa CHEGADA. Quantas vezes a moeda deu CARA?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "12",
    "solution": "Solution:\n\n12",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26218,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUma estratégia com um número muito grande - Carlos escreveu em seguida todos os números de 1 a 60 :\n\n$$\n1234567891011121314 \\cdots 57585960\n$$\n\nDepois ele riscou 100 algarismos de modo que o número formado com os algarismos que não foram riscados fôsse o maior possível, sem mudar a ordem inicial de como os algarismos foram escritos. Qual é esse número?",
    "options": [],
    "answer": "9999785960",
    "solution": "Solution:\n\n9999785960 .",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26219,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNúmero ímpar - Se $n$ é um número inteiro qualquer, qual das seguintes opções é um número ímpar?\n(a) $n^{2}-n+2$\n(b) $n^{2}+n+2$\n(c) $n^{2}+n+5$\n(d) $n^{2}+5$\n(e) $n^{3}+5$",
    "options": [],
    "answer": "c",
    "solution": "Solution:\n\nLembremos que:\n- $n$ e $n^{2}$ têm a mesma paridade: $(\\text{par})^{2}=$ par e (ímpar $)^{2}=$ ímpar;\n- a soma ou diferença de números de mesma paridade é um número par: (par $\\pm$ par = par e ímpar $\\pm$ ímpar = par).\n\nSolução 1: Observemos que $n^{2}+n$ e $n^{2}-n$ são soma e diferença de dois números que sempre têm a mesma paridade, logo estes números sempre serão pares. Portanto $n^{2}+n+5=\\left(n^{2}+n\\right)+5$ é soma de um par e um ímpar, logo sempre será ímpar para todo valor inteiro de $n$. A opção correta é (c).\n\n\nSolução 2: Observemos que $n^{2}+n=n(n+1)$ e $n^{2}-n=n(n-1)$ são produtos de dois números consecutivos, logo estes números são pares. Portanto, $n^{2}+n+5=\\left(n^{2}+n\\right)+5$ é a soma de um par com um ímpar, assim este número é ímpar para todo valor inteiro de $n$.\nObservemos que (a) e (b) são sempre números pares para qualquer valor de $n$, enquanto que (d) e (e) podem ser pares ou ímpares, dependendo se $n$ é ímpar ou par.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26220,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm terreno compartilhado - Três amigas compraram um terreno quadrado e querem reparti-lo como indicado na figura, por que em $A$ se encontra uma fonte de água. Elas querem também que as áreas das três partes sejam iguais. Onde devem estar os pontos $M$ (sobre $BC$) e $N$ (sobre $CD$)?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "BM = DN = 2/3 of the side length (equivalently, MC = CN = 1/3 of the side length).",
    "solution": "Solution:\n\nComo as áreas de $\\triangle ABM$ e $\\triangle ADN$ são iguais e $AB = AD$, temos então\n$$\n\\frac{BM \\times AB}{2} = \\frac{ND \\times AD}{2} \\Longrightarrow BM = DN\n$$\n\n![](attached_image_2.png)\n\nAssim, a figura $AMCN$ é simétrica com respeito à diagonal $AC$. Portanto, a área do $\\triangle ACN$ é a metade da área do $\\triangle ADN$. Agora, como esses triângulos têm a mesma altura, então $DN = 2NC$ e, pela simetria, temos que $BM = 2MC$. Concluímos que $BM$ é $2/3$ do lado do quadrado, o mesmo ocorrendo com $DN$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26221,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm subconjunto - O conjunto $\\{1,2,3, \\ldots, 3000\\}$ contém um subconjunto de 2000 elementos tal que nenhum elemento é o dobro do outro?",
    "options": [],
    "answer": "No",
    "solution": "Solution:\n\nVamos construir o subconjunto pedido da seguinte forma:\n- ele contém todos os números ímpares: $1,3,5, \\ldots, 2999$. Aqui já temos uma lista com 1500 números.\n- o conjunto não pode conter os números que são da forma $2 \\times$ (número ímpar),\n- o conjunto pode conter os números que são da forma $4 \\times$ (número ímpar), isto é,\n$$\n\\underbrace{4 \\times 1}_{4}, \\underbrace{4 \\times 3}_{12}, \\underbrace{4 \\times 5}_{20}, \\ldots, \\underbrace{4 \\times 749}_{2996}\n$$\nessa lista tem 749 números e nenhum é o dôbro do outro. Além disso, nenhum deles é o dôbro de um número ímpar.\nLogo, o nosso conjunto já possui $1500+749=2249$ elemento; assim qualquer subconjunto dele com 2000 elementos satisfaz as condições pedidas.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26222,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nProduto de três números - No diagrama abaixo cada círculo representa um algarismo. Preencha o diagrama colocando em cada círculo um dos algarismos de 0 a 9, utilizando cada algarismo uma única vez.\n\n$$\n\\text{○} \\times \\text{○○} \\times \\text{○○○} = \\text{○○○○}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSejam $a, b, c, \\ldots$ os números em cada círculo como indicado abaixo.\n\n$$(a) \\times (b)(c) \\times (d)(e)(f) = (g)(h)(i)(j)$$\n\nTemos que $a$, $c$ e $f$ não podem ser zero, pois $0 \\times x = 0$.\n\nMas, o produto dos três números é um número de 4 algarismos, assim, $a b d < 10$ e portanto os números que aparecem em dito produto são $1,2,3$ ou $1,2,4$. Observemos que a segunda é impossível porque o mínimo produto que podemos obter neste caso é\n\n$$\n1 \\times 23 \\times 456 = 10488\n$$\n\nassim $a b d = 6$ e o produto é maior do que 6000. Por outra parte $a$ não pode ser 2 ou 3 porque nesse caso o mínimo valor que tem o produto é\n\n$$\n2 \\times 14 \\times 356 = 9968\n$$\n\ne os outros produtos ficam maiores do que 10000. Portanto $a = 1$.\n\nContinuando essa análise, obtemos a solução:\n\n$$\n(1) \\times (2) \\times (3)(4) = 8(9)(7)(0)\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26223,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO peso de um número é a soma de seus algarismos. Qual é o menor número que pesa 2000?",
    "options": [],
    "answer": "3*10^222 - 1",
    "solution": "Solution:\n\nObserve que os números $189$, $8307$ e $99$ têm todos peso $18$, e que $99$ é o menor número que pesa $18$. Note que: para aumentar o peso de um número e minimizar o número é preciso que o número seja composto do maior número possível de algarismos $9$. Por outro lado, podemos dizer que o $0$ está eliminado dos algarismos a ser considerados porque ele aumenta o número sem aumentar o peso.\n\nTemos que $2000 = 9 \\times 222 + 2$. Logo, o número procurado tem então $222$ algarismos $9$, e um algarismo $2$ ou dois algarismos $1$. Eliminamos o caso dos números com dois algarismos $1$ porque eles têm $224$ algarismos, e logo são maiores do que os números que possuem o algarismo $2$ e têm $223$ algarismos. Finalmente, o número procurado tem $222$ algarismos $9$ e um $2$. Logo esse número é $299\\ldots 999$, com $222$ algarismos $9$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26224,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO colar - Um colar é composto de pérolas grandes e pérolas pequenas, num total de menos do que 500 pérolas.\n\ni. Se substituirmos $70\\%$ das pérolas grandes por pequenas, o peso do colar diminui de $60\\%$.\n\nii. Se substituirmos $60\\%$ das pérolas pequenas por grandes, o peso do colar aumenta de $70\\%$.\n\nQuantas pérolas tem o colar?",
    "options": [],
    "answer": "85, 170, 255, 340, 425",
    "solution": "Solution:\n\nSejam $n$ o número de pérolas grandes, $p$ o número de pérolas pequenas, $a$ o peso de uma pérola grande e $b$ o de uma pérola pequena. Com essa notação temos:\n\n- número total de pérolas no colar $= p + n$. Logo: $n + p < 500$\n- peso das pérolas grandes $= n \\times a$\n- peso das pérolas pequenas $= p \\times b$\n- peso total do colar $= p b + n a$\n\nPara equacionar o problema, vamos equacionar antes as duas hipóteses:\n\ni. Ao substituirmos $70\\%$ das pérolas grandes pelas pequenas, o colar fica composto como\n$$\n\\underbrace{30\\% \\times n}_{\\text{grandes}} + \\underbrace{p + 70\\% \\times n}_{\\text{pequenas}} = 0,3 n + (p + 0,7 n)\n$$\ne seu peso fica sendo\n$$\n\\underbrace{0,3 n \\times a}_{\\text{peso das grandes}} + \\underbrace{(p + 0,7 n) \\times b}_{\\text{peso das pequenas}} = \\underbrace{0,4(n a + p b)}_{40\\% \\text{ do peso inicial}}\n$$\n\nii. Analogamente, temos que ao substituirmos $60\\%$ das pérolas pequenas pelas grandes, o colar fica composto como\n$$\n\\underbrace{n + 60\\% \\times p}_{\\text{grandes}} + \\underbrace{40\\% \\times p}_{\\text{pequenas}} = (n + 0,6 p) + 0,4 p\n$$\ne seu peso fica sendo\n$$\n\\underbrace{(n + 0,6 p) \\times a}_{\\text{peso das grandes}} + \\underbrace{0,4 p \\times b}_{\\text{peso das pequenas}} = \\underbrace{1,7(n a + p b)}_{170\\% \\text{ do peso inicial}}\n$$\n\nTemos, então, o sistema:\n$$\n\\left\\{\\begin{array}{l}\n0,3 n a + 0,7 n b + p b = 0,4(n a + p b) \\\\\nn a + 0,6 p a + 0,4 p b = 1,7(n a + p b)\n\\end{array}\\right.\n$$\n\nPara resolvê-lo, começamos eliminando as incógnitas $a$ e $b$, escrevendo o sistema na seguinte forma:\n$$\n\\left\\{\\begin{aligned}\n\\frac{7 n + 6 p}{n} & = \\frac{a}{b} \\\\\n\\frac{-13 p}{7 n - 6 p} & = \\frac{a}{b}\n\\end{aligned}\\right.\n$$\nSegue que\n$$\n\\frac{7 n + 6 p}{n} = \\frac{-13 p}{7 n - 6 p} \\Longrightarrow 36 p^2 - 13 p n - 49 n^2 = 0\n$$\nPara fatorar essa expressão, escrevemos\n$$\n-13 p n = 36 p n - 49 p n\n$$\ne temos:\n$$\n\\begin{aligned}\n36 p^2 - 13 p n - 49 n^2 & = 36 p^2 + 36 p n - 49 p n - 49 n^2 \\\\\n& = p(36 p - 49 n) + n(36 p - 49 n) \\\\\n& = (36 p - 49 n)(p + n)\n\\end{aligned}\n$$\nFinalmente:\n$$\n(36 p - 49 n)(p + n) = 0\n$$\nObtemos $36 p = 49 n$, e como $p$ e $n$ são inteiros positivos, segue que $n$ é múltiplo de $36$ e $p$ de $49$. Assim temos: $n = 36 k$ e $p = 49 k'$, onde $k$ e $k'$ são inteiros maiores do que $1$. Logo,\n$$\n36 \\times 49 k' = 49 \\times 36 k \\Longrightarrow k = k'\n$$\nPortanto, $n = 36 k$ e $p = 49 k$. Deduzimos que $n + p = 85 k$. Como $n + p < 500$, o colar só pode ter: $85, 170, 255, 340$ ou $425$ pérolas.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26225,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDividir um cubo - Se dividirmos um cubo de $1~\\mathrm{m}$ de aresta em cubinhos de $1~\\mathrm{mm}$ de aresta, que altura terá uma coluna formada por todos os cubinhos, dispostos sucessivamente um em cima do outro?\n\n(a) $1~\\mathrm{m}$\n(b) $1~\\mathrm{km}$\n(c) $10~\\mathrm{km}$\n(d) $100~\\mathrm{km}$\n(e) $1000~\\mathrm{km}$",
    "options": [],
    "answer": "e",
    "solution": "Solution:\n\nConvertendo metros em milímetros temos: $1~\\mathrm{m} = 1000~\\mathrm{mm}$. Assim, o cubo ficou dividido em $1000 \\times 1000 = 10^{6}$ cubinhos de lado $1~\\mathrm{mm}$ cada um. Colocando-se lado a lado os $10^{6}$ cubinhos, teremos uma coluna de comprimento\n$$\n1000 \\times 1000 = 10^{6}~\\mathrm{mm} = 10^{6} \\times 10^{-3}~\\mathrm{m} = 10^{3}~\\mathrm{m} = 1~\\mathrm{km}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26226,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUma distração - Em vez de multiplicar certo número por $6$, Julia se distraiu e dividiu o número por $6$. O erro cometido por Julia foi de aproximadamente\n\n(a) $100\\%$\n(b) $97\\%$\n(c) $83\\%$\n(d) $17\\%$\n(e) $3\\%$",
    "options": [],
    "answer": "b",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26227,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n(a) Adriano escolheu secretamente cinco números $a$, $b$, $c$, $d$ e $e$ e informou a Bruna os dez números $24, 28, 30, 30, 32, 34, 36, 36, 40$ e $42$ obtidos pelo cálculo de todas as somas de dois números dentre os cinco escolhidos.\nO objetivo de Bruna é descobrir $a$, $b$, $c$, $d$, $e$. Bruna pode alcançar seu objetivo?\n\n(b) Adriano escolheu secretamente quatro números $m$, $n$, $p$ e $q$ e informou a Carlos os seis números $10, 20, 22, 24, 26$ e $36$ obtidos pelo cálculo de todas as somas de dois números dentre os quatro escolhidos.\nO objetivo de Carlos é descobrir $m$, $n$, $p$ e $q$. Ele pode alcançar seu objetivo?",
    "options": [],
    "answer": "(a) Yes: a=11, b=13, c=17, d=19, e=23. (b) No; for instance, {3,7,17,19} and {4,6,16,20} produce the same pairwise sums.",
    "solution": "Solution:\n(a) Suponha que $a \\leqslant b \\leqslant c \\leqslant d \\leqslant e$. Logo a menor soma é $a+b$ e a maior soma é $d+e$. A segunda menor é $a+c$ e a segunda maior é $c+e$. Assim, temos o sistema\n$$\n\\left\\{\\begin{array}{l}\na+b=24 \\\\\na+c=28 \\\\\nc+e=40 \\\\\nd+e=42\n\\end{array}\\right.\n$$\nPor outro lado, cada número é utilizado em quatro somas e então\n$$\n\\begin{gathered}\na+b+c+d+e= \\\\\n\\frac{24+28+30+30+32+34+36+36+40+42}{4}=83\n\\end{gathered}\n$$\nAssim,\n$c=(a+b+c+d+e)-(a+b)-(d+e)=83-24-42=17$.\nLogo,\n$$\n\\begin{aligned}\n& a=28-c=11 \\\\\n& b=24-a=13 \\\\\n& e=40-c=23 \\\\\n& d=42-e=19\n\\end{aligned}\n$$\n\n(b) Observe que os números $3, 7, 17$ e $19$ geram as somas $10, 20, 22, 24, 26$ e $36$ e o mesmo acontece com os números $4, 6, 16$ e $20$. Carlos não alcançará seu objetivo!",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26228,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nUma das diagonais de um quadrilátero inscritível é um diâmetro de seu círculo circunscrito.\n![](attached_image_1.png)\nVerifique que as interseções $E$ e $F$ das retas perpendiculares por $A$ e $C$, respectivamente, à reta $BD$ satisfazem\n$$\nDE = BF\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nEstenda as perpendiculares $CF$ e $AE$ para a segunda interseção com o círculo nos pontos $H$ e $G$, respectivamente. Como $AG \\parallel CH$ e $AC$ é um diâmetro, segue que $AGCH$ é um retângulo. Seja $M$ o ponto médio de $EF$. Por ser uma corda da circunferência, temos $OM$ perpendicular a $BD$. Usando que $EF$ é perpendicular aos lados do retângulo, temos $EM = FM$ e daí\n$$\nDE = DM - ME = BM - MF = BF\n$$\n![](attached_image_2.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26229,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTodos os vértices do pentágono $A B C D E$ estão sobre um mesmo círculo. Se $\\angle D A C=50^{\\circ}$, determine $\\angle A B C+\\angle A E D$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "230°",
    "solution": "Solution:\n\nComo ângulos inscritos associados a um mesmo arco são iguais, temos $\\angle D A C=\\angle D B C$. Além disto, sabendo que a soma dos ângulos opostos de um quadrilátero inscritível é $180^{\\circ}$, segue que\n$$\n\\begin{aligned}\n\\angle A B C+\\angle A E D & = (\\angle A B D+\\angle A E D)+\\angle D B C \\\\\n& = 180^{\\circ}+\\angle D A C \\\\\n& = 180^{\\circ}+50^{\\circ} \\\\\n& = 230^{\\circ}\n\\end{aligned}\n$$\n\n![](attached_image_2.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26230,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSeja $ABCD$ um retângulo com $AB = \\sqrt{3}$. Se $\\angle ACD = 75^{\\circ}$, calcule o comprimento de $AC$ e o valor de $\\cos 75^{\\circ}$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "AC = 2√(6 + 3√3); cos 75° = 1/(2√(2 + √3))",
    "solution": "Solution:\n\n![](attached_image_2.png)\nConsidere os pontos $F$ e $E$ sobre o lado $BC$ de modo que $FC = FA$ e $EF = AE$. Então, pelo Teorema do Ângulo Externo, temos\n$$\n\\angle BEA = 2 \\cdot \\angle EFA = 4 \\cdot \\angle FCA = 4 \\cdot (90^{\\circ} - \\angle ACD) = 60^{\\circ}\n$$\nDaí, $\\frac{AB}{AE} = \\sen 60^{\\circ} = \\frac{\\sqrt{3}}{2}$ e $\\frac{AB}{BE} = \\tg 60^{\\circ}$, ou seja, $AE = 2$ e $BE = 1$. Além disto, $EF = AE = 2$. Pelo Teorema de Pitágoras no triângulo $BFA$, temos\n$$\n\\begin{aligned}\nAF & = \\sqrt{AB^2 + BF^2} \\\\\n   & = \\sqrt{(\\sqrt{3})^2 + 3^2} \\\\\n   & = 2\\sqrt{3}\n\\end{aligned}\n$$\nPortanto, $FC = AF = 2\\sqrt{3}$. Finalmente, aplicando o Teorema de Pitágoras no triângulo $ABC$, temos\n$$\n\\begin{aligned}\nAC & = \\sqrt{AB^2 + BC^2} \\\\\n   & = \\sqrt{(\\sqrt{3})^2 + (3 + 2\\sqrt{3})^2} \\\\\n   & = \\sqrt{24 + 12\\sqrt{3}} \\\\\n   & = 2\\sqrt{6 + 3\\sqrt{3}}\n\\end{aligned}\n$$\nConsequentemente, temos\n$$\n\\cos 75^{\\circ} = \\frac{CD}{AC} = \\frac{\\sqrt{3}}{2\\sqrt{6 + 3\\sqrt{3}}} = \\frac{1}{2\\sqrt{2 + \\sqrt{3}}}",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26231,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nRosa tem um papagaio que faz contas de um modo estranho. Cada vez que Rosa diz dois números ele faz a mesma conta. Por exemplo:\n- se Rosa diz \"4 e 2\" o papagaio responde \"12\";\n- se Rosa diz \"5 e 3\" o papagaio responde \"12\";\n- se Rosa diz \"3 e 5\" o papagaio responde \"14\";\n- se Rosa diz \"9 e 7\" o papagaio responde \"24\";\n- se Rosa diz \"0 e 0\" o papagaio responde \"1\".\nSe Rosa diz \"1 e 8\" o que responde o papagaio?",
    "options": [],
    "answer": "19",
    "solution": "Solution:\n\n19.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26232,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm número surpreendente é um número divisível por $9$, de nove algarismos diferentes, nenhum deles igual a $0$ tal que:\n\na. O número formado pelos $2$ primeiros algarismos é divisível por $2$;\nb. O número formado pelos $3$ primeiros algarismos é divisível por $3$;\nc. O número formado pelos $4$ primeiros algarismos é divisível por $4$;\nd. O número formado pelos $5$ primeiros algarismos é divisível por $5$;\ne. O número formado pelos $6$ primeiros algarismos é divisível por $6$;\nf. O número formado pelos $7$ primeiros algarismos é divisível por $7$;\ng. O número formado pelos $8$ primeiros algarismos é divisível por $8$;\n\nQual é esse número?",
    "options": [],
    "answer": "381654729",
    "solution": "Solution:\n\n381654729",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26233,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNo cabeleireiro - Três clientes estão no cabeleireiro pagando cada um a sua conta no caixa.\n- o primeiro cliente paga o mesmo montante que há no caixa e retira 10 reais de troco;\n- o segundo cliente efetua a mesma operação que o primeiro;\n- o terceiro cliente efetua a mesma operação que os dois primeiros.\nEncontre o montante que estava inicialmente no caixa, sabendo que ao fim das 3 operações o caixa ficou zerado.",
    "options": [],
    "answer": "8.75 reais",
    "solution": "Solution:\n\nSeja $x$ o montante inicial no caixa. Esse montante mais o que os 3 clientes pagaram nos dará o caixa zerado.\n\n- O 1º cliente paga $x-10$. Depois do primeiro cliente, há $x + (x-10) = 2x - 10$ reais no caixa.\n\n- O 2º cliente paga $(2x-10) - 10 = 2x - 20$. Depois do 2º cliente, há $(2x-10) + (2x-20) = 4x - 30$ no caixa.\n\n- O 3º cliente paga $(4x-30) - 10 = 4x - 40$. Depois do 3º cliente, há $(4x-30) + (4x-40) = 8x - 70$ no caixa, que sabemos ser igual a $0$.\n\nLogo, $8x = 70$ e obtemos $x = 8{,}75$ reais.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26234,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n(a) $1678^{2}-1677^{2}$\n\n(b) $1001^{2}+1000^{2}$\n\n(c) $19999^{2}$\n\n(d) $2001^{2}+2002^{2}+2003^{2}$",
    "options": [],
    "answer": "(a) 3355; (b) 2002001; (c) 399960001; (d) 12024014",
    "solution": "Solution:\n\n(a) Como $a^{2}-b^{2}=(a+b)(a-b)$, temos\n$$\n1678^{2}-1677^{2}=(1678+1677)(1678-1677)=3355\n$$\n\n(b) Como $(a+b)^{2}=a^{2}+2ab+b^{2}$, temos\n$$\n\\begin{aligned}\n1001^{2}+1000^{2} & =(1000+1)^{2}+1000^{2}=1000^{2}+2000+1+1000^{2}= \\\\\n& =2 \\times 1000^{2}+2001=2002001\n\\end{aligned}\n$$\n\n(c) Como $(a-b)^{2}=a^{2}-2ab+b^{2}$, temos\n$$\n\\begin{aligned}\n19999^{2} & =(20000-1)^{2}=\\left(2 \\times 10^{4}\\right)^{2}-4 \\times 10^{4}+1= \\\\\n& =4 \\times 10^{8}-4 \\times 10^{4}+1=399960001\n\\end{aligned}\n$$\n\n(d) Colocando em função de $2000$, temos\n$$\n\\begin{aligned}\n2001^{2}+2002^{2}+2003^{2} & =(2000+1)^{2}+(2000+2)^{2}+(2000+3)^{2}= \\\\\n& =3 \\times 2000^{2}+12 \\times 2000+14=12024014\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26235,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that if $a < b$ are in the interval $[0, \\pi/2]$ then $a - \\sin a < b - \\sin b$. Is this true for $a < b$ in the interval $[\\pi, 3\\pi/2]$?",
    "options": [],
    "answer": "Yes",
    "solution": "We have $\\sin b - \\sin a = 2 \\sin \\frac{b-a}{2} \\cos \\frac{b+a}{2} = 2 \\sin \\frac{b-a}{2} < 2(b-a)/2 = b-a$.\n\nThe second case is trivial because both $x$ and $-\\sin x$ are increasing in the interval $[\\pi, 3\\pi/2]$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26236,
    "subject": "Mathematics (Multi-modal)",
    "question": "The remainder on dividing the polynomial $p(x)$ by $x^2 - (a+b)x + ab$ (where $a$ and $b$ are unequal) is $mx + n$. Find the coefficients $m, n$ in terms of $a, b$. Find $m, n$ for the case $p(x) = x^{200}$ divided by $x^2 - x - 2$ and show that they are integers.",
    "options": [],
    "answer": "General: m = (p(a) − p(b)) / (a − b), n = (a p(b) − b p(a)) / (a − b). Specific (p(x) = x^200, divisor x^2 − x − 2): m = (2^200 − 1) / 3, n = (2^200 + 2) / 3, which are integers.",
    "solution": "Let $p(x) = q(x)(x-a)(x-b) + mx + n$. So putting $x = a, b$ we get $p(a) = ma + n$, $p(b) = mb + n$. Solving, $m = \\frac{p(a)-p(b)}{a-b}$, $n = \\frac{p(b)a-p(a)b}{a-b}$.\n\nIn the case given $a = 2$, $b = -1$, so $m = \\frac{2^{200}-1}{3}$, $n = \\frac{2^{200}+2}{3}$. Note that $2 \\equiv -1 \\pmod{3}$, so $2^{200} \\equiv 1 \\pmod{3}$ and hence $2^{200} - 1$ is a multiple of $3$, so $m$ is an integer. $n = m + 1$, so $n$ is also an integer.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 26237,
    "subject": "Mathematics (Multi-modal)",
    "question": "The vertex $C$ of the triangle $ABC$ is allowed to vary along a line parallel to $AB$. Find the locus of the orthocenter.",
    "options": [],
    "answer": "The locus is the entire parabola b y = a^2 − x^2, in coordinates with A = (−a, 0), B = (a, 0), and C moving along y = b.",
    "solution": "Take axes so that $A = (-a, 0)$, $B = (a, 0)$ and $C = (k, b)$. Then the orthocenter lies on the line $x = k$. The line $AC$ has gradient $\\frac{b}{k+a}$, so the perpendicular has gradient $-\\frac{k+a}{b}$. Hence the altitude from $B$ has equation $y + \\frac{(x-a)(k+a)}{b} = 0$. So the intersection is $x = k, y = -\\frac{(k-a)(k+a)}{b}$. So the locus is all or part of the parabola $by = a^2 - x^2$. But we can get an orthocenter with any x-coordinate (by taking $C$ to have the same x-coordinate), so we can get all points on the parabola.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 26238,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that the number of positive integer solutions to\n$$\nx_1 + 2^3 x_2 + 3^3 x_3 + \\cdots + 10^3 x_{10} = 3025 \\quad (*)\n$$\nequals the number of non-negative integer solutions to the equation\n$$\ny_1 + 2^3 y_2 + 3^3 y_3 + \\cdots + 10^3 y_{10} = 0\n$$\nHence show that (*) has a unique solution in positive integers and find it.",
    "options": [],
    "answer": "x1 = x2 = ... = x10 = 1",
    "solution": "We have $1^3 + 2^3 + \\cdots + 10^3 = 3025$. Now $x_i$ is a positive integer solution to $(*)$ iff $y_i = x_i - 1$ are all non-negative and satisfy $(y_1+1)+2^3(y_2+1)+3^3(y_3+1)+\\cdots+10^3(y_{10}+1) = 3025$ and hence $y_1+2^3y_2+3^3y_3+\\cdots+10^3y_{10} = 0$. But that clearly has the unique solution $y_i = 0$, so the unique solution to $(*)$ is $x_i = 1$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 26239,
    "subject": "Mathematics (Multi-modal)",
    "question": "(i) $ABCD$ is a square with side $1$. $M$ is the midpoint of $AB$, and $N$ is the midpoint of $BC$. The lines $CM$ and $DN$ meet at $I$. Find the area of the triangle $CIN$.\n\n(ii) The midpoints of the sides $AB$, $BC$, $CD$, $DA$ of the parallelogram $ABCD$ are $M$, $N$, $P$, $Q$ respectively. Each midpoint is joined to the two vertices not on its side. Show that the area outside the resulting 8-pointed star is $2/5$ the area of the parallelogram.\n\n(iii) $ABC$ is a triangle with $CA = CB$ and centroid $G$. Show that the area of $AGB$ is $1/3$ of the area of $ABC$.\n\n(iv) Is (ii) true for all convex quadrilaterals $ABCD$?",
    "options": [],
    "answer": "(i) 1/20; (ii) 2/5 of the area of the parallelogram; (iii) 1/3 of the area of ABC; (iv) No, it is not true for all convex quadrilaterals.",
    "solution": "(i) $CIN$ is similar to $DCN$, so area $CIN = \\left(\\frac{CN}{DN}\\right)^2$ area $DCN = \\left(\\frac{\\frac{1}{2}}{\\sqrt{5/2}}\\right)^2 \\cdot \\frac{1}{4} =$\n\n$\\frac{1}{20}$\n\n![](attached_image_1.png)\n\n(ii) If we stretch the plane parallel to one of sides of the square then all areas are increased by the same factor and hence the ratio $\\frac{\\text{area CIN}}{\\text{area ABCD}}$ is unchanged. If we now shear parallel to one of the sides, areas are unchanged, so the ratio $\\frac{\\text{area CIN}}{\\text{area ABCD}}$ remains $\\frac{1}{20}$. Thus the result holds for parallelograms. The area outside the star is made up of 8 small triangles, each area $\\frac{1}{20}$, so it is $\\frac{2}{5}$.\n\n![](attached_image_2.png)\n\n(iii) Let the median be $AM$. Then $AGB$ and $ABC$ have the same base $AB$, so $\\frac{\\text{area } AGB}{\\text{area } ABC} = \\frac{GM}{AM} = \\frac{1}{3}$. This is trivial, but this is a hint for part (iv).\n\n(iv) If we take $A$ and $B$ close together, then we get the same figure as in (iii) and so the ratio tends to $\\frac{1}{3}$. Hence the $\\frac{2}{5}$ result is not true for all convex quadrilaterals.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26240,
    "subject": "Mathematics (Multi-modal)",
    "question": "Box $A$ contains black balls and box $B$ contains white balls. Take a certain number of balls from $A$ and place them in $B$. Then take the same number of balls from $B$ and place them in $A$. Is the number of white balls in $A$ then greater, equal to, or less than the number of black balls in $B$?",
    "options": [],
    "answer": "equal",
    "solution": "Suppose we move $n$ balls from $A$ to $B$, then $x$ white balls from $B$ and $n-x$ black balls from $B$. That leaves $x$ black balls in $B$. So both numbers equal $x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26241,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that for any positive integer $n > 2$ we can find $n$ distinct positive integers such that the sum of their reciprocals is $1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We have $\\frac{1}{2} + \\frac{1}{3} + \\frac{1}{6} = 1$.\n\nNow we have $\\frac{1}{2} + \\frac{1}{4} + \\frac{1}{8} + \\cdots + \\frac{1}{2^n} + \\frac{1}{2^n} = 1$, replace the second $\\frac{1}{2^n}$ by $\\left(\\frac{1}{2} + \\frac{1}{3} + \\frac{1}{6}\\right) \\frac{1}{2^n}$ to get $n + 3$ terms ($n \\ge 1$):\n\n$\\frac{1}{2} + \\frac{1}{4} + \\frac{1}{8} + \\cdots + \\frac{1}{2^{n+1}} + \\frac{1}{3 \\cdot 2^n} + \\frac{1}{3 \\cdot 2^{n+1}} = 1$.\n\nFor example, we have $\\frac{1}{2} + \\frac{1}{4} + \\frac{1}{6} + \\frac{1}{12} = 1$ and $\\frac{1}{2} + \\frac{1}{4} + \\frac{1}{8} + \\frac{1}{12} + \\frac{1}{24} = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26242,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a triangle $ABC$ and a point $P_0$ on the side $AB$. Construct points $P_i$, $Q_i$, $R_i$ as follows: $Q_i$ is the foot of the perpendicular from $P_i$ to $BC$, $R_i$ is the foot of the perpendicular from $Q_i$ to $AC$ and $P_i$ is the foot of the perpendicular from $R_{i-1}$ to $AB$. Show that the points $P_i$ converge to a point $P$ on $AB$ and show how to construct $P$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "It is clear from the diagram that\n$$\nQ_n Q_{n+1} = P_n P_{n+1} \\cos \\angle B\n$$\n$$\nR_n R_{n+1} = Q_n Q_{n+1} \\cos \\angle C\n$$\n$$\nP_n P_{n+1} = R_{n-1} R_n \\cos \\angle A\n$$\nHence $P_n P_{n+1} = k^n P_0 P_1$, where $|k| = |\\cos A \\cos B \\cos C| < 1$. If we take the direction $A$ to $B$ as positive, then the signed distance $P_n P_{n+1}$ may be positive or negative, but the series $1 + |k| + |k|^2 + |k|^3 + \\cdots$ converges to $\\frac{1}{1-|k|}$, so $P_0 P_1 + P_1 P_2 + P_2 P_3 + \\cdots$ is absolutely convergent and hence convergent. So the points $P_i$ converge to a point $P$ on the line $AB$.\n\n![](attached_image_1.png)\nTake any point $P'$ on $AB$. Take $Q'$ as the foot of the perpendicular from $P'$ to $BC$. Now take $R'$ as the intersection of the lines through $P'$ perpendicular to $AB$ and through $Q'$ perpendicular to $AC$. Now $P'Q'R'$ is similar to the desired triangle $PQR$. Since $B, P', P$ are collinear and $B, Q', Q$ are collinear, it follows that $B, R', R$ must be collinear. Thus extend $BR'$ to meet $AC$ at\n\n$R$. It is now straightforward to construct $Q$, then $P$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26243,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given 5 points of a sphere of radius $r$, show that two of the points are a distance less than or equal to $r\\sqrt{2}$ apart.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose the result is false so that we can find 5 points with the distance between any two greater than $r\\sqrt{2}$. Then the angle subtended by any two at the center of the sphere is greater than $90^\\circ$. Take one of the points to be at the north pole. Then the other four must all be south of the equator. Two must have longitude differing by at most $90^\\circ$.\n\nIt is now fairly obvious that these two points subtend an angle at most $90^\\circ$ at the center. To prove it we may take rectangular coordinates with origin at the center of the sphere so that both points have all coordinates non-negative. Suppose one is $(a, b, c)$ and the other $(A, B, C)$. Then since both lie on the sphere $a^2 + b^2 + c^2 = A^2 + B^2 + C^2 = r^2$, and the square of the distance between them is $(a - A)^2 + (b - B)^2 + (c - C)^2 \\leq (a^2 + b^2 + c^2) + (A^2 + B^2 + C^2) = 2r^2$, so the distance is at most $r\\sqrt{2}$, as required.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 26244,
    "subject": "Mathematics (Multi-modal)",
    "question": "For which $k$ does the system $x^2 - y^2 = 0$, $(x - k)^2 + y^2 = 1$ have exactly\n(1) two, (2) three real solutions?",
    "options": [],
    "answer": "(1) Two solutions: k = ±√2. (2) Three solutions: k = ±1.",
    "solution": "We have $(x - k)^2 + x^2 = 1$, so $2x^2 - 2k x + k^2 - 1 = 0$. This has 0, 1 or 2 real solutions according as $k^2 > 2$, $k^2 = 2$ or $k^2 < 2$.\n\n$k = \\sqrt{2}$ gives $x = \\frac{1}{\\sqrt{2}}$, $y = \\frac{1}{\\sqrt{2}}$ or $y = -\\frac{1}{\\sqrt{2}}$, so there are two solutions to the original set. Similarly for $k = -\\sqrt{2}$.\n\nIf $|k| < \\sqrt{2}$, then $x = \\frac{k \\pm \\sqrt{2 - k^2}}{2}$, $y = \\pm x$. That gives 4 solutions unless one of the values of $x$ is 0, in which case we get 3 solutions. So $k = 1$ gives $(x, y) = (0, 0)$ or $(1, 1)$ or $(1, -1)$. $k = -1$ gives $(x, y) = (0, 0)$ or $(-1, 1)$ or $(-1, -1)$.\n\nIf $|k| > \\sqrt{2}$, there are no solutions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26245,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that there are at least 3 and at most 4 powers of $2$ with $m$ digits. For which $m$ are there $4$?",
    "options": [],
    "answer": "There are at least 3 and at most 4 powers of two with m digits. There are 4 precisely for those m such that there exists an integer n with (m−1)/log10 2 ≤ n < m/log10 2 − 3 (equivalently, an integer lies between (m−1)/log10 2 and m/log10 2 − 3).",
    "solution": "Take $n$ to be the smallest integer such that $2^n \\ge 10^{m-1}$. Then $2^{n-1} < 10^{m-1}$, so $2^{n+2} < 8 \\cdot 10^{m-1} < 10^m$. So $2^n$, $2^{n+1}$ and $2^{n+2}$ all have $m$ digits. Thus there are at least $3$ powers of $2$ with $m$ digits.\n\n$2^{n-1} \\ge 5 \\cdot 10^{m-2}$ (otherwise $2^n < 10^{m-1}$). Hence $2^{n+4} = 32 \\cdot 5 \\cdot 10^{m-2} > 10^m$, so $2^{n+4}$ has more than $m$ digits. Thus there are at most $4$ powers of $2$ with $m$ digits.\n\nThere are $4$ if there is an integer between $\\frac{m-1}{\\log_{10} 2}$ and $\\frac{m}{\\log_{10} 2} - 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26246,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a sheet of paper and the use of a rule, compass and pencil, show how to draw a straight line that passes through two given points, if the length of the ruler and the maximum opening of the compass are both less than half the distance between the two points. You may not fold the paper.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note that we can draw an arbitrarily long line through a given point by repeatedly extending a short line. We can also find the midpoint of an arbitrary line segment. For suppose the segment is $PQ$. Take a distance $k$ which is less than the maximum opening of the compass and less than the length of the ruler. Starting at $P$ and using the compasses, mark off $q$ distances of $k$ leaving a final distance of $r < k$ to $Q$. Now bisect the final segment of $r$ as usual and a segment length $k$. Then mark off $q$ distances of $k/2$ and one of $r/2$ from $P$ to get the midpoint.\n\nSo suppose the points given are $A$ and $B$. Take any lines through $A$ and $B$ meeting at $C$. Let $M_1, N_1$ be the midpoints of $AC, BC$ respectively. Then take $M_2, N_2$ as the midpoints of $M_1C, N_1C$ respectively, and so on until we get $M_n N_n < k$. We can now join $M_n$ and $N_n$ to get a line parallel to the desired line $AB$. That allows us to draw a short line through $A$ in the right direction. We mark off a point $X$ on $AC$ with $AX = M_n C$, then draw circles center $A$ radius $M_n N_n$ and center $X$ radius $CN_n$ to intersect at a point $Y$ with $AXY$ congruent to $M_n C N_n$ and hence $Y$ on $AB$. Now extend $AY$ to get $AB$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26247,
    "subject": "Mathematics (Multi-modal)",
    "question": "A graph has 100 points. Given any four points, there is one joined to the other three. Show that one point must be joined to all 99 other points. What is the smallest number possible of such points (that are joined to all the others)?",
    "options": [],
    "answer": "97",
    "solution": "Suppose that no point is joined to all the others. Then given any point $X$ we can find $Y$ not joined to $X$. So take arbitrary $A$ and $C$. Then take $B$ not joined to $A$ and $D$ not joined to $C$. Then the four points $A, B, C, D$ do not meet the required condition. Contradiction.\n\nSo find $X_1$ joined to all the other 99 points. Now repeat the argument for the other 99 points, that gives a point $X_2$ joined to the other 98. But it is also joined to $X_1$, so it is joined to all other 99 points. Now repeat for the other 98 points and so on. The last time we can repeat is when we have already found $X_1, X_2, \\dots, X_{96}$ leaving four points. We can now take $X_{97}$ joined to the other three and hence to all other 99. Thus we can get at least 97 points each joined to all points except itself.",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 26248,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two thieves stole a container of $8$ liters of wine. How can they divide it into two parts of $4$ liters each if all they have is a $3$ liter container and a $5$ liter container? Consider the general case of dividing $m+n$ liters into two equal amounts, given a container of $m$ liters and a container of $n$ liters (where $m$ and $n$ are positive integers). Show that it is possible iff $m+n$ is even and $\\frac{m+n}{2}$ is divisible by $\\gcd(m, n)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Call the containers $L_8$, $L_5$, $L_3$. Fill $L_5$ from $L_8$, then fill $L_3$ from $L_5$, leaving $2$ in $L_5$. Empty $L_3$ into $L_8$. Empty $L_5$ into $L_3$ (so now $L_8$ has $6$, $L_5$ has $0$, $L_3$ has $2$). Fill $L_5$ from $L_8$. Fill $L_3$ from $L_5$. Empty $L_3$ into $L_8$. Now $L_5$ and $L_8$ each contain $4$.\n\nNow consider the general case. It is an easy induction that the amount in each container is always a multiple of $\\gcd(m, n)$. Use induction on the number of steps, and note that the only possible move is to replace $a, b$ by $D, a+b-D$, where $D$ is one of $0, m, n, m+n$. So it is certainly a necessary condition that $\\frac{m+n}{2}$ is divisible by $\\gcd(m, n)$. In particular, it must be an integer and so $m+n$ must be even. So it remains to show that if $m+n$ is even and $\\frac{m+n}{2}$ is a multiple of $\\gcd(m, n)$ then we can get $\\frac{m+n}{2}$ into $L_m$.\nIf $m = n$, then that is trivial. So assume $m > n$. Put $d = m-n$. Now suppose that after some moves we have got $k$ in the $L_n$ and the rest $m+n-k$ in the $L_{m+n}$. Fill $L_m$ from $L_{m+n}$, then fill $L_n$ from $L_m$. That gives $m-(n-k) = k+d$ in $L_m$. Now $k+d = qn+r$ for some $0 \\le r < n$. Repeatedly (for more precisely $q$ times) fill $L_n$ from $L_m$ and empty it into $L_{m+n}$, finally pour the remainder of $r$ from $L_m$ into $L_n$. So starting with all the wine in $L_{m+n}$ (i.e. $k=0$), and iterating this process we get $[hd]$ in $L_n$ where $[hd]$ denotes the residue of $hd$ mod $n$.\nNow we may put $\\frac{m+n}{2} = Qn + R$, where $0 \\le R < n$. Since $n$ and $\\frac{m+n}{2}$ are multiples of $\\gcd(m, n)$, so is $R$. But $\\gcd(m, n) = \\gcd(d, n)$. So $R$ is a multiple of $\\gcd(d, n)$. But that means we can write $R = hd - h'n$ for some\n\n(1981)",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26249,
    "subject": "Mathematics (Multi-modal)",
    "question": "The centers of the faces of a cube form a regular octahedron of volume $V$. Through each vertex of the cube we may take the plane perpendicular to the long diagonal from the vertex. These planes also form a regular octahedron. Show that its volume is $27V$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the cube have side $k$. $A$, $B$ are two adjacent vertices of the small octahedron, and $AX = BX = k/2$ and $\\angle AXB = 90^\\circ$, so $AB = k/\\sqrt{2}$.\n\n![](attached_image_1.png)\n\nThe large octahedron has the vertices of the cube at the center of its faces.\nThe line joining the centers of $OPS$ and $OQR$ is parallel to $PQ$ and $2/3$ the length. But it is also $k\\sqrt{2}$, so the side of the large octahedron is $(3/2)k\\sqrt{2} = 3k/\\sqrt{2}$ or $3$ times the side of the small octahedron. Hence the volume of the large octahedron is $3^3 = 27$ times the volume of the small.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 26250,
    "subject": "Mathematics (Multi-modal)",
    "question": "The angles of the triangle $ABC$ satisfy $\\frac{\\angle A}{\\angle C} = \\frac{\\angle B}{\\angle A} = 2$. The incenter is $O$. $K$, $L$ are the excenters of the excircles opposite $B$ and $A$ respectively. Show that triangles $ABC$ and $OKL$ are similar.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "$\\angle AOC = 180^\\circ - \\angle A/2 - \\angle C/2$. But $\\angle KAO = \\angle KCO = 90^\\circ$, so $\\angle AKC = \\angle A/2 + \\angle C/2$. So considering triangle $AKL$, $\\angle ALK = 90^\\circ - \\angle A/2 - \\angle C/2 = \\angle B/2$. Similarly, $\\angle BLC = \\angle B/2 + \\angle C/2$, so $\\angle BKL = \\angle A/2$.\n\n![](attached_image_1.png)\n\nNow we use the given facts that $\\angle C = \\angle A/2$ and $\\angle A = \\angle B/2$. So $\\angle OKL = \\angle BKL = \\angle C$, and $\\angle OLK = \\angle ALK = \\angle A$. Hence triangles $OKL$ and $BCA$ are similar.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26251,
    "subject": "Mathematics (Multi-modal)",
    "question": "Any positive integer $n$ can be written in the form $n = 2^b(2c + 1)$. We call $2c + 1$ the *odd part of* $n$. Given an odd integer $n > 0$, define the sequence $a_0, a_1, a_2, \\ldots$ as follows: $a_0 = 2^n - 1$, $a_{k+1}$ is the odd part of $3a_k + 1$. Find $a_n$.",
    "options": [],
    "answer": "(3^n - 1)/2",
    "solution": "An induction shows that $a_k = 3^k 2^{n-k} - 1$ for $k \\le n-1$. It is certainly true for $k=0$. Suppose it is true for $k < n-1$. Then $3a_k + 1 = 3^{k+1} 2^{n-k} - 2$. Since $n-k > 1$, the odd part is $3^{k+1} 2^{n-(k+1)} - 1$, so the result is true for $k+1$. That gets us as far as $a_{n-1} = 3^{n-1} 2 - 1$.\n\nNow we want the odd part of $2(3^n - 1)$. Certainly $3^n - 1$ is even. We have $3^n - 1 \\equiv (-1)^n - 1 \\equiv 2 \\pmod 4$ for $n$ odd, so for $n$ odd it is not divisible by 4. Hence for $n$ odd we have $a_n = \\frac{3^{n-1}}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26252,
    "subject": "Mathematics (Multi-modal)",
    "question": "$S$ is a $(k+1) \\times (k+1)$ array of lattice points. How many squares have their vertices in $S$?",
    "options": [],
    "answer": "k(k+1)^2(k+2)/12",
    "solution": "The key is to consider how many squares have their vertices on the perimeter of a given $(n+1) \\times (n+1)$ array whose sides are parallel to the sides of the array.\n\n![](attached_image_1.png)\n\nThe diagram shows that there are $n$ such squares. There are $(k+1-n)^2$ such arrays. So the total number of squares is $k \\cdot 1^2 + (k-1)^2 \\cdot 2 + \\cdots + 1 \\cdot k^2 = \\sum_{1}^{k}(k+1-i)i^2 = \\frac{(k+1)k(k+1)(2k+1)}{6} - \\frac{k^2(k+1)^2}{4} = \\frac{k(k+1)2(k+2)}{12}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26253,
    "subject": "Mathematics (Multi-modal)",
    "question": "Three numbered tiles are arranged in a tray as shown:\n\n| 1 | 2 |\n|---|---|\n| 3 |   |\n\nShow that we cannot interchange the $1$ and the $3$ by a sequence of moves where we slide a tile to the adjacent vacant space.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Write down the order of the tiles reading clockwise around the perimeter, starting at $1$. We get $123$ and no move changes that, so we will always get $123$ after any sequence of moves. But the desired arrangement would give $132$, so it is not possible.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26254,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show how to construct a line segment length $\\sqrt[4]{a^4 + b^4}$ given segments of length $a$ and $b$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We show first how to get the square and the square root. Take $ABC$ with $\\angle A = 90^\\circ$, altitude $AD$ of length $a$ and $BD = 1$. Then by similar triangles $\\frac{BC}{AB} = \\frac{AB}{BD} \\iff BC = \\frac{AB^2}{BD} = AB^2 = a^2 + 1$. Hence $CD = a^2$.\n\n![](attached_image_1.png)\n\nConversely, we can take $BD = 1$, $CD = a$ and then construct $A$ (as the intersection of the perpendicular at $D$ and the circle with diameter $BC$) to obtain $AD$ of length $\\sqrt{a}$.\n\nSo now given $a, b$ construct $a^2, b^2$. Then take a right-angled triangle with those lengths as its shorter sides and the hypotenuse is $\\sqrt{a^4 + b^4}$. Finally, take the square root to get the required length.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26255,
    "subject": "Mathematics (Multi-modal)",
    "question": "Five spheres of radius $r$ are inside a right circular cone. Four of the spheres lie on the base of the cone. Each touches two of the others and the sloping sides of the cone. The fifth sphere touches each of the other four and also the sloping sides of the cone. Find the volume of the cone.",
    "options": [],
    "answer": "pi * r^3 * (2*sqrt(2) + 1)^3 / 3",
    "solution": "The left-hand diagram shows the four spheres on the base. Evidently $AC = 2r\\sqrt{2}$.\n\nThe right-hand diagram shows a vertical section through $A$, $C$ and the center $O$ of the top sphere. $P$ is the apex of the cone and $QR$ is a diameter of its base. Evidently $PQR$ is similar to $OAC$ and its sides are parallel and at distance $r$ outside the corresponding sides of $OAC$.\n\n![](attached_image_1.png)\n\nAlso $AOC$ is congruent to $ABC$, so $\\angle AOC = 90^\\circ$. Hence $\\angle QPR = 90^\\circ$ also and so $OP = r\\sqrt{2}$. The altitude from $O$ in $AOC$ has length $AC/2 = r\\sqrt{2}$. Hence the altitude from $P$ in $PQR$ has length $OP + r\\sqrt{2} + r = (2\\sqrt{2} + 1)r$. Thus the radius of the base of the cone is also $(2\\sqrt{2} + 1)r$ and its volume is $\\frac{\\pi r^3 (2\\sqrt{2}+1)^3}{3}$.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 26256,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that there are only finitely many solutions to\n$$\n\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} = \\frac{1}{1983}\n$$\nin positive integers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose without loss of generality that $a \\leq b \\leq c$. We must have $a \\leq 3 \\cdot 1983$, otherwise $\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} < \\frac{1}{1983}$. So there are only finitely many possible values for $a$.\n\nNow consider the number of solutions for fixed $a$. We have $\\frac{1}{b} + \\frac{1}{c} = \\frac{1}{1983} - \\frac{1}{a} = k$. Now we must have $b \\leq \\frac{2}{k}$, otherwise $\\frac{1}{b} + \\frac{1}{c} < k$. So there are only finitely many possible values for $b$. The number $c$ is fixed once $a$ and $b$ are fixed, so we have shown that there are only finitely many solutions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26257,
    "subject": "Mathematics (Multi-modal)",
    "question": "An equilateral triangle $ABC$ has side $a$. A square is constructed on the outside of each side of the triangle. A right regular pyramid with sloping side $a$ is placed on each square. These pyramids are rotated about the sides of the triangle so that the apex of each pyramid comes to a common point above the triangle. Show that when this has been done the other vertices of the bases of the pyramids (apart from the vertices of the triangle) form a regular hexagon.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The key observation is that the midpoints of the edges of a regular tetrahedron form the vertices of a regular octahedron. So we can place a tetrahedron (side $a$) with its base on a plane next to an octahedron (side $a$) with its base on the same plane and with a face of each coinciding. Now the pyramid is just half an octahedron, so evidently if we rotate it from its initial position so that its apex coincides with the apex $O$ of a regular tetrahedron (whose base is $ABC$), then one face of the pyramid is parallel to the base of the tetrahedron. Then the result follows immediately.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26258,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that $1 + \\frac{1}{2} + \\frac{1}{3} + \\cdots + \\frac{1}{n}$ is not an integer for $n > 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $2^m$ be the highest power of $2$ that does not exceed $n$. Then none of the other positive integers less than or equal to $n$ are divisible by $2^m$. Let $k = \\text{lcm}(1, 2, \\dots, n)$. Now write each of the terms $1, \\frac{1}{2}, \\dots, \\frac{1}{n}$ as fractions with denominator $k$. All will have even numerators except $\\frac{1}{2^m}$ which will have an odd numerator. Thus their sum is a fraction $\\frac{h}{k}$ with $h$ odd, so it cannot be an integer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26259,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that it is possible to color each point of a circle red or blue so that no right-angled triangle inscribed in the circle has its vertices all the same color.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Use any coloring with all pairs at opposite ends of a diameter having opposite colors. For example, take $AB$ to be a diameter. Color $A$ red, $B$ blue, all points in the interior of one arc $AB$ red and all points in the interior of the other blue.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26260,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that $1 \\le n^{1/n} \\le 2$ for all positive integers $n$. Find the smallest $k$ such that $1 \\le n^{1/n} \\le k$ for all positive integers $n$.",
    "options": [],
    "answer": "3^{1/3}",
    "solution": "If $k < 1$, then $k^n < 1$, so $n^{1/n} \\ge 1$. We have $2^1 > 1$, and $2^{n+1} - 2^n = 2^n > 1 = (n+1) - n$, so a trivial induction shows that $2^n > n$. Hence $2 > n^{1/n}$.\n\nNote that $3^2 > 2^3$, so taking 6th roots, $3^{1/3} > \\sqrt{2}$. Note also that $4^{1/4} = \\sqrt{2}$.\n\nWe show that $n^{1/n} < \\sqrt{2}$ for $n > 4$. This is another trivial induction. We have $2^{5/2} = 4\\sqrt{2} > 5$ and $2^{6/2} = 8 > 6$, so it is true for $n = 5$ and $n = 6$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26261,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that the maximum number of spheres of radius $1$ that can be placed touching a fixed sphere of radius $1$ so that no pair of spheres has an interior point in common is between $12$ and $14$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We can place $6$ spheres with their centers coplanar with the fixed sphere. Then we can place $3$ more above and $3$ more below as shown above. Thus $12$ can be achieved.\n\n![](attached_image_1.png)\n\nTake $O$ to be the center of the fixed sphere. Another sphere touching it blocks off a conical solid angle as shown. The angle between the center line of the cone and the surface is $30^\\circ$. It's not hard to show that for an angle $\\theta$ the solid angle is $2\\pi(1 - \\cos\\theta)$, so for $\\theta = 30^\\circ$, it is $\\pi(2 - \\sqrt{3})$. Thus we can have at most $\\frac{4\\pi}{\\pi(2-\\sqrt{3})} = 4(2+\\sqrt{3}) < 15$ spheres touching the fixed sphere. Hence at most $14$.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 26262,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all solutions in positive integers to $(n+1)^k - 1 = n!$.",
    "options": [],
    "answer": "(n, k) = (1, 1), (2, 1), (4, 2)",
    "solution": "The solutions are $n = 1, k = 1$; $n = 2, k = 1$; $n = 4, k = 2$.\nIt is easy to check that the solutions above are the only solutions for $n \\le 4$. So assume $n > 4$. So $n! + 1 > n + 1$, so $k > 1$. If $n$ is odd, then $n + 1$ is even, but $n! + 1$ is odd, so there are no solutions. So $n$ is even. Hence $n$ is composite. So $n$ divides $(n-1)!$. But using the binomial theorem we have $(n+1)^k - 1 = n^k + \\binom{k}{1}n^{k-1} + \\cdots + \\binom{k}{k-2}n^2 + kn$. Hence $(n-1)! = n(n^{k-2} + n^{k-3} + \\cdots + \\binom{k}{k-2}) + k$. Hence $n$ divides $k$. But that means $k \\ge n$, and $(n+1)^n > n! + 1$. So there are no other solutions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26263,
    "subject": "Mathematics (Multi-modal)",
    "question": "Each day 289 students are divided into 17 groups of 17. No two students are ever in the same group more than once. What is the largest number of days that this can be done?",
    "options": [],
    "answer": "18",
    "solution": "Each day student $X$ is with 16 different students. But there are only 288 students available, so it can be done for at most $288/16 = 18$ days.\n\nWe now show explicitly how it can be done for 18 days. Label the students as $(m, n)$, where $0 \\le m, n \\le 16$. Now for $0 \\le d, g \\le 16$, take group $g$ on day $d$ to be $(0, g)$, $(1, g+d)$, $(2, g+2d)$, $\\ldots$, $(16, g+16d)$, where all arithmetic is done mod 17. On the final day, take the groups as\n$$\n\\begin{array}{l}\n(0, 0), (0, 1), \\dots, (0, 16) \\\\\n(1, 0), (1, 1), \\dots, (1, 16) \\\\\n(2, 0), (2, 1), \\dots, (2, 16) \\\\\n\\vdots \\\\\n(16, 0), (16, 1), \\dots, (16, 16)\n\\end{array}\n$$\n\nNote first that for any given day the groups include all the students. That is obvious for the last day. Suppose we want to find the group for student $(a, b)$ on day $d$. Take $g = b - ad$ mod 17, then $(a, b) = (a, g + ad)$, so $(a, b)$ is in group $g$. Since the groups include just 289 students and each student is included at least once, it follows that each student is included exactly once.\n\nNow consider the pair $(a, b), (A, B)$. If $a = A$, then they are together on the last day. So suppose $a \\ne A$. Take $d = (b - B)(a - A)^{-1}$ mod 17 and take $g = b - ad$ mod 17. Then $(a, b) = (a, g + da)$ and $(A, B) = (A, g + dA)$, so they are together in group $g$ on day $d$. But there are only $18 \\cdot 17 \\cdot \\binom{17}{2} = \\binom{289}{2}$ pairs available and we have just shown that each occurs at least once. Hence each occurs at most once, as required.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 26264,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a regular dodecahedron of side $a$. Take two pairs of opposite faces: $E, E'$ and $F, F'$. For the pair $E, E'$ take the line joining the centers of the faces and take points $A$ and $C$ on the line each a distance $m$ outside one of the faces. Similarly, take $B$ and $D$ on the line joining the centers of $F, F'$ each a distance $m$ outside one of the faces. Show that $ABCD$ is a rectangle and find the ratio of its side lengths.",
    "options": [],
    "answer": "(1+sqrt(5))/2",
    "solution": "The centers of the faces of a regular dodecahedron form a regular icosahedron. Let $P, P'$ be two opposite vertices of a regular icosahedron. Then 5 of the remaining vertices are adjacent to $P$ and the other 5 are adjacent to $P'$. So we may label the other pair of opposite vertices $Q$ and $Q'$, where $Q$ is adjacent to $P$ and hence $Q'$ is adjacent to $P'$.\n\n![](attached_image_1.png)\n\nLet the side of the icosahedron be $k$. Then $PQP'Q'$ is obviously a rectangle and one pair of sides has length $k$. The other side is the diagonal of the regular pentagon $XYP'ZQ$ of side $k$ and hence has length $\\left(\\frac{1+\\sqrt{5}}{2}\\right)k$. $ABCD$ is similar to $PQP'Q'$ and so has the same ratio $\\frac{1+\\sqrt{5}}{2}$ for its side lengths.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26265,
    "subject": "Mathematics (Multi-modal)",
    "question": "$ABC$ is a triangle with $\\angle A = 90^\\circ$. For a point $D$ on the side $BC$, the feet of the perpendiculars to $AB$ and $AC$ are $E$ and $F$. For which point $D$ is $EF$ a minimum?",
    "options": [],
    "answer": "The foot of the perpendicular from A to BC.",
    "solution": "The foot of the perpendicular from $A$. $AEDF$ is a rectangle, so $EF = AD$. $AD$ is obviously a minimum for the altitude.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26266,
    "subject": "Mathematics (Multi-modal)",
    "question": "$ABCD$ is any convex quadrilateral. Squares center $E, F, G, H$ are constructed on the outside of the edges $AB, BC, CD$ and $DA$ respectively. Show that $EG$ and $FH$ are equal and perpendicular.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider $A, B, C, D$ to be the points $a, b, c, d$ in the complex plane. Then $E, F, G, H$ are the points $\\frac{a+b}{2} + i\\frac{b-a}{2}$, $\\frac{b+c}{2} + i\\frac{c-b}{2}$, $\\frac{c+d}{2} + i\\frac{d-c}{2}$, $\\frac{d+a}{2} + i\\frac{a-d}{2}$. Hence vector $\\vec{GE}$ is represented by the complex number $h = \\frac{a+b-c-d}{2} + i\\frac{-a+b+c-d}{2}$ and the vector $\\vec{FH}$ by the $k = \\frac{a-b-c+d}{2} + i\\frac{a+b-c-d}{2}$. But $k = ih$, so $EG$ and $FH$ are equal and perpendicular.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 26267,
    "subject": "Mathematics (Multi-modal)",
    "question": "$a$, $b$, $c$, $d$ are integers with $ad \\neq bc$. Show that $\\frac{1}{(ax+b)(cx+d)}$ can be written in the form $\\frac{r}{ax+b} + \\frac{s}{cx+d}$. Find the sum\n$$\n\\frac{1}{1 \\cdot 4} + \\frac{1}{4 \\cdot 7} + \\frac{1}{7 \\cdot 10} + \\dots + \\frac{1}{2998 \\cdot 3001}\n$$",
    "options": [],
    "answer": "1000/3001",
    "solution": "$$\n\\frac{a}{ax+b} - \\frac{c}{cx+d} = \\frac{ad-bc}{(ax+b)(cx+d)}, \\text{ so we can take } r = \\frac{a}{ad-bc}, s = -\\frac{c}{ad-bc}.\n$$\n\n$$\n\\sum_{k=0}^{999} \\frac{1}{(3k+1)(3k+4)} = \\frac{1}{3} \\sum_{k=0}^{999} \\left( \\frac{1}{3k+1} - \\frac{1}{3k+4} \\right) = \\frac{1}{3} \\left( 1 - \\frac{1}{3001} \\right) = \\frac{1000}{3001}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26268,
    "subject": "Mathematics (Multi-modal)",
    "question": "A convex quadrilateral is inscribed in a circle of radius $1$. Show that its perimeter minus the sum of its two diagonals lies between $0$ and $2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the quadrilateral be $ABCD$. We have $AB + BC > AC$ and $CD + DA > AC$, so the perimeter is bigger than $2 \\cdot AC$ and, similarly, bigger than $2 \\cdot BD$. Hence it's bigger than $AC + BD$, and the perimeter minus the sum of the diagonals is bigger than zero. Notice that this inequality is sharp: just consider a rectangle with two sides next to zero.\n\n![](attached_image_1.png)\n\nNow, $\\alpha + \\beta + \\gamma + \\delta = 180^\\circ$ and, since the radius of the circle is $1$, $AB = 2 \\sin \\gamma$, $BC = 2 \\sin \\beta$, $CD = 2 \\sin \\alpha$, $AD = 2 \\sin \\delta$, $AC = 2 \\sin(\\beta + \\gamma) = 2 \\sin(\\alpha + \\delta) = \\sin(\\beta + \\gamma) + \\sin(\\alpha + \\delta)$ and $BD = 2 \\sin(\\alpha + \\beta) = 2 \\sin(\\gamma + \\delta) = \\sin(\\alpha + \\beta) + \\sin(\\gamma + \\delta)$, so the difference required is\n$$\n2(\\sin \\alpha + \\sin \\beta + \\sin \\gamma + \\sin \\delta) - \\sin(\\alpha + \\delta) - \\sin(\\beta + \\gamma) - \\sin(\\alpha + \\beta) - \\sin(\\gamma + \\delta)\n$$\nNotice that $\\sin(\\alpha + \\gamma)$ and $\\sin(\\beta + \\delta)$ are “missing”. So we will prove a stronger result, that is,\n$$\n\\begin{aligned}\nE &= 2(\\sin \\alpha + \\sin \\beta + \\sin \\gamma + \\sin \\delta) \\\\\n&\\quad - \\sin(\\alpha + \\delta) - \\sin(\\beta + \\gamma) - \\sin(\\alpha + \\beta) - \\sin(\\gamma + \\delta) - \\sin(\\alpha + \\gamma) - \\sin(\\beta + \\delta) < 0 \\\\\n\\implies \\quad & 2(\\sin \\alpha + \\sin \\beta + \\sin \\gamma + \\sin \\delta) - \\sin(\\alpha + \\delta) - \\sin(\\beta + \\gamma) - \\sin(\\alpha + \\beta) - \\sin(\\gamma + \\delta) \\\\\n&< \\sin(\\alpha + \\gamma) + \\sin(\\beta + \\delta) \\le 2\n\\end{aligned}\n$$\nFirst notice that\n$$\n\\begin{aligned}\n\\sin(\\alpha + \\beta) + \\sin(\\alpha + \\gamma) &= 2 \\sin \\left( \\alpha + \\frac{\\beta + \\gamma}{2} \\right) \\cos \\left( \\frac{\\beta - \\gamma}{2} \\right) \\\\\n&= 2 \\cos \\left( \\frac{\\alpha - \\delta}{2} \\right) \\cos \\left( \\frac{\\beta - \\gamma}{2} \\right)\n\\end{aligned}\n$$\nso\n$$\n\\begin{aligned}\n& \\sin \\beta + \\sin \\gamma - \\sin(\\alpha + \\beta) - \\sin(\\alpha + \\gamma) \\\\\n&= 2 \\sin \\left( \\frac{\\beta + \\gamma}{2} \\right) \\cos \\left( \\frac{\\beta - \\gamma}{2} \\right) - 2 \\cos \\left( \\frac{\\alpha - \\delta}{2} \\right) \\cos \\left( \\frac{\\beta - \\gamma}{2} \\right) \\\\\n&= 2 \\cos \\left( \\frac{\\beta - \\gamma}{2} \\right) \\left( \\cos \\left( \\frac{\\alpha + \\delta}{2} \\right) - \\cos \\left( \\frac{\\alpha - \\delta}{2} \\right) \\right) \\\\\n&= -4 \\cos \\left( \\frac{\\beta - \\gamma}{2} \\right) \\sin \\frac{\\alpha}{2} \\sin \\frac{\\delta}{2}\n\\end{aligned}\n$$\nSumming three analogous equations, we obtain\n$$\n\\begin{aligned}\n& 2(\\sin \\alpha + \\sin \\beta + \\sin \\gamma) \\\\\n&\\quad - \\sin(\\alpha + \\delta) - \\sin(\\beta + \\gamma) - \\sin(\\alpha + \\beta) - \\sin(\\gamma + \\delta) - \\sin(\\alpha + \\gamma) - \\sin(\\beta + \\delta) \\\\\n&= -4 \\sin \\frac{\\delta}{2} \\left( \\cos \\left( \\frac{\\beta - \\gamma}{2} \\right) \\sin \\frac{\\alpha}{2} + \\cos \\left( \\frac{\\alpha - \\gamma}{2} \\right) \\sin \\frac{\\beta}{2} + \\cos \\left( \\frac{\\beta - \\alpha}{2} \\right) \\sin \\frac{\\gamma}{2} \\right) \\\\\n&= -4 \\sin \\frac{\\delta}{2} \\left( \\sin \\left( \\frac{\\alpha + \\beta - \\gamma}{2} \\right) + \\sin \\left( \\frac{\\alpha - \\beta + \\gamma}{2} \\right) + \\sin \\left( \\frac{-\\alpha + \\beta + \\gamma}{2} \\right) \\right)\n\\end{aligned}\n$$\nFinally,\n$$\n\\begin{align*}\nE &= 2\\sin\\delta - 4\\sin\\frac{\\delta}{2}\\left(\\sin\\left(\\frac{\\alpha+\\beta-\\gamma}{2}\\right) + \\sin\\left(\\frac{\\alpha-\\beta+\\gamma}{2}\\right) + \\sin\\left(\\frac{-\\alpha+\\beta+\\gamma}{2}\\right)\\right) \\\\\n&= -4\\sin\\frac{\\delta}{2}\\left(-\\cos\\frac{\\delta}{2} + \\sin\\left(\\frac{\\alpha+\\beta-\\gamma}{2}\\right) + \\sin\\left(\\frac{\\alpha-\\beta+\\gamma}{2}\\right) + \\sin\\left(\\frac{-\\alpha+\\beta+\\gamma}{2}\\right)\\right) \\\\\n&= -4\\sin\\frac{\\delta}{2}\\left(-\\sin\\left(\\frac{\\alpha+\\beta+\\gamma}{2}\\right) + \\sin\\left(\\frac{\\alpha+\\beta-\\gamma}{2}\\right) + 2\\cos\\left(\\frac{\\alpha-\\beta}{2}\\right)\\sin\\frac{\\gamma}{2}\\right) \\\\\n&= -8\\sin\\frac{\\delta}{2}\\left(-\\cos\\left(\\frac{\\alpha+\\beta}{2}\\right)\\sin\\frac{\\gamma}{2} + \\cos\\left(\\frac{\\alpha-\\beta}{2}\\right)\\sin\\frac{\\gamma}{2}\\right) \\\\\n&= -8\\sin\\frac{\\gamma}{2}\\sin\\frac{\\delta}{2}\\left(-\\cos\\left(\\frac{\\alpha+\\beta}{2}\\right) + \\cos\\left(\\frac{\\alpha-\\beta}{2}\\right)\\right) \\\\\n&= -16\\sin\\frac{\\alpha}{2}\\sin\\frac{\\beta}{2}\\sin\\frac{\\gamma}{2}\\sin\\frac{\\delta}{2} < 0\n\\end{align*}\n$$\nbecause $0 < \\frac{\\alpha}{2}, \\frac{\\beta}{2}, \\frac{\\gamma}{2}, \\frac{\\delta}{2} < 90^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26269,
    "subject": "Mathematics (Multi-modal)",
    "question": "$a$, $b$, $c$, $d$ are integers. Show that $x^2 + a x + b = y^2 + c y + d$ has infinitely many integer solutions iff $a^2 - 4b = c^2 - 4d$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $a^2 - 4b = c^2 - 4d$, then $a^2 \\equiv c^2 \\pmod{4}$, so $a$ and $c$ have the same parity. So if we take any integer $x$ and then $y$ to be the integer $x - \\frac{a-c}{2}$ we have $x - \\frac{a}{2} = y - \\frac{c}{2}$ and hence $(x - \\frac{a}{2})^2 - \\frac{a^2-4b}{4} = (y - \\frac{c}{2})^2 - \\frac{c^2-4d}{4} \\iff x^2 + a x + b = y^2 + c y + d$. Thus the equation has infinitely many integer solutions.\n\nConversely, suppose $k = (a^2 - 4b) - (c^2 - 4d) \\neq 0$, then we have $(2x - a)^2 - (2y - c)^2 = k$, so $(2x + 2y - a - c)(2x - 2y - a + c) = k$. But $k$ has only finitely many factorizations, so there are only finitely many possible values for the pair $(2x + 2y - a - c, 2x - 2y - a + c)$ and hence for the pair $(x, y)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26270,
    "subject": "Mathematics (Multi-modal)",
    "question": "$A$, $B$ are real numbers. Find a necessary and sufficient condition for $A x + B [x] = A y + B [y]$ to have no solutions except $x = y$.",
    "options": [],
    "answer": "A ≠ 0 and B/A ∉ (-2, 0) (equivalently, A ≠ 0 and B/A ≤ −2 or B/A ≥ 0).",
    "solution": "If $A = 0$, then we have $B \\lfloor x \\rfloor = B \\lfloor y \\rfloor$ which obviously has infinitely many solutions with $x \\neq y$. So assume $A \\neq 0$. Then we can write the equation as $\\frac{B}{A}(\\lfloor x \\rfloor - \\lfloor y \\rfloor) = y - x$. If $x \\neq y$, we can assume $x < y$. We cannot have $\\lfloor x \\rfloor = \\lfloor y \\rfloor$, since $y - x > 0$, so $\\lfloor x \\rfloor < \\lfloor y \\rfloor$. Write $x = n - \\epsilon$, $y = n + d + \\delta$, where $n$ is any integer, $d$ is a non-negative integer, $\\delta \\in [0, 1)$ and $\\epsilon \\in (0, 1]$. So $\\frac{B}{A} = -\\frac{d+\\nu}{d+1}$, where $d$ is a non-negative integer and $\\nu = \\delta + \\epsilon \\in (0, 2)$.\n\nWe note first that $-2 < -\\frac{d+\\nu}{d+1} < 0$. So if there are solutions with $x \\neq y$ and $A \\neq 0$, then $\\frac{B}{A}$ must belong to the open interval $(-2, 0)$.\n\nConversely, by taking $d = 0$ and suitable $\\nu$ it is clear that $\\frac{B}{A}$ can take any value in $(-2, 0)$. Thus a necessary and sufficient condition for solutions with $x \\neq y$ is $A = 0$ or $\\frac{B}{A} \\in (-2, 0)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26271,
    "subject": "Mathematics (Multi-modal)",
    "question": "A ball moves endlessly on a circular billiard table. When it hits the edge it is reflected. Show that if it passes through a point on the table three times, then it passes through it infinitely many times.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose $AB$ and $BC$ are two successive chords of the ball's path. Then by the reflection law $\\angle ABO = \\angle OBC$. But $OAB$ and $OBC$ are isosceles and so $\\angle AOB = \\angle BOC$. Hence $AB = BC$. So every chord of the path is the same length $d$.\n\nWe now claim that through any given point $P$ inside the circle there are at most two chords length $d$. Let $AB$ and $CD$ be a chord containing $P$, with $AP = a$ and $CP = b$. The power of $P$ with respect to the circle is $PA \\cdot PB = PC \\cdot PD \\iff a(d-a) = b(d-b) \\iff a = b$ or $a+b = d$. This means that $P$ always divides the chords containing it in two segments of fixed lengths $a$ and $d-a$. Now if three chords passes through $P$, the circle with center $P$ and radius $a$ would cut the circle of the billiard table three times, a contradiction.\n\nThus if the path passes through $P$ more than twice, then on two occasions it must be moving along the same chord $AB$. That implies that $\\angle AOB$ is a rational multiple of $2\\pi$ and hence the path will traverse $AB$ repeatedly.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 26272,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the number of ways that a positive integer $n$ can be represented as a sum of one or more consecutive positive integers.",
    "options": [],
    "answer": "the number of odd divisors of 2n",
    "solution": "The sum of $k$ consecutive integers is $(m+1) + (m+2) + \\dots + (m+k) = \\frac{k(2m+k+1)}{2}$. So we require $k(2m+k+1) = 2n$. Note that $k$ and $2m+k+1$ have opposite parity and that $k < 2m+k+1$.\n\nNow suppose $2n = ab$ with $a$ odd. Then $b$ must be even. So $a$ and $b$ cannot be equal. Take $k$ to be the smaller, then put $m = \\frac{|a-b-1|}{2}$ and we have a solution. So the total number of solutions is just the number of odd factors of $2n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26273,
    "subject": "Mathematics (Multi-modal)",
    "question": "The Poincaré plane is a half-plane bounded by a line $R$. The lines are taken to be (1) the half-lines perpendicular to $R$, and (2) the semicircles with center on $R$. Show that given any line $L$ and any point $P$ not on $L$, there are infinitely many lines through $P$ which do not intersect $L$. Show that if $ABC$ is a triangle, then the sum of its angles lies in the interval $(0, \\pi)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\nFor the first part there are three cases to consider: $L$ a semicircle and $P$ inside it, $L$ a semicircle and $P$ outside it, and $L$ a perpendicular half-line. The diagram above shows the first case. Take $L$ to have radius $R$. The semicircle through $P$ with the same center $O$ as $L$ does not intersect $L$. Suppose its radius is $R - k$. Now take a semicircle through $P$ with center $X$ such that $OX < \\frac{k}{2}$. Then its radius $XP < OP + OX < R - \\frac{k}{2}$, and $\\min(XA, XB) > R - \\frac{k}{2}$, so the semicircle lies entirely inside $L$. Thus we have an infinity of possible semicircles through $P$.\n\nAn exactly similar argument works for $P$ outside $L$.\n\n![](attached_image_2.png)\nFor the third case, just take the center sufficiently far from $P$ on the same side of $L$ as $P$.\n\nOn to the second part. Step 1 is to show that if we fix $C$ and allow $B$ to vary along a line $l$, then $\\angle x$ increases as $B$ moves down ($O$ is confined to the line and $OB = OC$). In fact, if $M$ is the midpoint of $BC$ then $M$ belongs to the line parallel to $l$ and passes through a point whose distance to $l$ is half the distance from $C$ to $l$. Moreover, the quadrilateral $OPBM$ is cyclic, so $\\angle OBM = \\angle OPM$, and $\\angle OPM$ decreases as $B$ moves down. This means that $\\angle x = 90^\\circ - \\angle OPM$ increases as $B$ moves down.\n\n![](attached_image_3.png)\nStep 2 is to show that $\\angle b + \\angle c + \\angle x = 180^\\circ$. Let the tangents to the arc passing through $B$ and $C$ intersect in $D$. So $\\angle BDC = \\angle b + \\angle c$ and, by looking at the internal angles of the quadrilateral $OBDC$, $\\angle x + 90^\\circ + 90^\\circ + \\angle b + \\angle c = 360^\\circ \\iff \\angle b + \\angle c + \\angle x = 180^\\circ$.\n\n![](attached_image_4.png)\nNow consider the special case where $AB$ is a straight line. Using the previous results one can show by some angle chasing that the sum of the angles in the triangle is $180^\\circ - (\\angle O' - \\angle O) < 180^\\circ$. Finally, use this special case to deduce the other cases (by dividing the general triangle into two parts which fall into the special case).",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26274,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all 10 digit numbers $a_0a_1\\ldots a_9$ such that, for each $k$, $a_k$ is the number of times that the digit $k$ appears in the number.",
    "options": [],
    "answer": "6210001000",
    "solution": "Note that $a_0 + a_1 + \\cdots + a_9$ is equal to the sum of number of 0s, 1s, ..., 9s, which is, in turn, equal to total number of digits, 10. In other words, the sum of the digits is 10. In particular, if $i > 5$, then $a_i = 0$ or 1 (because if $a_i = 2$, then we have at least 2 digits bigger than 5 with sum bigger than 10).\n\nSuppose all $a_i \\le 5$. Then in particular $a_0 \\le 5$, so there are at least 5 non-zero digits, and hence at least 4 of $a_1, a_2, \\dots, a_9$ are non-zero. But that means that $a_1 + 2a_2 + 3a_3 + \\cdots + 9a_9 \\ge 1 + 2 + 3 + 4$ with equality iff $a_1 = a_2 = a_3 = a_4 = 1$ and $a_5 = a_6 = a_7 = a_8 = a_9 = 0$. But we must have equality because $a_1 + 2a_2 + 3a_3 + \\cdots + 9a_9$ is the sum of the digits which is 10. However it is not possible to have $a_1 = a_2 = a_3 = a_4 = 1$ and $a_5 = a_6 = a_7 = a_8 = a_9 = 0$ because then there are four 1s, but $a_4 = 1$.\n\nSo suppose some $a_i \\ge 6$. Then $a_6 \\ge 1$. But $a_6 \\le 1$, so $a_6 = 1$. Hence $a_1 > 0$. We cannot have $a_1 = 1$, because then there are at least two 1s, so $a_1 \\le 2$. We cannot have $a_1 = 6$, because then there are six 1s and one 6, total 12, whereas all the digits only total 10. So $a_1$ must be some other digit $k > 1$. But we now have at least six 0s, at least two 1s, a $k$ and a 6, which is at least\n10. So it must be exactly 10. Hence $a_0 = 6$ and $a_1 = 2$. So the number must be 6210001000.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26275,
    "subject": "Mathematics (Multi-modal)",
    "question": "A number is written in each square of a chessboard, so that each number not on the border is the mean of the 4 neighboring numbers. Show that if the largest number is $N$, then there is a number equal to $N$ in the border squares.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Take the leftmost $N$. Suppose it is not in the border. Then it must be the mean of the 4 neighboring numbers. Hence each of the 4 neighboring numbers must also be $N$, but one of them is to the left of $N$. Contradiction.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 26276,
    "subject": "Mathematics (Multi-modal)",
    "question": "$p(x_1, x_2, \\dots, x_n)$ is a polynomial with integer coefficients. For each positive integer $r$, $k(r)$ is the number of $n$-tuples $(a_1, a_2, \\dots, a_n)$ such that $0 \\le a_i \\le r - 1$ and $p(a_1, a_2, \\dots, a_n)$ is prime to $r$. Show that if $u$ and $v$ are coprime then $k(u \\cdot v) = k(u) \\cdot k(v)$, and if $p$ is prime then $k(p^s) = p^{n(s-1)}k(p)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First observe that if $a_i \\equiv b_i \\pmod r$, $i = 1, 2, \\dots, n$ then $p(a_1, a_2, \\dots, a_n) \\equiv p(b_1, b_2, \\dots, b_n) \\pmod r$. If $u$ and $v$ are coprime then given $b_1, b_2, \\dots, b_n$ and $c_1, c_2, \\dots, c_n$ with $0 \\le b_i < u$ and $0 \\le c_i < v$ by the chinese remainder theorem there exist unique integers $a_1, a_2, \\dots, a_n$ with $0 \\le a_i < uv$ such that $a_i \\equiv b_i \\pmod u$ and $a_i \\equiv c_i \\pmod v$, which imply $p(a_1, a_2, \\dots, a_n) \\equiv p(b_1, b_2, \\dots, b_n) \\pmod u$, $p(a_1, a_2, \\dots, a_n) \\equiv p(c_1, c_2, \\dots, c_n) \\pmod v$. Thus\n$$\n\\gcd(p(a_1, a_2, \\dots, a_n), uv) = 1 \\iff \\begin{cases} \\gcd(p(a_1, a_2, \\dots, a_n), u) = 1 \\\\ \\gcd(p(a_1, a_2, \\dots, a_n), v) = 1 \\end{cases}\n$$\nand $k(uv) = k(u) \\cdot k(v)$.\nFor $p$ prime $\\gcd(p(a_1, a_2, \\dots, a_n), p^s) = 1 \\iff \\gcd(p(a_1, a_2, \\dots, a_n), p) = 1$. Divide each $a_i$ by $p$, obtaining quotient $q_i$ and remainder $r_i$. Reducing modulo $p$, we obtain $\\gcd(p(a_1, a_2, \\dots, a_n), p) = 1 \\iff \\gcd(p(r_1, r_2, \\dots, r_n), p) = 1$. Given $r_i$ there are $p^s/p = p^{s-1}$ possibilities for $q_i$ such that $0 \\le a_i < p^s$. Hence $k(p^s) = (p^{s-1})^n k(p) = p^{n(s-1)} k(p)$, as required.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26277,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a point $p$ inside a convex polyhedron $P$. Show that there is a face $F$ of $P$ such that the foot of the perpendicular from $p$ to $F$ lies in the interior of $F$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $F$ be the face of $P$ closer to $p$ and $p'$ be the projection of $p$ in the plane of $F$. Suppose $p'$ lies outside of $F$. The line through $p$ and $p'$ cuts $P$ in two points $A$ and $B$. Let $A$ be the point between $p$ and $p'$. $A$ belongs to a face different from $F$ and the distance from $p$ to $A$ is less than the distance between $p$ and $F$, which contradicts the minimality of $F$. So $p'$ lies inside $F$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26278,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two players play alternately. The first player is given a pair of positive integers $(x_1, y_1)$. Each player must replace the pair $(x_n, y_n)$ that he is given by a pair of non-negative integers $(x_{n+1}, y_{n+1})$ such that $x_{n+1} = \\min(x_n, y_n)$ and $y_{n+1} = \\max(x_n, y_n) - k \\cdot x_{n+1}$ for some positive integer $k$. The first player to pass on a pair with $y_{n+1} = 0$ wins. Find for which values of $\\frac{x_1}{y_1}$ the first player has a winning strategy.",
    "options": [],
    "answer": "Let phi = (1 + sqrt(5)) / 2. The first player has a winning strategy if and only if x1 = y1 or max(x1, y1) > phi * min(x1, y1). Equivalently, in terms of r = x1 / y1: the first player wins exactly when r < 1/phi, or r = 1, or r > phi.",
    "solution": "Note first that draws are not possible so any position $(x, y)$ is either a win or a loss for the player receiving it. Let $\\phi$ be the positive root of $t^2 - t - 1 = 0$, so $\\phi = \\frac{1+\\sqrt{5}}{2}$. Let $m = \\min(x, y)$, $M = \\max(x, y)$. We show that a player receiving $(x, y)$ wins if and only if $m = M$ or $M > \\phi m$.\n\nIf $m \\neq M$ and $M < \\phi m$, then the player must pass on $(m, M-m) \\neq (m, 0)$ and $m > \\phi(M-m)$. So it is sufficient to show that a player receiving a position with $M > \\phi m$ can either win or pass back a position with $\\min m'$ and $\\max M'$ such that $M' < \\phi m'$ and $M' \\neq m'$.\n\nIf $M > \\phi m$, and $M$ is a multiple of $m$, then the player can pass on $(m, 0)$ and win. So assume $M > \\phi m$ and $M = qm + r$ with $0 < r < m$. If $q \\ge 2$, then the player can choose whether to pass on $(m, r)$ or $(m, m+r)$. If $(m, r)$ is a losing position, then he wins by passing that. If it is a winning position, then\n\nhe passes $(m, m+r)$ to the other player. The other player is now forced to pass back $(m, r)$, which is a winning position. So we may assume $q = 1$. Now the player passes $(m, r)$. We claim that $m < \\phi r$ and $m \\neq r$. Certainly $m \\le r$. But $m+r > \\phi m$, so $(\\phi-1)m < r$, so $m < \\phi r$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26279,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given points $A_1 = (x_1, y_1, z_1)$, $A_2 = (x_2, y_2, z_2)$, ..., $A_n = (x_n, y_n, z_n)$ let $P = (x, y, z)$ be the point which minimizes $\\sum_{i=1}^n (|x - x_i| + |y - y_i| + |z - z_i|)$. Give an example (for each $n > 4$) of points $A_i$ for which the point $P$ lies outside the convex hull of the points $A_i$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose $n$ is $3m$. Take the points to be\n$$\n(1, 0, 0), (2, 0, 0), \\dots, (m, 0, 0), \\\\\n(0, 1, 0), (0, 2, 0), \\dots, (0, m, 0), \\\\\n(0, 0, 1), (0, 0, 2), \\dots, (0, 0, m)\n$$\nTake $P$ to be $(x, y, z)$. Then the sum is $2m|x| + |x-1| + |x-2| + \\dots + |x-m|$ plus similar terms in $y$ and $z$. Evidently we can minimize for $x, y, z$ separately. Looking at the terms in $x$, it is clear that if we increase $|x|$ by $k > 0$, then we increase $2m|x|$ by $2mk$. At most we can reduce the other terms by $mk$, so we increase the sum by at least $mk$. Similarly, if we reduce $|x|$ by $k > 0$, we reduce the sum by at least $mk$. So to minimize the sum we must take $|x| = 0$. Similarly for $y$ and $z$. Hence $P$ is at the origin which is outside the convex hull.\nIf $n$ is $3m-1$ or $3m-2$, then we drop $(0, 0, m)$ and/or $(0, m, 0)$. Then the argument is strengthened for the $y$ and $z$ coordinates. For the $x$ coordinate we have at worst $(2m-2)|x| + |x| + |x-1| + \\dots + |x-m|$. If $n > 6$, then $m > 2$, then $2m-2 > m$ and the same argument works. If $n = 5$, then we have $3|x| + |x| + |x-1|$ for the $x$-coordinate, and the same argument works.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26280,
    "subject": "Mathematics (Multi-modal)",
    "question": "$A$ and $B$ wish to divide a cake into two pieces. Each wants the largest piece he can get. The cake is a triangular prism with the triangular faces horizontal. $A$ chooses a point $P$ on the top face. $B$ then chooses a vertical plane through the point $P$ to divide the cake. $B$ chooses which piece to take. Which point $P$ should $A$ choose in order to secure as large a slice as possible?",
    "options": [],
    "answer": "Choose the centroid of the triangular face (this guarantees at least 4/9 of the cake).",
    "solution": "Given any triangle $ABC$, we have to find the point $P$ inside the triangle, such that any line through $P$ divides the triangle into two pieces which are as equal as possible. We take $P$ to be the centroid $G$. We show first that in that case the smallest piece is at least $\\frac{4}{9}$ of the total area. If we take a line through $G$ parallel to one of the sides, then the triangular piece has area $\\frac{4}{9}$ of the total area.\n\n![](attached_image_1.png)\n\nTake any other line through $G$, say $RS$ as shown above. We claim that $RG > GS$. Take $S'$ to be the point obtained by rotating $S$ through $180^\\circ$ about $G$. Then $GSE$ and $GS'D$ are congruent and $SE$ is parallel to $S'D$. So $R$ cannot coincide with $S'$ (because the lines $SE$ and $RD$ are not parallel, they meet at $A$). If $GR < GS'$, then $RD$ and $SE$ will meet on the wrong side of $DE$. So we must have $GR > GS'$ and hence $GR > GS$. The triangles $GRD$ and $GSE$ have the same height ($GD \\sin \\angle DGR$), so area $GRD >$ area $GSE$. Hence area $ARS >$ area $ADE = \\frac{4}{9}$ area $ABC$.\n\nLet $N$ be the midpoint of $AC$. As we move $R$ towards $B$, $S$ moves towards $A$. When $R$ reaches $B$, $S$ reaches $N$. So $S$ lies between $E$ and $N$. Now consider the triangles $BGR$, $NGS$. $BGR$ has the larger base, because $BG = 2GN$, and the larger height because $RG > SG$, so $RG \\sin \\angle RGB > SG \\sin \\angle SGN$. So area $BGR >$ area $NGS$ and hence area $ABN >$ area $ARS$. So area $ARS < \\frac{1}{2}$ area $ABC$. Thus $ARS$ is the smaller piece but its area is bigger than $\\frac{4}{9}$ area $ABC$. It remains to show that no other choice of $P$ is better than $G$.\n\n![](attached_image_2.png)\n\nTake lines through $G$ parallel to the sides. That gives three overlapping triangles which cover $ABC$. If $P$ is not at $G$, then it must lie inside at least one of these triangles, say $ADE$. Now take a line $D'E'$ through $P$ parallel to $DE$. Then area $AD'E' < $ area $ADE = \\frac{4}{9}$ area $ABC$, so $P$ is a worse choice than $G$ (from $A$'s point of view).",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26281,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all primes which can be written both as a sum of two primes and as a difference of two primes.",
    "options": [],
    "answer": "5",
    "solution": "The smallest prime is $2$, so it cannot be written as the sum of two primes. All other primes $p$ are odd, so if $p$ can be written as the sum of two primes and the difference of two primes, then in each case one of the primes must be $2$. Thus we have $p = q - 2 = r + 2$ for some primes $q, r$.\nIf $r \\equiv 1 \\pmod{3}$, then $p$ is a multiple of $3$, so $p$ must be $3$. But then $r = 1$, which is not a prime. So there are no solutions with $r \\equiv 1 \\pmod{3}$. If $r \\equiv 2 \\pmod{3}$, then $q$ is a multiple of $3$, so $q$ must be $3$, but then $p = 1$, which is not a prime. So there are no solutions with $r \\equiv 2 \\pmod{3}$. Hence $r$ must be a multiple of $3$. But $r$ is prime, so $r$ must be $3$. Hence $q = 7$ and $p = 5$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26282,
    "subject": "Mathematics (Multi-modal)",
    "question": "$P$ is a fixed point in the plane. $A$, $B$, $C$ are points such that $PA = 3$, $PB = 5$, $PC = 7$ and the area $ABC$ is as large as possible. Show that $P$ must be the orthocenter of $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider all points $A'$ such that $PA' = 3$. They lie on a circle with center $P$. The area of $A'BC$ is $\\frac{BC}{2}$ times the distance of $A'$ from $BC$. That distance is maximal for $A'P$ perpendicular to $BC$ (because the distance is the distance of $P$ from $BC$ is $PA' \\sin \\theta$, where $\\theta$ is the angle between $A'P$ and $BC$). Hence $AP$ must be perpendicular to $BC$. Similarly $BP$ must be perpendicular to $AC$, so $P$ must be the orthocenter.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26283,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $N$ be the natural numbers and $N' = N \\setminus \\{0\\}$. Find all functions $f: N' \\to N$ such that $f(xy) = f(x) + f(y)$, $f(30) = 0$ and $f(x) = 0$ for all $x \\equiv 7 \\pmod{10}$.",
    "options": [],
    "answer": "f(x) = 0 for all positive integers x",
    "solution": "$f(30) = f(2) + f(3) + f(5) = 0$ and $f(n)$ is non-negative, so $f(2) = f(3) = f(5) = 0$. For any positive integer $n$ not divisible by $2$ or $5$ we can find a positive integer $m$ such that $mn \\equiv 7 \\pmod{10}$. But then $f(mn) = 0$, so $f(n) = 0$.\nIt is a trivial induction that $f(2^a 5^b n) = f(5^b n) = f(n)$, so $f$ is identically zero.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26284,
    "subject": "Mathematics (Multi-modal)",
    "question": "A figure on a computer screen shows $n$ points on a sphere, no four coplanar. Some pairs of points are joined by segments. Each segment is colored red or blue. For each point there is a key that switches the colors of all segments with that point as endpoint. For every three points there is a sequence of key presses that makes the three segments between them red. Show that it is possible to make all the segments on the screen red. Find the smallest number of key presses that can turn all the segments red, starting from the worst case.",
    "options": [],
    "answer": "floor(n/2)",
    "solution": "Consider three of the $n$ points. The parity of the number of blue segments of the triangle with these points as vertices doesn't change while switching the keys. Since it is possible to make all three segments red, the number of blue segments in each triangle is even.\n\nLet $P$ be one of the $n$ points. Let $A$ be the set of points connected to $P$ by red segments and $B$ be the set of points connected to $P$ by blue segments. Let $A_1, A_2 \\in A$. So $PA_1$ and $PA_2$ are both red and thus $A_1A_2$ is red. Now consider $B_1, B_2 \\in B$. Then $PB_1$ and $PB_2$ are both blue and thus $B_1B_2$ is red. Finally, consider $A \\in A$ and $B \\in B$. $PA$ is red and $PB$ is blue, so $AB$ is blue. Put $P$ in $A$. All this reasoning shows that segments in the same set are red and segments connecting points in different sets are blue.\n\nSwitching all points in set $A$ will make all segments red. Indeed, all segments in $A$ will change twice, one time from each of its edges, all segments connecting points from $\\mathcal{A}$ and $\\mathcal{B}$ will change once, turning from blue to red and segments in $\\mathcal{B}$ won't change. This proves the first part.\n\nFor the second part, notice first that one needs to switch each point at most once. Let $|\\mathcal{A}| = k$ and $|\\mathcal{B}| = n-k$. If we switch $a$ points from $\\mathcal{A}$ and $b$ points from $\\mathcal{B}$ we change at most $a(n-k) + bk$ blue segments. Suppose without loss of generality that $k \\le n-k \\iff k \\le \\lfloor \\frac{n}{2} \\rfloor$. Then $k(n-k) \\le a(n-k) + bk \\le a(n-k) + b(n-k) \\implies k \\le a+b$. So the number of key presses is at most $k$ and, in the worst case, $\\lfloor \\frac{n}{2} \\rfloor$. This number is needed to make all segments red if $|\\mathcal{A}| = \\lfloor \\frac{n}{2} \\rfloor$.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 26285,
    "subject": "Mathematics (Multi-modal)",
    "question": "The triangle with vertices $(0, 0)$, $(0, 1)$, $(2, 0)$ is repeatedly reflected in the three lines $AB$, $BC$, $CA$ where $A$ is $(0, 0)$, $B$ is $(3, 0)$, $C$ is $(0, 3)$. Show that one of the images has vertices $(24, 36)$, $(24, 37)$ and $(26, 36)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $R$ be reflection in $BC$: $(x, y) \\rightarrow (3 - y, 3 - x)$. Let $S$ be reflection in $AC$: $(x, y) \\rightarrow (-x, y)$. Let $T$ be reflection in $AB$: $(x, y) \\rightarrow (x, -y)$. Let $RS$ denote the reflection $R$ followed by the reflection $S$. Then we have $Q = STR$ is $(x, y) \\rightarrow (y + 3, x + 3)$. So $Q^2$ is $(x, y) \\rightarrow (x + 6, y + 6)$, and $Q^{10}$ is $(x, y) \\rightarrow (x + 30, y + 30)$.\n$P = TRS$ is $(x, y) \\rightarrow (x, -y) \\rightarrow (y+3, -x+3) \\rightarrow (-y-3, -x+3)$. So $P^2$ is $(x, y) \\rightarrow (x-6, y+6)$. Hence $Q^{10}P^2$ is $(x, y) \\rightarrow (x+24, y+36)$, which is the required translation.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26286,
    "subject": "Mathematics (Multi-modal)",
    "question": "$n$ is a positive integer such that $\\frac{n(n+1)}{3}$ is a square. Show that $n$ is a multiple of $3$, and $n+1$ and $\\frac{n}{3}$ are squares.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose $\\frac{n(n+1)}{3} = m^2$, so $n(n+1) = 3m^2$. $n$ and $n+1$ are coprime so either\n\n(1) $n = 3a^2$ and $n+1 = b^2$, or\n\n(2) $n = a^2$ and $n+1 = 3b^2$.\n\nBut (2) implies that $3b^2 - 1$ is a square, whereas squares must be $0$ or $1$ mod $3$. So we must have (1).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26287,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $Z$ be the set of integers. $f: Z \\rightarrow Z$ is defined by $f(n) = n - 10$ for $n > 100$ and $f(n) = f(f(n+11))$ for $n = 100$. Find the set of possible values of $f$.",
    "options": [],
    "answer": "all integers greater than or equal to 91",
    "solution": "We show that $f(n) = 91$ for $n \\le 100$. For $n = 100, 99, \\dots, 90$, we have $f(n) = f(f(n+11)) = f(n+11-10) = f(n+1)$. But $f(101) = 91$, so $f(n) = 91$ for $n = 100, 99, \\dots, 90$. Now for $n = 89, 88, \\dots, 1, 0, -1, \\dots$ we use induction. We have $f(n) = f(f(n+11)) = f(91)$ which, by induction, is 91. So the range of $f$ is all the integers greater than or equal to 91.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26288,
    "subject": "Mathematics (Multi-modal)",
    "question": "$A$ and $B$ play a game. Each has $10$ tokens numbered from $1$ to $10$. The board is two rows of squares. The first row is numbered $1$ to $1492$ and the second row is numbered $1$ to $1989$. On the $n$th turn, $A$ places his token number $n$ on any empty square in either row and $B$ places his token on any empty square in the other row. $B$ wins if the order of the tokens is the same in the two rows, otherwise $A$ wins. Which player has a winning strategy? Suppose each player has $k$ tokens, numbered from $1$ to $k$. Who has the winning strategy? What if both rows are all the integers? Or both all the rationals?",
    "options": [],
    "answer": "With ten tokens on rows of lengths one thousand four hundred ninety two and one thousand nine hundred eighty nine, player B has a winning strategy. For k tokens on this board, B wins if k is at most ten and A wins if k is greater than ten. If both rows are all integers, B has a winning strategy. If both rows are all rationals, B has a winning strategy.",
    "solution": "Note that $B$ will lose if he does not space his tokens widely enough. Call the rows $R$ and $S$, so that $S^n$ denotes the number $n$ in row $S$. Suppose $A$ plays $1$ on $R5$, then $2$ on $R6$. If $B$ plays $1$ on $S5$ and $2$ on $S7$, then he loses, because $A$ swaps rows and plays $3$ on $S6$. However, $B$ does not need to make this mistake, so $A$ can only force a win by always playing in the longer row, until $B$ runs out of space to match him in the shorter row. So $A$ places each token as centrally as possible in one the shortest gap. $B$ must match him. The table\n\n| after | 1989 | 1492 |\n|-------|------|------|\n| 1     | 994  | 745  |\n| 2     | 496  | 372  |\n| 3     | 247  | 185  |\n| 4     | 123  | 92   |\n| 5     | 61   | 45   |\n| 6     | 30   | 22   |\n| 7     | 14   | 10   |\n| 8     | 6    | 4    |\n| 9     | 2    | 1    |\n\nEvidently $B$ can always play the first $10$ tokens. But $A$ can fit both tokens $10$ and $11$ into his gap of $2$, whereas $B$ can only fit token $10$. So $B$ loses for more than $10$ tokens, and wins for at most $10$ tokens.\n\nThe only way $B$ can lose is if he runs out of space to put his tokens. Clearly he cannot run out of space with the rationals, because there is always a rational between any two given rationals. Similarly, he should not run out of space with all the integers. He just copies $A$ in the other row, so that the tokens numbered $k$ are always placed on the same number in each row.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26289,
    "subject": "Mathematics (Multi-modal)",
    "question": "The circumcenter of a tetrahedron lies inside the tetrahedron. Show that at least one of its edges is at least as long as the edge of a regular tetrahedron with the same circumsphere.\n(1989)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "There are two main steps in the proof: first, we'll prove a tridimensional analogous to the famous $R \\ge 2r$ inequality; then, we'll prove the bidimensional analogous to the problem. Combining these two facts and other arguments will lead us to a solution.\n\nThe first step is proving that if $R$ and $r$ are the circumradius and the inradius of the circumscribed and inscribed spheres of a tetrahedron $T$ then $R \\ge 3r$.\n\nTo do this, consider the tetrahedron $T'$ whose vertices are the centroids of the faces of $T$. Then $T'$ is similar to $T$, with ratio $1 : 3$ (the faces of $T'$ are parallel to the faces of $T$ and the ratio of the altitudes is $1 : 3$). The circumradius of $T'$, which is $\\frac{R}{3}$, must be bigger than $r$, because the smallest sphere through four points, one in each face of $T$, is the inscribed sphere. So $R \\ge 3r$.\n\nNow we prove that if $ABC$ is not obtusangle and has all sides smaller than the side $l$ of an equilateral triangle then the circumradius $R$ of $ABC$ is less than the circumradius $R'$ of the equilateral triangle. Indeed, by the sine law, if $a$ is the greater side of $ABC$ then $60^{\\circ} \\le \\angle A \\le 90^{\\circ}$ and $2R = \\frac{a}{\\sin \\angle A} < \\frac{l}{\\sin 60^{\\circ}} = 2R'$. Let $T$ be a tetrahedron and $U$ be the regular tetrahedron with edge length $a$ and same circumradius $R$ as $T$. Let $O$ be the circumcenter of both $T$ and $U$ and suppose $ABC$ is the face nearer to $O$. Since $O$ is inside $T$, by problem 2 of 1987 its projection $O'$, which is the circumcenter of $ABC$, lies inside $ABC$, so $ABC$ is not obtusangle. Thus its circumradius $R'$ is less than the circumradius $R''$ of the equilateral triangle with side $a$.\n\nIf $d_1 \\le d_2 \\le d_3 \\le d_4$ are the distances from $O$ to each face ($d_1$ being the distance $OO'$ from $O$ to $ABC$) and $S_1, S_2, S_3, S_4$ the corresponding areas of the faces, then the volume $V$ of $T$ is such that\n$$\n3V = d_1S_1 + d_2S_2 + d_3S_3 + d_4S_4 = r(S_1 + S_2 + S_3 + S_4) \\\\ \\implies r = \\frac{S_1d_1 + S_2d_2 + S_3d_3 + S_4d_4}{S_1 + S_2 + S_3 + S_4},\n$$\nthat is, $r$ is the weighted mean of $d_1, d_2, d_3, d_4$ (the weights are the $S_i$'s). This means that the smallest value $d_1$ is not bigger than $r$, that is, $d_1 \\le r$.\n\nThe triangle formed by $O$, its projection $O'$ and $A$ is right-angled in $O'$. So $d_1^2 = R^2 - R'^2$ and then $R^2 - R'^2 \\le r^2 \\iff R'^2 \\ge R^2 - r^2$. But, by looking at $U$, $a = \\sqrt{3}R'' = \\frac{2}{3}\\sqrt{6}R \\implies R''^2 = \\frac{8}{9}R^2$. But $R'' > R'$, so $\\frac{8}{9}R^2 > R^2 - r^2 \\iff R < 3r$, a contradiction. So one of the sides of $ABC$ does not exceed $a$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 26290,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that a convex polyhedron with an odd number of faces has at least one face with an even number of edges.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Count the number $N$ of pairs $(F, e)$, where $F$ is a face and $e$ an edge belonging to $F$. Each edge belongs to two faces, so $N$ is even. Hence the number of faces with an odd number of edges must be even. The total number of faces is odd, so the number of faces with an even number of edges must be odd. In particular, there is at least one.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 26291,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that there are infinitely many positive integer solutions to $a^3 + 1990b^3 = c^4$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Take, for instance, $a = b = 1991k^4$ and $c = 1991k^3$ (there are lots of other solutions; can you find them all?).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26292,
    "subject": "Mathematics (Multi-modal)",
    "question": "Each face of a tetrahedron is a triangle with sides $a$, $b$, $c$ and the tetrahedron has circumradius $1$. Find $a^2 + b^2 + c^2$.",
    "options": [],
    "answer": "8",
    "solution": "Opposite edges of the tetrahedron have the same length. Let $AB$, $CD$ have length $a$. Take parallel planes through $AB$ and $CD$. Project $A$, $B$ onto the other plane, and $C$, $D$ onto the other plane. Then we get $8$ points at the vertices of a rectangular block. If the sides have lengths $x$, $y$, $z$, then we have $a^2 = y^2 + z^2$, $b^2 = x^2 + z^2$, $c^2 = x^2 + y^2$. A long diagonal of the block has length $\\sqrt{x^2 + y^2 + z^2}$ and is a diameter of the circumsphere. So $x^2 + y^2 + z^2 = 4$. Hence $a^2 + b^2 + c^2 = 2(x^2 + y^2 + z^2) = 8$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26293,
    "subject": "Mathematics (Multi-modal)",
    "question": "$ABCD$ is a convex quadrilateral. $E, F, G, H$ are the midpoints of sides $AB, BC, CD, DA$ respectively. Find the point $P$ such that area $PHAE = \\text{area } PEBF = \\text{area } PFCG = \\text{area } PGDH$.",
    "options": [],
    "answer": "Let M and N be the midpoints of AC and BD. Take P as the intersection of the line through M parallel to BD and the line through N parallel to AC. For this P, the four areas are equal (each equals one quarter of the area of ABCD).",
    "solution": "Let $M$ be the midpoint of $AC$, $N$ the midpoint of $BD$. Take $P$ so that $PM$ is parallel to $BD$ and $PN$ is parallel to $AC$. Now $EM$ is a midline of $\\triangle ABC$, so $\\text{area } AEM = \\frac{\\text{area } ABC}{4}$. Similarly, $\\text{area } AHM = \\frac{\\text{area } ADC}{4}$. So $\\text{area } MHAE = \\frac{\\text{area } ABCD}{4}$. But $MP$ is parallel to $EH$, so $\\text{area } MEH = \\text{area } PEH$. Hence\n\n$$\n\\begin{align*}\n\\text{area } PHAE &= \\text{area } PEH + \\text{area } AEH \\\\\n&= \\text{area } MEH + \\text{area } AEH \\\\\n&= \\text{area } MHAE = \\frac{\\text{area } ABCD}{4}\n\\end{align*}\n$$\n\nSimilarly for $\\text{area } PEBF$ and the other quadrilaterals.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26294,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given that $f(x) = \\frac{ax+b}{cx+d}$, $f(0) \\neq 0$, $f(f(0)) \\neq 0$. Let $F(x) = \\underbrace{f(\\dots(f(x)\\dots))}_{n \\text{ times}}$. If $F(0) = 0$, show that $F(x) = x$ for all $x$ where the expression is defined.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Put $f^k(x) = \\underbrace{f(\\dots(f(x))\\dots)}_{k \\text{ times}}$. Then we have $f^k(x) = \\frac{a_k x + b_k}{c_k x + d_k}$. From\n$f^{k+1}(x) = f(f^k(x))$ we get:\n$$\n\\begin{align*} \na_{k+1} &= a a_k + b c_k \\\nb_{k+1} &= a b_k + b d_k \\\nc_{k+1} &= c a_k + d c_k \\\nd_{k+1} &= c b_k + d d_k \n\\end{align*}\n$$\nBut we also have $f^{k+1}(x) = f^k(f(x))$, so\n$$\n\\begin{align*} \na_{k+1} &= a a_k + c b_k \\\nb_{k+1} &= b a_k + d b_k \\\nc_{k+1} &= a c_k + c d_k \\\nd_{k+1} &= b c_k + d d_k \n\\end{align*}\n$$\nComparing, we get $b c_k = c b_k$ and $(a-d) b_k = b(a_k - d_k)$.\nIf $f^n(0) = 0$, then $b_n = 0$, so $b c_n = 0$. We are given that $f(0) \\neq 0$, so $b \\neq 0$. Hence $c_n = 0$. Also $b(a_n - d_n) = 0$ and hence $a_n = d_n$. Hence $f^n(x) = x$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 26295,
    "subject": "Mathematics (Multi-modal)",
    "question": "At a party every woman dances with at least one man, and no man dances with every woman. Show that there are men $M$ and $M'$ and women $W$ and $W'$ such that $M$ dances with $W$, $M'$ dances with $W'$, but $M$ does not dance with $W'$, and $M'$ does not dance with $W$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $M$ be one of the men who dance with the maximal number of women, $W'$ one of the women he doesn't dance with, and $M'$ one of the men $W'$ dances with. If $M'$ were to dance with every woman that $M$ dances with, then the maximality of the number of women that $M$ dances with would be contradicted, so there is a woman $W$ that dances with $M$ but not with $M'$, and we're done.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26296,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given $k > 0$, the sequence $a_1, a_2, a_3, \\dots$ is defined by its first two members and $a_{n+2} = a_{n+1} + \\frac{k}{n} a_n$. For which $k$ can we write an $a_n$ as a polynomial in $n$? For which $k$ can we write $\\frac{a_{n+1}}{a_n} = \\frac{p(n)}{q(n)}$, where $p$ and $q$ are polynomials?",
    "options": [],
    "answer": "First: k must be a nonnegative integer. Second: k must be an integer.",
    "solution": "Rewrite the equation as $\\frac{n(a_{n+2}-a_{n+1})}{a_n} = k$. If $a_n$ is a polynomial in $n$, say $a_n = r(n) = \\sum_{i=0}^{d} r_i n^i$, then the limit $\\lim_{n \\to \\infty} \\frac{n(a_{n+2}-a_{n+1})}{a_n}$ must exist and be equal to $k$. But\n$$\n\\begin{align*}\n& \\lim_{n \\to \\infty} \\frac{n(a_{n+2} - a_{n+1})}{a_n} \\\\\n&= \\lim_{n \\to \\infty} \\frac{n \\left( \\sum_{i=0}^{d} r_i (n+2)^i - \\sum_{i=0}^{d} r_i (n+1)^i \\right)}{\\sum_{i=0}^{d} r_i n^i} \\\\\n&= \\lim_{n \\to \\infty} \\frac{n \\sum_{i=0}^{d} r_i ((n+2)^i - (n+1)^i)}{\\sum_{i=0}^{d} r_i n^i} \\\\\n&= \\lim_{n \\to \\infty} \\frac{n \\sum_{i=0}^{d} r_i ((n+2)^{i-1} + (n+2)^{i-2}(n+1) + \\dots + (n+1)^{i-1})}{\\sum_{i=0}^{d} r_i n^i} \\\\\n&= \\lim_{n \\to \\infty} \\frac{n^d \\sum_{i=0}^{d} r_i n^{i-d} \\left( \\left(1+\\frac{2}{n}\\right)^{i-1} + \\left(1+\\frac{2}{n}\\right)^{i-2} \\left(1+\\frac{1}{n}\\right) + \\dots + \\left(1+\\frac{1}{n}\\right)^{i-1} \\right)}{n^d \\sum_{i=0}^{d} r_i n^{i-d}} \\\\\n&= \\lim_{n \\to \\infty} \\frac{r_d \\left( \\left(1+\\frac{2}{n}\\right)^{d-1} + \\left(1+\\frac{2}{n}\\right)^{d-2} \\left(1+\\frac{1}{n}\\right) + \\dots + \\left(1+\\frac{1}{n}\\right)^{d-1} \\right)}{r_d} = d\n\\end{align*}\n$$\nSo if $a_n$ is a polynomial in $n$ then $k$ is the degree of $a_n$ and is a nonnegative integer.\n\nNow we need to show an example of polynomial $r_k(n)$ for each nonnegative integer $k$. For $k = 0$ and $k = 1$, one may consider $r_0(n) = 0$ and $r_1(n) = n$. For $k \\ge 2$, let's find a polynomial\n$$\nr(n) = n^k + r_{k-1} n^{k-1} + \\cdots + r_1 n + r_0\n$$\nsuch that $n(r(n+2) - r(n+1)) = kr(n)$ for all $n$.\nExpanding $(n + 2)^i$ and $(n + 1)^i$ via binomial theorem, we obtain the system of equations\n$$\n\\begin{align*}\n& 3 \\binom{k}{k-2} + \\binom{k-1}{k-2} r_{k-1} = k r_{k-1} \\\\\n& 7 \\binom{k}{k-3} + 3 \\binom{k-1}{k-3} r_{k-1} + \\binom{k-2}{k-3} r_{k-2} = k r_{k-2} \\\\\n& \\qquad \\dots \\\\\n& (2^j - 1) \\binom{k}{k-j} + (2^{j-1} - 1) \\binom{k-1}{k-j} r_{k-1} + \\dots + \\binom{k-j+1}{k-j} r_{k-j+1} = k r_{k-j+1} \\\\\n& \\qquad \\dots\n\\end{align*}\n$$\nSince $\\binom{k-j+1}{k-j} = k-j+1 \\neq k$ it is possible to find $r_{k-j+1}$ in each equation and the system is possible, so there is such a polynomial.\n\nFor the second part, rewrite the given equation as $n\\left(\\frac{a_{n+2}}{a_{n+1}} - 1\\right) = k \\frac{a_n}{a_{n+1}}$. If there exists polynomials $p$ and $q$ such that $\\frac{a_{n+1}}{a_n} = \\frac{p(n)}{q(n)}$, then\n$$\nn \\left( \\frac{p(n+1)}{q(n+1)} - 1 \\right) = k \\frac{q(n)}{p(n)} \\iff n(p(n+1) - q(n+1))p(n) = kq(n+1)q(n)\n$$\nWe may suppose without loss of generality that $p(n)$ and $q(n)$ don't have common factors. Let $m$ and $m'$ be the degrees of $p$ and $q$, respectively. If $m = m'$ then the right hand side has degree $2m$ and the degree of $p(n+1) - q(n+1)$ is $m-1$. This means that the leading coefficients of $p$ and $q$ are equal. Since $p(n)$ and $q(n)$ don't have common factors, $p(n)$ divides $q(n+1)$, which in this case means $p(n) = q(n+1)$. Substituting yields $n(q(n+2) - q(n+1)) = kq(n)$, and $q$ satisfies the first part of the problem, and thus $k$ must be a nonnegative integer.\n\nIf $m \\neq m'$ by checking degrees we must have $1 + \\max(m, m') + m = 2m' \\Rightarrow 2m' \\le 1 + m + m' \\iff m < m' \\Rightarrow 1 + m' + m = 2m' \\iff m' = m + 1$. The polynomials $q(n+1)$ and $p(n+1) - q(n+1)$ don't have common factors, so $q(n)$ divides $p(n+1) - q(n+1)$. These two polynomials have both degree $m'$ and opposite leading coefficients, so $p(n+1) = q(n+1) - q(n)$. Substituting yields the polynomial identity $n(q(n) - q(n-1)) = -kq(n+1)$. Now if $r(n) = q(1-n)$ and $m = -n$ then $n(r(m+2) - r(m+1)) = -kr(m)$ and we reduce the problem to the first part again for $-k$ instead of $k$. So, in this case, $k$ must be a nonpositive integer.\n\nConversely, it's not hard to obtain $p(n)$ and $q(n)$ from the above case, so the answer for the second part is $k$ integer.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 26297,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that there is a number of the form $199\\dots 91$ (with $n$ 9s) with $n > 2$ which is divisible by $1991$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "There are many ways to solve the problem, using, for instance, Euler-Fermat theorem. But one student, André Reys Leal, obtained a clever, simple solution: consider all numbers of the form $1999\\dots 91$ with more than two nines. If one of them is multiple of $1991$ we are done. If not, since there are infinite of them there are two that are the same modulo $1991$. By subtracting them, we obtain $1999\\dots 98000\\dots 0$, which is a multiple of $1991$. We may ignore all rightmost zeros, but we do keep three of them, obtaining $1999\\dots 98000$, which is still a multiple of $1991$. Sum $1991$ to it and we are done.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26298,
    "subject": "Mathematics (Multi-modal)",
    "question": "$P_0 = (1,0), P_1 = (1,1), P_2 = (0,1), P_3 = (0,0)$. $P_{n+4}$ is the midpoint of $P_n P_{n+1}$. $Q_n$ is the quadrilateral $P_n P_{n+1} P_{n+2} P_{n+3}$. $A_n$ is the interior of $Q_n$. Find $\\bigcap_{n \\ge 0} A_n$.",
    "options": [],
    "answer": "(3/7, 4/7)",
    "solution": "Interpreting the points $P_k$ as vectors, we have $P_{n+4} = \\frac{P_n + P_{n+1}}{2}$, which is a linear homogeneous recursion. Its characteristic polynomial is $2x^4 - x - 1 = (x - 1)(2x^3 + 2x^2 + 2x + 1)$. So let $\\alpha, \\beta, \\gamma$ be the roots of $f(x) = 2x^3 + 2x^2 + 2x + 1$, so that $P_n = Q_0 + Q_1 \\alpha^n + Q_2 \\beta^n + Q_3 \\gamma^n$, $Q_0, Q_1, Q_2, Q_3$ being constant vectors.\n\nWe have $\\alpha^2 + \\beta^2 + \\gamma^2 = (\\alpha + \\beta + \\gamma)^2 - 2(\\alpha\\beta + \\beta\\gamma + \\gamma\\alpha) = \\left(-\\frac{2}{2}\\right)^2 - 2 \\cdot \\frac{2}{2} = -1 < 0$, so $f$ has one real root and two conjugate complex roots. Suppose $\\alpha$ is real. Since $f(-1) = -1 < 0$ and $f\\left(-\\frac{1}{2}\\right) = \\frac{1}{4} > 0$, $-1 < \\alpha < -\\frac{1}{2}$ and $|\\beta| = |\\gamma| = \\sqrt{| -\\frac{1}{2} |} < 1$, so all roots of $f$ have modulus smaller than 1 and so $P_n$ tends to $Q_0$ as $n$ goes to infinity. This means that $Q_n$ tends to $Q_0$ and thus the intersection of all $Q_n$ is $Q_0$. To find $Q_0$, notice that\n\n$$\n\\left| \n\\begin{array}{l}\nP_0 = Q_0 + Q_1 + Q_2 + Q_3 \\\\\n2P_1 = 2Q_0 + Q_1 \\cdot 2\\alpha + Q_2 \\cdot 2\\beta + Q_3 \\cdot 2\\gamma \\\\\n2P_2 = 2Q_0 + Q_1 \\cdot 2\\alpha^2 + Q_2 \\cdot 2\\beta^2 + Q_3 \\cdot 2\\gamma^2 \\\\\n2P_3 = 2Q_0 + Q_1 \\cdot 2\\alpha^3 + Q_2 \\cdot 2\\beta^3 + Q_3 \\cdot 2\\gamma^3 \\\\\n\\Rightarrow P_0 + 2P_1 + 2P_2 + 2P_3 = 7Q_0 + f(\\alpha)Q_1 + f(\\beta)Q_2 + f(\\gamma)Q_3 = 7Q_0 \\\\\n\\iff Q_0 = \\frac{P_0 + 2P_1 + 2P_2 + 2P_3}{7} = \\left(\\frac{3}{7}, \\frac{4}{7}\\right)\n\\end{array}\n\\right.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26299,
    "subject": "Mathematics (Multi-modal)",
    "question": "The polynomial $x^3 + px + q$ has three distinct real roots. Show that $p < 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The sum of the squares of the roots $\\alpha, \\beta, \\gamma$ is $s_2 = \\alpha^2 + \\beta^2 + \\gamma^2 = (\\alpha + \\beta + \\gamma)^2 - 2(\\alpha\\beta + \\beta\\gamma + \\gamma\\alpha) = 0^2 - 2p$. Since the roots are distinct, $s_2 > 0 \\iff p < 0$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 26300,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that there is a positive integer $n$ such that the first 1992 digits of $n^{1992}$ are 1.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We only need to prove that there are positive integers $n$, $k$ such that\n$$\n\\underbrace{11 \\dots 11}_{1992} \\cdot 10^k \\le n^{1992} < \\underbrace{11 \\dots 1}_{1991} 2 \\cdot 10^k \\\\\n\\iff \\sqrt[1992]{11 \\dots 11} \\cdot 10^t \\le n < \\sqrt[1992]{11 \\dots 12} \\cdot 10^t\n$$\nin which $k = 1992t$.\nSuppose the decimal expansions of $\\sqrt[1992]{11 \\dots 11}$ and $\\sqrt[1992]{11 \\dots 12}$ differ in the $t$th position. So\n$$\nn = \\left\\lceil \\sqrt[1992]{\\underbrace{11 \\dots 11}_{1992}} \\cdot 10^t \\right\\rceil + 1\n$$\nsuffices.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26301,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given positive real numbers $x_1, x_2, \\dots, x_n$ find the polygon $A_0A_1 \\dots A_n$ with $A_0A_1 = x_1$, $A_1A_2 = x_2$, $\\dots$, $A_{n-1}A_n = x_n$ which has the greatest area.",
    "options": [],
    "answer": "The unique maximizer is the convex polygon whose vertices lie on a semicircle with endpoints at the first and last vertices, so that each interior vertex subtends a right angle with these endpoints (that is, it is inscribed in a semicircle with diameter A0An).",
    "solution": "The answer is the convex polygon inscribed in a semicircle of diameter $A_0A_n$, that is, $\\angle A_0A_iA_n = 90^\\circ$ for all $i = 1, 2, \\dots, n-1$.\n\nFirst of all, the polygon must be convex because if it isn't the case one can obtain a polygon of greater area by reflecting the sides of the polygon inside of its convex hull.\n\nConsider an arbitrary vertex $A_i$ and fix the polygons $A_0A_1 \\dots A_i$, $A_iA_{i+1} \\dots A_n$; in particular, $A_0A_i$ and $A_iA_n$ are fixed. Now the area of $A_0A_1A_2 \\dots A_n$ is the sum of the areas of the fixed polygons $A_0A_1 \\dots A_i$, $A_iA_{i+1} \\dots A_n$ and the triangle $A_0A_iA_n$, which attains its maximum if and only if $\\angle A_0A_iA_n = 90^\\circ$, since the area of the triangle is $\\frac{1}{2}A_iA_0 \\cdot A_iA_n \\sin \\angle A_0A_iA_n$ and $A_iA_0$ and $A_iA_n$ are both fixed.\n\nIt remains to show that such polygon is unique. Indeed, if $A_0A_n = 2r$ and $O$ is the midpoint of $A_0A_n$ and center of the semicircle, then $\\angle A_iOA_{i+1} = 2 \\arcsin \\frac{x_i+x_{i+1}}{2r}$. But $\\sum_{i=1}^n 2 \\arcsin \\frac{x_i+x_{i+1}}{2r} = \\pi$ and $\\arcsin$ is increasing in $[0, 1]$, so $r$ is unique and hence the polygon is unique.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26302,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $d(n)$ be the number of positive divisors of $n$. Show that\n$$\nn\\left(\\frac{1}{2} + \\frac{1}{3} + \\dots + \\frac{1}{n}\\right) \\le d(1) + d(2) + \\dots + d(n) \\le n\\left(1 + \\frac{1}{2} + \\frac{1}{3} + \\dots + \\frac{1}{n}\\right)\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let's count the number $N$ of pairs $(j, k)$ such that $j$ divides $k$ and $1 \\le j, k \\le n$. Fixing $j$, we must count the multiples of $j$, which is $\\lfloor \\frac{n}{j} \\rfloor$. So\n$$\nN = \\lfloor \\frac{n}{1} \\rfloor + \\lfloor \\frac{n}{2} \\rfloor + \\lfloor \\frac{n}{3} \\rfloor + \\dots + \\lfloor \\frac{n}{n} \\rfloor\n$$\nFixing $k$, we must count the number of divisors of $k$, which is $d(k)$ and then $N = d(1) + d(2) + \\dots + d(n)$. So\n$$\nd(1) + d(2) + \\dots + d(n) = \\lfloor \\frac{n}{1} \\rfloor + \\lfloor \\frac{n}{2} \\rfloor + \\lfloor \\frac{n}{3} \\rfloor + \\dots + \\lfloor \\frac{n}{n} \\rfloor\n$$\nand since $x - 1 < \\lfloor x \\rfloor \\le x$\n$$\n\\left(\\frac{n}{1} - 1\\right) + \\left(\\frac{n}{2} - 1\\right) + \\dots + \\left(\\frac{n}{n} - 1\\right) < d(1) + d(2) + \\dots + d(n) \\le \\frac{n}{1} + \\frac{n}{2} + \\dots + \\frac{n}{n}\n$$\n$$\nn \\left( \\frac{1}{2} + \\frac{1}{3} + \\dots + \\frac{1}{n} \\right) \\le d(1) + d(2) + \\dots + d(n) \\le n \\left( 1 + \\frac{1}{2} + \\frac{1}{3} + \\dots + \\frac{1}{n} \\right)\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26303,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a set of $n$ elements, find the largest number of subsets such that no subset is contained in any other.",
    "options": [],
    "answer": "binomial(n, floor(n/2))",
    "solution": "Let $A$ have $n$ elements. The number of sequences of chains\n$$\nS_1 \\subset S_2 \\subset \\dots \\subset S_{n-1} \\subset A\n$$\nsuch that $|S_i| = i$ is $n!$: for each permutation $(a_1, a_2, \\dots, a_n)$ of elements from $A$ let $S_i = S_{i-1} \\cup \\{a_i\\}$.\n\nFix a subset $B$ of $A$. If $|B| = k$, then the chains containing $B$\n$$\nS_1 \\subset S_2 \\subset \\dots \\subset B \\subset \\dots S_{n-1} \\subset A\n$$\nis $k!(n-k)!$, because there are $k!$ choices for $S_1, \\dots, S_{k-1}$ and $(n-k)!$ choices for $S_{k+1}, \\dots, S_{n-1}$.\nIf $B$ and $B'$ are such that $B \\not\\subset B'$ and $B' \\not\\subset B$, then every pair of chains, one containing $B$ and other containing $B'$, are disjoint. So if $B_1, B_2, \\dots, B_m$ are subsets such that $B_i \\not\\subset B_j$ and $B_j \\not\\subset B_i$ then, denoting $|B_i| = m_i$,\n$$\n\\sum_{i=1}^{m} m_i!(n - m_i)! \\le n! \\iff \\sum_{i=1}^{n} \\frac{1}{\\binom{n}{m_i}} \\le 1\n$$\nThe maximum value of $\\binom{n}{m}$ is $\\binom{n}{\\lfloor \\frac{n}{2} \\rfloor}$. Hence\n$$\n\\sum_{i=1}^{m} \\frac{1}{\\binom{n}{m_i}} \\le 1 \\implies \\sum_{i=1}^{m} \\frac{1}{\\binom{n}{\\lfloor \\frac{n}{2} \\rfloor}} \\le 1 \\iff m \\le \\binom{n}{\\lfloor \\frac{n}{2} \\rfloor}\n$$\n\nOne example with $m = \\binom{n}{\\lfloor \\frac{n}{2} \\rfloor}$ is considering all subsets with $\\lfloor \\frac{n}{2} \\rfloor$ elements.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26304,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all solutions in positive integers to $n^a + n^b = n^c$.",
    "options": [],
    "answer": "All solutions are n = 2 with a = b and c = a + 1.",
    "solution": "We must have $n > 1$, $c > a$ and $c > b$. Suppose without loss of generality that $a \\le b$. Dividing by $n^a$ we get $1 + n^{b-a} = n^{c-a}$. But $c - a > b - a$, so $n^{c-a} \\ge n \\cdot n^{b-a}$. So we must have $n = 2$, $b - a = 0$ and $c - a = 1$. It is easy to check that that gives a solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26305,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a chess tournament each player plays every other player once. A player gets 1 point for a win, $\\frac{1}{2}$ point for a draw and 0 for a loss. Both men and women played in the tournament and each player scored the same total of points against women as against men. Show that the total number of players must be a square.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $m$ and $w$ be the number of men and women in the tournament. Each game assigns a total of 1 point to the players, the total of points assigned to games between two men and two women are $\\binom{m}{2}$ and $\\binom{w}{2}$. Since each player scored the same total of points against men as against women and the total of games played is $\\binom{m+w}{2}$,\n$$\n2 \\left( \\binom{m}{2} + \\binom{w}{2} \\right) = \\binom{m+w}{2} \\iff m+w = (m-w)^2,\n$$\nthat is, the total number of players, $m+w$, is a perfect square.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26306,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all $n > 5$ for which it is possible to find a convex polyhedron with all $n$ faces congruent such that each face has another face parallel to it.",
    "options": [],
    "answer": "All even integers greater than 5",
    "solution": "Since for each face there is another parallel to it, $n$ must be even. For $n = 4k$, attach two congruent regular pyramids with $2k$ lateral faces by its bases. For $n = 4k + 2$, attach two congruent regular pyramids with $2k + 1$ lateral faces and twist one of them. Possible coordinates for the vertices of this solid are $(\\cos \\frac{\\pi j}{2k+1}, \\sin \\frac{\\pi j}{2k+1}, (-1)^j(1 - \\cos \\frac{\\pi}{2k+1}))$, $j = 1, 2, \\dots, 4k + 1$ and $(0, 0, \\pm (1 + \\cos \\frac{\\pi}{2k+1}))$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26307,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sequence $a_1, a_2, a_3, \\ldots$ is defined by $a_1 = 8$, $a_2 = 18$, $a_{n+2} = a_{n+1} \\cdot a_n$.\nFind all terms which are perfect squares.",
    "options": [],
    "answer": "All terms with indices divisible by 3 (a3, a6, a9, ...).",
    "solution": "We have $a_n = 2^{b_n} 3^{c_n}$, where $b_{n+2} = b_n + b_{n+1}$, and $c_{n+2} = c_n + c_{n+1}$.\n\n$c_1$ and $c_2$ are even, so all $c_n$ are even.\n\n$b_1$ and $b_2$ are odd, so $b_n$ is even for $n$ multiple of $3$.\n\nWe need both $b_n$ and $c_n$ even for $a_n$ to be a perfect square.\n\nSo the answer is all $n$ multiple of $3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26308,
    "subject": "Mathematics (Multi-modal)",
    "question": "A real number with absolute value less than $1$ is written in each cell of an $n \\times n$ array, so that the sum of the numbers in each $2 \\times 2$ square is zero. Show that for $n$ odd the sum of all the numbers is less than $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider a $L$-shaped piece formed by a square and two neighbouring squares.\nThe sum of its numbers is the opposite of the number that completes the $L$\nto a $2 \\times 2$ square, so the sum of numbers in any $L$ is less than $1$.\nWe proceed by induction on $n$. For $n=3$, divide the array in four regions: a $2 \\times 2$ square, an $L$ and two single cells.\n![](attached_image_1.png)\nThe sum is less than $3$, because of the single cells and the $L$.\nFor bigger odd $n$, divide the array in a square of side $n-2$, several squares of side $2$ and a $L$:\n![](attached_image_2.png)\nThe sum of the square of side $n-2$ is less than $n-2$ by induction hypothesis and there are only an $L$ and a single square adding to less than $2$. So the sum is less than $n-2+2=n$ and the proof is complete.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26309,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a circle and its center $O$, a point $A$ inside the circle and a distance $h$, construct a triangle $BAC$ with $\\angle A = 90^\\circ$, $B$ and $C$ on the circle and the altitude from $A$ with length $h$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $H$ on $BC$ such that $AH = h$. Since $BAC$ is a right angle, $BH \\cdot CH = h^2$.\n\nBut the power of $H$ with respect to the given circle is $HB \\cdot HC = R^2 - OH^2$.\n\nSo $OH = \\sqrt{R^2 - h^2}$ is determined and $H$ is the intersection of the circle with center $O$ and radius $\\sqrt{R^2 - h^2}$ and the circle with center $A$ and radius $h$. So $H$ is determined. To determine $B$ and $C$, it is enough to trace a perpendicular to $AH$ passing through $H$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26310,
    "subject": "Mathematics (Multi-modal)",
    "question": "$ABCD$ is a convex quadrilateral with $\\angle BAC = 30^\\circ$, $\\angle CAD = 20^\\circ$, $\\angle ABD = 50^\\circ$, $\\angle DBC = 30^\\circ$. If the diagonals intersect at $P$, show that $PC = PD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We have $\\angle ADB = 80^\\circ$, $\\angle ACB = 70^\\circ$. Applying the sine rule repeatedly we have $PD = \\frac{\\sin 20^\\circ}{\\sin 80^\\circ} PA$, $PC = \\frac{\\sin 30^\\circ}{\\sin 70^\\circ} PB = \\frac{\\sin 30^\\circ}{\\sin 70^\\circ} \\frac{\\sin 30^\\circ}{\\sin 50^\\circ} PA$. So we have to show that $\\sin 30^\\circ \\sin 30^\\circ \\sin 80^\\circ = \\sin 70^\\circ \\sin 50^\\circ \\sin 20^\\circ$. Indeed we have $\\sin 80^\\circ = 2 \\sin 40^\\circ \\cos 40^\\circ = 4 \\sin 20^\\circ \\cos 20^\\circ \\cos 40^\\circ = 4 \\sin 20^\\circ \\sin 70^\\circ \\sin 50^\\circ$ and $\\sin 30^\\circ \\sin 30^\\circ = \\frac{1}{4}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26311,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find a real-valued function $f(x)$ on the non-negative reals such that $f(0) = 0$, and $f(2x + 1) = 3f(x) + 5$ for all $x$.",
    "options": [],
    "answer": "One example is: define f(x)=0 for x in [0,1), and for each integer n ≥ 1 define f(x) = 5(3^n − 1)/2 for x in [2^n − 1, 2^{n+1} − 1).",
    "solution": "Let $I_n$ be the interval $[2^n - 1, 2^{n+1} - 1)$ for $n = 0, 1, 2, \\dots$. Then the $I_n$ are disjoint and cover the non-negative reals. Also $x \\to 2x + 1$ maps $I_n$ onto $I_{n+1}$. Thus $f$ is determined by the values it takes on $I_0 = [0, 1)$. These can be arbitrary, but the simplest is to take $f(x) = 0$ on $[0, 1)$. Then we get $f(x) = 5$ on $[1, 3)$, $20$ on $[3, 7)$, $65$ on $[7, 15)$ and by a simple induction $\\frac{5(3^n-1)}{2}$ on $I_n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26312,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P_1P_2...P_n$ be a polygon inscribed in a circle and $Q$ a point outside the plane of the circle. Let $\\beta_i$ be the plane perpendicular to $QP_i$ and passing through $P_i$, $i = 1, 2, ..., n$. Prove that the intersection of all planes $\\beta_i$ is a point.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $S$ be the sphere that contains the circumcircle of the polygon and $Q$ and let $Q'$ be the point diametrically opposite to $Q$ in $S$. Then $\\angle Q'P_iQ = 90^\\circ$, that is, $Q'P_i$ is perpendicular to $QP_i$ and since $\\beta_i$ contains all lines perpendicular to $QP_i$ passing through $P_i$, $Q'$ belongs to $\\beta_i$.\n\nIt remains to prove that the intersection of the planes is not a line. Let $t$ be the intersection of $\\beta_1$ and $\\beta_2$. It is a line, since the intersection is non-empty. This line is perpendicular to both $QP_1$ and $QP_2$, so it is perpendicular to plane $QP_1P_2$. Analogously, the intersection $u$ of $\\beta_2$ and $\\beta_3$ is perpendicular to the plane $QP_2P_3$. Since the planes $QP_1P_2$ and $QP_2P_3$ are secant in $QP_2$, $t \\neq u$ and the intersection of all planes is the intersection of $t$ and $u$, which is a single point.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26313,
    "subject": "Mathematics (Multi-modal)",
    "question": "The edges of a cube are labeled from $1$ to $12$ in an arbitrary manner. Show that it is not possible to get the sum of the edges at each vertex the same. Show that we can get eight vertices with the same sum if one of the labels is changed to $13$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Each edge contributes to the total sum twice, one for each of its vertices. So if each vertex has sum $v$, the sum of all numbers is $8v = 2(1+2+\\cdots+12) = 4 \\cdot 39 \\Rightarrow v = \\frac{39}{2}$, which can't be possible.\n\nThe following diagram shows a solution for sums equal to $13$.\n\n![](attached_image_1.png)",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 26314,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given any convex polygon, show that there are three consecutive vertices such that the polygon lies inside the circle through them.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $V_1$ and $V_2$ be consecutive vertices of the polygon. Let $V_k$ be a vertex such that $\\angle V_1 V_k V_2$ is minimal. Since all angles $\\angle V_1 V_j V_2 \\ge \\angle V_1 V_k V_2$ for all $j \\ne k$ of the polygon, then all vertices are contained in the circumcircle of the triangle $V_1 V_2 V_k$.\n\nIf $k=3$, we are done. Otherwise, consider the arc $V_2V_k$ that doesn't contain $V_1$ and let $V_\\ell$ a point in the region between the arc $V_2V_k$ and the chord $V_2V_k$ such that $\\angle V_2V_\\ell V_k$ is minimal. Notice that the circumcircle of $V_2V_\\ell V_k$ still contains the polygon and that the number of vertices between 1 and $k$ is less than the number of vertices between 2 and $k$. Repeat the procedure exchanging $V_2$ and $V_k$ by $V_\\ell$ and $V_k$. We obtain again a circle with less vertices between the edges of the triangle. Continuing in this fashion we eventually obtain three consecutive vertices.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26315,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that if the positive real numbers $a$, $b$ satisfy $a^3 = a + 1$ and $b^6 = b + 3a$, then $a > b$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "$a^3 = a + 1 > 1$, so $a > 1$, so $b^6 = b + 3a > 3$, so $b > 1$.\n$a^6 - b^6 = (a+1)^2 - (b+3a) = (a-1)^2 + (a-b) > a-b$. But $a^6 - b^6 = (a-b)(a^5 + a^4b + a^3b^2 + a^2b^3 + ab^4 + b^5)$, so if $b > a$, then $b^6 - a^6 \\ge 6(b-a)$ and hence $a^6 - b^6 < a - b$. Contradiction. Obviously if $a=b$, then $a^6 = b^6$ and so $a^6 - b^6$ is not greater than $a-b$. Hence we must have $a > b$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26316,
    "subject": "Mathematics (Multi-modal)",
    "question": "Call a super-integer an infinite sequence of decimal digits: $\\ldots d_n \\ldots d_3 d_2 d_1$. Given two super-integers $\\ldots c_n \\ldots c_3 c_2 c_1$ and $\\ldots d_n \\ldots d_3 d_2 d_1$, their product $\\ldots p_n \\ldots p_3 p_2 p_1$ is formed by taking $p_n \\ldots p_3 p_2 p_1$ to be the last $n$ digits of the product $c_n \\ldots c_3 c_2 c_1$ and $d_n \\ldots d_3 d_2 d_1$. Can we find two non-zero super-integers with zero product? A zero super-integer has all its digits zero.",
    "options": [],
    "answer": "Yes",
    "solution": "The answer is yes. In fact, there exist two sequences $x = \\ldots a_n \\ldots$ and $y = \\ldots b_n \\ldots$ such that, for every $k \\ge 1$, $2^k$ divides $a_k \\ldots a_2 a_1$ and $5^k$ divides $b_k \\ldots b_2 b_1$; then $10^k$ divides $a_k \\ldots a_2 a_1 \\times b_k \\ldots b_2 b_1$ and $xy = 0$.\n\nConsider first $x$. We proceed by induction. For $k=1$ choose $a_1=2$. Suppose $a_m \\ldots a_2 a_1$ is a multiple of $2^m$ and let $c_m$ be $a_m \\ldots a_2 a_1$ divided by $2^m$. Then $a_{m+1} a_m \\ldots a_2 a_1 = 10^m \\cdot a_{m+1} + a_m \\ldots a_2 a_1 = 2^m (5^m \\cdot a_{m+1} + c_m)$ and we may choose $a_{m+1}=0$ or $1$ according to the parity of $c_m$, so $5^m \\cdot a_{m+1} + c_m$ is even and, consequently, $a_{m+1} a_m \\ldots a_2 a_1$ is a multiple of $2^{m+1}$. This completes the induction.\n\nThe induction for $y$ is not that different. Indeed, choose $b_1=5$ and suppose $b_m \\ldots b_2 b_1 = 5^m \\cdot d_m$. Then $b_{m+1} b_m \\ldots b_2 b_1 = 10^m \\cdot b_{m+1} + b_m \\ldots b_2 b_1 = 5^m (2^m \\cdot b_{m+1} + d_m)$ and it's always possible to choose $b_{m+1} \\in \\{0, 1, 2, 3, 4\\}$ such that $2^m \\cdot b_{m+1} \\equiv -d_m \\pmod{5}$, since $2^m$ admits inverse modulo $5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26317,
    "subject": "Mathematics (Multi-modal)",
    "question": "A triangle has semi-perimeter $s$, circumradius $R$ and inradius $r$. Show that it is right-angled iff $2R = s - r$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $a \\ge b \\ge c$ be the sides of the triangle. If the triangle is right-angled, then\n$r = \\frac{b+c-a}{2}$, $2R = a$ and $s - r = \\frac{a+b+c}{2} - \\frac{b+c-a}{2} = a = 2R$.\n\nIf $2R = s - r$, consider\n$$\nE = (2R-a)(2R-b)(2R-c) = 8R^3 - 4R^2(a+b+c) + 2R(ab+bc+ca) - abc\n$$\nLet $\\Delta$ be the area of the triangle. We have $a+b+c = 2s$ and $R = \\frac{abc}{4\\Delta} \\iff abc = 4R\\Delta$. Finally, since $\\Delta = sr$, $abc = 4Rsr$ and\n$$\n\\begin{align*}\nsr &= \\Delta = \\sqrt{s(s-a)(s-b)(s-c)} \\\\\n\\iff sr^2 &= s^3 - (a+b+c)s^2 + (ab+bc+ca)s - abc \\\\\n\\iff ab + bc + ca &= r^2 - s^2 + 2s \\cdot s + 4Rr = r^2 + s^2 + 4Rr\n\\end{align*}\n$$\nThus\n$$\n\\begin{align*}\nE &= 8R^3 - 4R^2 \\cdot 2s + 2R(r^2 + s^2 + 4Rr) - 4Rsr \\\\\n&= 2R(4R^2 - 4Rs + r^2 + s^2 + 4Rr - 2sr) \\\\\n&= 2R(4R^2 + (r-s)^2 + 4Rr - 4Rs) \\\\\n&= 2R(4R^2 + 4R^2 + 4Rr - 4Rs) \\\\\n&= 8R^2(2R - s + r) = 0\n\\end{align*}\n$$\nand so one of the sides of the triangle is equal to $2R$, completing the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26318,
    "subject": "Mathematics (Multi-modal)",
    "question": "$ABCD$ is a quadrilateral with a circumcircle center $O$ and an inscribed circle center $I$. The diagonals intersect at $S$. Show that if two of $O$, $I$, $S$ coincide, then it must be a square.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $S = O$ then $AC = BD = 2R$, where $R$ is the circumradius of $ABCD$ and $S$ is the midpoint of both $AC$ and $BD$. This means that $ABCD$ is an inscribed parallelogram. The sum of its opposite angles, which are congruent, is $180^\\circ$, so all angles of $ABCD$ are right, that is, $ABCD$ is a rectangle. But there is an inscribed circle of radius $r$ in $ABCD$, so all sides are equal to $2r$ and so $ABCD$ is a square.\n\nIf $S = I$ then the diagonals $AC$ and $BD$ are bisectors of the angles of $ABCD$, which means that every side subtends the same arc of the circumcircle as its neighbouring sides. This means that the four vertices $A$, $B$, $C$, $D$ divide the circle in four equal parts, so $ABCD$ is a square.\n\nIf $I = O$ then the quadrilateral can be partitioned in eight congruent right triangles with hypotenuse equal to $R$ and one of the legs equal to $r$; the angle between $R$ and $r$ in those right triangles has always vertex on $O$, so all central angles $\\angle AOB$, $\\angle BOC$, $\\angle COD$ and $\\angle DOA$ are equal and again $A$, $B$, $C$ and $D$ divide the circumcircle in four equal arcs, leading again to $ABCD$ being a square.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26319,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all real-valued functions on the positive integers such that $f(x+1019) = f(x)$ for all integer $x$, and $f(xy) = f(x)f(y)$ for all integers $x, y$.",
    "options": [],
    "answer": "All solutions f: N -> R are exactly the following four functions (with p = 1019):\n1) f(n) = 0 for all n.\n2) f(n) = 1 for all n.\n3) The principal character modulo p: f(n) = 0 if p divides n, and f(n) = 1 otherwise.\n4) The quadratic character modulo p: f(n) = 0 if p divides n; if not, then f(n) = 1 when n is a quadratic residue modulo p and f(n) = -1 otherwise (equivalently, f(n) = (n/p), the Legendre symbol).",
    "solution": "The condition $f(x + 1019) = f(x)$ allows to make all computations modulo $1019$, which is a prime number. Hence the problem asks the number of multiplicative functions in $\\mathbb{Z}/1019\\mathbb{Z}$.\n\nTwo such functions are $f(x) \\equiv 1 \\pmod{1019}$ and $f(x) \\equiv 0 \\pmod{1019}$. From now on suppose $f$ is different from these two functions.\n\nSubstituting $y = 0$, $f(0) = f(x)f(0) \\Leftrightarrow f(0) = 0$; substituting $y = 1$, $f(x) = f(x)f(1) \\Leftrightarrow f(1) = 1$. By Euler-Fermat's theorem, $x^{\\phi(1019)} \\equiv 1 \\pmod{1019}$ so $f(x)^{\\phi(1019)} = f(x^{\\phi(1019)}) = f(1) = 1 \\Leftrightarrow f(x) = \\pm 1$.\n\nNow let $g$ be a primitive root of $1019$ (that is, $g$ such that the least positive integer $d$ such that $g^d \\equiv 1 \\pmod{1019}$ is $\\phi(1019)$). So every $x$ in $\\mathbb{Z}/1019\\mathbb{Z}$ can be written as $g^k$ for some $k$. Then $f(x) = (f(g))^k$. If $f(g) = 1$ then $f(x) = 0$ if $1019$ divides $x$ and $1$ otherwise; if $f(g) = -1$ then $f(x) = 0$ if $1019$ divides $x$, $f(x) = 1$ if $x$ is a quadratic residue modulo $1019$ (such residues are $1, g^2, g^4, \\dots, g^{1016}$) and $f(x) = -1$ otherwise.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26320,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p(n)$ be the largest prime which divides $n$. Show that there are infinitely many positive integers $n$ such that $p(n) < p(n + 1) < p(n + 2)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $q$ be an odd prime and take $n + 1 = q^{2^k}$. Then $p(q^{2^k}) = q$. Since $\\gcd(q^{2^k} + 1, q^{2^l} + 1) = 2$ for $k \\neq l$ (indeed, if $d$ is such $\\gcd$ and $k < l$,\n\n$$q^{2k} \\equiv -1 \\pmod d \\implies q^{2l} \\equiv 1 \\pmod d \\iff d \\mid 2$$\n\n$p(q^{2k} + 1)$ can be arbitrarily large. So let $k$ be the least integer value such that $p(q^{2k} + 1) > q$. Hence all prime divisors of $q^{2t} + 1$, $t < k$, are smaller than $q$. Since $q^{2k} - 1 = (q-1)(q+1)(q^2+1)\\cdots(q^{2k-1}+1)$ and $q-1 < q$, all prime divisors of $q^{2k} - 1$ are smaller than $q$, so $p(q^{2k} - 1) < q$. So $p(q^{2k} - 1) < p(q^{2k}) < p(q^{2k} + 1)$ and, since there are infinite prime numbers, the result follows.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26321,
    "subject": "Mathematics (Multi-modal)",
    "question": "A regular tetrahedron has side $L$. What is the smallest $x$ such that the tetrahedron can be passed through a loop of twine of length $x$?",
    "options": [],
    "answer": "2L",
    "solution": "The answer is $2L$. Consider the following net of the tetrahedron:\n\n![](attached_image_1.png)\n\nLet $P$ be a point of one of the edges of the tetrahedron. The loop will pass through $P$ some time. But $P'$ on the net coincides with $P$ on the tetrahedron, so by the triangular inequality the loop must be at least $PP' = 2L$ long.\n\nIt is possible to maintain the loop on the net parallel to $AA'$, so the least value is $2L$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26322,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that the $n$th root of a rational (for $n$ a positive integer) cannot be a root of the polynomial $x^5 - x^4 - 4x^3 + 4x^2 + 2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First we prove that $P(x) = x^5 - x^4 - 4x^3 + 4x^2 + 2$ is irreducible in $\\mathbb{Z}$. Its possible rational roots are $\\pm 1, \\pm 2$. Since $P(x) = x^2(x-1)(x-2)(x+2)+2$ it is clear that $\\pm 2$ and $1$ are not roots and $P(-1) \\neq 0$ too. So if $P(x)$ is reducible then it must be written as a product of two nonlinear polynomials. Looking $P(x)$ modulo $2$ we have $P(x) = x^4(x-1)$. Since there is unique factoring in polynomials in $\\mathbb{Z}/2\\mathbb{Z}$ then if $P(x)$ is reducible then $P(x) = (x^k + 2P_1(x))(x^l(x-1) + 2P_2(x))$, $P_1(x)$ and $P_2(x)$ with integer coefficients and $k, l > 0$. But then $P(0) = 4P_1(x)P_2(x) \\iff P_1(x)P_2(x) = \\frac{1}{2}$, which is not possible. So $P(x)$ is irreducible.\n\nLet $\\alpha$ be a root of $P(x)$. Then $P(x)$ is the minimal polynomial of $\\alpha$, and if $\\alpha^n = r$, $r$ rational, then $P(x)$ divides $Q(x) = x^n - r$. But looking modulo $2$ again we have $x^5 \\equiv x^4 \\pmod{P(x)}$, so $x^n \\equiv x^4 \\pmod{P(x)}$ and so $\\alpha$ is a root of $x^4 - r$. This is a contradiction, since the minimal polynomial of $\\alpha$ has degree $5$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 26323,
    "subject": "Mathematics (Multi-modal)",
    "question": "$X$ has $n$ elements. $\\mathcal{F}$ is a family of subsets of $X$ each with three elements, such that any two of the subsets have at most one element in common. Show that there is a subset of $X$ with at least $\\lfloor\\sqrt{2n}\\rfloor$ members which does not contain any members of $\\mathcal{F}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $Y$ be a maximal subset of $X$ in the sense that if one adjoins another element from $X$ in $Y$ then it will contain a subset from $\\mathcal{F}$. Define $f: X \\setminus Y \\to \\binom{Y}{2}$, where $\\binom{Y}{2}$ is the family of 2-subsets from $Y$, with $f(x) = A$ if $A \\cup \\{x\\}$ is one of the sets from $\\mathcal{F}$; if there is more than one set, choose any of them. $f$ is an injective function, because if $f(x_1) = f(x_2) = B$ then $B \\cup \\{x_1\\}$ and $B \\cup \\{x_2\\}$ would be in $\\mathcal{F}$ and these two sets would have more than one element in their intersection. Thus, by the injective principle,\n$$\n|X \\setminus Y| \\le \\binom{|Y|}{2} \\iff n - |Y| \\le \\frac{|Y|(|Y| - 1)}{2} \\iff |Y| \\ge -\\frac{1}{2} + \\sqrt{2n + \\frac{1}{4}}\n$$\nLet $\\lfloor\\sqrt{2n}\\rfloor = k$. We need to prove that $-\\frac{1}{2} + \\sqrt{2n + \\frac{1}{4}} > k - 1$ in order to prove that $|Y| > k - 1 \\iff |Y| \\ge k$. But this is only a small computation:\n$$-\\frac{1}{2} + \\sqrt{2n + \\frac{1}{4}} > k - 1 \\iff 2n + \\frac{1}{4} > k^2 - k + \\frac{1}{4} \\iff \\sqrt{2n} > \\sqrt{k^2 - k},$$\nwhich is true because $\\sqrt{2n} \\ge k > \\sqrt{k^2 - k}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26324,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that the equation $x^2 + y^2 + z^2 = 3xyz$ has infinitely many solutions in positive integers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We can regard $x^2 - 3yz \\cdot x + (y^2 + z^2) = 0$ as a quadratic in $x$. So if $1 \\le x \\le y \\le z$ is a solution, then so is $y, z, 3yz - x$.\n\nAlso we have $3y \\ge 3$, so $3yz - x \\ge 3yz - z \\ge 2z > z$. So $1 \\le y \\le z < 3yz - x$, and the new solution has larger largest element.\n\nSo starting with the solution $x = y = z = 1$ and repeating, we get infinitely many solutions.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 26325,
    "subject": "Mathematics (Multi-modal)",
    "question": "Does there exist a set of $n > 2$ points in the plane such that no three are collinear and the circumcenter of any three points of the set is also in the set?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "No, it's not possible. Let $S$ be the set of $n$ points. Consider the smallest circumcircle $\\omega$ of all triangles determined by the $n$ points. Let $A, B, C \\in S$ be three points on $\\omega$ and let $O$ be its center. If one of the central angles $\\angle AOB$, $\\angle BOC$, $\\angle COA$ is less than $120^\\circ$ then the circumcircle of the correspondent triangle is smaller than $\\omega$. Since the sum of the three central angles is $360^\\circ$, $\\angle AOB = \\angle BOC = \\angle COA = 120^\\circ$. The circumcenters of $AOB$, $BOC$ and $COA$ are the midpoints of the arcs $AB$, $BC$, $CA$, and then there are six points from $S$ on $\\omega$ forming a regular hexagon. Take $O$ and two of these points and obtain a circumcircle smaller than $\\omega$, which is a contradiction. So there is no such set.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26326,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f(n)$ be the smallest number of $1$s needed to represent the positive integer $n$ using only $1$s, $+$ signs, $\\times$ signs and brackets. For example, you could represent $80$ with $13$ $1$s as follows: $(1+1+1+1+1) \\times (1+1+1+1) \\times (1+1+1+1)$. Show that $3\\log_3 n \\le f(n) \\le 5\\log_3 n$ for $n > 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The upper bound follows by induction. Let $S_m$ be the statement that $f(n) \\le 5\\log_3 n$ for $n \\le 3m+2$. We have $2 = 1+1$, $3 = 1+1+1$, $4 = 1+1+1+1$, $5 = 1+1+1+1+1$, so $f(2) = 2$, $f(3) = 3$, $f(4) = 4$, $f(5) = 5$. We have $1 = \\log_3 3 < \\log_3 4 < \\log_3 5$, so $f(3) < 5\\log_3 3$, $f(4) < 5\\log_3 4$ and $f(5) < 5\\log_3 5$. Also $2^4 = 16 > 3^2$, so $2 < \\log_3 2^4 = 4\\log_3 2$. Hence $5\\log_3 2 > 4\\log_3 2 > 2 = f(2)$. So $S_1$ is true. Suppose $S_m$ is true. We have $3m+3 = (1+1+1) \\times (m+1)$, so $f(3m+3) \\le 3+f(m+1) = 3+5\\log_3(m+1) < 5(1+\\log_3(m+1)) = 5\\log_3(3m+3)$. Similarly, since $3m+4 = 1+(3m+3)$, we have $f(3m+4) \\le 4+f(m+1) < 5\\log_3(3m+3) < 5\\log_3(3m+4)$. Similarly, $f(3m+5) \\le 5+f(m+1) \\le 5+5\\log_3(m+1) = 5\\log_3(3m+3) < 5\\log_3(3m+5)$. So $S_{m+1}$ is true. So $S_m$ is true for all $m$.\n\nInduction also works for the lower bound. In any expression for $n$ using $1$s we either have $n = a + b$, with further expressions for $a$ and $b$, or we have $n = a \\cdot b$ with further expressions for $a$ and $b$. In the first case if $a = 1$, we have to show that $1 + \\log_3 b \\ge \\log_3(b+1)$, but that is almost obvious since $1 + \\log_3 b = \\log_3(3b) > \\log_3(b+1)$. If $a, b > 1$, then we have to show that $\\log_3 a + \\log_3 b \\ge \\log_3(a+b)$, which is again obvious, since if $a \\ge b \\ge 2$ we have $\\log_3 a + \\log_3 b = \\log_3(ab) \\ge \\log_3(2a) \\ge \\log_3(a+b)$. In the second case we can assume that $a, b > 1$ (since there is no benefit in the expression if $a = 1$ or $b = 1$), so we have to show that $\\log_3 a + \\log_3 b \\ge \\log_3(ab)$ which is true (we have equality). So all we need are the starting cases. For $n = 2$, it is clear that $f(2) = 2$, but $8 < 9$, so $\\log_3 8 < \\log_3 9$ or $3\\log_3 2 < 2 = f(2)$, as required.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26327,
    "subject": "Mathematics (Multi-modal)",
    "question": "$ABC$ is acute-angled. $D$ is a variable point on the side $BC$. $O_1$ is the circumcenter of $ABD$, $O_2$ is the circumcenter of $ACD$, and $O$ is the circumcenter of $AO_1O_2$. Find the locus of $O$.",
    "options": [],
    "answer": "The locus of O is the open segment of the perpendicular bisector of AO′ between the circumcenters of triangles AO′B and AO′C.",
    "solution": "Let $O'$ be the circumcenter of $ABC$ and $M, N, P$ be the respective midpoints of $AB, AC, AD$. Notice that $O', M$ and $N$ are fixed points and, by Thales theorem, $P$ is a variable point on the segment $MN$. Then $O_1$ is the intersection of the perpendicular bisectors of $AB$ and $AD$, $O_2$ is the intersection of the perpendicular bisectors of $AD$ and $AC$ and, similarly, $O'$ is the intersection of the perpendicular bisectors of $AB$ and $AC$.\n\n![](attached_image_1.png)\n\nSince $\\angle ANO_2 = \\angle APO_2 = 90^\\circ$, $APNO_2$ is cyclic; analogously, $AMO_1P$ is cyclic as well. So $\\angle O_2AN = \\angle O_2PN = \\angle O_1PM = \\angle O_1AM$. Hence $\\angle O_1AO_2 = \\angle BAC$.\n\nQuadrilateral $AMO'N$ is cyclic as well, so $\\angle MO'N = 180^\\circ - \\angle MAN = 180^\\circ - \\angle BAC$. Thus $\\angle O_1O'O_2 + \\angle O_1AO_2 = 180^\\circ$, which implies that $AO_1O'O_2$ is cyclic. Its circumcircle is $O$, and so $O$ lies on the perpendicular bisector of $AO'$, which is fixed. So the locus is a segment contained in such perpendicular bisector.\n\nIt remains to find the vertices of this segment. When $D$ tends to $B$, the perpendicular bisectors of $AD$ and $AB$ tends to coincide, so $O_2$ and $O'$ tends to coincide and $O$ tends to be the circumcenter of $AO'B$; analogously, when $D$ tends to $C$, $O$ tends to be the circumcenter of $AO'C$. So the locus is the open segment with vertices on the circumcenters of $AO'B$ and $AO'C$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26328,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $n$ boys $B_1, B_2, \\dots, B_n$ and $n$ girls $G_1, G_2, \\dots, G_n$. Each boy ranks the girls in order of preference, and each girl ranks the boys in order of preference. Show that we can arrange the boys and girls into $n$ pairs so that we cannot find a boy and a girl who prefer each other to their partners. For example if $(B_1, G_3)$ and $(B_4, G_7)$ are two of the pairs, then it must not be the case that $B_4$ prefers $G_3$ to $G_7$ and $G_3$ prefers $B_4$ to $B_1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "In round $k$ each unpaired boy $B$ in turn proposes to the girl $G$ he ranks highest amongst those he has not yet proposed to. If $G$ is not yet paired, or if she prefers $B$ to the boy she is currently paired with, then we pair $B$ and $G$ (and remove any existing pair for $G$). We claim that this algorithm terminates and gives a stable pairing.\n\nFirst we must establish that it terminates. Each boy can propose to at most $n$ girls and in each round at least one boy must make a proposal, so it must terminate in $n^2$ rounds or less.\n\nSecond we must establish that it results in every boy and girl being paired off. Once a girl is paired, she always remains paired, although not necessarily to the same boy. Suppose girl $G$ ends up unpaired. Since there are equal numbers of boys and girls, a boy $B$ must also end up unpaired. By the end $B$ must have proposed to every girl, including $G$. Hence $G$ must have been paired at the end of that round. Contradiction.\n\nFinally, we must establish that the pairing is stable. Suppose not, so that we end up with pairs $(B, G)$ and $(B', G')$, where $B$ prefers $G'$ to $G$ and $G'$ prefers $B$ to $B'$. Since $B$ prefers $G'$ to $G$ he must have proposed to $G'$ before $G$. At the end of that round either $B$ and $G'$ were paired, or $G'$ was paired to $B''$ whom she preferred to $B$. In the first case, we have a contradiction because a girl can only change her pairing in favor of someone she prefers. So if she ends up paired to $B'$ having earlier been paired to $B$, then she must prefer $B'$ to $B$. Contradiction. In the second case, $G'$ must prefer $B''$ to $B$ and $B'$ to $B''$. Again a contradiction. So the pairing is stable.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26329,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p(x)$ be the polynomial $x^3 + 14x^2 - 2x + 1$. Let $p^{(n)}(x)$ denote $p(p^{(n-1)}(x))$. Show that there is an integer $N$ such that $p^{(N)}(x) - x$ is divisible by $101$ for all integers $x$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We have $p(x) - p(y) = (x^3 + 14x^2 - 2x + 1) - (y^3 + 14y^2 - 2y + 1) = (x - y)(x^2 + xy + y^2 + 14x + 14y - 2)$. Since $101$ is prime, $p(x) \\equiv p(y) \\pmod{101} \\Leftrightarrow x \\equiv y \\pmod{101}$ or $x^2 + xy + y^2 + 14x + 14y - 2 \\equiv 0 \\pmod{101}$.\n\nCompleting squares, we have\n$$\n\\begin{aligned}\nx^2 + xy + y^2 + 14x + 14y - 2 &\\equiv 0 \\pmod{101} \\\\\n&\\Leftrightarrow (2x + y + 14)^2 + 3y^2 + 28y - 2 \\equiv 0 \\pmod{101} \\\\\n&\\Leftrightarrow (2x + y + 14)^2 \\equiv -3(y - 29)^2 \\pmod{101} \\quad (*)\n\\end{aligned}\n$$\nBy the quadratic reciprocity law, $\\left(\\frac{-3}{101}\\right) \\cdot \\left(\\frac{101}{-3}\\right) = (-1)^{\\frac{101-1}{2} \\cdot \\frac{-3-1}{2}} = 1 \\Leftrightarrow \\left(\\frac{-3}{101}\\right) = \\left(\\frac{2}{3}\\right) = -1$, so $-3$ is not a quadratic residue modulo $101$. Thus\n$$ (*) \\Leftrightarrow 2x + y + 14 \\equiv y - 29 \\equiv 0 \\pmod{101} \\Leftrightarrow x \\equiv y \\equiv 29 \\pmod{101} $$\nThus $p(x) \\equiv p(y) \\pmod{101} \\Leftrightarrow x \\equiv y \\pmod{101}$, which means that $p(x)$ admits an inverse function.\n\nTo finish the problem, fix $x$ and consider $x, p(x), p^{(2)}(x), \\dots \\pmod{101}$. Since there are infinite numbers and $101$ remainders, there are $m$ and $n$ such that $m > n$ and $p^{(m)}(x) = p^{(n)}(x) \\pmod{101} \\Leftrightarrow p^{(m-n)} \\equiv x \\pmod{101}$. So for each $k \\pmod{101}$ there is a positive integer $n_k$ such that $p^{(n_k)}(k) \\equiv k \\pmod{101}$. Choose $N = \\text{lcm}(n_1, n_2, \\dots, n_{101})$ and we are done.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26330,
    "subject": "Mathematics (Multi-modal)",
    "question": "$A$ is a set of $n$ non-negative integers. We say it has property $P$ if the set $\\{x + y \\mid x, y \\in A\\}$ has $\\frac{n(n+1)}{2}$ elements. We call the largest element of $A$ minus the smallest element the *diameter* of $A$. Let $f(n)$ be the smallest diameter of any set $A$ with property $P$. Show that $\\frac{n^2}{4} \\le f(n) < n^3$.\n(If you still have time, try to improve these bounds. For example, try to prove that $f(p) < 2p^2$ for all primes $p$.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $a_1 < a_2 < \\dots < a_n$ be the elements of $A$. The diameter of $A$ is $d(A) = a_n - a_1$. The minimum and maximum of $A + A = \\{x + y \\mid x, y \\in A\\}$ are $a_1 + a_1$ and $a_n + a_n$. Since $A + A$ has $\\frac{n(n+1)}{2}$ elements, $2a_n - 2a_1 + 1 \\ge \\frac{n(n+1)}{2} \\iff d(A) \\ge \\frac{n^2}{4} + \\frac{n-2}{4} \\ge \\frac{n^2}{4}$.\n\nTo prove the upper bound, we need to construct a set $A$ with property $P$ such that $d(A) < n^3$. We do this recursively. Start with $A = \\{0, 1\\}$ and we will adjoin elements to $A$. Suppose we already have $a_1 = 0 < a_2 = 1 < \\dots < a_n$ in $A$ such that $\\{a_1, a_2, \\dots, a_n\\}$ has property $P$. Adjoining $a_{n+1}$ will generate $n+1$ more sums: $a_1 + a_{n+1}, a_2 + a_{n+1}, \\dots, a_{n+1} + a_{n+1}$. All these sums must be different from any $a_i + a_j$, $1 \\le i \\le j \\le n$. Moreover, $a_{n+1}$ must be different from all numbers $a_i$, $1 \\le i \\le n$. This means that $a_{n+1} \\notin \\{a_i + a_j - a_k \\mid 1 \\le i \\le j \\le k\\} \\cup \\left\\{\\frac{a_i+a_j}{2} \\mid 1 \\le i,j \\le n\\right\\} = B$. This set has no more than $n^3 + \\frac{n(n+1)}{2}$ elements. So we may choose $a_{n+1}$ as the smaller number not in $B$, and so $a_{n+1} \\le n^3 + \\frac{n(n+1)}{2} < (n+1)^3$. Thus\n$d(A) = a_{n+1} - a_1 = a_{n+1} < (n+1)^3$.\n\nNow we are going to prove the “bonus” part of the problem. Let $p$ be a prime and consider $A = \\{k + 2p \\cdot (k^2 \\bmod p) \\mid 0 \\le k \\le p-1\\}$. We have\n$d(A) \\le (p-1)+2p(p-1) = 2p^2-p-1 < 2p^2$. Moreover, $r+2p(r^2 \\bmod p)+$\n$s+2p(s^2 \\bmod p) = t+2p(t^2 \\bmod p)+u+2p(u^2 \\bmod p) \\iff r+s=t+u$\nand $r^2 \\bmod p + s^2 \\bmod p = t^2 \\bmod p + u^2 \\bmod p$. So $r-t=u-s$ and\n$r^2-t^2 \\equiv u^2-s^2 \\pmod p \\iff (r-t)(r+t) \\equiv (u-s)(u+s) \\pmod p$\n$\\iff r-t \\equiv u-s \\equiv 0 \\pmod p$ or $r+t \\equiv u+s \\pmod p$, hence\n$r=t$ and $s=u$ or $r=u$ and $s=t$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26331,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $R$ be the set of real numbers. Show that there are no functions $f, g: R \\to R$ such that $g(f(x)) = x^3$ and $f(g(x)) = x^2$ for all $x$. Let $S$ be the set of all real numbers greater than $1$. Show that there are functions $f, g: S \\to S$ satisfying the condition above.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First of all, since $g(f(x)) = x^3$ and $x^3$ is injective, $f$ is injective. Indeed,\n$f(x) = f(y) \\implies g(f(x)) = g(f(y)) \\iff x^3 = y^3 \\iff x = y$. Plugging\n$f(x)$ in $f(g(x)) = x^2$, we obtain $f(g(f(x))) = f(x)^2 \\iff f(x^3) = f(x)^2$. If\n$f, g: R \\to R$, consider $x = -1, 0, 1$ (i.e., the roots of $x^3 = x$): $f(-1) = f(-1)^2$,\n$f(0) = f(0)^2$ and $f(1) = f(1)^2$, so $\\{f(-1), f(0), f(1)\\} \\subset \\{0, 1\\}$, which\ncontradicts the fact that $f$ is injective. So in this case there are no such\nfunctions $f, g$.\n\nIf $f, g: S \\to S$, consider $f(x^3) = (f(x))^2$ and $g(f(g(x))) = (g(x))^3 \\iff$\n$g(x^2) = (g(x))^3$. Applying these two identities $n$ times we obtain $f(x^{3^n}) =$\n$(f(x))^{2^n}$ and $g(x^{2^n}) = (g(x))^{3^n}$. In particular, $f(2^{3^n}) = (f(2))^{2^n}$ and\n$g(2^{2^n}) = (g(2))^{3^n}$. Extend these identities for all $n$ real. We have $x =$\n$2^{3^n} \\iff n = \\log_3 2 \\log_2 x$ and $x = 2^{2^n} \\iff n = \\log_2 \\log_2 x$. This means\nthat $f(x) = a^{2^{\\log_3 2 \\log_2 x}}$ and $g(x) = b^{2^{\\log_2 2 \\log_2 x}}$. For the sake of simplicity, write all\nequations in base $2$: $f(x) = 2^{2^{\\log_3 2 \\log_2 \\log_2 x+k}}$ and $g(x) = 2^{2^{\\log_2 3 \\log_2 \\log_2 x+\\ell}}$.\nNow substitute in the original equations:\n$$\n\\begin{align*}\nf(g(x)) &= f(2^{2^{\\log_2 3 \\log_2 \\log_2 x + \\ell}}) = 2^{2^{\\log_3 2 \\log_2 \\log_2 2^{2^{\\log_2 3 \\log_2 \\log_2 x + \\ell}}}+ k} = 2^{2^{\\log_3 2 \\cdot (\\log_2 3 \\log_2 \\log_2 x + \\ell) + k}} \\\\\n&= 2^{2^{\\log_2 \\log_2 x + \\ell} \\log_3 2 + k} = 2^{2^{\\log_2 \\log_2 x} \\cdot 2^{\\ell} \\log_3 2 + k} = (2^{2^{\\log_2 \\log_2 x}})^{2^{\\ell} \\log_3 2 + k} = x^{2^{\\ell} \\log_3 2 + k}\n\\end{align*}\n$$\n$$\n\\text{so } 2^{\\ell \\log_3 2+k} = 2 \\iff \\ell \\log_3 2 + k = 1.\n$$\n$$\n\\begin{align*}\ng(f(x)) &= g(2^{2^{\\log_3 2 \\log_2 \\log_2 x+k}}) = 2^{2^{\\log_2 3 \\log_2 \\log_2 2^{2^{\\log_3 2 \\log_2 \\log_2 x+k}}}+ \\ell} = 2^{2^{\\log_2 3 \\cdot (\\log_3 2 \\log_2 \\log_2 x+k) + \\ell}} \\\\\n&= 2^{2^{\\log_2 \\log_2 x+k} \\log_2 3+\\ell} = 2^{2^{\\log_2 \\log_2 x} \\cdot 2^k \\log_2 3+\\ell} = (2^{2^{\\log_2 \\log_2 x}})^{2^k \\log_2 3+\\ell} = x^{2^k \\log_2 3+\\ell}\n\\end{align*}\n$$\n$$\n\\text{so } 2^{k \\log_2 3 + \\ell} = 3 \\iff k \\log_2 3 + \\ell = \\log_2 3.\n$$\nIt is clear that both $f$ and $g$ are well defined in $S$ and that it's possible to\nchoose $k$ and $\\ell$ (for example, $k=1$ and $\\ell=0$), so such functions do exist.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26332,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $F_n$ be the Fibonacci sequence $F_1 = F_2 = 1$, $F_{n+2} = F_{n+1} + F_n$. Put $V_n = \\sqrt{F_n^2 + F_{n+2}^2}$. Show that $V_n$, $V_{n+1}$, $V_{n+2}$ are the sides of a triangle of area $1/2$.",
    "options": [],
    "answer": "1/2",
    "solution": "Let $A = (F_{n+4}, 0)$, $B = (0, F_{n+2})$ and $C = (F_{n+3}, F_n)$. Notice that\n$$\nAB = \\sqrt{F_{n+4}^2 + F_{n+2}^2} = V_{n+2}, \\quad BC = \\sqrt{F_{n+3}^2 + (F_{n+2} - F_n)^2} = \\sqrt{F_{n+3}^2 + F_{n+1}^2} = V_{n+1}\n$$\nand $CA = \\sqrt{(F_{n+4} - F_{n+3})^2 + F_n^2} = \\sqrt{F_{n+2}^2 + F_n^2} = V_n$. Its area equals $\\frac{|D|}{2}$, where\n$$\n\\begin{aligned}\nD &= \\begin{vmatrix} F_{n+4} & 0 & 1 \\\\ 0 & F_{n+2} & 1 \\\\ F_{n+3} & F_n & 1 \\end{vmatrix} = F_{n+4}F_{n+2} - F_{n+3}F_{n+2} - F_nF_{n+4} \\\\\n&= F_{n+4}(F_{n+2} - F_n) - F_{n+3}F_{n+2} = F_{n+4}F_{n+1} - F_{n+3}F_{n+2} \\\\\n&= \\begin{vmatrix} F_{n+1} & F_{n+2} \\\\ F_{n+3} & F_{n+4} \\end{vmatrix}\n\\end{aligned}\n$$\nLet $A_{n+1} = \\begin{pmatrix} F_{n+1} & F_{n+2} \\\\ F_{n+3} & F_{n+4} \\end{pmatrix}$. An easy induction proves that $A_n = \\begin{pmatrix} 0 & 1 \\\\ 1 & 2 \\end{pmatrix}$.\n$(\\begin{pmatrix} 0 & 1 \\\\ 1 & 1 \\end{pmatrix})^n$. Indeed, $A_0 = \\begin{pmatrix} F_0 & F_1 \\\\ F_2 & F_3 \\end{pmatrix} = \\begin{pmatrix} 0 & 1 \\\\ 1 & 2 \\end{pmatrix} \\cdot \\begin{pmatrix} 0 & 1 \\\\ 1 & 1 \\end{pmatrix}^0$ and\n$$\n\\begin{aligned}\nA_{n+1} \\cdot \\begin{pmatrix} 0 & 1 \\\\ 1 & 1 \\end{pmatrix} &= \\begin{pmatrix} F_{n+1} & F_{n+2} \\\\ F_{n+3} & F_{n+4} \\end{pmatrix} \\begin{pmatrix} 0 & 1 \\\\ 1 & 1 \\end{pmatrix} = \\begin{pmatrix} F_{n+2} & F_{n+1} + F_{n+2} \\\\ F_{n+4} & F_{n+3} + F_{n+4} \\end{pmatrix} \\\\\n&= \\begin{pmatrix} F_{n+2} & F_{n+3} \\\\ F_{n+4} & F_{n+5} \\end{pmatrix} = A_{n+2}\n\\end{aligned}\n$$\nHence\n$$\nD = |A_{n+1}| = \\begin{vmatrix} 0 & 1 \\\\ 1 & 2 \\end{vmatrix} \\cdot \\begin{vmatrix} 0 & 1 \\\\ 1 & 1 \\end{vmatrix}^{n+1} = (0 \\cdot 2 - 1 \\cdot 1) \\cdot (0 \\cdot 1 - 1 \\cdot 1)^{n+1} = (-1)^{n+2}\n$$\nand area $ABC = \\frac{|D|}{2} = \\frac{1}{2}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26333,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $c$ be a rational. Let $f(x) = x^2 + c$. Define $f^{(0)}(x) = x$, $f^{(n+1)}(x) = f(f^{(n)}(x))$. Show that there are only finitely many rationals $x$ such that the sequence $f^{(0)}(x), f^{(1)}(x), f^{(2)}(x), \\dots$ takes only finitely many values.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For sake of simplicity, call $x$ a *periodic number* if $f^{(0)}(x)$, $f^{(1)}(x)$, $f^{(2)}(x)$, $\\dots$ takes finitely many values.\n\nIf $|x| > |c| + 1$ then $x^2 - |x| = |x|(|x| - 1) > (|c| + 1)|c| \\ge |c| \\implies x^2 - |c| > |x|$, so $|f(x)| = |x^2 + c| \\ge x^2 - |c| > |x|$, that is, $|f^{(n+1)}(x)| > |f^{(n)}(x)| > \\dots > |x|$ and $f^{(0)}(x)$, $f^{(1)}(x)$, $f^{(2)}(x)$, $\\dots$ takes infinitely many values. So if $x$ is periodic then $|x| \\le |c| + 1$, that is, all periodic numbers lie in the interval $[-(|c| + 1), |c| + 1]$.\n\nLet $c = \\frac{r}{s}$, $x = \\frac{y}{z}$ and $f(x) = \\frac{u}{v}$, $\\gcd(r, s) = \\gcd(y, z) = \\gcd(u, v) = 1$, $s, z, v > 0$. So\n$$\n\\frac{u}{v} = \\left(\\frac{y}{z}\\right)^2 + \\frac{r}{s} \\iff y^2 s v = z^2 (u s - r v)\n$$\nSince $\\gcd(y, z) = 1$, $z^2$ divides $s v$, so $s v \\ge z^2 \\iff v \\ge \\frac{z^2}{s}$. If $\\frac{z^2}{s} > z \\iff z > s$ then the denominator of $x$ is less than the denominator of $f(x)$ and consequently the denominator of $f^{(n)}(x)$ is less than the denominator $f^{(n+1)}(x)$, so $x$ is not a periodic point.\n\nSo all rational periodic points of $f$ lie in the interval $[-(|c|+1), |c|+1]$ and have denominator not greater than the denominator of $c$. Thus the number of rational periodic points of $f$ is finite.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26334,
    "subject": "Mathematics (Multi-modal)",
    "question": "$f$ is a map on the plane such that two points a distance $1$ apart are always taken to two points a distance $1$ apart. Show that for any $d$, $f$ takes two points a distance $d$ apart to two points a distance $d$ apart.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Observe first that the images of equilateral triangles of side $1$ are equilateral triangles of side $1$. Consider two equilateral triangles $ABC$ and $A'B'C$ with side $1$ and a common side. Notice that $AA' = \\sqrt{3}$ and $f(A)f(A') = 0$ or $\\sqrt{3}$. If $f(A) = f(A')$, then consider $B$ such that $AB = 1$ and $A'B = \\sqrt{3}$, so $f(A)f(B) = 1 \\Leftrightarrow f(A')f(B) = 1$, contradiction. So $f(A)f(A') = \\sqrt{3}$ and all points $\\sqrt{3}$ apart are taken to two points $\\sqrt{3}$ apart.\n\nSo any triangular lattice is taken to a triangular lattice. In particular, any triangle with sides $1$, $\\sqrt{n^2-n+1}$ and $\\sqrt{n^2+n+1}$ are preserved by $f$.\n\n![](attached_image_1.png)\n\nConsider two different triangles $ABC$ and $AB'C$ such that $B'C = BC = 1$, $AB = AB' = \\sqrt{n^2-n+1}$ and $AC = \\sqrt{n^2+n+1}$. Notice that $BB' = \\sqrt{\\frac{3}{n^2+n+1}}$. Let $\\epsilon_n = \\sqrt{\\frac{3}{n^2+n+1}}$. Again, $f(B)f(B') = 0$ or $\\epsilon_n$. If $B = A_0$ and $B' = A_1$ and $f(B) = f(B')$, let $k$ be an integer such that $k\\epsilon_n < 1 \\le (k+1)\\epsilon_n$ and $A_2, A_3, \\dots, A_{k+1}$ points such that $A_iA_{i+1} = \\epsilon_n$, $i = 0, 1, 2, \\dots, k$, and $A_0A_{k+1} = 1$. We have $f(A_0)f(A_{k+1}) = 1$, so by the triangle inequality,\n\n$$\n1 = f(A_0A_{k+1}) \\le f(A_0)f(A_1) + f(A_1)f(A_2) + \\dots + f(A_k)f(A_{k+1}) \\\\\n\\le f(A_0)f(A_1) + k\\epsilon_n < 1,\n$$\n\nwhich is a contradiction. So $f(A_0)f(A_1) = \\epsilon_n$ for all points $A_0, A_1$, $\\epsilon_n$ apart. Moreover, if $AB = m\\epsilon_n$, $m$ positive integer, $f(A)f(B) = m\\epsilon_n$.\n\nNow let $X$ and $Y$ be two arbitrary points and suppose that $f(X)f(Y) \\ne XY$. Choose $n$ such that $|XY - f(X)f(Y)| > 4\\epsilon_n$ and $P$ such that $PX = m\\epsilon_n$, $m$ integer, and $PY = 2\\epsilon_n$. Then $f(X)f(P) = XP = m\\epsilon_n$, $f(Y)f(P) = 2\\epsilon_n$ and $|f(X)f(Y) - XY| \\le |f(X)f(Y) - f(P)f(X)| + |f(P)f(X) - XY| =$\n\n$|f(X)f(Y) - f(P)f(X)| + |PX - XY| \\le f(Y)f(P) + PY = 4\\epsilon_n$, contradiction.\n\nSo $f(X)f(Y) = XY$ for all points $X$, $Y$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26335,
    "subject": "Mathematics (Multi-modal)",
    "question": "15 positive integers smaller than $1998$ are relatively prime (no pair has a common factor larger than $1$). Show that at least one of them must be prime.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose there are $15$ such integers, none prime. If we list the primes in order, the $15$th is $p_{15} = 47$. Now let $p(n)$ be the smallest prime dividing $n$. Take $N$ to be that integer $n$ among the $15$ which has the largest $p(n)$. Then since $N$ is not prime and $p(n)$ is its smallest prime factor, we must have $N \\geq p(n)^2$. Since all $15$ integers are relatively prime, they must all have different $p(n)$s. Hence $p(n) = p_{15}$, so $N \\geq p_{15}^2 > 1998$. Contradiction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26336,
    "subject": "Mathematics (Multi-modal)",
    "question": "$ABC$ is a triangle. $D$ is the midpoint of $AB$, $E$ is a point on the side $BC$ such that $BE = 2EC$ and $\\angle ADC = \\angle BAE$. Find $\\angle BAC$.",
    "options": [],
    "answer": "90°",
    "solution": "Let $F$ be the midpoint of $BE$. Thus $DF$ and $AE$ are parallel. Let $M$ be the intersection point of $AE$ and $CD$.\n\n![](attached_image_1.png)\n\n$E$ is the midpoint of $CF$. Since $ME$ is parallel to $DF$, by Thales theorem $DM = MC$. But triangle $ADM$ is isosceles, so $AM = DM = MC$ and triangle $AMC$ is isosceles as well. Let $\\alpha = \\angle ADC = \\angle BAE$. Then $\\angle AMC = 2\\alpha$ and $\\angle MAC = 90^\\circ - \\alpha$ and, consequently, $\\angle BAC = \\angle DAM + \\angle MAC = \\alpha + 90^\\circ - \\alpha = 90^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26337,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two players play a game as follows. There are $n > 1$ rounds and $d \\ge 1$ is fixed. In the first round A picks a positive integer $m_1$, then B picks a positive integer $n_1 \\ne m_1$. In round $k$ (for $k = 2, \\dots, n$), A picks an integer $m_k$ such that $m_{k-1} < m_k \\le m_{k-1} + d$. Then B picks an integer $n_k$ such that $n_{k-1} < n_k \\le n_{k-1} + d$. A gets $\\gcd(m_k, n_{k-1})$ points and B gets $\\gcd(m_k, n_k)$ points. After $n$ rounds, A wins if he has at least as many points as B, otherwise he loses. For each $n, d$ which player has a winning strategy?",
    "options": [],
    "answer": "B",
    "solution": "$B$ has a winning strategy. Let $s = N!$ for $N$ sufficiently large (so $N!$ is divisible by all $m_i$'s). $B$ can compute $N$ after $A$ makes his first choice, because he knows that the biggest number that $A$ can choose in any round is at most $m_1 + (n-1)d$, so $N$ may be any number bigger than $m_1 + (n-1)d$. $B$ always chooses $n_k = m_k + s$. Then $B$ wins $\\gcd(m_k, m_k + s) = m_k$ points and $A$ wins $\\gcd(m_{k-1}, s+m_k) = \\gcd(m_{k-1}, m_k) \\le m_{k-1} < m_k$ points. Since $B$ wins more points than $A$ in every round, $B$ wins the game.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26338,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two players play a game as follows. The first player chooses two non-zero integers $A$ and $B$. The second player forms a quadratic with $A$, $B$ and $1998$ as coefficients (in any order). The first player wins iff the equation has two distinct rational roots. Show that the first player can always win.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Choose $A$ and $B$ such that $A + B + 1998 = 0$. Then $1$ is a root of the quadratic equation no matter how the second player arranges the coefficients.\n\nThe other root is also rational, because the product of the roots, the quotient of two of the rational coefficients $A$, $B$, $1998$, is rational.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 26339,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $N = \\{0, 1, 2, 3, \\dots\\}$. Find all functions $f: N \\to N$ which satisfy $f(2f(n)) = n + 1998$ for all $n$.",
    "options": [],
    "answer": "no function exists",
    "solution": "First of all, $f$ is injective: indeed, $f(x) = f(y) \\implies f(2f(x)) = f(2f(y)) \\iff x + 1998 = y + 1998 \\iff x = y$.\n\nMoreover, $f$ takes all integer values bigger than $1997$: if $m \\ge 1998$ then $m = 1998 + k$ for some $k \\ge 0$ and $f(2f(k)) = k+1998 = m$.\n\nNotice that the values of $m$ such that $f(m) \\ge 1998$ are all even; since $f$ is injective, $f(x)$ can't be bigger than $1997$ for every $x$ odd. This means that all odd numbers are taken by $f$ to numbers not exceeding $1998$. By the pigeon hole principle, there are two different odd numbers $x$ and $y$ such that $f(x) = f(y)$, contradiction. So there is no such function.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26340,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two mathematicians, lost in Berlin, arrived on the corner of Barbarossa street with Martin Luther street and need to arrive on the corner of Meininger street with Martin Luther street. Unfortunately they don't know which direction to go along Martin Luther Street to reach Meininger Street nor how far it is, so they must go forwards and backwards along Martin Luther street until they arrive at their destination. What is the smallest value for a positive integer $K$ so that they can be sure that if there are $N$ blocks between Barbarossa street and Meininger street then they can arrive at their destination by walking no more than $KN$ blocks (no matter what $N$ turns out to be)? The mathematicians always walk together.",
    "options": [],
    "answer": "9",
    "solution": "Since the mathematicians didn't know which side and at what distance their destination was, they should adopt the following strategy: walk $a_1$ blocks on one side (say, left), then go back to the starting point and walk $a_2$ blocks to the right, then go back again and walk $a_3$ blocks to the left, and so on, with $a_1 < a_3 < a_5 < \\dots$ and $a_2 < a_4 < a_6 < \\dots$ until they find their destination. The worst cases happen when the destination is $a_{2k+1} + 1$ blocks to the left or $a_{2k} + 1$ blocks to the right.\nIn these cases, there are $a_n + 1$ blocks between the starting point and the destination and the mathematicians walk a total of $2a_1 + 2a_2 + \\dots + 2a_n + 2a_{n+1} + a_n + 1$ blocks until they reach their destination. So $2a_1 + 2a_2 + \\dots + 2a_n + 2a_{n+1} + a_n \\le k(a_n + 1)$ or $s_{n+1} \\le \\left(\\frac{k-1}{2}\\right)(a_n + 1)$ for all naturals $n$, where $s_t = a_1 + a_2 + \\dots + a_t$.\nFor $k=9$ there are strategies satisfying the conditions of the problem: for example, consider $a_m = 2^m$. Notice that $s_{n+1} = 2^1 + 2^2 + \\dots + 2^{n+1} = 2^{n+2} - 2 < 4(2^n + 1) = \\left(\\frac{k-1}{2}\\right)(a_n + 1)$ for all naturals $n$.\nNow we prove that $k$ cannot be less than 9. If $k < 9$ then $c = \\frac{k-1}{2} < 4$. Since $a_n = s_n - s_{n-1}$ we have $s_{n+1} \\le c(s_n - s_{n-1} + 1)$ for all naturals $n$. Let $u_n = s_n - c$ so that $u_{n+1} \\le c(u_n - u_{n-1})$. Since $c < 4$, $u_n > 0$ for all $n \\ge 3$ and if $v_n = \\frac{u_{n+1}}{u_n}$ for $n \\ge 3$, $v_n \\le c\\left(1 - \\frac{1}{v_{n-1}}\\right)$ for all $n \\ge 4$ and $v_n - v_{n-1} \\le c\\left(1 - \\frac{1}{v_{n-1}}\\right) - v_{n-1} = \\frac{cv_{n-1} - c - v_{n-1}^2 - 1}{v_{n-1}} = \\frac{c^2/4 - c - (c/2 - v_{n-1})^2}{v_{n-1}} \\le \\frac{c(c-4)}{4v_{n-1}} < 0$, from which $v_n < v_{n-1}$ for all $n \\ge 4$. On the other hand, for all $n \\ge 4$ we have $v_n - v_{n-1} \\le \\frac{c(c-4)}{v_{n-1}} \\implies v_n - v_{n-1} \\le \\frac{c(c-4)}{4v_3}$ for $n \\ge 4$ and so $v_n \\le v_3 + \\frac{(n-3)c(c-4)}{4v_3}$ for all $n \\ge 4$, a contradiction because the right hand side is negative for $n > 3 + \\frac{4v_3^2}{c(4-c)}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26341,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCDE$ be a regular pentagon such that the star $ACEBD$ has area $1$. Let $P$ be the point of intersection of $AC$ and $BE$ and $Q$ be the point of intersection of $BD$ and $CE$. Find the area of $APQD$.",
    "options": [],
    "answer": "1/2",
    "solution": "Let $AD$ meet $EC$ and $EB$ at $R$ and $S$ respectively. $PQDR$ is a parallelogram, so the area of $PQD$ equals the area of $RQD$, which by its turn equals the area of $ERS$. Therefore the area of $APQD$ equals the area of $APDRES$, which is half of the area of the whole star.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26342,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that there is at least one nonzero digit between the 1,000,000th and the 3,000,000th decimal digits of $\\sqrt{2}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us suppose that all digits between the 1,000,000th and the 3,000,000th decimal digits of $\\sqrt{2}$ are zeros. Then\n$$\n\\begin{aligned}\n\\sqrt{2} &= \\frac{n}{10^{10^6}} + \\epsilon, && n \\in \\mathbb{Z},\\ n < 2 \\cdot 10^{10^6} \\text{ and } 0 < \\epsilon < 10^{-3 \\cdot 10^6} \\\\\n\\Leftrightarrow 2 \\cdot 10^{2 \\cdot 10^6} - n^2 &= 2n\\epsilon 10^{10^6} + \\epsilon^2 \\cdot 10^{2 \\cdot 10^6}\n\\end{aligned}\n$$\nThe left hand side is a positive integer. The right hand side is smaller than 1 nevertheless. The result thus follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26343,
    "subject": "Mathematics (Multi-modal)",
    "question": "$n$ tokens are to be placed on the squares of a $10 \\times 10$ board such that no 4 tokens be the vertices of a rectangle with sides parallel to the sides of the board. Find the greatest value of $n$ for which this is possible.",
    "options": [],
    "answer": "34",
    "solution": "Let $A_i \\subset \\{1, 2, \\dots, 10\\}$ be the set of the positions of the tokens in the $i$$-$th line of the board, $1 \\le i \\le 10$. The problem is equivalent to finding $A_1, A_2, \\dots, A_{10}$ such that $|A_i \\cap A_j| \\le 1$ for $i \\ne j$ and $|A_1| + |A_2| + \\dots + |A_{10}|$ is maximum.\n\nLet $k_i$ be $|A_i|$. The $\\binom{k_i}{2}$ subsets of $A_i$ with 2 elements must not be contained in any other $A_j, j \\ne i$. Hence\n$$\n\\sum_{1 \\le i \\le 10} \\binom{k_i}{2} \\le \\binom{10}{2} \\Leftrightarrow \\sum_{1 \\le i \\le 10} (2k_i - 1)^2 \\le 370\n$$\nBy Cauchy's inequality,\n$$\n\\begin{aligned}\n& \\sum_{1 \\le i \\le 10} 1^2 \\cdot \\sum_{1 \\le i \\le 10} (2k_i - 1)^2 \\ge \\left( \\sum_{1 \\le i \\le 10} (2k_i - 1) \\right)^2 \\\\\n\\Rightarrow & \\sum_{1 \\le i \\le 10} (2k_i - 1) \\le \\sqrt{10 \\cdot 370} \\\\\n\\Leftrightarrow & \\sum_{1 \\le i \\le 10} k_i \\le 35\n\\end{aligned}\n$$\nThe equality holds if and only if 5 of the $k_i$'s equal 4 and the other 5 equal 3. In this case, $\\sum_{1 \\le i \\le 10} \\binom{k_i}{2} = \\binom{10}{2}$ and hence each subset of $\\{1, 2, \\dots, 10\\}$ with 2 elements should be in exactly one $A_i$.\n\nTherefore if it were possible to construct an instance with 35 tokens, each element of $\\{1, 2, \\dots, 10\\}$ would either be in 3 subsets with 4 elements or\nin 1 subset with 4 elements and 3 subsets with 3 elements. Since there are 5 subsets with 4 elements, there must be elements which belong to 3 subsets with 4 elements. We may thus suppose wlog that $A_1 = \\{1, 2, 3, 4\\}$, $A_2 = \\{1, 5, 6, 7\\}$, $A_3 = \\{1, 8, 9, 10\\}$. However any other subset with 4 elements would be contained in $\\{2, 3, \\dots, 10\\}$ and therefore its intersection with one of $A_1$, $A_2$ or $A_3$ would have at least 2 elements. We conclude that it is impossible to have $\\sum_{1\\le i\\le 10} k_i = 35$.\n\nOn the other hand, there exist instances with 34 tokens:\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26344,
    "subject": "Mathematics (Multi-modal)",
    "question": "Planet *Zork* is spherical and has many towns. For each town there is a corresponding antipodal town (i.e. symmetric in relation to the centre of the planet).\nThere are roads connecting pairs of towns in Zork. If there is a road connecting towns $P$ and $Q$ then there is also a road connecting towns $P'$ and $Q'$, where $P'$ is the antipode of $P$ and $Q'$ is the antipode of $Q$. Besides the roads do not cross each other and for any given two towns $P$ and $Q$ it is possible to travel from $P$ to $Q$ through some sequence of roads.\nThe prices of *Kriptonita* in *Urghs* (the planetary currency) in two towns connected by a road differ by no more than 100 *Urghs*. Prove that there exist two antipodal towns such that the prices of *Kriptonita* in these towns differ by no more than 100 *Urghs*.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $a_i \\ge b_i$ be the prices of Kriptonita in the antipodal towns $A_i$ and $B_i$ respectively. Suppose that the prices differ by more than 100 Urghs at all antipodal towns. Thus $a_i - b_i > 100$ for each $i$.\nIn a sequence of roads connecting $A_0$ to $B_0$, we can find a road connecting $A_j$ and $B_k$ for some $j, k$. Hence there exists a road connecting the antipodes $B_j$ and $A_k$ of $A_j$ and $B_k$.\nIf $j = k$ there exists a road connecting $A_j$ and $B_j$, which implies $a_j - b_j \\le 100$, contradiction.\nIf $j \\ne k$ then\n$$\n\\left| \\begin{array}{l} a_j - b_k \\le 100 \\\\ a_k - b_j \\le 100 \\end{array} \\right. \\implies a_j + a_k - b_j - b_k \\le 200\n$$\nOn the other hand, we have\n$$\n\\left| \\begin{array}{l} a_j - b_j > 100 \\\\ a_k - b_k > 100 \\end{array} \\right. \\implies a_j + a_k - b_j - b_k > 200,\n$$\nwhich is a contradiction.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26345,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $n$ football teams in *Tumbólia*.\nA championship is to be organized in which each team plays against every other exactly once. Every match takes place on Sundays and one team must not play more than once in the same day.\nFind the least integer $m$ for which it's possible to set up a championship lasting $m$ Sundays.",
    "options": [],
    "answer": "n-1 if n is even; n if n is odd",
    "solution": "First case: $n$ is even.\nLet $T_1, T_2, \\dots, T_n$ be the teams. Each team shall play $n-1$ times on at least $n-1$ Sundays. Let us show that $n-1$ Sundays are sufficient to set up the championship.\n\nTeams $T_i$ and $T_j$ play on the $d_{ij}$-th Sunday. Define $d_{ij}$ by the following rules:\n$$\nd_{ij} = \\begin{cases} i+j \\pmod{n-1} & \\text{if } i, j \\neq n \\\\ 2i \\pmod{n-1} & \\text{if } i \\neq n \\text{ and } j = n \\\\ 2j \\pmod{n-1} & \\text{if } i = n \\text{ and } j \\neq n \\end{cases}\n$$\nLet us show that $d_{ij} = d_{ik}$ implies $j = k$. If $i = n$, we have $2j \\equiv 2k \\pmod{n-1} \\Leftrightarrow j \\equiv k \\pmod{n-1}$. Since $j, k \\le n-1$, we must have $j = k$. If $i \\ne n$ and $j, k \\ne n$, $i+j \\equiv i+k \\pmod{n-1} \\Leftrightarrow j \\equiv k \\pmod{n-1}$, which implies $j = k$. If $i \\ne n$ and $j = n$ and $k \\ne n$, we have $2i \\equiv i+k \\pmod{n-1} \\Leftrightarrow i \\equiv k \\pmod{n-1}$, which implies $i = k$, absurd. Hence $d_{ij} = d_{ik} \\Leftrightarrow j = k$.\nThus $n-1$ Sundays are sufficient to set up the championship.\n\nSecond case: $n$ is odd.\nEach team must play $n-1$ times on at least $n-1$ Sundays. However on each Sunday at least one team does not play. Hence $m \\ge n$.\nLet us prove that $n$ Sundays are sufficient. Let $T_0$ be a \"dummy\" team. If $T_i$ plays against $T_0$ on a given Sunday, it means that $T_i$ doesn't play on that Sunday. Thus we have $n+1$ teams. As $n+1$ is even, as shown above, we know that $(n+1)-1 = n$ Sundays are sufficient.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26346,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a triangle $ABC$, explain how to construct with ruler and compass a triangle $A'B'C'$ of minimum area such that $C' \\in AC$, $A' \\in AB$, $B' \\in BC$ and $\\angle B'A'C' = \\angle BAC$, $\\angle A'C'B' = \\angle ACB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\angle A$, $\\angle B$ and $\\angle C$ be the angles of $\\triangle ABC$ and $\\angle A'$, $\\angle B'$ and $\\angle C'$ be the angles of $\\triangle A'B'C'$. Thus $\\angle A = \\angle A'$ and $\\angle C = \\angle C'$.\n\nThe circumcircle of $\\triangle AA'C'$ meets the circumcircle of $\\triangle CC'B'$ at $D \\neq C'$. Thus $\\angle A'DC' = \\pi - \\angle A$ and $\\angle C'DB' = \\pi - \\angle C$. Moreover, $\\angle A'DB' = 2\\pi - (\\pi - \\angle A) - (\\pi - \\angle C) = \\pi - \\angle B$. Hence the circumcircle of $\\triangle BB'A'$ passes through $D$.\n\nIt is easy to see that $\\angle DAA' = \\angle DC'A' = \\alpha$ and $\\angle DA'C' = \\angle DAC' = \\alpha$. Since $\\angle A = \\angle A'$ we conclude that $\\angle DA'B' = \\alpha$. Furthermore $\\angle DBB' = \\alpha$ as $BB'DA'$ is cyclic. Analogously $\\angle DCC' = \\alpha$.\n\nHence $D$ is a fixed point since it satisfies\n$$\n\\angle DAB = \\angle DBC = \\angle DCA\n$$\nand consequently does not depend on $A'$, $B'$ and $C'$. Its construction is shown below.\n\nAs the angles $\\angle A'DB'$, $\\angle B'DC'$ and $\\angle C'DA'$ are constant, the minimum area of $\\triangle A'B'C'$ occurs when the lengths of $DA'$, $DB'$ and $DC'$ are minimum. Thus the lines $DA'$, $DB'$ and $DC'$ ought to be respectively orthogonal to the sides $AB$, $BC$ and $CA$.\n\nConstruction of point $D$\n\nLet $E$ be the intersection point of the perpendicular bisector of $AB$ and the perpendicular to $BC$ through $B$. Thus the circle with center $E$ and radius $EA$ touches $BC$ at $B$. Therefore $\\angle XAB = \\angle XBC$ for each point $X$ in the shorter arc $AB$.\n\nSimilarly, define $F$ as the intersection point of the perpendicular bisector of $BC$ and the perpendicular to $CA$ through $C$. The circle with center $F$ and radius $FB$ touches $CA$ at $C$. Therefore $\\angle XBC = \\angle XCA$ for each point $X$ on the shorter arc $BC$.\n\nThe point $D$ is the intersection point of these two arcs. It is easy to see that the point $D$ is unique.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26347,
    "subject": "Mathematics (Multi-modal)",
    "question": "Line *r* passes through the corner $A$ of a sheet of paper and makes an angle $\\alpha$ with the horizontal border, as shown in figure 1. In order to divide $\\alpha$ into three equal parts we proceed as follows:\n\na) initially we mark two points $B$ and $C$ on the vertical border such that $AB = BC$; through $B$ we draw a line $s$ parallel to the border (figure 2);\n\nb) after that, we fold the sheet so as to make $C$ coincide with a point $C'$ on the line $r$ and $A$ with a point $A'$ on line $s$ (figure 3); we call $B'$ the point which coincides with $B$.\n\n![](attached_image_1.png)\nFigure 1\n![](attached_image_2.png)\nFigure 2\n![](attached_image_3.png)\nFigure 3\n\nShow that lines $AA'$ and $AB'$ divides angle $\\alpha$ into three equal parts.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $P$ and $X$ be the points determined by the crease in the lower horizontal border of the sheet and on the line $s$, respectively. Let $\\beta = \\angle PAA'$. Since $AP = AP'$, $\\angle PA'A = \\angle AA'B = \\beta$, thus $\\angle PA'X = 2\\beta$. Moreover $\\triangle APX \\cong \\triangle A'PX$, so that $\\angle XAA' = \\angle XAP - \\beta = \\beta$.\n\nNow observe that $\\angle BXA = 2\\beta$ and, since $A'P \\parallel B'P$, $\\angle B'XA' = 2\\beta$ also. Thus $A, X$ and $B'$ are collinear. Therefore $AB'$ is a median and height of $\\triangle C'AA'$, so $AB'$ is also an angle bisector of $\\angle C'AA'$. Thus $\\angle C'AB' = \\angle B'AA' = \\angle A'AP = \\beta$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26348,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\sigma(n)$ be the sum of all positive divisors of $n$, where $n$ is a positive integer (for instance, $\\sigma(6) = 12$ and $\\sigma(11) = 12$). We say that $n$ is *almost perfect* if $\\sigma(n) = 2n - 1$ (for instance, 4 is almost perfect since $\\sigma(4) = 7$). Let $n \\mod k$ be the remainder of the division of $n$ by $k$ and $s(n) = \\sum_{1 \\le k \\le n} n \\mod k$ (for instance, $s(6) = 0 + 0 + 0 + 2 + 1 + 0 = 3$ and $s(11) = 0 + 1 + 2 + 3 + 1 + 5 + 4 + 3 + 2 + 1 + 0 = 22$). Prove that $n$ is almost perfect if and only if $s(n) = s(n-1)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "$$\n\\begin{align*}\ns(n) = s(n-1) &\\iff \\sum_{1 \\le k \\le n} n \\bmod k = \\sum_{1 \\le k \\le n-1} (n-1) \\bmod k \\\\\n&\\iff \\sum_{1 \\le k \\le n} n \\bmod k = \\sum_{1 \\le k \\le n-1} ((n \\bmod k) - 1) + \\sum_{\\substack{1 \\le k \\le n-1 \\\\ k \\nmid n}} k \\\\\n&\\iff \\sum_{1 \\le k \\le n-1} 1 = \\sigma(n) - n \\\\\n&\\iff 2n - 1 = \\sigma(n)\n\\end{align*}\n$$",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 26349,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f: \\mathbb{Z}_{+}^{*} \\rightarrow \\mathbb{Z}_{+}^{*}$ be a function defined as follows: given $n$, we write $n = 2^a \\cdot (2b+1)$, with $a$ and $b$ non-negative integers, and define $f(n) = a^2 + a + 1$.\n\nDetermine the least positive integer $n$ such that $f(1) + f(2) + \\cdots + f(n) \\ge 123456$.",
    "options": [],
    "answer": "24710",
    "solution": "Let $S_k = f(1) + f(2) + f(3) + \\dots + f(2^k)$. Then\n$$\nS_k = f(1) + f(2) + \\dots + f(2^{k-1}) + f(1) + f(2) + \\dots + f(2^{k-1} - 1) + f(2^k)\n$$\n$$\n\\Leftrightarrow S_k = 2S_{k-1} - f(2^{k-1}) + f(2^k)\n$$\n$$\n\\Leftrightarrow S_k = 2(S_{k-1} + k)\n$$\nIf $n = (a_t a_{t-1} \\dots a_0)_2$ is the binary expansion of $n$, it is easy to see that $f(1) + \\dots + f(n) = \\sum a_i \\cdot S_i$. Also, by induction, $S_k > S_{k-1} + \\dots + S_0$, so that in order to obtain the least $n$ satisfying $f(1) + \\dots + f(n) \\ge 123456$, we have to choose\n$$\nS_{14} + S_{13} + S_7 + S_2 + S_1 = 123456,\n$$\nthat is, $n = 2^{14} + 2^{13} + 2^7 + 2^2 + 2^1 = 24710$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26350,
    "subject": "Mathematics (Multi-modal)",
    "question": "Providence Ave. has infinitely many traffic lights, all equally spaced and synchronized. The distance between any two consecutive ones is $1500$ m. The traffic lights stay green $1.5$ minute, red $1$ minute, then green again $1.5$ minute, and so on. Suppose that a car is passing through Providence Ave. at constant speed, equal to $v$ m/s. For which values for $v$ is it possible for the car to pass through an arbitrarily large number of traffic lights without stopping?",
    "options": [],
    "answer": "All speeds of the form 20/k meters per second for positive integers k.",
    "solution": "Suppose that at instant $0$ the traffic lights turn green and the car passes through the first light at instant $t_0 \\ge 0$ (time is measured in seconds). The traffic lights will stay green between the time instants $150k + 90$ and $150(k+1)$, for each integer $k$. The car will pass through the lights at the instants $t_0 + \\frac{1500}{v} r$, for each nonnegative integer $r$.\n\nTherefore, a necessary and sufficient condition for a non-stopping trip is that, for every integer $r$, $\\frac{t_0}{150} + \\frac{10}{v}$ equals an integer plus a number between $0$ and $\\frac{3}{5}$. This is clearly possible if $\\frac{10}{v}$ is an integer (with any $t_0$ between $0$ and $90$) or if $\\frac{10}{v}$ is half of an odd integer (with any $t_0$ between $0$ and $15$).\n\nLet's show that these are the only possible situations. If $\\frac{10}{v}$ is irrational then by Kronecker theorem the fractional part of $\\frac{t_0}{150} + \\frac{10}{v}$ is dense in $(0, 1)$ and thus admits values between $\\frac{3}{5}$ and $1$. If $\\frac{10}{v}$, say, $\\frac{10}{v} = \\frac{p}{q}$, $\\gcd(p, q) = 1$, then the fractional part of $\\frac{10}{v}$ is $\\frac{pr}{q}$ takes all values of the form $\\frac{s}{q}$, $0 \\le s \\le q-1$ (for instance, consider $r = s \\cdot p^{-1} \\mod q$). So consecutive values of the fractional part of $\\frac{t_0}{150} + \\frac{10}{v}$ are $\\frac{1}{q}$ apart; the fractional part of $\\frac{t_0}{150} + \\frac{10}{v}$ takes values between $\\frac{3}{5}$ and $1$ if $\\frac{1}{r} \\le 1 - \\frac{3}{5} \\iff r \\ge 3$. Thus $r \\le 2$, that is, $\\frac{10}{v}$ is an integer or half of an odd integer.\n\nWe conclude that the possible values for the speed of the car are $\\frac{20}{k}$ m/s, for every positive integer $k$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26351,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $X$ be the set of all sequences $a = (a_1, a_2, \\dots, a_{2000})$ such that $a_i \\in \\{0, 1, 2\\}$ for $1 \\le i \\le 1000$ and $a_i \\in \\{0, 1\\}$ for $1001 \\le i \\le 2000$. Given $a, b$ in $X$, we define the distance $d(a, b)$ between $a$ and $b$ as the number of values for $i$, $1 \\le i \\le 2000$, for which $a_i \\ne b_i$. Find the number of distance-preserving functions $f: X \\to X$, i.e., such that $d(f(a), f(b)) = d(a, b)$ for all $a$ and $b$ in $X$.",
    "options": [],
    "answer": "1000!^2 * 12^{1000}",
    "solution": "Let $a = f(0, 0, 0, \\dots, 0)$, $b = f(1, 0, 0, \\dots, 0)$ and $c = f(2, 0, 0, \\dots, 0)$, say $a = (a_1, a_2, \\dots, a_{2000})$, $b = (b_1, b_2, \\dots, b_{2000})$ and $c = (c_1, c_2, \\dots, c_{2000})$. It's easy to conclude that there exists a unique $1 \\le i_0 \\le 2000$ such that $a, b$ and $c$ differ at position $i_0$, i.e., $a_{i_0} \\ne b_{i_0} \\ne c_{i_0} \\ne a_{i_0}$. This also gives $i_0 \\le 1000$. Another easy argument shows that if $x_0 = f(0, d_2, d_3, \\dots, d_{2000}) = (x_1, x_2, \\dots, x_{2000})$, then $x_{i_0} = a_{i_0}$ and also if $x_1 = f(1, e_2, e_3, \\dots, e_{2000}) = (y_1, y_2, \\dots, y_{2000})$, then $y_{i_0} = b_{i_0}$.\nA similar argument shows the following: let\n$$\na_t = f(0, 0, \\dots, 0, \\dots, 0) = (a_1, a_2, \\dots, a_{2000}),\n$$\n$$\nb_t = f(0, 0, \\dots, 1, \\dots, 0) = (b_1, b_2, \\dots, b_{2000}),\n$$\n$$\nc_t = f(0, 0, \\dots, 2, \\dots, 0) = (c_1, c_2, \\dots, c_{2000}),\n$$\nwhere $t \\le 1000$ is the position of 1 and 2, and $x = f(x_1, x_2, \\dots, x_{2000}) = (y_1, y_2, \\dots, y_{2000})$. Therefore $y_{i_t} = a_{i_t}$ if $x_t = 0$ ($i_t$ is the position that changes from $a_t$ to $b_t$), $y_{i_t} = b_{i_t}$ if $x_t = 1$ and $y_{i_t} = c_{i_t}$ if $x_t = 2$.\nNow, it's clear that $(i_1, i_2, \\dots, i_{1000})$ is a permutation of $(1, 2, \\dots, 1000)$. On the other hand, if $a_t = f(0, 0, \\dots, 0, \\dots, 0) = (a_1, a_2, \\dots, a_{2000})$ and $b_t = f(0, 0, \\dots, 1, \\dots, 0) = (b_1, b_2, \\dots, b_{2000})$, where $t \\ge 1000$ is the position of 1, there is a unique $j \\ge 1000$ such that $a_j \\ne b_j$. Denoting this $j$ by $i_j$, if $x = f(x_1, x_2, \\dots, x_{2000}) = (y_1, y_2, \\dots, y_{2000})$, one can show in a similar way that $y_{i_t} = a_{i_t}$ if $x_t = 0$ and $y_{i_t} = b_{i_t}$ if $x_t = 1$. Finally, in order to count the number of different functions $f: X \\to X$, it suffices to choose $(i_1, i_2, \\dots, i_{1000})$, which is a permutation of $(1, 2, \\dots, 1000)$, $(i_{1001}, i_{1002}, \\dots, i_{2000})$, which is a permutation of $(1001, 1002, \\dots, 2000)$, and $(a_{i_t}, b_{i_t}, c_{i_t})$ for $1 \\le t \\le 1000$ and $(a_{i_t}, b_{i_t})$ for $1001 \\le t \\le 2000$.\nThis gives that the total number of functions is $1000!^2 \\times 12^{1000}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26352,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $C$ be a wooden cube. For each pair $(X, Y)$ of vertices of $C$, we cut $C$ through the plane orthogonal to $\\overline{XY}$ and passing through its midpoint. Into how many pieces is the cube divided?",
    "options": [],
    "answer": "96",
    "solution": "Let's call the plane orthogonal to and passing through the midpoint of a segment $\\overline{AB}$ the \"medial plane\" of $\\overline{AB}$. There exist three different kinds of medial planes: the medial planes of the edges of $C$ (there are 3 different such planes); the medial planes of the diagonals of the faces of $C$ (there are 6 of such planes); the medial planes of the diagonals of $C$ (3 more planes). They divide $C$ into triangular pyramids, all of them having the center of $C$ as vertex. Moreover they divide each face of $C$ into triangles, that serve as bases for the pyramids. It's easy to see that each face is divided in exactly 16 triangles, so that the total number of pieces is $16 \\times 6 = 96$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26353,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that\n$$\n(a+b)(a+c) \\ge 2\\sqrt{abc(a+b+c)}\n$$\nfor all positive real numbers $a$, $b$ and $c$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By AM-GM,\n$$\n(a+b)(a+c) = bc + a(a+b+c) \\geq 2\\sqrt{bc \\cdot a(a+b+c)}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26354,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a given integer $a_0 > 1$ one defines a sequence $(a_n)_{n \\ge 0}$ as follows: for each $k \\ge 0$, define $a_{k+1}$ as the smallest integer greater than $a_k$ such that $\\text{gcd}(a_{k+1}, a_0 a_1 \\cdots a_k) = 1$. Determine all values $a_0$ for which all terms $a_k$ are primes or powers of primes.",
    "options": [],
    "answer": "a0 ∈ {2, 3, 4, 7, 8}",
    "solution": "We show first that if (1) $a_n$ is prime, (2) all terms $a_i$ for $i < n$ are primes or prime powers, (3) given any prime $p < a_n$, we have $p \\mid a_i$ for some $i < n$, then $a_i$ is prime for all $i > n$. For suppose $a_n < k < q$, where $q$ is the next larger prime than $a_n$. Then $k$ lies between two consecutive primes, so it must be composite. So it must have a prime factor less than $a_n$. By (3) that must divide $a_i$ for some $i < n$. Hence $a_{n+1} \\neq k$. On the other hand $q$ is obviously coprime to all $a_i$ for $i \\le n$, so $a_{n+1} = q$.\n\n* $a_0 = 2$ gives all terms primes.\n* $a_0 = 3$ gives $a_1 = 4$, $a_2 = 5$ and hence all higher terms prime.\n* $a_0 = 4$ gives $a_1 = 5$, $a_2 = 7$, $a_3 = 9$, $a_4 = 11$, and hence all higher terms prime.\n* $a_0 = 5$ gives $a_1 = 6$, which is not a prime or prime power.\n* $a_0 = 6$ is not a prime or prime power.\n* $a_0 = 7$ gives $a_1 = 8$, $a_2 = 9$, $a_3 = 11$, $a_4 = 13$, $a_5 = 17$, $a_6 = 19$, $a_7 = 23$, $a_8 = 25$, $a_9 = 29$, and hence all higher terms prime.\n* $a_0 = 8$ gives 9, 11, 13, 17, 19, 23, 25, 29, 31, 37, 41, 43, 47, 49, 53 and hence all higher terms prime.\n* $a_0 = 9$ gives $a_1 = 10$, which is not a prime or prime power.\n\nNow suppose $a_0 > 9$. We show first that there are no solutions to $3^k = 2^h \\pm 1$ for $k > 2$. Consider $3^k = 2^h+1$. If $k > 2$, then $h > 3$, so $3^k - 1 = (4-1)^k - 1$ is divisible by 4, so $k$ must be even. Put $k = 2m$. Then $2^h = (3^m+1)(3^m-1)$. But $3^m - 1$ and $3^m + 1$ cannot both be powers of 2 for $m > 1$, so there are no solutions for $k > 2$. Consider $3^k = 2^h - 1$. Again $2 = (3-1)$, so $h$ must be even. Put $h = 2m$, then $3^k = (2^m-1)(2^m+1)$. But $2^m - 1$ and $2^m + 1$ cannot both be powers of 3 for $m > 1$. So there are no solutions for $k > 2$.\n\nNow if $a_0 = 6n$, then $a_0$ is obviously not a prime power. If $a_0 = 6n - 1$, then $a_1 = 6n$, which is not a prime power. If $a_0 = 6n + 1$, then $a_1 = 6n + 2$, which is divisible by 2. So if it is a prime power, then it must be a power of 2. But $a_2 = 6n + 3$, which is divisible by 3, so it can only be a prime power if it is a power of 3. We have just shown that is impossible for $a_0 > 9$. Thus\n$a_0 = 6n+1$ fails for $a_0 > 9$. Similarly, if $a_0 = 6n+2$, then $a_0$ must be a power of 2, and $a_1 = 6n+3$ must be a power of 3, which is impossible for $a_0 > 9$. Similarly, if $a_0 = 6n+3$, $a_1 = 6n+4$ and again they cannot both be prime powers.\n\nWe claim that there are no solutions to $3^k = 2^h + 5$ for $k > 2$. Putting $3 = 4-1$, we see that $k$ must be even. Putting $2 = 3-1$, we see that $h$ must be even. Put $k = 2K$, $h = 2H$, then $3^{2K}-2^{2H} = 5$, so $(3^K+2^H)(3^K-2^H) = 5$. But that is impossible for $K > 2$ and hence $k > 4$. It is easy to see that $3^4-5$ and $3^3-5$ are not powers of 2. If $a_0 = 6n-2$ then $a_1 = 6n-1$, $a_2 = 6n+1$, $a_3 = 6n+3$. Then $a_0$ must be a power of 2 and $a_3$ must be a power of 3. But we have just shown that is impossible for $a_0 > 9$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26355,
    "subject": "Mathematics (Multi-modal)",
    "question": "Initially, a calculator displays number $1$. An operation consists in pressing either key $\\sin$ or key $\\cos$, which calculates respectively the sine and cosine with the arguments in radians. After performing $2001$ operations, what is the greatest possible value that can be achieved?",
    "options": [],
    "answer": "cos(sin(sin(…sin(1)…))) with 2000 applications of sin followed by cos",
    "solution": "Obviously the value does not exceed $1$. We cannot get close to $1$ with $\\sin$. So the final press must be $\\cos$ and we want the previous value to be as close to $0$ as possible. We cannot get close to $0$ with $\\cos$, so the $2000$th press must be $\\sin$. Thus we want the previous value to be as close to $0$ as possible. So, by a simple induction, all presses except the last must be $\\sin$.\n\nPut $a_1 = \\sin 1$. Put $a_{n+1} = \\sin a_n$ for $n = 1, \\dots, 1999$ and put $a_{2001} = \\cos a_{2000}$. The best value is $a_{2001}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26356,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a convex quadrilateral, the *altitude* relative to a side is defined to be the line perpendicular to this side through the midpoint of the opposite side. Prove that the four altitudes have a common point if and only if the quadrilateral is cyclic, that is, if and only if, there exists a circle which contains its four vertices.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider the following\n**Lemma.** Let $ABCD$ be a convex quadrilateral. The lines that connect the midpoints of opposite sides meet in $E$ and the perpendicular bisectors of opposite sides $AB$ and $CD$ meet in $O$. Then the altitudes relative to $AB$ and $CD$ meet in a point $O'$ symmetric to $O$ with respect to $E$.\n\n![](attached_image_1.png)\n\n*Proof.* It's well known that the quadrilateral whose vertices are the midpoints of $ABCD$ is a parallelogram. So $E$ is the midpoint of both its diagonals, in particular $MN$. But since the pairs of lines $OM$, $NO'$ and $ON$, $MO'$ are parallel, $MO'NO$ is also a parallelogram, so the diagonals $MN$ and $OO'$ meet in their respective midpoints. This means that $E$ is the midpoint of $OO'$ and we're done.\n\nLet $O_1$ and $O_2$ be the intersection of the perpendicular bisectors of $AB$, $CD$ and $AD$, $BC$, respectively and $O_1'$ and $O_2'$ be the intersection points of the altitudes relative to $AB$, $CD$ and $AD$, $BC$ respectively. The four altitudes have a common point if and only if $O_1' = O_2'$, which by the lemma is equivalent to $O_1 = O_2$, which is, in turn, equivalent to $ABCD$ being cyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26357,
    "subject": "Mathematics (Multi-modal)",
    "question": "One has a long row of glasses and $n$ stones in the central glass (glass $0$). The following movements are allowed:\n\n• Movement $A$\n![](attached_image_1.png)\nIf there is at least one stone in glass $i$ and at least one in glass $i+1$, one may make one stone in glass $i+1$ jump to glass $i-1$, eliminating one stone in glass $i$.\n\n• Movement *B*\n![](attached_image_2.png)\nIf there are at least two stones in glass *i* one may make a stone in glass *i* jump to glass *i* + 2 and make another stone jump to glass *i* - 1.\n\nProve the following: performing movements A and B for a sufficiently long time, one shall always obtain a configuration from which neither movements are allowed anymore. Besides this final configuration does not depend on the choices of movements during the process.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assign $x'$ to a stone in the $i$th glass. Let's consider what happens to the sum of the number assigned to all stones with each movement.\n\nMovement *A* replaces $x^i + x^{i+1}$ with $x^{i-1}$; movement *B* replaces $2x^i$ with $x^{i-1} + x^{i+2}$. So the difference in the sum is either $x^{i-1}(1 - x - x^2)$ or $x^{i-1}(1 + x^3 - 2x) = x^{i-1}(1 - x - x^2)(1 - x)$.\n\nChoose $x$ so that $1-x-x^2=0$, say $x = \\phi^{-1} = \\frac{\\sqrt{5}-1}{2}$. Then the sum $S$ of the numbers assigned to the stones always stays the same. There is a number $x^k$ such that $S < x^k$, so there is a limit to the leftmost stone, so it won't move after a while. Now consider the sum without this stone; the leftmost stone (that is, the second leftmost stone overall) won't move after a while and so on. So the configuration will remain constant at some time, that is, there won't be any two stones in the same glass nor two stones in two consecutive glasses.\n\nNow notice that the representation of $S$ in base $\\phi^{-1}$ (all digits equal to one, no two consecutive ones) is unique: suppose on the contrary that there are two such representations and let $k$ be the first position from left to right in which those representations differ. This means that $\\phi^{-k}$ is equal to the sum of some numbers of the form $\\phi^{-m}$ for $m > k + 1$. But $\\sum_{m=k+2}^{\\infty} \\phi^{-m} = \\frac{\\phi^{-k-2}}{1-\\phi^{-1}} = \\phi^{-k}$, so this isn't possible. So the final configuration is unique.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 26358,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that there is a set of $2002$ distinct positive integers such that the sum of one or more elements of the set is never a square, cube, or higher power.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $p$ be a prime and $A = \\{p, 2p, 3p, \\dots, 2002p\\}$. The sum of any quantity of numbers from $A$ is at most $p + 2p + \\dots + 2002p = 1001 \\cdot 2003p$. Choose any $p > 1001 \\cdot 2003$ and we are done, because every sum of numbers from $A$ is a multiple of $p$ but not of $p^2$, and cannot be a perfect power.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26359,
    "subject": "Mathematics (Multi-modal)",
    "question": "$ABCD$ is a cyclic quadrilateral and $M$ a point on the side $CD$ such that $ADM$ and $ABCM$ have the same area and the same perimeter. Show that two sides of $ABCD$ have the same length.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\nSince the perimeter of $ADM$ and $ABCM$ are equal and share the side $AM$, $a + b + m = n + c \\iff n - m = a + b - c$. Since they also have the same area, the area of $ADM$ is half the area of $ABCD$. If $\\angle ABC = \\theta$, then $\\angle CDA = 180^\\circ - \\theta$ and\n$$\n\\begin{aligned}\n\\frac{c n \\sin(180^\\circ - \\theta)}{2} &= \\frac{1}{2} \\cdot \\frac{a b + c(m + n)}{2} &&\\iff 2 c n = a b + c m + c n \\\\\n\\iff c(n - m) &= a b &&\\iff c(a + b - c) = a b \\\\\n\\iff (c - a)(c - b) &= 0 &&\\iff c = a \\text{ or } c = b\n\\end{aligned}\n$$\nand we are done.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26360,
    "subject": "Mathematics (Multi-modal)",
    "question": "The squares of an $m \\times n$ board are labeled from $1$ to $mn$ so that the squares labeled $i$ and $i+1$ always have a side in common. Show that for some $k$ the squares $k$ and $k+3$ have a side in common.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider the center of the $mn$ unit squares and connect two centers if the numbers assigned to the correspondent squares are consecutive. Then we obtain a path with $mn-1$ unit segments. The $mn$ centers determine a lattice with $(m-1)(n-1)$ unit squares. Squares with numbers $k$ and $k+3$ correspond to a lattice square with three unit segments:\n\n![](attached_image_1.png)\n\nSo we need to prove that there is a lattice square with at least $3$ segments. Suppose the contrary, so the number of segments is at most $2(m-1)(n-1)$, in which border segments are counted once and the interior segments are counted twice. The path uses at most $2(m-1)+2(n-1)-1 = 2m+2n-5$ segments on the border. So there are at least $mn-1-(2m+2n-5) = mn-2m-2n+4$ interior segments. Hence the path uses $2m+2n-5+2(mn-2m-2n+4) = 2mn-2m-2n+3 > 2(m-1)(n-1)$ lattice segments (counting repetitions), a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26361,
    "subject": "Mathematics (Multi-modal)",
    "question": "For any non-empty subset $A$ of $\\{1, 2, \\dots, n\\}$ define $f(A)$ as the largest element of $A$ minus the smallest element of $A$. Find $\\sum f(A)$ where the sum is taken over all non-empty subsets of $\\{1, 2, \\dots, n\\}$.",
    "options": [],
    "answer": "(n - 3) * 2^n + n + 3",
    "solution": "Let $m$ and $M$ be the sum of the minima and maxima of all subsets. Since the diameter of a set is the difference between its maximum and its minimum, the desired sum is $M - m$. We may include unitary subsets, since their minima and maxima coincide.\n\nThe number $k$, $1 \\le k \\le n$, is the minimum of all subsets of the form $\\{k\\} \\cup A$, where $A \\subset \\{k+1, k+2, \\dots, n\\}$. So $k$ is the minimum of $2^{n-k}$ subsets. Hence\n$$\nm = \\sum_{k=1}^{n} k \\cdot 2^{n-k} = \\sum_{k=0}^{n-1} (n-k) \\cdot 2^k.\n$$\nNow we count the number of subsets with diameter $k$. Let $a$ be the minimum of a subset. The maximum is $a+k$. Since $a+k \\le n$, one may choose $a$ in $n-k$ ways. Since there are $k-1$ numbers between $a$ and $a+k$, there are $(n-k) \\cdot 2^{k-1}$ subsets with diameters $k$. Since there are $2^n-n-1$ non-empty and non-unitary subsets,\n$$\n\\begin{align*}\n\\sum_{k=1}^{n-1} (n-k) \\cdot 2^{k-1} &= 2^n - n - 1 \\\\\n\\iff \\sum_{k=1}^{n-1} (n-k) \\cdot 2^k &= 2^{n+1} - 2n - 2 \\\\\n\\iff \\sum_{k=0}^{n-1} (n-k) \\cdot 2^k &= 2^{n+1} - 2n - 2 + n = 2^{n+1} - n - 2\n\\end{align*}\n$$\nand $m = 2^{n+1} - n - 2$.\n\nIn order to compute $M$, notice that the association $A = \\{a_1, a_2, \\dots, a_m\\} \\to f(A) = \\{n+1-a_1, n+1-a_2, \\dots, n+1-a_m\\}$ is clearly a bijection that transform minima $\\ell$ in maxima $n+1-\\ell$. Since there are $2^n-1$ non-empty subsets, $M+m = (n+1)(2^n-1) \\iff M-m = (n+1)(2^n-1)-2(2^{n+1}-n-2) = (n-3) \\cdot 2^n + n + 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26362,
    "subject": "Mathematics (Multi-modal)",
    "question": "A finite collection of squares has total area $4$. Show that they can be arranged to cover a square of side $1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the sides of the squares be equal to $a_i$ for $i = 1, 2, \\ldots, N$ ($N$ is the number of squares).\n\nIf some $a_k > 1$ then the $k$th square will cover the unit square. Now let's assume the case when $a_i < 1$ for all $i$.\n\nEvery number $a_i$ must satisfy $2^{-k_i} \\le a_i < 2^{-k_i+1}$ for some integer $k_i$. Now let's decrease every $i$th square to the square with side $2^{-k_i}$. Then its area would decrease by at most $4$ times. Therefore the area of all squares will be greater than $1$.\n\nNow let's prove we can tile the unit square fully with the new squares. Let's divide the unit square into $4$ squares of side $\\frac{1}{2}$. First place the squares with side $\\frac{1}{2}$, if they exist. Then on the non-tiled squares with side $\\frac{1}{2}$, if they exist, place the squares with side $\\frac{1}{4}$, if they exist, dividing each non-tiled square with side $\\frac{1}{2}$ into $4$ equal squares. We will continue this procedure for $k = 3, 4, \\ldots$ by placing squares with side $\\frac{1}{2^k}$ on the non-tiled squares with side $\\frac{1}{2^{k-1}}$, in turn dividing them into $4$ equal squares.\n\nFinally, because the sum of areas of squares is greater than $1$, then on some step, we will cover the square. Then increasing the $i$th square to the square with side $a_i$, we will get the tiling of the unit square with the given squares.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26363,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that we cannot form more than $4096$ binary sequences of length $24$ so that any two differ in at least $8$ positions.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $p$, $q$ be two binary sequences. Define $d(p, q)$ as the number of positions $p$ and $q$ differ.\n\n**Lemma.** $d$ satisfies the triangle inequality, that is, $d(p, q) \\le d(p, r) + d(r, q)$ for all binary sequences $p$, $q$, $r$.\n\n*Proof.* $p$ and $r$ differ in $d(p, r)$ positions, so they coincide in $24 - d(p, r)$ positions. Of these positions, in at most $d(q, r)$ positions $q$ and $r$ differ, so $p$, $q$ and $r$ coincide in at least $24 - d(p, r) - d(q, r)$ positions. So $d(p, q) \\le d(p, r) + d(r, q)$.\n\nDefine the *sphere* with center $p$ as the set of binary sequences $q$ such that $d(p, q) \\le 4$. Let $S$ be a set of binary sequences of length $24$ so that any two differ in at least $8$ positions and $p_1$, $p_2$ two binary sequences from $S$. If the spheres with centers $p_1$ and $p_2$ have an intersection, then $p_1$ and $p_2$ are at distance at most $4 + 4 = 8$. So $d(p_1, p_2) = 8$ and for all sequences $q$ in the intersection $d(p_1, q) = d(p_2, q) = 4$. Each sequence $q$ belongs to at most $24 / 4 = 6$ spheres with center in $S$, because if $q$ belongs to more spheres then there are two sequences $q_1$ and $q_2$ such that they differ with $q$ in the same positions. This means that $d(q_1, q_2) < d(q_1, q) + d(q, q_2) = 8$, contradiction.\n\nThe number of sequences in each sphere is $\\binom{24}{0} + \\binom{24}{1} + \\binom{24}{2} + \\binom{24}{3} + \\binom{24}{4}$. Since each pair of spheres can intersect and each sequence belongs to at most $6$ spheres,\n$$\n|S| \\cdot \\left( \\binom{24}{0} + \\binom{24}{1} + \\binom{24}{2} + \\binom{24}{3} + \\frac{1}{6} \\binom{24}{4} \\right) \\le 2^{24} \\\\\n\\iff |S| \\cdot 4096 \\le 2^{24} \\iff |S| \\le 4096\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26364,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest positive prime that divides $n^2 + 5n + 23$ for some integer $n$.",
    "options": [],
    "answer": "17",
    "solution": "Let $f(n) = n^2 + 5n + 23$. For $n = -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4$ we get $f(n) = 47, 37, 29, 23, 19, 17, 17, 19, 23, 29, 37, 47, 59$. Observe that $f(-n-5) = f(n)$, so this only involves calculating 7 terms. None of these are divisible by any of $2, 3, 5, 7, 11, 13$. But all possible values mod $13$ are assumed by $f(n)$ for any 13 consecutive integers, similarly for 11, 7 etc. So $f(n)$ is not divisible by $2, 3, 5, 7, 11, 13$ for any $n$. On the other hand $f(-2) = 17$. So the answer is $17$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26365,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $S$ be a set with $n$ elements. Take a positive integer $k$. Let $A_1, A_2, \\dots, A_k$ be any distinct subsets of $S$. For each $i$ take $B_i = A_i$ or $B_i = S - A_i$. Find the smallest $k$ such that we can always choose $B_i$ so that $\\bigcup_{1 \\le i \\le k} B_i = S$.",
    "options": [],
    "answer": "the smallest k such that 2^k > n (equivalently, floor(log2 n) + 1)",
    "solution": "The $2^k$ sets $C_1 \\cap C_2 \\cap \\dots \\cap C_k$, where $C_i = A_i$ or $C_i = S - A_i$, are all disjoint. If $2^k > n$, it follows that one of them must be empty. Hence its complement (which is $D_1 \\cup D_2 \\cup \\dots \\cup D_k$, where $D_i = S - A_i$), is $S$.\n\nOn the other hand if $2^k = n$, then we can choose $k$ subsets $A_i$ such that all $2^k$ intersections are non-empty. That means all possible unions are incomplete. Thus the answer is the smallest $k$ such that $2^k > n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26366,
    "subject": "Mathematics (Multi-modal)",
    "question": "$ABCD$ is a rhombus. Take points $E, F, G, H$ on sides $AB, BC, CD, DA$ respectively so that $EF$ and $GH$ are tangent to the incircle of $ABCD$. Show that $EH$ and $FG$ are parallel.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $O$ be the center of the incircle. We show first that $AE \\cdot CF = AO^2$. Let $\\angle AOE = \\theta$. Then if $OX$ is the perpendicular from $O$ to $AB$, we have $\\angle AOX = B/2$ and hence $\\angle XOE = \\theta - B/2$. If $EF$ touches the circle at $Y$, $\\angle EOY = \\angle XOE$, so $\\angle BEF = 2(\\theta - B/2) = 2\\theta - B$. Hence $\\angle BFE = 180^\\circ - 2\\theta$. Hence $\\angle CFE = 2\\theta$. So $\\angle CFO = \\theta$. So we have established that $\\angle AOE = \\angle CFO$. But $\\angle EAO = \\angle OCF$ (since $ABCD$ is a rhombus), so $AOE$ and $CFO$ are similar. Hence $AE/AO = CO/CF \\Leftrightarrow AE \\cdot CF = AO^2$.\n\nSimilarly, $AH \\cdot CG = AO^2$. Hence $AH/AE = CF/CG$. So $AHE$ and $CFG$ are similar. So $\\angle AEH = \\angle CGF$, so $EH$ and $FG$ are parallel.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26367,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a circle and a point $A$ inside the circle, but not at its center. Find points $B, C, D$ on the circle which maximize the area of the quadrilateral $ABCD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If we fix the length $BD$, then we obviously maximize area $BDA$ by taking the distance of $A$ from $BD$ as large as possible and hence by taking $BD$ perpendicular to $AO$ and on the opposite side of $O$ to $A$. We maximize $BCD$ by taking $C$ the midpoint of the arc $BD$. So suppose the radius of the circle is $R$, $OA = d$ and we take $BD = 2x$. We find area $ABCD = x(R + d)$, which is maximized by maximizing $x$. Thus we take $BD$ to be the diameter perpendicular to $AO$ and $C$ the point of the circle on the ray $AO$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26368,
    "subject": "Mathematics (Multi-modal)",
    "question": "$f(x)$ is a real-valued function defined on the positive reals such that\n(1) if $x < y$, then $f(x) < f(y)$;\n(2) $f\\left(\\frac{2xy}{x+y}\\right) = \\frac{f(x)+f(y)}{2}$ for all $x$.\nShow that $f(x) < 0$ for some value of $x$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Put $x_n = 1/n$, $y_n = f(x_n)$. We have $\\frac{2x_{n-1}x_{n+1}}{x_{n-1} + x_{n+1}} = x_n$, so $y_n = \\frac{y_{n-1} + y_{n+1}}{2}$, or $y_n - y_{n+1} = y_{n-1} - y_n$. Now $1/2 < 1$, so $y_2 < y_1$.\n\nPut $y_1 - y_2 = d > 0$. Then $y_n - y_{n+1} = d$ for all $n$. Hence $y_{n+1} = y_1 - nd$. So $y_n$ is negative for sufficiently large $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26369,
    "subject": "Mathematics (Multi-modal)",
    "question": "A graph $G$ with $n$ vertices is called *great* if we can label each vertex with a different positive integer not exceeding $\\left\\lfloor \\frac{n^2}{4} \\right\\rfloor$ and find a set of non-negative integers $D$ so that there is an edge between two vertices if and only if the difference between their labels is in $D$. Show that if $n$ is sufficiently large we can always find a graph with $n$ vertices which is not great.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let's count the number of ordered pairs $(f, D)$, where $f$ is a labeling of vertices and $D$ a possible set of differences of labels of vertices joined by an edge.\nNote that an ordered pair $(f, D)$ determines at most one great graph, since the labeling $f$ determine the vertices and the set of differences $D$ determine the edges of the graph (we join two vertices if and only if the difference between their labels is in $D$), unless there is some contradiction among the differences. In the latter case $(f, D)$ determines no great graph.\nThere are $\\frac{\\left\\lfloor \\frac{n^2}{4} \\right\\rfloor!}{\\left(\\left\\lfloor \\frac{n^2}{4} \\right\\rfloor - n\\right)!} \\le \\left(\\frac{n^2}{4}\\right)^n$ labelings and since $D \\subset \\{1, 2, \\dots, \\left\\lfloor \\frac{n^2}{4} \\right\\rfloor\\}$, there are $2^{\\left\\lfloor \\frac{n^2}{4} \\right\\rfloor} \\le 2^{\\frac{n^2}{4}}$ possibilities for $D$. Thus the number of ordered pairs (and, henceforth, the number of great graphs with $n$ vertices) is not greater than $\\left(\\frac{n^2}{4}\\right)^n \\cdot 2^{\\frac{n^2}{4}}$.\nBut the number of graphs with $n$ vertices is $2^{\\binom{n}{2}} = 2^{\\frac{n(n-1)}{2}}$. Therefore it suffices to show that for sufficiently large $n$\n$$\n\\left(\\frac{n^2}{4}\\right)^n \\cdot 2^{\\frac{n^2}{4}} < 2^{\\frac{n(n-1)}{2}}\n$$\nTaking logarithms on both sides,\n$$\nn \\cdot \\log_2 \\frac{n^2}{4} + \\frac{n^2}{4} < \\frac{n^2}{2} - \\frac{n}{2} \\iff 2 \\log_2 \\left(\\frac{n}{2}\\right) < \\frac{n}{4} - \\frac{1}{2}\n$$\nwhich is true for sufficiently large $n$, since log grows slower than any polynomial with degree at least 1 and positive leader coefficient (it is true, indeed, for $n > 33$).",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 26370,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a convex quadrilateral. Prove that the incircles of the triangles $ABC$, $BCD$, $CDA$ and $DAB$ have a point in common if, and only if, $ABCD$ is a rhombus.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $ABCD$ is a rhombus the incircle touch the respective triangle in the midpoint of the diagonals, which belongs to all four circles.\n\nNow suppose the circles have a common point $P$. Since the incircle are contained in its triangle and the intersection of the triangles $ABC$, $BCD$, $CDA$, $DAB$ is the intersection of the diagonals, $P$ must coincide with $O$.\n\nLet's prove that the diagonals are perpendicular. Suppose that it's not the case. So we may suppose without loss of generality that $\\angle AOB = \\angle COD > 90^\\circ$. Let $l_1, l_2, l_3$ and $l_4$ be the incenters of $ABC$, $BCD$, $CDA$ and $DAB$, respectively. Thus $l_1$ and $l_4$ are in the interior of $\\angle ABC$ and $l_2$ and $l_3$ are in the interior of $\\angle COD$.\n\n![](attached_image_1.png)\n\nLet $E$ and $F$ be the intersection of $AI_4$, $BI_1$ and $CI_2$, $DI_3$, respectively. So $\\angle BEA > \\angle AOB > 90^\\circ \\Rightarrow 180^\\circ - \\frac{\\angle DAB}{2} - \\frac{\\angle ABC}{2} > 90^\\circ \\Leftrightarrow \\angle DAB + \\angle ABC < 180^\\circ$. Using a similar argument on $F$ we obtain $\\angle BCD + \\angle CDA < 180^\\circ$. Summing these two inequalities, we obtain that the sum of the internal angles of the quadrilateral $ABCD$ is less than $360^\\circ$, contradiction. So $AC$ and $BD$ are perpendicular and, moreover, the four incenters $l_1, l_2, l_3$ and $l_4$ lie in the diagonals. This means that the bisectors and altitudes from vertices $A, B, C, D$ with respect to the triangles $DAB$, $ABC$, $BCD$, $CDA$ coincide and hence these triangles are isosceles. Thus $DA = AB = BC = CD$ and $ABCD$ is a rhombus.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26371,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all values of $n$ such that it is possible to divide a triangle in $n$ smaller triangles such that there are not three collinear vertices and such that each vertex belongs to the same number of segments.",
    "options": [],
    "answer": "3, 7, 19",
    "solution": "Consider the planar graph which vertices are the vertices of the triangles and edges are the sides of the triangles. Let $V$, $E$ and $F$ be the number of vertices, edges and faces of such graph and $d$ be the degree of each vertex. Note that $n = F - 1$. Then $E = \\frac{3F}{2} = \\frac{dV}{2} \\iff V = \\frac{2E}{d}$ and $F = \\frac{2E}{3}$. By Euler's theorem, $V - E + F = 2 \\iff \\frac{1}{d} + \\frac{1}{3} = \\frac{1}{2E} + \\frac{1}{2} > \\frac{1}{2} \\implies \\frac{1}{d} > \\frac{1}{6} \\iff d < 6$. It's not hard to see that $d \\ge 3$ and that each value for $d$ determines $V$, $E$ and $F$. Indeed $d = 3 \\implies F = 4$; $d = 4 \\implies F = 8$; $d = 5 \\implies F = 20$. The following examples show that the possible values for $n$ are indeed $3$, $7$ and $19$.\n![](attached_image_1.png)\n![](attached_image_2.png)\n![](attached_image_3.png)",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 26372,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x_1, x_2, \\dots, x_{2004}$ be a sequence of integer numbers such that $x_{k+3} = x_{k+2} + x_k x_{k+1}$, $1 \\le k \\le 2001$. Is it possible that more than half of the elements are negative?",
    "options": [],
    "answer": "Yes",
    "solution": "The answer is yes. For instance, consider $x_0 = -1$, $x_1 = -n$ and $x_2 = -n^2$, $n$ sufficiently large. All $x_i$'s are polynomials in $n$ and, for all large $n$ its sign is equal to the sign of the coefficient of the term of greatest degree. Let $a(x_n)$ such term. Then\n$$\n\\begin{aligned}\na(x_0) &= -1, & a(x_1) &= -n, & a(x_2) &= -n^2, & a(x_3) &= -n^3, \\\\\na(x_4) &= n^3, & a(x_5) &= n^4, & a(x_6) &= -n^5, & a(x_7) &= n^7\n\\end{aligned}\n$$\nLet's prove by induction that, for $k \\ge 5$, $a(x_k) = a(x_{k-2})a(x_{k-3})$, which is equivalent to prove that the degree of $a(x_{k-1})$ is less than the degree of $a(x_{k-2}x_{k-3})$. This is true for $k = 5, 6, 7$. Suppose that it is true for all $k$ less than $m$. So the degree of $a(x_{m-1}) = a(x_{m-3})a(x_{m-4})$ is less than the degree of $a(x_{m-2})a(x_{m-3})$ because by the induction hypothesis the degree of $a(x_l)$ is less than the degree of $a(x_{l-1})$ and we're done.\nSo the sign of $x_k$ is equal to the sign of $x_{k-2}x_{k-3}$ for $k \\ge 5$. Thus the signs of the sequence $x_k$ are\n$$\n-, -, -, -, +, +, -, +, -, -, -, +, +, -, +, +, -, +, -, -, -, +, +, -, +, -, \\dots\n$$\nwhich are composed of several cycles of the form $-, -, -, +, +, -, +$. Since 4 out of the 7 terms in each cycle are negative, the result follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26373,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider all the ways of writing exactly ten times each one of the numbers $0, 1, 2, 3, \\ldots, 9$ in the squares of a $10 \\times 10$ board.\nFind the greatest integer $n$ with the property that there is always a row or a column with $n$ different numbers.",
    "options": [],
    "answer": "4",
    "solution": "Let's count in two ways the number of ordered pairs $(d, l)$, where $d$ is a digit and $l$ is a row or column containing $d$. For simplicity, let a *line* be a row or a column. Since there are $10$ occurrences of $d$, they are present in at least $7$ lines (the intersections of the rows and columns must cover all ten numbers). So the number of pairs are at least $7 \\cdot 10 = 70$. Since there are $10 + 10 = 20$ lines, one line must contain at least $\\lfloor \\frac{70-1}{20} \\rfloor + 1 = 4$ different numbers. The\n\nfollowing example shows that the answer is indeed $4$:\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26374,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the sequence $(a_n)_{n \\in \\mathbb{N}}$ with $a_0 = a_1 = a_2 = a_3 = 1$ and $a_n a_{n-4} = a_{n-1} a_{n-3} + a_{n-2}^2$, $n \\ge 4$. Prove that all the terms of this sequence are integer numbers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let's prove by induction that both $a_n$ is an integer and $\\gcd(a_n, a_{n-1}) = \\gcd(a_n, a_{n-2}) = \\gcd(a_n, a_{n-3}) = 1$. It is certainly true for $n = 3$ and direct substitutions show that it is true for $n \\le 7$. Suppose that it's true for $n \\le k$. First we prove that $a_{k+1}$ is an integer by showing that $a_{k-3} \\mid a_k a_{k-2} + a_{k-1}^2$.\nIn fact,\n$$\n\\begin{aligned}\na_{k-2}a_k + a_{k-1}^2 &= \\frac{a_{k-5}a_{k-3} + a_{k-4}^2}{a_{k-6}} \\cdot \\frac{a_{k-3}a_{k-1} + a_{k-2}^2}{a_{k-4}} + \\left( \\frac{a_{k-4}a_{k-2} + a_{k-3}^2}{a_{k-5}} \\right)^2 \\\\\n&= \\frac{a_{k-5}^2(a_{k-5}a_{k-3} + a_{k-4}^2) \\cdot (a_{k-3}a_{k-1} + a_{k-2}^2) + a_{k-4}a_{k-6}(a_{k-4}a_{k-2} + a_{k-3}^2)^2}{a_{k-4}a_{k-6}a_{k-5}^2}\n\\end{aligned}\n$$\nReducing modulo $a_{k-3}$ and noting that $\\gcd(a_{k-3}, a_{k-4}) = \\gcd(a_{k-3}, a_{k-5}) = \\gcd(a_{k-3}, a_{k-6}) = 1$ by induction hypothesis, this is equivalent to prove that\n$$\na_{k-2}^2 a_{k-4}^2 (a_{k-5}^2 + a_{k-4} a_{k-6}) \\equiv 0 \\pmod{a_{k-3}}\n$$\nBut $a_{k-5}^2 + a_{k-4}a_{k-6} = a_{k-3}a_{k-7}$, so $a_{k-3} \\mid a_k a_{k-2} + a_{k-1}^2$.\nNow the $\\gcd$ part. Notice that\n$$\n\\begin{aligned}\n\\gcd(a_{k+1}, a_k) &= \\gcd\\left(\\frac{a_k a_{k-2} + a_{k-1}^2}{a_{k-4}}, a_k\\right) \\\\\n&\\le \\gcd(a_k a_{k-2} + a_{k-1}^2, a_k) = \\gcd(a_{k-1}^2, a_k) = 1\n\\end{aligned}\n$$\nwhere the last equality follows from the induction hypothesis. The other equalities follow similarly:\n$$\n\\gcd(a_{k+1}, a_{k-1}) \\le \\gcd(a_k a_{k-2} + a_{k-1}^2, a_{k-1}) = \\gcd(a_2 a_{k-2}, a_{k-1}) = 1\n$$\n$$\n\\gcd(a_{k+1}, a_{k-2}) \\le \\gcd(a_k a_{k-2} + a_{k-1}^2, a_{k-2}) = \\gcd(a_{k-1}^2, a_{k-1}) = 1\n$$\nand the induction step is complete.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26375,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ and $b$ be real numbers. Define $f_{a,b}: \\mathbb{R}^2 \\to \\mathbb{R}^2$ by $f_{a,b}(x, y) = (a - b y - x^2, x)$. If $P = (x, y) \\in \\mathbb{R}^2$, define $f_{a,b}^0(P) = P$ and $f_{a,b}^{k+1}(P) = f_{a,b}(f_{a,b}^k(P))$ for all nonnegative integers $k$.\n\nThe set $\\operatorname{per}(a, b)$ of the *periodic points* of $f_{a,b}$ is the set of points $P \\in \\mathbb{R}^2$ such that $f_{a,b}^n(P) = P$ for some positive integer $n$.\n\nFix $b$. Prove that the set $A_b = \\{a \\in \\mathbb{R} \\mid \\operatorname{per}(a, b) \\neq \\emptyset\\}$ admits a minimum. Find this minimum.",
    "options": [],
    "answer": "-((b-1)^2)/4",
    "solution": "Let the *orbit* $\\operatorname{orb}(P)$ of $P = (x_0, y_0)$ be the least $n > 0$ such that $f_{a,b}^n(P) = P$.\n\nAlso, let $f_{a,b}^k(P) = (x_k, y_k)$. We have $(x_{k+1}, y_{k+1}) = (a - b y_k - x_k^2, x_k)$, that is,\n$y_{k+1} = x_k$ and, consequently, $x_{k+1} = a - b x_{k-1} - x_k^2$.\n\nSum this relation over $k$, $k$ from $0$ to $\\operatorname{orb}(P) - 1$. Setting $\\operatorname{orb}(P) = m$, $S_1 = x_0 + x_1 + \\cdots + x_{m-1}$ and $S_2 = x_0^2 + x_1^2 + \\cdots + x_{m-1}^2$ for simplicity, we obtain $S_1 = m a - b S_1 - S_2$ which implies\n$$\nS_2 = m a - (b-1) S_1.\n$$\nBut it is known by Cauchy, for example, that $S_2 \\geq S_1^2/m$.\n\nSo\n$$\nm a - (b-1) S_1 \\geq \\frac{S_1^2}{m} \\iff \\frac{S_1^2}{m} + (b-1) S_1 - m a \\leq 0,\n$$\nwhich has solution iff\n$$(b-1)^2 - 4 \\cdot \\frac{1}{m} \\cdot (-m a) \\geq 0 \\iff a \\geq -\\frac{(b-1)^2}{4}.$$\n\nThus $A_b = \\left[-\\frac{(b-1)^2}{4}, +\\infty\\right[$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26376,
    "subject": "Mathematics (Multi-modal)",
    "question": "We have four charged batteries, four uncharged batteries and a radio which needs two charged batteries to work.\nSuppose we don't know which batteries are charged and which ones are uncharged. Find the least number of attempts sufficient to make sure the radio will work. An attempt consists in putting two batteries in the radio and check if the radio works or not.",
    "options": [],
    "answer": "7",
    "solution": "Let's generalize this problem to $2n$ batteries, $n$ of them charged. The number of charged batteries needed to turn on the radio is still two.\n\nOne can see that the order of the attempts doesn't matter and that a set of attempts is always successful if and only if in every set of $n$ batteries there is an attempt with two of these batteries, that is, no set of $n$ batteries is \"pairwise untested\". Indeed, if there is a set $S$ of $n$ pairwise untested batteries, you can be unlucky enough: the working batteries might be these in $S$. Conversely, if every set of $n$ batteries has two tested batteries, we will choose at some moment two charged batteries and the radio will work.\n\nSo consider a graph $G$ in which each battery is a vertex and we connect two batteries iff we **do not** test these batteries. The number of attempts is the number of disconnected pairs, that is $\\binom{2n}{2} - |E|$, where $E$ is the set of edges of $G$. Since we want to minimize the number of attempts, we must maximize $|E|$, that is, we need to have the maximum number of edges. But the only restriction is that we don't have a set $S$ of $n$ pairwise untested batteries, which speaking \"graph-wise\", is the same as $G$ not having a $n$-clique. But, by Turán's Theorem, the graph with the maximum number of edges without a $n$-clique is the most balanced $(n-1)$-partite graph. In this case, the partition sets of vertices of $G$ must contain $3$, $3$, $2$, $2$, ..., $2$ vertices (where we have $n-3$ twos). So the least number of attempts is the number of disconnected pairs of vertices, that is, $2 \\cdot 3 + (n-3) = n + 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26377,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nVeja a seguir a página do calendário de abril de 2005:\n\n| $\\mathrm{D}$ | $\\mathrm{S}$ | $\\mathrm{T}$ | $\\mathrm{Q}$ | $\\mathrm{Q}$ | $\\mathrm{S}$ | $\\mathrm{S}$ |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n|  |  |  |  |  | 1 | 2 |\n| 3 | 4 | 5 | 6 | 7 | 8 | 9 |\n| 10 | 11 | 12 | 13 | 14 | 15 | 16 |\n| 17 | 18 | 19 | 20 | 21 | 22 | 23 |\n| 24 | 25 | 26 | 27 | 28 | 29 | 30 |\n\nQual mês de 2005 ou de 2006 terá uma página igual?",
    "options": [],
    "answer": "Setembro de 2006",
    "solution": "Solution:\nSetembro de 2006",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26378,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle. The internal bisector of $\\angle B$ meets $AC$ in $P$ and $I$ is the incenter of $ABC$. Prove that if $AP + AB = CB$, then $API$ is an isosceles triangle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Draw $PP'$ parallel to $IA$ so that $P'$ is on line $AB$. Then $\\triangle PAP'$ is isosceles, which implies that $BC = AB + AP = AB + AP' = BP'$. This then implies that $\\triangle P'BC$ is isosceles, which in turn implies that, since $P$ is on the angle bisector of $\\angle B$, $P'PC$ is also isosceles, with $PP' = PC$. It then follows, using similarity of triangles and the angle bisector theorem, that\n$$\n\\frac{IA}{PP'} = \\frac{BA}{BP'} = \\frac{BA}{BC} = \\frac{AP}{PC} = \\frac{AP}{PP'}\n$$\nfrom which $IA = AP$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26379,
    "subject": "Mathematics (Multi-modal)",
    "question": "A natural number is a palindrome when one obtains the same number when writing its digits in reverse order. For example, $481184$, $131$ and $2$ are palindromes.\nDetermine all pairs $(m, n)$ of positive integers such that $\\overbrace{111\\ldots1}^{m \\text{ ones}} \\times \\overbrace{111\\ldots1}^{n \\text{ ones}}$ is a palindrome.",
    "options": [],
    "answer": "All pairs with min(m, n) ≤ 9; equivalently, at least one of m or n is at most nine.",
    "solution": "Note that $N = \\overbrace{111\\ldots1}^{m \\text{ ones}} \\times \\overbrace{111\\ldots1}^{n \\text{ ones}}$ has exactly $m+n-1$ digits, since $\\overbrace{111\\ldots1}^{m \\text{ ones}} \\times \\overbrace{111\\ldots1}^{n \\text{ ones}} < 2 \\cdot 10^{m-1} \\times 2 \\cdot 10^{n-1} = 4 \\cdot 10^{m+n-2}$. If $m, n > 9$, then considering the tenth leftmost digit, there would be a carry, thus the number consisting of the first nine digits of $N$ is bigger than the number formed by the last nine digits of $N$, in reversed order. Then if $m$ and $n$ are bigger than $9$ the number $N$ is not a palindrome.\n\nIf one of the numbers $m$, $n$ does not exceed $9$ then there won't be a carry and thus the number $N$ is palindrome.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26380,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be an integer, $n \\ge 3$. Let $f(n)$ be the largest number of isosceles triangles whose vertices belong to some set of $n$ points in the plane without three colinear points. Prove that there exists positive real constants $a$ and $b$ such that $an^2 < f(n) < bn^2$ for every integer $n$, $n \\ge 3$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First consider $n-1$ points in the circumference of a circle and its center. Since any two points in the circumference and the center determine an isosceles triangle, the total number of isosceles triangles in this set of points is at least $(n-1)(n-2)/2 > en^2$ for some small $e$.\n\nIn another hand, let's consider any set of $n$ points in the plane, no three collinear. There are $n(n-1)/2$ choices for two distinct points among the $n$. For each, say $A$ and $B$, there are at most two points $M$ among the given ones such that $AMB$ is isosceles at $M$, because such points $M$ belong to the perpendicular bisector of $AB$ and no three of the points can belong to this line. Thus the total number of isosceles triangles is at most $n(n-1) < n^2$. Therefore $f(n) < n^2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26381,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the smallest real number $C$ such that the inequality\n$$\nC(x_1^{2005} + x_2^{2005} + x_3^{2005} + x_4^{2005} + x_5^{2005}) \\geq x_1 x_2 x_3 x_4 x_5 (x_1^{125} + x_2^{125} + x_3^{125} + x_4^{125} + x_5^{125})^{16}\n$$\nholds for all positive real numbers $x_1, x_2, x_3, x_4, x_5$.",
    "options": [],
    "answer": "5^{15}",
    "solution": "We have\n$$\n5 (x_1^{2005} + x_2^{2005} + x_3^{2005} + x_4^{2005} + x_5^{2005}) \\geq (x_1^5 + x_2^5 + x_3^5 + x_4^5 + x_5^5) (x_1^{2000} + x_2^{2000} + x_3^{2000} + x_4^{2000} + x_5^{2000})\n$$\nby Chebyshev. Also,\n$$\nx_1^5 + x_2^5 + x_3^5 + x_4^5 + x_5^5 \\geq 5x_1x_2x_3x_4x_5\n$$\nby AM-GM and\n$$\n\\frac{x_1^{2000} + x_2^{2000} + x_3^{2000} + x_4^{2000} + x_5^{2000}}{5} \\geq \\left( \\frac{x_1^{125} + x_2^{125} + x_3^{125} + x_4^{125} + x_5^{125}}{5} \\right)^{16}\n$$\nCombining these inequalities gives $C \\leq 5^{15}$. But substituting $x_1 = x_2 = x_3 = x_4 = x_5 = 1$ gives $C \\geq 5^{15}$. Thus $C = 5^{15}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 26382,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that\n$$\nf(x)f(y) + f(x) = 2f(x) + xy\n$$\nfor every real $x, y$.",
    "options": [],
    "answer": "f(x) = x + 1",
    "solution": "First, $f$ is bijective because, plugging $x = 1$, we get $f(f(y) + f(1)) = 2f(1) + x$; $x + 2f(1)$ spans all real numbers (so $f$ is surjective) and $f(x) = f(y) \\iff f(x) + f(1) = f(y) + f(1) \\iff f(f(x) + f(1)) = f(f(y) + f(1)) \\iff 2f(1) + x = 2f(1) + y \\iff x = y$ (so $f$ is injective).\nChoose $x \\neq 0$ and $y$ such that $2f(x) + xy = f(x) \\iff y = -\\frac{f(x)}{x}$. By injectivity,\n$$\nx f\\left(-\\frac{f(x)}{x}\\right) + f(x) = x \\iff f\\left(-\\frac{f(x)}{x}\\right) + \\frac{f(x)}{x} = 1 \\quad (*)\n$$\nNow, set $x = y = 0$: $f(f(0)) = 2f(0)$. Let $a = f(0)$, so that $f(a) = 2a$. If $a = 0$, setting $y = 0$ we get $f(f(x)) = 2f(x)$ and by surjectivity $f(x) = 2x$ for all $x$. This function doesn't work (by testing), so $a$ can't be 0. So $a \\neq 0$ and plugging $x = a$ in $(*)$, we obtain (recall that $f(a) = 2a$):\n$$\nf(-2) + 2 = 1 \\iff f(-2) = -1\n$$\nNow set $y = 0$: $f(f(x) - x) = 2(f(x) - x)$. Now we find all real numbers $k$ such that $f(k) = 2k$. If there is only one possible value for $k$, we are almost done because $f(x) - x = k \\iff f(x) = x + k$ and by testing the function, $f(x) = x + 1$.\nLet $m$ be such that $f(m) = k$ (it exists because $f$ is surjective). Plug $x = m$ and $y = 0$: recalling that $f(0) = a$,\n$$\nf(mf(0) + f(m)) = 2f(m) \\iff f(ma + k) = 2k = f(k)\n$$\nBy injectivity,\n$$\nma + k = k \\iff ma = 0 \\iff m = 0\n$$\nbecause we already proved that $a \\neq 0$. So there is only one value for $k$, namely, $k = f(0) = a$. Then, we're done: the only function is $f(x) = x + 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26383,
    "subject": "Mathematics (Multi-modal)",
    "question": "A square is contained in a cube when all of its points are in the faces or in the interior of the cube. Determine the biggest $l$ such that there exists a square of side $l$ contained in a cube with edge $1$.",
    "options": [],
    "answer": "3√2/4",
    "solution": "The answer is $\\frac{3\\sqrt{2}}{4}$. Just consider the points $(\\frac{3}{4}, 0, 0)$, $(0, \\frac{3}{4}, 0)$, $(1, \\frac{1}{4}, 1)$ and $(\\frac{1}{4}, 1, 1)$.\n\nNow suppose that there is a square with side $\\ell > \\frac{3\\sqrt{2}}{4}$ inside the cube. First, it's not hard to prove that we can suppose without loss of generality that the centers of the cube and of the square coincide. If it isn't the case, consider the three planes that cut the cube in two blocks with dimensions $\\frac{1}{2}, 1, 1$. Now focus on only one of the planes, which we'll call $\\alpha$. If the center of the square is not in $\\alpha$, it is inside one of the two blocks defined by the plane. Next, draw a plane $\\beta$ passing through the center of the square and parallel to $\\alpha$. It cuts the square in two congruent pieces and the cube in two blocks, one smaller than the other. Half of the square is inside the smaller block, so if we perform a translation that transforms $\\beta$ in $\\alpha$ there wouldn't be any problems, that is, the square won't go outside the cube. Repeating this procedure two more times with the two other planes, the center of the square will eventually coincide with the center of the cube.\n\nThe next step is to draw a sphere $S$ with center on the center of the square (which now coincides with the center of the cube) and radius $\\frac{\\ell\\sqrt{2}}{2} > \\frac{3}{4}$. Note that all vertices of the square will lie on the surface of $S$ and that two opposite vertices of the square will be antipodal on $S$. Then we know where are the vertices of the square: each is in one of the $8$ regions determined by the intersection of the surface of $S$ and the cube. One vertex is in one of these regions, say $R$; the other one lies on the opposite region, say $T$.\n\nNow consider a block containing two neighbour regions. This block has dimensions $1, x$ and $x$. Recalling that $S$ has radius bigger than $\\frac{3}{4}$, then $x$ is less than $\\frac{1}{2} - \\sqrt{(\\frac{3}{4})^2 - (\\frac{\\ell\\sqrt{2}}{2})^2} = \\frac{1}{4}$. But then the maximum distance between any two points inside the block is less than $\\sqrt{1^2 + 2x^2} = \\frac{3\\sqrt{2}}{4}$, so it is impossible to have two vertices of the square in two neighbour regions. But $R$ eliminates three regions and $T$ eliminate the other three remaining regions. The other two vertices of the square won't have any region to go, contradiction.\n\nThus the biggest square contained in the cube has side $\\frac{3\\sqrt{2}}{4}$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 26384,
    "subject": "Mathematics (Multi-modal)",
    "question": "A positive integer is **bold** iff it has $8$ positive divisors that sum up to $3240$. For example, $2006$ is bold because its $8$ positive divisors, $1$, $2$, $17$, $34$, $59$, $118$, $1003$ and $2006$, sum up to $3240$. Find the smallest positive bold number.",
    "options": [],
    "answer": "1614",
    "solution": "Let $n = p_1^{\\alpha_1} \\cdots p_k^{\\alpha_k}$. Then $8 = (\\alpha_1 + 1) \\cdots (\\alpha_k + 1)$, and $3240 = \\frac{p_1^{\\alpha_1+1}-1}{p_1-1} \\cdots \\frac{p_k^{\\alpha_k+1}-1}{p_k-1}$. Hence there are three cases:\n(a) $8 = \\alpha_1 + 1 \\implies n = p^7$.\n(b) $8 = (\\alpha_1 + 1)(\\alpha_2 + 1) \\implies n = p_1 p_2^3$.\n(c) $8 = (\\alpha_1 + 1)(\\alpha_2 + 1)(\\alpha_3 + 1) \\implies n = p_1 p_2 p_3$.\nThen we check $\\sigma(n)$:\n(a) $1 + p^2 + \\cdots + p^7 = 3240$. We have $2 < p < 5$, so $p = 3$. But substituting yields no solution.\n(b) $(p_1 + 1)(p_2^3 + p_2^2 + p_2 + 1) = 3240 \\iff (p_1 + 1)(p_2 + 1)(p_2^2 + 1) = 3240$. First note that if $q$ is an odd prime then, by Fermat's theorem, $p_2^2 + 1 \\equiv 0 \\pmod{q} \\implies 1 \\equiv p_2^{q-1} \\equiv (-1)^{\\frac{q-1}{2}} \\pmod{q} \\implies q \\equiv 1 \\pmod{q}$. Since $3240 = 2^3 \\cdot 3^4 \\cdot 5$, the only primes that can divide $p_2^2 + 1$ are $2$ and $5$. This leaves the possibilities $p_2 = 2$ and $p_2 = 3$, none of which yield a solution.\n\n(c) $(p_1+1)(p_2+1)(p_3+1) = 3240$. Let's consider some cases.\n(c.1) One of the primes $p_i$ is $2$. Suppose that $p_1 = 2$. Then $(p_2+1)(p_3+1) = 1080 \\iff (\\frac{p_2+1}{2})(\\frac{p_3+1}{2}) = 270$.\nLet $x = \\frac{p_2+1}{2}$ and $y = \\frac{p_3+1}{2}$. Then $xy = 270$ is fixed and we want to minimize $2p_2p_3 = 2(2x-1)(2y-1) = 8 \\cdot 270 - 4(x+y) + 2$, that is, we want to maximize $x+y$. This happens when $|x-y|$ is maximum. Since $p_2$ and $p_3$ are primes, the optimal values for $x$ and $y$ are $x=2$ and $y=135$, that is, $p_2 = 3$ and $p_3 = 269$, leading to the minimal solution $n = 1614$.\n(c.2) All primes $p_i$ are odd. Then $(\\frac{p_1+1}{2})(\\frac{p_2+1}{2})(\\frac{p_3+1}{2}) = 405 = 3^3 \\cdot 5$. Then one prime, say $p_1$, is equal to $2 \\cdot 3^k \\cdot 5 - 1 = 10 \\cdot 3^k - 1$; and since $\\frac{p_i+1}{2}$ cannot be equal to $1$, then they must have at least one factor $3$ for $i=2,3$; there are only three factors $3$, so $k=0$ or $k=1$. $k=0$ is not possible; $k=1$ yields $p_1 = 29$ and $(\\frac{p_2+1}{2})(\\frac{p_3+1}{2}) = 27 \\implies p_2 = 5$ and $p_3 = 17$, leading to $n = 2465$.\nHence the smallest value for $n$ is $1614$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26385,
    "subject": "Mathematics (Multi-modal)",
    "question": "We have four charged batteries, four uncharged batteries and a radio which needs two charged batteries to work.\nSuppose we don't know which batteries are charged and which ones are uncharged. Find the least number of attempts sufficient to make sure the radio will work. An attempt consists in putting two batteries in the radio and check if the radio works or not.",
    "options": [],
    "answer": "7",
    "solution": "Let's generalize this problem to $2n$ batteries, $n$ of them charged. The number of charged batteries needed is still two.\nOne can see that the order of the attempts doesn't matter and that a set of attempts is always successful if and only if in every set of $n$ batteries there is an attempt with two of these batteries, that is, no set of $n$ batteries is \"pairwise untested\". Indeed, if there is a set $S$ of $n$ pairwise untested batteries, you can be unlucky enough: the working batteries might be these in $S$. Conversely, if every set of $n$ batteries has two tested batteries, we will choose at some moment two charged batteries and the radio will work.\nSo consider a graph $G$ in which each battery is a vertex and we connect two batteries iff we **do not** test these batteries. The number of attempts is the number of disconnected pairs, that is $(2^n) - |E|$, where $E$ is the set of edges of $G$. Since we want to minimize the number of attempts, we must maximize $|E|$, that\n\nis, we need to have the maximum number of edges. But the only restriction is that we don't have a set S of n pairwise untested batteries, which speaking \"graph-wise\", is the same as G not having a n-clique. But, by Túran's Theorem, the graph with the maximum number of edges without a n-clique is the most balanced $(n - 1)$-partite graph. In this case, the partition sets of vertices of $G$ must contain $3, 3, 2, 2, \\ldots, 2$ vertices (where we have $n - 3$ twos). So the least number of attempts is the number of disconnected pairs of vertices, that is, $2 \\cdot 3 + (n - 3) = n + 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26386,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle and $F$ its Fermat point, that is, the interior point of $ABC$ such that $\\angle AFB = \\angle BFC = \\angle CFA = 120^\\circ$. For each one of triangles $ABF$, $BCF$ and $CAF$, draw its Euler line, that is, the line connecting its circumcenter and its centroid.\nProve that these three lines pass through one common point.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First we'll prove the following well-known\n**Lemma.** Let $ABP$, $BCM$ and $CAN$ be the equilateral triangles constructed externally to triangle $ABC$. The lines $AM$, $BN$ and $CP$ concur on the Fermat point of $ABC$.\n*Proof.* Let $F$ be the intersection point of $BN$ and $CP$. Since triangles $BAN$ and $PAC$ are congruent, $\\angle APE = \\angle ABF$. So the quadrilateral $APBF$ is cyclic, and thus $\\angle AEP = \\angle ABP = 60^\\circ$ and $\\angle BFP = \\angle BAP = 60^\\circ$. Hence $\\angle AFB = \\angle BFC = 120^\\circ$, which proves that $F$ is the Fermat point of triangle $ABC$. Analogously, $AM$ and $BN$ pass through the same point $F$, and the lemma is proved.\n\nNow, let $O_A$, $O_B$ and $O_C$ be the centers of these equilateral triangles (and also the circumcenters of triangles $BCF$, $ACF$ and $ABF$, respectively); $G_A$, $G_B$ and $G_C$ are the centroids of $BCF$, $ACF$ and $ABF$, respectively. It is immediate that the Euler lines $O_A G_A$, $O_B G_B$, $O_C G_C$ are parallel to $AF$, $BF$ and $CF$, respectively.\n\nNow consider the homothety with center $F$ and ratio $3/2$, which transforms $G_A$, $G_B$ and $G_C$ into the midpoints of $BC$, $CA$ and $AB$, respectively. This homothety also transforms the Euler lines of $BCF$, $ACF$ and $ABF$ into the lines $l_A$, $l_B$ and $l_C$, parallel to $AF$, $BF$ and $CF$ and passing through the midpoints of $BC$, $CA$ and $AB$, respectively. Hence these Euler lines are concurrent iff $l_A$, $l_B$ and $l_C$ are.\n\nFinally, consider the homothety with center $G$ (centroid of $ABC$) and ratio $-2$. The midpoints of $BC$, $CA$ and $AB$ are transformed into the vertices $A$, $B$ and $C$, respectively, so $l_A$, $l_B$ and $l_C$ are transformed into $AF$, $BF$ and $CF$, respectively. Since the latter are concurrent at $F$, it follows that $l_A$, $l_B$ and $l_C$ are too, and we are done.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26387,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P$ be a convex $2006$-gon. The $1003$ diagonals connecting opposite vertices and the $1003$ lines connecting the midpoints of opposite sides are concurrent, that is, all $2006$ lines have a common point. Prove that the opposite sides of $P$ are parallel and congruent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $A_1A_2\\ldots A_{2006}$ be a polygon such that $A_1A_{1004}$, $A_2A_{1005}$, $\\ldots$, $A_{1003}A_{2006}$ and $M_1M_{1004}$, $M_2M_{1005}$, $\\ldots$, $M_{1003}M_{2006}$ are concurrent at $O$, where $M_1, M_2, \\ldots, M_{2006}$ are the midpoints of $A_1A_2, A_2A_3, \\ldots, A_{2006}A_1$, respectively.\n\nNote that $A_1A_{1004}$, $A_2A_{1005}$, $M_1M_{1004}$ are concurrent, so $A_1A_2 \\parallel A_{1004}A_{1005}$. Similarly, we obtain that any opposite sides are parallel.\n\nNow we have by applying the ratio theorem repetitively, we obtain\n$$\n\\frac{OA_1}{OA_{1004}} = \\frac{OA_2}{OA_{1005}} = \\dots = \\frac{OA_{1003}}{OA_{2006}} = \\frac{OA_{1004}}{OA_1} = \\frac{OA_{1005}}{OA_2} = \\dots = \\frac{OA_{2006}}{OA_{1003}}\n$$\nThis implies that $O$ is the midpoint of every main diagonal. Therefore, all opposite sides are congruent and parallel.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26388,
    "subject": "Mathematics (Multi-modal)",
    "question": "Professor Piraldo takes part in soccer matches with a lot of goals and judges a match in his own peculiar way. A match with score of $m$ goals to $n$ goals, $m \\ge n$, is *tough* when $m \\le f(n)$, where $f(n)$ is defined by $f(0) = 0$ and, for $n \\ge 1$, $f(n) = 2n - f(r) + r$, where $r$ is the largest integer such that $r < n$ and $f(r) \\le n$.\nLet $\\phi = \\frac{1+\\sqrt{5}}{2}$. Prove that a match with score of $m$ goals to $n$, $m \\ge n$, is tough if $m \\le \\phi n$ and is not tough if $m \\ge \\phi n + 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First note that if $n$ is written in the Fibonacci basis (as a sum of distinct Fibonacci numbers containing no neighbors), then the representation of $f(n)$ is just that of $n$ with a 0 in the end. The proof goes by induction: it is true for $n=0$. Suppose $n > 0$. Then $r$ will be such that $f(r) = n$, if the last digit of the Fibonacci of $n$ is zero, or $f(r) = n-1$, if it is one. Then $f(n) = n+r+n-f(r)$. If the last digit of the Fibonacci of $n$ is zero, then $r$ is obtained by deleting the last digit from $n$. So $f(n) = n+r$ is the sum of the Fibonacci numbers used in the representation of $n$ and its respective antecessors in the Fibonacci sequence, resulting in the subsequent Fibonacci numbers, which is exactly $n$ with a zero at its right. If the last digit of the Fibonacci of $n$ is zero, then $r$ is still obtained by deleting the last digit from $n$, but $f(n) = n+1+r$. Note that $n-1$ ends with a zero, so summing $r$ will do the same thing as the preceding case; and we're substituting the rightmost digit one with 2, which corresponds to a rightmost 10.\n\nNow, express $n$ and $f(n)$ by using the closed form for Fibonacci numbers $F_k = \\frac{1}{\\sqrt{5}}(\\phi^k - (-\\phi)^{-k})$, and note that we cannot use two consecutive Fibonacci numbers in base Fibonacci, so $|f(n) - \\phi \\cdot nZ| < \\phi^{-1}(1 + \\phi^{-2} + \\phi^{-4} + \\dots) = \\frac{\\phi^{-1}}{1-\\phi^{-2}} = 1$, that is, $f(n) - 1 < \\phi n < f(n) + 1$. Thus if $m \\le \\phi n$ then $m < f(n) + 1 \\iff m \\le f(n)$ and the match is tough; if $m \\ge \\phi n + 1$ then $m > f(n) - 1 + 1 = f(n)$ and the match is not tough.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26389,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given positive integers $a$, $c$ and integer $b$, prove that there exists a positive integer $x$ such that\n$$\na^x + x \\equiv b \\pmod{c},\n$$\nthat is, there exists a positive integer $x$ such that $c$ is a divisor of $a^x + x - b$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\ell$ be the length of the cycle of the sequence $a, a^2, a^3, \\dots$ mod $c$. Thus\n$$\na^{\\ell + \\ell k} \\equiv a^\\ell \\pmod{c} \\quad (*)\n$$\nfor all positive integer $k$ and $\\ell$ large enough.\n\nLet $d = \\gcd(\\ell, c)$. The multiples of $\\ell$ mod $c$ are the multiples of $d$. Let's prove that if $c > 1$ then $d < c$. Since $d$ divides $c$, $d \\leq c$.\nSuppose that $d = c$. So $c$ divides $\\ell$. Note that it's not possible that two equal remainders appear in the same cycle, because $a' \\equiv a' \\pmod{c}$ implies $a'^c \\equiv a'^c \\pmod{c}$, that is, $i-j$ is multiple of $\\ell$, contradiction. Hence the length of the cycle of $a, a^2, a^3, \\dots$ mod $c$ does not exceed $c$. Thus, if $c$ divides $\\ell$, $c = \\ell$. This means that $c$ divides some $a^n$ and also every bigger power of $a$. Hence, $\\ell = 1$ and $c = 1$.\nThe problem now goes by induction on $c$. It is obvious for $c = 1$. Suppose that $c > 1$ and that the result is true for every positive integer less than $c$. Then it is true for $d$, because $d < c$. By the induction hypothesis, there are sufficiently large $n_0, n_1, n_2, \\dots, n_{d-1}$ such that\n$$\na^{n_i} + n_i \\equiv i \\pmod{d}\n$$\nfor every $i = 0, 1, 2, \\dots, d-1$.\nLet $b = qd + r$, with $0 \\leq r < d$. So, from $a^{n_r} + n_r = r + md$ and $(*)$,\n$$\na^{n_r + \\ell k} + (n_r + \\ell k) \\equiv a^{n_r} + (n_r + \\ell k) \\equiv r + md + \\ell k \\pmod{c} \\quad (**)\n$$\nBut, when $k$ varies, $\\ell k \\pmod{c}$ goes through all multiples of $d$. Thus there exists $k$ such that $\\ell k \\equiv (q-m)d \\pmod{c}$. Plugging it in $(**)$,\n$$\na^{n_r + \\ell k} + (n_r + \\ell k) \\equiv r + md + (q - m)d \\pmod{c} \\\\\n\\iff a^{n_r + \\ell k} + (n_r + \\ell k) \\equiv r + qd \\equiv b \\pmod{c}\n$$\nHence, we can take $x = n_r + \\ell k$ and the induction step is complete.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26390,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUma cidade ainda não tem iluminação elétrica e todos usam velas à noite. Na casa de João, usa-se uma vela por noite, sem queimá-la totalmente. Com os tocos de quatro destas velas, é possível fazer uma nova vela. Durante quantas noites João poderá iluminar sua casa com 43 velas?\nA) 43\nB) 53\nC) 56\nD) 57\nE) 60",
    "options": [],
    "answer": "D",
    "solution": "Solution:\n\n(D) De 43 velas obtém-se 43 tocos. Como $43=4 \\times 10+3$, com esses 43 tocos se pode fazer 10 velas e guardar 3 tocos. Dessas 10 velas, obtemos 10 tocos que, com os 3 que sobraram, dão 13. Sendo $13=4 \\times 3+1$, fazemos então 3 velas com 12 tocos, sobrando 1 toco. Depois de usar estas 3 velas, teremos um total de 4 tocos, que nos dá 1 vela extra. No total, obtemos $43+10+3+1=57$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26391,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nResolva a equação $\\frac{|x-1|}{x^{2}}=6$.",
    "options": [],
    "answer": "x = -1/2 or x = 1/3",
    "solution": "Solution:\n\n1. $1^{\\circ}$ caso: $x \\geq 1$\n\nNesse caso $x-1 \\geq 0$, donde $|x-1|=x-1$. A equação toma a forma $\\frac{x-1}{x^{2}}=6$ ou $6 x^{2}-x+1=0$. Esta equação não tem raízes reais porque $\\Delta=(-1)^{2}-4 \\times 6 \\times 1=1-24$ é negativo. Logo, não temos soluções maiores ou iguais a 1.\n\n2. caso: $x<1$\n\nNesse caso $x-1<0$, donde $|x-1|=-(x-1)=1-x$. A equação toma a forma $\\frac{1-x}{x^{2}}=6$ ou $6 x^{2}+x-1=0$. Resolvendo esta equação temos:\n\n$$\nx=\\frac{-1 \\pm \\sqrt{1-4 \\times 6 \\times(-1)}}{2 \\times 6}=\\frac{-1 \\pm \\sqrt{25}}{12}=\\frac{-1 \\pm 5}{12}\n$$\n\n$\\Rightarrow x=-\\frac{1}{2}$ e $x=\\frac{1}{3}$.\n\nComo essas duas raízes são menores que 1, elas são as raízes da equação do enunciado.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26392,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nQual é o maior dos números?\n(A) $1000+0,01$\n(B) $1000 \\times 0,01$\n(C) $1000 / 0,01$\n(D) $0,01 / 1000$\n(E) $1000-0,01$",
    "options": [],
    "answer": "C",
    "solution": "Solution:\n\nTemos: $1000+0,01=1000,01$ ; $1000 \\times 0,01=1000 \\times \\frac{1}{100}=10$;\n$$\n\\frac{1000}{0,01}=\\frac{1000}{\\frac{1}{100}}=1000 \\times 100=100000\n$$\n$\\frac{0,01}{1000}$ é o inverso de $\\frac{1000}{0,01}$, logo, de (C) temos que $\\frac{1000}{0,01}=\\frac{1}{100000}=0,00001$. Agora, $1000-0,01$ é menor do que $1000$ (não é preciso efetuar o cálculo para obter esta conclusão). Portanto, o maior número é $\\frac{1000}{0,01}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26393,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUma loja de sabonetes realiza uma promoção com o anúncio \"Compre um e leve outro pela metade do preço\". Outra promoção que a loja poderia fazer oferecendo o mesmo desconto percentual é:\nA) \"Leve dois e pague um\"\nB) \"Leve três e pague um\"\nC) \"Leve três e pague dois\"\nD) \"Leve quatro e pague três\"\nE) \"Leve cinco e pague quatro\"",
    "options": [],
    "answer": "D",
    "solution": "Solution:\n\n(D) Pela promoção, quem levar 2 unidades paga pelo preço de 1,5 unidade, logo quem levar 4 unidades paga pelo preço de 3 unidades, ou seja, leva quatro e paga três.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26394,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n1) Os valores positivos de $x$ para os quais $(x-1)(x-2)(x-3)<0$ formam o conjunto:\n$(1,3)$ $(2,3)$\n$(0,3)$\n$(0,1) \\cup (2,3)$\n$(1,2)$",
    "options": [],
    "answer": "(0,1) \\cup (2,3)",
    "solution": "Solution:\n\n1. Para que um produto de três fatores seja negativo, devemos ter dois fatores positivos e um fator negativo, ou os três negativos.\n$(+)(+)(-) = -$ \\ $(+)(-)(+) = -$ \\ $(-)(+)(+) = -$ \\ $(-)(-)(-) = -$\n\nAs possibilidades são:\n\n1) $\\underbrace{(x-1)}_{+}(\\underbrace{x-2}_{-})(\\underbrace{x-3}_{-}) \\Rightarrow x>1, x>2 \\text{ e } x<3$.\n\n![](attached_image_1.png)\n\nNesse caso, a solução é $2 < x < 3$.\n\n2) $\\underbrace{(x-1)}_{+}(\\underbrace{x-2}_{-})(\\underbrace{x-3)}_{+} \\Rightarrow x>1, x<2 \\text{ e } x>3$.\n\nNesse caso temos $1 < x < 2$ e $x > 3$, o que não é possível. Logo, esse caso não pode ocorrer.\n\n![](attached_image_2.png)\n\n3) $\\underbrace{(x-1)}_{-} \\underbrace{x-2)}_{+} \\underbrace{(x-3)}_{+} \\Rightarrow x<1, x>2 \\text{ e } x>3$.\n\nNesse caso temos $x<1, x>2$ e $x>3$, o que não é possível.\n\n![](attached_image_3.png)\n\n4) $\\underbrace{(x-1)}_{-}(\\underbrace{x-2}_{-})(\\underbrace{x-3)}_{-}) \\Rightarrow x<1, x<2 \\text{ e } x<3$.\n\n![](attached_image_4.png)\n\nNesse caso, a solução é $x < 1$. Logo, a solução são todos os números reais $x$ tais que $x < 1$ ou $2 < x < 3$; ou seja, a união de dois intervalos: $(0,1) \\cup (2,3)$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26395,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa figura ao lado $ABCD$ é um retângulo e $ABE$ e $CDF$ são triângulos retângulos. A área do triângulo $ABE$ é $150~\\mathrm{cm}^2$ e os segmentos $AE$ e $DF$ medem, respectivamente, $15~\\mathrm{cm}$ e $24~\\mathrm{cm}$. Qual o comprimento do segmento $CF$?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "7",
    "solution": "Solution:\n\nO segmento $CF$, que queremos calcular, é um cateto do triângulo retângulo $CDF$. O teorema de Pitágoras, aplicado a este triângulo, diz que $CD^2 = CF^2 + FD^2 = CF^2 + 24^2$ e daí tiramos $CF^2 = CD^2 - 24^2$. Ou seja, para achar $CF$ basta conhecer $CD$. Como os lados opostos de um retângulo (e, mais geralmente, de um paralelogramo) são iguais, temos $CD = AB$, e nosso objetivo passa a ser o cálculo de $AB$.\n\nPara isso, olhemos para o triângulo $ABE$; sua área é $\\frac{AE \\times BE}{2} = \\frac{15 \\times BE}{2} = 150$, donde tiramos $BE = 20$. O teorema de Pitágoras aplicado a este triângulo nos dá $AB^2 = AE^2 + BE^2 = 15^2 + 20^2 = 625 = 25^2$, donde $AB = 25$.\n\nLogo $CD = AB = 25$ e, de acordo com nossa observação anterior, temos $CF^2 = CD^2 - 24^2 = 25^2 - 24^2 = (25 + 24)(25 - 24) = 49$. Obtemos então $CF = 7$.\n\nNotamos que a solução independe da medida dos lados $AD$ e $BE$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26396,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nQual a menor das raízes da equação $2(x-3 \\sqrt{5})(x-5 \\sqrt{3})=0$ ?",
    "options": [],
    "answer": "3√5",
    "solution": "Solution:\nA equação já é dada na forma fatorada $a(x-m)(x-n)=0$, logo as raízes são $m=3 \\sqrt{5}$ e $n=5 \\sqrt{3}$. Devemos decidir qual delas é a maior.\n\nSabemos que $\\sqrt{5}<2,3$ e $1,7<\\sqrt{3}$, logo $3 \\sqrt{5}<3 \\times 2,3=6,9$ e $5 \\sqrt{3}>5 \\times 1,7=8,5$. Como $6,9$ é menor do que $8,5$, concluímos que $3 \\sqrt{5}$ é menor do que $5 \\sqrt{3}$.\n\n![](attached_image_1.png)\n\nComparar $3 \\sqrt{5}$ e $5 \\sqrt{3}$ é o mesmo que comparar $(3 \\sqrt{5})^{2}$ e $(5 \\sqrt{3})^{2}$. Assim, $(3 \\sqrt{5})^{2}=9 \\times 5=45<75=25 \\times 3=(5 \\sqrt{3})^{2}$. Logo, $3 \\sqrt{5}$ é menor do que $5 \\sqrt{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26397,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n1) Determine o valor de $(666666666)^2-(333333333)^2$.",
    "options": [],
    "answer": "333333332666666667",
    "solution": "Solution:\n\n1. Usando a fatoração $x^2-y^2=(x-y)(x+y)$, obtemos:\n$$\n\\begin{aligned}\n666\\,666\\,666^2 - 333\\,333\\,333^2 &= (666\\,666\\,666 - 333\\,333\\,333)(666\\,666\\,666 + 333\\,333\\,333) \\\\\n&= 333\\,333\\,333 \\times 999\\,999\\,999 \\\\\n&= 333\\,333\\,333 \\times (1\\,000\\,000\\,000 - 1) \\\\\n&= 333\\,333\\,333\\,000\\,000\\,000 - 333\\,333\\,333 \\\\\n&= 333\\,333\\,332\\,666\\,666\\,667\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26398,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nQual é o maior fator primo de $2006$?",
    "options": [],
    "answer": "59",
    "solution": "Solution:\n\nA decomposição de $2006$ em fatores primos é $2006 = 2 \\times 17 \\times 59$. Logo, o maior fator primo de $2006$ é $59$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26399,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nQuantos são os pares diferentes de inteiros positivos $(a, b)$ tais que $a+b \\leq 100$ e $\\frac{a+\\frac{1}{b}}{\\frac{1}{a}+b}=13$?",
    "options": [],
    "answer": "7",
    "solution": "Solution:\n\nTemos: $13=\\frac{a+\\frac{1}{b}}{\\frac{1}{a}+b}=\\frac{\\frac{a b+1}{b}}{\\frac{1+a b}{a}}=\\frac{a}{b}$. Logo, $a=13 b$ e como $a+b \\leq 100$ segue que $14 b \\leq 100 \\Rightarrow b \\leq 7,14$. Como $b$ é inteiro devemos ter $b \\leq 7$. Logo os pares são em número de 7, a saber:\n$$\n(13,1),\\ (26,2),\\ (39,3),\\ (52,4),\\ (65,5),\\ (78,6),\\ (91,7)\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26400,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPara fabricar 9 discos de papelão circulares para o Carnaval usam-se folhas quadradas de $10~\\mathrm{cm}$ de lado como indicado na figura. Qual a área do papel não aproveitado?\n![](attached_image_1.png)\n\n(A) $25~\\mathrm{cm}^2$\n(B) $22,5~\\mathrm{cm}^2$\n(C) $21,5~\\mathrm{cm}^2$\n(D) $21~\\mathrm{cm}^2$\n(E) $22~\\mathrm{cm}^2$",
    "options": [],
    "answer": "C",
    "solution": "Solution:\n\nLembre que a área de um círculo é $\\pi r^{2}$, onde $r$ é o raio do círculo. Se $r$ é o raio dos círculos da figura, então a área pedida é:\n\n\\[\n\\underbrace{\\text{área do quadrado}}_{10 \\times 10 = 100} - \\underbrace{\\text{área dos 9 círculos}}_{9 \\times \\pi r^{2}} = 100 - 9 \\times \\pi \\times \\left(\\frac{5}{3}\\right)^{2} = 100 - 25\\pi\n\\]\n\nA área do círculo de raio $r$ é $\\pi r^{2}$.\n\nUsando a aproximação $\\pi \\approx 3,14$, obtemos:\n\n\\[\n100 - 25\\pi \\approx 100 - 25 \\times 3,14 = 21,5~\\mathrm{cm}^{2}\n\\]",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26401,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEncontre o produto: $\\left(1-\\frac{1}{4}\\right)\\left(1-\\frac{1}{9}\\right)\\left(1-\\frac{1}{16}\\right) \\ldots\\left(1-\\frac{1}{225}\\right)$.\n\n(A) $\\frac{10}{125}$\n(B) $\\frac{5}{9}$\n(C) $\\frac{3}{5}$\n(D) $\\frac{8}{15}$\n(E) $\\frac{1}{120}$",
    "options": [],
    "answer": "D",
    "solution": "Solution:\n\n(D) Cada um dos fatores é da forma \"diferença de quadrados\", isto é $a^{2}-b^{2}$, onde $a=1$\n\n$\\left(1-\\frac{1}{4}\\right)\\left(1-\\frac{1}{9}\\right)\\left(1-\\frac{1}{16}\\right) \\ldots\\left(1-\\frac{1}{225}\\right)=\\left(1-\\frac{1}{2^{2}}\\right)\\left(1-\\frac{1}{3^{2}}\\right)\\left(1-\\frac{1}{4^{2}}\\right) \\ldots\\left(1-\\frac{1}{15^{2}}\\right)$\n\nUsando a fatoração $a^{2}-b^{2}=(a-b)(a+b)$, obtemos:\n\n$\\left(1-\\frac{1}{4}\\right)\\left(1-\\frac{1}{9}\\right)\\left(1-\\frac{1}{16}\\right) \\ldots\\left(1-\\frac{1}{225}\\right)=\\left(1-\\frac{1}{2}\\right)\\left(1+\\frac{1}{2}\\right)\\left(1-\\frac{1}{3}\\right)\\left(1+\\frac{1}{3}\\right)\\left(1-\\frac{1}{4}\\right)\\left(1+\\frac{1}{4}\\right) \\ldots\\left(1-\\frac{1}{15}\\right)\\left(1+\\frac{1}{15}\\right)$\n\n$=\\frac{1}{2} \\times \\frac{3}{2} \\times \\frac{2}{3} \\times \\frac{4}{3} \\times \\frac{3}{4} \\times \\frac{5}{4} \\times \\ldots \\times \\frac{15}{14} \\times \\frac{14}{15} \\times \\frac{16}{15}$\n\n$=\\frac{1}{2} \\times \\frac{16}{15}=\\frac{8}{15}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26402,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEm 1998, a população do Canadá era de 30,3 milhões. Qual das opções abaixo representa a população do Canadá em 1998?\nA) 30300000\nB) 303000000\nC) 30300\nD) 303000\nE) 30300000000",
    "options": [],
    "answer": "A",
    "solution": "Solution:\n\nTemos que 1 milhão $= 1000000$. Logo, 30,3 milhões $= 30,3 \\times 1000000 = 30300000$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26403,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n1) Qual é o maior dos números?\n(A) $2 \\times 0 \\times 2006$\n(B) $2 \\times 0+6$\n(C) $2+0 \\times 2006$\n(D) $2 \\times(0+6)$\n(E) $2006 \\times 0+0 \\times 6$",
    "options": [],
    "answer": "D",
    "solution": "Solution:\n\n(D) Lembre que se num produto um dos fatores é zero, então o produto também é zero. Temos:\n$2 \\times 0 \\times 2006=0$\n$2 \\times 0+6=0+6=6$\n$2+0 \\times 2006=2+0=2$\n$2 \\times(0+6)=2 \\times 6=12$\n$2006 \\times 0+0 \\times 6=0+0=0$\nLogo, o maior é $2 \\times(0+6)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26404,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDetermine o valor de $123456123456 \\div 1000001 =$.",
    "options": [],
    "answer": "123456",
    "solution": "Solution:\n\nÉ claro que com números tão grandes, a questão não pretende que se efetue a divisão. Para resolvê-la vamos usar alguns truques aritméticos:\n$$\n\\begin{aligned}\n& 123456123456 = 123456000000 + 123456 = 123456 \\times 1000000 + 123456 = \\\\\n& = 123456 \\times (1000000 + 1) = 123456 \\times 1000001\n\\end{aligned}\n$$\nLogo, $123456123456 \\div 1000001 = 123456$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26405,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSe $m$ e $n$ são inteiros maiores do que zero com $m < n$, definimos $m \\nabla n$ como a soma dos inteiros entre $m$ e $n$, incluindo $m$ e $n$. Por exemplo, $5 \\nabla 8 = 5 + 6 + 7 + 8 = 26$.\n\nEntão o valor de $\\frac{22 \\nabla 26}{4 \\nabla 6}$ é:\n\nA) 4\nB) 6\nC) 8\nD) 10\nE) 12",
    "options": [],
    "answer": "C",
    "solution": "Solution:\n\nDe acordo com a definição de $\\nabla$, temos\n$$\n\\frac{22 \\nabla 26}{4 \\nabla 6} = \\frac{22 + 23 + 24 + 25 + 26}{4 + 5 + 6} = \\frac{120}{15} = 8.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26406,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSe eu der duas barras de chocolate para Tião, ele me empresta sua bicicleta por 3 horas. Se eu lhe der 12 bombons, ele me empresta a bicicleta por 2 horas. Amanhã, eu lhe darei uma barra de chocolate e 3 bombons. Por quantas horas ele me emprestará a bicicleta?\n(A) $1 / 2$\n(B) 1\n(C) 2\n(D) 3\n(E) 4",
    "options": [],
    "answer": "C",
    "solution": "Solution:\n\n$\\left\\{\\begin{array}{l}2 \\text{ barras } \\xrightarrow{\\text{ corresponde }} 3 \\text{ horas } \\\\ 12 \\text{ bombons } \\xrightarrow{\\text{ corresponde }} 2 \\text{ horas }\\end{array}\\right.$\n\nLogo,\n$\\left\\{\\begin{array}{l}1 \\text{ barra } \\xrightarrow{\\text{ corresponde }} 1,5 \\text{ horas } \\\\ 3 \\text{ bombons } \\xrightarrow{\\text{ corresponde }} 0,5 \\text{ horas }\\end{array}\\right.$\n\nLogo, Tião me emprestará a bicicleta por $1,5+0,5=2$ horas",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26407,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUma professora tem 237 balas para dar a seus 31 alunos. Qual é o número mínimo de balas a mais que ela precisa conseguir para que todos os alunos recebam a mesma quantidade de balas, sem sobrar nenhuma?\nA) 11\nB) 20\nC) 21\nD) 31\nE) 41",
    "options": [],
    "answer": "A",
    "solution": "Solution:\n\nO algoritmo de divisão de Euclides nos dá $237 = 7 \\times 31 + 20$; logo 237 não é divisível por 31. Isso quer dizer que a professora realmente vai ter que comprar mais balas para que todos os alunos recebam o mesmo número de balas. De acordo com o enunciado, devemos então adicionar à expressão $7 \\times 31 + 20$ o menor inteiro positivo $x$ tal que $7 \\times 31 + 20 + x$ seja múltiplo de 31. Como $x = 31 - 20 = 11$, basta que a professora compre 11 balas.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26408,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nQuantos são os números inteiros $x$ tais que $-5 < x-1 \\leq 5$?\n(A)\n(B) 9\n(C) 10\n(D) 11\n(E) 12",
    "options": [],
    "answer": "C",
    "solution": "Solution:\n\nSomando 1 a todos os membros das duas desigualdades temos $-5+1 < x-1+1 \\leq 5+1 \\Rightarrow -4 < x \\leq 6$.\n\nOs valores inteiros de $x$ que satisfazem as duas desigualdades são: $-3, -2, -1, 0, 1, 2, 3, 4, 5, 6$.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26409,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO quadrado abaixo é chamado quadrado mágico, porque a soma dos números de cada linha, de cada coluna e de cada diagonal é sempre a mesma. Neste caso essa soma é $15$.\n\n| 4 | 9 | 2 |\n| :--- | :--- | :--- |\n| 3 | 5 | 7 |\n| 8 | 1 | 6 |\n\nComplete os cinco números que faltam no quadrado abaixo para que ele seja um quadrado mágico.\n\n| -12 |   | -4 |\n| :--- | :--- | :--- |\n|   | 0 |   |\n| 4 |   |   |",
    "options": [],
    "answer": "-12  16  -4\n 8   0  -8\n 4  -16  12",
    "solution": "Solution:\n\nComo a soma dos números de uma diagonal é $4+0+(-4)=0$, este deve ser o valor da soma dos números de cada linha, coluna e diagonal.\n\nAssim, obtemos de imediato os números que faltam nas casas em cinza no primeiro tabuleiro: $16$, $8$ e $12$, pois $(-12)+16+(-4)=0$ (na primeira linha), $(-12)+8+4=0$ (na primeira coluna) e $(-12)+0+12=0$ (na diagonal).\n\n| -12 | 16 | -4 |\n| :---: | :---: | :---: |\n| 8 | 0 | -8 |\n| 4 | -16 | 12 |\n\nAgora, o número que falta na segunda linha do tabuleiro é $-8$, porque $8+0+(-8)=0$. Para a terceira linha, obtemos $-16$, pois $4+(-16)+12=0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26410,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSete equipes, divididas em dois grupos, participaram do torneio de futebol do meu bairro.\nO grupo 1 foi formado pelas equipes Avaqui, Botágua e Corinense.\nO grupo 2 foi formado pelas equipes Dinossauros, Esquisitos, Flurinthians e Guaraná.\nNa primeira rodada do torneio, cada equipe enfrentou cada uma das equipes do seu grupo exatamente uma vez.\nNa segunda rodada do torneio, cada equipe enfrentou cada uma das equipes do outro grupo exatamente uma vez.\n\na) Quantas partidas foram disputadas na primeira rodada no grupo 1?\nb) Quantas partidas foram disputadas na primeira rodada no grupo 2?\nc) Quantas partidas foram disputadas na segunda rodada?",
    "options": [],
    "answer": "a) 3; b) 6; c) 12",
    "solution": "Solution:\n\na) Foram disputadas 3 partidas que são: $A \\times B$, $B \\times C$, $C \\times A$.\n\nb) Foram disputadas 6 partidas que são: $D \\times E$, $D \\times F$, $D \\times G$, $E \\times F$, $E \\times G$, $F \\times G$\n\nc) Na segunda rodada, cada equipe do grupo 1 joga 4 partidas; uma com cada equipe do grupo 2. Como o grupo 1 tem 3 equipes, o total de partidas será $3 \\times 4 = 12$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26411,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSe $x + |x| + y = 5$ e $x + |y| - y = 6$ então $x + y$ é:\n(A) -1\n(B) 11\n(C) $\\frac{9}{5}$\n(D) 1\n(E) -11",
    "options": [],
    "answer": "C",
    "solution": "Solution:\n\nSe $x < 0$, então $|x| = -x$ e da $1^{a}$ equação temos $x + (-x) + y = 5 \\Rightarrow y = 5$. Substituindo esse valor na $2^{a}$ equação obtemos $x = 6$, o que não é possível pois estamos supondo $x < 0$. Logo, não há solução para $x < 0$.\n\nSe $y \\geq 0$, então $|y| = y$ e da $2^{a}$ equação segue que $x = 6$. Substituindo esse valor na $1^{a}$ equação encontramos $y = -7$, o que não é possível porque estamos supondo que $y$ é positivo.\n\nConcluímos que não há solução para $y \\geq 0$ e $x < 0$. Logo, $y < 0$ e $x \\geq 0$, e as equações são: $2x + y = 5$ e $x - 2y = 6$. Resolvendo obtemos $x = \\frac{16}{5}$ e $y = -\\frac{7}{5}$. Portanto, $x + y = \\frac{9}{5}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26412,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUma fábrica embala 8 latas de palmito em caixas de papelão cúbicas de $20~\\mathrm{cm}$ de lado. Estas caixas são colocadas, sem deixar espaços vazios, em caixotes de madeira de $80~\\mathrm{cm}$ de largura por $120~\\mathrm{cm}$ de comprimento por $60~\\mathrm{cm}$ de altura. Qual o número máximo de latas de palmito em cada caixote?\nA) 576\nB) 4608\nC) 2304\nD) 720\nE) 144",
    "options": [],
    "answer": "A",
    "solution": "Solution:\n\nEm cada caixote de madeira de dimensões $a \\times b \\times c$ cabem, empilhados regularmente, $\\frac{a}{l} \\times \\frac{b}{l} \\times \\frac{c}{l}$ cubos de lado $l$. No nosso caso, $a=60$, $b=80$, $c=120$ e $l=20$. Como $60$, $80$ e $120$ são múltiplos de $20$, podemos encher o caixote sem deixar espaços com $\\frac{60}{20} \\times \\frac{80}{20} \\times \\frac{120}{20} = 72$ caixas de papelão cúbicas de $20~\\mathrm{cm}$ de cada lado. Logo, em cada caixote cabem $72 \\times 8 = 576$ latas de palmito.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26413,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNum triângulo retângulo, definimos o cosseno de seus ângulos agudos $\\alpha$ por: $\\cos \\alpha=\\frac{\\text{cateto adjacente}}{\\text{hipotenusa}}=\\frac{c}{a}$.\n\n![](attached_image_1.png)\n\nO triângulo retângulo da figura tem cateto $OA=1$. Escreva em ordem crescente os cossenos dos ângulos de $25^{\\circ}, 41^{\\circ}$ e $58^{\\circ}$\n\n![](attached_image_2.png)",
    "options": [],
    "answer": "cos 58° < cos 41° < cos 25°",
    "solution": "Solution:\n\nDe acordo com a definição de cosseno, temos: $\\cos 25^{\\circ}=\\frac{1}{OM}$, $\\cos 41^{\\circ}=\\frac{1}{ON}$ e $\\cos 58^{\\circ}=\\frac{1}{OB}$. Na figura, vemos que $OM<ON<OB$, logo $\\cos 58^{\\circ}<\\cos 41^{\\circ}<\\cos 25^{\\circ}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26414,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUsando apenas os dígitos $1,2,3,4$ e $5$, Peri construiu uma sequência da seguinte forma: um $1$, dois $2$, três $3$, quatro $4$, cinco $5$, seis $1$, sete $2$ e assim por diante; abaixo vemos os primeiros termos desta sequência:\n\n$1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,1,1,1,1,1,1,2,2,2, \\ldots$\n\nQual é o $100^{\\circ}$ termo dessa sequência?",
    "options": [],
    "answer": "4",
    "solution": "Solution:\n\nAgrupamos a sequência em blocos numerados consecutivamente, cada bloco formado pelos termos iguais consecutivos, como mostrado a seguir.\n$$\n\\underbrace{1}_{\\text{bloco 1}}, \\underbrace{2,2}_{\\text{bloco 2}}, \\underbrace{3,3,3}_{\\text{bloco 3}}, \\underbrace{4,4,4,4}_{\\text{bloco 4}}, \\underbrace{5,5,5,5,5}_{\\text{bloco 5}}, \\underbrace{1,1,1,1,1,1}_{\\text{bloco 6}}, \\underbrace{2,2,2,2,2,2,2}_{\\text{bloco 7}}, \\underbrace{3,3,3,3,3,3,3,3}_{\\text{bloco 8}}, \\ldots, \\underbrace{k, k, k, \\ldots, k}_{\\substack{\\text{bloco } n \\\\ k \\in \\{1,2,3,4,5\\}}}\n$$\nObserve que a numeração de cada bloco coincide com o número de termos que ele contém: o bloco $1$ tem $1$ termo, o bloco $2$ tem $2$ termos, o bloco $3$ tem $3$ termos, ..., o bloco $n$ tem $n$ termos. A posição na sequência do último termo de cada bloco é obtida somando todos os números de $1$ até o número atribuído ao bloco. Por exemplo:\n- O último $3$ do bloco $8$ é o $36^{\\circ}$ termo da sequência porque $1+2+3+4+5+6+7+8=36$.\n- O último $1$ do bloco $11$ é o $66^{\\circ}$ termo da sequência porque $1+2+3+4+5+6+7+8+9+10+11=66$.\n\nEm geral, o último elemento do bloco $n$ está na posição $1+2+3+\\ldots+n$. Para calcular o valor desta soma, lembramos que $1,2,3, \\ldots, n$ é uma progressão aritmética de razão $1$, termo inicial $a_{1}=1$ e $n$-ésimo termo $a_{n}=n$; a soma de seus $n$ primeiros termos é\n$$\n1+2+\\cdots+n=\\frac{n\\left(a_{1}+a_{n}\\right)}{2}=\\frac{n(n+1)}{2}\n$$\n\nPrecisamos agora descobrir em qual bloco se encontra o $100^{\\circ}$ termo da sequência. Suponhamos que ele esteja no bloco $n$; então sua posição é no máximo a do último termo deste bloco. Como ele não está no bloco $n+1$, concluímos que $n$ é o menor inteiro tal que\n$$\n100 \\leq \\frac{n(n+1)}{2}\n$$\nou seja\n$$\n200 \\leq n(n+1)\n$$\nPara determinar o valor de $n$ devemos resolver essa inequação e escolher entre suas soluções o menor número inteiro. Como a expressão é bastante simples, é mais fácil resolvê-la por tentativas. Fazendo isso, vemos que $n=14$; de fato, $13 \\times (13+1)=182<200$ e $14 \\times (14+1)=210>200$. Logo o $100^{\\circ}$ termo da sequência está no bloco $14$. Os números que aparecem nos blocos se repetem de $5$ em $5$ blocos na ordem $1, 2, 3, 4, 5$. Como $14=5 \\times 2+4$, o bloco $14$ é formado pelo número $4$. Assim, o $100^{\\circ}$ termo da sequência é $4$.\n\nComentário: A resolução acima da inequação $200 \\leq n(n+1)$ apesar de correta, não serviria se o problema pedisse, por exemplo, para determinar o $10.000^{\\circ}$ termo da sequência. Neste caso, teríamos que lidar com a inequação $20.000 \\leq n(n+1)$, e é claro que achar sua menor solução inteira por tentativas não funciona (a não ser com muita, muita sorte!). Por isso vamos resolvê-la como seria feito para um caso qualquer.\nPrimeiro escrevemos $200 \\leq n(n+1)$ como $n^{2}+n-200 \\geq 0$, o que nos leva ao estudo de sinal da função quadrática $f(x)=x^{2}+x-200$. As raízes de $f(x)$ são $x_{1}=\\frac{-1-\\sqrt{1+800}}{2}$, que é negativa, e $x_{2}=\\frac{-1+\\sqrt{1+800}}{2}$, que é aproximadamente $13,6$. Como $f(x) \\geq 0$ para $x \\leq x_{1}$ e $x \\geq x_{2}$, segue que o $n$ que estamos procurando é o menor inteiro que é maior ou igual a $x_{2}$, ou seja, $n=14$.\n\nAgora, se quiséssemos determinar o $10.000^{\\circ}$ termo da sequência repetindo procedimento acima, encontraríamos $x_{2}=\\frac{-1+\\sqrt{1+80000}}{2}$, que é aproximadamente $140,9$. Logo $n=141$ e o $10.000^{\\circ}$ termo da sequência está no $141^{\\circ}$ bloco. Como $141=28 \\times 5+1$ segue que o $10.000^{\\circ}$ termo é $1$.\n\n![](attached_image_1.png)\n\n![](attached_image_2.png)",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26415,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nQuantas frações da forma $\\frac{n}{n+1}$ são menores do que $\\frac{7}{9}$, sabendo que $n$ é um número inteiro positivo?\nA) 1\nB) 2\nC) 3\nD) 4\nE) 5",
    "options": [],
    "answer": "C",
    "solution": "Solution:\n\nAs frações da forma $\\frac{n}{n+1}$, com $n$ inteiro positivo são:\n$$\n\\underbrace{\\frac{1}{2}}_{n=1} ; \\quad \\underbrace{\\frac{2}{3}}_{n=2} ; \\quad \\underbrace{\\frac{3}{4}}_{n=3} ; \\quad \\underbrace{\\frac{4}{5}}_{n=4} ; \\quad \\underbrace{\\frac{5}{6}}_{n=5} \\cdots\n$$\nObserve que esta sequência de frações é crescente, isto é: $\\frac{1}{2}<\\frac{2}{3}<\\frac{3}{4}<\\frac{4}{5}<\\ldots$\n\nPara comparar cada uma dessas frações com $\\frac{7}{9}$ precisamos igualar os denominadores. Temos:\n$$\n\\frac{1}{2}=\\frac{9}{18}<\\frac{14}{18}=\\frac{7}{9} \\quad ; \\quad \\frac{2}{3}=\\frac{6}{9}<\\frac{7}{9} \\quad ; \\quad \\frac{3}{4}=\\frac{27}{36}<\\frac{28}{36}=\\frac{7}{9} \\quad ; \\quad \\frac{4}{5}=\\frac{36}{45}>\\frac{35}{45}=\\frac{7}{9}\n$$\nLogo, $4/5$ é maior do que $7/9$, e como a sequência é crescente, a partir de $4/5$ todas as frações desta sequência são maiores do que $7/9$. Assim, as frações da forma $\\frac{n}{n+1}$ menores do que $\\frac{7}{9}$ são $\\frac{1}{2}, \\frac{2}{3}, \\frac{3}{4}$. Portanto, a resposta é 3.\n\n\nSolução 2 - Transformando em números decimais temos: $\\frac{7}{9}=0,777\\ldots$ e $\\frac{1}{2}=0,5$; $\\frac{2}{3}=0,666\\ldots$; $\\frac{3}{4}=0,75$; $\\frac{4}{5}=0,8$; $\\frac{5}{6}=0,8333\\ldots$\nLogo, a sequência é crescente e apenas $\\frac{1}{2}=0,5$; $\\frac{2}{3}=0,666\\ldots$; $\\frac{3}{4}=0,75$ são menores do que $\\frac{7}{9}=0,777\\ldots$\n\n\nSolução 3 - Se $\\frac{n}{n+1}<\\frac{7}{9}$, então $\\frac{n}{n+1}-\\frac{7}{9}<0 \\Rightarrow \\frac{9n-7(n+1)}{9(n+1)}=\\frac{2n-7}{9(n+1)}<0$. Como $9(n+1)>0$, devemos ter $2n-7<0$, isto é $n<\\frac{7}{2}=3,5$. Logo, $n=1,2,3$ e portanto, as frações são $\\frac{1}{2}, \\frac{2}{3}$ e $\\frac{3}{4}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26416,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nEntre 1986 e 1989, a moeda do nosso país era o cruzado ($\\mathrm{Cz\\ \\$}$). De lá para cá, tivemos o cruzado novo, o cruzeiro, o cruzeiro novo e, hoje, temos o real. Para comparar valores do tempo do cruzado e de hoje, os economistas calcularam que 1 real equivale a 2.750.000.000 cruzados.\n\nImagine que a moeda não tivesse mudado e que João, que ganha hoje 640 reais por mês, tivesse que receber seu salário em notas de 1 cruzado cada uma. Se uma pilha de 100 notas de 1 cruzado tem $1{,}5\\ \\mathrm{~cm}$ de altura, qual seria a altura do salário do João?\n\nA) $26{,}4\\ \\mathrm{~km}$\nB) $264\\ \\mathrm{~km}$\nC) $26400\\ \\mathrm{~km}$\nD) $264000\\ \\mathrm{~km}$\nE) $2640000\\ \\mathrm{~km}$",
    "options": [],
    "answer": "D",
    "solution": "Solution:\nO enunciado diz que 1 real $= 275 \\times 10^{7}$ cruzados. O salário de João é 640 reais, o que é equivalente a $640 \\times 275 \\times 10^{7} = 176.000 \\times 10^{7} = 176 \\times 10^{10}$ cruzados. O número de pilhas de 100 notas que se podem fazer com este número de notas de 1 cruzado é $\\frac{176 \\times 10^{10}}{10^{2}} = 176 \\times 10^{8}$.\n\nComo cada uma destas pilhas tem altura $1{,}5\\ \\mathrm{~cm}$, a altura de todas elas é $1{,}5 \\times 176 \\times 10^{8} = 264 \\times 10^{8}\\ \\mathrm{~cm}$.\n\nLembramos agora que $1\\ \\mathrm{~km} = 1000\\ \\mathrm{~m} = 10^{3}\\ \\mathrm{~m}$ e $1\\ \\mathrm{~m} = 100\\ \\mathrm{~cm} = 10^{2}\\ \\mathrm{~cm}$, donde $1\\ \\mathrm{~km} = 10^{3} \\times 10^{2} = 10^{5}\\ \\mathrm{~cm}$. Logo uma pilha de $264 \\times 10^{8}\\ \\mathrm{~cm}$ tem $\\frac{264 \\times 10^{8}}{10^{5}} = 264 \\times 10^{3} = 264.000\\ \\mathrm{~km}$ de altura.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26417,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nQual o maior número de 6 algarismos que se pode encontrar suprimindo-se 9 algarismos do número $778157260669103$ sem mudar a ordem dos algarismos?\n\n(A) $778152$\n(B) $781569$\n(C) $879103$\n(D) $986103$\n(E) $987776$",
    "options": [],
    "answer": "C",
    "solution": "Solution:\n\nSolução 1. Para que seja o maior possível, o número deve começar com o maior algarismo. Para termos 6 algarismos sem mudar a ordem, o maior é $8$ depois $7$, faltam agora $4$ algarismos para completar o número, escolhemos $9103$. Logo, o número é $879103$ ($77$-$8793$).\n\nSolução 2. As opções D e E não servem, pois a ordem foi alterada, já nas opções A, B e C, não. O maior número entre as opções $\\{A\\}, \\{B\\}$ e C é C.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26418,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSe um arco de $60^{\\circ}$ num círculo I tem o mesmo comprimento que um arco de $45^{\\circ}$ num círculo II, então a razão entre a área do círculo I com a do círculo II é:\n(A) $16/9$\n(B) $9/16$\n(C) $4/3$\n(D) $3/4$\n(E) $6/9$",
    "options": [],
    "answer": "B",
    "solution": "Solution:\n\nComo o arco de $60^{\\circ}$ do círculo I tem o mesmo comprimento que o arco de $45^{\\circ}$ no círculo II, concluímos que o raio do círculo I é menor que o do círculo II. Denotemos por $r$ e $R$ os raios dos círculos I e II respectivamente.\n\nNo círculo I o comprimento do arco de $60^{\\circ}$ é igual a $1/6$ de seu comprimento total, ou seja, $\\frac{2\\pi r}{6} = \\frac{\\pi r}{3}$.\n\nAnalogamente, no círculo II o comprimento do arco de $45^{\\circ}$ é igual a $1/8$ de seu comprimento total, ou seja, $\\frac{2\\pi R}{8} = \\frac{\\pi R}{4}$.\n\nLogo, $\\frac{\\pi r}{3} = \\frac{\\pi R}{4} \\Rightarrow \\frac{r}{R} = \\frac{3}{4}$.\n\nFinalmente temos:\n\n![](attached_image_1.png)\n\n$$\n\\frac{\\text{área do círculo I}}{\\text{área do círculo II}} = \\frac{\\pi r^{2}}{\\pi R^{2}} = \\left(\\frac{r}{R}\\right)^{2} = \\left(\\frac{3}{4}\\right)^{2} = \\frac{9}{16}\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26419,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDetermine quais afirmações são verdadeiras:\n(A) $|-108|>100$\n(B) $|5-13|=|5|-|13|$\n(C) $|2-9|=9-2$\n(D) $|a^{2}+5|=a^{2}+5$\n(E) $|-6 a|=6|a|$",
    "options": [],
    "answer": "A, C, D, E",
    "solution": "Solution:\n\n(A) $|-108|=108>100$, verdadeira\n\n(B) $|5-13|=|-8|=8$ e $|5|-|13|=5-13=-8$, falsa.\n\n(C) $|2-9|=-(2-9)=9-2$ porque $2-9<0$, verdadeira.\n\n(D) $|a^{2}+5|=a^{2}+5$ porque $a^{2}+5>0$ para qualquer valor de $a$, verdadeira.\n\n$|x|=\\begin{cases}x \\text{ se } x \\geq 0 \\\\ -x \\text{ se } x<0\\end{cases}$\n\n(E) $|-6 a|=|-6| \\times|a|=6|a|$, verdadeira.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26420,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO preço de uma corrida de táxi é $R\\$ 2,50$ fixos (\"bandeirada\"), mais $R\\$ 0,10$ por cada 100 metros rodados. Tenho apenas $R\\$ 10,00$ no bolso. Logo, tenho dinheiro para uma corrida de até:\nA) $2,5~\\mathrm{km}$\nB) $5,0~\\mathrm{km}$\nC) $7,5~\\mathrm{km}$\nD) $10,0~\\mathrm{km}$\nE) $12,5~\\mathrm{km}$",
    "options": [],
    "answer": "C",
    "solution": "Solution:\n\nComo a bandeirada é fixa, temos $10,00 - 2,50 = 7,50$ reais a serem gastos apenas com os metros rodados. Cada trecho de 100 metros rodado custa $R\\$ 0,10$, então com $R\\$ 7,50$ posso fazer uma corrida de $\\frac{7,50}{0,10} = \\frac{750}{10} = 75$ trechos de 100 metros cada um, ou seja $75 \\times 100 = 7500$ metros. Como 1 quilômetro tem 1000 metros, segue que com $R\\$ 10,00$ posso pagar uma corrida de até 7500 metros $= \\frac{7500}{1000}$ quilômetros $= 7,5$ quilômetros.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26421,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUma bola de futebol é feita com 32 peças de couro. Dessas peças 12 são pentágonos regulares idênticos e as outras 20 são hexágonos, também regulares e idênticos. Os lados dos pentágonos são iguais aos lados dos hexágonos. Para unir dois lados de duas dessas peças é necessária uma costura. Quantas são as costuras necessárias para fazer uma bola?\nA) 60\nB) 64\nC) 90\nD) 120\nE) 180",
    "options": [],
    "answer": "C",
    "solution": "Solution:\n\nSe somarmos os números de lados de todos os polígonos (20 hexágonos e 12 pentágonos) que compõem a superfície da bola, obteremos um valor que é duas vezes o número de costuras, pois cada costura é lado comum de exatamente dois polígonos. Assim, temos que $2 \\times$ (número de costuras) $= 12 \\times 5 + 20 \\times 6 = 180$, donde o número de costuras é $90$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26422,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$2-2\\{2-2[2-2(4-2)]\\}$ é igual a:\n(A) 0\n(B) 2\n(C) -2\n(D) 4\n(E) -10",
    "options": [],
    "answer": "E",
    "solution": "Solution:\n\n(E) As ordens de prioridade para resolver uma expressão são:\n$$\n\\begin{aligned}\n& \\underbrace{\\text{parênteses}}_{1^{\\circ}} \\rightarrow \\underbrace{\\text{colchete}}_{2^{\\circ}} \\rightarrow \\underbrace{\\text{chaves}}_{3^{\\circ}} \\text{ e } \\underbrace{\\text{multiplicações e divisões}}_{1^{\\circ}} \\rightarrow \\underbrace{\\text{somas e subtrações}}_{2^{\\circ}} \\\\\n& \\quad 2-2\\{2-2[2-2(\\underbrace{4-2}_{2})]\\}=2-2\\{2-2[2-\\underbrace{2 \\times 2}_{4}]\\}=2-2\\{2-2[\\underbrace{2-4}_{-2}]\\}=\\end{aligned}\n$$\nTemos:\n$$\n=2-2\\{2-\\underbrace{2 \\times(-2)}_{-4}\\}=2-2\\{2-(-4)\\}=2-2\\{\\underbrace{2+4}_{6}\\}=2-\\underbrace{2 \\times 6}_{12}=2-12=-10\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26423,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO símbolo $\\odot$ representa uma operação especial com números. Veja alguns exemplos $2 \\odot 4=10, 3 \\odot 8=27, \\quad 4 \\odot 27=112, \\quad 5 \\odot 1=10$. Quanto vale $4 \\odot(8 \\odot 7)$ ?\n(A) 19\n(B) 39\n(C) 120\n(D) 240\n(E) 260",
    "options": [],
    "answer": "E",
    "solution": "Solution:\n\nTemos que descobrir qual é a regra dessa operação. Note que $2 \\odot 4=10=2 \\times 4+2$, $3 \\odot 8=27=3 \\times 8+3$, $4 \\odot 27=112=4 \\times 27+4$, $5 \\odot 1=10=5 \\times 1+5$.\n\nPodemos concluir que a regra que define a operação $\\odot$ é $a \\odot b = a \\times b + a$.\n\nAssim, temos:\n\n$4 \\odot (8 \\odot 7) = 4 \\odot (8 \\times 7 + 8) = 4 \\odot 64 = 4 \\times 64 + 4 = 260$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26424,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nQuais os valores de $x$ que satisfazem $\\frac{1}{x-2}<4$?\n\n(A) $x<\\frac{9}{4}$\n\n(B) $x>2$\n\n(C) $2<x<\\frac{9}{4}$\n\n(D) $x<-2$\n\n(E) $x<2$ ou $x>\\frac{9}{4}$",
    "options": [],
    "answer": "(E)",
    "solution": "Solution:\n\n$\\frac{1}{x-2}<4 \\Rightarrow \\frac{1}{x-2}-4<0 \\Rightarrow \\frac{1-4(x-2)}{x-2}=\\frac{9-4x}{x-2}<0$\n\n$1^\\circ$ caso : $9-4x>0$ e $x-2<0$ :\n$9-4x>0 \\Rightarrow x<\\frac{9}{4}$ e $x-2<0 \\Rightarrow x<2$.\nComo $2<\\frac{9}{4}$ a solução são todos os números $x$ menores que 2, isto é $x<2$.\n\n$2^\\circ$ caso : $9-4x<0$ e $x-2>0$ :\n$9-4x<0 \\Rightarrow x>\\frac{9}{4}$ e $x-2>0 \\Rightarrow x>2$\nComo $2<\\frac{9}{4}$ a solução são todos os números $x$ maiores que 9/4, isto é $x>\\frac{9}{4}$.\n\nLogo, a solução da inequação é $x<2$ ou $x>\\frac{9}{4}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26425,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nToda vez que Joãozinho vai ao cinema, ele toma 2 refrigerantes. Ele gastou toda a sua mesada de $R\\$ 50,00$ indo ao cinema 6 vezes e tomando um total de 20 refrigerantes, incluindo os que ele tomou quando foi ao cinema. Se Joãozinho tivesse tomado só um refrigerante cada vez que foi ao cinema, com essa economia ele poderia ter ido ao cinema mais uma vez, tomando um refrigerante também nessa ocasião. A respeito do preço do ingresso no cinema e preço do refrigerante, podemos afirmar que:\nA) o preço do ingresso é o triplo do preço do refrigerante.\nB) o preço do ingresso é o quádruplo do preço do refrigerante.\nC) o preço do ingresso é o quíntuplo do preço do refrigerante.\nD) o ingresso é $R \\$ 6,00$ mais caro que o refrigerante.\nE) o ingresso é $R \\$ 5,00$ mais caro que o refrigerante",
    "options": [],
    "answer": "C",
    "solution": "Solution:\n\n(C) A economia teria sido equivalente a 6 refrigerantes, permitindo a Joãozinho mais um cinema e mais um refrigerante. Logo o ingresso do cinema é 5 vezes o valor do refrigerante.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26426,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa figura temos $\\hat{B}=50^{\\circ}$, $AD$ e $CD$ são as bissetrizes dos ângulos $\\hat{A}$ e $\\hat{C}$ respectivamente.\nQual a medida do ângulo $A \\hat{D} C$?\nA) $90^{\\circ}$\nB) $100^{\\circ}$\nC) $115^{\\circ}$\nD) $122.5^{\\circ}$\nE) $125^{\\circ}$\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "C",
    "solution": "Solution:\n\nNesta questão, usaremos o seguinte importante teorema da Geometria Plana:\nTeorema: A soma dos ângulos internos de um triângulo é $180^{\\circ}$.\n\nDo teorema acima temos $\\hat{A}+\\hat{B}+\\hat{C}=180^{\\circ}$, e como $\\hat{B}=50^{\\circ}$, segue que\n\n![](attached_image_2.png)\n\n$$\n\\hat{A}+50^{\\circ}+\\hat{C}=180^{\\circ} \\Rightarrow \\hat{A}+\\hat{C}=130^{\\circ}\n$$\n\nAplicando agora o teorema ao triângulo $ADC$, obtemos:\n$$\n\\frac{\\hat{A}}{2}+\\frac{\\hat{C}}{2}+A \\hat{D} C=180^{\\circ}\n$$\nComo $\\frac{\\hat{A}}{2}+\\frac{\\hat{C}}{2}=\\frac{\\hat{A}+\\hat{C}}{2}=\\frac{130^{\\circ}}{2}=65^{\\circ}$, concluímos da igualdade acima que $A \\hat{D} C=180^{\\circ}-65^{\\circ}=115^{\\circ}$.\n\n![](attached_image_2.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26427,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm artesão começa a trabalhar às 8 h e produz 6 braceletes a cada vinte minutos; já seu auxiliar começa a trabalhar uma hora depois e produz 8 braceletes do mesmo tipo a cada meia hora. O artesão pára de trabalhar às $12~\\mathrm{h}$, mas avisa ao seu auxiliar que este deverá continuar trabalhando até produzir o mesmo que ele. A que horas o auxiliar irá parar?\nA) $12~\\mathrm{h}$\nB) $12~\\mathrm{h}~30~\\mathrm{min}$\nC) $13~\\mathrm{h}$\nD) $13~\\mathrm{h}~30~\\mathrm{min}$\nE) $14~\\mathrm{h}~30~\\mathrm{min}$",
    "options": [],
    "answer": "D",
    "solution": "Solution:\n\nO artesão produz 6 braceletes a cada 20 minutos. Como 1 hora $=60$ minutos $=3 \\times 20$ minutos, o artesão produz $6 \\times 3=18$ braceletes em 1 hora. Como ele trabalhou $12$ horas $-8$ horas $=4$ horas, o número de braceletes feitos pelo artesão é $18 \\times 4=72$.\n\nO auxiliar produz 8 braceletes a cada meia-hora, portanto em 1 hora ele produz 16 braceletes. Para produzir 72 braceletes ele precisará de $\\frac{72}{16}=4,5$ horas $=4$ horas e 30 minutos. Como ele inicia seu trabalho às 9 horas, ele terminará seu trabalho às 9 horas $+4$ horas $+30$ minutos $=13$ horas e 30 minutos.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26428,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUma certa máquina é capaz de produzir $8$ réguas em cada minuto. Quantas réguas esta máquina consegue produzir em $15$ minutos?\nA) $104$\nB) $110$\nC) $112$\nD) $128$\nE) $120$",
    "options": [],
    "answer": "E",
    "solution": "Solution:\n\nSe a máquina produz $8$ réguas em $1$ minuto, em $15$ minutos ela produzirá $8 \\times 15 = 120$ réguas.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26429,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nHá $1002$ balas de banana e $1002$ balas de maçã numa caixa. Lara tira, sem olhar o sabor, duas balas da caixa. Se $q$ é a probabilidade das duas balas serem de sabores diferentes e $p$ é a probabilidade das duas balas serem do mesmo sabor, qual o valor de $q-p$?\n\nA) $0$\nB) $1/2004$\nC) $1/2003$\nD) $2/2003$\nE) $1/1001$",
    "options": [],
    "answer": "C",
    "solution": "Solution:\n\nA primeira bala pode ser de qualquer sabor; para fixar idéias suponhamos que seja de banana. Depois que esta bala é retirada, sobram $1002 + 1001$ balas na caixa — no nosso caso, $1002$ de maçã e $1001$ de banana.\n\nA probabilidade $q$ de que a segunda bala seja diferente (no nosso exemplo, de maçã) é\n$$\nq = \\frac{1002}{2003}\n$$\nA probabilidade $p$ de que a segunda bala seja igual (no nosso exemplo, de banana) é\n$$\np = \\frac{1001}{2003}\n$$\nA diferença $q - p$ é, portanto,\n$$\nq - p = \\frac{1002}{2003} - \\frac{1001}{2003} = \\frac{1}{2003}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26430,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSe $x > 0$, $y > 0$, $x > y$ e $z \\neq 0$, então a única opção errada é:\n(A) $x + z > y + z$\n(B) $x - z > y - z$\n(C) $x z > y z$\n(D) $\\frac{x}{z^{2}} > \\frac{y}{z^{2}}$\n(E) $x z^{2} > y z^{2}$",
    "options": [],
    "answer": "C",
    "solution": "Solution:\nNessa questão usaremos as propriedades das desigualdades.\nPodemos somar o mesmo número a ambos os membros de uma desigualdade sem alterar o sinal, temos: $x > y \\Rightarrow \\left\\{\\begin{array}{l} x + z > y + z \\text{ (somando $z$ a ambos os membros)} \\\\ x - z > y - z \\text{ (somando $-z$ a ambos os membros)} \\end{array}\\right.$\n(A) e (B) corretas\n\nA opção (C) é falsa porque $z$ pode ser negativo, por exemplo: $x = 5$, $y = 3$ e $z = -2$ temos:\n$5 > 3$, no entanto $\\underbrace{5 \\times (-2)}_{x z} = -10 < -6 = \\underbrace{3 \\times (-2)}_{y z}$.\n\nComo $z \\neq 0$ então $z^{2} > 0$ e $\\frac{1}{z^{2}} > 0$, logo as opções (D) e (E) estão corretas porque foram obtidas multiplicando-se ambos os membros de $x > y$ por um número positivo; em (E) por $z^{2}$ e em (D) por $\\frac{1}{z^{2}}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26431,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa figura, os três círculos são concêntricos, e as áreas do menor círculo e do maior anel (em cinza) são iguais. O raio do menor círculo é $5~\\mathrm{cm}$ e do maior $13~\\mathrm{cm}$. Qual o raio do círculo intermediário?\n\n(A) 12\n(B) 11\n(C) $10 \\sqrt{65}$\n(D) $5 \\sqrt{3}$\n(E) $12 \\sqrt{2}$\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "A",
    "solution": "Solution:\n\nA área do menor círculo é $5^2 \\pi = 25\\pi~\\mathrm{cm}^2$ e do maior é $13^2 \\pi = 169\\pi~\\mathrm{cm}^2$. Seja $r$ o raio do círculo intermediário, então a área do maior anel é $169\\pi - \\pi r^2$. Logo,\n$$\n169\\pi - \\pi r^2 = 25\\pi \\Rightarrow r^2 = 169 - 25 = 144,\n$$\ndonde $r = 12~\\mathrm{cm}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26432,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSe $n$ é um número natural e $\\frac{n}{24}$ é um número entre $\\frac{1}{6}$ e $\\frac{1}{4}$, então $n$ é igual a:\n\n(A) 5\n(B) 6\n(C) 7\n(D) 8\n(E) 9",
    "options": [],
    "answer": "A",
    "solution": "Solution:\n\nComo $\\frac{1}{6} = \\frac{4}{24}$ e $\\frac{1}{4} = \\frac{6}{24}$, então $n$ só pode ser igual a $5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26433,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOs ramais de uma central telefônica têm apenas 2 algarismos, de $00$ a $99$. Nem todos os ramais estão em uso. Trocando a ordem de dois algarismos de um ramal em uso, ou se obtém o mesmo número ou um número de um ramal que não está em uso. O maior número possível de ramais em uso é:\n(A) Menos que $45$\n(B) $45$\n(C) entre $45$ e $55$\n(D) mais que $55$\n(E) $55$",
    "options": [],
    "answer": "E",
    "solution": "Solution:\n\n(i) Os dois algarismos são iguais ($00, 11, 22, 33, 44, 55, 66, 77, 88$, e $99$), esses são em número de $10$.\n\n(ii) Os dois algarismos são distintos, nesse caso temos $10 \\times 9 = 90$ números, e metade deles podem ser usados.\n\nLogo, temos no máximo $10 + 45 = 55$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26434,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n![](attached_image_1.png)\n\nSeja $v$ a soma das áreas das regiões pertencentes unicamente aos três discos pequenos (em cinza claro), e seja $w$ a área da região interior unicamente ao maior disco (em cinza escuro). Os diâmetros dos círculos são 6, 4, 4 e 2. Qual das igualdades abaixo é verdadeira?\n\n(A) $3 v=\\pi w$\n(B) $3 v=2 w$\n(C) $v=w$\n(D) $\\pi v=3 w$\n(E) $\\pi v=w$",
    "options": [],
    "answer": "C",
    "solution": "Solution:\n\n(C) Os raios dos três discos menores são $1$, $2$ e $2$; e do disco maior $3$.\n\nDenotemos por $b$ a área em branco, temos:\n$$\n\\begin{aligned}\n& v = 9\\pi - b \\\\\n& \\text{de raio } 3 \\\\\n& w = 9\\pi - b\n\\end{aligned}\n$$\nLogo, $v = w$.\n\nA área do disco de raio $r$ é $\\pi r^{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26435,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm atleta corre $5000~\\mathrm{m}$ por semana em uma quadra de esportes, que tem uma pista curta e outra longa. Em uma semana ele treinou seis dias, sendo que a cada dia correu uma vez na pista longa e duas na pista curta. Na semana seguinte ele treinou sete dias, sendo que a cada dia correu uma vez em cada pista. Podemos então afirmar que:\nA) a pista longa é três vezes maior que a curta.\nB) a pista longa é quatro vezes maior que a curta.\nC) a pista longa é cinco vezes maior que a curta.\nD) a pista longa é $600~\\mathrm{m}$ mais longa que a curta.\nE) a pista longa é $500~\\mathrm{m}$ mais longa que a curta.",
    "options": [],
    "answer": "C",
    "solution": "Solution:\n\nDenotemos por $x$ e $y$ os comprimentos das pistas longa e curta, respectivamente.\n\nNuma semana, ele corre $6(x+2y)$ e na outra $7(x+y)$. Como, em cada semana, ele corre os mesmos $5000$ metros, temos:\n$$\n6(x+2y) = 7(x+y)\n$$\nSegue que $6x + 12y = 7x + 7y$, e portanto, $5y = x$. Assim, o comprimento da pista longa é cinco vezes o da pista curta.\n\n\nNa semana em que Joãozinho treinou sete dias, ele correu uma pista longa a mais e cinco pistas curtas a menos do que a semana em que ele treinou apenas seis dias. Como a distância corrida foi a mesma nas duas semanas, concluímos que o comprimento da pista longa é igual ao comprimento de cinco pistas curtas.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26436,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNuma certa povoação africana vivem 800 mulheres. Delas, $3\\%$ usam apenas um brinco; das restantes, metade usa dois brincos e a outra metade, nenhum. Qual o número total de brincos usados por todas as mulheres?\nA) 776\nB) 788\nC) 800\nD) 812\nE) 824",
    "options": [],
    "answer": "C",
    "solution": "Solution:\n\nDo enunciado temos que o número de mulheres que usam apenas um brinco é $0,03 \\times 800 = 24$. Restam $800 - 24 = 776$, das quais $388$ usam dois brincos e $388$ não usam brincos. Logo, o número total de brincos usados por todas as mulheres é: $24 + 388 \\times 2 = 800$.\nSolution:\n\nSe cada mulher com dois brincos der um dos seus a uma das que não têm brincos, todas as $800$ mulheres ficarão com um único brinco. Logo, o número de brincos é igual ao de mulheres, ou seja, $800$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26437,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSe $\\frac{\\sqrt{x}}{\\sqrt{y}}=5$ então $\\frac{x+y}{2y}$ é igual a:\n(A) $5/2$\n(B) $3\\sqrt{2}$\n(C) $13y$\n(D) $\\frac{25y}{2}$\n(E) $13$",
    "options": [],
    "answer": "E",
    "solution": "Solution:\n\nElevando ao quadrado ambos os membros de $\\frac{\\sqrt{x}}{\\sqrt{y}}=5$, obtemos $\\frac{x}{y}=25$. Agora,\n$$\n\\frac{x+y}{2y} = \\frac{1}{2} \\times \\frac{x+y}{y} = \\frac{1}{2} \\times \\left(\\frac{x}{y} + \\frac{y}{y}\\right) = \\frac{1}{2} \\times \\left(\\frac{x}{y} + 1\\right) = \\frac{1}{2} \\times (25 + 1) = 13.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26438,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nQuais os valores de $x$ que satisfazem $\\frac{1}{x-2}<4$?\n\n(A) $x>\\frac{3}{4}$\n(B) $x>2$\n(C) $\\frac{3}{4}<x<2$\n(D) $x<2$\n(E) todos os valores de $x$.",
    "options": [],
    "answer": "C",
    "solution": "Solution:\n\nTemos: $\\frac{1}{x-2}<4 \\Rightarrow \\frac{1}{x-2}-4<0 \\Rightarrow \\frac{1-4(x-2)}{x-2}<0 \\Rightarrow \\frac{3-4x}{x-2}<0$.\n\nPara que uma fração seja negativa, o numerador e o denominador têm que ter sinais trocados.\n\n$1^\\circ$ caso: $3-4x>0$ e $x-2<0$.\n\n$3-4x>0 \\Rightarrow x<\\frac{3}{4}$ e $x-2<0 \\Rightarrow x<2$, o que é impossível.\n\n$2^\\circ$ caso: $3-4x<0$ e $x-2>0$.\n\n$3-4x<0 \\Rightarrow x>\\frac{3}{4}$ e $x-2<0 \\Rightarrow x<2$. Logo, a resposta é $\\frac{3}{4}<x<2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26439,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSeja $n=9867$. Se você calculasse $n^{3}-n^{2}$, encontraria um número cujo algarismo das unidades é:\nA) 0\nB) 2\nC) 4\nD) 6\nE) 8",
    "options": [],
    "answer": "C",
    "solution": "Solution:\n\nSolução 1: O algarismo final de $9867^{3}$ é o mesmo que o de $7^{3}=343$, isto é, 3; o algarismo final de $9867^{2}$ é o mesmo que o de $7^{2}=49$, isto é, 9. Se de um número terminado em 3 subtraímos outro terminado em 9, o algarismo final do resultado é 4.\n\nComentário: Observe que:\nalgarismo das unidades de $\\left(9867^{3}-9867^{2}\\right)=$ algarismo das unidades de $\\left(7^{3}-7^{2}\\right)$\n\nSolução 2: $n^{3}-n^{2}=n^{2}(n-1)$. Assim, $n^{2}=(9867)^{2}$ termina em 9 e $n-1=9866$ em 6. Como, $9 \\times 6=54$, o algarismo final do resultado é 4.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26440,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSe dois lados de um triângulo medem $5~\\mathrm{cm}$ e $7~\\mathrm{cm}$, então o terceiro lado não pode medir:\n(A) $11~\\mathrm{cm}$\n(B) $10~\\mathrm{cm}$\n(C) $6~\\mathrm{cm}$\n(D) $3~\\mathrm{cm}$\n(E) $1~\\mathrm{cm}$",
    "options": [],
    "answer": "E",
    "solution": "Solution:\n\nLembre que num triângulo a soma de dois lados quaisquer tem que ser maior que o terceiro lado. Como $1+5$ não é maior do que $7$, o terceiro lado não pode ser $1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26441,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAndré, Bruno, Celina e Dalva ganharam juntos 21 medalhas num concurso. André foi o que mais ganhou medalhas, Bruno ganhou o dobro de Celina e Dalva 3 a mais que Bruno. Quantas medalhas cada um pode ter ganhado?",
    "options": [],
    "answer": "Possible distributions: André 8, Bruno 4, Celina 2, Dalva 7; André 13, Bruno 2, Celina 1, Dalva 5; André 18, Bruno 0, Celina 0, Dalva 3.",
    "solution": "Solution:\n\nDenotemos por $A$, $B$, $C$ e $D$ o número de medalhas ganhas por André, Bruno, Celina e Dalva respectivamente, então $A+B+C+D=21$. Agora, temos:\n- \"Bruno ganhou o dobro de Celina\" $\\Rightarrow B=2C$\n- \"Dalva 3 a mais que Bruno\": $\\Rightarrow D=B+3$\n\nDaí obtemos:\n$$\nA + B + C + D = 21\n$$\nSubstituindo $B=2C$ e $D=B+3$:\n$$\nA + B + C + (B+3) = 21 \\\\\nA + 2B + C + 3 = 21 \\\\\nA + 2B + C = 18\n$$\nMas $B=2C$, então $C = B/2$:\n$$\nA + 2B + (B/2) = 18 \\\\\nA + (5B)/2 = 18\n$$\nMultiplicando ambos os lados por 2:\n$$\n2A + 5B = 36\n$$\nComo $A$ e $B$ são inteiros, temos as seguintes possibilidades para $A$ e $B$:\n\n| $A$ | $B$ | $2A + 5B$ |\n| :--- | :--- | :--- |\n| 3 | 6 | $2 \\times 3 + 5 \\times 6 = 36$ |\n| 8 | 4 | $2 \\times 8 + 5 \\times 4 = 36$ |\n| 13 | 2 | $2 \\times 13 + 5 \\times 2 = 36$ |\n| 18 | 0 | $2 \\times 18 + 5 \\times 0 = 36$ |\n\nComo André foi o que mais recebeu medalhas, a solução $A=3$ e $B=6$ não serve.\n\nAgora usando as condições $C = B/2$ e $D = B+3$, obtemos as seguintes possibilidades de medalhas para cada um deles:\n\n| André | Bruno | Celina | Dalva | Total |\n| :---: | :---: | :---: | :---: | :---: |\n| 8 | 4 | 2 | 7 | 21 |\n| 13 | 2 | 1 | 5 | 21 |\n| 18 | 0 | 0 | 3 | 21 |\n\nPortanto, as possíveis distribuições de medalhas são:\n- André: 8, Bruno: 4, Celina: 2, Dalva: 7\n- André: 13, Bruno: 2, Celina: 1, Dalva: 5\n- André: 18, Bruno: 0, Celina: 0, Dalva: 3",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26442,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nQuantas soluções inteiras e positivas satisfazem a dupla inequação $2000<\\sqrt{n(n+1)}<2005$?\n\n(A) 1\n(B) 2\n(C) 3\n(D) 4\n(E) 5",
    "options": [],
    "answer": "E",
    "solution": "Solution:\n\nComo os números que aparecem são todos positivos, podemos elevá-los ao quadrado mantendo os sinais, isto é: $2000^{2}<n(n+1)<2005^{2}$. Observe que $n$ e $n+1$ são inteiros consecutivos. Logo, temos as seguintes opções:\n$$\n\\begin{aligned}\n& 2000^{2}<2000 \\times 2001<2005^{2} \\\\\n& 2000^{2}<2001 \\times 2002<2005^{2} \\\\\n& 2000^{2}<2002 \\times 2003<2005^{2} \\\\\n& 2000^{2}<2003 \\times 2004<2005^{2} \\\\\n& 2000^{2}<2004 \\times 2005<2005^{2}\n\\end{aligned}\n$$\nLogo, temos 5 possibilidades para $n$: $2000$, $2001$, $2002$, $2003$ e $2004$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26443,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLuíza, Maria, Antônio e Júlio são irmãos. Dois deles têm a mesma altura. Sabe-se que:\n- Luíza é maior que Antônio\n- Maria é menor que Luíza\n- Antônio é maior do que Júlio\n- Júlio é menor do que Maria.\nQuais deles têm a mesma altura?\nA) Maria e Júlio\nB) Júlio e Luíza\nC) Antônio e Luíza\nD) Antônio e Júlio\nE) Antônio e Maria",
    "options": [],
    "answer": "E",
    "solution": "Solution:\n\nUsaremos a notação $a < b$ que significa que $a$ é menor do que $b$, ou equivalentemente, $b$ é maior do que $a$. Assim, $a < b < c$ significa que $a$ é menor do que $b$ e $b$ é menor do que $c$.\n\nPara simplificar, vamos denotar a altura de cada um dos irmãos pela letra inicial de seu nome.\n\nDo enunciado temos:\n\n(i) $L$ maior que $A$ ou, equivalentemente, $A$ menor que $L$ ($A < L$)\n\n(ii) $M$ menor que $L$ ($M < L$)\n\n(iii) $A$ maior que $J$ ou, equivalentemente, $J$ menor que $A$ ($J < A$)\n\n(iv) $J$ menor que $M$ ($J < M$)\n\nDe (i) e (iii) segue que: $J < A < L$. Portanto, os irmãos de mesma altura não estão entre Júlio, Antônio e Luíza.\n\nDe (ii) e (iv) segue que: $J < M < L$. Portanto, os irmãos de mesma altura não estão entre Júlio, Maria e Luíza.\n\nLogo, a única opção é que Antônio e Maria tenham a mesma altura.\n\n\nSolution 2:\n\nPelo enunciado, as opções A, C e D não ocorrem. Como Luíza é maior do que Antônio e Antônio é maior do que Júlio, temos que Luíza é maior do que Júlio. Logo, a opção correta é (E).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26444,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSinal de um produto e sinal de um quociente: $a$, $b$, $c$ e $d$ são quatro números não nulos tais que os quocientes $\\frac{a}{5}$, $\\frac{-b}{7 a}$, $\\frac{11}{a b c}$, $\\frac{-18}{a b c d}$ são positivos. Determine os sinais de $a$, $b$, $c$ e $d$.",
    "options": [],
    "answer": "a > 0, b < 0, c < 0, d < 0",
    "solution": "Solution:\n\n- $\\frac{a}{5} > 0 \\Rightarrow a > 0$\n\n- Temos $a > 0 \\Rightarrow 7a > 0$, logo: $\\frac{-b}{7a} > 0 \\Rightarrow -b > 0 \\Rightarrow b < 0$\n\n- $\\frac{11}{a b c} > 0 \\Rightarrow a b c > 0$\n\n- $\\frac{-18}{a b c d} > 0 \\Rightarrow a b c d < 0$, como $a b c > 0$ segue que $d < 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26445,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa lista de frações, no quadro ao lado, temos:\n- 2 frações cuja soma é $\\frac{5}{2}$\n- 2 frações cuja diferença é $\\frac{5}{2}$\n- 2 frações cujo produto é $\\frac{5}{2}$\n- 2 frações cujo quociente é $\\frac{5}{2}$\nEncontre a fração que está sobrando.\n\n| $\\frac{5}{4}$ | $\\frac{17}{6}$ | $\\frac{-5}{4}$ | $\\frac{10}{7}$ | $\\frac{2}{3}$ |\n| :--- | :--- | :--- | :--- | :--- |\n$\\begin{array}{llll}\\frac{14}{8} & \\frac{-1}{3} & \\frac{5}{3} & \\frac{-3}{2}\\end{array}$",
    "options": [],
    "answer": "-3/2",
    "solution": "Solution:\n\na. 2 frações cuja diferença é $\\frac{5}{2}$:\n$\\frac{5}{4} - \\left(-\\frac{5}{4}\\right) = \\frac{5}{4} + \\frac{5}{4} = \\frac{10}{4} = \\frac{5}{2}$\n\nb. 2 frações cujo produto é $\\frac{5}{2}$:\n$\\frac{10}{7} \\times \\frac{14}{8} = \\frac{10}{7} \\times \\frac{7}{4} = \\frac{10}{4} = \\frac{5}{2}$\n\nc. 2 frações cuja soma é $\\frac{5}{2}$:\n$\\frac{17}{6} + \\left(-\\frac{1}{3}\\right) = \\frac{17}{6} - \\frac{1}{3} = \\frac{17}{6} - \\frac{2}{6} = \\frac{15}{6} = \\frac{5}{2}$\n\nd. 2 frações cujo quociente é $\\frac{5}{2}$:\n$\\frac{5}{3} \\div \\frac{2}{3} = \\frac{5}{3} \\times \\frac{3}{2} = \\frac{5}{2}$\n\nLogo, a fração que está sobrando é $-\\frac{3}{2}$.\n\n![](attached_image_1.png)\n\n![](attached_image_2.png)\n\n![](attached_image_3.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26446,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO quociente de $50^{50}$ por $25^{25}$ é igual a:\nA) $25^{25}$\nB) $10^{25}$\nC) $100^{25}$\nD) $2^{25}$\nE) $2 \\times 25^{25}$",
    "options": [],
    "answer": "C",
    "solution": "Solution:\n\nSolução 1:\n$$\n\\frac{50^{50}}{25^{25}} = \\frac{\\left(2 \\times 5^{2}\\right)^{50}}{\\left(5^{2}\\right)^{25}} = \\frac{2^{50} \\times 5^{100}}{5^{50}} = 2^{50} \\times 5^{50} = \\left(2^{2} \\times 5^{2}\\right)^{25} = 100^{25}\n$$\n\nSolução 2:\n$$\n\\frac{50^{50}}{25^{25}} = \\frac{(2 \\times 25)^{50}}{25^{25}} = \\frac{2^{50} \\times 25^{50}}{25^{25}} = 2^{50} \\times 25^{25} = (2^{2} \\times 25)^{25} = 100^{25}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26447,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nQuantos números entre $1$ e $601$ são múltiplos de $3$ ou múltiplos de $4$?\n\nA) $100$\nB) $150$\nC) $250$\nD) $300$\nE) $430$",
    "options": [],
    "answer": "D",
    "solution": "Solution:\n\nPara achar o número de múltiplos de $3$ compreendidos de $1$ a $601$, basta usar o algoritmo da divisão e escrever $601 = 200 \\times 3 + 1$. Isso mostra que $3 \\times 1, 3 \\times 2, \\ldots, 3 \\times 200$ são os múltiplos de $3$ de $1$ a $601$, ou seja, temos $200$ destes múltiplos. Do mesmo modo vemos que existem $150$ múltiplos de $4$ de $1$ a $601$.\n\nNesse total $\\underbrace{200}_{\\begin{array}{c}\\text{múltiplos} \\\\ \\text{de } 3\\end{array}} + \\underbrace{150}_{\\begin{array}{c}\\text{múltiplos} \\\\ \\text{de } 4\\end{array}} = 350$, alguns números aparecem contados duas vezes, pois são múltiplos de $3$ e de $4$ ao mesmo tempo. Por exemplo: $12, 36$ e $60$ foram incluídos nos $200$ múltiplos de $3$ e também nos $150$ múltiplos de $4$. Lembre que os múltiplos de $3$ e de $4$ são também múltiplos de $12$. O mesmo argumento usado acima mostra que temos $50$ múltiplos de $12$ de $1$ a $601$. Logo o número de múltiplos de $3$ ou $4$ de $1$ a $601$ é $350 - 50 = 300$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26448,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCapitu tem cem cartões numerados de $1$ a $100$. Todos os cartões têm uma face amarela e a outra vermelha, e o número de cada cartão está escrito em ambas as faces. Os cartões foram colocados sobre uma mesa, todos com a face vermelha voltada para cima. Capitu virou todos os cartões de número par e depois todos os cartões de número múltiplo de $3$, colocando-os com a face amarela voltada para cima. Quantos cartões ficaram com a face vermelha para cima?",
    "options": [],
    "answer": "33",
    "solution": "Solution:\n\nCapitu virou, em primeiro lugar, os $50$ cartões pares; após isto, ficaram então na mesa os $50$ cartões pares com a face amarela para cima e os $50$ cartões ímpares com a face vermelha para cima. Ao virar agora os múltiplos de $3$, ela virou apenas os múltiplos de $3$ ímpares, que são $3, 9, 15, 21, 27, 33, 39, 45, 51, 57, 63, 69, 75, 81, 87, 93$ e $99$. Assim, temos $17$ múltiplos de $3$ que são ímpares. Logo, Capitu virou para cima a face amarela de $50+17=67$ cartões. Sobraram com a face vermelha para cima $100-67=33$ cartões.\nSolution:\n\nCapitu virou, em primeiro lugar, os $50$ cartões pares; após isto, ficaram então na mesa os $50$ cartões pares com a face amarela para cima e os $50$ cartões ímpares com a face vermelha para cima. Ao virar então os cartões múltiplos de $3$, Capitu fez o seguinte:\n\n- Entre os cartões pares: Ela virou os que eram também múltiplos de $3$. Um número que é múltiplo de $2$ e de $3$ também é múltiplo de $6$. Como $100=16 \\times 6+4$, concluímos que Capitu virou $16$ cartões entre os cartões pares. Estes cartões voltaram a ficar com a face vermelha para cima, ficando os outros $34$ com a face amarela para cima.\n\n- Entre os cartões ímpares: Como $100=33 \\times 3+1$, segue que o número total de cartões (pares e ímpares) múltiplos de $3$ é $33$. Como vimos acima, entre estes cartões $16$ são pares, logo $17$ são ímpares. Assim, Capitu virou $17$ cartões ímpares, e estes cartões passaram a ter a face amarela para cima, enquanto que os outros $33$ continuaram com a face vermelha para cima.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26449,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCorrendo com velocidade de $10~\\mathrm{km}/\\mathrm{h}$, João completa uma certa distância em 6 minutos. A qual velocidade ele pode completar a mesma distância em 8 minutos?\n(A) $7{,}5~\\mathrm{km}/\\mathrm{h}$\n(B) $7{,}75~\\mathrm{km}/\\mathrm{h}$\n(C) $8~\\mathrm{km}/\\mathrm{h}$\n(D) $8{,}25~\\mathrm{km}/\\mathrm{h}$\n(E) $8{,}5~\\mathrm{km}/\\mathrm{h}$",
    "options": [],
    "answer": "A",
    "solution": "Solution:\n\n6 minutos é $1/10$ da hora, logo a distância corrida em 6 minutos é $10:10 = 1~\\mathrm{km}$. Como, espaço $=$ velocidade $\\times$ tempo, temos $1~\\mathrm{km} = v \\times 8~\\mathrm{min} \\Rightarrow v = 1~\\mathrm{km}/8~\\mathrm{min}$ (onde $v$ é a velocidade). Logo, João corre $1~\\mathrm{km}$ em 8 minutos, precisamos determinar essa velocidade em horas.\n\n| $8~\\mathrm{min}$ | corresponde a | $1~\\mathrm{km}$ |\n| :--- | :--- | :--- |\n| $4~\\mathrm{min}$ | corresponde a | $0{,}5~\\mathrm{km}$ |\n| $60~\\mathrm{min}$ | corresponde a | $0{,}5 \\times 15~\\mathrm{km} = 7{,}5~\\mathrm{km}$ |\n\nLogo, a velocidade é $7{,}5~\\mathrm{km}/\\mathrm{h}$.\n\n\nSolução 2 : Podemos usar diretamente a seguinte Regra de Três:\n\n| Velocidade <br> em $\\mathrm{km}/\\mathrm{h}$ | Tempo <br> em horas |  |\n| :---: | :---: | :---: |\n| 10 | $\\longrightarrow$ | $\\frac{6}{60}$ |\n| $x$ | $\\longrightarrow \\frac{8}{60}$ |  |\n\nVelocidade e tempo são grandezas inversamente proporcionais (aumentando a velocidade diminui o tempo), logo:\n$$\n\\frac{x}{10} = \\frac{\\frac{6}{60}}{\\frac{8}{60}} = \\frac{6}{8} \\Rightarrow x = \\frac{60}{8} = \\frac{15}{2} = 7{,}5~\\mathrm{km}/\\mathrm{h}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26450,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA menor raiz da equação $\\frac{|x-1|}{x^{2}}=6$ é:\n(A) $-\\frac{1}{3}$\n(B) $-\\frac{1}{2}$\n(C) $\\frac{1}{3}$\n(D) $\\frac{1}{4}$\n(E) $\\frac{3}{2}$",
    "options": [],
    "answer": "B",
    "solution": "Solution:\nCaso 1: $x \\geq 1$\n\nNesse caso, $x-1 \\geq 0$, donde $|x-1|=x-1$. A equação toma a forma $\\frac{x-1}{x^{2}}=6$ ou $6x^{2}-x+1=0$. Essa equação não tem raízes reais porque $\\Delta = (-1)^{2} - 4 \\times 6 \\times 1 = 1 - 24$ é negativo.\n\n\nCaso 2: $x < 1$\n\nNesse caso, $x-1 < 0$, donde $|x-1|=-(x-1)=1-x$. A equação toma a forma $\\frac{1-x}{x^{2}}=6$ ou $6x^{2}+x-1=0$. Resolvendo essa equação temos:\n\n$$\nx = \\frac{-1 \\pm \\sqrt{1-4 \\times 6 \\times (-1)}}{2 \\times 6} = \\frac{-1 \\pm \\sqrt{25}}{12} = \\frac{-1 \\pm 5}{12}\n$$\n\nPortanto, $x = -\\frac{1}{2}$ e $x = \\frac{1}{3}$. Como essas duas raízes são menores que $1$, elas são as raízes da equação do enunciado. A menor delas é $x = -\\frac{1}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26451,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNuma sequência, cada termo, a partir do terceiro, é a soma dos dois termos imediatamente anteriores, o segundo termo é 1 e o quinto termo é 2005. Qual é o sexto termo?\n\nA) 3002\nB) 3008\nC) 3010\nD) 4002\nE) 5004",
    "options": [],
    "answer": "B",
    "solution": "Solution:\n\nSeja $x$ o primeiro termo. Como o segundo termo é $1$ e, a partir do terceiro, cada termo é a soma dos dois anteriores, temos:\n\n- terceiro termo: $1 + x$;\n- quarto termo: $1 + (1 + x) = 2 + x$;\n- quinto termo: $(1 + x) + (2 + x) = 3 + 2x$;\n- sexto termo: $(2 + x) + (3 + 2x) = 5 + 3x$.\n\nComo o quinto termo é $2005$, temos $3 + 2x = 2005$, donde $x = 1001$; logo o sexto termo é $5 + 3 \\times 1001 = 3008$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26452,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO gráfico da parábola $y = x^{2} - 5x + 9$ é rodado de $180^{\\circ}$ em torno da origem. Qual é a equação da nova parábola?\nA) $y = x^{2} + 5x + 9$\nB) $y = x^{2} - 5x - 9$\nC) $y = -x^{2} + 5x - 9$\nD) $y = -x^{2} - 5x + 9$\nE) $y = -x^{2} - 5x - 9$",
    "options": [],
    "answer": "E",
    "solution": "Solution:\n\nUma rotação de $180^{\\circ}$ também é conhecida como meia-volta. Neste problema, temos uma meia-volta em torno da origem. O desenho ao lado ilustra o que esta meia-volta faz com as coordenadas dos pontos do plano. Por exemplo, o ponto $A^{\\prime}$ é o resultado da meia-volta aplicada ao ponto $A$; em outras palavras, $A^{\\prime}$ é onde o ponto $A$ vai parar após a meia-volta. Do mesmo modo, $B^{\\prime}$ é onde $B$ vai parar após a meia volta. É fácil ver que na passagem de $A$ para $A^{\\prime}$ as coordenadas trocam de sinal. Deste modo, vemos que uma meia-volta em torno da origem leva um ponto qualquer $(x, y)$ no ponto $(-x, -y)$.\n\n![](attached_image_1.png)\n\nTemos: $(a, b)$ pertence à nova parábola $\\Leftrightarrow (-a, -b)$ pertence à parábola $y = x^{2} - 5x + 9 \\Leftrightarrow -b = a^{2} + 5a + 9 \\Leftrightarrow b = -a^{2} - 5a - 9$. Logo a equação da nova parábola é $y = -x^{2} - 5x - 9$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26453,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAna, Bento e Lucas participam de um concurso que consta de 20 perguntas com a seguinte regra:\n- cada resposta certa ganha 5 pontos,\n- cada resposta errada perde 3 pontos,\n- cada resposta em branco perde 2 pontos.\n\nVeja os resultados na tabela a seguir:\n\n|  | Número de respostas certas | Número de respostas erradas | Número de respostas em branco |\n| :--- | :---: | :---: | :---: |\n| Ana | 12 | 4 | 4 |\n| Bento | 13 | 7 | 0 |\n| Lucas | 12 | 3 | 5 |\n\nEscrevendo os nomes dos três em ordem decrescente de classificação no concurso, encontramos:\nA) Ana, Bento, Lucas\nB) Lucas, Bento, Ana\nC) Ana, Lucas, Bento\nD) Lucas, Ana, Bento\nE) Bento, Lucas, Ana",
    "options": [],
    "answer": "E",
    "solution": "Solution:\n\nO número de pontos de cada um deles é:\n\nAna: $5 \\times 12 + (-3) \\times 4 + (-2) \\times 4 = 60 - 12 - 8 = 40$\n\nBento: $5 \\times 13 + (-3) \\times 7 + (-2) \\times 0 = 65 - 21 = 44$\n\nLucas: $5 \\times 12 + (-3) \\times 3 + (-2) \\times 5 = 60 - 9 - 10 = 41$\n\nLogo, Bento foi o mais bem classificado, seguido de Lucas e depois de Ana.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26454,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nResolva geometricamente as equações:\n\na. $|x-5|=2$\n\nb. $|x+3|=1$\n\nc. $|3x-7|=9$\n\nd. $|x+2|=|x-5|$",
    "options": [],
    "answer": "{'a': [3, 7], 'b': [-4, -2], 'c': [-2.0, 16.0], 'c_simplified': [-0.6666666667, 5.3333333333], 'c_exact': ['-2/3', '16/3'], 'd': '3/2'}",
    "solution": "Solution:\n\n## Interpretação geométrica\n$|a-b|=$ distância entre $a$ e $b$\n\na. $|x-5|=2 \\Leftrightarrow$ números cuja distância ao $5$ é $2$.\nLogo as raízes são $3$ e $7$.\n\n![](attached_image_1.png)\n\nb. $|x+3|=1 \\Leftrightarrow$ números cuja distância ao $-3$ é $1$.\nLogo as raízes são $-4$ e $-2$.\n\n![](attached_image_2.png)\n\nc. Fazendo a mudança de variável $y=3x$, a equação toma a forma $|y-7|=9 \\Leftrightarrow$ números cuja distância ao $7$ é $9$.\nLogo as raízes são $y=-2$ e $y=16$.\nDestrocando a variável temos $3x=-2$ e $3x=16$, e obtemos raízes da equação: $x=-\\frac{2}{3}$ e $x=\\frac{16}{3}$.\n\n![](attached_image_3.png)\n\nd. As raízes da equação $|x+2|=|x-5|$ são os números equidistantes de $-2$ e de $5$. Esses números só podem estar entre $-2$ e $5$.\nLogo, a solução é $x=1,5$\n\n![](attached_image_4.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26455,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO limite de peso que um caminhão pode transportar corresponde a $50$ sacos de areia ou $400$ tijolos. Se este caminhão já contém $32$ sacos de areia, quantos tijolos, no máximo, ele ainda pode carregar?\nA) $132$\nB) $144$\nC) $146$\nD) $148$\nE) $152$",
    "options": [],
    "answer": "B",
    "solution": "Solution:\n\nO enunciado mostra que o peso de $1$ saco de areia é o mesmo que o de $8$ tijolos. Se no caminhão já há $32$ sacos de areia, ele pode carregar ainda $18$ sacos, o que equivale $18 \\times 8 = 144$ tijolos.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26456,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm ônibus, um trem e um avião partem no mesmo horário da cidade $A$ para a cidade $B$. Se eu tomar o ônibus cuja velocidade média é $100~\\mathrm{km}/\\mathrm{h}$, chegarei à cidade $B$ às 20 horas. Se eu tomar o trem, cuja velocidade média é $300~\\mathrm{km}/\\mathrm{h}$, chegarei à cidade $B$ às 14 horas. Qual será o horário de chegada do avião se sua velocidade média é de $900~\\mathrm{km}/\\mathrm{h}$?",
    "options": [],
    "answer": "12:00",
    "solution": "Solution:\n\nSeja $d$ a distância entre as duas cidades e $h$ o horário de partida comum do ônibus, do trem e do avião. Como, distância $=$ velocidade $\\times$ tempo, temos:\n$$\nd = 100 \\times (20 - h)\n$$\ne\n$$\nd = 300 \\times (14 - h)\n$$\nLogo,\n$$\n100 \\times (20 - h) = 300 \\times (14 - h)\n$$\nDonde\n$$\n100 \\times 20 - 100h = 300 \\times 14 - 300h\n$$\n$$\n2000 - 100h = 4200 - 300h\n$$\n$$\n200h = 2200\n$$\n$$\nh = 11\n$$\nPortanto, a distância entre as duas cidades é\n$$\nd = 100 \\times (20 - 11) = 900~\\mathrm{km}\n$$\nLogo, o avião gasta\n$$\n\\frac{900}{900} = 1~\\text{hora}\n$$\nda cidade $A$ à cidade $B$, e portanto ele chega às 12 horas.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26457,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEncontre os algarismos que estão faltando sobre cada um dos traços:\n\na) $\\frac{126}{8\\_} = \\frac{21}{\\_}$;\n\nb) $\\frac{\\_\\_8}{33\\_} = \\frac{4}{5}$",
    "options": [],
    "answer": "a) 84 and 14; b) 268 and 335",
    "solution": "Solution:\n\na) Observe que $126 \\div 6 = 21$, logo, o numerador $126$ foi dividido por $6$ para obter o numerador $21$ da outra fração. Logo, o denominador $8\\_$ também é divisível por $6$. O único número da forma $8\\_$ que é divisível por $6$ é $84$, e $84 \\div 6 = 14$. Podemos então completar as frações:\n\n$$\n\\frac{126}{8\\underline{4}} \\xrightarrow{\\stackrel{\\div 6}{\\div 6}} \\frac{21}{1\\underline{4}}\n$$\n\nb) Note que $33\\_$ deve ser múltiplo de $5$, logo só pode ser $330$ ou $335$. Temos como $\\frac{4}{5} = 0,8$, segue que $\\frac{\\_\\_8}{330} = 0,8$ ou $\\frac{\\_\\_8}{335} = 0,8$. Temos $330 \\times 0,8 = 264$ e $335 \\times 0,8 = 268$, segue que $\\_\\_8 = 268$ e $33\\_ = 335$. Podemos completar as frações: $\\frac{268}{335} = \\frac{4}{5}$. Note que $\\frac{268}{335} = \\frac{268 \\div 67}{335 \\div 67} = \\frac{4}{5}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26458,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm ponto $P$ está no centro de um quadrado com $10~\\mathrm{cm}$ de lado. Quantos pontos da borda do quadrado estão a uma distância de $6~\\mathrm{cm}$ de $P$?\n\nA) 1\nB) 2\nC) 4\nD) 6\nE) 8",
    "options": [],
    "answer": "E",
    "solution": "Solution:\n\nOs pontos que estão a $6~\\mathrm{cm}$ de distância do ponto $P$ formam uma circunferência de centro $P$ e raio $R=6~\\mathrm{cm}$. Se $D$ denota a diagonal do quadrado, do teorema de Pitágoras temos\n$$\nD=\\sqrt{10^{2}+10^{2}}=\\sqrt{2 \\times 10^{2}}=10 \\sqrt{2}\n$$\nA circunferência de raio $L/2=5$ tangencia o quadrado em 4 pontos.\n\n![](attached_image_1.png)\n\nA circunferência de raio $D/2$ toca o quadrado em 4 pontos (os vértices do quadrado).\nTemos: $L=10$, $R=6$ e $D=10 \\sqrt{2}$, logo $\\underbrace{5}_{L / 2}<\\underbrace{6}_{R}<\\underbrace{5 \\sqrt{2}}_{D / 2}$. (Observe que $1,2<\\sqrt{2}$, logo $5 \\times 1,2<5 \\times \\sqrt{2}$ e portanto, $6<5 \\sqrt{2}$)\n\nAssim, a circunferência de raio $R=6$ está “entre\" as duas circunferências de raios $5$ e $5 \\sqrt{2}$.\nLogo, ela corta o quadrado em 8 pontos.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26459,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nVocê possui apenas palitos com $6~\\mathrm{cm}$ e $7~\\mathrm{cm}$ de comprimento. O número mínimo de palitos que você precisa para cobrir com esses palitos um segmento de reta com 2 metros é:\nA) 29\nB) 30\nC) 31\nD) 32\nE) 33",
    "options": [],
    "answer": "A",
    "solution": "Solution:\n\nA quantidade utilizada de palitos é mínima quando o número de palitos de $7~\\mathrm{cm}$ utilizado é o maior possível. Dividindo $200$ por $7$ obtemos $200 = 28 \\times 7 + 4$. Como $200 = 26 \\times 7 + 18 = 26 \\times 7 + 3 \\times 6$, usando $26$ palitos de $7~\\mathrm{cm}$ e $3$ palitos de $6~\\mathrm{cm}$ obtemos o que queríamos. Logo, o número mínimo de palitos é $26 + 3 = 29$.\n\nComentário: Observe que a solução equivale a encontrar números inteiros $x$ e $y$ tais que $200 = \\underset{\\text{múltiplo de } 7}{7y} + \\underset{\\text{múltiplo de } 6}{6x}$ e $y$ seja o maior possível, onde $y =$ número de palitos de $7~\\mathrm{cm}$ e $x =$ número de palitos de $6~\\mathrm{cm}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26460,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSe $x$, $y$ e $z$ são números inteiros positivos tais que $x y z = 240$, $x y + z = 46$ e $x + y z = 64$, qual é o valor de $x + y + z$?\n\nA) 19\nB) 20\nC) 21\nD) 24\nE) 36",
    "options": [],
    "answer": "B",
    "solution": "Solution:\n\nSolução 1:\nDe $x y z = 240$ segue que $x y = \\frac{240}{z}$; substituindo em $x y + z = 46$ obtemos $\\frac{240}{z} + z = 46$, ou seja, $z^{2} - 46z + 240 = 0$. As raízes desta equação são números cuja soma é $46$ e cujo produto é $240$, ou seja, as raízes são $6$ e $40$. Logo, $z = 6$ ou $z = 40$ (I).\n\nDo mesmo modo, a substituição de $y z = \\frac{240}{x}$ em $x + y z = 64$ nos leva a $x = 4$ ou $x = 60$ (II).\n\nDe $x y z = 240$, segue que $y = \\frac{240}{x z}$. Como $y$ é um número inteiro, então $x z$ é um divisor de $240$. Segue de (I) e (II) que as possibilidades para $x z$ são:\n\n![](attached_image_1.png)\n\nVemos que só podemos ter $x = 4$ e $z = 6$, pois em qualquer outro caso o produto $x z$ não é um divisor de $240$. Segue que $y = \\frac{240}{x z} = \\frac{240}{4 \\times 6} = 10$, donde $x + y + z = 4 + 10 + 6 = 20$.\n\n![](attached_image_1.png)\n\n\nSolução 2:\nSomando $x y + z = 46$ e $x + y z = 64$, obtemos:\n$$\nx y + z + x + y z = (x + z) + y(x + z) = (x + z)(y + 1) = 110\n$$\ne vemos que $y + 1$ é um divisor de $110$. Logo, temos as possibilidades $y + 1 = 1, 2, 5, 10, 11, 22, 55$ e $110$, ou seja, $y = 0, 1, 4, 9, 10, 21, 54$ e $109$. Por outro lado, $y$ é um divisor de $240$ porque $x y z = 240$, além disso $y$ é positivo, que nos deixa com as possibilidades $y = 1, 4$ e $10$.\n\nSe $y = 1$ então\n\\[\n\\begin{cases}\n(x + z)(y + 1) = 110 \\Rightarrow x + z = 55 \\\\\nx y + z = 46 \\Rightarrow x + z = 46\n\\end{cases}\n\\]\no que não é possível. Logo $y \\neq 1$.\n\nSe $y = 4$ então\n\\[\n\\begin{cases}\n(x + z)(y + 1) = 110 \\Rightarrow x + z = 22 \\\\\nx y z = 240 \\Rightarrow x z = 60\n\\end{cases}\n\\]\ne podemos verificar (por exemplo, com uma lista de divisores de $60$ ou então resolvendo a equação $w^{2} - 22w + 60 = 0$) que não há valores inteiros positivos de $x$ e $z$ que verifiquem estas duas condições. Logo $y \\neq 4$.\n\nSe $y = 10$ então\n\\[\n\\begin{cases}\n(x + z)(y + 1) = 110 \\Rightarrow x + z = 10 \\\\\nx y z = 240 \\Rightarrow x z = 24\n\\end{cases}\n\\]\ndonde concluímos que $x = 4$ e $z = 6$. Finalmente, temos $x + y + z = 4 + 10 + 6 = 20$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26461,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO perímetro de um retângulo é $100~\\mathrm{cm}$ e a diagonal mede $x~\\mathrm{cm}$. Qual é a área do retângulo em função de $x$?\nA) $625-x^{2}$\nB) $625-\\frac{x^{2}}{2}$\nC) $1250-\\frac{x^{2}}{2}$\nD) $250-\\frac{x^{2}}{2}$\nE) $2500-\\frac{x^{2}}{2}$",
    "options": [],
    "answer": "C",
    "solution": "Solution:\n\nSolução 1:\nComo o perímetro do retângulo é $100$, seu semiperímetro é $50$. Como o semiperímetro de um retângulo é a soma do comprimento com a largura, concluímos que esses são da forma $a$ e $50-a$. A área de um retângulo é o produto do comprimento pela largura. No nosso caso, esta área é $(50-a) \\cdot a = 50a - a^{2}$.\n\nPelo teorema de Pitágoras, temos $x^{2} = (50-a)^{2} + a^{2}$, ou seja,\n$$\nx^{2} = 2500 - 100a + 2a^{2} = 2500 - 2(50a - a^{2})\n$$\nLogo,\n$$\n50a - a^{2} = \\frac{1}{2}(2500 - x^{2})\n$$\ne obtemos a expressão da área do retângulo em função de $x$.\n\n![](attached_image_1.png)\n\nSolução 2:\nÁrea do retângulo de medidas $a$ e $b$ é $A = ab$. Como $a + b = 50$, temos\n$$(a + b)^{2} = a^{2} + b^{2} + 2ab = 50^{2}$$\nPelo Teorema de Pitágoras, $x^{2} = a^{2} + b^{2}$, assim,\n$$x^{2} + 2A = 2500$$\nPortanto,\n$$A = \\frac{1}{2}(2500 - x^{2})$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26462,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSe $x > 5$, então qual dos números abaixo é o menor?\n\n(A) $\\dfrac{5}{x}$\n\n(B) $\\dfrac{5}{x+1}$\n\n(C) $\\dfrac{5}{x-1}$\n\n(D) $\\dfrac{x}{5}$\n\n(E) $\\dfrac{x+1}{5}$",
    "options": [],
    "answer": "B",
    "solution": "Solution:\n\nSolução 1:\nComo a questão tem uma única resposta, ela é válida para qualquer valor de $x$. Podemos então escolher um valor para $x$, por exemplo $x = 10$. Temos:\n\n$\\dfrac{5}{x} = \\dfrac{5}{10}$\n\n$\\dfrac{5}{x+1} = \\dfrac{5}{11}$\n\n$\\dfrac{5}{x-1} = \\dfrac{5}{9}$\n\n$\\dfrac{x}{5} = \\dfrac{10}{5}$\n\n$\\dfrac{x+1}{5} = \\dfrac{11}{5}$\n\nVemos que $x / 5$ e $(x+1) / 5$ são maiores que $1$, logo estão excluídos porque as outras três opções são menores que $1$. Como $5/10$, $5/11$ e $5/9$ têm o mesmo numerador, o menor é o que tiver maior denominador, que é $5/11$, ou seja, $\\dfrac{5}{x+1}$.\n\n\nSolução 2:\nSe $x > 5$, então $\\dfrac{5}{x}$, $\\dfrac{5}{x+1}$ e $\\dfrac{5}{x-1}$ são menores do que $1$ e $\\dfrac{x}{5}$ e $\\dfrac{x+1}{5}$ são maiores do que $1$. Logo, as opções D e E estão excluídas. Como $\\dfrac{5}{x}$, $\\dfrac{5}{x+1}$ e $\\dfrac{5}{x-1}$ têm o mesmo numerador, o menor é o que tem maior denominador, que é $\\dfrac{5}{x+1}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26463,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nGeni é cliente de uma companhia telefônica que oferece o seguinte plano:\n- tarifa mensal fixa de $R\\$ 18,00$\n- gratuidade em 10 horas de ligações por mês\n- $R\\$ 0,03$ por cada minuto que exceder às 10 horas.\nEm janeiro, Geni usou seu telefone por 15 horas e 17 minutos, e em fevereiro por 9 horas e 55 minutos. Qual a despesa de Geni com telefone nesses dois meses?\nA) $R\\$ 45,51$\nB) $R\\$ 131,10$\nC) $R\\$ 455,10$\nD) $R\\$ 13,11$\nE) $R\\$ 4,55$",
    "options": [],
    "answer": "A",
    "solution": "Solution:\nVejamos a despesa em janeiro. Como 10 horas são gratuitas e Geni usou seu telefone por 15 horas e 17 minutos, ela deve pagar o custo de apenas 5 horas e 17 minutos mais a tarifa fixa mensal de 18 reais. Como o preço é dado em minutos, vamos reduzir a minutos o tempo a pagar. Sabemos que 1 hora $=60$ minutos, portanto 5 horas $=5 \\times 60=300$ minutos. Logo, $5\\ \\mathrm{h}\\ 17\\ \\mathrm{m}=300+17=317\\ \\mathrm{m}$. Portanto, a conta telefônica de Geni em janeiro foi:\n$$\n18+317 \\times 0,03=18+9,51=27,51 \\text{ reais. }\n$$\nEm fevereiro, Geni usou seu telefone menos do que 10 horas, portanto neste mês ela só precisa pagar a tarifa fixa mensal de 18 reais. Logo, a despesa de Geni com telefone nesses dois meses foi:\n$$\n27,51+18=45,51 \\text{ reais. }\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26464,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEm um quente dia de verão, 64 crianças comeram, cada uma, um sorvete pela manhã e outro à tarde. Os sorvetes eram de 4 sabores: abacaxi, banana, chocolate e doce de leite. A tabela abaixo mostra quantas crianças consumiram um destes sabores pela manhã e outro à tarde; por exemplo, o número 7 na tabela indica que 7 crianças tomaram sorvete de banana pela manhã e de chocolate à tarde.\n\n|        | TARDE    |           |           |            |                   |\n| :---:  | :---:    | :---:     | :---:     | :---:      | :---:             |\n|        |          | Abacaxi   | Banana    | Chocolate  | Doce de leite     |\n| $\\mathbf{MANH\\tilde{A}}$ | Abacaxi   | 1         | 8         | 0          | 3                 |\n|        | Banana    | 6         | 2         | 7          | 5                 |\n|        | Chocolate | 3         | 3         | 0          | 5                 |\n|        | Doce de leite | 2     | 9         | 9          | 1                 |\n\nQuantas crianças tomaram sorvetes de sabores diferentes neste dia?\nA) 58\nB) 59\nC) 60\nD) 61\nE) 62",
    "options": [],
    "answer": "C",
    "solution": "Solution:\n\nVamos primeiro analisar a informação contida na diagonal da tabela indicada pelos números dentro dos quadradinhos.\n\n|        | TARDE    |           |           |            |                   |\n| :---:  | :---:    | :---:     | :---:     | :---:      | :---:             |\n|        |          | Abacaxi   | Banana    | Chocolate  | Doce de leite     |\n| $M$    | Abacaxi   | 1         | 8         | 0          | 3                 |\n|        | Banana    | 6         | 2         | 7          | 5                 |\n|        | Chocolate | 3         | 3         | 0          | 5                 |\n|        | Doce de leite | 2     | 9         | 9          | 1                 |\n\nEsses números indicam quantos foram as crianças que tomaram sorvetes com o mesmo sabor pela manhã e pela tarde: 1 tomou sorvetes de abacaxi, 2 de banana, 0 de chocolate e 1 de doce de leite. Todos os outros estudantes comeram sorvetes de sabores diferentes pela manhã e à tarde; estes são em número de $64-(1+2+0+1)=60$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26465,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nQuais dos números abaixo são negativos?\n$10 - 3 \\sqrt{11}$;\n$3 \\sqrt{11} - 10$;\n$18 - 5 \\sqrt{13}$;\n$51 - 10 \\sqrt{26}$;\n$10 \\sqrt{26} - 51$.",
    "options": [],
    "answer": "3\\sqrt{11} - 10; 10\\sqrt{26} - 51; 18 - 5\\sqrt{13}",
    "solution": "Solution:\n\nComo $100 > 99$ então $\\underbrace{\\sqrt{100}}_{10} > \\underbrace{\\sqrt{99}}_{3 \\sqrt{11}}$. Logo, $10 - 3 \\sqrt{11} > 0$ e $3 \\sqrt{11} - 10 < 0$.\n\nAnalogamente: $2601 > 2600 \\Rightarrow \\underbrace{\\sqrt{2601}}_{51} > \\underbrace{\\sqrt{2600}}_{10 \\sqrt{26}}$.\nAssim, $51 - 10 \\sqrt{26} > 0$ e $10 \\sqrt{26} - 51 < 0$.\n\nFinalmente, $324 < 325 \\Rightarrow \\underbrace{\\sqrt{324}}_{18} < \\underbrace{\\sqrt{325}}_{5 \\sqrt{13}} \\Rightarrow 18 - 5 \\sqrt{13} < 0$.\n\nOs números negativos são $3 \\sqrt{11} - 10$, $10 \\sqrt{26} - 51$ e $18 - 5 \\sqrt{13}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26466,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSe $2\\left(2^{2x}\\right) = 4^{x} + 64$, então $x$ é igual a:\nA) -2\nB) -1\nC) 1\nD) 2\nE) 3",
    "options": [],
    "answer": "E",
    "solution": "Solution:\n\nSolução 1: Notamos que os termos do lado direito da equação dada podem ser escritos como potências de 2; de fato, $4^{x} = \\left(2^{2}\\right)^{x} = 2^{2x}$ e $64 = 2^{6}$. Desse modo, a equação se torna $2\\left(2^{2x}\\right) = 2^{2x} + 2^{6}$. Temos então $2\\left(2^{2x}\\right) - 2^{2x} = 2^{6}$, donde $2^{2x}(2-1) = 2^{6}$, ou seja $2^{2x} = 2^{6}$. Logo $2x = 6$ e segue que $x = 3$.\n\n\nSolução 2: $2\\left(4^{x}\\right) = 4^{x} + 4^{3} \\Rightarrow 4^{x} = 4^{3} \\Rightarrow x = 3$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26467,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA figura mostra a marca de uma empresa, formada por dois círculos concêntricos e outros quatro círculos de mesmo raio, cada um deles tangente a dois dos outros e aos dois círculos concêntricos. O raio do círculo menor mede $1~\\mathrm{cm}$. Qual é, em centímetros, o raio do círculo maior?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "3 + 2√2",
    "solution": "Solution:\n\nSeja $r$ o raio das quatro circunferências iguais. Ligando os centros $A$ e $B$ de duas destas circunferências ao centro $O$ das circunferências concêntricas, obtemos o triângulo $OAB$ como na figura ao lado. Lembrando que a reta que une os centros de duas circunferências tangentes passa pelo ponto de tangência, vemos que $OA = OB = 1 + r$ e $AB = 2r$. Lembrando também que o triângulo $OAB$ é retângulo em $O$, o teorema de Pitágoras nos diz então que $(2r)^2 = (1 + r)^2 + (1 + r)^2$, ou seja, $4r^2 = 2r^2 + 4r + 2$. Logo $r^2 - 2r - 1 = 0$; daqui tiramos $r = \\frac{2 \\pm 2\\sqrt{2}}{2} = 1 \\pm \\sqrt{2}$. Como $1 - \\sqrt{2}$ é negativo,\n\n![](attached_image_2.png)\n\ndescartamos esta raiz e obtemos $r = 1 + \\sqrt{2}$. Segue que o raio da circunferência maior é $1 + 2r = 3 + 2\\sqrt{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26468,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSabendo-se que $0,333\\ldots=\\frac{1}{3}$, qual é a fração irredutível equivalente a $0,1333\\ldots$ ?\nA) $\\frac{1}{13}$\nB) $\\frac{1}{15}$\nC) $\\frac{1}{30}$\nD) $\\frac{2}{15}$\nE) $\\frac{1333}{10000}$",
    "options": [],
    "answer": "D",
    "solution": "Solution:\n\nSolução 1 - Usando o dado da questão temos: $0,1333\\ldots=\\frac{1,333\\ldots}{10}=\\frac{1+0,333\\ldots}{10}=\\frac{1}{10}\\left(1+\\frac{1}{3}\\right)=\\frac{1}{10} \\times \\frac{4}{3}=\\frac{2}{15}$.\n\n\nSolução 2 - Usando a regra que fornece a geratriz de uma dízima periódica, temos: $0,1333\\ldots=\\frac{13-1}{90}=\\frac{12}{90}=\\frac{2}{15}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26469,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nQuantos números entre $10$ e $13000$, quando lidos da esquerda para a direita, são formados por algarismos consecutivos e em ordem crescente? Por exemplo, $456$ é um desses números, mas $7890$ não é.\n\nA) $10$\nB) $13$\nC) $18$\nD) $22$\nE) $25$",
    "options": [],
    "answer": "D",
    "solution": "Solution:\n\nOs números em questão são:\n- com $2$ algarismos: $12, 23, 34, 45, \\ldots, 89$ ($8$ números),\n- com $3$ algarismos: $123, 234, 345, \\ldots, 789$ ($7$ números),\n- com $4$ algarismos: $1234, 2345, \\ldots, 6789$ ($6$ números)\ne, por fim,\n- com $5$ algarismos: $12345$, um total de $8+7+6+1=22$ números.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26470,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUma cerca de arame reta tem 12 postes igualmente espaçados. A distância entre o terceiro e o sexto poste é de $3{,}3$ m. Qual é a distância entre o primeiro e o último poste?\nA) $8,4~\\mathrm{m}$\nB) $12,1~\\mathrm{m}$\nC) $9,9~m$\nD) $13,2~m$\nE) $9,075~\\mathrm{m}$",
    "options": [],
    "answer": "B",
    "solution": "Solution:\n\nA distância entre dois postes consecutivos é $\\frac{3,3~m}{3} = 1,1~m$, donde a distância entre o primeiro e o último poste é $11 \\times 1,1~\\mathrm{m} = 12,1~\\mathrm{m}$\n\n![](attached_image_1.png)",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26471,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPara encher de água um tanque em forma de um bloco retangular de $300~\\mathrm{cm}$ de comprimento, $50~\\mathrm{cm}$ de largura e $36~\\mathrm{cm}$ de altura, um homem\n![](attached_image_1.png)\n\nutiliza um balde cilíndrico, de $30~\\mathrm{cm}$ de diâmetro em sua base e $48~\\mathrm{cm}$ de altura, para pegar água numa fonte. Cada vez que ele vai à fonte, ele enche $\\frac{4}{5}$ do balde e no caminho derrama $10\\%$ do seu conteúdo. Estando o tanque inicialmente vazio, quantas viagens à fonte o homem terá que fazer para que a água no tanque chegue a $\\frac{3}{4}$ de sua altura?",
    "options": [],
    "answer": "17",
    "solution": "Solution:\n\nNesta solução todas as medidas de volume são dadas em $\\mathrm{cm}^3$.\n\nO volume $V$ do balde é dado pela fórmula habitual do volume de um cilindro, ou seja, $V = $ área da base $\\times$ altura. A base do balde é um círculo de diâmetro $30~\\mathrm{cm}$; seu raio é então $r = 15~\\mathrm{cm}$ e sua área é $\\pi r^2 = 225\\pi~\\mathrm{cm}^2$. Logo $V = 48 \\times 225\\pi = 10.800\\pi$.\n\nA cada viagem, o volume de água que o homem coloca no balde é $\\frac{4}{5}$ de $V$, e deste volume ele perde $10\\%$.\n\nLogo, resta no balde $90\\%$ de $\\frac{4}{5}$ de $V$, isto é, $\\frac{9}{10} \\times \\frac{4}{5} V = \\frac{18}{25} V = 0,72 V = 0,72 \\times 10800 \\pi = 7776 \\pi$; esta é a quantidade de água que ele coloca no tanque em cada viagem, que denotaremos por $B$.\n\nO volume de $\\frac{3}{4}$ do tanque é $T = \\frac{3}{4} \\times 300 \\times 36 \\times 50 = 405.000$.\n\nLogo, o número de baldes necessários para atingir esse volume é $\\frac{405000}{B} = \\frac{405000}{7776 \\pi} = \\frac{625}{12 \\pi}$. Usando a aproximação $3,14$ para o número $\\pi$ temos $\\frac{625}{12 \\pi} \\approx \\frac{625}{12 \\times 3,14} \\approx 16,587$.\n\nLogo o homem necessitará 16 baldes mais $0,587$ de um balde. Concluímos que o homem deverá fazer 17 viagens.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26472,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA pista de um autódromo tem $20~\\mathrm{km}$ de comprimento e forma circular. Os pontos marcados na pista são: $A$, que é o ponto de partida, $B$ que dista $5~\\mathrm{km}$ de $A$ no sentido do percurso, $C$ que dista $3~\\mathrm{km}$ de $B$ no sentido do percurso, $D$ que dista $4~\\mathrm{km}$ de $C$ no sentido do percurso e $E$ que dista $5~\\mathrm{km}$ de $D$ no sentido do percurso. Um carro que parte de $A$ e pára após percorrer $367~\\mathrm{km}$ estará mais próxima de qual dos 5 pontos?\n(A) $A$\n(B) $B$\n(C) $C$\n(D) $D$\n(E) $E$",
    "options": [],
    "answer": "C",
    "solution": "Solution:\n\n(C) Vamos marcar os 4 pontos a partir de $A$.\nComo o comprimento é de $20~\\mathrm{km}$, o comprimento de cada um dos 4 quadrantes é $5~\\mathrm{km}$. Podemos então marcar os pontos. Como $367 = 18 \\times 20 + 7$, o carro deu 18 voltas completas e percorreu mais $7~\\mathrm{km}$ a partir de $A$. Logo, ele passa $2~\\mathrm{km}$ após $B$, o que significa que ele pára $1~\\mathrm{km}$ de $C$. Portanto, $C$ é o ponto mais próximo.\n\n![](attached_image_1.png)",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26473,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAs vizinhas Elza, Sueli, Patrícia, Heloísa e Cláudia chegam juntas do trabalho e começam a subir as escadas do prédio de 5 andares onde moram. Cada uma mora num andar diferente. Heloísa chega a seu andar depois de Elza, mas antes de Cláudia. Quando Sueli chega ao seu andar, Heloísa ainda tem 2 andares para subir, e o mesmo ocorre a Patrícia quando Elza chega ao seu andar. Sueli não mora no $1^{\\circ}$ andar. Em qual andar mora cada uma delas?",
    "options": [],
    "answer": "Elza on the 1st floor, Sueli on the 2nd floor, Patrícia on the 3rd floor, Heloísa on the 4th floor, and Cláudia on the 5th floor.",
    "solution": "Solution:\n\nVejamos as informações dadas no enunciado:\n\"Heloísa chega a seu andar depois de Elza, mas antes de Cláudia\".\n$$\n\\Rightarrow \\text{Heloísa mora acima de Elza e abaixo de Cláudia.}\n$$\n\"Quando Sueli chega ao seu andar, Heloísa ainda tem 2 andares para subir, e o mesmo ocorre a Patrícia quando Elza chega ao seu andar\".\n$$\n\\Longrightarrow \\text{Heloísa mora dois andares acima de Sueli e Patrícia dois andares acima de Elza.}\n$$\nSueli não mora no $1^{\\circ}$ andar e Heloísa mora 2 andares acima de Sueli, logo temos as seguintes possibilidades ao lado.\n\n|               | $1^{\\circ}$ andar | $2^{\\circ}$ andar | $3^{\\circ}$ andar | $4^{\\circ}$ andar | $5^{\\circ}$ andar |\n|:-------------:|:-----------------:|:-----------------:|:-----------------:|:-----------------:|:-----------------:|\n| Possibilidade |                   | Sueli             |                   | Heloísa           |                   |\n| Possibilidade |                   |                   | Sueli             |                   | Heloísa           |\n\nComo Cláudia mora acima de Heloísa, então Heloísa não pode morar no último andar que é o $5^{\\circ}$ andar. Logo, Sueli mora no $2^{\\circ}$ andar, Heloísa no $4^{\\circ}$ e Cláudia só pode morar no $5^{\\circ}$.\n\n|               | $1^{\\circ}$ andar | $2^{\\circ}$ andar | $3^{\\circ}$ andar | $4^{\\circ}$ andar | $5^{\\circ}$ andar |\n|:-------------:|:-----------------:|:-----------------:|:-----------------:|:-----------------:|:-----------------:|\n|               |                   | Sueli             |                   | Heloísa           | Cláudia           |\n\nFinalmente, Patrícia mora dois andares acima de Elza, logo Elza mora no $1^{\\circ}$ andar e Patrícia no $3^{\\circ}$ andar.\n\n|               | $1^{\\circ}$ andar | $2^{\\circ}$ andar | $3^{\\circ}$ andar | $4^{\\circ}$ andar | $5^{\\circ}$ andar |\n|:-------------:|:-----------------:|:-----------------:|:-----------------:|:-----------------:|:-----------------:|\n|               | Elza              | Sueli             | Patrícia          | Heloísa           | Cláudia           |",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26474,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUma mesa quadrada tem 1 metro de lado. Qual o menor diâmetro de uma toalha redonda que cubra completamente o tampo da mesa?\n\n(A) 1\n(B) 1,5\n(C) 2\n(D) $\\sqrt{2}$\n(E) $\\sqrt{3}$",
    "options": [],
    "answer": "D",
    "solution": "Solution:\n\nPara que a toalha cubra inteiramente a mesa e que tenha o menor diâmetro possível, o quadrado deve estar inscrito no círculo. A diagonal do quadrado é o diâmetro do círculo, logo pelo Teorema de Pitágoras, temos:\n$$d^{2} = 1^{2} + 1^{2} \\Rightarrow d = \\sqrt{2}$$\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26475,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n| Seleção | Jogos | V | E | D | GM | GS | P |\n| :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| Dinamarca | 3 | 2 | 1 | 0 | 5 | 2 | 7 |\n| Senegal | 3 | 1 | 2 | 0 | 5 | 4 | $?$ |\n| Uruguai | 3 | 0 | 2 | 1 | 4 | $\\boldsymbol\\{?\\}$ | 2 |\n| França | 3 | 0 | 1 | 2 | 0 | 3 | 1 |\nUtilize as informações abaixo para resolver as duas próximas questões:\nA tabela ao lado mostra o desempenho das seleções do grupo A da Copa do Mundo de 2002:\n# Legenda: \nV - vitórias, E - empates, D - derrotas, GM - Gols Marcados, GS - Gols Sofridos, P - Pontos.\nNuma partida de futebol, a equipe vencedora ganha 3 pontos, em caso de empate as duas ganham 1 ponto e a perdedora não ganha nem perde pontos.\n5) Quantos pontos obteve a seleção do Senegal?\nA) 3\nB) 4\nC) 5\nD) 6\nE) 7",
    "options": [],
    "answer": "C",
    "solution": "Solution:\n\n5. (C) Segundo as condições da copa, uma vitória vale 3 pontos, um empate vale 1 ponto e quem sofre uma derrota não pontua. Como Senegal teve uma vitória e dois empates, ele somou: $1 \\times 3 + 2 \\times 1 = 5$ pontos.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26476,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAs retas $r$ e $s$ são paralelas, encontre $x$ e $y$:\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "x = 40°, y = 100°",
    "solution": "Solution:\n\nTemos $80^{\\circ} + y = 180^{\\circ} \\Rightarrow y = 100^{\\circ}$.\n\nComo as retas $r$ e $s$ são paralelas, segue que, $60^{\\circ} + x + 80^{\\circ} = 180^{\\circ}$, donde $x = 40^{\\circ}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26477,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nVeja as promoções de dois supermercados:\n\n| Supermercado A | Supermercado B |\n| :---: | :---: |\n| 6 latas de 3 litros do  \n sorvete QUENTE | Sorvete QUENTE - lata de 3  \n litros |\n| $R\\$ 24,00$ | 4 latas - só $R\\$ 14,00$ |\n\nJoana quer comprar 12 latas de sorvete para a festa de seu aniversário. Em qual supermercado ela deve comprar?\n\nA) No A, pois economizará $R \\$ 7,00$ em relação ao $B$.\nB) No A, pois economizará $R \\$ 6,00$ em relação ao B.\nC) No B, pois economizará $R \\$ 8,00$ em relação ao $A$.\nD) No B, pois economizará $R \\$ 6,00$ em relação ao $A$.\nE) Tanto faz, porque o preço é o mesmo nos dois supermercados.",
    "options": [],
    "answer": "D",
    "solution": "Solution:\n\n(D) Se comprar no supermercado $A$, Joana gastará $2 \\times R\\$ 24,00 = R\\$ 48,00$.\nSe comprar no supermercado $B$, ela gastará $3 \\times R\\$ 14,00 = R\\$ 42,00$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26478,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCamila e Lara têm, cada uma, um tabuleiro $4 \\times 4$, inicialmente ambos em branco. Com estes tabuleiros elas fazem uma brincadeira do seguinte modo:\n- Camila, escondida de Lara, pinta algumas casas de seu tabuleiro, de preto;\n- Ainda em seu tabuleiro, Camila escreve em cada casa o número de casas vizinhas que estão pintadas de preto (duas casas distintas são vizinhas se possuem um lado ou um vértice em comum);\n- Camila copia os números escritos em seu tabuleiro no tabuleiro de Lara;\n- Lara deve adivinhar, a partir dos números escritos em seu tabuleiro, quantas são as casas pretas do tabuleiro de Camila.\n\nPor exemplo, se Camila pintou seu tabuleiro assim\n\n![](attached_image_1.png)\n\nentão ela vai colocar os números no tabuleiro de Lara do seguinte modo:\n\n| 1 | 1 | 3 | 1 |\n| :--- | :--- | :--- | :--- |\n| 2 | 3 | 2 | 2 |\n| 1 | 3 | 1 | 1 |\n| 1 | 2 | 0 | 0 |\n\nSe o tabuleiro de Lara tem os números\n\n| 1 | 2 | 1 | 1 |\n| :--- | :--- | :--- | :--- |\n| 0 | 2 | 1 | 2 |\n| 2 | 3 | 3 | 1 |\n| 1 | 0 | 2 | 1 |\n\nquantas foram as casas que Camila pintou?\n\nA) 3\nB) 4\nC) 5\nD) 6\nE) 7",
    "options": [],
    "answer": "B",
    "solution": "Solution:\n\nNotamos primeiro que se uma casa tem o algarismo $0$, então nenhuma das casas vizinhas pode estar pintada. Logo as casas marcadas com um $\\mathrm{X}$ na figura ao lado não foram pintadas:\n\nConsideremos agora a casa do canto superior direito, na qual aparece o número $1$. Ela tem $3$ vizinhas, e já sabemos que duas delas não foram pintadas; logo, a vizinha que sobra (a casa imediatamente abaixo) foi pintada.\n\n![](attached_image_2.png)\n\n|   |   |   |   |\n| :--- | :--- | :--- | :--- |\n|   |   |   |   |\n|   |   |   |   |\n| 1 |   |   | 1 |$\\Rightarrow$| $\\mathrm{X}$ | $\\mathrm{X}$ |   |   |\n|   | $\\mathrm{X}$ |   |   |\n| $\\mathrm{X}$ | $\\mathrm{X}$ | $\\mathrm{X}$ |   |\n| $\\mathrm{X}$ |   | $\\mathrm{X}$ |   |\n\nPodemos aplicar o mesmo argumento às casas do canto inferior esquerdo e do canto inferior direito.\n\nOlhamos agora para o $2$ na última linha. Como esta casa já tem duas vizinhas pintadas, todas suas outras vizinhas não foram pintadas:\n\n![](attached_image_3.png)\n![](attached_image_4.png)\n![](attached_image_5.png)\n\nArgumento idêntico se aplica à casa da segunda linha e terceira coluna, pois nela aparece um $1$ e já temos uma de suas vizinhas pintadas. Logo, as suas outras $3$ vizinhas não foram pintadas.\n\nFinalmente, usamos o $3$ que aparece na casa da terceira linha e terceira coluna; esta casa já tem $2$ vizinhas pintadas, logo deve haver mais uma de suas vizinhas pintada. Esta vizinha só pode ser a casa em branco na figura acima, e podemos completar a tabela:\n\nConcluímos que o número de casas pintadas é $4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26479,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA maior raiz da equação $(x-37)^2-169=0$ é:\nA) 39\nB) 43\nC) 47\nD) 50\nE) 53",
    "options": [],
    "answer": "D",
    "solution": "Solution:\nSolução 1: Usando a fatoração $a^2-b^2=(a-b)(a+b)$ : $(x-37)^2-13^2=0 \\Leftrightarrow (x-37-13)(x-37+13)=0 \\Leftrightarrow (x-50)(x-24)=0$.\nLogo, as raízes são 24 e 50.\n\n\nSolução 2: Extraindo a raiz quadrada em ambos os lados:\n$(x-37)^2=13^2 \\Leftrightarrow x-37=13$ ou $x-37=-13$. Assim, $x=50$ ou $x=24$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26480,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSe $x + y = 8$ e $x y = 15$, qual é o valor de $x^{2} + 6 x y + y^{2}$?\n\nA) 64\nB) 109\nC) 120\nD) 124\nE) 154",
    "options": [],
    "answer": "D",
    "solution": "Solution:\n\nUsando a identidade $(x + y)^{2} = x^{2} + 2 x y + y^{2}$, temos\n\n$x^{2} + 6 x y + y^{2} = \\left(x^{2} + 2 x y + y^{2}\\right) + 4 x y = (x + y)^{2} + 4 x y = 8^{2} + 4 \\times 15 = 124$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26481,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO quadrado $STUV$ é formado de um quadrado limitado por 4 retângulos iguais. O perímetro de cada retângulo é $40~\\mathrm{cm}$. Qual é a área, em $\\mathrm{cm}^2$, do quadrado $STUV$?\n\n![](attached_image_1.png)\n\n(A) 400\n(B) 200\n(C) 160\n(D) 100",
    "options": [],
    "answer": "A",
    "solution": "Solution:\n\nDenotemos por $C$ e $L$, o comprimento e a largura respectivamente de cada um dos quatro retângulos. O perímetro de cada retângulo é $2(C+L)$. Então, $2 \\times (C+L) = 40 \\Rightarrow C+L = 20$. Observe na figura que o lado do quadrado $STUV$ é $C+L$, e portanto sua área é $A = (C+L)^2 = 20^2 = 400~\\mathrm{cm}^2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26482,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nArtur quer desenhar uma \"espiral\" de 4 metros de comprimento formada de segmentos de reta. Ele já traçou 7 segmentos, como mostra a figura. Quantos segmentos ainda faltam traçar?\n(A) 28\n(B) 30\n(C) 24\n(D) 32\n(E) 36\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "D",
    "solution": "Solution:\n\nA figura mostra que a \"espiral\" é formada de segmentos cujos comprimentos formam uma sequência finita da forma $1, 1, 2, 2, 3, 3, 4, 4, \\ldots, n, n$ (se os dois últimos segmentos da espiral têm o mesmo comprimento) ou da forma $1, 1, 2, 2, 3, 3, 4, 4, \\ldots, n, n, n+1$ (se os dois últimos segmentos da espiral têm comprimentos diferentes). Como o comprimento total é $4\\ \\mathrm{m} = 400\\ \\mathrm{cm}$, devemos ter:\n\n![](attached_image_2.png)\n\n\\[\n\\left\\{\n\\begin{array}{l}\n1+1+2+2+3+3+\\ldots+n+n=400 \\Rightarrow 2 \\times (1+2+3+\\ldots+n)=400 \\\\\n\\mathrm{ou} \\\\\n1+1+2+2+3+3+\\ldots+n+n+n+1=400 \\Rightarrow 2 \\times (1+2+3+\\ldots+n)+n+1=400\n\\end{array}\n\\right.\n\\]\n\nA soma dos $n$ primeiros números naturais é $1+2+3+\\ldots+n=\\frac{n(n+1)}{2}$, logo temos:\n\n\\[\n\\left\\{\n\\begin{array}{l}\n2 \\times \\frac{n(n+1)}{2}=400 \\Rightarrow n(n+1)=400 \\\\\n\\text{ou} \\\\\n2 \\times \\frac{n(n+1)}{2}+n+1=400 \\Rightarrow n(n+1)+n+1=400 \\Rightarrow (n+1)^2=400\n\\end{array}\n\\right.\n\\]\n\nNão existem dois números naturais consecutivos cujo produto seja $400$, logo, a equação $n(n+1)=400$ não tem solução. De $(n+1)^2=400$ segue que $n+1=20$. Portanto, o último segmento da espiral tem $20\\ \\mathrm{cm}$ e o penúltimo $19\\ \\mathrm{cm}$. Os comprimentos dos segmentos da espiral formam a sequência de números $1, 1, 2, 2, 3, 3, 4, 4, \\ldots, 19, 19, 20$.\n\nPortanto, são $19 \\times 2 + 1 = 39$ segmentos. Como $7$ já foram traçados, faltam $32$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26483,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAndré treina para a maratona dando voltas em torno de uma pista circular de raio $100~\\mathrm{m}$. Para percorrer aproximadamente $42~\\mathrm{km}$, o número de voltas que André precisa dar está entre:\nA) $1$ e $10$\nB) $10$ e $50$\nC) $50$ e $100$\nD) $100$ e $500$\nE) $500$ e $1000$",
    "options": [],
    "answer": "C",
    "solution": "Solution:\n\nO comprimento de uma circunferência de raio $r$ é $2 \\pi r$. Assim, em cada volta, André percorre $2 \\pi \\times 100~\\mathrm{m} = 200 \\pi~\\mathrm{m}$. Logo, o número de voltas que André precisa dar é $\\frac{42000}{200 \\pi} = \\frac{210}{\\pi}$.\n\nPodemos agora finalizar o problema de duas maneiras:\n\n1a) A aproximação de $\\pi$ até a segunda casa decimal é $3,14$. Daí, $\\frac{210}{\\pi} \\approx \\frac{210}{3,14} \\approx 66,878 \\approx 66,88$. Como $66,88$ está entre $50$ e $100$, a opção correta é C.\n\n2a) Como $3 < \\pi < 4$ segue que $\\frac{1}{\\pi} < \\frac{1}{3}$ e $\\frac{1}{4} < \\frac{1}{\\pi}$. Multiplicando ambos os lados dessas desigualdades por $210$ obtemos:\n$$\n\\frac{210}{\\pi} < \\frac{210}{3} = 70 \\text{ e } \\frac{210}{4} < \\frac{210}{\\pi}\n$$\nComo $\\frac{210}{4} = 52,5$, concluímos que André deve dar entre $53$ e $70$ voltas na pista para percorrer $42000~\\mathrm{m}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26484,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDois espelhos formam um ângulo de $30^{\\circ}$ no ponto $V$. Um raio de luz parte de um ponto $S$ paralelamente a um dos espelhos e é refletido pelo outro espelho no ponto $A$, como mostra a figura. Depois de uma certa quantidade de reflexões, o raio retorna a $S$.\nSe $A S$ e $A V$ têm ambos 1 metro, qual o comprimento em metros do trajeto percorrido pelo raio de luz?\n\n![](attached_image_1.png)\nA) 2\nB) $2+\\sqrt{3}$\nC) $1+\\sqrt{2}+\\sqrt{3}$\nD) $\\sqrt{2}(1+\\sqrt{3})$\nE) $5 \\sqrt{3}$",
    "options": [],
    "answer": "B",
    "solution": "Solution:\n\nVamos acompanhar o trajeto do raio de luz a partir do ponto $S$. Para isso, lembramos a propriedade básica da reflexão de um raio de luz em um espelho: o ângulo de reflexão é igual ao ângulo de incidência. Por exemplo, na figura ao lado, os ângulos $a$ e $b$ são iguais, bem como $d$ e $e$. Notamos que temos na figura as paralelas $A S$ e $B V$ cortadas pela transversal $A B$, daí segue que:\n- $a = 30^{\\circ} = b$,\n- $a + b + c = 180^{\\circ}$, donde $c = 120^{\\circ}$.\n- $c + d = 180^{\\circ}$, donde $d = 60^{\\circ} = e$.\nComo a soma dos ângulos internos do triângulo $B C V$ é $180^{\\circ}$, segue que $f = 90^{\\circ}$. Isso quer dizer que o nosso raio de luz, ao atingir $C$, será refletido sobre si mesmo e fará então o caminho inverso.\n$C \\rightarrow B \\rightarrow A \\rightarrow S$. Desse modo, o trajeto completo do raio será $S \\rightarrow A \\rightarrow B \\rightarrow C \\rightarrow B \\rightarrow A \\rightarrow S$.\n\n![](attached_image_2.png)\n\nDesse modo, o comprimento do trajeto do raio até retornar a $S$ é duas vezes a soma dos comprimentos dos segmentos $A S$, $A B$ e $B C$. O enunciado nos diz que $A S = 1~\\text{m}$. Falta calcular $A B$ e $B C$. Para isso, olhamos para o triângulo $A B C$. Ele é um triângulo retângulo com ângulos de $30^{\\circ}$ e $60^{\\circ}$. Sabemos que em um tal triângulo o cateto oposto ao ângulo de $30^{\\circ}$ tem comprimento igual à metade do comprimento da hipotenusa (exercício). No nosso caso, temos $B C = \\frac{1}{2} A B$.\nNotamos agora que os triângulos $A B C$ e $V B C$ são congruentes, pois são triângulos retângulos ($f = 90^{\\circ}$) com ângulos iguais ($b = 30^{\\circ}$) e um cateto comum ($B C$), o que nos mostra que $A C = \\frac{1}{2}~\\text{m}$. Pondo $A B = x$ temos $B C = \\frac{1}{2} x$, e o teorema de Pitágoras aplicado ao triângulo $A B C$ nos dá $x^{2} = \\left(\\frac{1}{2}\\right)^{2} + \\left(\\frac{1}{2} x\\right)^{2}$; simplificando, obtemos $\\frac{3}{4} x^{2} = \\frac{1}{4}$, donde $x = \\frac{1}{\\sqrt{3}} = \\frac{\\sqrt{3}}{3}$. Desse modo, temos o comprimento do trajeto do raio de luz:\n$$\n2(S A + A B + B C) = 2\\left(1 + \\frac{\\sqrt{3}}{3} + \\frac{1}{2} \\frac{\\sqrt{3}}{3}\\right) = (2 + \\sqrt{3})~\\text{m}\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26485,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm padeiro quer gastar toda sua farinha para fazer pães. Trabalhando sozinho, ele conseguiria acabar com a farinha em 6 horas; com um ajudante, o mesmo poderia ser feito em 2 horas. O padeiro começou a trabalhar sozinho; depois de algum tempo, cansado, ele chamou seu ajudante e assim, após 150 minutos a farinha acabou. Quanto tempo o padeiro trabalhou sozinho?",
    "options": [],
    "answer": "45 minutes",
    "solution": "Solution:\n\nSeja $x$ a quantidade de farinha, em quilos, de que o padeiro dispõe. Trabalhando sozinho, ele usaria $\\frac{x}{6}$ quilos de farinha em 1 hora; trabalhando com seu ajudante, eles usariam $\\frac{x}{2}$ quilos de farinha em 1 hora. Seja $t$ o tempo, em horas, que o padeiro trabalhou sozinho. Como a farinha acaba em 150 minutos (2 horas e 30 minutos $=2,5$ horas), o tempo que ele trabalhou com seu ajudante foi $2,5-t$ horas.\n\nLogo, a quantidade gasta de farinha durante o tempo que o padeiro trabalhou sozinho é $\\frac{x}{6} \\times t$, e a quantidade gasta durante o tempo que o padeiro trabalhou com seu ajudante é $\\frac{x}{2} \\times (2,5-t)$. Como\n\n![](attached_image_1.png)\n\ntemos $x=\\frac{x}{6} t+\\frac{x}{2}(2,5-t)$. A quantidade de farinha que o padeiro tinha inicialmente era não nula, isto é $x \\neq 0$. Logo, podemos dividir ambos os membros por $x$ e encontramos $1=\\frac{t}{6}+\\frac{2,5-t}{2}$, portanto, $t=0,75$ horas $=0,75 \\times 60$ minutos $=45$ minutos.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26486,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA tabela mostra as temperaturas máximas e mínimas durante 5 dias seguidos em certa cidade. Em qual dia ocorreu o maior variação de temperatura?\n\n| Dia         | Temperatura máxima em $^{\\circ} \\mathrm{C}$ | Temperatura mínima em $^{\\circ} \\mathrm{C}$ |\n|:----------:|:------------------------------------------:|:-------------------------------------------:|\n| $2^{\\mathrm{a}}$-feira | 7 | -12 |\n| 3a-feira    | 0 | -11 |\n| $4^{\\mathrm{a}}$-feira | -2 | -15 |\n| 5a-feira    | 9 | -8 |\n| $6^{\\mathrm{a}}$-feira | 13 | -7 |",
    "options": [],
    "answer": "6a-feira",
    "solution": "Solution:\n\nA variação de temperatura é a diferença entre a máxima e a mínima. Temos:\n\n| Dia         | Temperatura máxima em $^{\\circ} \\mathrm{C}$ | Temperatura mínima em $^{\\circ} \\mathrm{C}$ | Variação |\n|:----------:|:------------------------------------------:|:-------------------------------------------:|:--------:|\n| 2a-feira    | 7  | -12 | $7-(-12)=7+12=19$ |\n| 3a-feira    | 0  | -11 | $0-(-11)=0+11=11$ |\n| 4-feira     | -2 | -15 | $-2-(-15)=-2+15=13$ |\n| 5a-feira    | 9  | -8  | $9-(-8)=9+8=17$ |\n| 6a-feira    | 13 | -7  | $13-(-7)=13+7=20$ |\n\nLogo, a maior variação ocorreu na $6^{a}$ feira.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26487,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUma certa máquina tem um visor, onde aparece um número inteiro $x$, e duas teclas $A$ e $B$. Quando se aperta a tecla $A$ o número do visor é substituído por $2x+1$. Quando se aperta a tecla $B$ o número do visor é substituído por $3x-1$.\nSe no visor está o número $5$, o maior número de dois algarismos que se pode obter apertando alguma sequência das teclas $A$ e $B$ é:\nA) 85\nB) 87\nC) 92\nD) 95\nE) 96",
    "options": [],
    "answer": "D",
    "solution": "Solution:\n\n(D) O diagrama a seguir mostra os resultados que podem ser obtidos a partir do número $5$ apertando-se cada uma das duas teclas.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26488,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLarissa e Jorge estão jogando com cartões numerados de $1$ a $6$ que devem ser colocados nas casas do tabuleiro abaixo de modo a formar um número de seis algarismos.\n\n![](attached_image_1.png)\n\nJorge coloca o primeiro cartão e a seguir as jogadas são alternadas entre os dois. O objetivo de Larissa é obter o maior número possível e o de Jorge é obter o menor número possível. Larissa tem os cartões com os algarismos $1, 3$ e $5$ e Jorge tem os cartões com os algarismos $2, 4$ e $6$. Se os dois jogadores forem espertos, qual o número que aparecerá ao final do jogo?\n\nA) $254361$\n\nB) $253416$\n\nC) $251634$\n\nD) $256134$\n\nE) $251346$",
    "options": [],
    "answer": "B",
    "solution": "Solution:\n\nA formação de um número de $6$ algarismos é ilustrada a seguir.\n\n| centena <br> de milhar | dezena de <br> milhar | unidade <br> de <br> milhar | centena | dezena | unidade |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n\nPara se obter o menor número possível, os menores algarismos devem estar o mais à esquerda possível (na casa do milhar); e para se obter o maior número possível os maiores algarismos devem também estar o mais à esquerda possível (na casa do milhar).\n\nJorge joga primeiro: Para obter o menor número possível, ele coloca o menor algarismo que ele possui, que é o $2$, na casa das centenas de milhar. Se ele não fizesse isso, Larissa colocaria seu $5$ nesta casa na próxima jogada, obtendo assim um número maior.\n\n| $2$ | dezena de <br> milhar | unidade <br> de <br> milhar | centena | dezena | unidade |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n\nAgora é a vez de Larissa: Para obter o maior número possível, ela coloca o maior algarismo que ela possui, que é o $5$, na casa das dezenas de milhar, pois a casa das centenas de milhar já está ocupada.\n\n| $2$ | $5$ | unidade <br> de <br> milhar | centena | dezena | unidade |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n\nJorge tem agora os algarismos $4$ e $6$, e Larissa $1$ e $3$. Logo, os algarismos de Larissa são menores do que os de Jorge, o que determina a estratégia de Jorge: ele deve tentar colocar seus algarismos o mais à direita possível, com o $6$ à direita do $4$. Por sua vez, Larissa deve tentar colocar seus algarismos o mais à esquerda possível, com o $3$ à esquerda do $1$. Jorge então coloca o $6$ na casa das unidades.\n\nJorge joga: Ele coloca o algarismo $6$ na casa das unidades.\n\n| $2$ | $5$ | unidade <br> de <br> milhar | centena | dezena | $6$ |\n| :--- | :--- | :--- | :--- | :--- | :--- |\n\nLarissa joga: Ela coloca seu $1$ na casa das dezenas.\n\n| $2$ | $5$ | unidade <br> de <br> milhar | centena | $1$ | $6$ |\n| :--- | :--- | :--- | :--- | :--- | :--- |\n\nAgora Jorge tem apenas o algarismo $4$ e Larissa o $3$. Ele então coloca o $4$ na casa das centenas, e Larissa coloca o $3$ na casa das unidades de milhar, acabando assim o jogo.\n\n| $2$ | $5$ | $3$ | $4$ | $1$ | $6$ |\n| :--- | :--- | :--- | :--- | :--- | :--- |\n\nLogo, o número final obtido se os dois jogadores forem espertos é $253416$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26489,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\na) Calcule as diferenças: $1-\\frac{1}{2}$; $\\frac{1}{2}-\\frac{1}{3}$; $\\frac{1}{3}-\\frac{1}{4}$; $\\frac{1}{4}-\\frac{1}{5}$; $\\frac{1}{5}-\\frac{1}{6}$\n\nb) Deduza de (a) o valor da soma: $\\frac{1}{2}+\\frac{1}{6}+\\frac{1}{12}+\\frac{1}{20}+\\frac{1}{30}$\n\nc) Calcule a soma: $\\frac{1}{2}+\\frac{1}{6}+\\frac{1}{12}+\\frac{1}{20}+\\frac{1}{30}+\\frac{1}{42}+\\cdots+\\frac{1}{999000}$",
    "options": [],
    "answer": "a) 1/2, 1/6, 1/12, 1/20, 1/30; b) 5/6; c) 999/1000",
    "solution": "Solution:\n\na)\n$1-\\frac{1}{2}=\\frac{1}{2}$\n\n$\\frac{1}{2}-\\frac{1}{3}=\\frac{1}{6}$\n\n$\\frac{1}{3}-\\frac{1}{4}=\\frac{1}{12}$\n\n$\\frac{1}{4}-\\frac{1}{5}=\\frac{1}{20}$\n\n$\\frac{1}{5}-\\frac{1}{6}=\\frac{1}{30}$\n\n\nb)\n$\\frac{1}{2}+\\frac{1}{6}+\\frac{1}{12}+\\frac{1}{20}+\\frac{1}{30}=1-\\frac{1}{2}+\\frac{1}{2}-\\frac{1}{3}+\\frac{1}{3}-\\frac{1}{4}+\\frac{1}{4}-\\frac{1}{5}+\\frac{1}{5}-\\frac{1}{6}=1-\\frac{1}{6}=\\frac{5}{6}$\n\n\nc)\nNote que os denominadores são produtos de números consecutivos, iniciando no 1:\n$\\frac{1}{2}+\\frac{1}{6}+\\frac{1}{12}+\\frac{1}{20}+\\frac{1}{30}=1-\\frac{1}{6}$\n\n![](attached_image_1.png)\n\nMas, geralmente, usando a decomposição de cada parcela como no item (a) podemos provar que:\n$$\n\\frac{1}{1 \\times 2}+\\frac{1}{2 \\times 3}+\\frac{1}{3 \\times 4}+\\frac{1}{4 \\times 5}+\\frac{1}{5 \\times 6}+\\frac{1}{6 \\times 7}+\\cdots+\\frac{1}{n \\times(n+1)}=1-\\frac{1}{n+1}\n$$\nLogo:\n$$\n\\frac{1}{2}+\\frac{1}{6}+\\frac{1}{12}+\\frac{1}{20}+\\frac{1}{30}+\\frac{1}{42}+\\cdots+\\frac{1}{999000}=1-\\frac{1}{1000}=\\frac{999}{1000}=0,999\n$$\n\n![](attached_image_1.png)\n\n![](attached_image_2.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26490,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIara possui $R\\$ 50,00$ para comprar copos que custam $R\\$ 2,50$ e pratos que custam $R\\$ 7,00$. Ela quer comprar no mínimo 4 pratos e 6 copos. O que ela pode comprar?",
    "options": [],
    "answer": "She can buy 4 plates and 8 cups, or 5 plates and 6 cups.",
    "solution": "Solution:\n\nSejam $c$ e $p$ o número de copos e pratos que Iara pode comprar. Logo seu gasto é $2,5c + 7p$. Ela só tem $R\\$ 50,00$, logo $2,5c + 7p \\leq 50$ (I).\n\nAlém disso, ela quer comprar no mínimo 4 pratos e 6 copos, logo $p \\geq 4$ e $c \\geq 6$ (II).\n\nDevemos encontrar dois números inteiros $c$ e $p$ (número de copos e pratos são números inteiros) que satisfaçam (I) e (II).\n\nSe ela comprar 4 pratos sobram $50 - 4 \\times 7 = 22$ reais para os copos. Como $22 = 8 \\times 2,50 + 2$, ela pode comprar 8 copos (sobrando-lhe $R\\$ 2,00$).\n\nSe ela comprar 5 pratos sobram $50 - 5 \\times 7 = 15$ reais para os copos. Como $15 = 6 \\times 2,50$, ela pode comprar 6 copos.\n\nSe ela comprar 6 pratos sobram $50 - 6 \\times 7 = 8$ reais para os copos, o que lhe permite comprar apenas 1 copo, que não é o que ela quer.\n\nLogo, Iara pode comprar 4 pratos e 8 copos, ou 5 pratos e 6 copos.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26491,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNuma corrida de carros, um piloto percorreu três trechos: um de $240~\\mathrm{km}$, um de $300~\\mathrm{km}$ e um de $400~\\mathrm{km}$. O piloto sabe que as velocidades médias nesses trechos foram $40~\\mathrm{km}/\\mathrm{h}$, $75~\\mathrm{km}/\\mathrm{h}$ e $80~\\mathrm{km}/\\mathrm{h}$, mas não se lembra qual dessas velocidades corresponde a cada um desses trechos. Podemos garantir que o tempo total em horas gasto pelo piloto para percorrer os três trechos foi:\nA) menor ou igual a 13 horas\nB) maior ou igual a 13 horas e menor ou igual a 16 horas\nC) maior ou igual a 16 horas e menor ou igual a 17 horas\nD) maior ou igual a 15 horas e menor ou igual a 18 horas\nE) maior ou igual a 18 horas",
    "options": [],
    "answer": "D",
    "solution": "Solution:\n\nO menor tempo de percurso é obtido quando se percorre o maior trecho com a maior velocidade e o menor trecho com a menor velocidade. Já o maior tempo é obtido quando se percorre o maior trecho com a menor velocidade e o menor trecho com a maior velocidade. Assim, o tempo total gasto pelo piloto nos três trechos é no mínimo $\\frac{240}{40} + \\frac{300}{75} + \\frac{400}{80} = 15$ horas e no máximo $\\frac{240}{80} + \\frac{300}{75} + \\frac{400}{40} = 17$ horas.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26492,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa figura estão indicadas em graus as medidas de alguns ângulos em função de $x$. Quanto vale $x$?\nA) $6^{\\circ}$\nB) $12^{\\circ}$\nC) $18^{\\circ}$\nD) $20^{\\circ}$\nE) $24^{\\circ}$\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "C",
    "solution": "Solution:\n\nCompletamos a figura marcando os ângulos $\\alpha$ e $\\beta$, lembrando que ângulos opostos pelo vértice são iguais. Como a soma dos ângulos internos de um triângulo é $180^{\\circ}$, podemos escrever as três igualdades abaixo, uma para cada um dos triângulos da figura:\n$$\n\\begin{aligned}\n& \\alpha+7x=180^{\\circ} \\\\\n& \\beta+8x=180^{\\circ} \\\\\n& \\alpha+\\beta+5x=180^{\\circ}\n\\end{aligned}\n$$\nLogo,\n![](attached_image_2.png)\n$$\n(\\alpha+7x)+(\\beta+8x)-(\\alpha+\\beta+5x)=180^{\\circ}+180^{\\circ}-180^{\\circ}=180^{\\circ}\n$$\ne como\n$$\n(\\alpha+7x)+(\\beta+8x)-(\\alpha+\\beta+5x)=\\alpha+7x+\\beta+8x-\\alpha-\\beta-5x=10x\n$$\nsegue que $10x=180^{\\circ}$, donde $x=18^{\\circ}$\n\n![](attached_image_2.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26493,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPaulo quer comprar um sorvete com 4 bolas em uma sorveteria que dispõe de três sabores: açaí, baunilha e cajá. De quantos modos diferentes ele pode fazer a compra?\nA) 6\nB) 9\nC) 12\nD) 15\nE) 18",
    "options": [],
    "answer": "D",
    "solution": "Solution:\n\nVamos denotar cada sabor de sorvete pela sua letra inicial:\n$a \\rightarrow$ açaí, $b \\rightarrow$ baunilha, $c \\rightarrow$ cajá\n\nPara enumerar todas as possibilidades de compra do sorvete com quatro bolas, devemos considerar os seguintes casos:\n\n- 4 bolas do mesmo sabor\n- 3 bolas do mesmo sabor e 1 de sabor diferente\n- 2 bolas de um mesmo sabor e 2 de outro sabor\n- 2 bolas de um mesmo sabor e as outras 2 dos outros dois sabores\n\nObtemos assim 15 modos de fazer a compra do sorvete.\n\n$aaac$ $aacc$ $bbac$\n$bbba$\n$b b b c$\n$ссса$\n$c c c b$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26494,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA função $f$ é dada pela tabela a seguir.\n\n| $\\boldsymbol{x}$ | 1 | 2 | 3 | 4 | 5 |\n| :---: | :--- | :--- | :--- | :--- | :--- |\n| $f(x)$ | 4 | 1 | 3 | 5 | 2 |\n\nPor exemplo, $f(2)=1$ e $f(4)=5$. Quanto vale $\\underbrace{f(f(f(f(\\ldots \\ldots . . f}_{2004 \\text{ veces }}(4) \\ldots .)))}$?\n\nA) 1\nB) 2\nC) 3\nD) 4\nE) 5",
    "options": [],
    "answer": "D",
    "solution": "Solution:\n\nDa tabela temos:\n\n$$\n\\begin{gathered}\nf(4)=5 \\, , \\, f(f(4))=f(5)=2 \\, , \\, f(f(f(4)))=f(f(5))=f(2)=1 \\, , \\\\\nf(f(f(f(4))))=f(f(f(5)))=f(f(2))=f(1)=4 \\\\\n\\text{Logo, } \\underbrace{f(f(f(f(4))))}_{\\text{4 vezes}}=4.\n\\end{gathered}\n$$\n\nComo $2004$ é múltiplo de $4$, segue que $\\underbrace{f(f(f(f(\\ldots f(4) \\ldots ))))}_{2004 \\text{ veces }}=4$.\n\nA sequência a seguir ilustra esta composição.\n\n$$\n\\underbrace{4 \\stackrel{f}{\\rightarrow} 5 \\stackrel{f}{\\rightarrow} 2 \\stackrel{f}{\\rightarrow} 1 \\stackrel{f}{\\rightarrow} 4}_{\\text{4 vezes}} \\underbrace{\\stackrel{f}{\\rightarrow} 5 \\stackrel{f}{\\rightarrow} 2 \\stackrel{f}{\\rightarrow} 1 \\stackrel{f}{\\rightarrow} 4}_{\\text{8 vezes}} \\rightarrow \\underbrace{5 \\stackrel{f}{\\rightarrow} 2 \\stackrel{f}{\\rightarrow} 1 \\stackrel{f}{\\rightarrow} 4}_{12 \\text{ vezes}} \\rightarrow \\ldots \\cdots \\underbrace{\\stackrel{f}{\\rightarrow} 5 \\stackrel{f}{\\rightarrow} 2 \\stackrel{f}{\\rightarrow} 1 \\stackrel{f}{\\rightarrow} 4}_{\\text{2004 vezes}}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26495,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nManoel testou sua pontaria lançando cinco flechas no alvo reticulado de quadrados de comprimento $1~\\mathrm{cm}$, ilustrado na figura. Uma flecha que acerta dentro do círculo menor conta 300 pontos; na região sombreada conta 100 pontos, entre a região sombreada e o círculo maior conta 50 pontos e fora do círculo maior não conta nada. As flechas de Manoel acertaram os pontos\n$$\nA=(1,-1),\\ B=\\left(\\frac{5}{2}, 1\\right),\\ C=(1,-4),\\ D=(-4,-4)\\ \\text{e}\n$$\n$E=(6,5)$.\n![](attached_image_1.png)\n\na. Marque na figura os pontos onde Manoel acertou suas flechas.\n\nb. Quantas flechas ele acertou no interior do menor círculo?\n\nc. Quantos pontos Manoel fez no total?",
    "options": [],
    "answer": "b: 1; c: 500",
    "solution": "Solution:\n\na. Marcamos os pontos, conforme mostra a figura:\n\n![](attached_image_1.png)\n\nb. No círculo menor temos apenas o ponto $A$. Portanto, Manoel acertou apenas uma vez neste círculo, o que lhe dá 300 pontos.\n\nc. Para calcular o total de pontos, observe que no ponto $B$ ele ganha 100 pontos, no $C$ ganha 50 pontos e no $D$ ganha 50 pontos. Já no ponto $E$, ele não ganha pontos, porque este está fora do alvo. Logo, o número total de pontos foi de $300+100+50+50=500$ pontos.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26496,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nQual é o menor número inteiro positivo $N$ tal que $\\frac{N}{3}$, $\\frac{N}{4}$, $\\frac{N}{5}$, $\\frac{N}{6}$ e $\\frac{N}{7}$ são números inteiros?\n\nA) 420\nB) 350\nC) 210\nD) 300\nE) 280",
    "options": [],
    "answer": "A",
    "solution": "Solution:\n\nPara que $\\frac{N}{3}$, $\\frac{N}{4}$, $\\frac{N}{5}$, $\\frac{N}{6}$ e $\\frac{N}{7}$ sejam números inteiros, $N$ deve ser múltiplo comum de $3, 4, 5, 6$ e $7$. Como queremos o menor $N$ possível, ele deve ser o menor múltiplo comum de $3, 4, 5, 6$ e $7$. Sendo o MMC entre $3, 4, 5, 6$ e $7$ igual a $420$, temos $N = 420$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26497,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSe $3$ e $\\frac{1}{3}$ são as raízes da equação $a x^{2}-6 x+c=0$, qual o valor de $a+c$?\n\nA) $1$\nB) $0$\nC) $-\\frac{9}{5}$\nD) $\\frac{18}{5}$\nE) $-5$",
    "options": [],
    "answer": "D",
    "solution": "Solution:\n\nSolução 1 - Como $3$ e $\\frac{1}{3}$ são raízes da equação $a x^{2}-6 x+c=0$ temos:\n$$\n9a - 18 + c = 0 \\Rightarrow 9a + c = 18 \\text{ e } \\frac{a}{9} - 2 + c = 0 \\Rightarrow \\frac{a}{9} + c = 2\n$$\nResolvendo o sistema\n$$\n\\left\\{\\begin{array}{l}\n9a + c = 18 \\\\\n\\frac{a}{9} + c = 2\n\\end{array}\\right.\n$$\nobtemos $a = c = \\frac{9}{5}$. Logo, $a + c = \\frac{18}{5}$.\n\n\nSolução 2 - Numa equação do $2^{\\circ}$ grau $a x^{2} + b x + c = 0$, a soma das raízes é $-\\frac{b}{a}$ e o produto $\\frac{c}{a}$. Logo: $3 + \\frac{1}{3} = \\frac{10}{3} = \\frac{6}{a} \\Rightarrow a = \\frac{9}{5}$ e $3 \\times \\frac{1}{3} = 1 = \\frac{c}{a} \\Rightarrow a = c$. Assim, $a + c = 2 \\times \\frac{9}{5} = \\frac{18}{5}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26498,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nQuantos dentre os números $-5,-4,-3,-2,-1,0,1,2,3$ satisfazem a desigualdade $-3 x^{2}<-14$ ?\n(A) 1\n(B) 2\n(C) 3\n(D) 4\n(E) 5",
    "options": [],
    "answer": "D",
    "solution": "Solution:\nSe $-3 x^{2}<-14$ então $3 x^{2}>14$ ou $x^{2} > \\frac{14}{3} = 4 \\frac{2}{3}$. Como estamos olhando apenas para valores inteiros de $x$, então $x^{2}$ também é inteiro. Sendo $x^{2} > 4 \\frac{2}{3}$, concluímos que $x^{2}$ é no mínimo $5$. Os números acima que satisfazem essa condição são $-5$, $-4$, $-3$ e $3$. Logo a resposta é $4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26499,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa figura abaixo temos dois quadrados. O maior tem lado $a+b$ e o menor lado $a$.\n\n![](attached_image_1.png)\n\nQual é a área da região em cinza?\nA) $b$\nB) $a+b$\nC) $a^{2}+2 a b$\nD) $b^{2}$\nE) $2 a b+b^{2}$",
    "options": [],
    "answer": "E",
    "solution": "Solution:\n\nSolução 1 - Usaremos que $(a+b)^{2}=a^{2}+2 a b+b^{2}$. Lembre que a área de um quadrado de lado $l$ é $l^{2}$. Note que a área da região cinza é a diferença entre as áreas do maior e do menor quadrado. O lado do maior é $a+b$, portanto sua área é $(a+b)^{2}$. Já o lado do menor é $a$, logo sua área é $a^{2}$. Concluímos que a área da região cinza é:\n$$(a+b)^{2}-a^{2}=a^{2}+2 a b+b^{2}-a^{2}=2 a b+b^{2}.$$\n\n\nSolução 2 - Lembre que a área de um retângulo é o produto da largura pelo comprimento.\nPodemos dividir a região cinza em dois retângulos, um da largura $b$ e comprimento $a$, e o outro de largura $b$ e comprimento $a+b$, como mostra a figura. A área em cinza é a soma das áreas desses dois retângulos, ou seja:\n$$a \\times b + b \\times (a+b) = a b + a b + b^{2} = 2 a b + b^{2}.$$ \nPortanto, a área solicitada é $2 a b + b^{2}$.\n\n![](attached_image_2.png)\n\n\nSolução 3 - A região cinza é formada por 2 retângulos de dimensões $a \\times b$ e um quadrado de lado $b$. Logo a sua área é: $2 \\times a b + b^{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26500,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEm um quadrado mágico, a soma dos 3 números de cada linha, coluna ou diagonal é sempre a mesma. A seguir temos um quadrado mágico, parcialmente preenchido.\n\n|   |   |   |\n| :--- | :--- | :--- |\n| 1 | 14 | $x$ |\n| 26 |   | 13 |\n\nQual é o valor de $x$?\nA) 20\nB) 22\nC) 23\nD) 25\nE) 27",
    "options": [],
    "answer": "E) 27",
    "solution": "Solution:\n\nSeja $y$ um dos números do quadrado mágico, conforme a figura. De acordo com a regra de quadrado mágico temos:\n$$\n\\underbrace{26+14+y}_{\\begin{array}{c}\\text{ soma dos números } \\\\ \\text{ da diagonal que contém } y\\end{array}} = \\underbrace{y+x+13}_{\\begin{array}{c}\\text{ soma dos números } \\\\ \\text{ da coluna que contém } y\\end{array}}\n$$\nSegue que $26+14 = x+13$, donde $x=27$.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26501,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO café, o bolo e o gato - Dez minutos antes de colocar o bolo no forno, eu coloquei meu gato do lado de fora da casa. O bolo deve cozinhar por 35 minutos, então eu coloquei o despertador para tocar 35 minutos, após colocar o bolo no forno. Imediatamente fiz um café para mim, o que me tomou 6 minutos. Três minutos antes de acabar de beber o café o gato entrou em casa. Isso foi 5 minutos antes do despertador tocar. O telefone tocou no meio do tempo entre eu acabar de fazer o café e o gato entrar em casa. Falei ao telefone por 5 minutos e desliguei. Eram 3h59min da tarde.\n\na) A que horas coloquei o gato fora de casa?\nb) Quantos minutos depois de colocar o gato fora de casa, o despertador tocou?\nc) Quanto tempo o gato estava fora de casa até o momento em que o telefone tocou?",
    "options": [],
    "answer": "a) 3:26; b) 45 minutes; c) 28 minutes",
    "solution": "Solution:\n\nVamos listar os eventos ocorridos e contar o tempo gasto em cada um. A primeira atividade foi colocar o gato fora de casa, logo nossa lista começa com essa atividade e o tempo é contado a partir dela.\n\n| Atividade                | Tempo depois que o gato foi posto fora de casa |\n| :---                     | :---                                           |\n| Gato fora de casa        | 0 minutos                                      |\n| Bolo no forno            | 10 minutos                                     |\n| Fazer o café             | $10+6=16$ minutos                              |\n| Despertador toca         | $35+10=45$ minutos                             |\n| Gato entra em casa       | $45-5=40$ minutos                              |\n| Acabar de tomar o café   | $40+3=43$ minutos                              |\n| Telefone toca            | $16+(40-16): 2=28$ minutos                     |\n| Desligar o telefone      | $28+5=33$ minutos                              |\n\nPodemos agora dar as respostas.\n\na) Às 3:59 horas desliguei o telefone, o que ocorreu 33 minutos depois de colocar o gato fora de casa. Logo a resposta é $3:59-0:33=3:26$.\n\nb) O despertador toca 45 minutos após colocar o gato fora de casa.\n\nc) 28 minutos\n\nPodemos saber exatamente a hora de cada atividade; veja na tabela a seguir.\n\n| Atividade                | Tempo depois que o gato foi posto fora de casa | Hora atual           |\n| :---                     | :---                                           | :---                |\n| Gato fora de casa        | 0 minutos                                      | $3:59-0:33=3:26$    |\n| Bolo no forno            | 10 minutos                                     | $3:26+0:10=3:36$    |\n| Fazer o café             | $10+6=16$ minutos                              | $3:26+0:16=3:42$    |\n| Despertador toca         | $35+10=45$ minutos                             | $3:26+0:45=4:11$    |\n| Gato entra em casa       | $45-5=40$ minutos                              | $3:26+0:40=4:06$    |\n| Acabar de tomar o café   | $40+3=43$ minutos                              | $3:26+0:43=4:09$    |\n| Telefone toca            | $16+(40-16): 2=28$ minutos                     | $3:26+0:28=3:54$    |\n| Desligar o telefone      | $28+5=33$ minutos                              | $3:59$              |",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26502,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEsmeralda escreveu em ordem crescente todos os números de $1$ a $999$, sem separá-los, formando o número mostrado a seguir: $12345678910111213\\ldots 997998999$. Nesse número, quantas vezes aparece o agrupamento \"21\", nesta ordem?",
    "options": [],
    "answer": "31",
    "solution": "Solution:\n\nVamos primeiro listar os números que têm o agrupamento $21$ no meio de sua representação decimal:\n$21, 121, 221, \\ldots, 921 \\rightarrow 10$ números\n$210, 211, \\ldots, 219 \\rightarrow 10$ números\n\nTemos também que contar os agrupamentos $21$ obtidos a partir de um par de números consecutivos tal que o primeiro termina com $2$ e o segundo começa com $1$, que são os seguintes $11$ casos:\n$12$-$13$, $102$-$103$, $112$-$113$, $122$-$123$, $132$-$133$, $142$-$143$, $152$-$153$, $162$-$163$, $172$-$173$, $182$-$183$, $192$-$193$\n\nTemos então um total de $11+20=31$ números.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26503,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA festa de aniversário de André tem menos do que 120 convidados. Para o jantar, ele pode dividir os convidados em mesas completas de 6 pessoas ou em mesas completas de 7 pessoas. Nos dois casos são necessárias mais do que 10 mesas e todos os convidados ficam em alguma mesa. Quantos são os convidados?",
    "options": [],
    "answer": "84",
    "solution": "Solution:\n\nComo podemos repartir o total de convidados em mesas de 6 ou 7, o número de convidados é um múltiplo de 6 e de 7. Como o menor múltiplo comum de 6 e 7 é $42$, podemos ter $42, 84, 126, \\ldots$ convidados. Como são menos do que $120$ convidados, só podemos ter $42$ ou $84$ convidados. Por outro lado, como são necessárias mais do que $10$ mesas, temos mais do que $60$ convidados. Logo, descartamos o $42$, e o número de convidados só pode ser $84$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26504,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPedro montou um quadrado com quatro das cinco peças abaixo. Qual é a peça que ele não usou?\n(a)\n![](attached_image_1.png)\n\n(d)\n![](attached_image_2.png)\n\n(b)\n![](attached_image_3.png)\n\n(c)\n![](attached_image_4.png)\n\n(e)\n![](attached_image_5.png)",
    "options": [],
    "answer": "b",
    "solution": "Solution:\n\nSolução 1 - Contando o total de quadrados nas peças.\nPara que seja possível montar o quadrado, o número total de quadradinhos deve ser um quadrado perfeito (Um número é um quadrado perfeito se ele é igual ao quadrado de um número inteiro. Por exemplo, $1, 9$ e $16$ são quadrados perfeitos pois $1=1^{2}$, $9=3^{2}$, $16=4^{2}$).\nContando o total de quadradinhos apresentados nas cinco opções de resposta, obtemos: $4+5+6+7+8=30$.\nPortanto, devemos eliminar uma peça de modo que o total de quadradinhos resultante seja um quadrado perfeito. A única possibilidade é a (b). De fato, eliminando (b), a soma fica sendo $25$ que é um quadrado perfeito, pois $25=5^{2}$.\n\n\nSolução 2 - Tentando montar o quadrado com 4 das cinco peças.\nNeste caso, conseguimos montar um quadrado com as peças $a, c, d$ e $e$, como na figura:\n\n![](attached_image_6.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26505,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDados $a$ e $b$ números reais seja $a \\diamond b = a^{2} - a b + b^{2}$. Quanto vale $1 \\diamond 0$?\nA) 1\nB) 0\nC) 2\nD) -2\nE) -1",
    "options": [],
    "answer": "A",
    "solution": "Solution:\n\nFazendo $a = 1$ e $b = 0$ em $a \\diamond b = a^{2} - a b + b^{2}$ obtemos:\n$$\n1 \\diamond 0 = 1^{2} - 1 \\times 0 + 0^{2} = 1\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26506,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nVocê já viu um truque numérico? Aqui vão os passos de um truque numérico:\n\n(I) Escolha um número qualquer.\n(II) Multiplique-o por 6.\n(III) Do resultado subtraia 21.\n(IV) Divida agora este novo resultado por 3.\n(V) Deste último resultado subtraia o dobro do número que você escolheu.\n\na. Experimente fazer esses cinco passos três vezes, iniciando cada vez com um número diferente. Qual foi o resultado de seu experimento?\n\nb. A seguir, usando a letra $x$ para representar o número que você pensou, mostre por que os resultados do item (a) não são apenas uma coincidência, mas sim um fato matemático.",
    "options": [],
    "answer": "-7",
    "solution": "Solution:\n\na.\nVamos fazer o experimento com os números $0$, $5$ e $-4$.\n\nPara $0$:\n\n- Escolha $0$.\n- Multiplique por $6$: $0 \\times 6 = 0$.\n- Subtraia $21$: $0 - 21 = -21$.\n- Divida por $3$: $-21 \\div 3 = -7$.\n- Subtraia o dobro do número escolhido: $-7 - 2 \\times 0 = -7$.\n\nPara $5$:\n\n- Escolha $5$.\n- Multiplique por $6$: $5 \\times 6 = 30$.\n- Subtraia $21$: $30 - 21 = 9$.\n- Divida por $3$: $9 \\div 3 = 3$.\n- Subtraia o dobro do número escolhido: $3 - 2 \\times 5 = 3 - 10 = -7$.\n\nPara $-4$:\n\n- Escolha $-4$.\n- Multiplique por $6$: $-4 \\times 6 = -24$.\n- Subtraia $21$: $-24 - 21 = -45$.\n- Divida por $3$: $-45 \\div 3 = -15$.\n- Subtraia o dobro do número escolhido: $-15 - 2 \\times (-4) = -15 + 8 = -7$.\n\nO resultado final é sempre $-7$.\n\nb.\nÉ razoável conjeturar então que para qualquer número escolhido o resultado final deste procedimento será sempre $-7$.\n\nSeja $x$ o número inicial. Temos então as operações:\n\n- Multiplique por $6$: $6x$\n- Subtraia $21$: $6x - 21$\n- Divida por $3$: $\\dfrac{6x - 21}{3} = 2x - 7$\n- Subtraia o dobro do número escolhido: $2x - 7 - 2x = -7$\n\nPortanto, o resultado será $-7$ qualquer que seja o número inicialmente escolhido.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26507,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSão dadas 4 moedas aparentemente iguais, das quais 3 são verdadeiras e por isso têm o mesmo peso; uma é falsa e por isso tem peso diferente. Não se sabe se a moeda falsa é mais leve ou mais pesada que as demais. Mostre que é possível determinar a moeda diferente empregando somente duas pesagens em uma balança de pratos. Observação: Neste tipo de balança podemos comparar os pesos colocados nos dois pratos, ou seja, a balança pode ficar equilibrada ou pender para o lado mais pesado.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSejam $A$, $B$, $C$ e $D$ as quatro moedas. Comparamos as moedas $A$ e $B$ na balança, colocando uma em cada prato. Dois casos podem ocorrer: a balança fica em equilíbrio ou a balança não fica em equilíbrio. Vamos analisar separadamente cada caso.\n\n$1^{\\circ}$ Caso: A balança fica equilibrada. Podemos concluir que $A$ e $B$ têm o mesmo peso, e logo são verdadeiras. Vamos então comparar $A$ com $C$. Para isso, mantemos $A$ na balança e colocamos $C$ no lugar de $B$. Se houver equilíbrio novamente, é porque $A$ e $C$ têm o mesmo peso e logo são verdadeiras. Portanto, $A$, $B$ e $C$ são verdadeiras, e a única opção é que $D$ seja falsa. Se não houver equilíbrio, $C$ será a moeda falsa.\n\n$2^{\\circ}$ Caso: A balança não fica equilibrada. Logo uma das duas moedas, $A$ ou $B$ será falsa. Substituímos $A$ por $C$ na balança. Se houver equilíbrio, $A$ será a moeda falsa. Se não houver equilíbrio, a moeda falsa será $B$.\n\nObserve que nos dois casos só utilizamos a balança duas vezes.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26508,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUma linha de ônibus possui $12$ paradas numa rua em linha reta. A distância entre duas paradas consecutivas é sempre a mesma. Sabe-se que a distância entre a terceira e a sexta paradas é $3300$ metros. Qual é a distância entre a primeira e a última parada?\n\nA) $8,4~\\mathrm{km}$\nB) $12,1~\\mathrm{km}$\nC) $9,9~\\mathrm{km}$\nD) $13,2~\\mathrm{km}$\nE) $9,075~\\mathrm{km}$",
    "options": [],
    "answer": "B",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nComo a distância entre a $3^a$ e a $6^a$ paradas é $3300~\\mathrm{m}$, então a distância entre duas paradas consecutivas é $3300 \\div 3 = 1100~\\mathrm{m}$.\n\nPortanto, a distância entre a primeira e a última paradas é $1100~\\mathrm{m} \\times 11 = 12100~\\mathrm{m}$. Como as opções da resposta são dadas em quilômetro, devemos reduzir $12100~\\mathrm{m}$ a quilômetro. Como $1~\\mathrm{km} = 1000~\\mathrm{m}$, temos $12100~\\mathrm{m} = 12,1~\\mathrm{km}$.\n\n![](attached_image_1.png)",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26509,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAdriano, Bruno, César e Daniel são quatro bons amigos. Daniel não tinha dinheiro, mas os outros tinham. Adriano deu a Daniel um quinto do seu dinheiro, Bruno deu um quarto do seu dinheiro e César deu um terço do seu dinheiro. Cada um deu a Daniel a mesma quantia. A quantia que Daniel possui agora representa que fração da quantia total que seus três amigos juntos possuíam inicialmente?\nA) $\\frac{1}{10}$\nB) $\\frac{1}{4}$\nC) $\\frac{1}{3}$\nD) $\\frac{2}{5}$\nE) $\\frac{1}{2}$",
    "options": [],
    "answer": "B",
    "solution": "Solution:\n\nSuponha que Daniel tenha recebido $x$ reais de cada um de seus amigos. Então, Adriano tinha, inicialmente, $5x$ reais, Bruno tinha $4x$ reais e César tinha $3x$ reais. Segue que o total de dinheiro dos três no início era de $5x + 4x + 3x = 12x$ reais. Como cada um de seus três amigos lhe deu $x$ reais, Daniel tem agora $3x$ reais, o que representa a quarta parte de $12x$. Logo, ele possui agora $\\frac{1}{4}$ da quantia que seus três amigos juntos possuíam inicialmente.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26510,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFrações inteiras - Quantos números inteiros positivos $n$ existem tais que $\\frac{2 n^{2}+4 n+18}{3 n+3}$ é um inteiro?",
    "options": [],
    "answer": "4",
    "solution": "Solution:\n\nComo\n$$\n\\frac{2 n^{2}+4 n+18}{3 n+3} = \\frac{2}{3} \\frac{\\left(n^{2}+2 n+1\\right)+8}{n+1} = \\frac{1}{3}\\left(2 n+2+\\frac{16}{n+1}\\right)\n$$\nsegue que $n+1$ tem que dividir $16$. Assim, $n$ tem que pertencer ao conjunto $\\{1,3,7,15\\}$. Em cada um destes casos temos\n\n| $n$ | $\\frac{2 n^{2}+4 n+18}{3 n+3}$ |\n| :---: | :---: |\n| 1 | 4 |\n| 3 | 4 |\n| 7 | 6 |\n| 15 | 11 |\n\nPortanto para os quatro valores de $n$, $1, 3, 7$ e $15$, tem-se que $\\frac{2 n^{2}+4 n+18}{3 n+3}$ é um inteiro.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26511,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNúmeros proporcionais - Se $\\frac{x}{y}=\\frac{3}{z}$, então $9 y^{2}$ é igual a:\n(a) $\\frac{x^{2}}{9}$\n(b) $x^{3} z$\n(c) $3 x^{2}$\n(d) $x^{2} z^{2}$\n(e) $\\frac{1}{9} x^{2} z^{2}$",
    "options": [],
    "answer": "d",
    "solution": "Solution:\n\nComo $\\frac{x}{y}=\\frac{3}{z}$, então $x z=3 y$. Elevando ao quadrado ambos os membros dessa igualdade obtemos $x^{2} z^{2}=9 y^{2}$. A opção correta é (d).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26512,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEquação cúbica - Sobre a equação $2007 x^{3}+2006 x^{2}+2005 x=0$ é certo afirmar:\n(a) Não possui raízes\n(b) Tem 3 raízes reais distintas\n(c) Tem 2 raízes iguais\n(d) Tem apenas uma raiz real\n(e) Tem 3 raízes positivas",
    "options": [],
    "answer": "d",
    "solution": "Solution:\n\nObservemos que $x=0$ é uma solução, logo as possibilidades (a) e (e) ficam descartadas. Agora só precisamos estudar as soluções de $2007 x^{2}+2006 x+2005=0$, que é uma equação do $2^{o}$ grau com discriminante\n$$\n\\begin{aligned}\n\\Delta & =2006^{2}-4 \\times 2007 \\times 2005=2006^{2}-4(2006+1)(2006-1) \\\\\n& =2006^{2}-4\\left(2006^{2}-1\\right)=-3 \\times 2006^{2}+4<0\n\\end{aligned}\n$$\nlogo não possui raízes reais. Portanto, a equação inicial tem uma única raiz real, e a opção correta é (d).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26513,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUma companhia tem um lucro de $6\\%$ nos primeiros $R\\$ 1000,00$ reais de venda diária, e $5\\%$ em todas as vendas que excedem $R\\$ 1000,00$ reais, nesse mesmo dia. Qual é o lucro dessa companhia num dia que as vendas alcançam $R\\$ 6000,00$ reais?\n\n(a) $R\\$ 250$\n(b) $R\\$ 300$\n(c) $\\$ 310$\n(d) $R\\$ 320$\n(e) $R\\$ 360$",
    "options": [],
    "answer": "c",
    "solution": "Solution:\n\nNos primeiros $R\\$ 1000$ reais a companhia tem lucro de $R\\$ 60$ reais, e para os $R\\$ 5000$ reais restantes tem lucro de $5000 \\times 5\\% = 250$ reais. Logo o lucro da empresa nesse dia é $R\\$ 310$ reais.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26514,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSeqüência triangular - Qual é o $21^{\\text{\\textordmasculine}}$ termo da seqüência\n$$\n1 ; 2+3 ; 4+5+6 ; 7+8+9+10 ; 11+12+13+14+15 ; \\ldots ?\n$$",
    "options": [],
    "answer": "4641",
    "solution": "Solution:\n\nObserve que o $21^{\\circ}$ termo é a soma de 21 números consecutivos. Tomando a primeira parcela de cada termo, isto é, $1, 2, 4, 7, 11, 16, \\ldots$, temos que\n$$\n\\begin{aligned}\n2 & = 1 + 1 \\\\\n4 & = 2 + 1 + 1 \\\\\n7 & = 3 + 2 + 1 + 1 \\\\\n11 & = 4 + 3 + 2 + 1 + 1 \\\\\n16 & = 5 + 4 + 3 + 2 + 1 + 1\n\\end{aligned}\n$$\nAssim, a primeira parcela do $21^{\\circ}$ termo é\n$$\n20 + 19 + \\cdots + 3 + 2 + 1 + 1 = \\frac{20 \\times 21}{2} + 1 = 211\n$$\ne o $21^{\\circ}$ termo é\n$$\n211 + 212 + \\cdots + 230 + 231 = \\frac{(211 + 231) \\times 21}{2} = 221 \\times 21 = 4641\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26515,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nQuatro prefeitos decidem construir uma rodovia circular que passe em suas cidades, entretanto, as quatro cidades não estão sobre um mesmo círculo. Eles contratam uma empresa para elaborar um projeto para a construção da rodovia circular eqüidistante das quatro cidades. Qual o maior número de projetos geograficamente distintos que a empresa elaborou?",
    "options": [],
    "answer": "7",
    "solution": "Solution:\n\nO número de rodovias é igual ao número de pontos que podem ser centros da circunferência formada pelas rodovias.\n\nObservemos por outra parte que podemos ter dois tipos de configuração. Na primeira configuração a circunferência divide o conjunto das 4 cidades em dois conjuntos: um conjunto com 3 cidades e outro com uma cidade. Na segunda configuração a circunferência divide o conjunto das cidades em dois conjuntos, cada um deles com 2 cidades.\n\nNas figuras abaixo está ilustrado um exemplo de cada uma destas configurações onde a circunferência contínua é a rodovia planejada.\n\n![](attached_image_1.png)\n\nNa primeira configuração temos que o centro da circunferência está na mesma distância das três cidades que ficam do mesmo lado da rodovia e assim o centro desta rodovia é o centro da circunferência circunscrita ao triângulo formado pelas três cidades. Logo, o número de rodovias é igual ao número de triângulos que podemos formar com as 4 cidades, isto é, 4 possíveis rodovias.\n\nNa segunda configuração, temos que o centro da circunferência formada pela rodovia está sobre a mediatriz das duas cidades que ficam na parte interna da rodovia e também sobre a mediatriz das duas cidades que ficam na parte externa da rodovia. Assim, o número de rodovias é igual ao número de formas de dividir o conjunto de 4 elementos em dois conjuntos com 2 elementos cada um, isto é 3 formas.\n\nLogo o número possível de projetos é $4+3=7$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26516,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEsportistas de uma escola - Em um grupo de $40$ estudantes, $20$ jogam futebol, $19$ jogam vôlei e $15$ jogam exatamente uns destes dois esportes. Quantos estudantes não praticam futebol e vôlei?\n(a) $7$\n(b) $5$\n(c) $13$\n(d) $9$\n(e) $10$",
    "options": [],
    "answer": "c",
    "solution": "Solution:\n\nDenotemos por $x$ o número de estudantes que praticam simultaneamente os dois esportes. Logo, temos que o número de estudantes que pratica somente futebol é $20-x$ e o que pratica somente vôlei é $19-x$. Portanto os estudantes que praticam exatamente um esporte são\n$$\n(20-x)+(19-x)=15\n$$\nSegue-se que $x=12$ e teremos que os estudantes que praticam algum esporte são\n$$\n20+(19-x)=27\n$$\nPortanto, os que não praticam esporte são $13$. A opção correta é (c).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26517,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO perfume de Rosa - Rosa ganhou um vidro de perfume no formato de um cilindro de $7~\\mathrm{cm}$ de raio da base e $10~\\mathrm{cm}$ de altura. Depois de duas semanas usando o perfume restou $0{,}45~l$ no vidro. Qual a fração que representa o volume que Rosa já usou?",
    "options": [],
    "answer": "(49π - 45)/(49π)",
    "solution": "Solution:\n\nO volume de um cilindro é o produto da área da base pela altura. Como o raio da base é $7~\\mathrm{cm}$, a área da base é: $\\pi \\times 7^{2}$, e então o volume do vidro é\n$$\n\\pi \\times 7^{2} \\times 10~\\mathrm{cm}^{3} = 490 \\pi~\\mathrm{cm}^{3} = \\frac{490 \\pi}{1000}~\\mathrm{dm}^{3} = 0,49 \\pi \\text{ litros }\n$$\nlembrando que $1000~\\mathrm{cm}^{3} = 1~\\mathrm{dm}^{3} = 1$ litro.\nDepois de duas semanas, restaram $0,45$ litros de perfume, então ela gastou $(0,49 \\pi - 0,45)$ litros. Portanto, a fração que representa o volume gasto é:\n$$\n\\frac{\\text{ volume gasto }}{\\text{ volume total }} = \\frac{0,49 \\pi - 0,45}{0,49 \\pi} = \\frac{49 \\pi - 45}{49 \\pi}\n$$",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 26518,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa campanha \"Vamos ao teatro\", 5 ingressos podem ser adquiridos pelo preço usual de 3 ingressos. Mário comprou 5 ingressos nessa campanha. A economia que Mário fez representa que percentual sobre o preço usual dos ingressos?\n(a) $20\\%$\n(b) $33 \\frac{1}{3} \\%$\n(c) $40\\%$\n(d) $60\\%$\n(e) $66 \\frac{2}{3} \\%$",
    "options": [],
    "answer": "(c)",
    "solution": "Solution:\n\nMário pagou 3 e levou 5, logo ele pagou apenas $\\frac{3}{5}$ do preço usual e portanto, economizou $\\frac{2}{5}$. Como $\\frac{2}{5}=\\frac{40}{100}$, a economia foi de $40\\%$. A opção correta é (c).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26519,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nQuais números naturais $m$ e $n$ satisfazem a $2^{n}+1=m^{2}$?",
    "options": [],
    "answer": "m=3, n=3",
    "solution": "Solution:\n\n$$\n2^{n}=m^{2}-1=(m+1)(m-1)\n$$\n\nTemos que $m-1$ e $m+1$ são potências de $2$ cuja diferença é $2$. Logo, a única solução possível é $m-1=2$ e $m+1=2^{2}$, donde $m=3$. Segue que $2^{n}+1=3^{2}$, e obtemos $n=3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26520,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFatoriais - Se $n$ é um número natural, denotamos por $n!$ o produto de todos os inteiros de 1 a $n$. Por exemplo: $5! = 1 \\times 2 \\times 3 \\times 4 \\times 5$ e $13! = 1 \\times 2 \\times 3 \\times 4 \\times 5 \\times \\ldots \\times 12 \\times 13$. Por convenção, $0! = 1$. Encontre três números inteiros diferentes $a$, $b$ e $c$, entre 0 e 9, tais que o número de três algarismos $abc$ é igual a $a! + b! + c!$.",
    "options": [],
    "answer": "145",
    "solution": "Solution:\n\nPrimeiramente observe que como o número tem 3 algarismos, então o maior dos algarismos tem que ser menor que ou igual a 6, já que $7! > 1000$. Como o número tem que ter 3 algarismos e $4! = 1 \\times 2 \\times 3 \\times 4 = 12$, então um dos algarismos tem que ser 5 ou 6, mas $6! = 720$ implicaria que a soma teria um algarismo maior ou igual a 7, logo o maior dos algarismos é 5. Por outra parte, $5! = 120$ e $5! + 4! + 3! = 120 + 24 + 6 = 150$, assim a soma dos fatoriais está entre 100 e 150, portanto o algarismo das centenas é 1. Por último, como $5! + 1! + 4! = 145$, então 145 é solução.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26521,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNuma folha de papel cabem 100 caracteres na largura e 100 na altura. Nessa folha são escritos sucessivamente os números $1, 2, 3, \\ldots$ com um espaço entre cada um. Quando no final de uma linha não há espaço para escrever um número este é escrito na linha seguinte. Qual é o ultimo número escrito na folha?",
    "options": [],
    "answer": "2220",
    "solution": "Solution:\n\nNa $1^\\text{a}$ linha escrevemos os números de $1$ a $9$, cada um seguido de um espaço, ocupando $18$ espaços, e sobram $82$ espaços. Cada número de $2$ algarismos mais um espaço ocupa $3$ lugares na linha. Como $82 = 27 \\times 3 + 1$, completamos a $1^\\text{a}$ linha com $27$ números de dois algarismos a partir do $10$. Logo o último número da primeira linha é $36$. Representando cada espaço por um traço, a $1^\\text{a}$ linha fica como\n$$\n\\underbrace{1-2-3-4-5-6-7-8-9}_{18} \\underbrace{10-\\cdots-36-}_{82}\n$$\nComo $100 = 33 \\times 3 + 1$, em cada linha podemos colocar $33$ números de $2$ algarismos, cada um seguido de um espaço, e no final da linha ainda sobra um espaço:\n$$\n2^\\text{a} \\operatorname{linha}: \\underbrace{37-38-\\cdots-69-}_{99}-\n$$\nNa $3^\\text{a}$ linha, colocamos de $70$ a $99$, ocupando $30 \\times 3 = 90$ espaços. Os $10$ espaços restantes ocupamos com dois números de $3$ algarismos:\n![](attached_image_1.png)\nAgora, em cada linha podemos colocar $100 \\div 4 = 25$ números de $3$ algarismos com seus respectivos espaços. De $102$ a $999$ inclusive temos $999 - 102 + 1 = 198$ números. Como $198 = 25 \\times 7 + 23$, ocupamos da $4^\\text{a}$ a $10^\\text{a}$ linha com os números de $3$ algarismos e ainda sobram $23$ espaços na $10^\\text{a}$ linha, que podemos ocupar com $23 \\div 5 = 4$ números de $4$ algarismos:\n$10^\\text{a}$ linha : $\\underbrace{--\\cdots-999-}_{67} \\underbrace{1000-1001-1002-}_{23}---$\nEm cada linha podemos colocar $100 \\div 5 = 20$ números de $4$ algarismos e seus respectivos espaços. Portanto, nas $90$ linhas restantes podemos colocar $90 \\times 20 = 1800$ números de $4$ algarismos. Começando com $1003$ chegaremos até o número $2802$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26522,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPara a escola de bicicleta - Cátia sai da escola todos os dias no mesmo horário e volta para casa de bicicleta. Quando ela pedala a $20~\\mathrm{km}/\\mathrm{h}$, ela chega em casa às $4:30$ horas da tarde. Se ela pedalar a $10~\\mathrm{km}/\\mathrm{h}$, ela chega em casa às $5:15$ horas da tarde. A qual velocidade ela deve pedalar para chegar em casa às $17:00$ horas?",
    "options": [],
    "answer": "12 km/h",
    "solution": "Solution:\n\nSeja $t$ o tempo que ela gasta pedalando a $20~\\mathrm{km}/\\mathrm{h}$. Pedalando a $10~\\mathrm{km}/\\mathrm{h}$, ela faz o percurso no dobro do tempo que pedalando a $20~\\mathrm{km}/\\mathrm{h}$, isto é, $2t$. No entanto, como ela demora 45 minutos a mais, temos:\n$$\n2t - t = 45 \\Longrightarrow t = 45~\\mathrm{min}\n$$\nLogo, diariamente ela sai da escola às\n$$\n4:30~h - 45~\\min = 3:45~h\n$$\ne o percurso até em casa é de\n$$\n45~\\mathrm{min} \\times 20~\\mathrm{km}/\\mathrm{h} = \\frac{3}{4} \\times 20 = 15~\\mathrm{km}\n$$\nPara percorrer $15~\\mathrm{km}$ em $5:00~h - 3:45~h = 1:15~h = \\frac{5}{4}~h$, ela deve manter uma velocidade de\n$$\n\\frac{15~\\mathrm{km}}{\\frac{5}{4}~h} = 12~\\mathrm{km}/h\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26523,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUma desigualdade - Os valores de $x$ que satisfazem $\\frac{1}{x-1}>1$ são:\n(a) $x<2$\n(b) $x>1$\n(c) $1<x<2$\n(d) $x<1$\n(e) $x>2$",
    "options": [],
    "answer": "c",
    "solution": "Solution:\n\nNote que o inverso de um número $b$ só é maior do que 1 quando $b$ for positivo e menor do que 1. Portanto,\n$$\n\\frac{1}{x-1}>1 \\Longleftrightarrow 0<x-1<1 \\Longleftrightarrow 1<x<2\n$$\nA opção correta é (c).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26524,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO caminho da pulga - Para percorrer um caminho reto de 10 metros de comprimento, uma pulga usa a seguinte estratégia: a cada dia ela percorre a metade do caminho que faltava no dia anterior. Portanto, no primeiro dia ela percorre 5 metros, no segundo 2,5 metros e assim por diante (o tamanho da pulga é desprezível).\n\na) Quantos metros ela terá percorrido ao final do sétimo dia? E do décimo?\n\nb) A partir de qual dia a pulga estará a menos de $0{,}001~\\mathrm{m}$ do final do caminho?",
    "options": [],
    "answer": "a) After 7 days: 10*(127/128) = 9.921875 m; after 10 days: 10*(1023/1024) = 9.990234375 m. b) From day 14 onward the flea is within 0.001 m of the end.",
    "solution": "Solution:\n\nNo $1^{\\mathrm{o}}$ pulo a pulga percorre $10 \\times \\frac{1}{2}$, no $2^{\\mathrm{o}}$, $10 \\times \\frac{1}{2^{2}}$, e assim por diante.\n\nOs pulos da pulga\n\n![](attached_image_1.png)\n\nDepois de 7 dias a pulga terá percorrido\n$$\n\\begin{aligned}\n& 10\\left(\\frac{1}{2}\\right)+10\\left(\\frac{1}{2}\\right)^{2}+10\\left(\\frac{1}{2}\\right)^{3}+10\\left(\\frac{1}{2}\\right)^{4}+10\\left(\\frac{1}{2}\\right)^{5}+10\\left(\\frac{1}{2}\\right)^{6}+10\\left(\\frac{1}{2}\\right)^{7}= \\\\\n& 10\\left[\\frac{1}{2}+\\left(\\frac{1}{2}\\right)^{2}+\\left(\\frac{1}{2}\\right)^{3}+\\left(\\frac{1}{2}\\right)^{4}+\\left(\\frac{1}{2}\\right)^{5}+\\left(\\frac{1}{2}\\right)^{6}+\\left(\\frac{1}{2}\\right)^{7}\\right]= \\\\\n& 10 \\times \\frac{2^{6}+2^{5}+2^{4}+2^{3}+2^{2}+2+1}{2^{7}}=10 \\times \\frac{127}{128} \\approx 9,9\n\\end{aligned}\n$$\nLogo em 7 dias ela percorreu, aproximadamente $9,9~\\mathrm{m}$. Em geral, depois de $n$ dias a pulga terá percorrido\n$$\n10\\left(\\frac{1}{2}+\\cdots+\\frac{1}{2^{n}}\\right)\n$$\nPara calcular a soma acima, note que $\\left(\\frac{1}{2}+\\cdots+\\frac{1}{2^{n}}\\right)$ é a soma dos $n$ termos de uma progressão geométrica cujo primeiro termo é $a_{1}=1 / 2$ e a razão é $q=1 / 2$. A fórmula para essa soma é:\n$$\nS_{n}=\\frac{a_{1}\\left(1-q^{n}\\right)}{1-q}=\\frac{1 / 2\\left(1-1 / 2^{n}\\right)}{1-1 / 2}=1-\\frac{1}{2^{n}}\n$$\nLogo, $10\\left(\\frac{1}{2}+\\cdots+\\frac{1}{2^{n}}\\right)=10\\left(1-\\frac{1}{2^{n}}\\right)$.\nPortanto, ao final do décimo dia a pulga terá percorrido $10\\left(1-\\frac{1}{2^{10}}\\right)~\\mathrm{m}$.\n\nA pulga estará a menos de $0{,}001~\\mathrm{m}$ do final do caminho, quando ela já tiver percorrido pelo menos $10-0{,}001=9{,}999$, ou seja quando\n$$\n10\\left(1-\\frac{1}{2^{n}}\\right) \\geq 9{,}999\n$$\nVamos determinar o menor valor de $n$ que satisfaz a desigualdade acima.\n$$\n\\begin{aligned}\n& 10\\left(1-\\frac{1}{2^{n}}\\right) \\geq 9{,}999 \\quad \\Longrightarrow 10-\\frac{10}{2^{n}} \\geq 9{,}999 \\quad \\Longrightarrow \\quad 10-9{,}999 \\geq \\frac{10}{2^{n}} \\\\\n& \\Longrightarrow 0{,}001 \\geq \\frac{10}{2^{n}} \\quad \\Longrightarrow 2^{n} \\geq \\frac{10}{0{,}001} \\quad \\Longrightarrow 2^{n} \\geq 10000\n\\end{aligned}\n$$\nAgora,\n$$\n2^{10}=2^{5} \\times 2^{5}=32 \\times 32=1024\n$$\nsegue que\n$$\n2^{13}=2^{10} \\times 2^{3}=1024 \\times 8=8192\n$$\nLogo, devemos ter $n=14$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26525,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA árvore de Emília cresce de acordo com a seguinte regra: após 2 semanas do aparecimento de um galho, esse mesmo galho produz um novo galho a cada semana, e o galho original continua a crescer. A árvore tem 5 galhos depois de 5 semanas, como mostra a figura. Quantos galhos, incluindo o galho principal, a árvore terá no final de 8 semanas?\n![](attached_image_1.png)",
    "options": [],
    "answer": "21",
    "solution": "Solution:\nDenotemos por $f_{n}$ o número de galhos da árvore depois de $n$ semanas. Como depois de duas semanas aparece um galho então $f_{2}=1$. Na seguinte semana este galho produz um novo galho, logo $f_{3}=2$. Pela regra, o número de galhos na $n+1$-ésima semana é igual ao número de galhos que existiam na $n$-ésima semana, mais os galhos novos. Mas, os galhos novos nascem dos galhos que têm pelo menos duas semanas, isto é, nasce um galho novo por cada galho que existia na semana $n-1$.\nAssim, temos que $f_{n+1}=f_{n}+f_{n-1}$. Logo:\n$$\n\\begin{aligned}\n& f_{4}=2+1=3 \\\\\n& f_{5}=3+2=5 \\\\\n& f_{6}=5+3=8 \\\\\n& f_{7}=8+5=13 \\\\\n& f_{8}=13+8=21\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26526,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA sala do Newton- Professor Newton dividiu seus alunos em grupos de $4$ e sobraram $2$. Ele dividiu seus alunos em grupos de $5$ e um aluno ficou de fora. Se $15$ alunos são mulheres e tem mais mulheres do que homens, o número de alunos homens é:\n(a) $7$\n(b) $8$\n(c) $9$\n(d) $10$\n(e) $11$",
    "options": [],
    "answer": "e",
    "solution": "Solution:\n\nComo o número de alunos homens é menor do que $15$ e das mulheres é $15$, temos\n$$\n15 < \\text{alunos homens} + \\text{alunas mulheres} < 15 + 15 = 30\n$$\nou seja: o número de alunos está entre $15$ e $30$.\n\nPor outro lado, quando dividimos por $4$ sobram $2$ alunos, então o número de alunos é par. Quando dividimos por $5$ sobra um, então o último algarismo do número é $1$ ou $6$, mas sendo par só pode ser $6$. Assim só temos dois possíveis valores: $16$ e $26$. Descartamos $16$ porque é divisível por $4$. Logo, a resposta é $26$.\nSolution:\n\nComo acima, o número de alunos está entre $15$ e $30$. Observemos que o número $6$ dividido por $4$ deixa resto $2$ e dividido por $5$ deixa resto $1$. Logo, se somamos a $6$ um múltiplo comum de $4$ e $5$, o número obtido também terá esta propriedade. O menor múltiplo comum de $4$ e $5$ é $20$, assim os possíveis valores para o número de alunos são $6, 26, 46, 66, \\ldots$ Dado que o número de alunos está entre $15$ e $30$ então a solução é $26$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26527,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO Riquinho distribuiu $R\\$ 1000,00$ reais entre os seus amigos: Antônio, Bernardo e Carlos da seguinte maneira: deu, sucessivamente, 1 real ao Antônio, 2 reais ao Bernardo, 3 reais ao Carlos, 4 reais ao Antônio, 5 reais ao Bernardo, etc. Quanto que o Bernardo recebeu?",
    "options": [],
    "answer": "345",
    "solution": "Solution:\n\nO dinheiro foi repartido em parcelas na forma\n$$\n1+2+3+\\cdots+n \\leq 1000\n$$\nComo $1+2+3+\\cdots+n$ é a soma $S_{n}$ dos $n$ primeiros números naturais a partir de $a_{1}=1$ temos:\n$$\nS_{n}=\\frac{\\left(a_{1}+a_{n}\\right) n}{2}=\\frac{(1+n) n}{2} \\leq 1000 \\Longrightarrow n^{2}+n-2000 \\leq 0\n$$\nTemos que\n$$\nn^{2}+n-2000<0 \\quad \\text{ para valores de } n \\text{ entre as raízes }\n$$\nComo a solução positiva de $n^{2}+n-2000=0$ é\n$$\nn=\\frac{-1+\\sqrt{1+8000}}{2} \\simeq 44,22\n$$\nentão $n \\leq 44$. Assim Bernardo recebeu\n$$\n2+5+8+11+\\cdots+44=\\frac{(44+2) \\cdot 15}{2}=23 \\cdot 15=345\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26528,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUma soma alternada - Se $S_{n}=1-2+3-4+5-6+\\ldots+(-1)^{n+1} n$, onde $n$ é um inteiro positivo, então $S_{1992}+S_{1993}$ é:\n(a) -2\n(b) -1\n(c) 0\n(d) 1\n(e) 2",
    "options": [],
    "answer": "d",
    "solution": "Solution:\n\nLembre que\n$$\n(-1)^{n+1}= \\begin{cases}1 & \\text{ se } n \\text{ é ímpar } \\\\ -1 & \\text{ se } n \\text{ é par }\\end{cases}\n$$\nObservemos que associando duas a duas parcelas consecutivas,\n$$\n(1-2)+(3-4)+(5-6)+\\cdots\n$$\nobtemos uma soma de $n$ parcelas todas iguais a $-1$. Logo,\n$$\nS_{1992}=\\underbrace{(1-2)+(3-4)+(5-6)+\\cdots+(1991-1992)}_{1992 \\div 2=996 \\text{ parcelas }}=(-1) \\times 996=-996.\n$$\n$$\nS_{1993}=(1-2)+(3-4)+(5-6)+\\cdots+(1991-1992)+1993=-996+1993=997.\n$$\nAssim, $S_{1992}+S_{1993}=-996+997=1$. A opção correta é (d).\nSolution:\n\nNote que\n$$\nS_{2 n}=\\underbrace{(1-2)+(3-4)+(5-6)+\\cdots+[2 n-(2 n+1)]}_{n \\text{ parcelas iguais a }-1}\n$$\nObtemos que $S_{2 n}=-n$ e $S_{2 n+1}=S_{2 n}+(2 n+1)=-n+2 n+1=n+1$. Logo, $S_{2 n}+S_{2 n+1}=1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26529,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO raio da circunferência - Um arco de circunferência mede $300^{\\circ}$ e o seu comprimento é $2~\\mathrm{km}$. Qual o número inteiro mais próximo da medida do raio em metros?\n\n(a) 157\n(b) 284\n(c) 382\n(d) 628\n(e) 764",
    "options": [],
    "answer": "c",
    "solution": "Solution:\n\nSe o raio é $r$ então o comprimento de um arco de $\\theta$ graus é $2 \\pi \\frac{\\theta}{360} r$. Assim, no problema dado, temos que\n$$\n2 \\pi \\frac{300}{360} r = 2000~m \\Longrightarrow r = 2000 \\times \\frac{3}{5 \\pi} \\simeq 382,17~m\n$$\nLogo, a opção correta é (c).\nSolution:\n\nComo a circunferência tem $360^{\\circ}$, um arco de $300^{\\circ}$ representa $\\frac{5}{6}$ da circunferência, logo, seu comprimento é $\\frac{5}{6}$ do comprimento da circunferência, isto é:\n$$\n\\frac{5}{6} \\times 2 \\pi r = 2000~m \\Longrightarrow r = \\frac{2000 \\times 6}{10 \\pi} = \\frac{1200}{\\pi} \\approx 382,17~m\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26530,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm teste vocacional - Foi feito um teste vocacional em 1000 estudantes de uma escola. A tabela a seguir apresenta os resultados por área de estudo e sexo.\n\n|             | Exatas | Humanas | Biológicas |\n|:-----------:|:------:|:-------:|:----------:|\n| Masculino   | 232    | 116     | 207        |\n| Feminino    | 112    | 153     | 180        |\n\nSe um aluno é escolhido ao acaso, determine a probabilidade de:\n\na) Ser da área de exatas.\n\nb) Ser da área de humanas, sendo do sexo masculino.\n\nc) Ser do sexo feminino, dado que é da área biológica.",
    "options": [],
    "answer": "a) 0.344; b) 116/555 ≈ 0.209; c) 180/387 ≈ 0.465",
    "solution": "Solution:\n\na) De exatas temos $232 + 112 = 344$ estudantes, logo a probabilidade de escolher ao acaso um aluno de exatas é $\\frac{344}{1000} = 0,344$.\n\nb) Como o número de estudantes do sexo masculino é $555$, temos que a probabilidade de ser da área de humanas é $\\frac{116}{555} = 0,209$.\n\nc) O número de estudantes da área biológica é $387$. Assim, a probabilidade de escolher um do sexo feminino é $\\frac{180}{387} = 0,465$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26531,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nRetângulo com dimensões inteiras - As diagonais de um retângulo medem $\\sqrt{1993}\\ \\mathrm{cm}$. Quais são suas dimensões, sabendo que elas são números inteiros?",
    "options": [],
    "answer": "43 and 12",
    "solution": "Solution:\n\nSe $a \\geq b$ são os comprimentos dos lados do retângulo, então pelo teorema de Pitágoras temos\n$$\na^{2}+b^{2}=1993\n$$\nComo $a^{2} \\geq b^{2}$, segue que\n$$\n2 a^{2} \\geq a^{2}+b^{2}=1993>a^{2}\n$$\nLogo,\n$$\n\\sqrt{1993}>a \\geq \\sqrt{996,5}\n$$\nAssim, $44 \\geq a \\geq 32$. Usando o fato que $a^{2}-(a-1)^{2}=2 a-1$ podemos completar a seguinte tabela, somando aos elementos da segunda coluna na linha $a-1$ o número $2 a-1$ para obter o elemento da segunda coluna na linha $a$.\n\n| $a$ | $b^{2}=1993-a^{2}$ |\n| :---: | :---: |\n| 44 | 57 |\n| 43 | 144 |\n| 42 | 229 |\n| $\\vdots$ | $\\vdots$ |\n\nAssim, temos que $a=43$ e $b=12$ é solução.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26532,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm jardim retangular - O retângulo $ABCD$ representa um terreno retangular cuja largura é $3/5$ do comprimento. O retângulo $ABEF$ representa um jardim retangular cuja largura é também $3/5$ do comprimento. Qual a razão entre a área do jardim e a área total do terreno?\n\n![](attached_image_1.png)\n\n(a) $30\\%$\n(b) $36\\%$\n(c) $40\\%$\n(d) $45\\%$\n(e) $50\\%$",
    "options": [],
    "answer": "b",
    "solution": "Solution:\n\nPelos dados do problema sabemos que\n$$\nAD = \\frac{5}{3} AB \\quad \\text{e} \\quad AB = \\frac{5}{3} AF\n$$\nLogo,\n$$\nAD = \\left(\\frac{5}{3}\\right)^2 AF = \\frac{25}{9} AF\n$$\nA área do terreno é $AB \\times AD$ e a área do jardim é $AB \\times AF$, portanto a razão entre as áreas é\n$$\n\\frac{AB \\times AF}{AB \\times AD} = \\frac{AF}{AD} = \\frac{AF}{\\left(\\frac{5}{3}\\right)^2 AF} = \\frac{9}{25} = 36\\%\n$$\nA opção correta é (b).",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26533,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNúmeros decrescentes - Escreva os números abaixo em ordem decrescente\n$$\n\\sqrt[5]{3}, \\quad 3^{-2 / 3}, \\quad 3^{-2}, \\quad\\left(\\frac{1}{3}\\right)^{3}, \\quad\\left(\\frac{1}{3}\\right)^{-1}\n$$",
    "options": [],
    "answer": "(1/3)^{-1}, √[5]{3}, 3^{-2/3}, 3^{-2}, (1/3)^3",
    "solution": "Solution:\n\nSabemos que\n- $3^{-2 / 3} = \\frac{1}{3^{2 / 3}} < 1$,\n- $3^{-2} = \\frac{1}{3^{2}} < 1$,\n- $\\left(\\frac{1}{3}\\right)^{3} = \\frac{1}{3^{3}} < 1$,\n- $1 < \\sqrt[5]{3} < 3$.\n\nSe $a, b$ e $c$ são não nulos e\n$$\na > b > c\n$$\nentão\n$$\n\\frac{1}{a} < \\frac{1}{b} < \\frac{1}{c}\n$$\nComo $3^{3} > 3^{2} > 3^{2 / 3}$ temos então :\n$$\n\\frac{1}{3^{3}} < \\frac{1}{3^{2}} < \\frac{1}{3^{2 / 3}} < 1 < \\sqrt[5]{3} < 3\n$$\nPortanto,\n$$\n\\left(\\frac{1}{3}\\right)^{3} < 3^{-2} < 3^{-2 / 3} < \\sqrt[5]{3} < \\left(\\frac{1}{3}\\right)^{-1}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26534,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEm um táxi podem se sentar um passageiro na frente e três atrás. De quantas maneiras podem se sentar os quatro passageiros se um deles quer ficar na janela?",
    "options": [],
    "answer": "18",
    "solution": "Solution:\n\nO passageiro que quer ficar na janela tem 3 possíveis lugares para se sentar, o seguinte pode-se sentar em qualquer lugar livre, logo tem 3 possíveis lugares, o seguinte dois possíveis lugares, e o último não tem escolha. Concluímos que o número de formas de se sentar é $3 \\times 3 \\times 2 = 18$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26535,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMúltiplos de 3 e quadrados perfeitos - Escreve-se em ordem crescente cada múltiplo de $3$ cuja soma com o número $1$ é um quadrado perfeito:\n$$\n3 ; \\quad 15 ; \\quad 24 ; 48 ; \\quad \\ldots\n$$\nQual é o múltiplo na posição $2006^{\\circ}$ ?",
    "options": [],
    "answer": "9060099",
    "solution": "Solution:\n\nChamemos $a$ um número qualquer da lista, então sabemos que:\n- $a$ é múltiplo de $3$\n- $a+1$ é um quadrado: $a+1=k^{2}$, sendo $k$ um número natural.\nAssim $a=k^{2}-1$, e logo\n$$\na=(k-1)(k+1)\n$$\nComo $a$ é divisível por $3$, então ou $k+1$ ou $k-1$ é divisível por $3$. Logo, $k$ não é divisível por $3$, portanto, $k$ tem que ser da forma $3n+1$ ou $3n+2$, ou seja, para cada valor de $n$ temos dois números que não são múltiplos de $3$.\nO número desta lista que está na posição $2006$ é $2006 \\times \\frac{3}{2}-1=3008$, e neste caso $a=3008^{2}-1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26536,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOs bombons misturados - Marta e Carmem ganharam, cada uma, muitos bombons. Elas misturaram os bombons e agora não sabem mais qual o número de bombons que cada uma ganhou. Vamos ajudá-las a descobrir os números sabendo que:\n- juntas ganharam 200 bombons;\n- cada número é múltiplo de 8;\n- Marta se lembra que ganhou menos de 100 bombons, mas mais do que $\\frac{4}{5}$ do que ganhou Carmem.",
    "options": [],
    "answer": "Marta 96, Carmem 104",
    "solution": "Solution:\n\nSejam $x$ o número de bombons que Marta ganhou e $y$ o que Carmem ganhou. Temos $x + y = 200$. Como $x < 100$ então $y \\geq 100$. Por outro lado, $x > \\frac{4}{5} y$ e $y \\geq 100$, concluímos que $x > \\frac{4}{5} \\times 100 = 80$. Logo, $x$ é um inteiro compreendido entre 80 e 100 e múltiplo de 8, logo, só pode ser 88 ou 96. Vamos decidir:\n\n- Se $x = 88$, então $y = 200 - 88 = 112$. Logo: $x > \\frac{4}{5} \\times 112 = 89,5$, o que não é possível.\n- Se $x = 96$, então $y = 200 - 96 = 104$ e $x > \\frac{4}{5} \\times 104 = 83,2$, o que é possível.\n\nLogo Marta ganhou 96 bombons e Carmem 104.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26537,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nJulia precisava calcular $6x$, mas por distração calculou $\\frac{x}{6}$. Qual foi o erro percentual cometido por Julia?",
    "options": [],
    "answer": "35/36 × 100% = 97.22%",
    "solution": "Solution:\n\nSeja $x$ o número. Julia tinha que obter $6x$ e com sua distração, obteve $\\frac{x}{6}$. Logo, seu erro foi de $6x - \\frac{x}{6} = \\frac{35x}{6}$. Portanto, em termos percentuais o erro foi de\n$$\n\\frac{\\frac{35x}{6}}{6x} = \\frac{35}{36} \\approx 0,9722 = 97,22\\%\n$$\nA opção correta é (b).\nSolution:\n\nSe $N$ é o valor que a Julia tinha que obter, então ela com seu erro encontrou $\\frac{N}{6}$. Assim, o erro absoluto cometido foi de $N - \\frac{N}{6} = \\frac{5N}{6}$. Portanto, o erro relativo foi de\n$$\n\\frac{5}{6} \\times 100\\% = 83,33\\%\n$$\n\nNo entanto, se o erro foi calcular $\\frac{N}{36}$ em vez de $N$, então o erro absoluto seria $N - \\frac{N}{36} = \\frac{35}{36}N$, e o erro relativo:\n$$\n\\frac{35}{36} \\times 100\\% = 97,22\\%\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26538,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMaria encomendou certo número de televisores a $R\\$ 1994,00$ cada um. Ela reparou que no total a pagar, não tem nem 0, nem 7, nem 8 e nem 9. Qual foi o menor número de televisores que ela encomendou?",
    "options": [],
    "answer": "56",
    "solution": "Solution:\n\nSe Maria comprou $n$ televisores, então ela gastou $1994n$, que é um múltiplo de $1994$ onde não aparecem os algarismos $0, 7, 8$ e $9$. Vamos tentar limitar o valor de $n$. Primeiro observe que\n$$\n1994 n = 2000 n - 6 n\n$$\ne também que se\n$$\n6 n < 300\n$$\nentão o número $2000 n - 6 n$ tem $7$ ou $8$ ou $9$ no algarismo das centenas (faça alguns exemplos, lembre-se que $2000 n$ termina com 3 zeros e depois convença-se). Assim devemos ter\n$$\n6 n \\geq 300, \\quad \\text{isto é } n \\geq 50\n$$\nObservemos que $50$ não pode ser porque o valor terminaria em $0$, logo $n \\geq 51$. Dado que $1994 \\times 51 = 101694$ temos que $n$ não pode ser $51$ e portanto $n = 51 + m$ com $m$ positivo. Agora como precisamos que o número não tenha $0$, assim $1994 m$ tem que eliminar o $0$ de $101694$, portanto $m \\geq 4$, mas $m = 4$ não é solução porque $1994 \\times 55$ termina em $0$. Se testamos $n = 56$ temos que\n$$\n1994 \\times 56 = 111664\n$$\né o número procurado.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26539,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAs cinco cartas abaixo estão sobre uma mesa, e cada uma tem um número numa face e uma letra na outra. Simone deve decidir se a seguinte frase é verdadeira: \"Se uma carta tem uma vogal numa face, então ela tem um número par na outra.\" Qual o menor número de cartas que ela precisa virar para decidir corretamente?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "3",
    "solution": "Solution:\n\n![](attached_image_2.png)\nEla não precisa virar a carta que tem o número $2$, porque sendo vogal ou consoante, ela cumpre a condição, de igual forma. Ela também não precisa virar a carta com a letra $M$. A carta que tem o número $3$ tem que ser virada, para comprovar que na outra face tem uma consoante, e também as cartas com a letra $A$ e a letra $E$ têm que ser viradas para verificar que os números na outra face são pares. Assim, ela precisa virar somente $3$ cartas.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26540,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOs cinco círculos - Cinco discos de mesmo raio estão dispostos como mostra a figura. Quatro centros são os vértices de um quadrado e três estão alinhados. Trace uma reta que divida a figura formada pelos 5 discos em duas partes de mesma área.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "The straight line through the square’s center and the center of the fifth disk.",
    "solution": "Solution:\n\nObservemos que qualquer linha que passe por $O$, o centro do quadrado $ABCD$, divide a área formada pelos círculos $\\mathcal{C}_1$, $\\mathcal{C}_2$, $\\mathcal{C}_3$ e $\\mathcal{C}_4$ na metade. Por outro lado, qualquer linha reta que passe por $F$ divide a área do círculo $\\mathcal{C}_5$ na metade. Assim, a reta procurada é a reta $FO$.\n\n![](attached_image_2.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26541,
    "subject": "Mathematics (Multi-modal)",
    "question": "A positive integer is *dapper* if at least one of its multiples begins with $2008$. For example, $7$ is dapper because $200858$ is a multiple of $7$ and begins with $2008$. Observe that $200858 = 28694 \\times 7$.\nProve that every positive integer is dapper.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $n$ be any positive integer. Choose $k$ to be an integer greater than the number of digits of $n$. So the interval $[2008 \\cdot 10^k, 2009 \\cdot 10^k[$, which contains $10^k > n$ integer consecutive numbers, has a multiple of $n$. So every positive integer $n$ is dapper.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26542,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $S$ be a set of $6n$ points in a line. Choose arbitrarily $4n$ of these points and paint them blue; the other $2n$ points are painted green. Prove that there exists a line segment that contains exactly $3n$ points from $S$, $2n$ of them blue and $n$ of them green.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $A_i$ be the segment with exactly $3n$ points from $S$, the leftmost being the $i$-th point from left to right of $S$, $i = 1, 2, \\dots, 3n + 1$. Define $f(i)$ as the number of blue points in $A_i$. We have to prove that $f(i) = 2n$ for some $i$.\n\nNotice that $|f(i + 1) - f(i)| \\le 1$ since $A_i$ and $A_{i+1}$ have $3n - 1$ common points and $f(1) + f(3n+1) = 4n$ because the disjoint segments $A_1$ and $A_{3n+1}$ cover $S$.\n\nIf $f(1) = 2n$ we are done. So suppose without loss of generality that $f(1) < 2n$. So $f(3n + 1) > 2n$ and, since $f(i)$ increases or decreases at most 1, there must exist a number $k$ such that $f(k) = 2n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26543,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a cyclic quadrilateral and $r$ and $s$ the lines obtained reflecting $AB$ with respect to the internal bisectors of $\\angle CAD$ and $\\angle CBD$, respectively. If $P$ is the intersection of $r$ and $s$ and $O$ is the center of the circumscribed circle of $ABCD$, prove that $OP$ is perpendicular to $CD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\nLet $E$ and $F$ be the intersections of $r$ and $s$ with the circumcircle of $ABCD$, respectively. Since $ABCD$ is cyclic and $BA$ and $BF$ are symmetric with respect to the bisector of $\\angle CBD$, $\\angle ACD = \\angle ABD = \\angle CBF = \\angle CAF$, hence $AF$ and $CD$ are parallel. Analogously, $BE$ and $CD$ are also parallel. Thus, since $CD$, $AF$ and $BE$ are parallel chords, its perpendicular bisectors coincide.\n\nThe intersection $P$ of $AE$ and $BF$ belong to this common perpendicular bisector because $\\angle BEA = \\angle EBF$. Thus this common perpendicular bisector is $OP$, which is perpendicular to $CD$ and, moreover, passes through its midpoint.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26544,
    "subject": "Mathematics (Multi-modal)",
    "question": "The venusian prophet Zabruberson sent to his pupils a 10000-letter word, each letter being $A$ or $E$: the *Zabrubic word*. Their pupils decided that, for $1 \\le k \\le 10000$, each word comprised of $k$ consecutive letters of the *Zabrubic* word is a *prophetic word* of length $k$. It is known that there are at most 7 prophetic words of length 3. Find the maximum number of prophetic words of length 10.",
    "options": [],
    "answer": "504",
    "solution": "Let $f(n)$ be the maximum number of prophetic words of length $n$. Obviously, $f(1) = 2$, $f(2) = 4$ and $f(3) = 7$. Moreover, since there are exactly $2^3 = 8$ 3-letter words, each letter being $A$ or $E$, there is a forbidden substring $W$ of length three. So we can estimate $f(n)$ from above: in fact, each of the $n$-letter words belongs to exactly one of three sets: the set of words whose last letter is different from the last letter of $W$; the set of words whose penultimate letter is different from the second letter of $W$ and whose last letter is equal to the last letter of $W$; the set of words whose third-to-last letter is different from the first letter of $W$ and whose two last letters are equal to the two last letters of $W$. There are exactly $f(n-1)$, $f(n-2)$ and $f(n-3)$ words in these sets, respectively (notice the last letter, two last letters and three last letters of the words in the respective sets are determined and that they don't impose any restriction on the other letters). So $f(n) \\le f(n-1)+f(n-2)+f(n-3)$. If $W = AAA$ it's not hard to prove that $f(n)$ is actually equal to $f(n-1) + f(n-2)+f(n-3)$. Just construct the words without a substring $AAA$ using all words that fit the description above. Substituting, we get $f(10) = 504$ and now it remains to prove that there is a Zabrubic word satisfying the conditions of the problem, but this isn't hard, too: first, concatenate all 504 10-letter prophetic words, separate them with an $E$ and complete the word with $10000 - 504 \\cdot 11$ $E$'s.\nSo the maximum number of 10-letter prophetic words is 504.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26545,
    "subject": "Mathematics (Multi-modal)",
    "question": "Emerald writes $2009^2$ integers in a $2009 \\times 2009$ table, one number in each entry of the table. She sums all the numbers in each row and in each column, obtaining 4018 sums. She notices that all sums are distinct. Is it possible that all such sums are perfect squares?",
    "options": [],
    "answer": "Yes",
    "solution": "Yes, it is. For instance, consider the table\n\n| $(3^1 - 2)^2$ | $3(2 \\cdot 3^1 - 1)$ | $0$ | $0$ | $\\ldots$ | $0$ |\n|---|---|---|---|---|---|\n| $0$ | $(3^2 - 2)^2$ | $3(2 \\cdot 3^2 - 1)$ | $0$ | $\\ldots$ | $0$ |\n| $0$ | $0$ | $(3^3 - 2)^2$ | $3(2 \\cdot 3^3 - 1)$ | $\\ldots$ | $0$ |\n| $\\ldots$ | $\\ldots$ | $\\ldots$ | $\\ldots$ | $\\ldots$ | $\\ldots$ |\n| $0$ | $0$ | $0$ | $0$ | $\\ldots$ | $(3^{2009} - 2)^2$ |\n\nThe sum of the numbers in each row except the last one is $(3^k - 2)^2 + 3(2 \\cdot 3^k - 1) = 3^{2k} + 2 \\cdot 3^k + 1 = (3^k + 1)^2$, $1 \\le k \\le 2008$, and the sum of the numbers in the last row is $(3^{2009} - 2)^2$.\n\nThe sum of the numbers in each column except the first one is $3(2 \\cdot 3^{k-1} - 1) + (3^k - 2)^2 = 3^{2k} - 2 \\cdot 3^k + 1 = (3^k - 1)^2$, $2 \\le k \\le 2009$, and the sum of the numbers in the first column is $1$.\n\nNote that all 4018 sums are distinct perfect squares, so we are done.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26546,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f: [0, 1] \\to [0, 1]$ be an increasing and derivable function that has an inverse $f^{-1}$.\nIf $\\int_0^1 f(x) dx = \\int_0^1 f^{-1}(x) dx$, prove that there exist two distinct real numbers $a$ and $b$, $0 \\le a < b \\le 1$, such that $f'(a) = f'(b) = 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First notice that, since $f$ is increasing and has an inverse, then it is bijective and hence $f(0) = 0$ and $f(1) = 1$ (if $f(0) > 0$ then $f$ would never be equal to $0$; the same applies if $f(1) < 1$). So, by applying the substitution $x = f(y)$, for which $\\frac{dx}{dy} = f'(y)$, and integration by parts,\n$$\n\\int f^{-1}(x) dx = \\int f^{-1}(f(y)) f'(y) dy = \\int y f'(y) dy = y f(y) - \\int f(y) dy\n$$\nThus\n$$\n\\int_0^1 f^{-1}(x) dx = \\left. y f(y) \\right|_0^1 - \\int_0^1 f(y) dy = f(1) - 0 - \\int_0^1 f(x) dx \\\\\n\\iff \\int_0^1 f(x) dx + \\int_0^1 f^{-1}(x) dx = 1\n$$\nand both definite integrals are equal to $\\frac{1}{2}$.\nSince $\\int_0^1 x dx = \\frac{1}{2}$ as well,\n$$\n\\int_0^1 (f(x) - x) dx = 0\n$$\nLet $g(x) = f(x) - x$. Then $g(0) = f(0) - 0 = 0$ and $g(1) = f(1) - 1 = 0$. Since $\\int_0^1 g(x) dx = 0$, there exist two real numbers $0 < t < u < 1$ such that $g(t) \\le 0$ and $g(u) \\ge 0$, and so there is $c \\in ]t, u[$ such that $g(c) = 0$. Then, by the mean value theorem (or Rolle's theorem) there exist $a \\in ]0, c[$ and $b \\in ]c, 1[$ such that $g'(a) = g'(b) = 0$. Since $g(x) = f(x) - x \\implies g'(x) = f'(x) - 1$ and so $f'(a) = f'(b) = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26547,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $N = \\{0, 1, 2, 3, \\dots\\}$. Given sets $A, B \\subset N$, for each positive integer $n$ denote $r(A, B, n)$ as the number of solutions to the equation $a + b = n$, $a \\in A, b \\in B$.\nProve that there exists $n_0 \\in N$ such that $r(A, B, n+1) > r(A, B, n)$ for all $n > n_0$ if and only if $N \\setminus A$ and $N \\setminus B$ are both finite.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First suppose that $r(A, B, n)$ is increasing for $n > n_0$. For the sake of simplicity, let $\\bar{A} = N \\setminus A$ and $\\bar{B} = N \\setminus B$. Then\n$$\nn+1 = r(N, N, n) = r(A, B, n) + r(\\bar{A}, B, n) + r(A, \\bar{B}, n) + r(\\bar{A}, \\bar{B}, n) \\\\\n\\iff r(\\bar{A}, B, n) + r(A, \\bar{B}, n) + r(\\bar{A}, \\bar{B}, n) = n+1 - r(A, B, n)\n$$\nSince $r(A, B, n)$ is increasing, there exists a constant $c$ such that $n+1 - r(A, B, n) \\le c$ for all $n$. In particular, for all $n$, $r(\\bar{A}, N, n) = r(\\bar{A}, B, n) + r(\\bar{A}, \\bar{B}, n) < c$, implying $|\\bar{A}| < c$. In fact, if $\\bar{A} = \\{a_1, a_2, a_3, \\dots, a_l, \\dots\\}$, $a_1 < a_2 < a_3 < \\dots$, for $a_k < n < a_{k+1}$, we have $r(\\bar{A}, N, n) = k$. Therefore if $\\bar{A}$ were infinite, $r(\\bar{A}, N, n)$ would be unbounded. Analogously $\\bar{B}$ must be finite as well.\n\nConversely, suppose $\\bar{A}$ and $\\bar{B}$ are both finite. For all $n > \\max(\\bar{A}) + \\max(\\bar{B})$, if $a \\in \\bar{A}$ then $n - a > \\max \\bar{B}$, so $n - a \\notin \\bar{B}$, hence $n - a \\in B$. In a similar fashion, if $b \\in \\bar{B}$ then $n - b \\notin \\bar{A}$, hence $n - b \\in A$. Thus\n$$\nr(A, B, n) = n + 1 - |\\bar{A}| - |\\bar{B}| < n + 2 - |\\bar{A}| - |\\bar{B}| = r(A, B, n + 1)\n$$\nSo $r(A, B, n)$ is increasing for $n > \\max(\\bar{A}) + \\max(\\bar{B})$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 26548,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $q = 2p+1$, $p, q > 0$ primes. Prove that there exists a multiple of $q$ whose digits sum in decimal base is positive and at most 3.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First notice that $10$ is a multiple of $5 = 2 \\cdot 2 + 1$, so we are done if $q$ and $10$ are not coprime. If $q$ and $10$ are coprime, since $q$ is prime, by Fermat's theorem $q \\mid 10^{q-1} - 1 \\Leftrightarrow q \\mid 10^{2p} - 1 \\Leftrightarrow q \\mid 10^p - 1$ or $q \\mid 10^p + 1$. If $q \\mid 10^p + 1$, we are done: the sum of the digits of $10^p + 1$ is $2$. From now on, suppose that $q \\mid 10^p - 1$.\n\nConsider the minimum positive integer $d$ such that $10^d \\equiv 1 \\pmod{q}$, that is, $d = \\text{ord}_q 10$. Then $d \\mid p$, that is, the possible values of $d$ are $1$ and $p$. If $d = 1$, $q \\mid 10^1 - 1 \\Leftrightarrow q = 3$, which is not acceptable. So $d = p$, which means that $1, 10, 10^2, \\dots, 10^{p-1}$ are all distinct mod $q$.\n\nNow let $A = \\{0\\} \\cup \\{10^k \\text{ mod } q : 0 \\le k < p\\}$ and $B = \\{q-1\\} \\cup \\{-1 - 10^k \\text{ mod } q : 0 \\le k < p\\}$. Since $10^k \\not\\equiv 0 \\pmod{q}$, $|A| = |B| = p+1$. This and the obvious fact that $|A \\cup B| \\le q = 2p+1$ implies $A \\cap B \\ne \\emptyset$.\n\nSo either there are integers $r, s$ such that $10^r \\equiv -1 - 10^s \\pmod{q}$, in which case $10^r + 10^s + 1$, whose digit sum is $3$, is a multiple of $q$, or $10^r \\equiv -1 \\pmod{q}$, in which $10^r + 1$, whose digit sum is $2$, is a multiple of $q$, or $0 \\equiv -1 - 10^s \\pmod{q}$, which is analogous to the previous case. So, in all cases, we are done.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26549,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$ be one of the intersection points of two circles with centers $X$ and $Y$. The tangent lines to these circles passing through $A$ meet the circles again at $B$ and $C$. Let $P$ be the point in the plane such that $PXAY$ is a parallelogram. Prove that $P$ is the circumcenter of the triangle $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The tangent lines passing through $A$ are perpendicular to $AX$ and $AY$, so $AX \\perp AC$ and $AY \\perp AB$. Since $AXPY$ is a parallelogram, $PX$ is parallel to $AY$, so $PX \\perp AB$. Since $AB$ is a chord of a circle with center $X$, $PX$ is the perpendicular bisector of $AB$. Analogously, $PY$ is the perpendicular bisector of $AC$. So $P$ is the intersection of the perpendicular bisectors of $AB$ and $AC$.\n\nand, therefore, circumcenter of the triangle $ABC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26550,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $2009$ pebbles in some points $(x, y)$ with both coordinates integer. An operation consists in choosing a point $(a, b)$ with four or more pebbles, removing four pebbles from $(a, b)$ and putting one pebble in each of the points\n$$\n(a, b - 1), (a, b + 1), (a - 1, b), (a + 1, b)\n$$\nShow that after a finite number of operations each point will necessarily have at most three pebbles. Prove that the final configuration doesn't depend on the order of the operations.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For each configuration that can be obtained, let $S$ be the sum of the squares of the coordinates of the pebbles. Each operation takes four pebbles at $(x, y)$ to one pebble at each of the points $(x-1, y)$, $(x+1, y)$, $(x, y-1)$, $(x, y+1)$, so each operation increases $S$ by\n$$(x-1)^2 + y^2 + (x+1)^2 + y^2 + x^2 + (y-1)^2 + x^2 + (y+1)^2 - 4(x^2 + y^2) = 4.$$ \nIt is easy to see that the baricenter $G$ of the pebbles doesn't change.\n\nLet $G$ be the graph whose vertices are the pebbles and we connect two vertices if the distance between the corresponding pebbles is less than $M$, $M > 2$ sufficiently large to make the graph connected. One can easily check that the graph remains connected after performing the operation: in fact, if the distance between two pebbles $a, b$ surpasses $M$, then one of the pebbles, say $a$, was moved; then one of the other three pebbles moved gets closer to $b$, and is at a distance at most $2$ from $a$, so there is still a path leading $a$ to $b$. This, added to the fact that $G$ is invariant, proves that the pebbles cannot go too far from $G$; indeed, all pebbles are contained in a disk with center $G$ and radius $2009M$.\n\nSince $S$ always increases by $4$, the number of operations cannot be arbitrarily large, because if it isn't the case, $S$ would go to infinity, and one of the pebbles' coordinates would go to infinity as well, which is not possible. Hence the number of operations is always finite.\n\nNow let's prove that the final configuration doesn't depend on the order of the operations. Suppose that it does and consider, of all possible configurations that can be obtained, one with $S$ maximum that can lead to more than one final configuration. This means that all configurations with higher values of $S$ can only lead to one final configuration. Then one can perform the operation at two distinct points $A$ and $B$, leading to configurations $A$ and $B$, respectively, that lead to different final configurations. Note that $A$ and $B$ have higher values of $S$, so each one leads to its only one final configuration, no matter what is the order of the following operations. But one can perform the operation at $B$ in $A$ and at $A$ in $B$, and these operations lead to the same configuration, which is a contradiction. So the final configuration doesn't depend on the order of the operations.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26551,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that there are no positive integers $x$ and $y$ such that $x^3 + y^3 = 2^{2009}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "One can verify that $x^3 \\equiv 0, 1, -1 \\pmod{7}$. Since $2^{2009} = (2^3)^{669} \\cdot 2^2 \\equiv 1^{669} \\cdot 4 \\equiv 4 \\pmod{7}$ and $x^3 + y^3 \\equiv -2, -1, 0, 1, 2 \\pmod{7}$, it is not possible that $x^3 + y^3 \\equiv 2^{2009} \\pmod{7}$, so the equation $x^3 + y^3 = 2^{2009}$ has no solutions.\n\n**Comment:** One can solve the problem without the aid of arithmetic mod $n$: let $x = d \\cdot x'$ and $y = d \\cdot y'$, where $d = \\gcd(x, y)$. So $x^3 + y^3 = d^3(x'^3 + y'^3) = d^3((x'+y')(x'^2 - x'y'+y'^2))$ and one can prove that $\\gcd(x'+y', x'^2-x'y'+y'^2) = \\gcd(x'+y', (x'+y')^2 - 3x'y') = \\gcd(x'+y', 3x'y')$ equals 1 or 3. Since $x^3+y^3$ is not divisible by 3, $\\gcd(x' + y', x'^2 - x'y' + y'^2) = 1$ and, considering that both $x' + y'$ and $x'^2 - x'y' + y'^2$ are powers of 2, $x' + y' = 1$, which is not possible because $x'$, $y'$ are positive integers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26552,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given positive integers $n$, $a_1$, $a_2$, $\\dots$, $a_n$, define $q_0 = 1$, $q_1 = a_1$ and $q_{k+1} = a_{k+1}q_k + q_{k-1}$, for $1 \\le k \\le n-1$.\nProve that, given $c > 1$, there exists $K > 0$ such that, for all $M > K$, there exist a positive integer $n$ and $a_1, a_2, \\dots, a_n \\in \\{1, 2\\}$ such that $M \\le q_n < c \\cdot M$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will choose two large positive integers $r$, $s$, and take $m = r + s$, $a_j = 1$ for $1 \\le j \\le r$ and $a_j = 2$ for $r + 1 \\le j \\le r + s = m$.\nWe have $q_{k+1} = q_k + q_{k-1}$, for $1 \\le k \\le r-1$, and so $q_j = F_{j+1}$, for $0 \\le j \\le r$, where\n$$\nF_j = \\frac{1}{\\sqrt{5}} \\left( \\left( \\frac{1+\\sqrt{5}}{2} \\right)^j - \\left( \\frac{1-\\sqrt{5}}{2} \\right)^j \\right) = \\frac{1+o(1)}{\\sqrt{5}} \\left( \\frac{1+\\sqrt{5}}{2} \\right)^j\n$$\nis the $j$-th term of Fibonacci's sequence, for $j \\ge 1$. So $q_j = \\frac{1+o(1)}{\\sqrt{5}} (\\frac{1+\\sqrt{5}}{2})^{j+1}$ for large $j$.\n\nOn the other hand, we have $q_{k+1} = 2q_k + q_{k-1}$ for $r \\le k \\le m-1$, and so $q_{r+j} = u_{j+1}q_r + u_j q_{r-1}$, where $(u_j)_{j \\ge 0}$ is the sequence given by $u_0 = 0$, $u_1 = 1$ and $u_{k+2} = 2u_{k+1} + u_k$, for $k \\ge 0$. Since\n$$\nu_k = \\frac{1}{2\\sqrt{2}}((1+\\sqrt{2})^k - (1-\\sqrt{2})^k) = \\frac{1+o(1)}{2\\sqrt{2}}(1+\\sqrt{2})^k, \\quad \\text{for } k \\ge 0,$$\nwe get\n$$\n\\begin{align*}\nq_{r+j} &= \\frac{1+o(1)}{2\\sqrt{2}}((1+\\sqrt{2})q_r + q_{r-1})(1+\\sqrt{2})^j \\\\\n&= \\frac{1+o(1)}{2\\sqrt{10}}\\left(1+\\sqrt{2}+\\frac{\\sqrt{5}-1}{2}\\right)(1+\\sqrt{2})^j\\left(\\frac{1+\\sqrt{5}}{2}\\right)^{r+1} \\\\\n&= (1+o(1))\\frac{4+\\sqrt{10}+\\sqrt{2}}{8\\sqrt{5}}(1+\\sqrt{2})^j\\left(\\frac{1+\\sqrt{5}}{2}\\right)^{r+1},\n\\end{align*}\n$$\nprovided that $j$ and $r$ are large.\n\nSince $\\log(1 + \\sqrt{2}) / \\log(\\frac{1+\\sqrt{5}}{2})$ is irrational, the result follows (by taking logarithms) from the elementary fact below:\nGiven $\\alpha$, $\\beta > 0$ such that $\\alpha/\\beta$ is irrational, $\\epsilon > 0$ and $r > 0$, there is $x_0 > 0$ such that, for every $x \\in \\mathbb{R}$, $x \\ge x_0$, there are positive integers $m$, $n \\ge r$ such that $|m\\alpha + n\\beta - x| < \\epsilon$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26553,
    "subject": "Mathematics (Multi-modal)",
    "question": "Carlitos has several pieces formed by four unit squares, shaped as an L:\n![](attached_image_1.png)\nHe assembles bigger figures with these pieces, making them share one or more sides of the little squares. In the following example, the figure in the left was assembled by two pieces sharing a unit side. The figures are not allowed to have holes.\n![](attached_image_2.png)\n![](attached_image_3.png)\n\na) Draw a figure with perimeter $14$.\n\nb) Describe how is it possible to obtain a figure with perimeter $2010$.\n\nc) Is it possible to obtain a figure with odd perimeter? Justify your answer.",
    "options": [],
    "answer": "a) One can arrange the pieces to obtain a figure with perimeter fourteen (for example, as in the provided sample drawing). b) Construct a rectangle of four by one thousand one using the pieces and remove a two by two corner notch; this yields perimeter two thousand ten. c) No; an odd perimeter is impossible because every such figure has even perimeter.",
    "solution": "a) For example,\n![](attached_image_4.png)\n(of course, there are other possibilities)\n\nb) For example,\n![](attached_image_5.png)\nwhich consists of a rectangle with dimensions $4 \\times 1001$ minus a square of side $2$.\n\nc) No, it's not possible, because $2009$ is odd. Each piece has perimeter $10$, which is even, and each junction between a piece and a figure adds to the perimeter $10$ minus twice the number of the common sides of the piece and the figure. Hence the perimeter of every figure is always even.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 26554,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve, in real numbers, the system\n$$\n\\begin{aligned}\nx + \\frac{1}{y} = y + \\frac{1}{z} = z + \\frac{1}{x} \\\\\nxyz = 1\n\\end{aligned}\n$$",
    "options": [],
    "answer": "(x,y,z) = (1,1,1) or (x,y,z) = (-(1+t)/t, t, -1/(1+t)) for any real t ≠ 0, −1, together with all cyclic permutations of this triple.",
    "solution": "Since $xyz = 1$,\n$$\n\\begin{aligned}\nx + \\frac{1}{y} &= y + \\frac{1}{z} \\iff xyz + z = y^2z + y \\\\\n\\iff 1 - y = -z(1 - y^2) \\iff y = 1 \\text{ or } z = -\\frac{1}{1+y}\n\\end{aligned}\n$$\nIf $y = 1$, the system reduces to $xz = 1 \\iff z = \\frac{1}{x}$ and\n$$\nx + 1 = 1 + \\frac{1}{z} = z + \\frac{1}{x} \\iff x + 1 = 1 + x = \\frac{2}{x} \\iff x = 1 \\text{ or } x = -2\n$$\nIf $z = -\\frac{1}{1+y}$, then $xyz = 1 \\iff x = -\\frac{1+y}{y}$ and\n$$\n-\\frac{1+y}{y} + \\frac{1}{y} = y - (1+y) = -\\frac{1}{1+y} - \\frac{y}{1+y}\n$$\n(Solutions)\n---\nwhich is true for every $y \\neq 0, -1$ (all three expressions are equal to $-1$).\nSo all solutions are $(1, 1, 1)$ and $(-\\frac{1+t}{t}, t, -\\frac{1}{1+t})$ and its cyclic analogous triples. Notice that $t=1$ yields the solution $(-2, 1, -\\frac{1}{2})$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26555,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $H$ be the hyperboloid $3x^2 + 3y^2 - z^2 - 1 = 0$.\n(a) Prove that every point $(x, y, z) \\in H$ belong to exactly two lines contained in $H$.\n(b) Prove that all lines contained in $H$ form the same angle with the plane $z = 0$, and find that angle.",
    "options": [],
    "answer": "60 degrees",
    "solution": "The solution is based on the following\n**Theorem.** Let $P = (x_0, y_0, 0)$ be a point in the xy-plane and $(a, b, 1)$ a vector perpendicular to the vector $v = (x_0, y_0, 0)$, that is, such that $a x_0 + b y_0 = 0$.\nThen the set of points obtained by rotating the line $P + t \\cdot v$, $t \\in \\mathbb{R}$, around the $z$-axis is the hyperboloid\n$$\nx^2 + y^2 - (a^2 + b^2)z^2 = x_0^2 + y_0^2.\n$$\nMoreover, each point of the hyperboloid belong to exactly one of the rotated lines.\n**Proof.** A generic point in the line $\\ell: P + t \\cdot v$ is $Q = (x_0 + a t, y_0 + b t, t)$, $t \\in \\mathbb{R}$. Since\n$$\n(x_0 + a t)^2 + (y_0 + b t)^2 - (a^2 + b^2) t^2 \\\\\n= x_0^2 + y_0^2 + 2(a x_0 + b y_0)t + a^2 t^2 + b^2 t^2 - (a^2 + b^2)t^2 = x_0^2 + y_0^2,\n$$\n$Q$ belongs to the aforementioned hyperboloid. Since rotating the line around the $z$-axis does not change the $z$-coordinate of $Q$ and the section of the hyperboloid in the plane $z = t$ is a circle, the image of $Q$ obtained by rotating $\\ell$ belongs to the hyperboloid as well.\nConversely, if $(x_1, y_1, z_1)$ belongs to the hyperboloid, then it belongs to the plane $z = z_1$, whose section in the hyperboloid has equation $x^2 + y^2 = (a^2 + b^2)z_1^2 + x_0^2 + y_0^2 = (a^2 + b^2)z_1^2 + x_0^2 + y_0^2 + 2a x_0 z_1 + 2b y_0 z_1 = (x_0 + a z_1)^2 + (y_0 + b z_1)^2$. So there exists a rotation $\\sigma$ around the $z$-axis that maps $(x_1, y_1, z_1)$ to the point $Q = (x_0 + a z_1, y_0 + b z_1, z_1)$ from $\\ell$. So $(x_1, y_1, z_1)$ can be obtained by applying the inverse rotation $\\sigma^{-1}$ to $\\ell$. Note that $\\sigma$ is unique, so only one of the rotated lines passes through $Q$.\nNow we can solve the problem. In this case, since the hyperboloid has equation $3x^2 + 3y^2 - z^2 - 1 = 0 \\iff x^2 + y^2 - \\frac{1}{3}z^2 = \\frac{1}{3}$ we can use $(x_0, y_0, 0) = (\\frac{\\sqrt{3}}{3}, 0, 0)$ and $(a, b, 1) = (0, \\frac{\\sqrt{3}}{3}, 1)$.\n\na. Since $(-a, -b, 1) = (0, -\\frac{\\sqrt{3}}{3}, 1)$, which is not parallel to $(a, b, 1)$, is also perpendicular to $(x_0, y_0, 0)$, the hyperboloid can be obtained rotating either the line $\\ell_1: (\\frac{\\sqrt{3}}{3}, 0, 0) + t(0, \\frac{\\sqrt{3}}{3}, 1)$ or $\\ell_2: (\\frac{\\sqrt{3}}{3}, 0, 0) + t(0, -\\frac{\\sqrt{3}}{3}, 1)$. So each point from $H$ belongs to at least two lines. We will prove that there are no other lines by showing that $H$ does not contain a line that is neither parallel to $(0, \\frac{\\sqrt{3}}{3}, 1)$ nor $(0, -\\frac{\\sqrt{3}}{3}, 1)$. Suppose it does. Since the sections of $H$ by horizontal planes are circles, this new line is not parallel to the $xy$-plane. So it contains a point in this plane, which we can suppose, without loss of generality, that is $(\\frac{\\sqrt{3}}{3}, 0, 0)$.\nSo suppose the line $r: (\\frac{\\sqrt{3}}{3}, 0, 0) + t(c, d, 1)$ is contained in $H$. So, for all $t \\in \\mathbb{R}$,\n$$\n\\begin{aligned}\n& 3 \\left( \\frac{\\sqrt{3}}{3} + t c \\right)^2 + 3 \\cdot (t d)^2 - t^2 - 1 = 0 \\\\\n\\Leftrightarrow & (3c^2 + 3d^2 - 1)t^2 + \\frac{2\\sqrt{3}}{3} t c = 0 \\\\\n\\Leftrightarrow & \\begin{cases} \\frac{2\\sqrt{3}}{3}c = 0 \\\\ 3c^2 + 3d^2 - 1 = 0 \\end{cases} \\quad \\Leftrightarrow \\quad \\begin{cases} c = 0 \\\\ d = \\pm \\frac{\\sqrt{3}}{3} \\end{cases}\n\\end{aligned}\n$$\nSo the support vector of $r$ should be $(0, \\frac{\\sqrt{3}}{3}, 1)$ or $(0, -\\frac{\\sqrt{3}}{3}, 1)$, and we are done.\n\nb. We need to compute the angle that the vectors $(0, \\pm\\frac{\\sqrt{3}}{3}, 1)$ define with the $xy$-plane: $\\arctan \\frac{1}{\\sqrt{3}/3} = 60^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26556,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that there exists a positive integer $n_0$ with the following property: for each integer $n \\ge n_0$ it is possible to partition a cube into $n$ smaller cubes.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider the following two operations, A and B: A consists in cutting a cube in 8 equal cubes of half its dimensions; B consists in cutting a cube in 27 equal cubes of one third of its dimensions. A and B increases the total number of cubes in 7 and 26, respectively, so one can obtain $1 + 7x + 26y$ cubes by performing A $x$ times and B $y$ times. Since $\\text{gcd}(7, 26) = 1$ and $x, y$ are non-negative integers, one can obtain any number of cubes bigger than $1 + 7 \\cdot 26 - 7 - 26$, so one can choose $n_0 = 1 + 7 \\cdot 26 - 7 - 26 + 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26557,
    "subject": "Mathematics (Multi-modal)",
    "question": "An ant crawls in plane as follows: it initially crawls 1 cm in either direction. Then, after each step, it turns $60^\\circ$ to the left or to the right and crawls 1 cm in the new direction.\n![](attached_image_1.png)\nIs it possible for it to return to its starting point in\n(a) 2008 steps?\n(b) 2009 steps?",
    "options": [],
    "answer": "a) Yes. b) No.",
    "solution": "Let $O$ be the starting point of the ant. Then the ant will walk on the following hexagonal lattice:\n![](attached_image_2.png)\nColor the vertices in the lattice alternatively in black and white, as shown in the diagram. Then each step leads the ant from a black point to a white point or vice-versa. Hence the ant can only come back to $O$ after an even number of steps, and then the answer to (b) is **no**.\n\nThe ant can do 6-step round paths (a regular hexagon) and 10-step round paths (a non-convex decagon obtained by joining two regular hexagons by a common side). Since $2008 = 6 \\cdot 334 + 4 = 6 \\cdot 332 + 2 \\cdot 10$, the ant can do 332 6-step round paths and 2 10-step round paths, coming back to $O$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26558,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{Z} \\to \\mathbb{Z}$ such that\n(i) $f(f(n)) = f(n+1)$ for all $n \\in \\mathbb{Z}$;\n(ii) $f(2009n + 2008) = 2009 \\cdot f(n)$ for all $n \\in \\mathbb{Z}$.",
    "options": [],
    "answer": "All solutions are: (1) f(n) = n + 1 for all integers n; (2) f(n) = 0 for all integers n; (3) f(n) = n + 1 for n < 0 and f(n) = 0 for n ≥ 0.",
    "solution": "If $f$ is injective, then $f(f(n)) = f(n+1)$ reduces to $f(n) = n+1$, one of the solutions, another solution is $f(n) = 0$ for all $n \\in \\mathbb{Z}$. So suppose $f$ is neither injective nor identically zero. So there are two integers $a < b$ such that $f(a) = f(b)$. By applying (i) we obtain $f(a+1) = f(f(a)) = f(f(b)) = f(b+1)$ and one can deduce by induction that $f(a+n) = f(b+n)$, i.e., $f(n)$ is periodic for $n \\ge a$ and therefore bounded.\n\nWe will prove that $f(n) = 0$ for $n \\ge -1$. Suppose the contrary and let $k \\ge n$ be such that $f(k) = M$ has maximum absolute value. By (ii), $|f(2009k + 2008)| = |2009f(k)| = 2009|M|$. So either $n > 2009k + 2008 \\ge 2009n + 2008 \\implies n < -1$ or $M = 0$. Since $n \\ge -1$, $M = 0$ and so $f(N) = 0$ for all $N \\ge n$, and in particular $f(n) = 0$.\n\nNow suppose that there exist $a < b$ such that $f(a) = f(b) \\ne 0$. Then, again by the argument above $f(n)$ is periodic for $n \\ge a$ and so $f(a) = f(b) = 0$, a contradiction. So if $f(a), f(b) \\ne 0$ then $f(a) \\ne f(b)$.\n\nNow we will prove that the set of negative integers $t$ such that $f(t) = 0$ is finite. Suppose it's not, so for every integer $k$ there is an integer $r$ such that $f(r) = 0$ and $r < k$. By applying (i) repeatedly just like before we prove that $f(n) = 0$ for all $n \\ge r$; in particular $f(k) = 0$, and the function is identically zero. So there is an maximum integer $c$ such that $f(n) \\ne 0$ for all $n \\le c$. Since $f$ is injective in $\\mathbb{Z} \\cap ]-\\infty, c[$, for $n < c$ we have $f(f(n)) = f(n+1) \\ne 0 \\implies f(n) = n+1$. Now if $n < -1$ we have $2009n+2008 < n$; for example, $2009c+2008 < c$. Then $f(2009c+2008) = 2009f(c) \\iff 2009c + 2009 = 2009f(c) \\iff f(c) = c+1$. Finally, from $2009(c+1) + 2008 < c+1 \\implies 2009(c+1) + 2008 \\le c$ we have $f(2009(c+1)+2008)-2009f(c+1) \\iff f(c+1) = c+2$. But $f(c+1) = 0$, so $c = -2$. So we have a third solution\n$$\nf(n) = \\begin{cases} n+1, & \\text{if } n < 0 \\\\ 0, & \\text{if } n \\ge 0 \\end{cases}\n$$\nIt can be easily verified that this solution also satisfies the required conditions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26559,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be a triangle and $O$ its circumcenter. Lines $AB$ and $AC$ meet the circumcircle of $OBC$ again in $B_1 \\neq B$ and $C_1 \\neq C$, respectively, lines $BA$ and $BC$ meet the circumcircle of $OAC$ again in $A_2 \\neq A$ and $C_2 \\neq C$, respectively, and lines $CA$ and $CB$ meet the circumcircle of $OAB$ in $A_3 \\neq A$ and $B_3 \\neq B$, respectively. Prove that lines $A_2A_3$, $B_1B_3$ and $C_1C_2$ have a common point.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "One can guess, by drawing a good diagram, that the common point is $O$ and that the lines $A_2A_3$, $B_1B_3$ and $C_1C_2$ are the perpendicular bisectors of the triangle $ABC$. So it is sufficient to prove that $A_2A_3$ is the perpendicular bisector of $BC$. For the sake of simplicity, let $\\angle A = \\angle BAC$, $\\angle B = \\angle ABC$ and $\\angle C = \\angle ACB$. Then, considering that the perpendicular bisector of $BC$ also bisects the angle $\\angle BOC$, it suffices to prove that the angle between either the lines $OA_2$, $OA_3$ and either the lines $OB$, $OC$ is $\\angle A$ or $180^\\circ - \\angle A$.\n\nBy the definition of $A_2$ and $A_3$, $O, A, A_2, C$ lies on the same circle, as well as $O, A, A_3, B$. Now, if $A$ and $A_2$ are opposite vertices in the quadrilateral $AOA_2C$ then $\\angle A_2OC = \\angle A_2AC$ and, since $A_2$ lies on $AB$, $\\angle A_2AC = \\angle A$ or $\\angle A_2AC = 180^\\circ - \\angle A$. Either way, we are done. If $A$ and $A_2$ are neighbors in the quadrilateral $AA_2OC$, then $\\angle A_2OC = 180^\\circ - \\angle A_2AC$ and we are done again. Finally, if $A$ and $A_2$ are neighbors in the quadrilateral $AA_2CO$, then $\\angle A_2OC = \\angle A_2AC$ and we are done. The result follows analogously to $O, A, A_3, B$, so all cases are covered.\nAnother solution can be found by applying an inversion with respect to the circumcircle of $ABC$. It can be shown that it maps $A_2$ to $A_3$, $B_1$ to $B_3$ and $C_1$ to $C_2$, proving directly that all lines $A_2A_3$, $B_1B_3$ and $C_1C_2$ pass through the center of inversion $O$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26560,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n > 3$ be a fixed integer and $x_1, x_2, \\dots, x_n$ be positive real numbers. Find, in terms of $n$, all possible real values of\n$$\n\\frac{x_1}{x_n + x_1 + x_2} + \\frac{x_2}{x_1 + x_2 + x_3} + \\dots + \\frac{x_{n-1}}{x_{n-2} + x_{n-1} + x_n} + \\frac{x_n}{x_{n-1} + x_n + x_1}\n$$",
    "options": [],
    "answer": "(1, floor(n/2))",
    "solution": "The answer is all real numbers in the interval $]1, \\lfloor n/2 \\rfloor[$. For simplicity, let\n$$\nE = \\frac{x_1}{x_n + x_1 + x_2} + \\frac{x_2}{x_1 + x_2 + x_3} + \\frac{x_3}{x_2 + x_3 + x_4} + \\dots + \\frac{x_n}{x_{n-1} + x_n + x_1}.\n$$\nLet's prove first the lower bound. Let $S = x_1 + x_2 + \\dots + x_n$. First notice that\n$$\nE > \\frac{x_1}{S} + \\frac{x_2}{S} + \\frac{x_3}{S} + \\dots + \\frac{x_n}{S} = \\frac{x_1 + x_2 + \\dots + x_n}{S} = 1.\n$$\nBy making, say, $x_i = \\epsilon^{i-1}$, we obtain $\\frac{x_i}{x_{i-1}+x_i+x_{i+1}} = \\frac{\\epsilon}{1+\\epsilon+\\epsilon^2}$ for $1 < i < n$, $\\frac{x_1}{x_n+x_1+x_2} = \\frac{1}{\\epsilon^{n-1}+1+\\epsilon}$ and $\\frac{x_n}{x_{n-1}+x_n+x_1} = \\frac{\\epsilon^{n-1}}{\\epsilon^{n-2}+\\epsilon^{n-1}+1}$. Thus, by making $\\epsilon$ very small we obtain $E$ arbitrarily close to 1.\n\nNow, for the upper bound, notice that, for $n$ even,\n$$\nE < \\frac{x_1}{x_1 + x_2} + \\frac{x_2}{x_1 + x_2} + \\frac{x_3}{x_3 + x_4} + \\frac{x_4}{x_3 + x_4} + \\dots + \\frac{x_n}{x_{n-1} + x_n} = \\frac{n}{2} = \\lfloor \\frac{n}{2} \\rfloor\n$$\nFor $n$ odd, suppose without loss of generality that the minimum of the denominators $x_n + x_1 + x_2, x_1 + x_2 + x_3, \\dots, x_{n-1} + x_n + x_1$ is $x_1 + x_2 + x_3$. Thus\n$$\n\\frac{x_1}{x_n + x_1 + x_2} + \\frac{x_2}{x_1 + x_2 + x_3} + \\frac{x_3}{x_2 + x_3 + x_4} \\le \\frac{x_1}{x_1 + x_2 + x_3} + \\frac{x_2}{x_1 + x_2 + x_3} + \\frac{x_3}{x_1 + x_2 + x_3} = 1\n$$\nThis implies\n$$\n\\begin{aligned}\nE < 1 &+ \\frac{x_4}{x_4+x_5} + \\frac{x_5}{x_4+x_5} + \\frac{x_6}{x_6+x_7} + \\frac{x_7}{x_6+x_7} + \\dots + \\frac{x_{n-1}}{x_{n-1}+x_n} + \\frac{x_n}{x_{n-1}+x_n} \\\\\n= 1 &+ \\frac{n-3}{2} = \\lfloor \\frac{n}{2} \\rfloor\n\\end{aligned}\n$$\nNow, to attain the upper bound, choose $x_{2k} = \\epsilon$ and $x_{2k-1} = 1$, $k = 1, 2, \\dots, \\lfloor n/2 \\rfloor$. For $n$ even, we have the summands $\\frac{x_{2k-1}}{x_{2k-2}+x_{2k-1}+x_{2k}} = \\frac{1}{1+2\\epsilon}$ and $\\frac{x_{2k}}{x_{2k-1}+x_{2k}+x_{2k+1}} = \\frac{\\epsilon}{2+\\epsilon}$ and $E$ gets arbitrarily close to $n/2$ as $\\epsilon$ gets small. For $n$ odd, notice that $x_1 = x_n = \\epsilon$, so $\\frac{x_{2k-1}}{x_{2k-2}+x_{2k-1}+x_{2k}} = \\frac{1}{1+2\\epsilon}$ and $\\frac{x_{2k}}{x_{2k-1}+x_{2k}+x_{2k+1}} = \\frac{\\epsilon}{2+\\epsilon}$ and $E$ gets arbitrarily close to $\\lfloor n/2 \\rfloor$ as $\\epsilon$ gets small.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26561,
    "subject": "Mathematics (Multi-modal)",
    "question": "For each positive integer $n$ let $f(n)$ be the number of products of integers bigger than $1$ whose result is at most $n$, i.e. $f(n)$ is the number of $k$-uples $(a_1, a_2, \\dots, a_k)$ where $k$ is a natural number, $a_i \\ge 2$ is an integer for all $i$ and $a_1 \\cdot a_2 \\cdot \\dots a_k \\le n$ (include, by convention, the empty $0$-uple $(\\cdot)$, whose product is $1$).\n\nThus, for example, $f(1) = 1$ because of the $0$-uple $(\\cdot)$ and $f(6) = 9$, because of the $0$-uple $(\\cdot)$, the $1$-uples $(2), (3), (4), (5)$ and $(6)$ and the $2$-uples $(2, 2), (2, 3)$ and $(3, 2)$.\n\nLet $\\alpha > 1$ such that $\\sum_{m=1}^{\\infty} \\frac{1}{m^{\\alpha}} = 2$.\n\na. Prove that there exists a constant $K > 0$ such that $f(n) \\le K \\cdot n^{\\alpha}$ for all positive integers $n$.\n\nb. Prove that there exists a constant $c > 0$ such that $f(n) \\ge c \\cdot n^{\\alpha}$ for all positive integers $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Extend the definition of $f$ to real numbers, that is, $f(x)$ is the number of $k$-uples whose product is at most $x$, $x \\in \\mathbb{R}$. If the last number in a $k$-uple is $m$, then we obtain a product that doesn't exceed $x/m$. Conversely, given a number $m$ and a product that doesn't exceed $x/m$ we obtain a $k$-uple whose last number is $m$. Adding the empty $0$-uple, we obtain\n$$\nf(x) = 1 + f\\left(\\frac{x}{2}\\right) + f\\left(\\frac{x}{3}\\right) + f\\left(\\frac{x}{4}\\right) + \\dots + f\\left(\\frac{x}{\\lfloor x \\rfloor}\\right)\n$$\n\na. Induct on $n$. We don't need to worry about all real numbers because $f(x) = f(\\lfloor x \\rfloor)$. Suppose $f(x) \\le K \\cdot x^{\\alpha}$, $\\alpha$ defined as in the problem statement, $x < n$. One can find a suitable $K$ for $n = 1$ (for instance, $K = 1$). Then\n$$\n\\begin{align*} \nf(x) &= f\\left(\\frac{x}{2}\\right) + f\\left(\\frac{x}{3}\\right) + f\\left(\\frac{x}{4}\\right) + \\dots + f\\left(\\frac{x}{\\lfloor x \\rfloor}\\right) \\\n&< K \\left(\\frac{x}{2}\\right)^{\\alpha} + K \\left(\\frac{x}{3}\\right)^{\\alpha} + K \\left(\\frac{x}{4}\\right)^{\\alpha} + \\dots \\\n&= K x^{\\alpha} \\left( \\left(\\frac{1}{2}\\right)^{\\alpha} + \\left(\\frac{1}{3}\\right)^{\\alpha} + \\left(\\frac{1}{4}\\right)^{\\alpha} + \\dots \\right) \\\n&= K x^{\\alpha} \\left( \\sum_{m=1}^{\\infty} \\frac{1}{m^{\\alpha}} - 1 \\right) = K x^{\\alpha} \n\\end{align*}\n$$\n\nb. Suppose that $f(n) \\ge c(k) \\cdot n^{\\alpha}$ for all $n < 2^k$. We will find a recurrence relation for $c(k)$ and prove that $c(k)$ tends to a positive constant. Let $c(1) = 2$ and, since $x < 2^{k+1} \\Rightarrow x/j < 2^k$ for $j \\ge 2$,\n$$\n\\begin{align*} \nf(x) &= 1 + f\\left(\\frac{x}{2}\\right) + f\\left(\\frac{x}{3}\\right) + f\\left(\\frac{x}{4}\\right) + \\dots + f\\left(\\frac{x}{\\lfloor x \\rfloor}\\right) \\\n&\\ge 1 + c(k) \\left( \\left(\\frac{x}{2}\\right)^{\\alpha} + \\left(\\frac{x}{3}\\right)^{\\alpha} + \\left(\\frac{x}{4}\\right)^{\\alpha} + \\dots + \\left(\\frac{x}{\\lfloor x \\rfloor}\\right)^{\\alpha} \\right) \n\\end{align*}\n$$\nSo we need\n$$\n\\begin{align*} \n& 1 + c(k)x^{\\alpha} \\left( \\left(\\frac{1}{2}\\right)^{\\alpha} + \\left(\\frac{1}{3}\\right)^{\\alpha} + \\left(\\frac{1}{4}\\right)^{\\alpha} + \\dots + \\left(\\frac{1}{\\lfloor x \\rfloor}\\right)^{\\alpha} \\right) \\ge c(k+1)x^{\\alpha} \\\n\\iff & 1 + c(k)x^{\\alpha} \\left( 2 - 1 - \\sum_{m>x} \\frac{1}{m^{\\alpha}} \\right) \\ge c(k+1)x^{\\alpha} \n\\end{align*}\n$$\nSince $m > x > 2^k$, by substituting every number between $2^t$ and $2^{t+1}$ by $2^t$ we obtain $\\sum_{m>x} \\frac{1}{m^{\\alpha}} < 2^k \\cdot \\frac{1}{2^{k\\alpha}} + 2^{k+1} \\cdot \\frac{1}{2^{(k+1)\\alpha}} + \\dots = \\sum_{t \\ge k} \\frac{1}{2^{(\\alpha-1)t}} = \\frac{1}{2^{(\\alpha-1)(k-1)}}$. So it's sufficient to have\n$$\nc(k+1) \\le x^{-\\alpha} + c(k) \\left(1 - \\frac{1}{2^{(k-1)(\\alpha-1)}(2^{\\alpha-1}-1)}\\right)\n$$\nSince $x < 2^{k+1}$, we can actually have\n$$\nc(k+1) \\le \\frac{1}{2^{(k+1)\\alpha}} + c(k) \\left(1 - \\frac{1}{2^{(k-1)(\\alpha-1)}(2^{\\alpha-1} - 1)}\\right)\n$$\nLet $\\beta = \\frac{1}{2^{\\alpha-1}-1}$. Let's try, then, to find constants $m_1, m_2$ such that $c(k) = m_1 + \\frac{m_2}{2^{k(\\alpha-1)}}$. Substituting, we find\n$$\nm_1 + \\frac{m_2}{2^{(k+1)(\\alpha-1)}} \\le \\frac{1}{2^{(k+1)\\alpha}} + \\left(m_1 + \\frac{m_2}{2^{k(\\alpha-1)}}\\right) \\left(1 - \\frac{\\beta}{2^{(k-1)(\\alpha-1)}}\\right) \\\\\n\\iff \\frac{\\beta m_2}{2^{(2k-1)(\\alpha-1)}} \\le \\frac{1}{2^{(k+1)\\alpha}} + \\frac{-m_2 + m_2 2^{\\alpha-1} - \\beta m_1 2^{2(\\alpha-1)}}{2^{(k+1)(\\alpha-1)}}\n$$\nWe can choose $m_1, m_2$ such that $-m_2 + m_2 2^{\\alpha-1} - \\beta m_1 2^{2(\\alpha-1)} = 0 \\iff 2^{2(\\alpha-1)}m_1 = (2^{\\alpha-1} - 1)^2 m_2$. Then the inequality reduces to\n$$\n\\frac{\\beta m_2}{2^{(2k-1)(\\alpha-1)}} \\le \\frac{1}{2^{(k+1)\\alpha}} \\iff \\frac{\\beta m_2}{2^{(k-1/2)(2\\alpha-2)}} \\le \\frac{1}{2^{(k+1)\\alpha}}\n$$\nwhich holds for many positive constants $\\beta, m_2$. So $c(k) = m_1 + \\frac{m_2}{2^{k(\\alpha-1)}}$ tends to a positive $m_1$ as $k$ goes to infinity, and, since $f(n) \\ge c(k) \\cdot 2^{\\alpha} > m_1 \\cdot 2^{\\alpha}$ we are done.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26562,
    "subject": "Mathematics (Multi-modal)",
    "question": "Compute\n$$\n\\int_{0}^{\\pi/4} \\frac{x}{(\\sin x + \\cos x) \\cos x} \\, dx.\n$$",
    "options": [],
    "answer": "π ln 2 / 8",
    "solution": "First notice the identity $\\sin x + \\cos x = \\sqrt{2}(\\cos x \\cos \\frac{\\pi}{4} + \\sin x \\sin \\frac{\\pi}{4}) = \\sqrt{2} \\cos(\\frac{\\pi}{4} - x)$. So\n$$\nI = \\int_{0}^{\\pi/4} \\frac{x}{(\\sin x + \\cos x) \\cos x} \\, dx = \\int_{0}^{\\pi/4} \\frac{x}{\\sqrt{2} \\cos(\\frac{\\pi}{4} - x) \\cos x} \\, dx\n$$\nNow consider the substitution $y = \\frac{\\pi}{4} - x$. So $dx = -dy$ and\n$$\n\\begin{aligned}\n\\int_{0}^{\\pi/4} \\frac{x}{\\sqrt{2} \\cos(\\frac{\\pi}{4} - x) \\cos x} \\, dx &= \\int_{\\pi/4}^{0} -\\frac{\\frac{\\pi}{4} - y}{\\sqrt{2} \\cos y \\cos(\\frac{\\pi}{4} - y)} \\, dy \\\\\n&= \\int_{0}^{\\pi/4} \\frac{\\frac{\\pi}{4} - y}{\\sqrt{2} \\cos y \\cos(\\frac{\\pi}{4} - y)} \\, dy \\\\\n= \\int_{0}^{\\pi/4} \\frac{\\frac{\\pi}{4} - x}{\\sqrt{2} \\cos(\\frac{\\pi}{4} - x) \\cos x} \\, dx\n\\end{aligned}\n$$\nSumming the two integrals, we find\n$$\n\\begin{aligned}\n2I &= \\int_{0}^{\\pi/4} \\frac{x}{\\sqrt{2} \\cos(\\frac{\\pi}{4} - x) \\cos x} \\, dx + \\int_{0}^{\\pi/4} \\frac{\\frac{\\pi}{4} - x}{\\sqrt{2} \\cos(\\frac{\\pi}{4} - x) \\cos x} \\, dx \\\\\n&= \\int_{0}^{\\pi/4} \\frac{\\frac{\\pi}{4} - x + x}{\\sqrt{2} \\cos(\\frac{\\pi}{4} - x) \\cos x} \\, dx \\\\\n= \\int_{0}^{\\pi/4} \\frac{\\frac{\\pi}{4}}{\\sqrt{2} \\cos(\\frac{\\pi}{4} - x) \\cos x} \\, dx\n\\end{aligned}\n$$\nSo $I = \\frac{\\pi}{8} \\int_{0}^{\\pi/4} \\frac{1}{(\\sin x + \\cos x) \\cos x} \\, dx$.\nLet $f(x) = \\ln(\\sin x + \\cos x)$ and $g(x) = \\ln(\\cos x)$. Notice that $f'(x) = \\frac{\\cos x - \\sin x}{\\sin x + \\cos x}$ and $g'(x) = -\\frac{\\sin x}{\\cos x} = -\\frac{\\sin^2 x + \\sin x \\cos x}{(\\sin x + \\cos x) \\cos x}$. So $f'(x) - g'(x) = \\frac{\\cos^2 x - \\sin x \\cos x + \\sin^2 x + \\sin x \\cos x}{(\\sin x + \\cos x) \\cos x} = \\frac{1}{(\\sin x + \\cos x) \\cos x}$. Hence\n$$\n\\begin{aligned}\nI &= \\frac{\\pi}{8} \\int_{0}^{\\pi/4} \\frac{1}{(\\sin x + \\cos x) \\cos x} \\, dx \\\\\n&= \\frac{\\pi}{8} \\left( f\\left(\\frac{\\pi}{4}\\right) - g\\left(\\frac{\\pi}{4}\\right) - (f(0) - g(0)) \\right) \\\\\n&= \\frac{\\pi}{8} \\left( \\ln(\\sqrt{2}) - \\ln\\left(\\frac{\\sqrt{2}}{2}\\right) - \\ln 1 + \\ln 1 \\right) = \\frac{\\pi \\ln 2}{8}\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26563,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f$ from real numbers to real numbers such that\n$$\nf(a + b) = f(ab)\n$$\nfor all irrationals $a, b$.",
    "options": [],
    "answer": "All constant functions: f(x) = k for all real x, where k is any real constant.",
    "solution": "Let $f(0) = k$. Plugging $a = \\sqrt{2}$ and $b = -\\sqrt{2}$ one obtain $f(\\sqrt{2} + (-\\sqrt{2})) = f(\\sqrt{2}(-\\sqrt{2})) \\iff f(-2) = k$.\nLet $\\alpha \\in \\mathbb{R} \\setminus \\mathbb{Q}$. Since the quadratic equation $x^2-\\alpha x-2=0$ has discriminant $\\Delta = \\alpha^2+8>0$, its roots have sum $\\alpha \\in \\mathbb{R} \\setminus \\mathbb{Q}$ and product $-2 \\in \\mathbb{Q}$, at least one of its roots $m$ is irrational; the other root, $n = -2/m$, is also irrational. Thus we can plug $m$ and $n$, obtaining $f(m+n) = f(mn) \\iff f(\\alpha) = f(-2) = k$.\nNow let $q \\in \\mathbb{Q}$. The quadratic equation $x^2 - qx - \\sqrt{2} = 0$ has discriminant $\\Delta = q^2 + 4\\sqrt{2} > 0$, sum of the roots $q \\in \\mathbb{Q}$ and product of the roots $-\\sqrt{2} \\in \\mathbb{R} \\setminus \\mathbb{Q}$, so one of its roots $r$ is irrational; the other root, $s = q-r$, is also irrational. We can plug $r$ and $s$, obtaining $f(r+s) = f(rs) \\iff f(q) = f(-\\sqrt{2}) = k$.\nSo all the functions are the constant functions $f(x) = k, k \\in \\mathbb{R}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26564,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDízima periódica - Sabendo que $0,333\\ldots=\\frac{1}{3}$, qual é a fração irredutível equivalente a $0,1333\\ldots$.\n\n(a) $\\frac{1}{13}$\n(b) $\\frac{1}{15}$\n(c) $\\frac{1}{30}$\n(d) $\\frac{2}{15}$\n(e) $\\frac{1333}{10000}$",
    "options": [],
    "answer": "d",
    "solution": "Solution:\n\nComo $\\frac{1}{3}=0,333\\ldots$, segue que\n$$\n0,1333\\ldots=0,333\\ldots-0,2=\\frac{1}{3}-\\frac{2}{10}=\\frac{1}{3}-\\frac{1}{5}=\\frac{2}{15}\n$$\n\nSolução 2: Usando simplesmente a regra que fornece a geratriz de uma dízima periódica, também podemos obter\n$$\n0,1333\\ldots=\\frac{13-1}{90}=\\frac{12}{90}=\\frac{2}{15}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26565,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO valor absoluto $|a|$ de um número $a$ qualquer é definido por\n\n$$\n|a|=\\left\\{\\begin{array}{cl}\na & \\text{ se } a>0 \\\\\n0 & \\text{ se } a=0 \\\\\n-a & \\text{ se } a<0\n\\end{array}\\right.\n$$\n\nPor exemplo, $|6|=6$, $|-4|=4$ e $|0|=0$. Quanto vale $N=|5|+|3-8|-|-4|$ ?\n\n(a) 4\n(b) -4\n(c) 14\n(d) -14\n(e) 6",
    "options": [],
    "answer": "e",
    "solution": "Solution:\n\nTemos: $|5|=5$, $|3-8|=|-5|=5$ e $|-4|=4$. Logo, $N=5+5-4=6$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26566,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO valor absoluto $|a|$ de um número $a$ qualquer é definido por\n$$\n|a|=\\left\\{\\begin{array}{cl}\na & \\text{ se } a>0 \\\\\n0 & \\text{ se } a=0 \\\\\n-a & \\text{ se } a<0\n\\end{array}\\right.\n$$\n\nPor exemplo, $|6|=6$, $|-4|=4$ e $|0|=0$. Quanto vale $N=|5|+|3-8|-|-4|$ ?\n(a) 4\n(b) -4\n(c) 14\n(d) -14\n(e) 6",
    "options": [],
    "answer": "e",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26567,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO triângulo de moedas - Um menino tentou alinhar 480 moedas em forma de um triângulo, com uma moeda na primeira linha, duas moedas na segunda linha, e assim por diante. Ao final da tentativa, sobraram 15 moedas. Quantas linhas tem esse triângulo?",
    "options": [],
    "answer": "30",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26568,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDuas tabelas - As linhas da primeira tabela dada são todas progressões aritméticas de uma mesma razão e as colunas dessa tabela são todas progressões aritméticas de uma mesma razão. Na segunda tabela dada foi utilizada a mesma lei de formação, mas alguém apagou alguns números deixando apenas três. Qual é o número que estava na posição indicada com\n\n| 5  | 8  | 11 | 14 | 17 |\n|:---:|:---:|:---:|:---:|:---:|\n| 12 | 15 | 18 | 21 | 24 |\n| 19 | 22 | 25 | 28 | 31 |\n| 26 | 29 | 32 | 35 | 38 |\n| 33 | 36 | 39 | 42 | 45 |\n\n|   |   | 39 |   |   |\n|---|---|---|---|---|\n|   |   |   |   |   |\n|   |   |   |   | 87|\n| 56|   |   |   |   |\n|   |   |   | $\\star$ |   |",
    "options": [],
    "answer": "102",
    "solution": "Solution:\n\nVemos que na primeira tabela cada linha é uma progressão aritmética de razão $3$ e cada coluna é uma progressão aritmética de razão $7$.\n\n| 5  | 8  | 11 | 14 | 17 |\n|:---:|:---:|:---:|:---:|:---:|\n| 12 | 15 | 18 | 21 | 24 |\n| 19 | 22 | 25 | 28 | 31 |\n| 26 | 29 | 32 | 35 | 38 |\n| 33 | 36 | 39 | 42 | 45 |\n\n|   |   | 39 |   |   |\n|---|---|---|---|---|\n|   |   |   |   |   |\n|   |   |   |   | 87|\n| 56|   |   |   |   |\n|   |   |   | $\\star$ |   |\n\nDigamos que na segunda tabela a razão das progressões aritméticas das linhas seja $a$ e a das colunas seja $b$. Assim, obtemos $39+2a+2b=87$ e $39-2a+3b=56$.\n\n| $39-2a$ | $39-a$ | 39 | $39+a$ | $39+2a$ |\n|:---:|:---:|:---:|:---:|:---:|\n| $39-2a+b$ |   |   |   | $39+2a+b$ |\n| $39-2a+2b$ |   |   |   | 87 |\n| 56 |   |   |   |   |\n|   |   |   | $\\star$ |   |\n\nSomando essas duas equações, resulta $78+5b=143$, donde $b=13$. Subtraindo as duas igualdades que anteriormente foram somadas, obtemos $4a-b=31$. Segue que $a=11$. Assim, o número que estava na posição $\\star$ era $39+a+4b=39+11+4\\times 13=102$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26569,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDuas tabelas - As linhas da primeira tabela dada são todas progressões aritméticas de uma mesma razão e as colunas dessa tabela são todas progressões aritméticas de uma mesma razão. Na segunda tabela dada foi utilizada a mesma lei de formação, mas alguém apagou alguns números deixando apenas três. Qual é o número que estava na posição indicada com\n\n| 5  | 8  | 11 | 14 | 17 |\n|:---:|:---:|:---:|:---:|:---:|\n| 12 | 15 | 18 | 21 | 24 |\n| 19 | 22 | 25 | 28 | 31 |\n| 26 | 29 | 32 | 35 | 38 |\n| 33 | 36 | 39 | 42 | 45 |\n\n|   |   | 39 |   |   |\n|---|---|---|---|---|\n|   |   |   |   |   |\n|   |   |   |   | 87|\n| 56|   |   |   |   |\n|   |   |   | $\\star$ |   |",
    "options": [],
    "answer": "102",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26570,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSete números inteiros positivos e consecutivos estão escritos em ordem crescente numa mesma linha. Determine se é possível colocar entre esses números cinco sinais de \"+\" e só um de \"=\" de tal modo que resulte uma igualdade.",
    "options": [],
    "answer": "Yes. The unique possibility is 9 + 10 + 11 + 12 = 13 + 14 + 15.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26571,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDuas partículas percorrem um caminho circular de $120~\\mathrm{m}$ de comprimento. A velocidade de uma delas é $2~\\mathrm{m}/\\mathrm{s}$ maior do que a da outra e ela completa cada volta num tempo que é 3 segundos inferior ao da outra. Qual é a velocidade de cada partícula?",
    "options": [],
    "answer": "8 m/s and 10 m/s",
    "solution": "Solution:\n\nDenotemos as partículas por $A$ e $B$ e seja $v$ a velocidade da partícula $B$. Supondo que $A$ seja a mais rápida, temos que $v+2$ é a velocidade de $A$. Assim, o tempo que $B$ demora para dar uma volta é $120 / v$ e o tempo que $A$ demora é $120 /(v+2)$. Como esse tempo é três segundos inferior ao de $B$, temos a equação básica\n$$\n\\frac{120}{v} - 3 = \\frac{120}{v+2}\n$$\nSimplificando, isso equivale a $v^{2} + 2v - 80 = 0$, cuja raiz positiva é\n$$\nv = \\frac{1}{2}[-2 + \\sqrt{4 + 320}] = -1 + \\sqrt{81} = 8\n$$\nPortanto, a velocidade da partícula mais lenta é $8~\\mathrm{m}/\\mathrm{s}$ e a da mais rápida é $10~\\mathrm{m}/\\mathrm{s}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26572,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm caminho triangular - Janete passeia por um caminho de forma triangular $\\triangle ABC$, com o lado $AB$ medindo $1992~\\mathrm{m}$. Ela gasta 24 minutos para percorrer esse lado $AB$ e, depois, com a mesma velocidade, ela percorre o outro lado $BC$ seguido da hipotenusa $CA$ em 2 horas e 46 minutos. Qual é o comprimento do lado $BC$?",
    "options": [],
    "answer": "6745 m",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26573,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm professor propõe 80 problemas a um aluno, informando que lhe atribuirá cinco pontos por problema resolvido corretamente e lhe descontará três pontos por problema não resolvido ou resolvido incorretamente. No final, o aluno fica com oito pontos. Quantos problemas ele resolveu corretamente?",
    "options": [],
    "answer": "31",
    "solution": "Solution:\n\nSeja $c$ o número de problemas resolvidos corretamente e seja $e$ a soma do número de problemas resolvidos incorretamente e de problemas não resolvidos. Logo, $c + e = 80$ e $5c - 3e$ é o número de pontos do aluno na avaliação. No caso presente,\n$$\n\\left\\{\\begin{aligned}\nc + e & = 80 \\\\\n5c - 3e & = 8\n\\end{aligned}\\right.\n$$\nResolvendo o sistema, encontramos $c = 31$ e $e = 49$. Logo, o aluno resolveu 31 problemas corretamente.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26574,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa figura dada, temos 16 pontos formando um reticulado quadrado e duas retas, $r$ e $s$, perpendiculares entre si.\n\n![](attached_image_1.png)\n\na) Quantos quadrados podemos construir, de tal maneira que seus vértices pertençam ao reticulado, porém nenhum de seus lados seja paralelo, nem à reta $r$, nem à reta $s$ ?\n\nb) Quantos triângulos retângulos isósceles podemos construir de tal maneira que seus vértices pertençam ao reticulado, porém nenhum de seus lados seja paralelo, nem à reta $r$, nem à reta $s$ ?",
    "options": [],
    "answer": "a) 5; b) 4",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26575,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSequência de algarismos - Todos os números naturais de $1$ em diante foram escritos consecutivamente, formando uma sequência de algarismos, como segue.\n\n$$\n1234567891011121314151617181920212223 \\ldots\n$$\n\nQual é o algarismo que aparece na posição de número $206788$ ?",
    "options": [],
    "answer": "7",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26576,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nColoque os números $663, 664, 665, 666, 667, 668, 669, 670$ e $671$, sem repetir, numa tabela $3 \\times 3$, de tal maneira que a soma em cada linha, em cada coluna e em cada diagonal seja $2001$. Caso isso não seja possível, justifique sua resposta.",
    "options": [],
    "answer": "Yes. One valid arrangement is: [[664, 669, 668], [671, 667, 663], [666, 665, 670]].",
    "solution": "Solution:\n\nVamos analisar se é possível dispor os números $663, 664, 665, 666, 667, 668, 669, 670$ e $671$ em uma tabela $3 \\times 3$ de modo que a soma de cada linha, coluna e diagonal seja $2001$.\n\nA soma de todos os números é:\n$$\n663 + 664 + 665 + 666 + 667 + 668 + 669 + 670 + 671 = (663 + 671) + (664 + 670) + (665 + 669) + (666 + 668) + 667 = (1334) + (1334) + (1334) + (1334) + 667 = 4 \\times 1334 + 667 = 5336 + 667 = 6003.\n$$\n\nEm um quadrado mágico $3 \\times 3$, a soma de todos os números é igual a $3$ vezes a soma de cada linha (ou coluna ou diagonal), pois há $3$ linhas:\n$$\n\\text{Soma total} = 3 \\times 2001 = 6003.\n$$\n\nPortanto, a soma total dos números é igual a $6003$, que é exatamente $3 \\times 2001$.\n\nAlém disso, em um quadrado mágico $3 \\times 3$, o número central deve ser a média dos nove números, ou seja:\n$$\n\\text{Número central} = \\frac{6003}{9} = 667.\n$$\n\nO número $667$ está entre os números dados, então é possível.\n\nAgora, vamos tentar construir o quadrado mágico. O quadrado mágico padrão de ordem $3$ com números de $1$ a $9$ é:\n\n| 2 | 7 | 6 |\n|---|---|---|\n| 9 | 5 | 1 |\n| 4 | 3 | 8 |\n\nA soma de cada linha, coluna e diagonal é $15$.\n\nPara transformar esse quadrado para os números dados, notamos que os números dados são $663$ a $671$, ou seja, $1$ a $9$ somados com $662$.\n\nAssim, basta somar $662$ a cada número do quadrado mágico padrão:\n\n| $2+662=664$ | $7+662=669$ | $6+662=668$ |\n|-------------|-------------|-------------|\n| $9+662=671$ | $5+662=667$ | $1+662=663$ |\n| $4+662=666$ | $3+662=665$ | $8+662=670$ |\n\nPortanto, a tabela é:\n\n| 664 | 669 | 668 |\n|-----|-----|-----|\n| 671 | 667 | 663 |\n| 666 | 665 | 670 |\n\nVerificando a soma de cada linha:\n- Primeira linha: $664 + 669 + 668 = 2001$\n- Segunda linha: $671 + 667 + 663 = 2001$\n- Terceira linha: $666 + 665 + 670 = 2001$\n\nColunas:\n- Primeira coluna: $664 + 671 + 666 = 2001$\n- Segunda coluna: $669 + 667 + 665 = 2001$\n- Terceira coluna: $668 + 663 + 670 = 2001$\n\nDiagonais:\n- Principal: $664 + 667 + 670 = 2001$\n- Secundária: $668 + 667 + 666 = 2001$\n\nPortanto, é possível montar a tabela conforme pedido.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26577,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nContando os zeros - Quantos zeros existem no final do número $9^{2007}+1$ ?",
    "options": [],
    "answer": "1",
    "solution": "Solution:\n\nInicialmente, verificamos como terminam as potências de $9$, ou seja, listamos os dois últimos algarismos, os da dezena e da unidade, das potências $9^{n}$, ordenadamente.\n\n| Se $n$ for | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| $9^{n}$ termina com | 01 | 09 | 81 | 29 | 61 | 49 | 41 | 69 | 21 | 89 | 01 | 09 | 81 |\n\nAssim, vemos que os dois últimos algarismos de $9^{10}$, $9^{11}$ e $9^{12}$ são os mesmos de $9^{0}$, $9^{1}$ e $9^{2}$. A partir $9^{10}$, os dois últimos algarismos das potências começam a se repetir, formando uma sequência periódica de período 10. Como $2007=10 \\times 200+7$ e os dois últimos algarismos de $9^{10 \\times 200}$ são 01, segue que os dois últimos algarismos de $9^{2007}$ são os dois últimos algarismos de $9^{7}$, ou seja, 69. Então, os dois últimos algarismos de $9^{2007}+1$ são iguais a $69+1=70$. Assim, existe um único zero no final do número $9^{2007}+1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26578,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCírculos dentro do quadrado - Dentro de um quadrado são colocados círculos, dois a dois disjuntos ou, então, tangentes externamente. Se o lado do quadrado mede $1~\\mathrm{cm}$, será possível colocar tantos desses círculos de tal modo que a soma de seus raios, em centímetros, seja maior do que $2008$?",
    "options": [],
    "answer": "No",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26579,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEncontre todos os números de oito algarismos formados somente com os algarismos $1$, $2$, $3$ e $4$, cada um deles duas vezes, tais que:\n\na. exista um único algarismo entre os dois algarismos $1$;\nb. existam dois algarismos entre os dois algarismos $2$;\nc. existam três algarismos entre os dois algarismos $3$;\nd. existam quatro algarismos entre os dois algarismos $4$.",
    "options": [],
    "answer": "41312432, 23421314",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26580,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOs algarismos $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$ e $9$ foram escritos (numa ordem desconhecida) ao redor de uma circunferência. Lendo esses algarismos de três em três no sentido horário, formam-se nove números de três algarismos. Determine a soma desses nove números.",
    "options": [],
    "answer": "4995",
    "solution": "Solution:\n\nConsidere os algarismos $a_1, a_2, \\ldots, a_9$ dispostos ao redor da circunferência, em ordem. Os nove números formados são:\n\n$N_1 = 100a_1 + 10a_2 + a_3$\n\n$N_2 = 100a_2 + 10a_3 + a_4$\n\n$N_3 = 100a_3 + 10a_4 + a_5$\n\n$\\ldots$\n\n$N_9 = 100a_9 + 10a_1 + a_2$\n\nA soma total é:\n\n$S = N_1 + N_2 + \\cdots + N_9$\n\nVamos somar todos os coeficientes de cada $a_i$.\n\nCada algarismo aparece três vezes em cada posição (centena, dezena, unidade):\n- $a_1$ aparece como centena em $N_1$, como dezena em $N_9$, como unidade em $N_8$.\n- O mesmo ocorre para os demais algarismos.\n\nPortanto, cada algarismo aparece uma vez em cada posição.\n\nAssim, a soma é:\n\n$S = 100(a_1 + \\cdots + a_9) + 10(a_1 + \\cdots + a_9) + (a_1 + \\cdots + a_9)$\n\nComo $a_1, \\ldots, a_9$ são os algarismos de $1$ a $9$, temos:\n\n$a_1 + \\cdots + a_9 = 1 + 2 + \\cdots + 9 = 45$\n\nLogo,\n\n$S = 100 \\times 45 + 10 \\times 45 + 45 = 4500 + 450 + 45 = 4995$\n\nPortanto, a soma dos nove números é $4995$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26581,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCinco peças de metal, confeccionadas, respectivamente, de ouro, prata, bronze, platina e níquel, foram colocadas em cinco cofres, numerados de $1$ a $5$. Cada cofre contém uma peça e o problema consiste em descobrir qual peça está em qual cofre. Na porta de cada cofre está escrita uma informação. Das cinco informações, quatro são falsas e a única que é verdadeira é a que aparece na porta do cofre que contém a peça de ouro. As informações nas portas dos cofres são as seguintes.\n\nCofre $1$: O ouro está no cofre $2$ ou $3$.\nCofre $2$: A prata está no cofre $1$.\nCofre $3$: O bronze não está aqui.\nCofre $4$: O níquel está no cofre cujo número é inferior, em uma unidade, ao que contém o ouro.\nCofre $5$: A platina está no cofre cujo número é superior, em uma unidade, ao que contém o bronze.",
    "options": [],
    "answer": "Chest 1: nickel; Chest 2: silver; Chest 3: bronze; Chest 4: platinum; Chest 5: gold.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26582,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoma de quadrados - Encontre três números, numa progressão aritmética de razão $2$, tais que a soma de seus quadrados seja um número formado de quatro algarismos iguais.",
    "options": [],
    "answer": "41, 43, 45",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26583,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCerto número deixa resto $1$ quando dividido por $3$, deixa resto $2$ quando dividido por $4$, deixa resto $3$ quando dividido por $5$ e deixa resto $4$ quando dividido por $6$. Qual é o menor número inteiro positivo que satisfaz essas propriedades?",
    "options": [],
    "answer": "58",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26584,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm código - Na expressão abaixo, cada letra corresponde a um algarismo, sendo que letras diferentes correspondem a algarismos diferentes. Determine esses algarismos.\n\n$$\n6 \\times A O B M E P = 7 \\times M E P A O B\n$$",
    "options": [],
    "answer": "A=5, O=3, B=8, M=4, E=6, P=1",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26585,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO triângulo $\\triangle ABC$ é equilátero e o ponto $P$ é tal que $PA = 3\\ \\mathrm{cm}$, $PB = 4\\ \\mathrm{cm}$ e $PC = 5\\ \\mathrm{cm}$. Calcule o comprimento dos lados do triângulo $\\triangle ABC$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "sqrt(25 + 12*sqrt(3)) cm",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26586,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAmigo oculto - Um grupo de cinco amigos decide brincar de amigo oculto, cada um compra um presente para seu amigo oculto. Pelas regras do jogo, cada um dá exatamente um presente e recebe exatamente um presente. De quantas maneiras podem os presentes ser distribuídos, de modo que ninguém dê presente para si mesmo?",
    "options": [],
    "answer": "44",
    "solution": "Solution:\n\nPrimeiramente observemos que o número de formas de distribuir os presentes sem nenhuma restrição é $5! = 5 \\times 4 \\times 3 \\times 2 \\times 1 = 120$. Daí temos que tirar os casos \"ruins\", isto é, os casos em que pelo menos um amigo tirou o seu próprio presente. Esses casos a eliminar podem ser listados pelo número de amigos que tiram seu próprio presente.\n\n- Os 5 amigos ficaram com seus próprios presentes. Só há uma possibilidade de acontecer isso.\n\n- Exatamente 4 amigos ficaram com seus próprios presentes. Isso não é possível.\n\n- Exatamente 3 amigos ficaram com seus próprios presentes. Nessa situação, os outros dois amigos trocam os presentes. Assim, escolhemos 3 pessoas dentre as 5, isto é, $\\frac{5 \\times 4 \\times 3}{3 \\times 2} = 10$ possibilidades.\n\n- Exatamente 2 amigos ficaram com seus próprios presentes. Nesse caso, escolhemos 2 pessoas dentre as 5, isto é, $\\frac{5 \\times 4}{2} = 10$ possibilidades. Os outros 3 amigos trocam os presentes entre si, dando um total de $10 \\times 2 = 20$ possibilidades.\n\n- Por último, exatamente um amigo ficou com seu próprio presente. Nesse caso, escolhemos uma pessoa dentre um total de 5, multiplicando pelo número de formas que os outros amigos não fiquem com seu presente, que são 9 maneiras. Logo, nessa situação, temos um total de $5 \\times 9 = 45$ possibilidades.\n\nPortanto, temos $120 - 45 - 20 - 10 - 1 = 44$ maneiras de distribuir os presentes sem que alguém fique com seu próprio presente.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26587,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nQuantos são os pares de números inteiros positivos $(x, y)$ tais que\n$$\n\\frac{x y}{x+y}=144 ?\n$$",
    "options": [],
    "answer": "45",
    "solution": "Solution:\nA equação dada é equivalente a $x y = 144(x + y) = 144 x + 144 y$, portanto, isolando $x$, obtemos $x = \\frac{144 y}{y - 144}$. Como $x$ e $y$ devem ser inteiros positivos, o denominador $y - 144$ deve ser um número inteiro positivo, digamos, $y - 144 = n$. Substituindo essa expressão no valor de $x$, obtemos\n$$\nx = \\frac{144(n + 144)}{n} = 144 + \\frac{144^2}{n}\n$$\nComo $x$ deve ser um número inteiro, $n$ deve ser um divisor de $144^2$. Sendo $144^2 = 12^4 = 2^8 \\cdot 3^4$, seus divisores são os números $d$ da forma $d = 2^a \\cdot 3^b$, com $0 \\leq a \\leq 8$ e $0 \\leq b \\leq 4$. Como há 9 valores possíveis para $a$ e 5 valores possíveis para $b$, concluímos que $144^2$ tem $9 \\times 5 = 45$ divisores.\nAssim, para cada divisor $n$ de $144^2$, obtemos uma solução\n$$\n(x, y) = \\left(144 + \\frac{144^2}{n}, n + 144\\right)\n$$\nda equação $\\frac{x y}{x + y} = 144$ dada. Portanto, essa equação possui 45 pares de números inteiros positivos $(x, y)$ que a satisfazem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26588,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nQuantos são os pares de números inteiros positivos $(x, y)$ tais que\n\n$$\n\\frac{x y}{x+y}=144 ?\n$$",
    "options": [],
    "answer": "45",
    "solution": "Solution:\n\nTemos a equação:\n$$\n\\frac{x y}{x+y} = 144\n$$\nMultiplicando ambos os lados por $x + y$:\n$$\nx y = 144(x + y)\n$$\nRearranjando:\n$$\nx y - 144x - 144y = 0\n$$\n$$\nx y - 144x - 144y + 144^2 = 144^2\n$$\n$$\n(x - 144)(y - 144) = 144^2\n$$\n\nAgora, como $x$ e $y$ são inteiros positivos, $x - 144$ e $y - 144$ são inteiros positivos maiores ou iguais a $1$.\n\nSeja $d$ um divisor positivo de $144^2$. Então, para cada divisor $d$, temos:\n$$\nx - 144 = d\n$$\n$$\ny - 144 = \\frac{144^2}{d}\n$$\nOu seja,\n$$\nx = d + 144\n$$\n$$\ny = \\frac{144^2}{d} + 144\n$$\n\nComo a ordem de $x$ e $y$ importa (pares ordenados), cada par $(d, \\frac{144^2}{d})$ gera um par $(x, y)$.\n\nO número de pares de inteiros positivos $(x, y)$ é igual ao número de divisores positivos de $144^2$.\n\nAgora, vamos calcular o número de divisores positivos de $144^2$.\n\nPrimeiro, fatoramos $144$:\n$$\n144 = 12^2 = (2^2 \\cdot 3)^2 = 2^4 \\cdot 3^2\n$$\nEntão,\n$$\n144^2 = (2^4 \\cdot 3^2)^2 = 2^8 \\cdot 3^4\n$$\nO número de divisores positivos de $2^8 \\cdot 3^4$ é:\n$$\n(8+1) \\times (4+1) = 9 \\times 5 = 45\n$$\n\nPortanto, existem $45$ pares de inteiros positivos $(x, y)$ que satisfazem a equação dada.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26589,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDeterminando uma sequência - Numa certa sequência de 80 números, qualquer termo, salvo as duas extremidades, é igual ao produto de seus termos vizinhos. O produto dos 40 primeiros termos da sequência é $8$ e o produto de todos os termos também é $8$. Determine os termos da sequência.",
    "options": [],
    "answer": "The sequence is periodic with period six: 1/64, sqrt(2)/32, 2 sqrt(2), 64, 16 sqrt(2), 1/(2 sqrt(2)), repeating.",
    "solution": "Solution:\n\nSeja a sequência $a_1, a_2, \\ldots, a_{80}$.\n\nDado que, para $2 \\leq k \\leq 79$, temos:\n$$\na_k = a_{k-1} \\cdot a_{k+1}\n$$\n\nIsso implica que:\n$$\na_{k+1} = \\frac{a_k}{a_{k-1}}\n$$\n\nPortanto, a sequência é recorrente e depende dos dois primeiros termos.\n\nVamos calcular os primeiros termos explicitamente:\n\n- $a_1$\n- $a_2$\n- $a_3 = \\frac{a_2}{a_1}$\n- $a_4 = \\frac{a_3}{a_2} = \\frac{a_2}{a_1 a_2} = \\frac{1}{a_1}$\n- $a_5 = \\frac{a_4}{a_3} = \\frac{1/a_1}{a_2/a_1} = \\frac{1}{a_2}$\n- $a_6 = \\frac{a_5}{a_4} = \\frac{1/a_2}{1/a_1} = \\frac{a_1}{a_2}$\n- $a_7 = \\frac{a_6}{a_5} = \\frac{a_1/a_2}{1/a_2} = a_1$\n- $a_8 = \\frac{a_7}{a_6} = \\frac{a_1}{a_1/a_2} = a_2$\n\nPortanto, a sequência é periódica de período $6$:\n$$\na_1,\\ a_2,\\ \\frac{a_2}{a_1},\\ \\frac{1}{a_1},\\ \\frac{1}{a_2},\\ \\frac{a_1}{a_2},\\ a_1,\\ a_2,\\ \\ldots\n$$\n\nVamos verificar o período:\n- $a_7 = a_1$\n- $a_8 = a_2$\n- $a_9 = \\frac{a_2}{a_1}$\n\nPortanto, a sequência repete a cada $6$ termos.\n\nComo $80 = 6 \\times 13 + 2$, temos $13$ períodos completos e mais dois termos.\n\nAgora, calculemos o produto dos $40$ primeiros termos.\n\nEm cada período de $6$ termos, o produto é:\n$$\nP_6 = a_1 \\cdot a_2 \\cdot \\frac{a_2}{a_1} \\cdot \\frac{1}{a_1} \\cdot \\frac{1}{a_2} \\cdot \\frac{a_1}{a_2}\n$$\nSimplificando:\n- $a_1$ cancela com $1/a_1$\n- $a_2$ cancela com $1/a_2$\n- $\\frac{a_2}{a_1}$ cancela com $\\frac{a_1}{a_2}$\n\nLogo,\n$$\nP_6 = 1\n$$\n\nO produto dos $6 \\times 6 = 36$ primeiros termos é $1$.\n\nRestam $4$ termos para completar $40$ termos: $a_{37}, a_{38}, a_{39}, a_{40}$.\n\nComo a sequência é periódica de $6$ termos, $a_{37} = a_1$, $a_{38} = a_2$, $a_{39} = \\frac{a_2}{a_1}$, $a_{40} = \\frac{1}{a_1}$.\n\nPortanto, o produto dos $40$ primeiros termos é:\n$$\nP_{40} = 1 \\cdot a_1 \\cdot a_2 \\cdot \\frac{a_2}{a_1} \\cdot \\frac{1}{a_1} = a_1 \\cdot a_2 \\cdot \\frac{a_2}{a_1} \\cdot \\frac{1}{a_1}\n$$\nSimplificando:\n- $a_1$ cancela com $1/a_1$\n- $a_2 \\cdot \\frac{a_2}{a_1} = \\frac{a_2^2}{a_1}$\n\nLogo,\n$$\nP_{40} = \\frac{a_2^2}{a_1^2}\n$$\n\nMas o enunciado diz que $P_{40} = 8$, então:\n$$\n\\frac{a_2^2}{a_1^2} = 8 \\implies \\frac{a_2}{a_1} = 2\\sqrt{2}\n$$\n\nAgora, o produto de todos os $80$ termos:\n- $80 = 6 \\times 13 + 2$\n- Produto dos $78$ primeiros termos: $P_{78} = 1$ (pois $6$ divide $78$)\n- Restam $a_{79}$ e $a_{80}$\n\n$a_{79} = \\frac{1}{a_2}$, $a_{80} = \\frac{a_1}{a_2}$\n\nPortanto,\n$$\nP_{80} = 1 \\cdot \\frac{1}{a_2} \\cdot \\frac{a_1}{a_2} = \\frac{a_1}{a_2^2}\n$$\n\nO enunciado diz que $P_{80} = 8$, então:\n$$\n\\frac{a_1}{a_2^2} = 8\n$$\n\nAgora, temos o sistema:\n\\begin{cases}\n\\frac{a_2}{a_1} = 2\\sqrt{2} \\\\\n\\frac{a_1}{a_2^2} = 8\n\\end{cases}\n\nDa primeira equação: $a_2 = 2\\sqrt{2} a_1$\n\nSubstituindo na segunda:\n$$\n\\frac{a_1}{(2\\sqrt{2} a_1)^2} = 8\n$$\n$$\n\\frac{a_1}{8 a_1^2} = 8\n$$\n$$\n\\frac{1}{8 a_1} = 8\n$$\n$$\na_1 = \\frac{1}{64}\n$$\n\nEntão,\n$$\na_2 = 2\\sqrt{2} \\cdot \\frac{1}{64} = \\frac{\\sqrt{2}}{32}\n$$\n\nPortanto, os termos da sequência são:\n$$\na_1 = \\frac{1}{64}\n$$\n$$\na_2 = \\frac{\\sqrt{2}}{32}\n$$\n$$\na_3 = \\frac{a_2}{a_1} = \\frac{\\sqrt{2}/32}{1/64} = 2\\sqrt{2}\n$$\n$$\na_4 = \\frac{1}{a_1} = 64\n$$\n$$\na_5 = \\frac{1}{a_2} = \\frac{32}{\\sqrt{2}} = 16\\sqrt{2}\n$$\n$$\na_6 = \\frac{a_1}{a_2} = \\frac{1/64}{\\sqrt{2}/32} = \\frac{1}{2\\sqrt{2}}\n$$\nE repete a cada $6$ termos.\n\n**Resposta:**\nA sequência é periódica de período $6$:\n$$\n\\frac{1}{64},\\ \\frac{\\sqrt{2}}{32},\\ 2\\sqrt{2},\\ 64,\\ 16\\sqrt{2},\\ \\frac{1}{2\\sqrt{2}},\\ \\frac{1}{64},\\ \\frac{\\sqrt{2}}{32},\\ \\ldots\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26590,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCarina está desenhando a planta de um jardim retangular que terá um de seus lados num muro reto de pedras. Ela comprou $140~\\mathrm{m}$ de cerca, em pedaços de $1~\\mathrm{m}$ cada um, para cercar os outros três lados. Ela não pode cortar esses pedaços e deve gastar todos eles.\n\n![](attached_image_1.png)\n\na. Se os dois lados vizinhos ao muro de pedra têm $40~\\mathrm{m}$ cada um, qual será o comprimento do terceiro lado?\n\nb. É possível que o maior dos lados a ser cercado tenha $85~\\mathrm{m}$? E $65~\\mathrm{m}$? Justifique.",
    "options": [],
    "answer": "a) 60 m. b) 85 m: not possible; 65 m: possible.",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26591,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm quadrilátero especial - Os ângulos $A \\widehat{B} C$ e $C \\widehat{D} A$ do quadrilátero $A B C D$ da figura são retos e os quatro lados do quadrilátero medem números inteiros que são todos distintos. Se $A D=7$ e $B C=11$, quanto medem os lados $A B$ e $C D$ ?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "AB = 3 and CD = 9, or AB = 17 and CD = 19",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26592,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrês quadrados - Dois quadrados, $ABCD$ com uma área de $30~\\mathrm{cm}^2$ e $FHIJ$ com uma área de $20~\\mathrm{cm}^2$, têm seus lados $AD$ e $HI$ sobre uma reta, conforme a figura. Se o ponto $E$ do segmento $AH$ for tal que $BEFG$ é um quadrado, calcule a área desse quadrado.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "50",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26593,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nBolinha de gude - Três amigos jogam uma partida de bolinha de gude, convencionando que o perdedor de cada rodada dobra as bolinhas dos outros jogadores, ou seja, ele dá aos outros dois um número tal de bolinhas que eles fiquem com o dobro do que tinham no início da rodada. O primeiro jogador perdeu a primeira rodada, o segundo jogador a segunda, o terceiro a terceira e todos terminaram com 64 bolinhas cada um. Com quantas bolinhas cada um dos três amigos começou essa partida?",
    "options": [],
    "answer": "First: 104, Second: 56, Third: 32",
    "solution": "Solution:\n\nDenotemos por $x$, $y$ e $z$ o número de bolinhas que cada um tinha no início da partida. Temos\n\n|   | $1^{o}$ | $2^{o}$ | $3^{o}$ |\n| :---: | :---: | :---: | :---: |\n| Início | $x$ | $y$ | $z$ |\n| $1^{a}$ rodada | $x-y-z$ | $2y$ | $2z$ |\n| $2^{a}$ rodada | $2(x-y-z)$ | $2y-2z-(x-y-z)$ | $4z$ |\n| $3^{a}$ rodada | $4(x-y-z)$ | $2(3y-x-z)$ | $4z-2(x-y-z)-(3y-x-z)$ |\n\nComo cada um terminou a partida com 64 bolinhas, segue que\n$$\n\\left\\{\\begin{aligned}\n4(x-y-z) & =64 \\\\\n2(3y-x-z) & =64 \\\\\n4z-2(x-y-z)-(3y-x-z) & =64\n\\end{aligned}\\right.\n$$\ndonde\n$$\n\\left\\{\\begin{aligned}\nx-y-z & =16 \\\\\n-x+3y-z & =32 \\\\\n-x-y+7z & =64\n\\end{aligned}\\right.\n$$\nPara resolver o sistema, somamos a primeira com a segunda equações e a primeira com a terceira, obtendo\n$$\n\\left\\{\\begin{aligned}\ny-z & =24 \\\\\n-y+3z & =40\n\\end{aligned}\\right.\n$$\nDaí, obtemos $z=32$ e $y=56$, portanto, $x=16+56+32=104$. Assim, o primeiro jogador começou a partida com 104 bolinhas, o segundo, com 56 e o terceiro, com 32.\n\n\n|   | $1^{o}$ | $2^{o}$ | $3^{o}$ |\n| :---: | :---: | :---: | :---: |\n| Início |  |  |  |\n| Após a 1-a rodada |  |  |  |\n| Após a 2a rodada |  |  |  |\n| Após a 3a rodada | 64 | 64 | 64 |\nSolution:\n\nPreenchemos a tabela \"de baixo para cima\", isto é, do final para o início do jogo. Começamos com 64 nas três casas finais.\n\nComo os dois primeiros jogadores dobraram a quantidade de bolinhas na terceira rodada,\n\n|   | $1^{o}$ | $2^{o}$ | $3^{o}$ |\n| :---: | :---: | :---: | :---: |\n| Início |  |  |  |\n| Após a 1- rodada |  |  |  |\n| Após a 2a rodada | 32 | 32 | 128 |\n| Após a 3a rodada | 64 | 64 | 64 |\n\ncada um tinha 32 bolinhas e o terceiro jogador deu 32 a cada um deles. Deduzimos que ele possuía $64+32+32=128$ bolinhas.\n\n|   | $1^{o}$ | $2^{o}$ | $3^{o}$ |\n| :---: | :---: | :---: | :---: |\n| Início |  |  |  |\n| Após a 1a rodada | 16 | $32+16+64=112$ | 64 |\n| Após a $2^{a}$ rodada | 32 | 32 | 128 |\n| Após a 3a rodada | 64 | 64 | 64 |\n\nQuem perdeu a segunda rodada foi o segundo jogador. Logo, a tabela era\n\nFinalmente,\n\n|   | $1^{o}$ | $2^{o}$ | $3^{o}$ |\n| :---: | :---: | :---: | :---: |\n| Início | $16+56+32=104$ | 56 | 32 |\n| Após a $1^{a}$ rodada | 16 | $32+16+64=112$ | 64 |\n| Após a 2a rodada | 32 | 32 | 128 |\n| Após a 3a rodada | 64 | 64 | 64 |\n\nAssim, o primeiro jogador começou a partida com 104 bolinhas, o segundo, com 56 e o terceiro, com 32.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26594,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCalcule o valor da soma\n\n$$\nS = \\frac{1}{1 \\cdot 2} + \\frac{1}{2 \\cdot 3} + \\frac{1}{3 \\cdot 4} + \\cdots + \\frac{1}{2006 \\cdot 2007} + \\frac{1}{2007 \\cdot 2008}\n$$",
    "options": [],
    "answer": "2007/2008",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26595,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCalcule o valor da soma\n$$\nS = \\frac{1}{1 \\cdot 2} + \\frac{1}{2 \\cdot 3} + \\frac{1}{3 \\cdot 4} + \\cdots + \\frac{1}{2006 \\cdot 2007} + \\frac{1}{2007 \\cdot 2008}\n$$",
    "options": [],
    "answer": "2007/2008",
    "solution": "Solution:\n\nInicialmente, observe que $\\frac{1}{k \\cdot (k+1)} = \\frac{1}{k} - \\frac{1}{k+1}$. Logo,\n$$\n\\frac{1}{1 \\cdot 2} = 1 - \\frac{1}{2}, \\quad \\frac{1}{2 \\cdot 3} = \\frac{1}{2} - \\frac{1}{3}, \\quad \\ldots, \\quad \\frac{1}{2007 \\cdot 2008} = \\frac{1}{2007} - \\frac{1}{2008}\n$$\nAssim, temos\n$$\nS = 1 - \\frac{1}{2} + \\frac{1}{2} - \\frac{1}{3} + \\frac{1}{3} - \\frac{1}{4} + \\ldots + \\frac{1}{2006} - \\frac{1}{2007} + \\frac{1}{2007} - \\frac{1}{2008}\n$$\ne, portanto, $S = 1 - \\frac{1}{2008} = \\frac{2007}{2008}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26596,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNo triângulo $\\triangle ABC$, $M$ é o ponto médio do lado $AC$, $D$ é um ponto do lado $BC$, tal que $AD$ é a bissetriz do ângulo $B\\widehat{A}C$, e $P$ é o ponto de interseção de $AD$ e $BM$. Sabendo que $AB = 10\\ \\mathrm{cm}$, $AC = 30\\ \\mathrm{cm}$ e a área do triângulo $\\triangle ABC$ mede $100\\ \\mathrm{cm}^2$, calcule a área do triângulo $\\triangle ABP$.",
    "options": [],
    "answer": "20",
    "solution": "Solution:\n\n$$\n\\frac{\\text{área}(\\triangle ABM)}{\\text{área}(\\triangle ABC)} = \\frac{\\frac{1}{2} AM \\cdot d}{\\frac{1}{2} AC \\cdot d} = \\frac{AM}{AC} = \\frac{1}{2}\n$$\nde modo que\n$$\n\\text{área}(\\triangle ABM) = \\frac{1}{2} \\text{área}(\\triangle ABC) = \\frac{1}{2} \\times 100 = 50\\ \\mathrm{cm}^2.\n$$\n![](attached_image_1.png)\nAnalogamente,\n$$\n\\frac{\\text{área}(\\triangle ABP)}{\\text{área}(\\triangle ABM)} = \\frac{BP}{BM}.\n$$\nPelo Teorema das Bissetrizes Internas,\n$$\n\\frac{BP}{PM} = \\frac{AB}{AM} = \\frac{10}{15} = \\frac{2}{3}\n$$\nportanto, $PM = \\frac{3}{2} BP$. Então obtemos\n$$\n\\frac{\\text{área}(\\triangle ABP)}{\\text{área}(\\triangle ABM)} = \\frac{BP}{BM} = \\frac{BP}{BP + PM} = \\frac{BP}{BP + \\frac{3}{2} BP} = \\frac{BP}{\\frac{5}{2} BP} = \\frac{2}{5}\n$$\nde modo que área $(\\triangle ABP) = \\frac{2}{5}$ área $(\\triangle ABM) = \\frac{2}{5} \\times 50 = 20\\ \\mathrm{cm}^2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26597,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nQuais são os dois últimos algarismos do número\n\n$$\n8 + 88 + 888 + \\cdots + \\overbrace{88 \\cdots 88}^{2008} ?\n$$",
    "options": [],
    "answer": "24",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26598,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOs vértices de um cubo são numerados de 1 a 8, de tal modo que uma das faces tem os vértices $\\{1,2,6,7\\}$ e as outras cinco têm os vértices $\\{1,4,6,8\\}$, $\\{1,2,5,8\\}$, $\\{2,3,5,7\\}$, $\\{3,4,6,7\\}$ e $\\{3,4,5,8\\}$. Qual é o número do vértice que está mais distante do vértice de número 6?\n\n(a) 1\n(b) 3\n(c) 4\n(d) 5\n(e) 7",
    "options": [],
    "answer": "d",
    "solution": "Solution:\n\nDesenhando o cubo e numerando seus vértices de acordo com o enunciado da questão, obtemos uma figura em que podemos ver que o vértice 5, por ser diametralmente oposto, é o mais distante do vértice 6.\n\n![](attached_image_1.png)\n\nO vértice 6 está nas faces $\\{1,2,6,7\\}$, $\\{1,4,6,8\\}$ e $\\{3,4,6,7\\}$. Como nessas faces só não aparece o 5, segue que este é o vértice diagonalmente oposto ao 6, ou seja, o 5 é o vértice mais distante do 6.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26599,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIdades múltiplas - Quando Isabel nasceu, sua mãe estava fazendo aniversário de 20 anos. Se Isabel e sua mãe viverem mais 100 anos, quantas vezes terão sido múltiplas as idades das duas?",
    "options": [],
    "answer": "6",
    "solution": "Solution:\n\nQuando Isabel tem $a$ anos, sua mãe tem $20+a$ anos. Se $a$ é divisor de $20+a$, então $(20+a)/a = (20/a) + 1$ é um número inteiro e, consequentemente, $20/a$ também é inteiro. Então, $a$ é um divisor de $20$ e, portanto, $a$ pode ser $1$, $2$, $4$, $5$, $10$ ou $20$. Assim, temos um total de $6$ vezes em que as idades das duas são múltiplos.\n\n| Isabel | 1 | 2 | 4 | 5 | 10 | 20 |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| Mãe | 21 | 22 | 24 | 25 | 30 | 40 |\n\nObserve que, depois dos $20$ anos de Isabel, nunca mais a idade da mãe será um múltiplo da idade de Isabel.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26600,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nAna tem um cubo de $10~\\mathrm{cm}$ de lado. Ela cortou o cubo em cubinhos de $1~\\mathrm{cm}$ de lado e, com esses cubinhos, ela brinca de formar outros blocos retangulares, mas sem que sobrem cubinhos. Por exemplo, ela formou um bloco de $10 \\times 20 \\times 5$. No total, quantos blocos diferentes ela pode construir com esses cubinhos, sem que sobre nenhum?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "19",
    "solution": "Solution:\nO volume do cubo é $10 \\times 10 \\times 10 = 1000~\\mathrm{cm}^3$. O volume $V$ de um bloco é o produto de suas três medidas, altura $(=a)$, largura $(=l)$ e comprimento $(=c)$.\n\nPara construir cada bloco, Ana deve usar todos os bloquinhos, portanto, o volume de cada bloco é\n$$\nV = \\text{altura} \\times \\text{largura} \\times \\text{comprimento} = a l c = 1000~\\mathrm{cm}^3\n$$\nAssim, precisamos saber de quantas maneiras podemos escrever $1000$ como o produto de três números inteiros positivos $a, l$ e $c$. Para isso, fatoramos $1000$, obtendo $a l c = 1000 = 2^3 \\times 5^3$.\n\nSolução 1: Podemos encontrar todos esses números listando as dimensões $a, l$ e $c$ dos blocos. Sem perda de generalidade, podemos supor que $a \\leq l \\leq c$. Então $a^3 \\leq a l c \\leq 1000$ e, portanto, $a \\leq 10$. Logo, $a = 1, 2, 4, 5, 8$ ou $10$. Mas se $a = 8$, então $l c = 125 = 5^3$ e, como $8 \\leq l \\leq c$, não há como obter medidas inteiras para $l$ e $c$. Assim, $a$ só pode ser $1, 2, 4, 5$ ou $10$. A tabela mostra as 19 possibilidades para esses blocos.\n\n| $a$ | $l$ | $c$ | $a$ | $l$ | $c$ |\n| ---: | ---: | ---: | ---: | ---: | ---: |\n|   | 1 | 1000 |   | 2 | 250 |\n|   | 2 | 500 |   | 4 | 125 |\n|   | 4 | 250 | 2 | 5 | 100 |\n| 1 | 5 | 200 |   | 10 | 50 |\n|   | 8 | 125 |   | 20 | 25 |\n|   | 10 | 100 |   | 5 | 40 |\n|   | 20 | 50 | 5 | 8 | 25 |\n|   | 25 | 40 |   | 10 | 20 |\n| 4 | 5 | 50 |   |   |   |\n|   | 10 | 25 | 10 | 10 | 10 |\n\nSolução 2: Podemos encontrar todos esses números listando as potências de 2 e 5, sem esquecer que uma das medidas pode ser 1 (no caso de potência 0). A tabela mostra as 19 possibilidades para esses blocos.\n\n| potência de 2 | potência de 5 | $a$ | $l$ | $c$ |\n| :---: | :---: | :---: | :---: | :---: |\n| 3 | 3 | 1 | 1 | $2^3 \\times 5^3$ |\n|   |   | 1 | $2^3$ | $5^3$ |\n| 1,2 | 3 | 1 | 2 | $2^2 \\times 5^3$ |\n|   |   | 1 | $2^2$ | $2 \\times 5^3$ |\n|   |   | 2 | $2^2$ | $5^3$ |\n| $1,1,1$ | 3 | 2 | 2 | $2 \\times 5^3$ |\n| 3 | 1,2 | 1 | $2^3 \\times 5$ | $5^2$ |\n|   |   | 1 | $2^3 \\times 5^2$ | 5 |\n|   |   | $2^3$ | 5 | $5^2$ |\n| 3 | $1,1,1$ | 5 | 5 | $2^3 \\times 5$ |\n| 1,2 | 1,2 | 1 | $2 \\times 5$ | $2^2 \\times 5^2$ |\n|   |   | 1 | $2 \\times 5^2$ | $2^2 \\times 5$ |\n|   |   | 2 | 5 | $2^2 \\times 5^2$ |\n|   |   | $2^2$ | $2 \\times 5$ | $5^2$ |\n|   |   | $2^2$ | $2 \\times 5^2$ | 5 |\n| 1,2 | $1,1,1$ | 2 | $2^2 \\times 5$ | $5^2$ |\n| $1,1,1$ | 1,2 | 5 | $2 \\times 5$ | $2^2 \\times 5$ |\n| $1,1,1$ | $1,1,1$ | $2 \\times 5$ | $2 \\times 5$ | $2 \\times 5^2$ |",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26601,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAna tem um cubo de $10~\\mathrm{cm}$ de lado. Ela cortou o cubo em cubinhos de $1~\\mathrm{cm}$ de lado e, com esses cubinhos, ela brinca de formar outros blocos retangulares, mas sem que sobrem cubinhos. Por exemplo, ela formou um bloco de $10 \\times 20 \\times 5$. No total, quantos blocos diferentes ela pode construir com esses cubinhos, sem que sobre nenhum?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "19",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26602,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nJoana escreveu os números de $1$ a $10000$ no quadro negro e, depois, apagou todos os múltiplos de $7$ e $11$. Qual foi o número que ficou na posição $2008$?",
    "options": [],
    "answer": "2573",
    "solution": "Solution:\n\nPrimeiro, vamos determinar quantos números de $1$ a $10000$ não são múltiplos de $7$ nem de $11$.\n\nO total de números é $10000$.\n\nNúmeros múltiplos de $7$: $\\left\\lfloor \\dfrac{10000}{7} \\right\\rfloor = 1428$.\n\nNúmeros múltiplos de $11$: $\\left\\lfloor \\dfrac{10000}{11} \\right\\rfloor = 909$.\n\nNúmeros múltiplos de $77$ (múltiplos de $7$ e $11$): $\\left\\lfloor \\dfrac{10000}{77} \\right\\rfloor = 129$.\n\nPelo Princípio da Inclusão-Exclusão, o total de números apagados é:\n\n$1428 + 909 - 129 = 2208$.\n\nPortanto, o total de números que sobraram é:\n\n$10000 - 2208 = 7792$.\n\nAgora, queremos saber qual número ficou na posição $2008$ (ou seja, o $2008$-ésimo número não apagado).\n\nSeja $n$ esse número. O número de inteiros de $1$ até $n$ que não são múltiplos de $7$ nem de $11$ deve ser $2008$.\n\nSeja $f(n)$ o número de inteiros de $1$ até $n$ que não são múltiplos de $7$ nem de $11$.\n\nTemos:\n\n$f(n) = n - \\left\\lfloor \\dfrac{n}{7} \\right\\rfloor - \\left\\lfloor \\dfrac{n}{11} \\right\\rfloor + \\left\\lfloor \\dfrac{n}{77} \\right\\rfloor$\n\nQueremos $f(n) = 2008$.\n\nVamos tentar encontrar $n$.\n\nAproximação:\n\nA cada $77$ números, $7 + 11 - 1 = 17$ são apagados ($11$ múltiplos de $7$, $7$ múltiplos de $11$, $1$ múltiplo de ambos), então $77 - 17 = 60$ ficam.\n\nPortanto, a cada $60$ números não apagados, percorremos $77$ números.\n\n$\\dfrac{2008}{60} \\approx 33.47$\n\nEntão, depois de $33$ blocos de $77$ números, temos $33 \\times 60 = 1980$ números não apagados, e já percorremos $33 \\times 77 = 2541$ números.\n\nFaltam $2008 - 1980 = 28$ números não apagados para chegar à posição $2008$.\n\nAgora, vamos olhar para os próximos números, de $2542$ em diante, e contar os que não são múltiplos de $7$ nem de $11$.\n\nVamos listar os números de $2542$ em diante e pular os múltiplos de $7$ ou $11$:\n\n$2542$ (não é múltiplo de $7$ nem $11$) $\\to$ $1981$\n$2543$ (não) $\\to$ $1982$\n$2544$ (não) $\\to$ $1983$\n$2545$ (não) $\\to$ $1984$\n$2546$ (não) $\\to$ $1985$\n$2547$ (não) $\\to$ $1986$\n$2548$ (não) $\\to$ $1987$\n$2549$ (não) $\\to$ $1988$\n$2550$ (não) $\\to$ $1989$\n$2551$ (não) $\\to$ $1990$\n$2552$ (não) $\\to$ $1991$\n$2553$ (não) $\\to$ $1992$\n$2554$ (não) $\\to$ $1993$\n$2555$ (não) $\\to$ $1994$\n$2556$ (não) $\\to$ $1995$\n$2557$ (não) $\\to$ $1996$\n$2558$ (não) $\\to$ $1997$\n$2559$ (não) $\\to$ $1998$\n$2560$ (não) $\\to$ $1999$\n$2561$ (não) $\\to$ $2000$\n$2562$ (não) $\\to$ $2001$\n$2563$ (não) $\\to$ $2002$\n$2564$ (não) $\\to$ $2003$\n$2565$ (não) $\\to$ $2004$\n$2566$ (não) $\\to$ $2005$\n$2567$ (não) $\\to$ $2006$\n$2568$ (não) $\\to$ $2007$\n$2569$ (não) $\\to$ $2008$\n\nAgora, precisamos verificar se algum desses números é múltiplo de $7$ ou $11$.\n\nVamos verificar de $2542$ em diante:\n\n$2548 \\div 7 = 364$ (exato), então $2548$ é múltiplo de $7$ e deve ser pulado.\n\n$2552 \\div 11 = 232$ (exato), então $2552$ é múltiplo de $11$ e deve ser pulado.\n\n$2555 \\div 11 = 232.27$ (não exato), $2555 \\div 7 = 365$ (não exato).\n\n$2556 \\div 7 = 365.14$ (não exato), $2556 \\div 11 = 232.36$ (não exato).\n\n$2557 \\div 7 = 365.28$ (não exato), $2557 \\div 11 = 232.45$ (não exato).\n\n$2559 \\div 11 = 232.63$ (não exato), $2559 \\div 7 = 365.57$ (não exato).\n\n$2561 \\div 7 = 365.86$ (não exato), $2561 \\div 11 = 232.82$ (não exato).\n\n$2563 \\div 7 = 366.14$ (não exato), $2563 \\div 11 = 233$ (exato), então $2563$ é múltiplo de $11$ e deve ser pulado.\n\n$2565 \\div 7 = 366.43$ (não exato), $2565 \\div 11 = 233.18$ (não exato).\n\n$2566 \\div 7 = 366.57$ (não exato), $2566 \\div 11 = 233.27$ (não exato).\n\n$2568 \\div 7 = 367$ (exato), então $2568$ é múltiplo de $7$ e deve ser pulado.\n\nAgora, vamos contar os números não apagados entre $2542$ e $2569$:\n\nDe $2542$ a $2569$ são $28$ números.\n\nMas precisamos subtrair os que são múltiplos de $7$ ou $11$:\n\nMúltiplos de $7$ nesse intervalo: $2548$, $2568$.\nMúltiplos de $11$ nesse intervalo: $2552$, $2563$.\n\nTotal de números apagados: $4$.\n\nPortanto, dos $28$ números, $28 - 4 = 24$ não são múltiplos de $7$ nem de $11$.\n\nOu seja, após $33$ blocos de $77$ números ($1980$ não apagados), precisamos de mais $28$ não apagados para chegar a $2008$.\n\nMas, como só há $24$ não apagados entre $2542$ e $2569$, precisamos ir além de $2569$.\n\n$1980 + 24 = 2004$ não apagados até $2569$.\n\nFaltam $2008 - 2004 = 4$ números não apagados.\n\nVamos continuar:\n\n$2570$ (não é múltiplo de $7$ nem $11$) $\\to$ $2005$\n$2571$ (não) $\\to$ $2006$\n$2572$ (não) $\\to$ $2007$\n$2573$ (não) $\\to$ $2008$\n\nVerificando:\n\n$2570 \\div 7 = 367.14$ (não exato), $2570 \\div 11 = 233.63$ (não exato).\n$2571 \\div 7 = 367.28$ (não exato), $2571 \\div 11 = 233.73$ (não exato).\n$2572 \\div 7 = 367.43$ (não exato), $2572 \\div 11 = 233.82$ (não exato).\n$2573 \\div 7 = 367.57$ (não exato), $2573 \\div 11 = 233.91$ (não exato).\n\nPortanto, o número que ficou na posição $2008$ é $2573$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26603,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nQual é o maior número possível de elementos de um subconjunto de $\\{1,2, \\ldots, 100\\}$ tal que nenhum de seus elementos seja um múltiplo de algum outro?",
    "options": [],
    "answer": "50",
    "solution": "Solution:\nInicialmente, observemos que o conjunto $\\{51,52,53, \\ldots, 100\\}$ tem 50 elementos e nenhum de seus elementos é múltiplo de outro. Assim, o subconjunto com o maior número de elementos e que satisfaz a propriedade exigida tem, no mínimo, 50 elementos.\n\nPara concluir que 50 é o maior número possível de elementos de um subconjunto que satisfaça a propriedade exigida, basta mostrar que todo subconjunto com mais de 50 elementos possui dois números múltiplos. Para isso, denotamos $B=\\{1,2,3, \\ldots, 100\\}$ e dividimos $B$ em 50 subconjuntos disjuntos, considerando os diversos subconjuntos de $B$ cujos elementos são do tipo número ímpar $\\times 2^{k}$, com $k$ natural. Como existem apenas 50 números ímpares entre 1 e 100, obtemos cinquenta subconjuntos dois a dois disjuntos construídos dessa forma, como segue.\n\n- $A_{1}=\\{1,2,4,8,16,32,64\\}=B \\cap\\left\\{1 \\times 2^{k} \\mid\\right.$ para algum $\\left.k=0,1,2,3, \\ldots\\right\\}$;\n- $A_{2}=\\{3,6,12,24,48,96\\}=B \\cap\\left\\{3 \\times 2^{k} \\mid\\right.$ para algum $\\left.k=0,1,2,3, \\ldots\\right\\}$;\n- $A_{3}=\\{5,10,20,40,80\\}=B \\cap\\left\\{5 \\times 2^{k} \\mid\\right.$ para algum $\\left.k=0,1,2,3, \\ldots\\right\\}$;\n- $A_{50}=\\{99\\}=B \\cap\\left\\{99 \\times 2^{k} \\mid\\right.$ para algum $\\left.k=0,1,2,3, \\ldots\\right\\}$.\n\nObserve que $B$ é a união desses cinquenta subconjuntos, isto é,\n$$\nB=\\{1,2, \\ldots, 100\\}=A_{1} \\cup A_{2} \\cup \\ldots \\cup A_{50}\n$$\ne que, se dois elementos de $B$ estiverem num mesmo subconjunto $A_{i}$, então um deles é múltiplo do outro. Assim, se um subconjunto $A$ de $B$ tiver mais do que 50 elementos, podemos afirmar que existem pelo menos dois elementos de $A$ num mesmo subconjunto $A_{i}$ e, portanto, um deles é múltiplo do outro. Isso prova que 50 é o número máximo de elementos de qualquer subconjunto de $B$ que não possua dois elementos tais que um deles seja múltiplo do outro.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26604,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDigite numa calculadora um número qualquer de três algarismos. Em seguida, digite o mesmo número obtendo, assim, um número de seis algarismos, da forma $abcabc$. Divida esse número por $7$, divida o resultado por $11$ e, finalmente, divida o número obtido por $13$. O que aconteceu? Por que você obteve esse resultado?",
    "options": [],
    "answer": "The final result is the original three-digit number.",
    "solution": "Solution:\n\nSeja o número de três algarismos $n = abc$. Ao digitá-lo duas vezes, obtemos o número de seis algarismos $abcabc$.\n\nEsse número pode ser escrito como:\n$$\nabcabc = 1000 \\times n + n = n \\times 1001\n$$\n\nAgora, dividimos $n \\times 1001$ por $7$, depois por $11$ e, finalmente, por $13$.\n\nNote que:\n$$\n1001 = 7 \\times 11 \\times 13\n$$\n\nPortanto,\n$$\nabcabc = n \\times 7 \\times 11 \\times 13\n$$\n\nAo dividir $abcabc$ por $7$, obtemos $n \\times 11 \\times 13$.\nDividindo por $11$, obtemos $n \\times 13$.\nDividindo por $13$, obtemos $n$.\n\nOu seja, ao final, o número obtido é exatamente o número de três algarismos que você digitou inicialmente.\n\nIsso acontece porque $1001$ é o produto de $7$, $11$ e $13$, e ao dividir sucessivamente por esses fatores, resta apenas o número original $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26605,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm galinheiro com $240~\\mathrm{m}^2$ de área deve abrigar galinhas e pintinhos, sendo desejável que haja um espaço livre de $4~\\mathrm{m}^2$ para cada galinha e $2~\\mathrm{m}^2$ para cada pintinho. Além disso, cada pintinho come $40~\\mathrm{g}$ de ração por dia e cada galinha come $160~\\mathrm{g}$ por dia, sendo permitido um gasto diário máximo de $8~\\mathrm{kg}$ de ração.\n\na. Represente algebricamente as condições do problema.\nb. Represente graficamente, no plano cartesiano $xOy$, as condições do problema.\nc. Esse galinheiro comporta 20 galinhas e 80 pintinhos? E 30 galinhas e 100 pintinhos?\nd. Qual é o número máximo de galinhas que podem ser colocadas no galinheiro, respeitando os espaços desejáveis e o gasto máximo de ração? E de pintinhos?",
    "options": [],
    "answer": "a) 4x + 2y ≤ 240; 160x + 40y ≤ 8000; x ≥ 0; y ≥ 0. b) Half-planes for 4x + 2y ≤ 240, 4x + y ≤ 200, x ≥ 0, y ≥ 0. c) 20 chickens and 80 chicks: yes; 30 chickens and 100 chicks: no. d) Maximum chickens: 50; maximum chicks: 120.",
    "solution": "Solution:\n\na.\nSeja $x$ o número de galinhas e $y$ o número de pintinhos.\n\nAs condições do problema são:\n\n- Área ocupada:\n  $$4x + 2y \\leq 240$$\n\n- Consumo de ração:\n  $$160x + 40y \\leq 8000$$\n  (pois $8~\\mathrm{kg} = 8000~\\mathrm{g}$)\n\n- Não negatividade:\n  $$x \\geq 0,\\quad y \\geq 0$$\n\n\nb.\nNo plano cartesiano $xOy$, as condições são representadas pelas regiões:\n\n- $4x + 2y \\leq 240$\n- $160x + 40y \\leq 8000$\n- $x \\geq 0$\n- $y \\geq 0$\n\nPodemos simplificar a segunda inequação:\n\n$$160x + 40y \\leq 8000$$\nDividindo ambos os lados por 40:\n$$4x + y \\leq 200$$\n\nPortanto, as regiões são:\n- $4x + 2y \\leq 240$\n- $4x + y \\leq 200$\n- $x \\geq 0$\n- $y \\geq 0$\n\n\nc.\nPara $x = 20$, $y = 80$:\n\n- Área: $4 \\times 20 + 2 \\times 80 = 80 + 160 = 240 \\leq 240$ (ok)\n- Ração: $160 \\times 20 + 40 \\times 80 = 3200 + 3200 = 6400 \\leq 8000$ (ok)\n\nPortanto, comporta 20 galinhas e 80 pintinhos.\n\nPara $x = 30$, $y = 100$:\n\n- Área: $4 \\times 30 + 2 \\times 100 = 120 + 200 = 320 > 240$ (não comporta)\n- Ração: $160 \\times 30 + 40 \\times 100 = 4800 + 4000 = 8800 > 8000$ (não comporta)\n\nPortanto, não comporta 30 galinhas e 100 pintinhos.\n\n\nd.\nNúmero máximo de galinhas ($y = 0$):\n\n- Área: $4x \\leq 240 \\implies x \\leq 60$\n- Ração: $160x \\leq 8000 \\implies x \\leq 50$\n\nLogo, o máximo é $x = 50$ galinhas.\n\nNúmero máximo de pintinhos ($x = 0$):\n\n- Área: $2y \\leq 240 \\implies y \\leq 120$\n- Ração: $40y \\leq 8000 \\implies y \\leq 200$\n\nLogo, o máximo é $y = 120$ pintinhos.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26606,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nQuinze minutos a mais - Dois carros partem, ao mesmo tempo, de uma cidade $A$ em direção a uma cidade $B$. Um deles viaja à velocidade constante de $60~\\mathrm{km}/\\mathrm{h}$ e o outro à velocidade constante de $70~\\mathrm{km}/\\mathrm{h}$. Se o carro mais rápido faz a viagem de $A$ a $B$ em $15$ minutos a menos do que o outro carro, qual é a distância entre essas duas cidades?",
    "options": [],
    "answer": "105 km",
    "solution": "Solution:\n\nSeja $d$ a distância entre as cidades $A$ e $B$.\n\nO tempo gasto pelo carro mais lento ($60~\\mathrm{km}/\\mathrm{h}$) é $t_1 = \\dfrac{d}{60}$ (em horas).\n\nO tempo gasto pelo carro mais rápido ($70~\\mathrm{km}/\\mathrm{h}$) é $t_2 = \\dfrac{d}{70}$ (em horas).\n\nSabemos que o carro mais rápido gasta $15$ minutos a menos:\n\n$$\nt_1 - t_2 = 15~\\text{minutos} = \\dfrac{1}{4}~\\text{hora}\n$$\n\nSubstituindo as expressões dos tempos:\n\n$$\n\\dfrac{d}{60} - \\dfrac{d}{70} = \\dfrac{1}{4}\n$$\n\nColocando em evidência $d$:\n\n$$\nd\\left(\\dfrac{1}{60} - \\dfrac{1}{70}\\right) = \\dfrac{1}{4}\n$$\n\nCalculando $\\dfrac{1}{60} - \\dfrac{1}{70}$:\n\n$$\n\\dfrac{1}{60} - \\dfrac{1}{70} = \\dfrac{70 - 60}{60 \\times 70} = \\dfrac{10}{4200} = \\dfrac{1}{420}\n$$\n\nPortanto:\n\n$$\nd \\cdot \\dfrac{1}{420} = \\dfrac{1}{4}\n$$\n\nMultiplicando ambos os lados por $420$:\n\n$$\nd = 420 \\times \\dfrac{1}{4} = 105\n$$\n\nLogo, a distância entre as cidades é $105~\\mathrm{km}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26607,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoma e inverte - Usando somente as duas operações \"+1=\\text{somar 1}\" e \" $-i=\\text{menos o inverso}$\", podemos formar várias sequências a partir de um número inicial. Por exemplo, iniciando com o número $3$, podemos formar a sequência\n$$\n3 \\xrightarrow{+1} 4 \\xrightarrow{+1} 5 \\xrightarrow{-i}-\\frac{1}{5} \\xrightarrow{+1} \\frac{4}{5} \\xrightarrow{-i}-\\frac{5}{4} \\xrightarrow{+1}-\\frac{1}{4} \\xrightarrow{+1} \\frac{3}{4} \\xrightarrow{-i}-\\frac{4}{3} .\n$$\nIniciando com $0$, com qual sequência obteremos novamente o $0$, usando apenas essas duas operações \"+1\" e \" $-i$ \"?",
    "options": [],
    "answer": "0 → +1 → 1 → −i → −1 → +1 → 0",
    "solution": "Solution:\n\nComo $0$ não é o inverso de número algum, qualquer sequência que comece e termine em $0$ deve ser dada por\n$$\n0 \\xrightarrow{+1} 1 \\longrightarrow \\cdots \\longrightarrow -1 \\xrightarrow{+1} 0.\n$$\nUma sequência dessas é a seguinte.\n$$\n\\begin{aligned}\n& 0 \\xrightarrow{+1} 1 \\xrightarrow{+1} 2 \\xrightarrow{+1} 3 \\xrightarrow{-i}-\\frac{1}{3} \\xrightarrow{+1} \\frac{2}{3} \\xrightarrow{+1} \\frac{5}{3} \\xrightarrow{+1} \\frac{8}{3} \\xrightarrow{-i}-\\frac{3}{8} \\xrightarrow{+1} \\frac{5}{8} \\xrightarrow{+1} \\frac{13}{8} \\xrightarrow{+1} \\frac{21}{8} \\\\\n& \\xrightarrow{-i}-\\frac{8}{21} \\xrightarrow{+1} \\frac{13}{21} \\xrightarrow{-i}-\\frac{21}{13} \\xrightarrow{+1}-\\frac{8}{13} \\xrightarrow{+1} \\frac{5}{13} \\xrightarrow{-i}-\\frac{13}{5} \\xrightarrow{+1}-\\frac{8}{5} \\xrightarrow{+1}-\\frac{3}{5} \\xrightarrow{+1} \\frac{2}{5} \\\\\n& \\xrightarrow{-i}-\\frac{5}{2} \\xrightarrow{+1}-\\frac{3}{2} \\xrightarrow{+1}-\\frac{1}{2} \\xrightarrow{+1} \\frac{1}{2} \\xrightarrow{-i}-2 \\xrightarrow{+1}-1 \\xrightarrow{+1} 0.\n\\end{aligned}\n$$\n\nUma outra sequência, bem mais curta e simples é, simplesmente,\n$$\n0 \\xrightarrow{+1} 1 \\xrightarrow{-i}-1 \\xrightarrow{+1} 0.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26608,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm triângulo isósceles tem uma base de $10~\\mathrm{cm}$ e dois lados iguais medindo $13~\\mathrm{cm}$. É possível cortar esse triângulo em dois outros triângulos de tal modo que, juntando esses triângulos de outra maneira obtenhamos um outro triângulo isósceles (evidentemente com a mesma área)?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Yes",
    "solution": "Solution:\n\nNum triângulo isósceles, a altura relativa à base coincide com a mediana. Traçando essa altura no triângulo dado, de base $10$, obtemos dois triângulos retângulos com catetos medindo $5$ e $h$ e hipotenusa $13$. Pelo Teorema de Pitágoras, temos $h^{2}+5^{2}=13^{2}$, donde $h^{2}=13^{2}-5^{2}=144$, portanto, $h=\\sqrt{144}=12$. Logo, a área do triângulo dado é\n$$\nA=\\frac{b \\times h}{2}=\\frac{10 \\times 12}{2}=60~\\mathrm{cm}^{2}\n$$\nAgora \"colamos\" os dois triângulos retângulos ao longo do cateto medindo $5$, obtendo um triângulo isósceles com $12+12=24~\\mathrm{cm}$ de base, lados de $13~\\mathrm{cm}$ e altura relativa à base igual a $5~\\mathrm{cm}$. Logo, esse novo triângulo isósceles também tem área igual a $\\frac{24 \\times 5}{2}=60~\\mathrm{cm}^{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26609,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm triângulo isósceles tem uma base de $10~\\mathrm{cm}$ e dois lados iguais medindo $13~\\mathrm{cm}$. É possível cortar esse triângulo em dois outros triângulos de tal modo que, juntando esses triângulos de outra maneira obtenhamos um outro triângulo isósceles (evidentemente com a mesma área)?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26610,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nClube de Matemática - Eu faço parte de um Clube de Matemática, onde tenho o mesmo número de colegas homens do que de colegas mulheres. Quando um garoto falta, três quartos da equipe são de meninas. Eu sou homem ou mulher? Quantas mulheres e quantos homens tem o clube?",
    "options": [],
    "answer": "Female; 3 women and 2 men",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26611,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDavi tem uma calculadora muito original, que efetua apenas duas operações, a adição usual $(+)$ e uma outra operação, denotada por $*$, que satisfaz\n\ni. $a * a = a$,\nii. $a * 0 = 2a$\niii. $(a * b) + (c * d) = (a + c) * (b + d)$,\n\npara quaisquer números inteiros $a$ e $b$. Quais são os resultados das operações $(2 * 3) + (0 * 3)$ e $1024 * 48$?",
    "options": [],
    "answer": "(2*3)+(0*3) = -2; 1024*48 = 2000",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26612,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDois irmãos - A diferença de idade entre dois irmãos é de três anos. Um ano atrás, a idade do pai desses irmãos era o dobro da soma das idades dos irmãos e, dentro de vinte anos, a idade do pai será a soma das idades desses dois filhos. Qual é a idade de cada um dos irmãos?",
    "options": [],
    "answer": "Younger sibling: 10 years; older sibling: 13 years",
    "solution": "Solution:\n\nSeja $x$ a idade do irmão mais novo e $x + 3$ a idade do irmão mais velho.\n\nSeja $p$ a idade do pai.\n\nUm ano atrás:\n- Idade do irmão mais novo: $x - 1$\n- Idade do irmão mais velho: $x + 2$\n- Idade do pai: $p - 1$\n\nA soma das idades dos irmãos, um ano atrás, era $(x - 1) + (x + 2) = 2x + 1$.\n\nO pai, um ano atrás, tinha o dobro da soma das idades dos filhos:\n$$\np - 1 = 2(2x + 1) = 4x + 2\n$$\nPortanto:\n$$\np = 4x + 3\n$$\n\nDaqui a 20 anos:\n- Idade do irmão mais novo: $x + 20$\n- Idade do irmão mais velho: $x + 3 + 20 = x + 23$\n- Idade do pai: $p + 20$\n\nA soma das idades dos irmãos, daqui a 20 anos, será:\n$$(x + 20) + (x + 23) = 2x + 43$$\n\nA idade do pai será igual à soma das idades dos filhos:\n$$\np + 20 = 2x + 43\n$$\n\nSubstituindo $p = 4x + 3$:\n$$\n4x + 3 + 20 = 2x + 43 \\\\\n4x + 23 = 2x + 43 \\\\\n4x - 2x = 43 - 23 \\\\\n2x = 20 \\\\\nx = 10\n$$\n\nPortanto:\n- Idade do irmão mais novo: $x = 10$ anos\n- Idade do irmão mais velho: $x + 3 = 13$ anos",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26613,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P(n)$ be a polynomial with real coefficients. Prove that there exist integers $n$ and $k$ such that $k$ has at most $n$ digits and at least $P(n)$ divisors.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $d$ be the degree of $P$, and consider $d+1$ distinct primes $p_1, p_2, \\dots, p_{d+1}$. Let $A = p_1p_2\\dots p_{d+1}$ and $k_N = A^N$. If $A$ has $r$ digits, then $k_N$ has at most $rN$ digits. On the other hand, $k_N$ has $(N+1)^{d+1}$ positive divisors. Since $P$ has degree $d$, $P(rN) < (N+1)^{d+1}$ for all sufficiently large $N$, which solves the problem.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 26614,
    "subject": "Mathematics (Multi-modal)",
    "question": "The cells of a $3 \\times 3$ table were numbered from $1$ to $9$, each number appearing exactly once. For each row the cell with the greatest number is colored red and the cell with the smallest number is colored green. Let $A$ be the smallest of the numbers in the red cells and $B$ the greatest of the numbers in the green cells.\n\na. Show a distribution of the numbers in the table such that $A - B = 4$.\n\nb. Show a distribution of the numbers in the table such that $A - B = -3$.\n\nc. Is it possible that $A = 4$ and $B = 3$?",
    "options": [],
    "answer": "a: Example arrangement with rows 7 4 1 / 8 5 2 / 9 6 3 gives A = 7 and B = 3, so A − B = 4.\nb: Example arrangement with rows 1 2 3 / 4 5 7 / 6 8 9 gives A = 3 and B = 6, so A − B = −3.\nc: Not possible (A = 4 and B = 3 cannot occur).",
    "solution": "a. For example,\n\n| 7 | 4 | 1 |\n|---|---|---|\n| 8 | 5 | 2 |\n| 9 | 6 | 3 |\n\nNotice that $A = \\min\\{7, 8, 9\\} = 7$ and $B = \\max\\{1, 2, 3\\} = 3$, so $A - B = 4$.\n\nb. For example,\n\n| 1 | 2 | 3 |\n|---|---|---|\n| 4 | 5 | 7 |\n| 6 | 8 | 9 |\n\nNotice that $A = \\min\\{3, 7, 9\\} = 3$ and $B = \\max\\{1, 4, 6\\} = 6$, so $A - B = -3$.\n\nc. No, it's not possible. Since $B = 3$ is the greatest of the numbers in green cells, which are the smallest ones in their respective rows, the numbers $1$, $2$ and $3$ must be on different rows. Now $A = 4$ means that $4$ is the greatest number of a row, but it could only happen if two of the numbers $1$, $2$, $3$ are on the same row as $4$, which cannot happen.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 26615,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a parallelogram and $\\Gamma$ the circumcircle of the triangle $ABD$. Lines $BC$ and $CD$ meet $\\Gamma$ at $E \\ne B$ and $F \\ne D$ respectively. Prove the circumcenter of the triangle $CEF$ lies on $\\Gamma$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider all angles oriented and modulo $180^\\circ$. Let $O$ be the center of the circle. In the cyclic pentagon $ABEDF$, $\\angle EBF = \\angle EDF = \\angle EDC = \\angle CED + \\angle DCE = \\angle BED + \\angle DCE = \\angle BAD + \\angle DCE = 2\\angle DCE$. This means that if $M$ is the midpoint of the arc $FE$ that does not contain $A$, $\\angle FME = \\angle EBF = 2\\angle DCE = 2\\angle FCE$. Since $ME = MF$, $M$ is the circumcenter of triangle $CEF$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26616,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be an integer and $n_1$ be one of its divisors. Let $A$ be a $n \\times n$ symmetric matrix defined by $a_{i,i} = 4$, $a_{i,i+1} = a_{i+1,i} = -1$ for all $i$ such that $1 \\le i \\le n-1$ and $i+1$ is not a multiple of $n_1$, $a_{i,i+n_1} = a_{i+n_1,i} = -1$ and $a_{i,j} = 0$ otherwise.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "See problem 3, grades 10–12.",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 26617,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOperação - Dados dois números reais $a$ e $b$, considere $a b = a^{2} - a b + b^{2}$. Quanto vale $1$?\n\n(a) 1\n(b) 0\n(c) 2\n(d) -2\n(e) -1",
    "options": [],
    "answer": "(a)",
    "solution": "Solution:\n\nFazendo $a = 1$ e $b = 0$ em $a b = a^{2} - a b + b^{2}$, obtemos $1 = 1^{2} - 1 \\times 0 + 0^{2} = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26618,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAda e Luisa treinam todos os dias, cada uma delas sempre com a mesma velocidade, para a grande corrida que vai acontecer no final do ano na escola. O treino começa num ponto $A$ e termina no ponto $B$, distantes $3000~\\mathrm{m}$. Elas partem no mesmo instante, mas quando Luisa termina a corrida, ainda faltam $120~\\mathrm{m}$ para Ada chegar ao ponto $B$. Ontem, Luisa deu uma chance para Ada: \"Partimos ao mesmo tempo, mas eu parto alguns metros antes do ponto A para chegarmos juntas.\" Quantos metros antes do ponto $A$ deve partir Luisa para chegar junto com Ada?",
    "options": [],
    "answer": "125",
    "solution": "Solution:\n\nSejam $v_A$ e $v_L$ as velocidades de Ada e Luisa, respectivamente. Ambas são constantes.\n\nQuando Luisa termina o percurso de $3000~\\mathrm{m}$, Ada ainda tem $120~\\mathrm{m}$ para percorrer, ou seja, Ada percorreu $3000 - 120 = 2880~\\mathrm{m}$ no mesmo tempo.\n\nAssim, o tempo que ambas levaram é o mesmo:\n\n$$\n\\frac{3000}{v_L} = \\frac{2880}{v_A}\n$$\n\nLogo,\n\n$$\nv_L = \\frac{3000}{2880} v_A = \\frac{25}{24} v_A\n$$\n\nAgora, para que cheguem juntas, Luisa deve partir $x$ metros antes do ponto $A$, ou seja, ela percorrerá $3000 + x$ metros, enquanto Ada percorre $3000$ metros.\n\nComo partem juntas e chegam juntas:\n\n$$\n\\frac{3000}{v_A} = \\frac{3000 + x}{v_L}\n$$\n\nSubstituindo $v_L$ por $\\frac{25}{24} v_A$:\n\n$$\n\\frac{3000}{v_A} = \\frac{3000 + x}{\\frac{25}{24} v_A}\n$$\n\nMultiplicando ambos os lados por $\\frac{25}{24} v_A$:\n\n$$\n\\frac{25}{24} \\cdot 3000 = 3000 + x\n$$\n\n$$\n\\frac{25 \\times 3000}{24} = 3000 + x\n$$\n\n$$\n3125 = 3000 + x\n$$\n\n$$\nx = 3125 - 3000 = 125~\\mathrm{m}\n$$\n\nPortanto, Luisa deve partir $125~\\mathrm{m}$ antes do ponto $A$ para chegar junto com Ada.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26619,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm triângulo tem vértices $A=(3,0)$, $B=(0,3)$ e $C$, onde $C$ está na reta de equação $x+y=7$. Qual é a área desse triângulo?",
    "options": [],
    "answer": "6",
    "solution": "Solution:\n\nObserve que a altura $h$ relativa ao lado $AB$ de todos os triângulos $\\triangle ABC$ que têm o vértice $C$ na reta $x+y=7$ é sempre a mesma, pois a reta $x+y=7$ é paralela à reta $x+y=3$ que passa por $A$ e $B$. Logo, todos esses triângulos têm a mesma área, a saber,\n$$\n\\frac{1}{2}(AB \\times h)\n$$\nResta, portanto, determinar $AB$ e $h$. Como $AB$ é a hipotenusa de um triângulo retângulo que tem os dois catetos iguais a $3$, segue do Teorema de Pitágoras que\n$$\nAB=\\sqrt{3^{2}+3^{2}}=\\sqrt{18}=3 \\sqrt{2}\n$$\nPor outro lado, $h$ é a distância entre as retas paralelas $x+y=3$ determinada pelos pontos $A$ e $B$ e $x+y=7$. A reta $x=y$ é perpendicular a essas retas paralelas e forma um ângulo de $45^{\\circ}$ com o eixo $Ox$. Sejam $M$ o pé da perpendicular à reta $x+y=7$ traçada a partir de $A$ e $D=(7,0)$ o ponto de corte da reta $x+y=7$ com o eixo $Ox$. O triângulo $\\triangle AMD$ assim formado é retângulo isósceles, com dois catetos iguais a $h$ e hipotenusa $7-3=4$. Pelo Teorema de Pitágoras segue que $4^{2}=h^{2}+h^{2}=2h^{2}$, ou seja, $h=\\sqrt{8}=2\\sqrt{2}$. Assim, a área procurada é dada por\n$$\n\\frac{1}{2}(3 \\sqrt{2} \\times 2 \\sqrt{2})=6\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26620,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUma companhia de eletricidade instalou um poste num terreno plano. Para fixar bem o poste, foram presos cabos no poste, a uma altura de $1{,}4$ metros do solo e a $2$ metros de distância do poste, sendo que um dos cabos mede $2{,}5$ metros, conforme a figura.\n\n\n\nUm professor de Matemática, após analisar estas medidas, afirmou que o poste não está perpendicular ao solo. Você acha que o professor está certo? Justifique sua resposta.",
    "options": [],
    "answer": "Yes. The professor is right; the pole is not perpendicular to the ground.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26621,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUma companhia de eletricidade instalou um poste num terreno plano. Para fixar bem o poste, foram presos cabos no poste, a uma altura de $1{,}4$ metros do solo e a $2$ metros de distância do poste, sendo que um dos cabos mede $2{,}5$ metros, conforme a figura.\n\n\n\nUm professor de Matemática, após analisar estas medidas, afirmou que o poste não está perpendicular ao solo. Você acha que o professor está certo? Justifique sua resposta.",
    "options": [],
    "answer": "Yes; the pole is not perpendicular to the ground.",
    "solution": "Solution:\n\nPara que o poste fique perpendicular ao solo, o ângulo em $A$ deve ser reto e, portanto, o triângulo $\\triangle ABC$ deve ser retângulo (ver figura). Nesse caso, os dados do problema dão que a hipotenusa mede $2{,}5\\ \\mathrm{m}$ e os catetos $1{,}4\\ \\mathrm{m}$ e $2\\ \\mathrm{m}$. Assim, pelo Teorema de Pitágoras teríamos $(2{,}5)^2 = (1{,}4)^2 + 2^2$.\n\n$$\n\\begin{array}{llll} \n& C & & \\\\\n& 2,5 & \\\\\n\\underset{-}{*} & & \\\\\nA & 2 & B\n\\end{array}\n$$\n\nEntretanto, $(1{,}4)^2 + 2^2 = 1{,}96 + 4 = 5{,}96$ e $(2{,}5)^2 = 6{,}25$. Logo, essas medidas não satisfazem o Teorema de Pitágoras e, portanto, o triângulo $\\triangle ABC$ não é retângulo. Assim, o ângulo em $A$ não é reto e, consequentemente, o poste não está perpendicular ao solo. Concluímos que o professor está certo.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26622,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n26. Festa de aniversário - A festa de aniversário de André tem menos do que 120 convidados. Para o jantar, ele pode dividir os convidados em mesas completas de seis pessoas ou em mesas completas de sete pessoas. Em ambos os casos, são necessárias mais do que 10 mesas e todos os convidados ficam em alguma mesa. Quantos são os convidados?\n\n27. Medida do cateto - Na figura dada, $ABCD$ é um retângulo e $\\triangle ABE$ e $\\triangle CDF$ são triângulos retângulos. A área do triângulo $\\triangle ABE$ é $150\\ \\mathrm{cm}^2$ e os segmentos $AE$ e $DF$ medem, respectivamente, $15$ e $24\\ \\mathrm{cm}$. Qual é o comprimento do segmento $CF$?",
    "options": [],
    "answer": "84",
    "solution": "Solution:\n\n26. Festa de aniversário - Como podemos repartir o total de convidados em mesas de 6 ou 7, o número de convidados é um múltiplo de 6 e de 7. Como o menor múltiplo comum de 6 e 7 é 42, podemos ter $42, 84, 126, \\ldots$ convidados. Como são menos do que 120 convidados, só podemos ter 42 ou 84 convidados. Por outro lado, como são necessárias mais do que 10 mesas, temos mais do que 60 convidados. Logo, descartamos o 42, e o número de convidados só pode ser 84.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26623,
    "subject": "Mathematics (Multi-modal)",
    "question": "What is the maximum area of the shadow cast by a unit cube? We understand “area of the shadow” of something as the area of its orthogonal projection in a given plane.",
    "options": [],
    "answer": "sqrt(3)",
    "solution": "Let $ABCD$ and $EFGH$ be two opposite faces, $AE$, $BF$, $CG$ and $DH$ being edges of the cube, and let $X'$ be the orthogonal projection of point $X$ onto the plane. Notice that $\\{A, G\\}$, $\\{B, H\\}$, $\\{C, E\\}$ and $\\{D, F\\}$ are pairs of opposite vertices. Suppose, without loss of generality, that $A'$ lies on the boundary of the projection of the cube. Then, considering the symmetry of the cube around the center of the cube, its symmetric point $G'$ lies on the boundary as well. Two of the three neighboring vertices of $A$ are going to be neighbors of $A'$ in the projection (unless, say, face *AEHD* projects onto a line; but in this case we consider a degenerate vertex inside this line). Suppose without loss that these neighbors are $B'$ and $D'$. So $E'$ is inside the projection. Again by symmetry $H'$ and $F'$ lie on the boundary of the projection and $C'$ lies inside the projection. Finally, since $\\overrightarrow{AE} = \\overrightarrow{BF} = \\overrightarrow{CG} = \\overrightarrow{DH}$, the projection of the cube is $A'D'H'G'F'B'$.\n\n![](attached_image_1.png)\n\nThe faces $ABCD$, $BCGF$ and $CDHG$ project onto the parallelograms (or line segments) $A'B'C'D$, $B'C'G'F$ and $C'D'H'G'$. Draw diagonals $B'D'$, $B'G'$ and $D'G'$. The area of the projection is then twice the area of the triangle $B'D'G'$, which is at most the area of triangle $BDG$. This triangle is equilateral with side $\\sqrt{2}$, so the desired maximum is $2\\frac{(\\sqrt{2})^2\\sqrt{3}}{4} = \\sqrt{3}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26624,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be an integer and $n_1$ be one of its divisors. Let $A$ be a $n \\times n$ symmetric matrix defined by $a_{i,i} = 4$, $a_{i,i+1} = a_{i+1,i} = -1$ for all $i$ such that $1 \\le i \\le n-1$ and $i+1$ is not a multiple of $n_1$, $a_{i,i+n_1} = a_{i+n_1,i} = -1$ and $a_{i,j} = 0$ otherwise.\nProve that $A$ has an inverse and that all of the entries in the inverse are positive.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let's find the inverse of $A/4 = I - X$, where all entries in $X$ are either $0$ or $1/4$. We will use the series\n$$\n(I - X)^{-1} = I + X + X^2 + X^3 + \\dots\n$$\nFirst let's prove that this series converges. It suffices to show that the maximum $M$ such that $\\|Xw\\| \\le M\\|w\\|$ for all column vectors $w$ of $X$ is less than $1$, so the sum of the entries always decrease by a factor smaller than $1$ if you multiply a vector by $X^2$; then we sum the series as $(I+X)(I+X^2+X^4+\\dots)$. Notice that every row of $X$ has at most four nonzero entries, all of which are equal to $1/4$. So if $w = (a_1, a_2, \\dots, a_n)$ then every entry of $Xw$ is of the form $\\frac{a_{r_1}+a_{r_2}+\\dots+a_{r_s}}{4}$, $s \\le 4$. By the Cauchy-Schwarz inequality, its square is at most $\\frac{s}{16}(a_{r_1}^2+a_{r_2}^2+\\dots+a_{r_s}^2) \\le \\frac{a_{r_1}^2+a_{r_2}^2+\\dots+a_{r_s}^2}{4}$, with equality if and only if $a_{r_1} = a_{r_2} = \\dots = a_{r_s}$ and $s=4$. Summing over all the rows, the sum of squares of the coordinates of $Xw$ is at most $a_1^2 + a_2^2 + \\dots + a_n^2 = \\|w\\|^2$, because all columns of $X$ have at most four nonzero entries. But equality would only happen if all entries $a_i$ are equal and $s=4$ always, which does not happen for, say, the first row. So $M < 1$ and the series converges.\n\nConsider the graph whose vertices are the numbers $v_1, v_2, \\dots, v_n$ and we connect $v_i$ and $v_j$ if and only if the entry $x_{ij}$ in $X$ is $1/4$. By the definition of the matrix $A$, this graph has a lattice-like configuration: it can be split in several paths $v_1v_2 \\dots v_{n_1-1}$, $v_{k_1}v_{k_1+1} \\dots v_{(k+1)n_1-1}$, $1 \\le k \\le \\frac{n}{2}-1$, $v_r v_{r+n_1} v_{r+2n_1} \\dots v_{r+n-n_1}$, $1 \\le r \\le n_1$. It is clear that this graph is connected. Since the entry $m_{ij}$ in $X^k$ is nonzero if and only if there exists a circuit from $i$ to $j$ with $k$ edges, for all $i, j$ there is $k$ such that the corresponding entry $m_{ij}$ in $X^k$ is nonzero. This proves that all entries in the inverse of $A$ is positive.",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 26625,
    "subject": "Mathematics (Multi-modal)",
    "question": "Arnold and Bernold play the following game in a $m \\times n$ board: Arnold chooses one of its cells and places a knight on it. Then Bernold and Arnold move the knight alternately, with the condition that the knight visits a cell at most once. The player who is unable to move the knight loses. Determine, in terms of $m$ and $n$, which player has the winning strategy.\n\n*Remark*: the knight always moves two cells in a row or a column and then one cell in the perpendicular direction.",
    "options": [],
    "answer": "Assuming the smaller side is at most the larger: if the smaller side equals two, then the first player wins if and only if the larger side is not divisible by four; if the smaller side is at least three, then the first player wins if and only if both side lengths are odd. Otherwise the second player wins.",
    "solution": "Suppose, without loss of generality, $m \\le n$. If $m = 2$, Arnold has winning strategy if and only if $n$ is not a multiple of $4$; for $m \\ge 3$, Arnold has winning strategy if and only if $m$ and $n$ are both odd.\n\nSuppose $m = 2$. If $4$ does not divide $n$, Arnold can win placing the knight on the first column if $n = 4k + 1$ and on the second column if $n = 4k + r$, $r = 2$ or $r = 3$. The knight should always be moved two columns ahead on each move, allowing exactly $2k$ moves.\nNow consider $n = 4k$. Divide the table into $2 \\times 4$ subtables, and pair the cells in such a way that it's always possible to move the knight between cells from each pair:\n\n| 1 | 2 | 3 | 4 |\n|---|---|---|---|\n| 3 | 4 | 1 | 2 |\n\nSince the whole table is divided into pairs, Bernold can always move, no matter where Arnold places the knight: it suffices to move the knight to the other cell in the pair. When Arnold plays, he will move the knight to a cell from another pair, and Bernold can repeat this strategy. So if $n = 4k$ Bernold has winning strategy.\nThis finishes the case $m = 2$. The case $m \\ge 3$ follows in a similar fashion, dividing the table into smaller subtables:\n\n| 1 | 2 | 3 | 4 |\n|---|---|---|---|\n| 3 | 6 | 1 | 5 |\n| 2 | 5 | 4 | 6 |\n\n| 1 | 2 | 3 | 4 | 5 | 6 |\n|---|---|---|---|---|---|\n| 3 | 4 | 1 | 7 | 8 | 9 |\n| 2 | 7 | 8 | 9 | 6 | 5 |\n\n| 1 | 2 | 3 |\n|---|---|---|\n| 4 | A | 1 |\n| 2 | 3 | 4 |\n\n| 1 | 2 | 3 | 4 | 5 |\n|---|---|---|---|---|\n| 3 | 4 | 6 | A | 7 |\n| 2 | 1 | 7 | 5 | 6 |\n\nThose tables prove that Bernold has winning strategy for tables $3 \\times n$, $n$ even and Arnold has winning strategy for tables $3 \\times n$, $n$ odd: divide the table into one $3 \\times 3$ or $3 \\times 5$ subtable and several $3 \\times 4$ subtables; it suffices to place the knight on the cell marked with an $A$ and then use the aforementioned Bernold's strategy.\nThe case $m = 4$ can be verified using several $4 \\times 2$ subtables if $n$ is even and one $4 \\times 3$ subtable and several $4 \\times 2$ subtables if $n$ is odd. This also proves that if either Arnold or Bernold has winning strategy for a $m \\times n$ table then he also has winning strategy for a $(m+4) \\times n$ table, $m \\ge 3$. So it suffices to solve the problem for $m \\in \\{3, 4, 5, 6\\}$.\nThe case $m = 5$ can be solved using the following subtables:\n\n| 1 | 2 | 3 | 4 | 5 | 6 |\n|---|---|---|---|---|---|\n| 3 | 4 | 1 | 7 | 8 | 9 |\n| 2 | 10 | 11 | 9 | 6 | 5 |\n| 12 | 13 | 14 | 15 | 7 | 8 |\n| 10 | 11 | 12 | 13 | 14 | 15 |\n\n| 1 | 2 | 3 | 4 | 5 |\n|---|---|---|---|---|\n| 3 | 4 | 5 | 6 | 7 |\n| 2 | 1 | 10 | 8 | 9 |\n| 11 | A | 12 | 7 | 6 |\n| 12 | 10 | 11 | 9 | 8 |\n\nNotice that Bernold has winning strategy for a $5 \\times 4$ table and Arnold has winning strategy for a $5 \\times 3$ table. Thus, in the case $5 \\times n$, $n$ even, we merge $5 \\times 4$ subtables if $n$ is a multiple of $4$ and one $5 \\times 6$ subtable and $5 \\times 4$ subtables if $n = 4k + 2$; if $n$ is odd, we merge several $5 \\times 4$ subtables to a $5 \\times 3$ or $5 \\times 5$ subtable, if $n = 4k + 3$ or $n = 4k + 1$, respectively.\nThe case $6 \\times n$ follows directly from the case $3 \\times n$ if $n$ is even (merge two $3 \\times n$ tables) and merging $6 \\times 4$ subtables to a $6 \\times 3$ or $6 \\times 5$ subtable, if $n = 4k + 3$ or $n = 4k + 1$, respectively. All the cases are covered.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26626,
    "subject": "Mathematics (Multi-modal)",
    "question": "Define the polynomials $\\binom{x}{j} = \\frac{x(x-1)(x-2)\\cdots(x-j+1)}{j!}$ for $j$ positive integer and $\\binom{x}{0} = 1$.\n\na. Prove that all non-nil polynomials can be written uniquely as a linear combination of such polynomials $\\binom{x}{n}$.\n\nb. Let $c(n,k)$ be the coefficient of $\\binom{x}{k}$ in $x^n$ (as described in the previous item). Compute\n$$\n\\frac{c(n, k) + c(n, k + 1)}{c(n + 1, k + 1)}.\n$$",
    "options": [],
    "answer": "1/(k+1)",
    "solution": "a. Induct on the degree $n$ of the polynomial $P(x) = \\sum_{k=0}^{n} a_k x^k$. It is immediate if $n=0$. Since the degree of $(x^k)$, $k < n$, is $k$, the coefficient on $(x^n)$ on $P(x)$ is $a_n \\cdot n!$. Now consider the polynomial $P(x) - a_n \\cdot n! (x^n)$. This is a polynomial with degree less than $n$, so $P(x) - a_n \\cdot n! (x^n)$ can be uniquely represented as a linear combination of the polynomials $(x^k)$, and the result follows.\n\nb. We have $x^n = \\sum_{k=0}^{n} c(n, k) \\binom{x}{k}$, so\n$$\n\\begin{align*} \nx^{n+1} &= \\sum_{k=0}^{n} c(n,k) \\binom{x}{k} x \\\n&= \\sum_{k=0}^{n} c(n,k) \\binom{x}{k} (x+1) - \\sum_{k=0}^{n} c(n,k) \\binom{x}{k} \\\n&= \\sum_{k=0}^{n} c(n,k) (k+1) \\binom{x+1}{k+1} - \\sum_{k=0}^{n} c(n,k) \\binom{x}{k} \\\n&= \\sum_{k=0}^{n} c(n,k) (k+1) \\left( \\binom{x}{k+1} + \\binom{x}{k} \\right) - \\sum_{k=0}^{n} c(n,k) \\binom{x}{k} \\\n&= \\sum_{k=0}^{n} (c(n,k-1)k + (k+1-1)c(n,k)) \\binom{x}{k} \\\n&= \\sum_{k=0}^{n} k(c(n,k-1) + c(n,k)) \\binom{x}{k} \n\\end{align*}\n$$\nin which we consider $c(n, -1) = 0$.\nThus $c(n+1, k) = k(c(n, k-1) + c(n, k)) \\iff \\frac{c(n,k)+c(n,k-1)}{c(n+1,k)} = \\frac{1}{k}$, and\n$\\frac{c(n,k+1)+c(n,k)}{c(n+1,k+1)} = \\frac{1}{k+1}$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 26627,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a quadrilateral with $\\angle ABC \\neq 90^\\circ$. Let $M$ and $N$ be the midpoints of $AD$ and $CD$, respectively. Prove that the lines perpendicular to $BC$ passing through $M$ and perpendicular to $AB$ passing through $N$ and $BD$ are concurrent if and only if the diagonals $BD$ and $AC$ are perpendicular.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider a homothety with center on $D$ that takes $M$ to $A$ and $N$ to $C$. So the perpendicular lines are mapped to the altitudes of the triangle $ABC$ relative to $A$ and $C$, and the intersection $P$ of the perpendicular lines is mapped to the orthocenter $H$ of triangle $ABC$.\n\n![](attached_image_1.png)\n\nNotice that the condition that the perpendicular lines and $BD$ are concurrent is equivalent to $B$, $P$ and $D$ being collinear. But the homothety implies that $D$, $P$ and $H$ are collinear, so the three lines are concurrent if and only if $BH$ and $BD$ coincide, that is, $BD \\perp AC$, since $BH \\perp AC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26628,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$ and $c$ be real numbers such that $a \\neq b$ and $a^2(b + c) = b^2(c + a) = 2010$. Compute $c^2(a + b)$.",
    "options": [],
    "answer": "2010",
    "solution": "Since $a \\neq b$, $a^2(b+c) = b^2(c+a) \\iff a^2b + a^2c - b^2c - ab^2 = 0 \\iff ab(a-b) + c(a-b)(a+b) = 0 \\iff ab+ca+bc = 0$.\n\nSo $(a-c)(ab+bc+ca) = 0 \\iff a^2b + abc + a^2c = abc + bc^2 + ac^2 \\iff c^2(a+b) = a^2(b+c) = 2010$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26629,
    "subject": "Mathematics (Multi-modal)",
    "question": "A positive integer $n$ is *clowny* if the number obtained by reversing its digits is greater than $n$. For example, $2009$ is clowny because $9002$ is greater than $2009$; however, $2010$ is not clowny because $0102 = 102$ is less than $2010$ and $3443$ is not clowny because it is equal to the number obtained by reversing its digits. How many four-digit numbers are clowny?",
    "options": [],
    "answer": "441",
    "solution": "Let $(abcd)$ be a four-digit number $a$, $b$, $c$, $d$ being its digits. So a number is clowny if and only if $(dcba) > (abcd)$. This means that either $d > a$ or $d = a$ and $c > b$. Notice that we cannot have both $d = a$ and $b = c$ because it would imply $(dcba) = (abcd)$.\n\nThere are $\\frac{9 \\cdot 8}{2} = 36$ choices in the first case (recall that $d > a > 0$) and $9 \\cdot \\frac{10 \\cdot 9}{2} = 405$ choices in the second case ($9$ choices for $a = d$ and $\\frac{10 \\cdot 9}{2}$ for $c > b \\ge 0$). So there are $36 + 405 = 441$ four-digit clowny numbers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26630,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nQual é a menor fração? Quantas frações da forma $\\frac{n}{n+1}$ são menores do que $\\frac{7}{9}$, sabendo que $n$ é um número inteiro positivo?\n\n(a) 1\n(b) 2\n(c) 3\n(d) 4\n(e) 5",
    "options": [],
    "answer": "c",
    "solution": "Solution:\n\nAs frações da forma $\\frac{n}{n+1}$, com $n$ inteiro positivo, são\n\n![](attached_image_1.png)\n\nObserve que temos $\\frac{1}{2}<\\frac{2}{3}<\\frac{3}{4}<\\frac{4}{5}<\\ldots$, ou seja, essa sequência de frações é crescente. Para comparar cada uma dessas frações com $\\frac{7}{9}$, precisamos igualar todos os denominadores, obtendo $\\frac{1}{2}=\\frac{9}{18}<\\frac{14}{18}=\\frac{7}{9}$, $\\frac{2}{3}=\\frac{6}{9}<\\frac{7}{9}$, $\\frac{3}{4}=\\frac{27}{36}<\\frac{28}{36}=\\frac{7}{9}$ e $\\frac{4}{5}=\\frac{36}{45}>\\frac{35}{45}=\\frac{7}{9}$. Logo, $\\frac{4}{5}$ é maior do que $\\frac{7}{9}$ e, como a sequência é crescente, a partir de $\\frac{4}{5}$, todas as frações dessa sequência são maiores do que $\\frac{7}{9}$. Assim, existem apenas três frações da forma $\\frac{n}{n+1}$ que são menores do que $\\frac{7}{9}$, a saber, $\\frac{1}{2}$, $\\frac{2}{3}$ e $\\frac{3}{4}$.\nSolution:\n\nTransformando tudo em números decimais, temos $\\frac{7}{9}=0,777\\ldots$ e $\\frac{1}{2}=0,5$, $\\frac{2}{3}=0,666\\ldots$, $\\frac{3}{4}=0,75$, $\\frac{4}{5}=0,8$, $\\frac{5}{6}=0,8333\\ldots$ Logo, a sequência é crescente e apenas $\\frac{1}{2}=0,5$, $\\frac{2}{3}=0,666\\ldots$ e $\\frac{3}{4}=0,75$ são menores do que $\\frac{7}{9}=0,777\\ldots$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26631,
    "subject": "Mathematics (Multi-modal)",
    "question": "a. Show a positive integer not greater than $1000$ with at least $20$ positive divisors.\n\nb. Does there exist a positive integer not greater than $11000$ with at least $200$ positive divisors?",
    "options": [],
    "answer": "a: 900; b: No",
    "solution": "a. For example, $900 = 2^2 \\cdot 3^2 \\cdot 5^2$, which has $(2+1) \\cdot (2+1) \\cdot (2+1) = 27$ positive divisors.\n\nb. No, there doesn't. Let $n$ be a number with at least $200$ divisors. If the $i$-th divisor is $d$, then the $i$-th to last divisor is $\\frac{n}{d}$. Let $m$ be the $100$th divisor. So $m \\ge 100$ and $\\frac{n}{m} > m \\iff n > m^2 = 10000$. Close enough, but how do we fix this? First notice that if $97$, $99$ and $100$ are all divisors of $n$, then $n \\ge \\text{lcm}(97, 99, 100) > 11000$. So the key observation is considering the $98$th, $99$th and $100$th divisors. Let $k, \\ell$ and $m$ be such divisors. Notice that if $m \\ge 105$ we are done because then $n > m^2 = 11025 > 11000$. So $98 \\le k < \\ell < m \\le 104$. But $\\text{gcd}(x, y) \\le |x - y|$ implies that $n \\ge \\text{lcm}(k, \\ell, m) \\ge \\frac{k \\cdot \\ell \\cdot m}{(\\ell-k)(m-\\ell)(m-k)} \\ge \\frac{98 \\cdot 99 \\cdot 100}{(104-98) \\cdot 3^2} > 11000$. Here we used the fact that if $x+y \\le 2t$ then $xy \\le t^2$ applied to $x = \\ell - k, y = m - \\ell$ and $t = 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26632,
    "subject": "Mathematics (Multi-modal)",
    "question": "For each finite subset $F$ of the space $R^3$, define $V_r(F)$ as the union of the open spheres with center on each point of $F$ and radius $r$. Prove that, for $0 < r < R$,\n$$\n\\mathrm{vol}(V_R(F)) \\le \\frac{R^3}{r^3} \\mathrm{vol}(V_r(F)).\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $F = \\{P_1, P_2, \\dots, P_n\\}$ and let $\\alpha_{ij}$ be the perpendicular plane bisector of $P_i$ and $P_j$ for $i \\neq j$. Those planes define $n$ convex regions $R_1, R_2, \\dots, R_n$, where $R_i$ is the intersection of the half-spaces determined by $\\alpha_{ij}$ that contain $P_i$. Finally, let $A_r(i)$ be the intersection of $R_i$ with the sphere with center $P_i$ and radius $r$. Thus, since the (disjoint) union of the regions $R_i$ is the whole space, $V_r(F)$ is the disjoint union of $A_r(1), A_r(2), \\dots, A_r(n)$.\n\nNow, for each point $P_i$, apply a homothety with center $P_i$ and ratio $r/R < 1$. It is clear that the image of $A_R(i)$ is contained in $A_r(i)$, since a sphere with radius $R$ is taken to a sphere with radius $r$ and the planes $\\alpha_{ij}$ are taken to planes $\\alpha'_{ij}$ closer to $P_i$. Thus $\\text{vol}(A_r(i)) \\ge (\\frac{r}{R})^3 \\text{vol}(A_R(i))$ and the result follows by summing up these inequalities for $i = 1, 2, \\dots, n$.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 26633,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $O$ be the intersection point of the diagonals of the cyclic quadrilateral $ABCD$. The circumcircles of triangles $AOB$ and $COD$ meet lines $BC$ and $AD$ again at $M$, $N$, $P$ and $Q$. Prove that the quadrilateral $MNPQ$ is inscribed in a circle with center $O$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider angles oriented modulo $180^\\circ$. Since $\\angle PCO = \\angle BCA = \\angle BDA = \\angle ODN$ are inscribed in the circumcircle of the triangle $OCD$, $OP = ON$. Analogously, $OM = OQ$.\n\n![](attached_image_1.png)\n\nNow, $\\angle QOP = \\angle CPO = \\angle CDO = \\angle CDB = \\angle CAB = \\angle OAB = \\angle OQB = \\angle OQP$, so $OP = OQ$, and analogously, $OM = ON$.\n\nSince $OM = ON = OP = OQ$, the result follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26634,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sidelengths and area of a triangle are all integer numbers. Find the minimum value of its area.",
    "options": [],
    "answer": "6",
    "solution": "The $3$–$4$–$5$ triangle has area $\\frac{3 \\cdot 4}{2} = 6$. We will prove that no other triangle with integer sidelengths and area has smaller area.\nLet $a$, $b$, $c$ be the sidelengths. Then its area is $S = \\sqrt{s(s-a)(s-b)(s-c)}$, where $s = \\frac{a+b+c}{2}$. Since the area is also an integer, $a+b+c$ is even, and $s$, $s-a$, $s-b$, $s-c$ are all integers.\nNow, notice that the triangle cannot be equilateral, since equilateral triangles with an integer side have irrational area. So, at least two of the three integer numbers $s-a$, $s-b$, $s-c$ are distinct and, since $s = (s-a) + (s-b) + (s-c) \\ge 1+1+2 = 4$, $S \\ge \\sqrt{4 \\cdot 2 \\cdot 2 \\cdot 1} = \\sqrt{8}$, so $S \\ge 3$.\nIf $S$ is odd, $s \\ge 5$ and $s-a$, $s-b$, $s-c$ are all odd, so $S \\ge \\sqrt{5 \\cdot 3 \\cdot 1 \\cdot 1} = \\sqrt{15}$, so $S \\ge 5$. The only relevant case is $S = 5$. But this would imply two of $s$, $s-a$, $s-b$, $s-c$ being equal to $5$, which is impossible.\nIf $S$ is even, the only relevant case is $S = 4$. But then all of $s$, $s-a$, $s-b$, $s-c$ are powers of two. So if $s > 4$ then $s \\ge 8$ and $s-a$, $s-b$, $s-c$ would be, in some order, $1$, $1$, $2$, which is not possible because $s = (s-a)+(s-b)+(s-c)$. If $s = 4$, the only possibility would be $s-a$, $s-b$, $s-c$ being $1$, $1$, $2$, which does not work either.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26635,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that if $10^{2n} + 8 \\cdot 10^n + 1$ has a prime factor of the form $60k + 7$, $k$ and $n$ both positive integers, then $n$ and $k$ are both even.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $p = 60k + 7$ be such a prime. Then $10^{2n} + 8 \\cdot 10^n + 1 \\equiv 0 \\pmod{p} \\iff (10^n - 1)^2 \\equiv -10^{n+1} \\pmod{p}$. Now suppose $n$ is odd. Then $(10^n - 1)^2 \\equiv -1 \\cdot (10^{(n+1)/2})^2 \\pmod{p}$, and $-1$ is a quadratic residue. But by the Euler criterion, $\\left(\\frac{-1}{p}\\right) = (-1)^{(p-1)/2} = -1$, a contradiction. So $n$ is even and, moreover, $\\left(\\frac{-10}{p}\\right) = 1 \\iff \\left(\\frac{2}{p}\\right)\\left(\\frac{5}{p}\\right) = -1 \\iff \\left(\\frac{2}{p}\\right) = -\\left(\\frac{5}{p}\\right)$.\nBy the quadratic reciprocity lemma, $\\left(\\frac{5}{p}\\right)\\left(\\frac{p}{5}\\right) = (-1)^{\\frac{5-1}{2} \\cdot \\frac{p-1}{2}} = 1 \\iff \\left(\\frac{5}{p}\\right) = \\left(\\frac{p}{5}\\right) = \\left(\\frac{2}{5}\\right) = -1$. So $\\left(\\frac{2}{p}\\right) = 1 \\iff (-1)^{\\frac{p^2-1}{8}} = 1$, so $\\frac{p^2-1}{8} = (30k+3)(15k+2)$ is even. Since $30k+3$ is odd, $15k+2$ is even, that is, $k$ is even.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26636,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $a, b$ such that\n$$\n3^a = 2b^2 + 1.\n$$",
    "options": [],
    "answer": "[[1, 1], [2, 2], [5, 11]]",
    "solution": "The only solutions are $(1, 1)$, $(2, 2)$ and $(5, 11)$.\n\nIf $a$ is even and greater than 2, the equation can be rewritten as $(3^{a/2} - 1) \\cdot (3^{a/2} + 1) = 2b^2$. But $\\gcd(3^{a/2} - 1, 3^{a/2} + 1) = \\gcd(3^{a/2} - 1, 2) = 2$, so $3^{a/2} + 1 = 4u^2$ and $3^{a/2} - 1 = 2v^2$ or $3^{a/2} + 1 = 2u^2$ and $3^{a/2} - 1 = 4v^2$.\n\nIn the former case, $3^{a/2} = (2v - 1)(2v + 1)$, and since $\\gcd(2v - 1, 2v + 1) = \\gcd(2v - 1, 2) = 1$, $2v - 1 = 1 \\iff v = 1$ and $a/2 = 1 \\iff a = 2$ and thus $b = 2$.\n\nIn the latter case, $3^{a/2} = 4v^2 + 1 \\implies 0 \\equiv v^2 + 1 \\pmod 3 \\iff v^2 \\equiv -1 \\pmod 3$, which is impossible.\n\nIf $a$ is odd, the equation is equivalent to $3 \\cdot (3^{(a-1)/2})^2 - 2b^2 = 1$. Let $c = 3^{(a-1)/2}$. Let's find the solutions to $3c^2 - 2b^2 = 1$ (*). Since\n$$\n(\\sqrt{3} + \\sqrt{2})(\\sqrt{3} - \\sqrt{2}) = 1 \\implies (\\sqrt{3} + \\sqrt{2})^{2k+1}(\\sqrt{3} - \\sqrt{2})^{2k+1} = 1\n$$\nand\n$$\n(\\sqrt{3} + \\sqrt{2})^{2k+1} = c_k\\sqrt{3} + b_k\\sqrt{2} \\quad \\text{y} \\quad (\\sqrt{3} - \\sqrt{2})^{2k+1} = c_k\\sqrt{3} - b_k\\sqrt{2} \\text{ (**)}\n$$\n$(c_k, b_k)$ are solutions to $(*)$, for integers $k \\ge 0$. Now suppose that $(\\alpha, \\beta)$ is a different solution. Then there exists an integer $k$ such that\n$$\n(\\sqrt{3} + \\sqrt{2})^{2k-1} < \\alpha\\sqrt{3} + \\beta\\sqrt{2} < (\\sqrt{3} + \\sqrt{2})^{2k+1} \\\\\n\\iff \\sqrt{3} + \\sqrt{2} < \\frac{\\alpha\\sqrt{3} + \\beta\\sqrt{2}}{(5 + 2\\sqrt{6})^{k-1}} < 9\\sqrt{3} + 11\\sqrt{2} \\\\\n\\iff \\sqrt{3} + \\sqrt{2} < (\\alpha\\sqrt{3} + \\beta\\sqrt{2}) \\cdot (5 - 2\\sqrt{6})^{k-1} < 9\\sqrt{3} + 11\\sqrt{2}\n$$\nIt's not hard to prove by induction that $(\\alpha\\sqrt{3} + \\beta\\sqrt{2}) \\cdot (5 - 2\\sqrt{6})^{k-1} = \\theta\\sqrt{3} + \\phi\\sqrt{2}$, $\\theta$ and $\\phi$ both integers. Moreover, $\\theta\\sqrt{3} + \\phi\\sqrt{2} > 1 > \\frac{1}{\\theta\\sqrt{3} + \\phi\\sqrt{2}} = \\theta\\sqrt{3} - \\phi\\sqrt{2} > 0$, so $\\theta\\sqrt{3} > \\phi\\sqrt{2} > 0$. Hence $(\\theta, \\phi)$ is a solution to $(*)$, with $1 < \\theta < 9$. But it can be verified that there are no solutions with $1 < \\theta < 9$, contradiction.\n\nNow suppose that $c > 9$, that is, $k > 1$. By $(**)$ and the binomial theorem,\n$$\nc = \\sum_{m=0}^{k} \\binom{2k+1}{2m+1} \\cdot 3^m \\cdot 2^{k-m} = (2k+1) \\cdot 2^k + \\binom{2k+1}{3} \\cdot 3 \\cdot 2^{k-1} + \\dots \\quad (***)\n$$\nThus 3 divides $2k+1$. Let $3^t$ be the largest power of 3 that divides $2k+1$. Since $\\binom{2k+1}{3} \\cdot 3 \\cdot 2^{k-1} = (2k+1) \\cdot k \\cdot (2k-1) \\cdot 2^{2k-1}$, the largest power of 3 that divides the second term of the right hand side of $(***)$ is also $3^t$ and, letting $3^s$ be the largest power of 3 that divides $2m+1$, the $m$-th term $\\binom{2k+1}{2m+1} \\cdot 3^m \\cdot 2^{k-m} = \\frac{2k+1}{2m+1} \\binom{2k}{2m} \\cdot 3^m \\cdot 2^{k-m}$ has at least $t+m-s \\ge t + \\frac{3^s+1}{2} - s$ factors 3 ($t$ from $2k+1$, $m$ from $3^m$, minus $s$ from $2m+1$). Since $3^s = (1+2)^s \\ge 1+2s \\iff \\frac{3^s+1}{2} - s \\ge 1$ for $s \\ge 0$, with equality if and only if $s=0$ or $s=1$, the $m$-th term has more than $t+1$ factors 3 except when $m = \\frac{3^s+1}{2}$. If $s=0$ we have $m=1 \\iff 2m+1=3$, a known solution; if $s=1$ we have $m=2$ which has no factors 3. Thus, since $k>1$, all terms starting from the third one have at least $t+2$ factors 3, and\n$$\nc = (2k+1) \\cdot 2^k + (2k+1) \\cdot k \\cdot (2k-1) \\cdot 2^{k-1} + 3^{t+2} \\cdot N \\\\\n\\iff 3^{(a-1)/2} = (2k+1) \\cdot 2^{k-1} \\cdot [(2k+1)(k-1) + 3] + 3^{t+2} \\cdot N\n$$\nNotice that since 3 divides $(2k+1)$ then it also divides $(2k+1)(k-1)+3$; moreover, 9 divides $(2k+1)(k-1)$, and then the largest power of 3 that divides $(2k+1)(k-1)+3$ is 3. Hence the largest power of 3 that divides $(2k+1) \\cdot 2^{k-1} \\cdot [(2k+1)(k-1)+3]$ is $t+1$ ($t$ factors from $2k+1$ and 1 from $(2k+1)(k-1)+3$). Finally,\n$$\n3^{(a-1)/2} = c = 3^{t+1}(1 + 3N),\n$$\nwhich is impossible because $N > 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26637,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nQual é o conjunto formado por todos os valores reais positivos de $x$ tais que $(x-1)(x-2)(x-3)<0$?\n\n(a) $(1,2)$\n(b) $(1,3)$\n(c) $(0,1) \\cup (2,3)$\n(d) $(0,3)$\n(e) $(0,2)$",
    "options": [],
    "answer": "(c)",
    "solution": "Solution:\n\nPara que um produto de três fatores seja negativo, devemos ter dois fatores positivos e um fator negativo, ou os três negativos. As possibilidades são as seguintes.\n$(+)(+)(-) = -$\n$(+)(-)(+) = -$\n$(-)(+)(+) = -$\n$(-)(-)(-) = -$\n\n1) $\\underbrace{(x-1)}_{+} \\underbrace{(x-2)}_{+} \\underbrace{(x-3)}_{-}$.\nIsso equivale a $x>1$, $x>2$ e $x<3$, ou seja, $2 < x < 3$.\n\n2) $\\underbrace{(x-1)}_{+} \\underbrace{(x-2)}_{-} \\underbrace{(x-3)}_{+}$.\nIsso equivale a $x>1$, $x<2$ e $x>3$, o que não é possível. Logo, não pode ocorrer esse caso.\n\n3) $\\underbrace{(x-1)}_{-} \\underbrace{(x-2)}_{+} \\underbrace{(x-3)}_{+}$.\nIsso equivale a $x<1$, $x>2$ e $x>3$, o que não é possível. Logo, não pode ocorrer esse caso.\n\n![](attached_image_1.png)\n\n4) $\\underbrace{(x-1)}_{-} \\underbrace{(x-2)}_{-} \\underbrace{(x-3)}_{-}$.\nIsso equivale a $x<1$, $x<2$ e $x<3$, ou seja, $x<1$.\n\n![](attached_image_2.png)\n\nLogo, os únicos reais $x$ satisfazendo $(x-1)(x-2)(x-3)<0$ são os reais $x$ tais que $x<1$ ou $2 < x < 3$. Assim, a união de intervalos $(0,1) \\cup (2,3)$ é o conjunto formado por todos os reais positivos de $x$ tais que $(x-1)(x-2)(x-3)<0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26638,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFração radical - Se $\\frac{\\sqrt{x}}{\\sqrt{y}}=5$, quanto é $\\frac{x+y}{2 y}$ ?\n\n(a) $\\frac{5}{2}$\n(b) $3 \\sqrt{2}$\n(c) $13 y$\n(d) $\\frac{25 y}{2}$\n(e) 13",
    "options": [],
    "answer": "e",
    "solution": "Solution:\n\nElevando ao quadrado ambos os membros de $\\frac{\\sqrt{x}}{\\sqrt{y}}=5$, obtemos $\\frac{x}{y}=25$. Assim,\n$$\n\\frac{x+y}{2 y}=\\frac{1}{2} \\times \\frac{x+y}{y}=\\frac{1}{2} \\times\\left(\\frac{x}{y}+\\frac{y}{y}\\right)=\\frac{1}{2} \\times\\left(\\frac{x}{y}+1\\right)=\\frac{1}{2} \\times(25+1)=13\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26639,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nQual é a soma? Se $x + |x| + y = 5$ e $x + |y| - y = 6$, qual é o valor da soma $x + y$?\n\n(a) $-1$\n(b) $11$\n(c) $\\frac{9}{5}$\n(d) $1$\n(e) $-11$",
    "options": [],
    "answer": "c",
    "solution": "Solution:\n\n1º Caso: Se $x \\leq 0$, então $|x| = -x$ e, pela primeira equação, temos $x + (-x) + y = 5$, ou seja, $y = 5$. Substituindo esse valor na segunda equação, obtemos $x = 6$, o que não é possível, pois estamos supondo $x \\leq 0$. Logo, não há solução nesse caso $x \\leq 0$.\n\n2º Caso: Se $y \\geq 0$, então $|y| = y$ e, pela segunda equação, temos\n$$\nx + y - y = 6\n$$\nou seja, $x = 6$. Substituindo esse valor na primeira equação, obtemos $y = -7$, o que não é possível, pois estamos supondo $y \\geq 0$.\n\n3º Caso: Se $x > 0$ e $y < 0$, então $|x| = x$ e $|y| = -y$. Pela primeira equação temos $2x + y = 5$ e, pela segunda, $x - 2y = 6$. Multiplicando $2x + y = 5$ por 2 e somando com $x - 2y = 6$, obtemos $5x = 16$, de modo que $x = 16/5$ e segue que $y = 5 - 2x = -7/5$. Assim, $x + y = 9/5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26640,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSendo $x>0$, $y>0$, $x>y$ e $z \\neq 0$, encontre a única desigualdade falsa.\n\n(a) $x+z>y+z$\n\n(b) $x-z>y-z$\n\n(c) $x z>y z$\n\n(d) $\\frac{x}{z^{2}}>\\frac{y}{z^{2}}$\n\n(e) $x z^{2}>y z^{2}$",
    "options": [],
    "answer": "(c)",
    "solution": "Solution:\n\nNessa questão usaremos as propriedades de desigualdades seguintes. Podemos somar o mesmo número a ambos os membros de uma desigualdade sem alterar seu sentido. Podemos multiplicar ambos os membros de uma desigualdade por um número positivo sem alterar seu sentido. Assim,\n$$\nx>y \\Rightarrow\\left\\{\\begin{array}{l}\nx+z>y+z \\text{ (somando } z \\text{ qualquer a ambos os lados)} \\\\\nx z>y z \\text{ (muliplicando por } z>0 \\text{ em ambos os lados)}\n\\end{array}\\right.\n$$\nLogo, (a) e (b) estão corretas, pois foi somado $z$ e $-z$ a ambos os membros, bem como (d) e (e), pois ambos os membros foram multiplicados por $1 / z^{2}$ e $z^{2}$, ambos positivos, já que $z \\neq 0$. A opção (c) é falsa, porque $z$ pode ser negativo. Por exemplo, se $x=5$, $y=3$ e $z=-2$, temos $5>3$ e, no entanto,\n$$\n\\underbrace{5 \\times (-2)}_{x z}=-10<-6=\\underbrace{3 \\times(-2)}_{y z}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26641,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA calculadora do Dodó tem uma tecla especial com o símbolo $\\mathrm{R}$. Se o visor mostra um número $x$ diferente de $2$, ao apertar $\\mathrm{R}$ aparece o valor de $\\frac{2x-3}{x-2}$.\n\na. Se o Dodó colocar $4$ no visor e apertar $\\mathrm{R}$, qual número vai aparecer?\n\nb. Dodó colocou um número no visor e, ao apertar $\\mathrm{R}$, apareceu o mesmo número. Quais são os números que ele pode ter colocado no visor?\n\nc. Dodó percebeu que, colocando o $4$ no visor e apertando $\\mathrm{R}$ duas vezes, aparece de novo o $4$; da mesma forma, colocando o $5$ e apertando $\\mathrm{R}$ duas vezes, aparece de novo o $5$. O mesmo vai acontecer para qualquer número diferente de $2$? Explique.",
    "options": [],
    "answer": "a) 5/2; b) 1 and 3; c) Yes. For any starting number other than two, pressing the key twice returns the original number (since f(f(x)) = x and f(x) is never two).",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26642,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nExiste um número quadrado perfeito formado apenas por algarismos 0 e 6 ?",
    "options": [],
    "answer": "Only zero; no positive perfect square is composed solely of the digits zero and six.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26643,
    "subject": "Mathematics (Multi-modal)",
    "question": "Emerald wrote on the blackboard all the integers from $1$ to $2011$. Then she erased all the even numbers.\n\na. How many numbers were left on the board?\n\nb. How many of the remaining numbers were written with only the digits $0$ and $1$?",
    "options": [],
    "answer": "a) 1006; b) 8",
    "solution": "a. The erased numbers were $2 = 2 \\cdot 1$, $4 = 2 \\cdot 2$, $\\ldots$, $2010 = 2 \\cdot 1005$. So $2011 - 1005 = 1006$ numbers were left on the board.\n\nb. We can list the numbers: they are $1$, $11$, $101$, $111$, $1001$, $1011$, $1101$, $1111$, a total of $8$.\n\n**OR** we can argue that the number is of the form (abc1), where $a$, $b$, $c$ are digits equal to either $0$ or $1$. Notice that the units digit must be $1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26644,
    "subject": "Mathematics (Multi-modal)",
    "question": "For each real number $t$, let $P_t(x) = x^3 - 12x + t$ and let\n$$\n\\Delta(t) = \\max\\{c \\in \\mathbb{R} \\mid P_t(c) = 0\\} - \\min\\{c \\in \\mathbb{R} \\mid P_t(c) = 0\\}\n$$\nthe difference between the largest and the smallest real roots of $P_t(x)$. Determine the range of values that $\\Delta(t)$ can assume as $t$ varies.",
    "options": [],
    "answer": "{0} ∪ [6, 4√3]",
    "solution": "Let $Q(x) = x^3 - 12x$. Then $Q'(x) = 3x^2 - 12$ has roots $-2$ and $2$, and $Q$ has a local minimum at $(2, -16)$ and a local maximum at $(-2, 16)$. So $Q(x) = -t$ has three (not necessarily distinct) real roots for $-16 \\le t \\le 16$ and one real root for $t < -16$ and $t > 16$, which means that $\\Delta(t) = 0$ for $t < -16$ or $t > 16$. So from now on we consider only $t \\in [-16, 16]$.\n\nLet $u \\le v \\le w$ be the roots. Since $P'_t(x) = 0 \\iff x = -2$ or $x = 2$ we have $u \\le -2 \\le v \\le 2 \\le w$. In particular, $-2 \\le v \\le 2$, and $v$ can assume any value in this interval: if $t = -16$, $v = -2$ and if $t = 16$, $v = 2$.\n\nBut we know that $u + v + w = 0$ and $uv + vw + uw = -12$, so $u + w = -v$ and $uw = -12 - v(u + w) = v^2 - 12$, so $(\\Delta(t))^2 = (w - u)^2 = (w + u)^2 - 4uw = 48 - 3v^2$, which lies in the range $[48 - 3 \\cdot 2^2, 48] = [36, 48]$. So $36 \\le (\\Delta(t))^2 \\le 48 \\iff 6 \\le \\Delta(t) \\le 4\\sqrt{3}$.\n\nSo the range of $\\Delta(t)$ is $\\{0\\} \\cup [6, 4\\sqrt{3}]$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26645,
    "subject": "Mathematics (Multi-modal)",
    "question": "We call a number *pal* if it doesn't have a zero digit and the sum of the squares of the digits is a perfect square. For example, $122$ and $34$ are *pal* but $304$ and $12$ are not *pal*. Prove that there exists a *pal* number with $n$ digits, $n > 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider the number $\\underbrace{55\\dots5}_{n \\text{ times}}$. The sum of the squares of its digits is $n \\cdot 5^2 = 25n$. We can exchange any two fives by one three and one four, so the sum of the squares decreases by $5^2$, until we run out of fives. So we can get any sum from $25 \\cdot \\lfloor n/2 \\rfloor$ and $25 \\cdot n$. So it suffices to show that there is an integer $k$ such that $\\frac{n}{2} \\le k^2 \\le n$. Choose $k$ such that $k^2 \\le n < (k+1)^2$. Suppose $k^2 < \\frac{n}{2}$. Then $n > 2k^2$, and $(k+1)^2 > n > 2k^2 \\implies (k+1)^2 \\ge 2k^2+2 \\iff k^2 - 2k + 1 \\le 0 \\iff (k-1)^2 \\le 0$, which is false except for $k=1$, or $2 < n < 4$, that is, $n=3$. But the statement of the problem itself gives an example with $n$ digits: $122$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26646,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUma grade de pontos com 10 linhas e 10 colunas é dada. Cada ponto é colorido de vermelho ou de azul. Sempre que dois pontos da mesma cor são vizinhos em uma mesma linha ou coluna, eles são ligados por um segmento da mesma cor dos pontos. Se dois pontos são vizinhos mas de cores diferentes, são ligados por um segmento verde. No total, existem 52 pontos vermelhos. Destes vermelhos, 2 estão nos cantos e outros 16 estão no bordo da grade. Os outros pontos vermelhos estão no interior da grade.\n![](attached_image_1.png)\nExistem 98 segmentos verdes. Determine o número de segmentos azuis.",
    "options": [],
    "answer": "37",
    "solution": "Solution:\n\nInicialmente, observe que existem 9 segmentos em cada linha e em cada coluna, de modo que existem $9 \\times 10 + 9 \\times 10 = 180$ segmentos no total.\nSeja $A$ o número de segmentos azuis e $V$ o número de segmentos vermelhos. Então $A + V + 98 = 180$, de modo que $A + V = 82$, já que existem 98 segmentos verdes.\n\nObserve que dos pontos vermelhos, só podem partir segmentos vermelhos ou verdes. Vamos contar o total de segmentos que partem dos pontos vermelhos. Neste total os segmentos verdes são contados exatamente uma vez e os segmentos vermelhos duas vezes, pois os segmentos vermelhos ligam dois pontos vermelhos.\n\nPartindo de um canto, existem 2 segmentos.\nDe um ponto sobre o bordo partem 3 segmentos.\nDe um ponto interior partem 4 segmentos.\n\nEntão, o número total de segmentos que partem dos vértices vermelhos é\n$$\n2 \\times 2 + 3 \\times 16 + 4 \\times 34 = 188\n$$\n\nMas como 98 segmentos que partem dos pontos vermelhos são os segmentos verdes, os restantes $188 - 98 = 90$ são vermelhos e foram contados duas vezes, de modo que $V = 45$.\n\nPortanto, $A = 82 - V = 37$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26647,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a regular $n$-gon inscribed in the unit circle. Compute the sum of the areas of all triangles determined by the vertices of the $n$-gon.",
    "options": [],
    "answer": "n^2/4 * cot(pi/n)",
    "solution": "First consider a triangle $ABC$ and its circumcenter $O$. Then the area of $ABC$ is $\\frac{R^2}{2}(\\sin 2\\angle A + \\sin 2\\angle B + \\sin 2\\angle C)$. Notice that if $\\angle B > 90^\\circ$ then $\\sin 2\\angle B < 0$.\n\n![](attached_image_1.png)\n\nSo the sum is equal to the sum of the areas of triangles $OA_iA_j$ with a plus sign or a minus sign, depending on the third vertex $A_k$ of the triangle $A_iA_jA_k$: if $A_k$ lies on the major arc $A_iA_j$ then we have a plus sign; else we have a minus sign (it won't matter if $A_iA_j$ is a diameter, because in that case the area of $OA_iA_j$ is zero).\n\nTherefore, if $A_iA_j$ subtend a minor arc of $k \\cdot \\frac{2\\pi}{n}$, $1 \\le k \\le \\lfloor n/2 \\rfloor$, the area of the triangle $OA_iA_j$ appears with a minus sign $k-1$ times and with a plus sign $n - (k-1) - 2 = n-k-1$ times. So it contributes with the sum $n-k-1 - (k-1) = n-2k$ times.\n\n![](attached_image_1.png)\n\n$$\nS = \\frac{n}{2} \\sum_{k=1}^{\\lfloor n/2 \\rfloor} (n-2k) \\sin k\\theta = \\frac{n^2}{2} \\sum_{k=1}^{\\lfloor n/2 \\rfloor} \\sin k\\theta - n \\sum_{k=1}^{\\lfloor n/2 \\rfloor} k \\sin k\\theta\n$$\n\nConsider the sums $S_1(\\theta) = \\sum_{k=1}^{\\lfloor n/2 \\rfloor} \\sin k\\theta$ and $S_2(\\theta) = \\sum_{k=1}^{\\lfloor n/2 \\rfloor} \\cos k\\theta \\implies S_2'(\\theta) = -\\sum_{k=1}^{\\lfloor n/2 \\rfloor} k \\sin k\\theta$. So we want to compute $\\frac{n^2}{2} S_1(\\theta) + n \\cdot S_2'(\\theta)$.\n\n$$\n\\begin{align*} \nS_2(\\theta) + iS_1(\\theta) &= \\sum_{k=1}^{\\lfloor n/2 \\rfloor} \\cos k\\theta + i \\sin k\\theta = \\sum_{k=1}^{\\lfloor n/2 \\rfloor} \\omega^k = \\omega \\cdot \\frac{\\omega^{\\lfloor n/2 \\rfloor} - 1}{\\omega - 1} \\\\ \n&= \\omega^{\\lfloor n/2 \\rfloor/2+1/2} \\frac{\\omega^{\\lfloor n/2 \\rfloor/2} - \\omega^{-\\lfloor n/2 \\rfloor/2}}{\\omega^{1/2} - \\omega^{-1/2}} \\\\ \n&= \\left( \\cos \\left( \\frac{(\\lfloor n/2 \\rfloor + 1)\\theta}{2} \\right) + i \\sin \\left( \\frac{(\\lfloor n/2 \\rfloor + 1)\\theta}{2} \\right) \\right) \\cdot \\frac{\\sin(\\lfloor n/2 \\rfloor \\theta/2)}{\\sin(\\theta/2)}, \n\\end{align*}\n$$\n\n$$\n\\begin{aligned}\nS_1 &= \\frac{\\sin\\left(\\frac{(\\lfloor n/2 \\rfloor + 1) \\theta}{2}\\right) \\sin\\left(\\frac{\\lfloor n/2 \\rfloor \\theta}{2}\\right)}{\\sin(\\theta/2)} = \\frac{\\cos(\\theta/2) - \\cos((\\lfloor n/2 \\rfloor + 1/2)\\theta)}{2 \\sin(\\theta/2)} \\\\\n&= \\frac{1}{2} \\cot(\\theta/2) - \\frac{\\cos((\\lfloor n/2 \\rfloor + 1/2)\\theta)}{2 \\sin(\\theta/2)} \\\\[1.5em]\nS_2 &= \\frac{\\cos\\left(\\frac{(\\lfloor n/2 \\rfloor + 1) \\theta}{2}\\right) \\sin\\left(\\frac{\\lfloor n/2 \\rfloor \\theta}{2}\\right)}{\\sin(\\theta/2)} = \\frac{\\sin(\\theta/2) + \\sin((\\lfloor n/2 \\rfloor + 1/2)\\theta)}{2 \\sin(\\theta/2)} \\\\\n&= \\frac{1}{2} + \\frac{\\sin((\\lfloor n/2 \\rfloor + 1/2)\\theta)}{2 \\sin(\\theta/2)}\n\\end{aligned}\n$$\n\n$$\n\\begin{aligned}\nS_2'(\\theta) &= \\frac{\\left(\\lfloor \\frac{n}{2} \\rfloor + \\frac{1}{2}\\right) \\cos\\left(\\left(\\lfloor \\frac{n}{2} \\rfloor + \\frac{1}{2}\\right) \\theta\\right) \\sin\\left(\\frac{\\theta}{2}\\right) - \\frac{1}{2} \\cos\\left(\\frac{\\theta}{2}\\right) \\sin\\left(\\left(\\lfloor \\frac{n}{2} \\rfloor + \\frac{1}{2}\\right) \\theta\\right)}{2 \\sin^2(\\theta/2)} \\\\\n&= \\frac{\\lfloor n/2 \\rfloor}{2} \\frac{\\cos((\\lfloor n/2 \\rfloor + 1/2)\\theta)}{\\sin(\\theta/2)} - \\frac{1}{4} \\frac{\\sin(\\lfloor n/2 \\rfloor \\theta)}{\\sin^2(\\theta/2)}\n\\end{aligned}\n$$\n\nSo the required sum is\n$$\n\\frac{n^2}{2} \\left( \\frac{1}{2} \\cot(\\theta/2) - \\frac{\\cos((\\lfloor n/2 \\rfloor + 1/2)\\theta)}{2 \\sin(\\theta/2)} \\right) + n \\left( \\frac{\\lfloor n/2 \\rfloor}{2} \\frac{\\cos((\\lfloor n/2 \\rfloor + 1/2)\\theta)}{\\sin(\\theta/2)} - \\frac{1}{4} \\frac{\\sin(\\lfloor n/2 \\rfloor \\theta)}{\\sin^2(\\theta/2)} \\right)\n$$\n\nIf $n$ is even, $\\lfloor n/2 \\rfloor = n/2$ and substituting $\\theta = \\frac{2\\pi}{n}$ the sum simplifies to\n$$\n\\frac{n^2}{4} \\cot \\frac{\\pi}{n}\n$$\nIf $n$ is odd, $\\lfloor n/2 \\rfloor = (n-1)/2$ and substituting $\\theta = \\frac{2\\pi}{n}$ the sum also simplifies to\n$$\n\\frac{n^2}{4} \\cot \\frac{\\pi}{n}\n$$\n\n$$\n\\frac{n^2}{4} \\cot \\frac{\\pi}{n}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26648,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a convex quadrilateral such that $AD = DC$, $AC = AB$ and $\\angle ADC = \\angle CAB$. Let $M$ and $N$ be the midpoints of $AD$ and $AB$. Prove that triangle $MNC$ is isosceles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $AD = CD$, $AB = AC$ and $\\angle ADC = \\angle BAC$, triangles $ADC$ and $BAC$ are similar by case SAS. Segments $CM$ and $CN$ are corresponding medians, so $\\frac{CM}{CN} = \\frac{CA}{CB}$ and $\\angle BCN = \\angle ACM \\iff \\angle BCN + \\angle NCA = \\angle ACM + \\angle NCA \\iff \\angle BCA = \\angle NCM$. Thus, again by case SAS, triangles $CMN$ and $CAB$ are similar, and therefore $CMN$ is an isosceles triangle with $CM = MN$.\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26649,
    "subject": "Mathematics (Multi-modal)",
    "question": "33 friends are collecting stickers for a 2011-sticker album. A distribution of stickers among the 33 friends is *incomplete* when there is a sticker that no friend has. Determine the least $m$ with the following property: every distribution of stickers among the 33 friends such that, for any two friends, there are at least $m$ stickers both don't have, is incomplete.",
    "options": [],
    "answer": "1890",
    "solution": "Since $2011 = 33 \\cdot 61 - 2$, consider the example where $S_i = \\{k \\in \\mathbb{Z} \\mid 61(i-1) < k \\le 61i\\}$ for $i = 1, 2, \\dots, 31$, $S_{32} = \\{k \\in \\mathbb{Z} \\mid 61 \\cdot 31 < k < 61 \\cdot 32\\}$ and $S_{33} = \\{k \\in \\mathbb{Z} \\mid 61 \\cdot 32 \\le k \\le 2011\\}$.\nNotice that $|S_i| = 61$ for $1 \\le i \\le 31$ and $|S_i| = 60$ for $i = 32$ and $i = 33$. Thus $|S_i \\cup S_j| \\le 2 \\cdot 61$, and therefore $m > 2011 - 2 \\cdot 61 = 1889$.\n\nNow we prove that the minimum value of $m$ is, in fact, $m = 1890$. First, notice that if $m = 1890$ then $|S_i \\cup S_j| \\le 2011 - 1890 = 121$. Suppose that $|S_1 \\cup S_2 \\cup S_3| > 181$. Then one of the sets, say $S_1$, has more than $181/3$ elements, that is, $|S_1| \\ge 61$. But $|(S_1 \\cup S_2 \\cup S_3) \\setminus (S_1 \\cup S_2)| > 181 - 121 \\iff |S_3 \\setminus (S_1 \\cup S_2)| > 60$. But $|S_3 \\cup S_1| = |S_3 \\setminus S_1| + |S_1| \\ge |S_3 \\setminus (S_1 \\cup S_2)| + |S_1| > 60 + 61 = 121$, contradiction. Hence $|S_1 \\cup S_2 \\cup S_3| \\le 181$.\nSo, $|S_1 \\cup S_2 \\cup \\dots \\cup S_{33}| \\le |S_1 \\cup S_2 \\cup S_3| + |S_4 \\cup S_5| + |S_6 \\cup S_7| + \\dots + |S_{32} + S_{33}| \\le 181 + 15 \\cdot 121 = 1996$, and there exists sixteen stickers that none of the 33 friends have.\nWe will prove that $m = 1890$ in another way. The example for $m = 1889$ is the same from the previous solution.\nAgain, number the stickers from 1 to 2011 and let $T_i$ be the set of the stickers that the friend $i$ does not have, $1 \\le i \\le 33$.\nConsider all pairs $(x, \\{i, j\\})$ such that $x \\in T_i \\cap T_j$. If for every sticker there is a friend that has it, that is, $T_1 \\cap T_2 \\cap \\dots \\cap T_{33} = \\emptyset$ then for each $x$ there exists $k$ such that $x \\notin T_k$. So each $x$ belongs to at most 32 sets $T_i$ and, hence, there exist at most $2011 \\cdot \\binom{32}{2}$ pairs $(x, \\{i, j\\})$. On the other hand, since $|T_i \\cap T_j| \\ge m$ there exists at least $m \\cdot \\binom{33}{2}$ pairs $(x, \\{i, j\\})$. Therefore, if $T_1 \\cap T_2 \\cap \\dots \\cap T_{33} = \\emptyset$ then\n$$\nm \\cdot \\binom{33}{2} \\le 2011 \\cdot \\binom{32}{2} \\iff m \\le 1890\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26650,
    "subject": "Mathematics (Multi-modal)",
    "question": "We have a red cube with sidelength $2$ cm. What is the minimum number of identical cubes that must be adjoined to the red cube in order to obtain a cube with volume $\\left(\\frac{12}{5}\\right)^3$ cm?",
    "options": [],
    "answer": "91",
    "solution": "The bigger cube has sidelength $\\frac{12}{5}$ cm, so the difference between the side-lengths is $\\frac{12}{5} - 2 = \\frac{2}{5}$ cm, that is, the red cubes should not have sidelength greater than this length. Cubes with sidelength $\\frac{2}{5}$ cm are the natural candidates, so we set a new unit $u = \\frac{2}{5}$ cm. Notice that the bigger cube should have sidelength $6u$ and the original cube must have sidelength $5u$. So we need $6^3 - 5^3 = 91$ red cubes.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 26651,
    "subject": "Mathematics (Multi-modal)",
    "question": "We call a number *pal* if it doesn't have a zero digit and the sum of the squares of the digits is a perfect square. For example, $2115522$ is pal (because $2^2 + 1^2 + 1^2 + 5^2 + 5^2 + 2^2 + 2^2 = 8^2$) but $304$ and $12$ are not pal.\n\na. What is the greatest two-digit pal number?\n\nb. Does there exist a $2011$-digit pal number?",
    "options": [],
    "answer": "a) 86; b) Yes",
    "solution": "a. First notice that $86$ is pal. Then it's not hard to check by hand that every number from $87$ to $99$ is not pal.\n\nb. The answer is *yes*. First consider the $2011$-digit number $11\\ldots1$. The sum of its digits is $2011$. The smallest perfect square greater than $2011$ is $45^2 = 2025$. Since $2025 - 2011 = 14$ and $14 = 2 \\cdot (2^2 - 1^2) + (3^2 - 1^2)$, we can exchange two $1$s by two $2$s and one $1$ by one $3$. So we obtain the pal number $11\\ldots123$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26652,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that, for all convex pentagons $P_1P_2P_3P_4P_5$ with area $1$, there are indices $i$ and $j$ (assume $P_6 = P_1$ and $P_7 = P_2$) such that:\n$$\n\\text{area } \\triangle P_i P_{i+1} P_{i+2} \\le \\frac{5 - \\sqrt{5}}{10} \\le \\text{area } \\triangle P_j P_{j+1} P_{j+2}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let's prove that there exists a triangle $P_j P_{j+1} P_{j+2}$ with area less than or equal to $\\alpha = \\frac{5-\\sqrt{5}}{10}$. Suppose that all triangles $P_j P_{j+1} P_{j+2}$ have area greater than $\\alpha$.\n\n![](attached_image_1.png)\n\nLet diagonals $P_1P_4$ and $P_3P_5$ meet at $Q$. Since $Q \\in P_3P_5$, $\\text{area } P_1P_2Q \\le \\max(\\text{area } P_1P_2P_5, \\text{area } P_1P_2P_3) < \\alpha$, so $\\text{area } P_1P_2P_4 = 1 - \\text{area } P_1P_4P_5 - \\text{area } P_2P_3P_4 > 1 - 2\\alpha$. Thus\n$$\n\\frac{P_1Q}{P_1P_4} = \\frac{\\text{area } P_1P_2Q}{\\text{area } P_1P_2P_4} < \\frac{\\alpha}{1 - 2\\alpha}\n$$\nWe also have $\\frac{P_1Q}{P_4Q} = \\frac{\\text{area } P_1P_3P_5}{\\text{area } P_3P_4P_5}$. Since $\\text{area } P_3P_4P_5 < \\alpha$ and $\\text{area } P_1P_3P_5 > 1 - 2\\alpha$,\n$$\n\\frac{P_1Q}{P_4Q} > \\frac{1 - 2\\alpha}{\\alpha} \\iff \\frac{P_1Q}{P_1P_4} > \\frac{1 - 2\\alpha}{1 - \\alpha}\n$$\nTherefore\n$$\n\\frac{1-2\\alpha}{1-\\alpha} < \\frac{P_1Q}{P_1P_4} < \\frac{\\alpha}{1-2\\alpha} \\implies 5\\alpha^2 - 5\\alpha + 1 < 0 \\iff \\frac{5-\\sqrt{5}}{10} < \\alpha < \\frac{5+\\sqrt{5}}{10},\n$$\ncontradiction.\n\nThe proof of the other inequality is analogous.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26653,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer and $A$ a subset of $\\mathbb{Z}/(n)$, the set of the integers modulo $n$, define $f(A) = \\min_{t \\in \\mathbb{Z}/(n)} |A \\cap (A+t)|$, where $A+t = \\{x+t, x \\in A\\} \\subset \\mathbb{Z}/(n)$. Define $g(n) = \\max\\{f(A); A \\subset \\mathbb{Z}/(n), |A| = \\lfloor n/2 \\rfloor\\}$.\n\na. Prove that $g(n) \\le \\lfloor n/4 \\rfloor - 1, \\forall n \\ge 1$.\n\nb. Prove that $g(n) = \\lfloor n/4 \\rfloor - 1$ for infinite values of $n \\ge 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a.\nLet $A = \\{a_1, a_2, \\dots, a_{\\lfloor n/2 \\rfloor}\\}$. Consider the sum\n$$\n\\sum_{t=0}^{n-1} |A \\cap (A+t)|\n$$\nNow each element $a \\in A$ appears in a set $A + t_i$ $|A| = \\lfloor n/2 \\rfloor$ times: choose $t_i = a - a_i$ for each $i = 1, 2, \\dots, \\lfloor n/2 \\rfloor$. So\n$$\n\\sum_{t=0}^{n-1} |A \\cap (A+t)| = \\left\\lfloor \\frac{n}{2} \\right\\rfloor^2\n$$\nand the average of $|A \\cap (A+t)|$ is\n$$\n\\frac{1}{n} \\left( \\left\\lfloor \\frac{n}{2} \\right\\rfloor \\right)^2 \\le \\frac{n}{4}.\n$$\nSince $|A \\cap (A+0)| = |A| > \\frac{n}{4}$ is above average, there is a $t$ such that $|A \\cap (A+t)|$ is below average, so\n$$\nf(A) \\le |A \\cap (A+t)| < \\frac{n}{4} \\implies f(A) \\le \\left\\lfloor \\frac{n}{4} \\right\\rfloor - 1.\n$$\nSo $g(n) \\le \\left\\lfloor \\frac{n}{4} \\right\\rfloor - 1$.\n\n\nb.\nLet $p \\equiv 3 \\pmod 4$ be a prime, and set\n$$\nA = (F_p^\\times)^2 \\pmod{p}\n$$\nWe have to show that\n$$\n|(A+t) \\cap A| \\ge \\left\\lfloor \\frac{p}{4} \\right\\rfloor - 1 \\quad \\text{for all } t \\in F_p\n$$\nSince this is clear for $t = \\bar{0}$, we henceforth assume that $t \\ne \\bar{0}$. From now on, all equalities are in $F_p$, that is, taken modulo $p$.\n\nWe have that $x^2 \\in A \\cap (A+t)$ if and only if there exists $y^2 \\in A$ such that\n$$\n\\begin{array}{l@{\\quad \\Longleftrightarrow \\quad}l}\nx^2 = y^2 + t & (x-y)(x+y) = t \\\\[1.5ex]\n\\left| \\begin{array}{l} x-y=u \\\\ x+y=u^{-1}t \\end{array} \\right. & \\text{for some } u \\in F_p^{\\times} \\\\[1.5ex]\n\\left| \\begin{array}{l} x = \\displaystyle\\frac{u+u^{-1}t}{2} \\\\ y = \\displaystyle\\frac{u^{-1}t-u}{2} \\end{array} \\right. & \\text{for some } u \\in F_p^{\\times}\n\\end{array}\n$$\n\n$$\nx^2 = \\frac{u^2 + 2t + u^{-2}t^2}{4}, \\quad u \\in F_p^{\\times},\n$$\nwith $x \\neq 0$ and $y \\neq 0$, that is,\n$$\n\\left| \\begin{array}{l} u + u^{-1}t \\neq 0 \\\\ u^{-1}t - u \\neq 0 \\end{array} \\right. \\iff u^2 \\neq \\pm t\n$$\n\n$$\n\\left(\\frac{\\pm t}{p}\\right) = \\left(\\frac{\\pm 1}{p}\\right) \\left(\\frac{t}{p}\\right) = \\pm \\left(\\frac{t}{p}\\right)\n$$\n\nOn the other hand, if $u^2 \\neq v^2$,\n$$\n\\frac{u^2 + 2t + u^{-2}t^2}{4} = \\frac{v^2 + 2t + v^{-2}t^2}{4} \\iff u^2 - v^2 = \\frac{t^2}{u^2} - \\frac{t^2}{v^2} \\iff u^2 v^2 = t^2\n$$\nHence, as $u^2$ runs over the non-zero quadratic residues, with exception of $\\pm t$, we obtain each $x^2 \\in A \\cap (A+t)$ twice (observe that the case $u^2 = v^2$ is excluded since $u^2 \\cdot u^2 = t^2 \\iff u^2 = \\pm t$). Therefore\n$$\n|A \\cap (A+t)| = \\frac{\\frac{p-1}{2} - 1}{2} = \\frac{p-3}{4} = \\left\\lfloor \\frac{p}{4} \\right\\rfloor - 1,\n$$",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 26654,
    "subject": "Mathematics (Multi-modal)",
    "question": "Emerald and Jade play the following game: Emerald writes a list with 2011 positive integers, but does not show it to Jade. Jade's goal is finding the product of the 2011 numbers in Emerald's list. In order to do so, she is allowed to ask Emerald the gcd or the lcm of any subset with at least two of the 2011 numbers (as, for instance, \"what is the gcd of the first, second, 10th and 2000th numbers from your list?\" or \"what is the lcm of all the numbers in your list?\"). Jade can make as many questions as she wants, but can only obtain her (correct) answers from Emerald after making all her questions (Emerald is generous and also says which answer corresponds to each question). Jade then can use any of the four elementary operations (add, subtract, multiply, divide) with Emerald's answers. Can Jade make a list of questions that guarantees that she can find the product of the 2011 numbers?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "She can obtain the product of any two numbers $a$ and $b$ by asking $\\gcd(a, b)$ and $\\text{lcm}(a, b)$, since $\\text{lcm}(a, b) \\cdot \\gcd(a, b) = ab$. The identity\n$$\nabc = \\frac{\\text{lcm}(a, b) \\cdot \\text{lcm}(a, c) \\cdot \\text{lcm}(b, c) \\cdot \\gcd(a, b, c)}{\\text{lcm}(a, b, c)}\n$$\nessentially finishes the proof, since the 2011 numbers can be divided into a set of three numbers and 1004 sets of two numbers.\n\nIt remains to prove the above identity. But this follows from the facts that $\\max(x, y)+\\max(x, z)+\\max(y, z)+\\min(x, y, z)-\\max(x, y, z) = x+y+z$, and if $p^{x_i} \\mid a_i$ then $p^{\\min\\{x_i\\}} \\mid \\gcd(a_1, a_2, \\dots, a_n)$ and $p^{\\max\\{x_i\\}} \\mid \\text{lcm}(a_1, a_2, \\dots, a_n)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26655,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the polynomial $f(x) = x^3 + x^2 - 4x + 1$.\n\na. Prove that if $r$ is a root of $f(x)$ then $r^2 + r - 3$ is also a root of $f(x)$.\n\nb. Let $\\alpha$, $\\beta$, $\\gamma$ be the three roots of $f(x)$, in some order. Determine all possible values of\n$$\n\\frac{\\alpha}{\\beta} + \\frac{\\beta}{\\gamma} + \\frac{\\gamma}{\\alpha}\n$$",
    "options": [],
    "answer": "3 and -10",
    "solution": "a. First notice that, since $r$ is a root, then $r^3 + r^2 - 4r + 1 = 0 \\iff r^2 + r - 3 = 1 - \\frac{1}{r}$. So we need to prove that\n$$\n\\begin{aligned}\n& \\left(1 - \\frac{1}{r}\\right)^3 + \\left(1 - \\frac{1}{r}\\right)^2 - 4\\left(1 - \\frac{1}{r}\\right) + 1 = 0 \\\\\n\\iff & 1 - \\frac{3}{r} + \\frac{3}{r^2} - \\frac{1}{r^3} + 1 - \\frac{2}{r} + \\frac{1}{r^2} - 4 + \\frac{4}{r} + 1 = 0 \\\\\n\\iff & -1 - \\frac{1}{r} + \\frac{4}{r^2} - \\frac{1}{r^3} = 0 \\\\\n\\iff & r^3 + r^2 - 4r + 1 = 0,\n\\end{aligned}\n$$\nwhich is true.\n\n\nb. Iterating $1 - \\frac{1}{r}$, we get $1 - \\frac{1}{1-\\frac{1}{r}} = \\frac{1}{1-r}$. It's not hard to see that $r, 1 - \\frac{1}{r} = \\frac{r-1}{r}$ and $\\frac{1}{1-r}$ are all distinct. In fact, if $r = 1 - \\frac{1}{r}$ then $r^2 - r + 1 = 0$, and $r^3 = -1$, so $r^2 - 4r = 0$, which is not true.\nSo there are two possible ways of computing $\\frac{\\alpha}{\\beta} + \\frac{\\beta}{\\gamma} + \\frac{\\gamma}{\\alpha}$:\n* $(\\alpha, \\beta, \\gamma) = (r, \\frac{r-1}{r}, \\frac{1}{1-r})$:\n$$ \n\\begin{aligned}\n\\frac{\\alpha}{\\beta} + \\frac{\\beta}{\\gamma} + \\frac{\\gamma}{\\alpha} &= \\frac{r^2}{r-1} - \\frac{(r-1)^2}{r} - \\frac{1}{r(r-1)} \\\\\n&= \\frac{r^3 - (r-1)^3 - 1}{r(r-1)} = \\frac{3r(r-1)}{r(r-1)} = 3\n\\end{aligned}\n$$\n\n* $(\\alpha, \\beta, \\gamma) = (r, \\frac{1}{1-r}, \\frac{r-1}{r})$:\n$$ \n\\begin{aligned}\n\\frac{\\alpha}{\\beta} + \\frac{\\beta}{\\gamma} + \\frac{\\gamma}{\\alpha} &= -r(r-1) - \\frac{r}{(r-1)^2} + \\frac{r-1}{r^2} \\\\\n&= \\frac{-r^6 + 3r^5 - 3r^4 + r^3 - 3r^2 + 3r - 1}{r^4 - 2r^3 + r^2}\n\\end{aligned}\n$$\n\nSince $-r^6 + 3r^5 - 3r^4 + r^3 - 3r^2 + 3r - 1 = (r^3 + r^2 - 4r + 1) \\cdot (-r^3 + 4r^2 - 11r + 29) - 80r^2 + 130r - 30 = -80r^2 + 130r - 30$ and $r^4 - 2r^3 + r^2 = (r^3 + r^2 - 4r - 1) \\cdot (r - 3) + 8r^2 - 13r + 3 = 8r^2 - 13r + 3$. So\n$$\n\\frac{\\alpha}{\\beta} + \\frac{\\beta}{\\gamma} + \\frac{\\gamma}{\\alpha} = \\frac{-80r^2 + 130r - 30}{8r^2 - 13r + 3} = -10.\n$$\n\nAnother way to solve this problem is realizing that if $(\\alpha, \\beta, \\gamma) = (r, \\frac{r-1}{r}, \\frac{1}{1-r})$ then the other sum is actually $\\frac{\\beta}{\\alpha} + \\frac{\\gamma}{\\beta} + \\frac{\\alpha}{\\gamma}$. By Vieta's formula, $\\sigma_1 = \\alpha+\\beta+\\gamma = -1$, $\\sigma_2 = \\alpha\\beta + \\beta\\gamma + \\gamma\\alpha = -4$ and $\\sigma_3 = \\alpha\\beta\\gamma = -1$.\n$$\n\\begin{aligned}\n\\frac{\\alpha}{\\beta} + \\frac{\\beta}{\\gamma} + \\frac{\\gamma}{\\alpha} + \\frac{\\beta}{\\alpha} + \\frac{\\gamma}{\\beta} + \\frac{\\alpha}{\\gamma} &= \\frac{\\alpha^2\\beta + \\alpha^2\\gamma + \\beta^2\\alpha + \\beta^2\\gamma + \\gamma^2\\alpha + \\gamma^2\\beta}{\\alpha\\beta\\gamma} \\\\\n&= \\frac{\\sigma_1\\sigma_2 - 3\\sigma_3}{\\sigma_3} = \\frac{(-1) \\cdot (-4) - 3(-1)}{-1} = -7,\n\\end{aligned}\n$$\nso\n$$\n\\frac{\\beta}{\\alpha} + \\frac{\\gamma}{\\beta} + \\frac{\\alpha}{\\gamma} = -7 - 3 = -10.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 26656,
    "subject": "Mathematics (Multi-modal)",
    "question": "Do there exist $2011$ positive integers $a_1 < a_2 < \\dots < a_{2011}$ such that $\\gcd(a_i, a_j) = a_j - a_i$ for any $i, j$ such that $1 \\le i < j \\le 2011$?",
    "options": [],
    "answer": "Yes",
    "solution": "The answer is yes and you can construct an example in several ways. The main observation is that $\\gcd(a_i, a_j) = a_j - a_i \\iff a_j - a_i \\mid a_i$. In fact, if $\\gcd(a_i, a_j) = a_j - a_i$ then $a_j - a_i \\mid a_i$ and, conversely, if $a_j - a_i \\mid a_i$ then $a_j - a_i \\mid a_i + (a_j - a_i) \\iff a_j - a_i \\mid a_j$, so $a_j - a_i \\mid \\gcd(a_i, a_j)$. But $\\gcd(a_i, a_j) \\mid a_i$ and $\\gcd(a_i, a_j) \\mid a_j$ implies $\\gcd(a_i, a_j) \\mid a_j - a_i$, so $\\gcd(a_i, a_j) = a_j - a_i$.\n\nOnce this fact is established, one can construct the sequence inductively as follows: first consider the two-term sequence $(1, 2)$. Now, given a sequence\n$(x_1, x_2, \\dots, x_{k-1})$ with $k-1$ terms such that $\\gcd(x_i, x_j) = x_j - x_i$, construct a new sequence adding $x_0$ to every term and putting $x_0$ at its beginning: $(x_0, x_1 + x_0, x_2 + x_0, \\dots, x_{k-1} + x_0)$. All we need to do is to find $x_0$. By the previous observation, we need $x_j - x_i \\mid x_i + x_0$ and $x_i \\mid x_0$. We already have that $x_j - x_i \\mid x_i$, so a good choice is $x_0 = \\operatorname{lcm}(x_1, x_2, \\dots, x_{k-1})$, because by definition $x_i \\mid x_0$ and, since $x_i \\mid x_0$ and $x_j - x_i \\mid x_i$, $x_j - x_i \\mid x_0$, so $x_j - x_i \\mid x_i + x_0$. So we obtained a new sequence with $k$ terms and the result follows by induction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26657,
    "subject": "Mathematics (Multi-modal)",
    "question": "Emerald wrote a list of positive integers. Renan noticed that each number in the list and any sum of any quantity of distinct numbers from the list were square-free (that is, not divisible by any perfect square except, of course, 1). What is the maximum quantity of numbers that Emerald's list can have?",
    "options": [],
    "answer": "3",
    "solution": "The smallest perfect square, apart from $1$, is $2^2 = 4$. So let $a_1, a_2, \\dots, a_k$ be the numbers on the list modulo $4$. We cannot have $a_i = 0$; also, there is at most one $a_i$ equal to $2$ and we cannot have $a_i = 1$ and $a_j = 3$ simultaneously.\n\nWe claim that among any four distinct numbers $a_1, a_2, a_3, a_4$ fulfilling the above properties there are three of them whose sum is a multiple of $4$. Indeed, there are two equal numbers, say $a_1, a_2$. We cannot have $a_1 = a_2 = 2$, so either $a_1 = a_2 = 1$ or $a_1 = a_2 = 3$. We can suppose wlog $a_1 = a_2 = 1$ (otherwise, reverse the signs of all four numbers modulo $4$). But since we also cannot have $a_j = 3$ and $a_3 = a_4 = 1$, one of $a_3, a_4$, say $a_3$, is $2$. But then $a_1 + a_2 + a_3 = 1 + 1 + 2 = 4$.\n\nSo the quantity of numbers is at most $3$. $5$, $13$ and $17$ is an example of a list with three numbers.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26658,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDetermine o maior múltiplo de $36$ que possui todos os algarismos pares e diferentes.",
    "options": [],
    "answer": "8640",
    "solution": "Solution:\nPara um número ser divisível por $36 = 4 \\times 9$, deve ser divisível por $4$ e por $9$. Assim, a soma dos algarismos do número $n$ procurado deve ser divisível por $9$.\n\nPor outro lado, como todos os algarismos são pares, a soma dos algarismos também é par. Assim, a soma dos algarismos é no mínimo $18$. Como $0 + 2 + 4 + 6 + 8 = 20$, o número $n$ deve ser formado pelos algarismos $0, 4, 6$ e $8$.\n\nO maior número que podemos formar com esses algarismos, sem repetir, é $8640$, o qual também é divisível por $4$, assegurando que este é o número procurado.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26659,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSejam\n$$\nR=3 \\times 9+4 \\times 10+5 \\times 11+\\cdots+2003 \\times 2009\n$$\n$$\nS=1 \\times 11+2 \\times 12+3 \\times 13+\\cdots+2001 \\times 2011\n$$\n(a) Qual é o maior número: $R$ ou $S$?\n(b) Calcule a diferença entre o maior e o menor.",
    "options": [],
    "answer": "R is larger; the difference is 32016",
    "solution": "Solution:\n(a) Cada parcela de $S$ é da forma $n \\times (n+10) = n^{2} + 10n$ e cada parcela de $R$ é da forma $(n+2) \\times (n+8) = n^{2} + 10n + 16$ com $n = \\{1, 2, \\ldots, 2001\\}$ em ambos os casos. Assim, para todo $n$, cada parcela de $R$ é maior que a correspondente em $S$, o que torna $R > S$.\n\n(b) A diferença entre as parcelas correspondentes é igual a\n$$\n\\left(n^{2} + 10n + 16\\right) - \\left(n^{2} + 10n\\right) = 16\n$$\nComo existem $2001$ parcelas, a diferença entre $R$ e $S$ é igual a $16 \\times 2001 = 32016$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26660,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm número $n$ de dois algarismos é dividido pela soma de seus algarismos, obtendo resto $r$.\n\na. Encontre um número $n$ tal que $r=0$.\nb. Mostre que $r$ não pode ser maior que 15.\nc. Mostre que para qualquer $r$ menor ou igual a 12, existe um $n$ que deixa resto $r$ ao dividi-lo pela soma de seus algarismos.",
    "options": [],
    "answer": "a) One example is n = 12 (remainder 0).\nb) The remainder cannot exceed 15; for instance, n = 79 gives remainder 15.\nc) For each r from 1 to 12 there exists a two-digit number: with digit sum 10, the numbers 19, 28, 37, 46, 55, 64, 73, 82, 91 yield remainders 9, 8, 7, 6, 5, 4, 3, 2, 1 respectively; and 65, 76, 87 yield remainders 10, 11, 12.",
    "solution": "Solution:\n\na. Existem vários exemplos onde o resto da divisão é 0, sendo o menor deles $n=12$.\n\nb. Denotemos por $S$ a soma dos algarismos de $n$.\nObservemos que $S \\leqslant 18$ e a igualdade somente acontece se $n=99$, mas neste caso o resto da divisão é 9.\nSe $S=17$, temos dois possíveis valores de $n=89$ e $98$, que quando divididos por 17 deixam respectivamente restos 4 e 13.\nNos números restantes, a soma dos algarismos é menor ou igual a 16. Assim, o resto deve ser menor ou igual a 15.\nO resto é igual a 15 se $n=79$. Verifique!\n\nc. Para terminar, basta mostrar um exemplo para cada resto entre 1 e 12. Se consideramos os números $19,28,37, \\ldots, 91$, em todos a soma de seus algarismos é 10 e os restos da divisão por 10 são respectivamente $9,8, \\ldots, 1$. Para os restos 10, 11 e 12, basta considerar os números 65, 76 e 87.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26661,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n(a) A soma de quatro inteiros positivos consecutivos pode ser um número primo? Justifique sua resposta.\n\n(b) A soma de três inteiros positivos consecutivos pode ser um número primo? Justifique sua resposta.",
    "options": [],
    "answer": "a) no; b) no",
    "solution": "Solution:\n\n(a) Seja $x$ o menor dos números. Então, a soma em questão é\n$$\nx + (x+1) + (x+2) + (x+3) = 4x + 6 = 2(x+3)\n$$\nEste número é par maior que $2$, portanto não pode ser um número primo.\n\n(b) Seja $y$ o menor dos números. Então, a soma em questão é\n$$\ny + (y+1) + (y+2) = 3y + 3 = 3(y+1)\n$$\nEste número é múltiplo de $3$ e maior que $3$, logo não pode ser um número primo.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26662,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nObserve que\n$$\n\\begin{gathered}\n1^{2}+2^{2}+(1 \\times 2)^{2}=3^{2} \\\\\n2^{2}+3^{2}+(2 \\times 3)^{2}=7^{2} \\\\\n3^{2}+4^{2}+(3 \\times 4)^{2}=13^{2}\n\\end{gathered}\n$$\nProve que se $a$ e $b$ são inteiros consecutivos então o número\n$$\na^{2}+b^{2}+(a b)^{2}\n$$\né um quadrado perfeito.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSuponha, sem perda de generalidade, que $b>a$, isto é, $b-a=1$. Então\n$$\n\\begin{gathered}\n(b-a)^{2}=1^{2} \\\\\nb^{2}-2 a b+a^{2}=1 \\\\\na^{2}+b^{2}=2 a b+1\n\\end{gathered}\n$$\nSomando $(a b)^{2}$ em cada lado da igualdade, temos\n$a^{2}+b^{2}+(a b)^{2}=(2 a b+1)+(a b)^{2}=(a b)^{2}+2(a b) \\cdot 1+1^{2}=(a b+1)^{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26663,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\na. Prove que o número $3999991$ não é primo.\n\nb. Prove que o número $1000343$ não é primo.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na.\nObserve que\n$$\n\\begin{aligned}\n3999991 &= 4000000 - 9 \\\\\n&= 4 \\cdot 10^{6} - 3^{2} \\\\\n&= \\left(2 \\cdot 10^{3}\\right)^{2} - 3^{2} \\\\\n&= \\left(2 \\cdot 10^{3} - 3\\right)\\left(2 \\cdot 10^{3} + 3\\right) = 1997 \\cdot 2003\n\\end{aligned}\n$$\ne portanto não é um número primo.\n\nb.\nObserve que\n$$\n\\begin{aligned}\n1000343 &= 10^{6} + 7^{3} \\\\\n&= \\left(10^{2}\\right)^{3} + 7^{3} \\\\\n&= \\left(10^{2} + 7\\right)\\left(\\left(10^{2}\\right)^{2} - 10^{2} \\cdot 7 + 7^{2}\\right) \\\\\n&= 107 \\cdot 9349\n\\end{aligned}\n$$\nportanto não é primo.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26664,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm conjunto de números é chamado trilegal se pode ser dividido em subconjuntos com três elementos de tal modo que um dos elementos seja a soma dos outros dois. Por exemplo, o conjunto $\\{1,2,3, \\ldots, 11,12\\}$ é trilegal pois pode ser dividido em $\\{1,5,6\\}$, $\\{2,9,11\\},\\{3,7,10\\}$ e $\\{4,8,12\\}$.\n\na. Mostre que $\\{1,2, \\ldots, 14,15\\}$ é trilegal.\n\nb. Mostre que $\\{1,2, \\ldots, 2010\\}$ não é trilegal.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na. Para a primeira parte basta encontrar uma distribuição em subconjuntos com três elementos, por exemplo\n$$\n\\{1,6,7\\},\\{2,12,14\\},\\{3,8,11\\},\\{4,9,13\\},\\{5,10,15\\}\n$$\n\nb. Observemos que se um conjunto de três elementos cumpre a condição de ser trilegal, então ele tem de ser da forma\n$$\n\\{\\text{par}, \\text{par}, \\text{par}\\}\n$$\nou\n$$\n\\{\\text{ímpar}, \\text{ímpar}, \\text{par}\\}\n$$\nSuponhamos que podemos dividir o conjunto em subconjuntos trilegais que têm $A$ conjuntos do primeiro tipo e $B$ conjuntos do segundo tipo. Como a quantidade de números ímpares menores que $2010$ é $1005$, devemos ter $2B=1005$, o que é contraditório.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26665,
    "subject": "Mathematics (Multi-modal)",
    "question": "Emerald writes the integers from $1$ to $9$ in a $3 \\times 3$ table, one number in each cell, each number appearing exactly once. Then she computes eight sums: the sums of three numbers on each row, the sums of the three numbers on each column and the sums of the three numbers on both diagonals.\n\na. Show a table such that exactly three of the eight sums are multiples of $3$.\n\nb. Is it possible that none of the eight sums is a multiple of $3$?",
    "options": [],
    "answer": "a. Example grid with rows: 1 2 3; 4 5 6; 8 9 7.\nb. No, it is not possible.",
    "solution": "a.\nFor instance,\n\n| 1 | 2 | 3 |\n|---|---|---|\n| 4 | 5 | 6 |\n| 8 | 9 | 7 |\n\nThe trick is to only adjust the last row. The usual order $7$, $8$, $9$ yields all sums to be multiple of $3$, so it's just a matter of rearranging them.\n\nb.\nNo, it's not possible. First, notice that the sum of three numbers $x$, $y$, $z$ is a multiple of $3$ iff $x \\equiv y \\equiv z \\pmod{3}$ or $x$, $y$, $z$ are $0$, $1$, $2$ mod $3$ in some order. Let $a$, $b$, $c$, $d$ be the numbers in the corner modulo $3$. So two of them are equal. We can suppose wlog that they are either $a = b$ or $a = d$. Also, let $x$ be the number in the central cell modulo $3$.\n\n| a | b |\n|---|---|\n| x |   |\n| c | d |\n\nIf $a = d$, then $x \\neq a$ and $x$ is equal to either $b$ or $c$. Suppose wlog $x = b \\neq a$. Then we have the following situation:\n\n| a | b |\n|---|---|\n|   | b |\n| c | a |\n\nLet $m$ be the other remainder (that is, $m \\neq a$ and $m \\neq b$). Then $m$ cannot be in the same line as $a$ and $b$. This leaves only one possibility:\n\n| a | b |\n|---|---|\n| m | b |\n| m | m | a |\n\nBut the remaining $a$ will necessarily yield a line with all three remainders. Now if $a = b$, then both $c$ and $d$ are different from $a$ (otherwise, we reduce the problem to the previous case). If $d \\neq c$, $a$, $c$, $d$ are the three distinct remainders, and we have no possibility for $x$. So $c = d$.\n\n| a | a |\n|---|---|\n|   | x |\n| c | c |\n\nBut this prevents the other remainder $m$ to appear in the middle row, leaving only two cells for three numbers, which is not possible.\nSo, in both cases, one of the sums is a multiple of $3$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 26666,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider 1000 points inside a square with sidelength 16. Prove that there is an equilateral triangle with sidelength $2\\sqrt{3}$ that covers at least 16 of those points.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $\\left(\\frac{16}{2\\sqrt{3}}\\right)^2 = \\frac{64}{3} = 21+\\frac{1}{3}$ lies between $4.5^2 = 20.25$ and $5^2$ and the altitude of the triangle is $\\frac{2\\sqrt{3}\\cdot\\sqrt{3}}{2} = 3$, we can cover a square with sidelength $16$ with $2 \\cdot 5 \\cdot \\lfloor\\frac{16}{3}\\rfloor = 60$ equilateral triangles. Since $\\lfloor\\frac{1000}{60}\\rfloor = 16$, by the pigeon hole principle there is an equilateral triangle that covers at least $17$ points.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26667,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle and $H$ is orthocenter. Let $D$ be the intersection of $BH$ and $AC$ and $E$ be the intersection of $CH$ and $AB$. The circumcircle of $ADE$ meets the circumcircle of $ABC$ at $F \\neq A$. Prove that the angle bisectors of $\\angle BFC$ and $\\angle BHC$ concur at a point on line $BC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\nBy the angle bisector theorem, it suffices to prove that $\\frac{BF}{FC} = \\frac{BH}{HC}$.\nWe have $\\angle EFB = 180^\\circ - \\angle FEA = 180^\\circ - \\angle FDA = \\angle FDC$ and $\\angle FBE = \\angle FBA = \\angle FCA = \\angle FCD$, so triangles $BEF$ and $CDF$ are similar. Thus\n$$\n\\frac{BF}{FC} = \\frac{BE}{CD} = \\frac{BH \\cos \\angle EBH}{CH \\cos \\angle DCH} = \\frac{BH \\cos(90^\\circ - \\angle BAC)}{CH \\cos(90^\\circ - \\angle BAC)} = \\frac{BH}{CH}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26668,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $u_1, \\dots, u_k \\in \\mathbb{R}^3$, denote by $C(u_1, \\dots, u_k)$ the cone generated by $u_1, \\dots, u_k$:\n$$\nC(u_1, \\dots, u_k) = \\{a_1 u_1 + \\dots + a_k u_k; a_1, \\dots, a_k \\in [0, +\\infty)\\}.\n$$\nLet $v_1, v_2, v_3, v_4$ points randomly and independently chosen from the unit sphere $x^2 + y^2 + z^2 = 1$.\n\na. What is the probability that $C(v_1, v_2, v_3, v_4) = \\mathbb{R}^3$?\n\nb. What is the probability that each of the vectors is needed to generate $C(v_1, v_2, v_3, v_4)$, i.e., that $C(v_1, v_2, v_3) \\neq C(v_1, v_2, v_3, v_4)$, $C(v_1, v_2, v_4) \\neq C(v_1, v_2, v_3, v_4)$, $C(v_1, v_3, v_4) \\neq C(v_1, v_2, v_3, v_4)$ and $C(v_2, v_3, v_4) \\neq C(v_1, v_2, v_3, v_4)$?",
    "options": [],
    "answer": "a: 1/8; b: 1/2",
    "solution": "a. The probability that the cone of the four vectors is proper is $\\frac{7}{8}$ so the probability that the cone is all of $\\mathbb{R}^3$ is $\\frac{1}{8}$.\nConstruct a vector $u_{12}$ that is normal to the plane spanned by $v_1$ and $v_2$ oriented so that $v_3 \\cdot u_{12} > 0$. Then the half-space $\\{w \\mid w \\cdot u_{12} \\ge 0\\}$ contains the cone generated by $\\{v_1, v_2, v_3\\}$. It follows that if $v_4 \\cdot u_{12} > 0$, the cone generated by all four vectors will be contained in the same half-space. So to keep the cone from being proper, we must assume that $v_4 \\cdot u_{12} < 0$.\n\nSimilarly, find $u_{13}$ orthogonal to $v_1$ and $v_3$ with $v_2 \\cdot u_{13} > 0$ and $u_{23}$ orthogonal to $v_2$ and $v_3$ with $v_1 \\cdot u_{13} > 0$. The cone is proper – contained in a half space – if and only if at least one of the three values $v_4 \\cdot u_{ij} > 0$. I further claim that if all three of those dot products are negative, then the cone covers all of space.\nIf $v_1, v_2$ and $v_3$ are fixed, the three signs of the dot products $v_4 \\cdot u_{ij}$ are not independent. But if we average over all choices of the vectors, then $-v_1$ occurs exactly as often as $+v_1$ and so on. We conclude that on average the three dot products in question are negative with probability $\\frac{1}{8}$.\n\n\nb. Given $v_1, v_2$ and $v_3$ then $v_4$ lies in the interior of the cone generated by those three if and only if $v_4 \\cdot u_{ij} > 0$ for all three such dot products. So there is a $\\frac{1}{8}$ chance that $v_4$ lies in $C(v_1, v_2, v_3)$. Similarly, there is a $\\frac{1}{8}$ chance that $v_2$ lies in $C(v_1, v_3, v_4)$. These two events are disjoint: only one vector can be in the interior of a triangle of the other three. So the probability we seek is the union of four disjoint events, each of probability $\\frac{1}{8}$, which gives a probability of $\\frac{1}{2}$.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 26669,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nJoãozinho escreveu os números de $1$ até $100000$ no quadro, depois foi trocando cada número pela soma de seus algarismos e repetiu este processo até obter uma lista de $100000$ números de um algarismo. Por exemplo, começando pelo número $7234$ obtemos $7+2+3+4=16$ e $1+6=7$.\n\na. Que número ficou no lugar do número $98765$?\n\nb. Quantas vezes aparece o número $8$ na lista final?\n\nc. Qual é o número que mais vezes se repete?",
    "options": [],
    "answer": "a: 8; b: 11111; c: 1",
    "solution": "Solution:\n\na. $98765 \\longrightarrow 9+8+7+6+5=35 \\longrightarrow 3+5=8$.\n\nb. Observemos que um número e a soma de seus algarismos deixam o mesmo resto quando divididos por $9$. Assim, depois de terminar todo o processo vamos obter uma lista da forma\n$$\n1,2,3,4,5,6,7,8,9,1,2,3,4, \\ldots, 8,9,1\n$$\nAssim até $99999$, cada um dos algarismos aparece $11111$ vezes, em particular o $8$ aparece $11111$ vezes.\n\nc. Do item anterior fica claro que o número que mais se repete é o $1$, pois aparece $11112$ vezes na lista.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26670,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $(x_n)_{n \\ge 0}$ be a sequence of integer numbers that fulfills a linear recursion of order $k$ for a fixed positive integer $k$, i.e., there exists real constant numbers $c_1, c_2, \\dots, c_k$ such that $x_{n+k} = \\sum_{r=1}^k c_r x_{n+k-r}$, $\\forall n \\ge 0$. Suppose $k$ is the minimum positive integer with this property. Prove that $c_j \\in \\mathbb{Z}$, for all $j$, $1 \\le j \\le k$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $(a_n^{(i)})$ be the sequence of integers obtained by shifting $a_n$ by $i$ positions, i.e., $a_n^{(i)} = a_{n+i}$. Then\n$$\n(a_n^{(0)}), (a_n^{(1)}), \\dots, (a_n^{(k)})\n$$\nis a basis for the $k$-dimensional $C$-vector space of sequences $(b_n)$ satisfying\n$$\nb_{n+k} = c_{k-1}b_{n+k-1} + \\dots + c_0b_n \\quad (n \\ge 0) \\quad (*)\n$$\nIn fact, any non-trivial $C$-linear relation among $(a_n^{(0)}, \\dots, a_n^{(k-1)})$ would\nimply that $k$ is not minimal.\n\nNow let\n$$\nf(x) = x^k - c_{k-1}x^{k-1} - \\dots - c_0\n$$\nand\n$\\alpha_1, \\dots, \\alpha_k$\nbe the roots of $f(x)$ (listed with multiplicity). First, observe that all coefficients $c_i$ are rational since they are solutions to the linear system with integer coefficients\n$$\n\\begin{pmatrix} a_0 & a_1 & \\dots & a_{k-1} \\\\ a_1 & a_2 & \\dots & a_k \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ a_{k-1} & a_k & \\dots & a_{2k-2} \\end{pmatrix} \\begin{pmatrix} c_0 \\\\ c_1 \\\\ \\vdots \\\\ c_{k-1} \\end{pmatrix} = \\begin{pmatrix} a_k \\\\ a_{k+1} \\\\ \\vdots \\\\ a_{2k-1} \\end{pmatrix}\n$$\n\n$$\nt_n = \\alpha_1^n + \\alpha_2^n + \\dots + \\alpha_k^n \\quad (n \\ge 0)\n$$\nThen all $t_n \\in \\mathbb{Q}$ (they are symmetric expression on the roots of $f(x) \\in \\mathbb{Q}[x]$) and the $t_n$ satisfy $(*)$, hence there are $r_i$ such that\n$$\nt_n = r_0 a_n^{(0)} + r_1 a_n^{(1)} + \\dots + r_{k-1} a_n^{(k-1)}\n$$\n\nTo sum up, if $d > 0$ is an integer such that $dr_i \\in \\mathbb{Z}$, $i = 0, 1, \\dots, k-1$, then\n$$\ndt_n \\in \\mathbb{Z}\n$$\nfor all $n$. Next, we show that this implies that the $\\alpha_i$ are algebraic integers, which in turn implies that all the $c_i$ are integers, finishing the problem.\nBy Newton's identities, we may write the elementary symmetric polynomials in $\\alpha^n$ as polynomials with rational coefficients of\n$$\n\\begin{align*} \nt_n &= \\alpha_1^n + \\dots + \\alpha_k^n \\\\ \nt_{2n} &= \\alpha_1^{2n} + \\dots + \\alpha_k^{2n} \\\\ &\\vdots \\\\ \nt_{kn} &= \\alpha_1^{kn} + \\dots + \\alpha_k^{kn} \\end{align*}\n$$\nTherefore the minimal polynomials of $\\alpha_i^n$ over $\\mathbb{Q}$ have coefficients with bounded denominators, independent of $n$ (they depend only on $d$ and $k$). Hence there exists an integer $\\Delta > 0$ such that $\\Delta \\cdot \\alpha_i^n$ are algebraic integers for all $n$. But that implies that the ring $\\mathbb{Z}[\\alpha_i]$ is contained in the finitely generated $\\mathbb{Z}$-module $\\frac{1}{\\Delta}\\mathcal{O}_{\\mathbb{Q}(\\alpha_i)}$, where $\\mathcal{O}_{\\mathbb{Q}(\\alpha_i)}$ is the ring of algebraic integers in the field $\\mathbb{Q}(\\alpha_i)$. Therefore $\\mathbb{Z}[\\alpha_i]$ itself is finitely generated as a $\\mathbb{Z}$-module (use the fact that $\\mathbb{Z}$ is noetherian or that $\\mathcal{O}_{\\mathbb{Q}(\\alpha_i)}$ is a free $\\mathbb{Z}$-module, together with the structure theorem of finitely generated modules over a PID). Hence $\\alpha_i$ is an algebraic integer, as required.\n\nAlternatively, suppose that $\\alpha_i$ is not an algebraic integer, so that it has prime ideal factorization\n$$\n(\\alpha_i) = p_1^{e_1} \\dots p_s^{e_s}\n$$\nwith $e_i < 0$ for some $i$, say $e_1$.\nThen\n$$\n(\\Delta \\alpha_i^n) = (\\Delta) \\cdot p_1^{ne_1} \\dots p_s^{ne_s}\n$$\nwould have a negative $p_1$-exponent for $n$ sufficiently large, contradicting the fact that $\\Delta \\cdot \\alpha_i^n$ is an algebraic integer for all $n \\ge 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26671,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_1, a_2, \\dots, a_{2011}$ be nonnegative reals with sum $\\frac{2011}{2}$. Prove that\n$$\n\\left| \\prod_{cyc} (a_n - a_{n+1}) \\right| = \\left| (a_1 - a_2)(a_2 - a_3) \\dots (a_{2011} - a_1) \\right| \\le \\frac{3\\sqrt{3}}{16}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "In what follows, indices are taken modulo $2011$ and $E = \\left| \\prod_{cyc} (a_n - a_{n+1}) \\right|$.\n\n**Lemma.** If $E$ is maximum, for every $i \\in \\{1, 2, \\dots, 2011\\}$, one of the numbers $a_{i-1}, a_i, a_{i+1}$ is zero.\n\n*Proof.* Suppose, by means of contradiction, that $E$ is maximum and there exists $a_i$ such that $a_{i-1}, a_i, a_{i+1}$ are all nonzero (that is, $a_{i-1}a_i a_{i+1} > 0$). Define $A = \\{a_i \\mid a_i > 0\\}$ and $B = \\{a_i \\mid a_{i-1}a_i a_{i+1} > 0\\}$. Then $B \\subset A$ and $B \\neq \\emptyset$. Let $a_k = \\min B$ and consider $a_{k-1}$ and $a_{k+1}$. We have the following cases:\n\n* $a_k < a_{k-1}$ and $a_k < a_{k+1}$. Let\n$$\na'_i = \\begin{cases} 0, & \\text{if } a_i = 0 \\text{ or } i = k \\\\ a_i + \\frac{a_k}{|A|-1}, & \\text{if } a_i > 0 \\text{ and } i \\neq k \\end{cases}\n$$\nThat is, we make $a_k$ be zero and distribute it among the remaining nonzero terms. So $|a_i - a_{i+1}|$ remains unchanged if $a_i, a_{i+1} \\in A$ and $k \\notin \\{i, i+1\\}$, or $a_i, a_{i+1} \\notin A$; increases from $|a_i - a_{i+1}| = \\max\\{a_i, a_{i+1}\\}$ to $\\max\\{a_i, a_{i+1}\\} + \\frac{a_k}{|A|-1}$ if $a_i \\notin A$ or $a_{i+1} \\notin A$, but not both; increases from $|a_{k \\pm 1} - a_k| = a_{k \\pm 1} - a_k$ to $a_{k \\pm 1} + \\frac{a_k}{|A|-1}$ if $k \\in \\{i, i+1\\}$.\n\n* $a_{k-1} < a_k < a_{k+1}$. This means that $a_{k-1} \\notin B$, and $a_k \\in B$, $a_{k-1} > 0$, that is, $a_{k-1} \\in A \\setminus B$, which means $a_{k-2} = 0$. In this case, we exchange $(a_{k-1}, a_k)$ for $(a'_{k-1}, a'_k) = (a_{k-1} + a_k, 0)$. Then $|a_i - a_{i+1}|$ remains unchanged for $i \\notin \\{k-2, k-1, k\\}$; for $i = k-2$ increases from $|a_{k-2} - a_{k-1}| = a_{k-1}$ to $|a_{k-2} - a'_{k-1}| = a_{k-1} + a_k$; for $i = k-1$ increases from $|a_{k-1} - a_k| = a_k - a_{k-1}$ to $|a'_{k-1} - a'_k| = a_{k-1} + a_k$; for $i = k$ increases from $|a_k - a_{k+1}| = a_{k+1} - a_k$ to $|a'_{k} - a_{k+1}| = a_{k+1}$.\n\n* $a_{k-1} > a_k > a_{k+1}$. Analogous to the previous case.\n\n* $a_k > a_{k-1}$ and $a_k > a_{k+1}$. This means $a_{k-1}, a_{k+1} \\in A \\setminus B$, that is, $a_{k-2} = a_{k+2} = 0$. In this case, exchange $(a_{k-1}, a_k, a_{k+1})$ for $(a'_{k-1}, a'_k, a'_{k+1}) = (a_{k-1} + a_k/2, 0, a_{k+1} + a_k/2)$. All differences $|a_i - a_{i+1}|$ remain unchanged except if $i \\in \\{k-2, k-1, k, k+1\\}$. The only change is $|(a_{k-2}-a_{k-1})(a_{k-1}-a_k)(a_k-a_{k+1})(a_{k+1}-a_{k+2})| = a_{k-1}(a_k-a_{k-1})(a_k-a_{k+1})a_{k+1}$ to $|(a_{k-2}-a'_{k-1})(a'_{k-1}-a'_k)(a'_k-a'_{k+1})(a'_{k+1}-a_{k+2})| = (a_{k-1}+a_k/2)^2(a_{k+1}+a_k/2)^2$. But\n$$ \n\\begin{aligned}\n& (a_{k-1} + a_k/2)^2 (a_{k+1} + a_k/2)^2 \\\\\n& = (a_{k-1}(a_{k-1} + a_k) + a_k^2/4)(a_{k+1}(a_{k+1} + a_k) + a_k^2/4) \\\\\n& > a_{k-1}(a_k + a_{k-1})(a_k + a_{k+1})a_{k+1} \\\\\n& > a_{k-1}(a_k - a_{k-1})(a_k - a_{k+1})a_{k+1}\n\\end{aligned}\n$$\n\nNow we only have groups with one or two consecutive nonzero variables. For a group $(0, a_k, 0)$, we obtain the product $|(a_{k-1} - a_k)(a_k - a_{k+1})| = a_k^2$; for a group $(0, a_k, a_{k+1}, 0)$, we obtain $|(a_{k-1} - a_k)(a_k - a_{k+1})(a_{k+1} - a_{k+2})| = a_k a_{k+1}|a_{k+1} - a_k|$. Notice that the groups can be interchanged, such that we can suppose wlog that all groups with two nonzero variables are contiguous.\n\n**Lemma.** If $E$ is maximum then there is exactly one group with two nonzero variables.\nSuppose, that there are at least two groups of nonzero variables $(0, a, b, 0)$ and $(0, c, d, 0)$. By the above remark, we can suppose wlog that the groups are consecutive, that is, it's $(0, a, b, 0, c, d, 0)$. Exchange these variables for $(0, a + b/2, 0, (b+c)/2, 0, d + c/2, 0)$. The product $abcd|(a-b)(c-d)|$ is exchanged for $(a+b/2)^2((b+c)/2)^2(d+c/2)^2$. But we already know that $(a+b/2)^2 > a|a-b|$, $(d+c/2)^2 > d|c-d|$ and, by AM-GM, $((b+c)/2)^2 \\ge bc$. Multiplying everything yields the lemma.\n\nCombining the two lemmas, we can suppose wlog that the nonzero variables are the ones with odd indices, that is, $a_1, a_3, \\dots, a_{2011}$. In this case, we obtain the product $a_1 a_{2011} |a_1 - a_{2011}| a_3^2 a_5^2 \\dots a_{2009}^2$, and we can optimize it locally.\nLet $a_1 + a_{2011} = s$ and suppose wlog $a_1 > a_{2011}$. Let $\\alpha, \\beta$ be positive real numbers to be determined. By AM-GM,\n$$\n\\begin{aligned}\na_1 a_{2011} (a_1 - a_{2011}) &= \\frac{1}{\\alpha\\beta} (\\alpha a_1) (\\beta a_{2011}) (a_1 - a_{2011}) \\\\\n&\\le \\frac{1}{\\alpha\\beta} \\left( \\frac{\\alpha a_1 + \\beta a_{2011} + (a_1 - a_{2011})}{3} \\right)^3 \\\\\n&= \\frac{1}{\\alpha\\beta} \\left( \\frac{(\\alpha + 1)a_1 + (\\beta - 1)a_{2011}}{3} \\right)^3\n\\end{aligned}\n$$\nSo we choose $\\alpha$ and $\\beta$ such that\n* we obtain $s$ in the end, that is, $\\alpha + 1 = \\beta - 1 \\Leftrightarrow \\beta - \\alpha = 2$;\n* the equality can occur, that is, $\\alpha a_1 = \\beta a_{2011} = a_1 - a_{2011} \\Leftrightarrow a_{2011} = (1-\\alpha)a_1$ and $a_1 = (\\beta+1)a_{2011}$, that is, $1 = (1-\\alpha)(\\beta+1) \\Leftrightarrow -\\alpha\\beta = \\alpha - \\beta = -2$.\nThus $-\\alpha$ and $\\beta$ are the roots of the quadratic $t^2 - 2t - 2 = 0$. Hence $\\alpha = \\sqrt{3} - 1$ and $\\beta = 1 + \\sqrt{3}$, and\n$$\n\\begin{aligned}\na_1 a_{2011} (a_1 - a_{2011}) &\\le \\frac{1}{\\alpha\\beta} \\left( \\frac{(\\alpha + 1)a_1 + (\\beta - 1)a_{2011}}{3} \\right)^3 \\\\\n&= \\frac{1}{2} \\left( \\frac{\\sqrt{3}(a_1 + a_{2011})}{3} \\right)^3 = \\frac{\\sqrt{3}}{18} s^3\n\\end{aligned}\n$$\n\nNow we optimize the rest. If $a_3 + a_5 + \\dots + a_{2009} = \\frac{2011}{2} - s$,\n$$\na_3^2 a_5^2 \\dots a_{2009}^2 \\le \\left( \\frac{a_3 + a_5 + \\dots + a_{2009}}{1004} \\right)^{2008} = \\left( \\frac{\\frac{2011}{2} - s}{1004} \\right)^{2008}\n$$\n\n\\begin{align*}\nE &= a_1 a_{2011} |a_1 - a_{2011}| a_3^2 a_5^2 \\dots a_{2009}^2 \\\\\n&\\le \\frac{\\sqrt{3}}{18} s^3 \\cdot \\left(\\frac{\\frac{2011}{2}-s}{1004}\\right)^{2008} = \\frac{\\sqrt{3}}{18\\gamma^3} (\\gamma s)^3 \\left(\\frac{\\frac{2011}{2}-s}{1004}\\right)^{2008} \\\\\n&\\le \\frac{\\sqrt{3}}{18\\gamma^3} \\left( \\frac{3\\gamma s + 2008 \\cdot \\frac{\\frac{2011}{2}-s}{1004}}{2011} \\right)^{2011} \\\\\n&= \\frac{\\sqrt{3}}{18\\gamma^3} \\left( \\frac{2011 + (3\\gamma - 2)s}{2011} \\right)^{2011}\n\\end{align*}",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26672,
    "subject": "Mathematics (Multi-modal)",
    "question": "For each positive integer $N$ with $2k$ digits, let $\\text{odd}(N)$ be the $k$-digit number obtained by writing the digits of odd order of $N$ and $\\text{even}(N)$ be the $k$-digit number obtained by writing the digits of even order of $N$. For example, $\\text{odd}(249035) = 405$ and $\\text{even}(249035) = 293$. Prove that there is no positive integer $N$ with $2k$ digits such that $N = \\text{odd}(N) \\cdot \\text{even}(N)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will prove by induction that $\\text{odd}(N) \\cdot \\text{even}(N) < N$ for all positive integers $N$ with $2k$ digits.\n\nIf $N = 10a + b$, $a, b \\in \\{0, 1, 2, \\dots, 9\\}$, $a \\neq 0$, $N = 10a + b > a \\cdot b + b \\ge a \\cdot b = \\text{even}(N) \\cdot \\text{odd}(N)$.\n\nNow suppose that $N$ has $2k > 2$ digits and that the claim is true for all numbers with $2k-2$ digits. Let $c$ and $d$ be the two leftmost digits of $N$, so that $N = c \\cdot 10^{2k-1} + d \\cdot 10^{2k-2} + N_0$, $N_0$ with $2k-2$ digits. Then $\\text{odd}(N) = d \\cdot 10^{k-1} + \\text{odd}(N_0)$ and $\\text{even}(N) = c \\cdot 10^{k-1} + \\text{even}(N_0)$. So we need to prove that\n$$\n\\begin{align*}\nc \\cdot 10^{2k-1} + d \\cdot 10^{2k-2} + N_0 &> (c \\cdot 10^{k-1} + \\text{even}(N_0)) \\cdot (d \\cdot 10^{k-1} + \\text{odd}(N_0)) \\\\\n\\iff c \\cdot 10^{2k-1} + d \\cdot 10^{2k-2} + N_0 &> cd \\cdot 10^{2k-2} + d \\cdot 10^{k-1} \\cdot \\text{even}(N_0) + c \\cdot 10^{k-1} \\cdot \\text{odd}(N_0) + \\text{odd}(N_0) \\cdot \\text{even}(N_0)\n\\end{align*}\n$$\nBut this is true, since both $\\text{odd}(N_0)$ and $\\text{even}(N_0)$ are less than $10^{k-1}$ and thus\n$$\n\\begin{align*}\nc \\cdot 10^{2k-1} &\\ge c(d+1) \\cdot 10^{2k-2} > cd \\cdot 10^{2k-2} + c \\cdot 10^{k-1} \\cdot \\text{odd}(N_0) \\\\\nd \\cdot 10^{2k-2} &> d \\cdot 10^{k-1} \\cdot \\text{even}(N_0) \\\\\nN_0 &> \\text{odd}(N_0) \\cdot \\text{even}(N_0)\n\\end{align*}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26673,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDe quantas formas é possível colorir as 6 faces de um cubo de preto ou branco? Duas colorações são iguais se é possível obter uma a partir da outra por uma rotação.",
    "options": [],
    "answer": "10",
    "solution": "Solution:\n\nObservemos que basta contar quantas colorações existem que têm exatamente 0, 1, 2 e 3 faces pretas, porque os outros casos são simétricos. Com uma ou nenhuma face preta existe uma única coloração para cada caso. Quando temos duas faces pretas temos duas possíveis colorações que são: quando estas faces são opostas e quando elas não são. Por último, com três faces pretas também temos dois casos: quando duas dessas faces pretas são opostas e quando não existem faces opostas de cor preta. Assim, no total temos $1+1+2+2+2+1+1=10$ possíveis colorações.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26674,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm professor e seus 30 alunos escreveram, cada um, os números de $1$ a $30$ em uma ordem qualquer. A seguir, o professor comparou as sequências. Um aluno ganha um ponto cada vez que um número aparece na mesma posição na sua sequência e na do professor. Ao final, observou-se que todos os alunos obtiveram quantidades diferentes de pontos. Mostre que a sequência de um aluno coincidiu com a sequência do professor.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nO número de acertos é um número entre $0$ e $30$ inclusive. Mas, observe que $29$ não pode ser obtido porque se $29$ números estão em posição certa, só há uma maneira de colocar o $30^{\\circ}$ número, que é em posição certa também.\n\nComo há $30$ alunos e $30$ possíveis resultados, $\\{0,1, \\ldots, 28,30\\}$, então um aluno escreveu exatamente a sequência do professor.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26675,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDez pontos são marcados ao redor de uma circunferência, como ilustra a figura.\n![](attached_image_1.png)\n(a) Quantas cordas podem ser formadas ligando dois quaisquer destes pontos? (Uma corda é um segmento de reta ligando dois pontos sobre uma circunferência.)\n\n(b) Quantos triângulos podem ser formados ligando três quaisquer destes pontos?",
    "options": [],
    "answer": "a) 45; b) 120",
    "solution": "Solution:\n(a) De cada ponto saem 9 cordas e temos 10 pontos. Mas cada corda é contada duas vezes (uma corda $AB$ é contada por sair de $A$ e por sair de $B$), assim temos $9 \\times 10 / 2 = 45$ cordas.\n\n(b) Cada corda é lado de 8 triângulos (basta escolher um ponto que não seja extremidade da corda escolhida) mas cada triângulo é contado três vezes (uma vez para cada corda). Como temos 45 cordas, então temos $8 \\times 45 / 3 = 120$ triângulos.\n\nContando Subconjuntos\n\nVamos resolver um problema mais geral em que temos $n$ pontos distribuídos na circunferência. Como cada corda está determinada por dois pontos, então precisamos contar de quantas formas podemos escolher 2 pontos entre os $n$.\n\nO primeiro ponto pode ser escolhido de $n$ formas, já o segundo pode ser escolhido de $n-1$ formas, pois ele deve ser distinto do primeiro selecionado. Assim temos $n(n-1)$ escolhas de pares ordenados, mas a ordem em que foram selecionados não importa, porque eles geram o mesmo subconjunto e assim o mesmo segmento. Portanto, o número de subconjuntos de dois pontos ou equivalentemente o número de segmentos é $n(n-1)/2$.\n\nSeguindo este raciocínio, encontrar todos os triângulos equivale a encontrar todos os subconjuntos de três pontos dentre os $n$ pontos. Assim, a escolha ordenada de três pontos pode ser realizada de $n(n-1)(n-2)$ maneiras, mas como a ordem não importa, então o subconjunto com três elementos $\\{a, b, c\\}$, está sendo contado seis vezes: $abc, acb, bac, bca, cab, cba$. Deste modo, o número de subconjuntos com três pontos, ou equivalentemente, o número de triângulos com vértices nos $n$ pontos é $n(n-1)(n-2)/6$.\n\nNo caso geral, se queremos saber quantos polígonos convexos com $k$ vértices existem (ou equivalentemente, quantos subconjuntos de $k$ pontos temos entre os $n$ pontos), a resposta é dada por $\\binom{n}{k}$ (lê-se \"$n$ escolhe $k$\"), que é calculado como\n$$\n\\binom{n}{k} = \\frac{n(n-1)(n-2) \\ldots (n-k+1)}{1 \\cdot 2 \\cdot 3 \\cdot \\ldots \\cdot k}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26676,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nCada uma das placas das bicicletas de Quixajuba contém três letras. A primeira letra é escolhida dentre os elementos do conjunto $\\mathcal{A}=\\{\\mathrm{G}, \\mathrm{H}, \\mathrm{L}, \\mathrm{P}, \\mathrm{R}\\}$, a segunda letra é escolhida dentre os elementos do conjunto $\\mathcal{B}=\\{\\mathrm{M}, \\mathrm{I}, \\mathrm{O}\\}$ e a terceira letra é escolhida dentre os elementos do conjunto $\\mathcal{C}=\\{\\mathrm{D}, \\mathrm{U}, \\mathrm{N}, \\mathrm{T}\\}$.\nDevido ao aumento no número de bicicletas da cidade, teve-se que expandir a quantidade de possibilidades de placas. Ficou determinado acrescentar duas novas letras a apenas um dos conjuntos ou uma letra nova a dois dos conjuntos.\nQual o maior número de novas placas que podem ser feitos, quando se acrescentam as duas novas letras?",
    "options": [],
    "answer": "40",
    "solution": "Solution:\nInicialmente, é possível fazer o emplacamento de $5 \\times 3 \\times 4 = 60$ bicicletas. Vamos analisar as duas situações possíveis:\n\n- Aumentamos duas letras num dos conjuntos. Com isso, podemos ter\n\n| $\\mathcal{A} \\times \\mathcal{B} \\times \\mathcal{C}$ | Número de Placas |\n| :---: | :---: |\n| $7 \\times 3 \\times 4$ | 84 |\n| $5 \\times 5 \\times 4$ | 100 |\n| $5 \\times 3 \\times 6$ | 90 |\n\nAssim, com a modificação mostrada, o número de novas placas é no máximo $100 - 60 = 40$.\n\n- Aumentar uma letra em dois dos conjuntos. Com isso, podemos ter\n\n| $\\mathcal{A} \\times \\mathcal{B} \\times \\mathcal{C}$ | Número de Placas |\n| :---: | :---: |\n| $6 \\times 4 \\times 4$ | 96 |\n| $6 \\times 3 \\times 5$ | 90 |\n| $5 \\times 4 \\times 5$ | 100 |\n\nNeste caso, o número de placas novas também é no máximo 40.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26677,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nNum torneio de tênis cada jogador passa para a rodada seguinte somente em caso de vitória. Se não for possível que sempre passe para a rodada seguinte um número par de jogadores, a organização do torneio decide quais rodadas determinados jogadores devem jogar. Por exemplo, um cabeça de chave pode, a critério dos organizadores, entrar na segunda rodada, ou passar da primeira para a terceira, de modo que o total de jogadores que participem de cada rodada seja par.\n\na. Considere um torneio de tênis com 64 jogadores. Quantas partidas são disputadas?\n\nb. E em um torneio com 2011 jogadores?",
    "options": [],
    "answer": "a) 63; b) 2010",
    "solution": "Solution:\n\na. Na primeira rodada são realizadas $32$ partidas, das quais $32$ jogadores passam para a fase seguinte. Depois são realizadas $16$ partidas, classificando $16$ para a rodada seguinte e assim por diante. Assim, o número de partidas do torneio é\n$$\n32 + 16 + 8 + 4 + 2 + 1 = 63\n$$\n\nb. Como em cada partida um jogador é eliminado, então o número de partidas é igual ao número de jogadores eliminados, isto é, $2011 - 1 = 2010$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26678,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPossuímos 32 pedras, todas com pesos diferentes. Descreva um processo para mostrar que podemos encontrar as duas pedras mais pesadas com 35 pesagens em uma balança de pratos.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nDividimos as pedras em 16 pares, pesamos cada par e pegamos as 16 mais pesadas. Repetimos o processo com as 16 pedras obtendo 8 pedras com oito pesagens a mais, 4 pedras com quatro pesagens, 2 pedras com 2 pesagens e a pedra mais pesada com a última pesagem.\n\nAté este momento foram usadas $16+8+4+2+1=31$ pesagens para encontrar a pedra mais pesada.\n\nA segunda pedra mais pesada deve ser uma das pedras que foi comparada com a pedra mais pesada, que foram 5 pedras no total. É claro que para descobrir a segunda pedra mais pesada devem ser registradas as comparações das pesagens anteriores para saber quais pedras foram comparadas com a pedra mais pesada.\n\nPara determinar a pedra mais pesada entre estas cinco pedras, precisamos de 4 pesagens porque cada vez que fazemos uma pesagem eliminamos a pedra mais leve. Portanto, precisamos de 35 pesagens para determinar as 2 pedras mais pesadas.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26679,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nQuantos números naturais de cinco algarismos têm o produto de seus algarismos igual a 2000?",
    "options": [],
    "answer": "30",
    "solution": "Solution:\nInicialmente, observe que $2000 = 2^{4} \\times 5^{3}$. Como os algarismos do número são menores que $10$, cada fator $5$ deve ser um algarismo desse número. Além disso, o produto dos outros algarismos deve ser $2^{4} = 16$. Assim, temos dois casos:\n\n- Os algarismos que faltam são $2$ e $8$. Nesse caso, existem cinco possibilidades para posicionarmos o $2$, quatro possibilidades para posicionarmos o $8$ e uma única possibilidade para posicionarmos cada $5$ que resta. Portanto, podemos formar $5 \\times 4 = 20$ números.\n\n- Os algarismos que faltam são $4$ e $4$. Nesse caso, podemos escolher dois lugares para os algarismos $4$ de $\\binom{5}{2} = 10$ modos (veja Contando Subconjuntos na página 118) e uma maneira de posicionarmos cada $5$ que resta. Portanto, podemos formar $10$ números.\n\nLogo, podem ser formados $20 + 10 = 30$ números.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26680,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNum tabuleiro $123 \\times 123$, cada casa é pintada de roxo ou azul de acordo com as seguintes condições:\n- Cada casa pintada de roxo que não está na borda do tabuleiro tem exatamente 5 casas azuis dentre suas 8 vizinhas.\n- Cada casa pintada de azul que não está na borda do tabuleiro tem exatamente 4 casas roxas dentre suas 8 vizinhas.\nNota: Duas casas são vizinhas se possuem um lado ou um vértice em comum.\n\na) Considere um tabuleiro $3 \\times 3$ dentro do tabuleiro $123 \\times 123$. Quantas casas de cada cor pode haver neste tabuleiro $3 \\times 3$ ?\n\nb) Calcule o número de casas pintadas de roxo no tabuleiro $123 \\times 123$.",
    "options": [],
    "answer": "(a) 4 purple and 5 blue; (b) 6724 purple",
    "solution": "Solution:\n\n(a) Observando um tabuleiro $3 \\times 3$, podemos claramente ver que seu centro não está na borda do tabuleiro. A casa do centro pode:\n- Estar pintada de roxo. Nesse caso, temos dentre suas 8 vizinhas, 5 azuis e 3 roxas. No total, há 4 casas roxas e 5 casas azuis nesse tabuleiro.\n- Estar pintada de azul. Nesse caso, temos dentre suas 8 vizinhas, 4 azuis e 4 roxas. No total, há 4 casas roxas e 5 casas azuis nesse tabuleiro.\n\n(b) Como em qualquer tabuleiro $3 \\times 3$ dentro do tabuleiro $123 \\times 123$ o número de casas azuis é 5 e o número de casas roxas é 4, podemos dividir o tabuleiro $123 \\times 123$ em tabuleiros menores $3 \\times 3$ conforme a figura 69.1. Deste modo, o tabuleiro é dividido em $\\left(\\frac{123}{3}\\right)^{2} = 41^{2} = 1681$ tabuleiros $3 \\times 3$. Como cada tabuleiro $3 \\times 3$ tem 4 casas roxas, então há no total $1681 \\times 4 = 6724$ casas roxas.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26681,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMontar a tabela de um torneio em que todas as $n$ equipes se enfrentam ao longo de $n-1$ rodadas (como, por exemplo, em cada turno do Brasileirão) é um problema matemático bastante elaborado e que possui vários métodos de solução. Nesta questão, vamos conhecer uma dessas abordagens.\nVamos considerar um torneio com 6 equipes. Associaremos os números 1, 2, 3, 4, 5 e $\\infty$ (infinito) a cada uma das equipes. A primeira rodada do torneio é $1 \\times \\infty, 2 \\times 5, 3 \\times 4$. Para montarmos a rodada $i$ somamos $i-1$ a cada número envolvido nas partidas da rodada inicial, considerando que\n- quando a soma ultrapassa 5, subtraímos 5 do resultado;\n- $\\infty$ adicionado a qualquer inteiro positivo é $\\infty$. Por exemplo, a segunda rodada será:\n$$\n\\begin{gathered}\n(1+1) \\times (\\infty+1), \\text{ isto é, } 2 \\times \\infty \\\\\n(2+1) \\times (5+1) \\text{, isto é, } 3 \\times 1 \\\\\n(3+1) \\times (4+1) \\text{, isto é, } 4 \\times 5\n\\end{gathered}\n$$\n\na. Determine as 3 rodadas restantes do torneio, seguindo o método descrito acima.\n\nb. A partir do procedimento mostrado, exiba as 7 rodadas de um torneio com 8 equipes.",
    "options": [],
    "answer": "a) For 6 teams (5 rounds):\n1) 1 vs ∞, 2 vs 5, 3 vs 4\n2) 2 vs ∞, 3 vs 1, 4 vs 5\n3) 3 vs ∞, 4 vs 2, 5 vs 1\n4) 4 vs ∞, 5 vs 3, 1 vs 2\n5) 5 vs ∞, 1 vs 4, 2 vs 3\n\nb) For 8 teams (7 rounds):\n1) 1 vs ∞, 2 vs 7, 3 vs 6, 4 vs 5\n2) 2 vs ∞, 3 vs 1, 4 vs 7, 5 vs 6\n3) 3 vs ∞, 4 vs 2, 5 vs 1, 6 vs 7\n4) 4 vs ∞, 5 vs 3, 6 vs 2, 7 vs 1\n5) 5 vs ∞, 6 vs 4, 7 vs 3, 1 vs 2\n6) 6 vs ∞, 7 vs 5, 1 vs 4, 2 vs 3\n7) 7 vs ∞, 1 vs 6, 2 vs 5, 3 vs 4",
    "solution": "Solution:\n\n(a)\n\n$$\n\\begin{gathered}\n\\left(\\begin{array}{c}\n1 \\times \\infty \\\\\n2 \\times 5 \\\\\n3 \\times 4\n\\end{array}\\right) \\rightarrow \\left(\\begin{array}{c}\n2 \\times \\infty \\\\\n3 \\times 1 \\\\\n4 \\times 5\n\\end{array}\\right) \\rightarrow \\left(\\begin{array}{c}\n3 \\times \\infty \\\\\n4 \\times 2 \\\\\n5 \\times 1\n\\end{array}\\right) \\rightarrow \\\\\n\\left(\\begin{array}{c}\n4 \\times \\infty \\\\\n5 \\times 3 \\\\\n1 \\times 2\n\\end{array}\\right) \\rightarrow \\left(\\begin{array}{c}\n5 \\times \\infty \\\\\n1 \\times 4 \\\\\n2 \\times 3\n\\end{array}\\right)\n\\end{gathered}\n$$\n\n(b)\n\n$$\n\\begin{gathered}\n\\left(\\begin{array}{c}\n1 \\times \\infty \\\\\n2 \\times 7 \\\\\n3 \\times 6 \\\\\n4 \\times 5\n\\end{array}\\right) \\rightarrow \\left(\\begin{array}{c}\n2 \\times \\infty \\\\\n3 \\times 1 \\\\\n4 \\times 7 \\\\\n5 \\times 6\n\\end{array}\\right) \\rightarrow \\left(\\begin{array}{c}\n3 \\times \\infty \\\\\n4 \\times 2 \\\\\n5 \\times 1 \\\\\n6 \\times 7\n\\end{array}\\right) \\rightarrow \\left(\\begin{array}{c}\n4 \\times \\infty \\\\\n5 \\times 3 \\\\\n6 \\times 2 \\\\\n7 \\times 1\n\\end{array}\\right) \\rightarrow \\\\\n\\left(\\begin{array}{c}\n5 \\times \\infty \\\\\n6 \\times 4 \\\\\n7 \\times 3 \\\\\n1 \\times 2\n\\end{array}\\right) \\rightarrow \\left(\\begin{array}{c}\n6 \\times \\infty \\\\\n7 \\times 5 \\\\\n1 \\times 4 \\\\\n2 \\times 3\n\\end{array}\\right) \\rightarrow \\left(\\begin{array}{c}\n7 \\times \\infty \\\\\n1 \\times 6 \\\\\n2 \\times 5 \\\\\n3 \\times 4\n\\end{array}\\right)\n\\end{gathered}\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26682,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDistribuímos nos vértices de um bloco retangular oito números dentre $1, 2, 3, 4, 5, 6, 7, 8, 9, 10$ de tal forma que a soma dos números de uma face qualquer seja igual a $18$.\n\na. Quais os números descartados na distribuição?\n\nb. Exiba uma possível distribuição.",
    "options": [],
    "answer": "a) The discarded numbers are 9 and 10. b) One valid placement using the numbers 1 through 8 is: bottom face corners in cycle 1, 8, 5, 4 and top face directly above them 7, 2, 3, 6 (so the vertical pairs are 1–7, 8–2, 5–3, 4–6). Each face then sums to 18.",
    "solution": "Solution:\n\n(a) Como o bloco possui seis faces, a soma dos números em todas as faces é $18 \\times 6 = 108$, mas o número atribuído a cada vértice é contado três vezes nesta soma. Portanto, a soma dos números distribuídos é $108/3 = 36$. Como a soma de todos os números de $1$ a $10$ é igual a $55$, a soma dos dois números descartados é $19$. Concluímos que os números descartados são $9$ e $10$.\n\n(b) Uma possível distribuição é exibida na figura 72.1.\n\n![](attached_image_1.png)\n\nFigura 72.1",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26683,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNuma corrida de São Paulo a Fortaleza participam quatro carros $A$, $B$, $C$, $D$ que largaram na seguinte ordem: primeiro $A$, segundo $B$, terceiro $C$ e por último $D$. Durante a corrida, $A$ e $B$ trocaram de posição (ultrapassaram um ao outro) 9 vezes e $B$ e $C$ trocaram de posição 8 vezes.\n\nPara saber em que ordem chegaram à Fortaleza, só é permitido fazer perguntas do tipo:\n\n\"Quantas vezes trocaram de posição os carros $X$ e $Y$?\"\n\nAntes de fazer uma pergunta se conhece a resposta da pergunta anterior. Formule três perguntas que permitam determinar a ordem em que os quatro terminaram a corrida.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nInicialmente, observe que se dois carros trocaram de posição um número par de vezes, eles terminaram na mesma ordem em que começaram e se trocaram de posição um número ímpar de vezes, eles terminaram na ordem inversa. Isto nos leva a concluir que $B$ terminou a corrida na frente de $A$ e de $C$.\n\nFazemos a primeira pergunta sobre os carros $A$ e $C$. De acordo com a resposta saberemos quem terminou na frente.\n\nSuponhamos que $A$ chegou na frente de $C$ (o outro caso é análogo). Falta determinar a posição de $D$, para a qual há quatro possibilidades (à frente de $B$, entre $B$ e $A$, entre $A$ e $C$ e atrás de $C$). Fazemos a segunda pergunta para $A$ e $D$ e dependendo de $D$ chegar na frente ou atrás de $A$, perguntamos para $B$ e $D$ ou $C$ e $D$, respectivamente. Com a última resposta descobriremos entre quais carros $D$ chegou, determinando a ordem de chegada.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26684,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA sequência de números $t_{1}, t_{2}, t_{3}, \\ldots$ está definida por\n$$\n\\left\\{\\begin{array}{l}\nt_{1}=2 \\\\\nt_{n+1}=\\frac{t_{n}-1}{t_{n}+1}\n\\end{array}\\right.\n$$\npara cada inteiro positivo $n$. Encontrar $t_{2011}$.",
    "options": [],
    "answer": "-1/2",
    "solution": "Solution:\n\nCalculemos os primeiros termos da sequência:\n$$\n\\begin{aligned}\n& t_{2}=\\frac{2-1}{2+1}=\\frac{1}{3} \\\\\n& t_{3}=\\frac{1/3-1}{1/3+1}=-\\frac{1}{2} \\\\\n& t_{4}=\\frac{-1/2-1}{-1/2+1}=-3 \\\\\n& t_{5}=\\frac{-3-1}{-3+1}=2\n\\end{aligned}\n$$\nAssim, os primeiros cinco termos da sequência são $2, 1/3, -1/2, -3$ e $2$. Observamos que a sequência se repete a cada 4 termos, isto é,\n$$\n2 = t_{1} = t_{5} = t_{9} = t_{13} = \\cdots = t_{2009}\n$$\nAssim, $t_{2010} = 1/3$ e $t_{2011} = -1/2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26685,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA progressão geométrica $121,242,484,968,1936, \\ldots$ possui três termos inteiros entre 200 e 1200.\n\na) Encontre uma progressão geométrica crescente que possui quatro termos inteiros entre 200 e 1200.\n\nb) Encontre uma progressão geométrica crescente que possui seis termos inteiros entre 200 e 1200.",
    "options": [],
    "answer": "a) 256, 384, 576, 864 (common ratio 3/2).\nb) 243, 324, 432, 576, 768, 1024 (common ratio 4/3).",
    "solution": "Solution:\n\nObservemos que para obter termos inteiros, a razão entre os termos inteiros deve ser um racional $\\frac{p}{q}$ e para obter mais de três termos a razão tem que ser menor do que 2, já que para uma progressão de razão maior ou igual a 2, com o primeiro termo maior ou igual a 200, o quarto termo é maior ou igual a $200 \\times 2^{3} = 1600$.\n\nComo a progressão geométrica é da forma $A \\cdot \\frac{p^{k}}{q^{k}}$, para essa expressão representar um número inteiro precisamos que $q$ divida $A$ \"muitas vezes\". Assim, possíveis valores de $A$ são as potências de $q$.\n\nPor exemplo, se $A = 2^{8} = 256$, $q = 2$ e $p = 3$, obtemos a sequência\n$$\n256, 384, 576, 864, 1296, \\ldots\n$$\ncom quatro termos inteiros entre 200 e 1200.\n\nPor outro lado, se $A = 3^{5} = 243$, $q = 3$ e $p = 4$ obtemos a sequência\n$$\n243, 324, 432, 576, 768, 1024, \\ldots\n$$\nque possui seis termos inteiros entre 200 e 1200.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26686,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPara um inteiro positivo $n$ considere a função\n$$\nf(n)=\\frac{4 n+\\sqrt{4 n^{2}-1}}{\\sqrt{2 n+1}+\\sqrt{2 n-1}}\n$$\n\nCalcule o valor de\n$$\nf(1)+f(2)+f(3)+\\cdots+f(40)\n$$",
    "options": [],
    "answer": "364",
    "solution": "Solution:\n\nSeja $a=\\sqrt{2 n+1}$ e $b=\\sqrt{2 n-1}$. Então $a b=\\sqrt{4 n^{2}-1}$, $a^{2}+b^{2}=4 n$ e $a^{2}-b^{2}=2$. Portanto,\n$$\nf(n)=\\frac{a^{2}+b^{2}+a b}{a+b}\n$$\nComo $a-b \\neq 0$, podemos escrever\n$f(n)=\\frac{a^{2}+b^{2}+a b}{a+b} \\cdot \\frac{a-b}{a-b}=\\frac{a^{3}-b^{3}}{a^{2}-b^{2}}=\\frac{(\\sqrt{2 n+1})^{3}-(\\sqrt{2 n-1})^{3}}{2}$.\nAssim,\n$$\n\\begin{gathered}\nf(1)+f(2)+\\cdots+f(40)= \\\\\n=\\frac{(\\sqrt{3})^{3}-(\\sqrt{1})^{3}}{2}+\\frac{(\\sqrt{5})^{3}-(\\sqrt{3})^{3}}{2}+\\cdots+\\frac{(\\sqrt{81})^{3}-(\\sqrt{79})^{3}}{2}= \\\\\n=\\frac{(\\sqrt{81})^{3}-(\\sqrt{1})^{3}}{2}=\\frac{729-1}{2}=364\n\\end{gathered}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26687,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSejam $a$ e $b$ números reais tais que existam números reais distintos $m$, $n$ e $p$, satisfazendo as igualdades abaixo:\n$$\n\\left\\{\\begin{array}{l}\nm^{3}+a m+b=0 \\\\\nn^{3}+a n+b=0 \\\\\np^{3}+a p+b=0\n\\end{array}\\right.\n$$\n\nMostre que $m+n+p=0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSubtraindo a segunda equação da primeira, obtemos\n$$\n\\begin{gathered}\nm^{3}-n^{3}+a m-a n=0 \\Longleftrightarrow \\\\\n(m-n)\\left(m^{2}+m n+n^{2}\\right)+a(m-n)=0 \\Longleftrightarrow \\\\\n(m-n)\\left(m^{2}+m n+n^{2}+a\\right)=0\n\\end{gathered}\n$$\ne como $m-n \\neq 0$, temos que $m^{2}+m n+n^{2}+a=0$. Subtraindo a terceira equação da primeira, obtemos de forma análoga $m^{2}+m p+p^{2}+a=0$.\nSubtraindo estas duas últimas relações encontradas, temos\n$$\n\\begin{aligned}\n& m n-m p+n^{2}-p^{2}=0 \\Longleftrightarrow \\\\\n& m(n-p)+(n+p)(n-p)=0 \\\\\n&(n-p)(m+n+p)=0\n\\end{aligned}\n$$\ne como $n-p \\neq 0$, concluímos finalmente que $m+n+p=0$.\n\n\nSegunda Solução: Considere o polinômio de terceiro grau $P(x)=x^{3}+0 x^{2}+a x+b$. As relações dadas no problema nos garantem que $m, n$ e $p$ são as raízes de $P$. Portanto, a soma das raízes dessa equação é $m+n+p=0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26688,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\na.\nMostre que a identidade abaixo é sempre verdadeira:\n$$\na^{n+1}+b^{n+1}=(a+b)\\left(a^{n}+b^{n}\\right)-a b\\left(a^{n-1}+b^{n-1}\\right)\n$$\n\nb.\nSejam $a$ e $b$ números reais tais que $a+b=1$ e $ab=-1$. Mostre que o número $a^{10}+b^{10}$ é inteiro, calculando seu valor.",
    "options": [],
    "answer": "123",
    "solution": "Solution:\n\na.\nObservemos que\n$$\n\\begin{aligned}\n(a+b)\\left(a^{n}+b^{n}\\right) & =a^{n+1}+a b^{n}+b a^{n}+b^{n+1}= \\\\\n& =a^{n+1}+b^{n+1}+a b\\left(a^{n-1}+b^{n-1}\\right)\n\\end{aligned}\n$$\ne a identidade segue.\n\nb.\nChamemos de $f_{n}=a^{n}+b^{n}$. Observe que $f_{1}=a+b=1$. Calculemos $f_{2}$ :\n$$\nf_{2}=a^{2}+b^{2}=(a+b)^{2}-2 a b=1^{2}-2 \\cdot(-1)=3\n$$\n\nPela identidade do item (a) temos que\n$$\na^{n+1}+b^{n+1}=(a+b)\\left(a^{n}+b^{n}\\right)-a b\\left(a^{n-1}+b^{n-1}\\right)\n$$\nou equivalentemente\n$$\nf_{n+1}=(a+b) f_{n}-a b f_{n-1}=f_{n}+f_{n-1}\n$$\nAssim,\n$$\n\\begin{aligned}\nf_{3} & =f_{2}+f_{1}=4 \\\\\nf_{4} & =f_{3}+f_{2}=7 \\\\\nf_{5} & =f_{4}+f_{3}=11 \\\\\nf_{6} & =f_{5}+f_{4}=18 \\\\\nf_{7} & =f_{6}+f_{5}=29 \\\\\nf_{8} & =f_{7}+f_{6}=47 \\\\\nf_{9} & =f_{8}+f_{7}=76 \\\\\nf_{10} & =f_{9}+f_{8}=123\n\\end{aligned}\n$$\nPortanto, $a^{10}+b^{10}=f_{10}=123$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26689,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n(a) Verifique a identidade\n$$\n(a+b+c)^{3}=a^{3}+b^{3}+c^{3}+3(a+b)(b+c)(c+a)\n$$\n\n(b) Resolva o sistema\n$$\n\\left\\{\\begin{array}{l}\nx+y+z=1 \\\\\nx^{2}+y^{2}+z^{2}=1 \\\\\nx^{3}+y^{3}+z^{3}=1\n\\end{array}\\right.\n$$",
    "options": [],
    "answer": "(0,0,1), (0,1,0), (1,0,0)",
    "solution": "Solution:\n\n(a) Vamos expandir $(a+b+c)^{3}$ como $[(a+b)+c]^{3}$.\n$$\n\\begin{aligned}\n{[(a+b)+c]^{3} } & =(a+b)^{3}+3(a+b) c(a+b+c)+c^{3} \\\\\n& =a^{3}+b^{3}+3 a b(a+b)+3(a+b) c(a+b+c)+c^{3} \\\\\n& =a^{3}+b^{3}+c^{3}+3(a+b)[a b+c(a+b+c)] \\\\\n& =a^{3}+b^{3}+c^{3}+3(a+b)\\left(c^{2}+c(a+b)+a b\\right) \\\\\n& =a^{3}+b^{3}+c^{3}+3(a+b)(b+c)(c+a)\n\\end{aligned}\n$$\n\n(b) Utilizando a identidade verificada no item (a), obtemos\n$$\n(x+y+z)^{3}=x^{3}+y^{3}+z^{3}+3(x+y)(y+z)(z+x)\n$$\nSubstituindo os valores de $x+y+z$ e $x^{3}+y^{3}+z^{3}$ chegamos a\n$$\n1^{3}=1+3(x+y)(y+z)(z+x)\n$$\ndonde $(x+y)(y+z)(z+x)=0$. Assim, $x=-y$ ou $y=-z$ ou $z=-x$. Como as soluções são simétricas, vamos supor $x=-y$. Logo, de $x+y+z=1$, obtemos $z=1$ e de $x^{2}+y^{2}+z^{2}=1$ obtemos $2 x^{2}=0$, ou $x=0$. Concluímos que as possíveis soluções são $(0,0,1),(0,1,0)$ e $(1,0,0)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26690,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\na. Determine $a$, $b$ e $c$ tais que a igualdade\n$$\n(n+2)^2 = a(n+1)^2 + b n^2 + c(n-1)^2\n$$\nseja verdadeira qualquer que seja o número $n$.\n\nb. Suponha que $x_1, x_2, \\ldots, x_7$ satisfazem o sistema\n$$\n\\left\\{\\begin{array}{l}\nx_1 + 4 x_2 + 9 x_3 + 16 x_4 + 25 x_5 + 36 x_6 + 49 x_7 = 1 \\\\\n4 x_1 + 9 x_2 + 16 x_3 + 25 x_4 + 36 x_5 + 49 x_6 + 64 x_7 = 12 \\\\\n9 x_1 + 16 x_2 + 25 x_3 + 36 x_4 + 49 x_5 + 64 x_6 + 81 x_7 = 123\n\\end{array}\\right.\n$$\n\nDetermine o valor de\n$$\n16 x_1 + 25 x_2 + 36 x_3 + 49 x_4 + 64 x_5 + 81 x_6 + 100 x_7\n$$",
    "options": [],
    "answer": "a=3, b=-3, c=1; desired sum = 334",
    "solution": "Solution:\n\na. Se um polinômio se anula para infinitos valores, então todos os seus coeficientes são nulos.\nExpandindo a igualdade temos\n$$\nn^2 + 4n + 4 = a(n^2 + 2n + 1) + b n^2 + c(n^2 - 2n + 1)\n$$\nAssim,\n$$\n(a + b + c - 1) n^2 + (2a - 2c - 4) n + (a + c - 4) = 0\n$$\nqualquer que seja o número $n$. Logo,\n$$\n\\left\\{\\begin{array}{l}\na + b + c - 1 = 0 \\\\\n2a - 2c - 4 = 0 \\\\\na + c - 4 = 0\n\\end{array}\\right.\n$$\nResolvendo o sistema encontramos $a = 3$, $c = 1$ e $b = -3$.\n\nb. Sejam\n$$\n\\begin{aligned}\n& S_1 = x_1 + 4 x_2 + 9 x_3 + 16 x_4 + 25 x_5 + 36 x_6 + 49 x_7 = 1 \\\\\n& S_2 = 4 x_1 + 9 x_2 + 16 x_3 + 25 x_4 + 36 x_5 + 49 x_6 + 64 x_7 = 12 \\\\\n& S_3 = 9 x_1 + 16 x_2 + 25 x_3 + 36 x_4 + 49 x_5 + 64 x_6 + 81 x_7 = 123\n\\end{aligned}\n$$\nPela identidade da parte (a), temos que\n$$\n\\begin{aligned}\n& 16 x_1 + 25 x_2 + 36 x_3 + 49 x_4 + 64 x_5 + 81 x_6 + 100 x_7 = \\\\\n& \\quad = 3 S_3 - 3 S_2 + S_1 = 3 \\cdot 123 - 3 \\cdot 12 + 1 = 334\n\\end{aligned}",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26691,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO quadrado de $13$ é $169$, que tem como algarismo das dezenas o número $6$. O quadrado de outro número tem como algarismo das dezenas o número $7$. Quais são os possíveis valores para o algarismo das unidades desse quadrado?",
    "options": [],
    "answer": "6",
    "solution": "Solution:\n\nSuponhamos que o número é $10a + b$, com $b$ um algarismo. Quando elevamos ao quadrado obtemos\n$$\n(10a + b)^2 = 100a^2 + 20ab + b^2\n$$\nque tem três parcelas: $100a^2$, $20ab$ e $b^2$.\nA primeira parcela termina em $00$, enquanto a segunda termina em um número par seguido por zero. Assim, para o algarismo das dezenas ser $7$, isto é, ímpar, é necessário que o algarismo das dezenas de $b^2$ seja ímpar, o que somente acontece quando $b = 4$ ou $b = 6$. Em cada um dos casos, $4^2 = 16$ e $6^2 = 36$, o algarismo das unidades do quadrado é $6$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26692,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSeja $n$ um número inteiro positivo. Para cada um dos inteiros $n+1$, \\ldots, $2n$ considere o seu maior divisor ímpar. Prove que a soma de todos estes divisores é igual a $n^{2}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nChamemos de $S_{n}$ a soma dos maiores divisores ímpares dos números $n+1, \\ldots, 2n$. Por cálculo direto temos que $S_{1}=1$, $S_{2}=3+1=4=2^{2}$, $S_{3}=1+5+3=9=3^{2}$ e $S_{4}=5+3+7+1=16=4^{2}$.\n\nSe queremos calcular $S_{n+1}$, que é a soma dos maiores divisores ímpares dos números\n$$\nn+2, n+3, \\ldots, 2n, 2n+1, 2(n+1)\n$$\ncomo $n+1$ e $2(n+1)$ têm os mesmos divisores ímpares, isto é equivalente a somar os maiores divisores ímpares de\n$$\nn+2, n+3, \\ldots, 2n+1, n+1\n$$\nque é igual a $S_{n} + (2n+1)$. Assim, $S_{n+1} = S_{n} + (2n+1)$.\n\nPortanto,\n$$\n\\begin{gathered}\nS_{2} - S_{1} = 3 \\\\\nS_{3} - S_{2} = 5 \\\\\n\\vdots \\\\\nS_{n} - S_{n-1} = 2n-1.\n\\end{gathered}\n$$\n\nSomando todas estas igualdades obtemos $S_{n} - S_{1} = 3 + 5 + \\cdots + (2n-1)$ e deste modo, $S_{n} = 1 + 3 + \\cdots + (2n-1) = n^{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26693,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCom os algarismos $a$, $b$ e $c$ construímos o número de três algarismos $abc$ e os números de dois algarismos $ab$, $bc$ e $ca$. Ache todos os possíveis valores de $a$, $b$ e $c$ tais que $$\\frac{abc + a + b + c}{ab + bc + ca}$$ seja um número inteiro.\n\nSugestão: Mostre que o denominador é sempre divisível por $11$ e que $a + c = 11$.",
    "options": [],
    "answer": "(a,b,c) = (5,1,6), (9,1,2), (6,4,5), (3,7,8), (5,7,6), (7,7,5), (9,7,2)",
    "solution": "Solution:\n\nObservemos que\n$$\nab + bc + ca = (10a + b) + (10b + c) + (10c + a) = 11(a + b + c)\n$$\nde forma que o denominador da fração é divisível por $11$. Como a fração é um inteiro, o numerador\n$$\nabc + a + b + c = (100a + 10b + c) + a + b + c = 101a + 11b + 2c\n$$\ntambém é divisível por $11$. Como\n$$\n101a + 11b + 2c = 11(9a + b) + 2(a + c)\n$$\nsegue que $a + c$ é divisível por $11$. Como $a$ e $c$ são algarismos e $a \\neq 0$, $1 \\leq a + c \\leq 18$, donde $a + c = 11$. Substituindo $c$ na expressão dada obtemos\n$$\n\\begin{aligned}\n\\frac{abc + a + b + c}{ab + bc + ca} &= \\frac{11(9a + b + 2)}{11(b + 11)} \\\\\n&= \\frac{9a + b + 2}{b + 11} \\\\\n&= \\frac{b + 11 + 9a - 9}{b + 11} \\\\\n&= 1 + \\frac{9(a - 1)}{b + 11}\n\\end{aligned}\n$$\ne $9(a - 1)/(b + 11)$ é um inteiro. O algarismo $a$ não pode ser $1$, porque $a + c = 11$. Observamos que se $b + 11$ não é divisível por $3$, teríamos que $b + 11$ divide $a - 1$ que é impossível, já que $b + 11 > a - 1$. Assim, $b + 11$ é igual a $12$, $15$ ou $18$. Então $b = 1, 4$ ou $7$.\n\n- Se $b = 1$, como $9(a - 1)/12 = 3(a - 1)/4$ é inteiro temos que $a = 5$ ou $9$ que gera os números $516$ e $912$.\n- Se $b = 4$, $9(a - 1)/15 = 3(a - 1)/5$ é inteiro e assim, $a = 6$ que gera o número $645$.\n- Se $b = 7$, $9(a - 1)/18 = (a - 1)/2$ e então $a = 3, 5, 7$ ou $9$ gerando os números $378$, $576$, $775$ e $972$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26694,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTio Mané tem duas caixas, uma com sete bolas distintas numeradas de $1$ a $7$ e outra com oito bolas distintas numeradas com todos os números primos menores que $20$. Ele sorteia uma bola de cada caixa.\n\nSugestão: Calcule a probabilidade do produto ser ímpar. Qual é a probabilidade de que o produto dos números das bolas sorteadas seja par?",
    "options": [],
    "answer": "1/2",
    "solution": "Solution:\n\nO produto dos números sorteados é ímpar somente se as duas bolas sorteadas têm números ímpares.\n\nA probabilidade de sortearmos da primeira caixa uma bola com número ímpar é $\\frac{4}{7}$ e a probabilidade de sortearmos uma bola ímpar da segunda caixa é $\\frac{7}{8}$, porque esta contém bolas com os números $\\{2,3,5,7,11,13,17,19\\}$.\n\nAssim, a probabilidade do produto dos números das caixas ser ímpar\n$$\n\\frac{4}{7} \\times \\frac{7}{8} = \\frac{1}{2}\n$$\nPortanto, a probabilidade do produto ser par é $1 - 1/2 = 1/2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26695,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nConsidere o conjunto $A=\\{1,2,3, \\ldots, 2011\\}$. Quantos subconjuntos de $A$ existem de modo que a soma de seus elementos seja 2023060?",
    "options": [],
    "answer": "4",
    "solution": "Solution:\n\nObserve que a soma $1+2+\\cdots+2011=\\frac{2011 \\times 2012}{2}=2023066$. Logo, para obtermos um subconjunto de $A$ que tenha para soma de seus elementos $2023060$, basta retirarmos de $A$ os elementos cuja soma é $6$. Os possíveis casos são:\n\n- Subconjuntos com um elemento: $\\{6\\}$.\n- Subconjuntos com dois elementos: $\\{2,4\\}$ e $\\{1,5\\}$.\n- Subconjuntos com três elementos: $\\{1,2,3\\}$.\n\nPortanto, há quatro subconjuntos de $A$ cuja soma de seus elementos é $6$ e por consequência também há quatro subconjuntos de $A$ cuja soma dos elementos é $2023060$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26696,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUma formiga se movimenta uma unidade por segundo sobre os pontos $0$, $1$ e $2$ da figura a seguir, começando do ponto $0$.\n![](attached_image_1.png)\n\na) Quais são os possíveis percursos da formiga até $3$ segundos?\nb) Quantos possíveis percursos pode fazer a formiga até $10$ segundos?",
    "options": [],
    "answer": "a) 0-1-0-1 and 0-1-2-1. b) 32",
    "solution": "Solution:\n\n(a) Até três segundos temos dois possíveis percursos: $0-1-0-1$ ou $0-1-2-1$.\n\n(b) Observemos que quando a formiga está nos pontos $0$ e $2$ ela somente tem uma possibilidade para caminhar no segundo seguinte, que é ir para $1$. Quando está em $1$ ela tem duas possibilidades no segundo seguinte, que é ir para $0$ ou $2$. Assim, nos segundos ímpares a formiga sempre está no $1$, enquanto nos segundos pares ela está no $0$ ou no $2$. Portanto, o número de caminhos possíveis depois de $10$ segundos é\n$$\n1 \\times 2 \\times 1 \\times 2 \\times 1 \\times 2 \\times 1 \\times 2 \\times 1 \\times 2 = 32\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26697,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTiago escreve todos os números de quatro algarismos não nulos distintos que possuem a mesma paridade. Qual a probabilidade de que, ao escolhermos um desses números, ele seja par?",
    "options": [],
    "answer": "1/6",
    "solution": "Solution:\n\nOs quatro algarismos escolhidos fazem parte dos conjuntos $A=\\{1,3,5,7,9\\}$ ou $B=\\{2,4,6,8\\}$.\n\nCom os elementos do conjunto $A$ temos 5 possibilidades para o primeiro algarismo, 4 para o segundo, 3 para o terceiro e 2 para o quarto, totalizando $5 \\times 4 \\times 3 \\times 2=120$ números com 4 algarismos distintos.\n\nJá com os elementos do conjunto $B$ temos 4 possibilidades para o primeiro algarismo, 3 para o segundo, 2 para o terceiro e 1 para o quarto, totalizando $4 \\times 3 \\times 2 \\times 1=24$ números com quatro algarismos distintos.\n\nAssim, é possível formar $120+24=144$ números. De todas as possibilidades calculadas, apenas as geradas pelo conjunto $B$ são números pares.\n\nPortanto, a probabilidade pedida é $\\dfrac{24}{144} = \\dfrac{1}{6}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26698,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nConsidere uma urna que contém uma bola preta, quatro bolas brancas e algumas bolas azuis. Uma bola é retirada ao acaso dessa urna, sua cor é observada e a bola é devolvida à urna. Em seguida, retira-se novamente, ao acaso, outra bola dessa urna. Para quais quantidades de bolas azuis, a probabilidade das duas bolas retiradas terem mesma cor vale $1 / 2$ ?",
    "options": [],
    "answer": "1 or 9",
    "solution": "Solution:\n\nChamemos de $n$ o número de bolas azuis da caixa. Quando retiramos as duas bolas, elas podem ser:\n- Duas bolas pretas. A probabilidade é\n$$\n\\frac{1}{n+5} \\times \\frac{1}{n+5} = \\left(\\frac{1}{n+5}\\right)^2\n$$\n- Duas bolas brancas. A probabilidade é\n$$\n\\frac{4}{n+5} \\times \\frac{4}{n+5} = \\left(\\frac{4}{n+5}\\right)^2\n$$\n- Duas bolas azuis. A probabilidade é\n$$\n\\frac{n}{n+5} \\times \\frac{n}{n+5} = \\left(\\frac{n}{n+5}\\right)^2\n$$\nLogo, a probabilidade das duas bolas serem da mesma cor é a soma das probabilidades individuais:\n$$\n\\left(\\frac{1}{n+5}\\right)^2 + \\left(\\frac{4}{n+5}\\right)^2 + \\left(\\frac{n}{n+5}\\right)^2 = \\frac{1 + 16 + n^2}{(n+5)^2} = \\frac{1}{2}\n$$\nSimplificando a igualdade obtemos que $n^2 - 10n + 9 = 0$, donde $n$ é igual a $1$ ou $9$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26699,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nConsidere um poliedro convexo com $100$ arestas. Todos os vértices foram aparados próximos a eles mesmos, usando uma faca plana afiada (isto foi feito de modo que os planos resultantes não se intersectassem no interior ou na fronteira do poliedro). Calcule para o poliedro resultante:\n(a) o número de vértices.\n(b) o número de arestas.\n\nQuantas faces tem este novo poliedro?",
    "options": [],
    "answer": "Vertices: 200; Edges: 300; Faces: 102",
    "solution": "Solution:\n\n(a) Quando realizamos os cortes, cada aresta antiga estará ligada a dois vértices novos, enquanto os vértices antigos desaparecem. Assim o novo poliedro tem $200$ vértices.\n\n(b) Quando realizamos um corte, de cada novo vértice surgem duas arestas novas (correspondentes a duas arestas consecutivas na nova face criada) e uma aresta antiga. Assim, de cada vértice do novo poliedro saem exatamente $3$ arestas. Deste modo, se somarmos a quantidade de arestas que partem de todos os vértices, encontraremos $3 \\times 200 = 600$. Este número corresponde ao dobro do número de arestas, pois cada uma foi contada em dois vértices. Logo, o número de arestas é $300$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26700,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nExistem bolas azuis e bolas vermelhas em uma caixa. A probabilidade de sortear duas bolas de cores diferentes, ao retirar duas bolas ao acaso, é $1/2$. Prove que o número de bolas na caixa é um quadrado perfeito.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSuponha que existam $a$ bolas azuis e $v$ bolas vermelhas na caixa.\n\n(1) O número de modos de escolher duas bolas de cores diferentes é $a v$.\n\n(2) O número de modos de escolher duas bolas quaisquer é $\\binom{a+v}{2}$.\n\n(3) De (1) e (2), a probabilidade de sortear duas bolas de cores diferentes é $\\dfrac{a v}{\\binom{a+v}{2}}$.\n\nPortanto,\n$$\n\\frac{a v}{\\binom{a+v}{2}} = \\frac{1}{2} \\Longleftrightarrow 2 a v = \\frac{(a+v)(a+v-1)}{2}\n$$\ndonde\n$$\n4 a v = (a+v)^2 - (a+v) \\Longleftrightarrow a+v = (a-v)^2\n$$\nLogo, a quantidade de bolas é um quadrado perfeito.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26701,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDez pontos são dados no plano e não existem três colineares. Quatro segmentos distintos ligando pares destes pontos são escolhidos ao acaso, mas todos com a mesma probabilidade. Qual é a probabilidade de três dos segmentos escolhidos formarem um triângulo?",
    "options": [],
    "answer": "16/473",
    "solution": "Solution:\n\nO número de possíveis segmentos entre os $10$ pontos é $\\binom{10}{2} = 45$ e o número de formas de escolher $4$ desses segmentos é $\\binom{45}{4}$.\n\nJá o número de formas de escolher $4$ segmentos de tal modo que três deles formem um triângulo é igual ao número de maneiras de escolher três vértices, que determinam os três segmentos do triângulo, multiplicado pelo número de formas de escolher o outro segmento, isto é $\\binom{10}{3} \\times (45-3)$. Portanto, a probabilidade de que três dos quatro segmentos formem um triângulo é\n$$\n\\frac{\\binom{10}{3} \\times 42}{\\binom{45}{4}} = \\frac{10 \\times 9 \\times 8 \\times 42 \\times 4!}{3! \\times 45 \\times 44 \\times 43 \\times 42} = \\frac{16}{473}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26702,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm poliedro convexo $\\mathcal{P}$ tem 26 vértices, 60 arestas e 36 faces. 24 faces são triangulares e 12 são quadriláteros. Uma diagonal espacial é um segmento de reta unindo dois vértices não pertencentes a uma mesma face. $\\mathcal{P}$ possui quantas diagonais espaciais?",
    "options": [],
    "answer": "241",
    "solution": "Solution:\n\nOs 26 vértices determinam exatamente $\\binom{26}{2} = 26 \\times 25 / 2 = 325$ segmentos. Destes segmentos, 60 são arestas e como cada quadrilátero tem duas diagonais, então temos $12 \\times 2 = 24$ diagonais que não são espaciais.\n\nPortanto, o número de diagonais espaciais é $325 - 60 - 24 = 241$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26703,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm torneio de futebol com $57$ times será disputado com as seguintes regras:\n\na. Nenhum jogo pode terminar empatado.\n\nb. O time que perder duas partidas será eliminado.\n\nc. O torneio termina quando sobrar apenas um time, que será o campeão.\n\nSe o time campeão perder uma vez, quantas partidas serão disputadas no torneio?",
    "options": [],
    "answer": "113",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26704,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nObserve que\n$$\n\\begin{aligned}\n& 1 \\times 2 \\times 3 \\times 4 + 1 = 5^{2} \\\\\n& 2 \\times 3 \\times 4 \\times 5 + 1 = 11^{2} \\\\\n& 3 \\times 4 \\times 5 \\times 6 + 1 = 19^{2}\n\\end{aligned}\n$$\nProve que o produto de quatro inteiros positivos consecutivos, aumentado em uma unidade, é um quadrado perfeito.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26705,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDe quantas formas é possível colorir as 12 arestas de um cubo de branco ou de preto? Duas colorações são iguais quando é possível obter uma a partir da outra por uma rotação.",
    "options": [],
    "answer": "218",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26706,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEm um torneio de xadrez cada jogador disputou uma partida com cada um dos demais participantes. A cada partida, havendo empate, cada jogador ganhou $1/2$ ponto; caso contrário, o vencedor ganhou $1$ ponto e o perdedor, $0$ ponto. Participaram homens e mulheres e cada participante conquistou o mesmo número de pontos contra homens que contra mulheres. Mostre que o número total de participantes é um quadrado perfeito.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26707,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCatarina tem 210 cartões numerados de 1 a 210.\n\na) Quantos desses cartões têm um número que é múltiplo de 3?\n\nb) Quantos desses cartões têm um número par que não é múltiplo de 3?\n\nc) Qual é o menor número de cartões que Catarina deve pegar, ao acaso, para ter certeza de que pelo menos dois deles tenham o número 2 ou o número 3 como divisor comum?",
    "options": [],
    "answer": "a) 70; b) 70; c) 73",
    "solution": "Solution:\n\na) Como $210 \\div 3 = 70$, existem 70 cartões cujos números são múltiplos de 3. Mais precisamente, esses cartões são os de número $3 = 1 \\times 3$, $6 = 2 \\times 3$, $9 = 3 \\times 3$, $12 = 4 \\times 3, \\ldots, 204 = 68 \\times 3, 207 = 69 \\times 3$ e $210 = 70 \\times 3$.\n\nb)\n$1^{a}$ solução: Um raciocínio idêntico ao do item a) mostra que existem $210 \\div 2 = 105$ cartões com números pares entre 1 e 210 (inclusive). Por outro lado, os números pares entre 1 e 210 que são múltiplos de 3 são $2 \\times 3, 4 \\times 3, 6 \\times 3, \\ldots, 68 \\times 3$ e $70 \\times 3$, em número de 35. Logo existem $105 - 35 = 70$ cartões com números pares que não são múltiplos de 3.\n\n$2^{\\mathrm{a}}$ solução: Há 105 cartões pares. Por outro lado, entre os 70 múltiplos de 3 há $70 \\div 2 = 35$ pares; logo $105 - 35 = 70$ são pares mas não múltiplos de 3.\n\n$3^{\\text{a}}$ solução: Retiram-se dos cartões os 70 cujos números são múltiplos de 3, restando assim $210 - 70 = 140$ cartões. Desses, metade são pares, pois entre dois múltiplos de 3 consecutivos um dos números é par e o outro ímpar; logo entre eles há $140 - 70 = 70$ cartões que não são nem pares nem múltiplos de 3.\n\n$4^{a}$ solução: Observamos que, entre 1 e 6, existem dois números pares que não são múltiplos de 3, a saber, 2 e 4. Do mesmo modo, entre 7 e 13 existem dois números pares que não são múltiplos de 3, a saber, 8 e 10. Esse padrão se repete a cada bloco de seis números consecutivos até chegar ao bloco de 205 a 210. Nesse último bloco, os números que não são múltiplos de 3 são 206 e 208. Temos assim $210 \\div 6 = 35$ blocos e, em cada um, dois números pares que não são múltiplos de 3, num total de $35 \\times 2 = 70$ números.\n\nc) As partes A, B e C da figura abaixo correspondem às conclusões dos itens anteriores. Restam, então, cartões com números que não são nem pares nem múltiplos de 3, correspondendo à parte $\\mathbf{D}$. Escolhendo um número na parte $\\mathbf{A}$, outro na parte $\\mathbf{B}$ (ou $\\mathbf{C}$) e 70 na parte $\\mathbf{D}$, vemos que é possível escolher 72 cartões de modo que quaisquer dois deles não contenham números que sejam simultaneamente pares ou múltiplos de 3. Por outro lado, ao escolher 73 cartões, os números de pelo menos três deles devem ficar fora da parte $\\mathbf{D}$, ou seja, devem pertencer às partes $\\mathbf{A}, \\mathbf{B}$ e $\\mathbf{C}$. Se dois desses números ficam na mesma parte, eles têm 2 ou 3 (ou mesmo ambos, no caso de ficarem na parte A) como divisor comum. Caso contrário, temos um na parte A e outro na parte B (ou C) que têm 3 (ou 2) como divisor comum. Logo Catarina deve pegar 73 cartões.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26708,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nO contrário de um número de dois algarismos, ambos diferentes de zero, é o número obtido trocando-se a ordem de seus algarismos. Por exemplo, o contrário de $25$ é $52$ e o contrário de $79$ é $97$. Qual dos números abaixo não é a soma de um número de dois algarismos com o seu contrário?\nA) $44$\nB) $99$\nC) $121$\nD) $165$\nE) $181$",
    "options": [],
    "answer": "E",
    "solution": "Solution:\nSeja $n$ um número de dois algarismos, sendo $a$ seu algarismo das dezenas e $b$ o das unidades; então $n = 10a + b$. Se $a$ e $b$ são ambos diferentes de zero, o contrário de $n$ é $10b + a$. Desse modo, a soma de $n$ e de seu contrário é:\n$$\n(10a + b) + (10b + a) = 11a + 11b = 11(a + b)\n$$\ne, portanto, a soma de um número com seu contrário é sempre um múltiplo de $11$. Basta agora notar que todas as opções são múltiplos de $11$, com a exceção de $181$.\n\nPode-se também verificar que as outras opções são todas somas de um número com seu contrário; de fato, $44 = 13 + 31$, $99 = 18 + 81$, $121 = 29 + 92$ e $165 = 69 + 96$.\n\nPara explicar como foram encontradas essas expressões, tomemos, como exemplo, $165 = 11 \\times 15$. O raciocínio inicial mostra que se escolhermos algarismos não nulos $a$ e $b$ de modo que sua soma seja $15$, então $165$ será a soma do número $10a + b$ e de seu contrário. Por exemplo, podemos tomar $a = 6$ e $b = 9$; para essa escolha obtemos a expressão $165 = 69 + 96$. Outras escolhas são possíveis; por exemplo, $a = 8$ e $b = 7$ leva a $165 = 87 + 78$. O mesmo raciocínio serve para as outras alternativas.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26709,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAlberto, Bernardo e Carlos disputaram uma corrida, na qual cada um deles correu com velocidade constante durante todo o percurso. Quando Alberto cruzou a linha de chegada, Bernardo e Carlos estavam 36 e 46 metros atrás dele, respectivamente. Quando Bernardo cruzou a linha de chegada, Carlos estava 16 metros atrás dele. Qual é o comprimento da pista?\nA) $96~\\mathrm{m}$\nB) $100~\\mathrm{m}$\nC) $120~\\mathrm{m}$\nD) $136~\\mathrm{m}$\nE) $144~\\mathrm{m}$",
    "options": [],
    "answer": "A",
    "solution": "Solution:\n\nSeja $x$ o comprimento em metros da pista. A distância entre Bernardo e Carlos era de 10 metros quando Alberto cruzou a linha de chegada, e era de 16 metros quando Bernardo cruzou a linha de chegada. Vemos assim que, durante o intervalo de tempo no qual Alberto e Bernardo completaram a corrida, Bernardo correu 36 metros enquanto Carlos correu 30; logo\n$$\n\\frac{\\text{velocidade de Carlos}}{\\text{velocidade de Bernardo}} = \\frac{30}{36} = \\frac{5}{6}\n$$\nComo Bernardo cruzou a linha de chegada 16 metros à frente de Carlos, temos a equação $\\frac{5}{6} = \\frac{x-16}{x}$, cuja solução é $x = 96$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26710,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUma calculadora esquisita tem apenas as teclas numéricas de 0 a 9 e duas teclas especiais $A$ e $B$. Quando a tecla $A$ é apertada, o número que aparece no visor é elevado ao quadrado; quando a tecla $B$ é apertada, soma-se 3 ao número que aparece no visor. Nessa calculadora é possível obter 22 a partir do 1 apertando as teclas $A$ e $B$ na ordem $B A B B$, como ilustrado abaixo:\n$$\n1 \\xrightarrow{B} 4 \\xrightarrow{A} 16 \\xrightarrow{B} 19 \\xrightarrow{B} 22\n$$\na) Com o 3 inicialmente no visor, qual o número que vai aparecer depois de apertar as teclas $A$ e $B$ na ordem $B B A B$ ?\nb) Mostre como obter 55 a partir do 1 usando as teclas $A$ e $B$.\nc) Explique porque não é possível obter 54 a partir do 2 usando as teclas $A$ e $B$.",
    "options": [],
    "answer": "a) 84; b) One example: B B A B B; c) Impossible to reach 54 from 2",
    "solution": "Solution:\n\na) A seguir vemos o que acontece quando começamos com o número 3 no visor e apertamos as teclas na ordem $B B A B$ :\n$$\n3 \\xrightarrow{B} 3+3=6 \\xrightarrow{B} 6+3=9 \\xrightarrow{A} 9^{2}=81 \\xrightarrow{B} 81+3=84\n$$\nLogo, o número que vai aparecer no visor é 84.\n\nb) Uma maneira é apertar as teclas na ordem $B B A B B$, como vemos a seguir:\n\n![](attached_image_1.png)\n\nOutra maneira é apertar a tecla $B$ dezoito vezes seguidas; ainda outra é $B A$ seguida de treze $B$'s.\n\nc)\n$1^{a}$ solução: Se o número que aparece no visor após apertar as teclas $A$ e $B$ algumas vezes não é um quadrado perfeito, a última tecla apertada foi necessariamente a tecla $B$. Desse modo, se o 54 aparece no visor, podemos reconstruir parcialmente a sequência das teclas apertadas até chegar a 54:\n\n![](attached_image_2.png)\n\nChegamos a 36, que é um quadrado perfeito. Aqui temos as possibilidades\n\n![](attached_image_3.png)\n\ne\n\n![](attached_image_4.png)\n\nComo 9 é um quadrado perfeito, essa última sequência nos dá também duas possibilidades, a saber,\n\n![](attached_image_5.png)\n\ne\n\n![](attached_image_6.png)\n\nVemos assim que é possível chegar a 54 a partir de 0 e 3, mas não a partir de 2.\n\n$2^{\\mathrm{a}}$ solução: Se um número inteiro $x$ não é múltiplo de 3 então:\n- $x+3$ não é múltiplo de 3.\nDe fato, se $x+3$ fosse múltiplo de 3, poderíamos escrever $x+3=3y$ para algum inteiro $y$ e então $x=3y-3=3(y-1)$ seria múltiplo de 3, absurdo.\n- $x^{2}$ não é múltiplo de 3.\nDe fato, os fatores primos de $x$ e $x^{2}$ são os mesmos; assim, se 3 não é fator primo de $x$ então também não será fator primo de $x^{2}$.\nAssim, começando com um número que não é múltiplo de 3 no visor, não é possível chegar a um múltiplo de 3 apertando as teclas $A$ e $B$. Como 2 não é múltiplo de 3 e $54=3 \\times 18$ é múltiplo de 3, concluímos que não se pode chegar a 54 a partir do 2.\n\n$3^{\\mathrm{a}}$ solução: Vamos tentar chegar a 54 a partir do 2. Como 54 não é múltiplo de 3, vemos que não é possível usar apenas a tecla $B$, ou seja, a tecla $A$ deve ser usada pelo menos uma vez. Por outro lado, a tecla $A$ só pode ser usada em números menores ou iguais a 7. Os números obtidos a partir do 2 que são menores ou iguais a 7 são $2,4=2^{2}, 5=2+3$ e $7=2^{2}+3$; seus quadrados são $4,16,25$ e 49. A partir de 16, 25 e 49 não podemos usar a tecla $A$ outra vez, e como nenhum desses números difere de 54 por um múltiplo de 3, vemos que a partir deles não é possível chegar a 54; o mesmo argumento se aplica ao 4 e a seu quadrado 16. Logo, não é possível obter 54 a partir do 2.\n\n$4^{\\mathrm{a}}$ solução: Notamos primeiro que começando do 2 e apertando apenas duas teclas quaisquer, o maior resultado possível é 24 (sequência $B A$), ou seja, não se chega ao 54. Vamos agora ver o que acontece quando o 2 está no visor e apertamos três teclas.\n\n| Sequência de teclas | Resultado |\n| :---: | :---: |\n| AAA | 256 |\n| AAB | 19 |\n| ABA | 49 |\n| BAA | 125 |\n| ABB | 10 |\n| BAB | 28 |\n| BBA | 64 |\n| BBB | 11 |\n\nPodemos eliminar as sequências $A A A, B A A$ e $B B A$ de nossas considerações, pois elas levam a resultados maiores que 54. Para chegar ao 54 a partir dos resultados das outras sequências, não podemos usar a tecla $A$, pois isso nos daria resultados maiores que 54. Por outro lado, a diferença entre 54 e qualquer dos números 19, 49, 10, 28 e 11 não é um múltiplo de 3, ou seja, também não podemos chegar ao 54 a partir desses números apenas com a tecla $B$. Logo, não é possível chegar ao 54 a partir do 2.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26711,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEstefânia tem cinco cartas marcadas com as letras $A, B, C, D$ e $E$, empilhadas nessa ordem de cima para baixo. Ela embaralha as cartas pegando as duas de cima e colocando-as, com a ordem trocada, embaixo da pilha. A figura mostra o que acontece nas duas primeiras vezes em que ela embaralha as cartas.\n\n![](attached_image_1.png)\n\nSe Estefânia embaralhar as cartas 74 vezes, qual carta estará no topo da pilha?\nA) $A$\nB) $B$\nC) $C$\nD) $D$\nE) $E$",
    "options": [],
    "answer": "E",
    "solution": "Solution:\n\nO leitor pode verificar que, se Estefânia embaralhar as cartas 6 vezes, elas voltarão à posição inicial. Como $74=12 \\times 6+2$, embaralhar as cartas 74 vezes tem o mesmo efeito que fazê-lo duas vezes, o que deixa a carta $E$ no topo da pilha.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26712,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEstefânia tem cinco cartas marcadas com as letras $A, B, C, D$ e $E$, empilhadas nessa ordem de cima para baixo. Ela embaralha as cartas pegando as duas de cima e colocando-as, com a ordem trocada, embaixo da pilha. A figura mostra o que acontece nas duas primeiras vezes em que ela embaralha as cartas.\n\n![](attached_image_1.png)\n\nSe Estefânia embaralhar as cartas 74 vezes, qual carta estará no topo da pilha?\nA) $A$\nB) $B$\nC) $C$\nD) $D$\nE) $E$",
    "options": [],
    "answer": "E",
    "solution": "Solution:\n\nO leitor pode verificar que, se Estefânia embaralhar as cartas 6 vezes, elas voltarão à posição inicial. Como $74=12 \\times 6+2$, embaralhar as cartas 74 vezes tem o mesmo efeito que fazê-lo duas vezes, o que deixa a carta $E$ no topo da pilha.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26713,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nQuatro times, entre os quais o Quixajuba, disputam um torneio de vôlei em que:\n- cada time joga contra cada um dos outros uma única vez;\n- qualquer partida termina com a vitória de um dos times;\n- em qualquer partida, os times têm a mesma probabilidade de ganhar;\n- ao final do torneio, os times são classificados em ordem pelo número de vitórias.\n\na) É possível que, ao final do torneio, todos os times tenham o mesmo número de vitórias? Por quê?\n\nb) Qual é a probabilidade de que o torneio termine com o Quixajuba isolado em primeiro lugar?\n\nc) Qual é a probabilidade de que o torneio termine com três times empatados em primeiro lugar?",
    "options": [],
    "answer": "a) No. b) 1/8. c) 1/8.",
    "solution": "Solution:\n\na) O número total de partidas disputadas no torneio é $3+2+1=6$. Como $6$ não é divisível por $4$, o torneio não pode acabar com os quatro times tendo o mesmo número de vitórias.\n\nb) 1ª solução: Para que o Quixajuba termine isolado em primeiro lugar, ele deve ganhar todas as suas partidas. De fato, se ele ganhar duas ou menos então os outros três times dividirão pelo menos quatro vitórias entre si, e assim algum deles deve ter pelo menos duas vitórias; nesse caso, o Quixajuba não seria o campeão isolado. Para cada um dos três jogos entre os outros times há duas possibilidades. Logo, o número de maneiras do Quixajuba terminar sozinho em primeiro lugar é $1 \\times 1 \\times 1 \\times 2 \\times 2 \\times 2 = 8$. Como há $2^6 = 64$ resultados possíveis para as seis partidas, a probabilidade de o Quixajuba ser o campeão isolado é $\\frac{8}{64} = \\frac{1}{8}$.\n\n2ª solução: Argumentamos como acima que o Quixajuba será o campeão isolado se e somente se ele vencer suas três partidas. Como a probabilidade de o Quixajuba ganhar um jogo contra qualquer dos outros times é $\\frac{1}{2}$, a probabilidade de ele ganhar suas três partidas é $\\frac{1}{2} \\times \\frac{1}{2} \\times \\frac{1}{2} = \\frac{1}{8}$.\n\nc) Suponhamos que os times sejam $A$, $B$, $C$ e $D$ e que o torneio termine com $D$ isolado em último lugar. Então $D$ perdeu todas suas partidas; de fato,\n- se $D$ tivesse ganho suas três partidas, teria terminado o torneio em primeiro lugar (como vimos no item anterior);\n- se $D$ tivesse ganho duas (ou uma) partidas, os outros times dividiriam quatro (ou cinco) vitórias entre si; neste caso, pelo menos um deles teria ganho no máximo uma partida e assim $D$ não teria ficado em último lugar isolado.\nLogo $A$, $B$ e $C$ dividem entre si as seis vitórias, ou seja, cada um deles ganhou duas vezes; uma contra $D$ e uma contra um dos outros. Para as partidas entre $A$, $B$ e $C$ temos apenas duas possibilidades: $A$ ganhou de $B$ que ganhou de $C$ que ganhou de $A$, ou $A$ ganhou de $C$ que ganhou de $B$ que ganhou de $A$. Em resumo, há apenas duas possibilidades para que $A$, $B$ e $C$ dividam a liderança, e neste caso $D$ acaba o torneio em último lugar isolado. Como qualquer um dos times pode acabar em último lugar isolado, enquanto os outros dividem a liderança, segue que o número de possibilidades para que isto aconteça é $4 \\times 2 = 8$. Por outro lado, o número total de possibilidades para os resultados das seis partidas é $2^6 = 64$. Logo a probabilidade de que três times dividam a liderança é $\\frac{8}{64} = \\frac{1}{8}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26714,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nJoão vai pintar figuras compostas por quadrados e triângulos. Cada quadrado pode ser pintado de azul, vermelho ou verde e cada triângulo de azul, vermelho ou amarelo, de modo que polígonos com um lado comum não tenham a mesma cor. Em cada um dos itens a seguir, determine de quantas maneiras João pode pintar a figura correspondente.\n![](attached_image_1.png)\na)\n![](attached_image_2.png)\nb)\n![](attached_image_3.png)\nc)",
    "options": [],
    "answer": "a = 7; b = 43; c = 667",
    "solution": "Solution:\na) Se João pintar o quadrado de azul, ele terá as escolhas vermelho e amarelo para o triângulo. Se ele pintar o quadrado de vermelho, ele terá as escolhas azul e amarelo para pintar o triângulo. Finalmente, se ele pintar o quadrado de verde, ele terá as escolhas azul, vermelho e amarelo para pintar o triângulo. Logo, ele pode pintar a figura de $2+2+3=7$ maneiras diferentes.\n\nb)\n1$^a$ solução: Se João escolher azul ou vermelho para o triângulo, cada um dos quadrados poderá ser pintado de duas cores; se ele escolher amarelo para o triângulo, cada quadrado poderá ser pintado de três cores. Logo, o número de maneiras diferentes de pintar essa figura é $2 \\times 2 \\times 2 \\times 2+1 \\times 3 \\times 3 \\times 3=43$.\n\n2$^a$ solução:\n- Se o quadrado de baixo é pintado de amarelo, o triângulo do meio pode ser pintado de verde, vermelho ou azul. No caso em que ele é pintado de verde, há com $3 \\times 3=9$ possibilidades para pintar os quadrados restantes. Já em cada um dos casos em que ele é pintado de vermelho ou azul; há $2 \\times 2=4$ modos de pintar os quadrados restantes. Assim, há, no total, $9+2 \\times 4=17$ possibilidades quando o quadrado inferior é pintado de amarelo.\n- Se o quadrado de baixo é pintado de vermelho, o triângulo do meio pode ser pintado de verde ou azul. No caso em que ele é pintado de verde, há, novamente, 9 possibilidades para pintar os quadrados restantes. No caso em que ele é pintado de azul, há 4 possibilidades para os quadrados restantes. Assim, há, no total, $9+4=13$ possibilidades quando o quadrado inferior é pintado de vermelho.\n- De maneira análoga, quando o quadrado de baixo é pintado de azul há, no total, $9+4=13$ possibilidades.\nPortanto, a figura pode ser colorida de $17+13+13=43$ modos distintos.\n\nc)\n1$^a$ solução: Se João escolher azul ou vermelho para o quadrado sombreado, os triângulos adjacentes poderão ser pintados de $2 \\times 2 \\times 2 \\times 2$ maneiras diferentes; em metade dessas maneiras o triângulo sombreado é azul ou vermelho, caso em que os quadrados adjacentes poderão ser pintados de $2 \\times 2$ maneiras e na outra metade ele é amarelo, quando os quadrados adjacentes poderão ser pintados de $3 \\times 3$ maneiras diferentes. Nesse caso, a figura poderá ser pintada de $2 \\times 2 \\times 2 \\times 2 \\times(1 \\times 2 \\times 2+1 \\times 3 \\times 3)=208$ maneiras diferentes.\n![](attached_image_4.png)\nSe ele escolher verde para o quadrado sombreado, os triângulos adjacentes poderão ser pintados de $3 \\times 3 \\times 3 \\times 3$ maneiras diferentes; em dois terços dessas maneiras o triângulo sombreado é azul ou vermelho, caso em que os quadrados adjacentes poderão ser pintados de $2 \\times 2$ maneiras e no terço restante ele é amarelo, quando os quadrados adjacentes poderão ser pintados de $3 \\times 3$ maneiras diferentes. Nesse caso, a figura poderá ser pintada de $3 \\times 3 \\times 3 \\times(2 \\times 2 \\times 2+1 \\times 3 \\times 3)=459$ maneiras diferentes. No total, a figura poderá ser pintada de $208+459=667$ maneiras diferentes.\n\n2$^a$ solução: Se João escolher azul ou vermelho para o triângulo sombreado, os quadrados adjacentes poderão ser pintados de $2 \\times 2 \\times 2$ maneiras diferentes; em metade dessas maneiras o quadrado sombreado é azul ou vermelho, caso em que os triângulos adjacentes poderão ser pintados de $2 \\times 2 \\times 2$ maneiras e na outra metade ele é amarelo, quando os quadrados adjacentes poderão ser pintados de $3 \\times 3 \\times 3$ maneiras diferentes. Nesse caso, a figura poderá ser pintada de $2 \\times 2 \\times 2 \\times(1 \\times 2 \\times 2 \\times 2+1 \\times 3 \\times 3 \\times 3)=280$ maneiras diferentes. Se ele escolher amarelo para o triângulo sombreado, os quadrados adjacentes poderão ser pintados de $3 \\times 3 \\times 3$ maneiras diferentes; em dois terços dessas maneiras o quadrado sombreado é azul ou vermelho, caso em que os triângulos adjacentes poderão ser pintados de $2 \\times 2 \\times 2$ maneiras e no terço restante ele é verde, quando os quadrados adjacentes poderão ser pintados de $3 \\times 3 \\times 3$ maneiras diferentes. Nesse caso, a figura poderá ser pintada de $3 \\times 3 \\times(2 \\times 2 \\times 2 \\times 2+1 \\times 3 \\times 3 \\times 3)=387$ maneiras diferentes. No total, a figura poderá ser pintada de $280+387=667$ maneiras diferentes.\n\n3$^{\\mathrm{a}}$ solução (usando a $2^a$ solução do item b)): Se o quadrado inferior é pintado de amarelo, os triângulos de baixo podem ser pintados de $3 \\times 3 \\times 3=27$ modos. Se ele é pintado de azul ou vermelho, os triângulos de baixo podem ser pintados de $2 \\times 2 \\times 2=8$ modos. Logo, o número total de possibilidades é $17 \\times 27+26 \\times 8=459+208=667$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26715,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAndré, Bianca, Carlos e Dalva querem sortear um livro entre si. Para isto, colocam 3 bolas brancas e 1 preta em uma caixa e combinam que, em ordem alfabética de seus nomes, cada um tirará uma bola, sem devolvê-la à caixa. Aquele que tirar a bola preta ganhará o livro.\na) Qual é a probabilidade de que André ganhe o livro?\nb) Qual é a probabilidade de que Dalva ganhe o livro?\nPara sortear outro livro entre eles, André sugeriu usar 2 bolas pretas e 6 brancas. Como antes, o primeiro que tirar uma bola preta ganhará o livro; se as primeiras quatro bolas saírem brancas, eles continuarão a retirar bolas, na mesma ordem. Nesse novo sorteio:\nc) Qual é a probabilidade de que André ganhe o livro?\nd) Qual é a probabilidade de que Dalva ganhe o livro?",
    "options": [],
    "answer": "a) 1/4, b) 1/4, c) 5/14, d) 1/7",
    "solution": "Solution:\n\na) Para André ganhar o livro ele deve retirar a bola preta. Como a caixa contém quatro bolas das quais apenas uma é preta, a probabilidade de ele retirar a bola preta é $\\frac{1}{4}$.\n\nUma outra solução aparece na $2^{\\mathrm{a}}$ solução do item $b$ ).\n\nb)\n1a solução: Para Dalva ganhar o livro, André, Bianca e Carlos devem retirar bolas brancas. Como inicialmente a caixa contém 3 bolas brancas, a probabilidade de André retirar uma bola branca é $\\frac{3}{4}$. Supondo que André tire uma bola branca, sobrarão na caixa 2 bolas brancas e 1 preta; assim, a probabilidade de Bianca tirar uma bola branca é $\\frac{2}{3}$. Do mesmo modo, se André e Bianca tirarem bolas brancas, a probabilidade de Carlos tirar uma bola branca será $\\frac{1}{2}$. Assim, a probabilidade de André, Carlos e Bianca tirarem bolas brancas é $\\frac{3}{4} \\times \\frac{2}{3} \\times \\frac{1}{2}=\\frac{1}{4}$, que é a probabilidade de Dalva ganhar o livro. Raciocínio semelhante mostra que a probabilidade de qualquer um dos amigos ganhar o livro é $\\frac{1}{4}$, ou seja, o sorteio é justo e a ordem em que eles retiram as bolas não tem importância. Para entender melhor isso, veja a seguinte solução.\n\n$2^{\\mathrm{a}}$ solução: Mantendo as regras do sorteio, vamos pintar uma bola branca de azul e outra de vermelho; temos então quatro bolas diferentes na caixa. O número de sorteios possíveis passa a ser $4 \\times 3 \\times 2 \\times 1=24$; desses, Dalva ganha o livro quando André, Bianca e Carlos ficam com as bolas branca, azul e vermelha, o que pode acontecer de $3 \\times 2 \\times 1=6$ maneiras diferentes. Logo, a probabilidade de Dalva ganhar o livro é $\\frac{6}{24}=\\frac{1}{4}$. Esse raciocínio se aplica a qualquer um dos amigos, justificando assim o comentário anterior sobre a justiça do sorteio.\n\nc)\n$1^{a}$ solução: André pode ganhar o livro de duas maneiras, a saber, quando a primeira bola retirada for preta ou então quando as quatro primeiras bolas retiradas forem brancas e a quinta preta. A probabilidade no primeiro caso é $\\frac{2}{8}=\\frac{1}{4}$ e no segundo é $\\frac{6}{8} \\times \\frac{5}{7} \\times \\frac{4}{6} \\times \\frac{3}{5} \\times \\frac{2}{4}=\\frac{3}{28}$. Assim, a probabilidade procurada é $\\frac{1}{4}+\\frac{3}{28}=\\frac{5}{14}$.\n\n$2^{a}$ solução: A probabilidade de que André ganhe o livro na primeira rodada, como visto acima, é $\\frac{1}{4}$. Para calcular a probabilidade de que ele ganhe o livro na segunda rodada vamos calcular os casos possíveis e os casos favoráveis. As primeiras cinco bolas podem ser sorteadas de $8 \\times 7 \\times 6 \\times 5 \\times 4$ maneiras. Para que André ganhe o livro na quinta bola, as quatro primeiras bolas devem ser brancas e a quinta preta, o que pode ocorrer de $6 \\times 5 \\times 4 \\times 3 \\times 2$ maneiras. Logo a probabilidade de que André ganhe o livro na quinta bola sorteada é $\\frac{6 \\times 5 \\times 4 \\times 3 \\times 2}{8 \\times 7 \\times 6 \\times 5 \\times 4}=\\frac{3}{28}$. Assim, a probabilidade procurada é $\\frac{1}{4}+\\frac{3}{28}=\\frac{5}{14}$.\n\nd)\n$1^{a}$ solução: Dalva só vai ganhar o livro no caso em que as três primeiras bolas sorteadas sejam brancas e a quarta preta; de fato, se as quatro primeiras bolas sorteadas forem brancas, sobrarão na caixa duas brancas e duas pretas e uma bola preta será retirada antes que chegue a sua vez. Assim, a probabilidade de que Dalva ganhe o livro é $\\frac{6}{8} \\times \\frac{5}{7} \\times \\frac{4}{6} \\times \\frac{2}{5}=\\frac{1}{7}=\\frac{2}{14}$.\n\n$2^{\\mathrm{a}}$ solução: Dalva só pode ganhar o livro no caso em que as três primeiras bolas sorteadas sejam brancas e a quarta preta. As quatro primeiras bolas podem ser sorteadas de $8 \\times 7 \\times 6 \\times 5$ modos. Para que Dalva ganhe o livro, as três primeiras devem ser brancas e a quarta preta, o que pode ocorrer de $6 \\times 5 \\times 4 \\times 2$ modos. Logo, a probabilidade de que Dalva ganhe o livro é $\\frac{6 \\times 5 \\times 4 \\times 2}{8 \\times 7 \\times 6 \\times 5}=\\frac{1}{7}=\\frac{2}{14}$.\n\nFica como exercício para o(a) leitor(a) mostrar que as probabilidades de Bianca e Carlos ganharem o livro são, respectivamente, $\\frac{4}{14} \\mathrm{e} \\frac{3}{14}$. O André foi bem esperto em propor esse novo sorteio!",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26716,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nUm número é enquadrado quando, ao ser somado com o número obtido invertendo a ordem de seus algarismos, o resultado é um quadrado perfeito. Por exemplo, $164$ e $461$ são enquadrados, pois $164+461 = 625=25^{2}$. Quantos são os números enquadrados entre $10$ e $100$?\nA) 5\nB) 6\nC) 8\nD) 9\nE) 10",
    "options": [],
    "answer": "C",
    "solution": "Solution:\nSeja $n$ um número enquadrado entre $10$ e $100$, $a$ seu algarismo das dezenas e $b$ seu algarismo das unidades; notamos que $1 \\leq a \\leq 9$ e $0 \\leq b \\leq 9$. Então $n=10a+b$ e o número obtido invertendo-se os algarismos de $n$ é $10b+a$. Como $n$ é enquadrado temos que $(10a+b)+(10b+a)=11a+11b=11(a+b)$ é um quadrado perfeito.\n\nNotamos primeiro que, se $b=0$, então não é possível que $11(a+b)$ seja um quadrado perfeito, já que $11a$ nunca é um quadrado perfeito para $a$ assumindo os valores de $1$ a $9$. Logo temos $b \\neq 0$. Com isso, vemos que $2 \\leq a+b \\leq 18$; dentre esses possíveis valores para $a+b$, o único que faz de $11(a+b)$ um quadrado perfeito é $11$. Logo $a+b=11$ e as possibilidades para $n$ são então $29$ e $92$, $38$ e $83$, $47$ e $74$ e $56$ e $65$, num total de $8$.\n\nObservação: podemos também chegar à conclusão de que $b \\neq 0$ verificando diretamente que $10, 20, 30, \\ldots, 90$ não são enquadrados.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26717,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO número $abcde$ tem cinco algarismos distintos e diferentes de zero, cada um deles representado por uma das letras $a, b, c, d, e$. Multiplicando-se este número por $4$ obtém-se um número de cinco algarismos $edcba$. Qual o valor de $a+b+c+d+e$?\n\nA) 22\nB) 23\nC) 24\nD) 25\nE) 27",
    "options": [],
    "answer": "E",
    "solution": "Solution:\n\nA multiplicação pode ser esquematizada como\n\n![](attached_image_1.png)\n\nA solução é baseada nas seguintes observações:\n\n- O algarismo $a$ só pode ser $1$ ou $2$, pois, se fosse $a \\geq 3$, então $4a$ seria um número de $2$ algarismos e portanto o número $edcba$ teria $6$ algarismos. Mas $a$ não pode ser $1$ pois $edcba$, sendo múltiplo de $4$, é par, donde seu último algarismo é par. Logo $a=2$.\n\n![](attached_image_2.png)\n\n- O algarismo $e$ só pode ser $8$ ou $9$, pois $2 \\times 4=8$ e $edcba$ tem apenas $5$ algarismos. No entanto, $e$ não pode ser $9$ porque $9 \\times 4$ termina em $6$ e não em $2$. Logo $e=8$.\n\n![](attached_image_3.png)\n\n- O algarismo $b$ só pode ser $1$ ou $2$, pois $4 \\times b$ tem que ser um número de apenas $1$ algarismo. Como $a=2$ e os cinco algarismos de $abcde$ são distintos, só podemos ter $b=1$.\n\n![](attached_image_4.png)\n\n- O algarismo $d$ só pode ser $2$ ou $7$, pois $4 \\times d+3$ é um número terminado em $1$. Como $a=2$ e os cinco algarismos de $abcde$ são distintos só podemos ter $d=7$.\n\n![](attached_image_5.png)\n\n- O algarismo $c$ só pode ser $9$, pois $4c+3$ é um número terminado em $c$.\n\n![](attached_image_6.png)\n\nLogo, a resposta é $8+7+9+1+2=27$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26718,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUma caixa contém 105 bolas pretas, 89 bolas cinzentas e 5 bolas brancas. Fora da caixa há bolas brancas em quantidade suficiente para efetuar repetidamente o seguinte procedimento, até que sobrem duas bolas na caixa:\n- retiram-se, sem olhar, duas bolas da caixa;\n- se as bolas retiradas forem de cores diferentes, a de cor mais escura é devolvida para a caixa;\n- caso contrário, descartam-se as bolas retiradas e coloca-se na caixa uma bola branca.\nSobre as cores das duas bolas que sobram, pode-se garantir que:\nA) as duas serão brancas.\nB) as duas serão cinzentas.\nC) as duas serão pretas.\nD) exatamente uma será preta.\nE) exatamente uma será cinzenta.",
    "options": [],
    "answer": "D",
    "solution": "Solution:\n\nQuando se retiram duas bolas pretas da caixa, elas não retornam; mas quando as bolas retiradas são uma preta e outra de cor distinta, a preta retorna. Isso mostra que o número de bolas pretas na caixa diminui de dois em dois. Como o número inicial de bolas pretas é ímpar, sempre haverá um número ímpar de bolas pretas na caixa; desse modo, exatamente uma das duas bolas que sobrar na caixa é preta.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26719,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEm uma caixa há $10$ bolas idênticas, numeradas de $1$ a $10$. O número de cada bola corresponde a um dos pontos da figura, os quais dividem a circunferência em $10$ partes iguais. Nos itens a seguir, considere que as bolas são retiradas ao acaso, uma a uma e sem reposição.\n\n![](attached_image_1.png)\n\na) Se forem retiradas duas bolas, qual é a probabilidade de que o segmento determinado pelos pontos correspondentes seja um diâmetro da circunferência?\n\nb) Se forem retiradas três bolas, qual é a probabilidade de que os pontos correspondentes sejam vértices de um triângulo retângulo?\n\nUm ângulo inscrito em uma circunferência é reto se e somente se o arco correspondente é uma semicircunferência.\n\nc) Se forem retiradas quatro bolas, qual é a probabilidade de que os pontos correspondentes sejam vértices de um retângulo?",
    "options": [],
    "answer": "a) 1/9, b) 1/3, c) 1/21",
    "solution": "Solution:\n\na) 1ª solução: O princípio multiplicativo mostra que o número de maneiras de retirar duas bolas, uma a uma, é $10 \\times 9 = 90$. Dessas retiradas, há dez para as quais o segmento determinado pelos pontos retirados é um diâmetro, a saber, $(1,6), (2,7), (3,8), (4,9), (5,10), (6,1), (7,2), (8,3), (9,4)$ e $(10,5)$. Logo, a probabilidade pedida é $\\frac{10}{90} = \\frac{1}{9}$.\n\n2ª solução: Retira-se uma bola qualquer. Das nove possibilidades de retirar outra bola, apenas uma determinará, junto com a primeira, um diâmetro. Logo, a probabilidade de retirar duas bolas que determinam um diâmetro é $\\frac{1}{9}$.\n\n3ª solução: É possível retirar duas bolas de $\\binom{10}{2} = 45$ maneiras diferentes. Dessas retiradas há cinco que determinam diâmetros; logo a probabilidade procurada é $\\frac{5}{45} = \\frac{1}{9}$.\n\nb) 1ª solução: O princípio multiplicativo mostra que o número de maneiras de retirar três bolas, uma a uma, é $10 \\times 9 \\times 8 = 720$. Para que uma retirada determine um triângulo retângulo, ela deve conter duas bolas $a$ e $b$ que determinam um diâmetro e uma terceira bola $x$ distinta dessas duas. Ordenando essas três bolas das $3! = 6$ maneiras possíveis, vemos que há seis retiradas que consistem dessas bolas. Como há cinco pares de bolas que determinam um diâmetro e a bola extra pode ser escolhida de oito maneiras diferentes, o número de retiradas que determinam um triângulo retângulo inscrito é $6 \\times 5 \\times 8 = 240$. Logo, a probabilidade procurada é $\\frac{240}{720} = \\frac{1}{3}$.\n\n2ª solução: Uma vez retiradas três bolas, podemos formar com elas três grupos de duas bolas. Observamos que se um desses grupos determina um diâmetro, então isso não pode acontecer para os outros dois grupos. Como cada grupo de duas bolas tem probabilidade $\\frac{1}{9}$ de determinar um diâmetro, a probabilidade procurada é então $\\frac{1}{9} + \\frac{1}{9} + \\frac{1}{9} = \\frac{1}{3}$.\n\n3ª solução: Há $\\binom{10}{3} = 120$ maneiras de escolher três bolas, ou seja, há $120$ triângulos inscritos com vértices nos vértices do decágono. Por outro lado, cada diâmetro determina oito triângulos retângulos inscritos, num total de $5 \\times 8 = 40$; ou seja, há $40$ escolhas de três bolas que determinam triângulos retângulos inscritos. A probabilidade procurada é então $\\frac{40}{120} = \\frac{1}{3}$.\n\nc) 1ª solução: O número de retiradas de quatro bolas é $10 \\times 9 \\times 8 \\times 7$ e cada uma dessas retiradas determina um quadrilátero inscrito. Por outro lado, as bolas de uma retirada que determina um retângulo inscrito devem determinar dois diâmetros. Há dez escolhas para a primeira bola de uma tal retirada e a bola diametralmente oposta pode então aparecer em qualquer uma das três posições seguintes; as outras duas bolas podem então ser escolhidas de oito maneiras diferentes, correspondentes aos quatro diâmetros ainda não determinados. Assim, as retiradas que determinam um triângulo retângulo são em número de $10 \\times 3 \\times 8$ e a probabilidade procurada é então $\\frac{10 \\times 3 \\times 8}{10 \\times 9 \\times 8 \\times 7} = \\frac{1}{21}$.\n\n2ª solução: Para que as quatro bolas retiradas determinem um retângulo, as três primeiras devem determinar um triângulo retângulo, o que acontece com probabilidade $\\frac{1}{3}$; uma vez isso feito, há uma única escolha para a quarta bola entre as sete remanescentes. Logo, a probabilidade procurada é $\\frac{1}{3} \\times \\frac{1}{7} = \\frac{1}{21}$.\n\n3ª solução: Há $\\binom{10}{4} = 210$ maneiras de escolher quatro bolas, ou seja, há $210$ quadriláteros inscritos com vértices nos vértices do decágono. Por outro lado, um retângulo inscrito é determinado por dois diâmetros, ou seja, há $\\binom{5}{2} = 10$ retângulos inscritos, correspondentes a dez escolhas de quatro bolas. Logo, a probabilidade procurada é $\\frac{10}{210} = \\frac{1}{21}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26720,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nJuliana quer dar a cada uma das 26 letras $A, B, C, D, \\ldots, W, X, Y, Z$ do alfabeto um valor numérico diferente de zero, de tal modo que $A \\times C = B$, $B \\times D = C$, $C \\times E = D$, e assim por diante, até $X \\times Z = Y$.\n\na) Se Juliana der a $A$ e $B$ os valores 5 e 7, respectivamente, quais serão os valores de $C, D$ e $E$?\n\nb) Mostre que $G = A$, quaisquer que sejam os valores que Juliana der para $A$ e $B$.\n\nc) Se Juliana der valores para $A$ e $B$ tais que $A \\times B = 2010$, qual será o valor do produto $A \\times B \\times C \\times D \\times \\cdots \\times W \\times X \\times Y \\times Z$?",
    "options": [],
    "answer": "a) C = 7/5, D = 1/5, E = 1/7. b) G = A. c) The total product equals 2010.",
    "solution": "Solution:\n\na) Substituindo $A = 5$ e $B = 7$ em $A \\times C = B$, temos $5 \\times C = 7$ e segue que $C = \\frac{7}{5}$. Podemos agora achar $D$ substituindo os valores de $B$ e $D$ em $B \\times D = C$; obtemos $7 \\times D = \\frac{7}{5}$ e então $D = \\frac{1}{5}$. Finalmente, de $C \\times E = D$ temos $\\frac{7}{5} \\times E = \\frac{1}{5}$ e vemos que $E = \\frac{1}{7}$.\n\nb) Multiplicando as expressões $A \\times C = B$ e $B \\times D = C$ obtemos $A \\times B \\times C \\times D = B \\times C$; como $B$ e $C$ são diferentes de 0, concluímos que $A \\times D = 1$, ou seja, $D = \\frac{1}{A}$. Do mesmo modo, multiplicando as expressões $B \\times D = C$ e $C \\times E = D$ obtemos $B \\times E = 1$, ou seja, $E = \\frac{1}{B}$. Repetindo esse raciocínio, vemos que cada letra a partir do $D$ é o inverso da letra que aparece três posições atrás dela; em particular, $G = \\frac{1}{D} = \\frac{1}{\\frac{1}{A}} = A$.\n\nc) O item anterior nos mostra que $C \\times D \\times E \\times F \\times G \\times H = C \\times D \\times E \\times \\frac{1}{C} \\times \\frac{1}{D} \\times \\frac{1}{E} = 1$; o mesmo raciocínio mostra que o produto de quaisquer seis letras consecutivas é igual a 1. Temos então:\n$$\nA \\times B \\times C \\ldots \\times Y \\times Z = A \\times B \\times (C \\times \\ldots \\times H) \\times (I \\times \\ldots \\times N) \\times (O \\times \\ldots \\times T) \\times (U \\times \\ldots \\times Z) = A \\times B = 2010\n$$\npois todos os produtos entre parênteses são produtos de seis letras consecutivas, logo são todos iguais a 1.\n\nObservação: Notamos que esse problema depende do fato de que, uma vez fixados os valores de $A$ e $B$, a sequência dos valores das letras do alfabeto é $A, B, \\frac{B}{A}, \\frac{1}{A}, \\frac{1}{B}, \\frac{A}{B}, A, B, \\frac{B}{A}, \\frac{1}{A}, \\frac{1}{B}, \\frac{A}{B}, A, B, \\ldots$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26721,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNo jogo Arrasta Um usa-se um tabuleiro quadriculado e peças redondas, uma preta e as outras brancas. Coloca-se uma peça em cada casa do tabuleiro, exceto em uma que é deixada vazia. Um movimento consiste em deslocar para a casa vazia a peça de uma casa adjacente. O jogo termina quando a peça preta chega ao canto superior direito do tabuleiro. Veja um exemplo de como terminar o Arrasta Um em quatro movimentos em um tabuleiro $2 \\times 2$.\n\n![](attached_image_1.png)\n\nposição inicial\n\n![](attached_image_2.png)\n\nEsta sequência de movimentos pode ser descrita por $(\\uparrow, \\leftarrow, \\downarrow, \\rightarrow)$.\n\na) Descreva como terminar o Arrasta Um em seis movimentos no tabuleiro $3 \\times 3$ abaixo.\n\n![](attached_image_3.png)\n\nb) Descreva como terminar o Arrasta Um em dez movimentos no tabuleiro $3 \\times 3$ abaixo.\n\n![](attached_image_4.png)\n\nc) Mostre que em um tabuleiro $n \\times n$, como na figura, é possível terminar o Arrasta Um em $6 n-8$ movimentos.\n\n![](attached_image_5.png)",
    "options": [],
    "answer": "a) (↓, ←, ↑, ←, ↓, →)\nb) (↑, ←, ↓, →), then the sequence from part (a), totaling 10 moves\nc) 6n − 8",
    "solution": "Solution:\n\na) A figura abaixo mostra que a sequência de seis movimentos $(\\downarrow, \\leftarrow, \\uparrow, \\leftarrow, \\downarrow, \\rightarrow)$ termina o jogo a partir da posição inicial dada.\n\n![](attached_image_6.png)\n\nb) A figura abaixo mostra que a sequência de quatro movimentos $(\\uparrow, \\leftarrow, \\downarrow, \\rightarrow)$ transforma a posição inicial dada na posição inicial do item a), a partir da qual é possível terminar o jogo em seis movimentos.\n\n![](attached_image_7.png)\n\nAssim, podemos terminar o jogo num total de $4+6=10$ movimentos.\n\nc) A ideia é fazer com que a peça preta se mova ao longo da diagonal do tabuleiro. Isso pode ser feito uma casa de cada vez usando primeiro os movimentos do exemplo do enunciado seguidos da repetição dos movimentos do item a). Abaixo ilustramos esse procedimento em um tabuleiro $4 \\times 4$.\n\n![](attached_image_8.png)\n\nEm geral, em um tabuleiro $n \\times n$, a peça preta deverá subir $n-1$ casas na diagonal. Pelo método indicado acima, pode-se subir a primeira delas em 4 movimentos e cada uma das $n-2$ restantes em 6 movimentos cada uma. Logo, pode-se acabar o jogo em $4+6(n-2)=6 n-8$ movimentos.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26722,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nConsidere uma pilha de cartas numeradas de $1$ a $104$. Um embaralhamento dessa pilha consiste em intercalar as $52$ cartas de cima com as $52$ de baixo, de modo que a carta que estava no topo fique em segundo lugar de cima para baixo. A figura mostra dois embaralhamentos seguidos a partir da situação inicial, na qual as cartas estão dispostas em ordem crescente de cima para baixo.\n\n![](attached_image_1.png)\n\na) Complete a tabela.\n\n| número de embaralhamentos <br> a partir da situação inicial | 1 | 2 | 3 | 4 | 5 | 6 |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| posição da carta de número 5 <br> a partir do topo da pilha | $10^{\\mathrm{a}}$ |  |  |  |  |  |\n\nb) Partindo da situação inicial, qual será a posição da carta de número $n$ após um embaralhamento?\n\nc) Partindo da situação inicial, ache duas cartas que trocam de lugar uma com a outra a cada embaralhamento.\n\nd) Um grupo de três cartas que trocam de lugar entre si a cada embaralhamento é chamado trio invariante. Partindo da situação inicial, encontre todos os trios invariantes.",
    "options": [],
    "answer": "a) Positions of card 5 after successive shuffles: 1→10th, 2→20th, 3→40th, 4→80th, 5→55th, 6→5th.\nb) From the initial state, after one shuffle the card originally in position n moves to 2n if n ≤ 52, and to 2n − 105 if n > 52.\nc) The two cards that swap positions at every shuffle are those initially at positions 35 and 70.\nd) The invariant trios are (15, 30, 60) and (45, 90, 75).",
    "solution": "Solution:\n\na) Vamos calcular a posição ocupada, após um embaralhamento, pela $n$-ésima carta da pilha. Há dois casos a considerar:\n\n1. Primeiro caso: $n \\leq 52$ (ou seja, a carta está na metade superior da pilha)\n\nNeste caso, após um embaralhamento, ficarão acima dela as primeiras $n$ cartas da metade inferior e as primeiras $n-1$ cartas da parte superior. Logo, sua posição na pilha passará a ser $n+(n-1)+1=2 n$.\n\n2. Segundo caso: $n>52$ (ou seja, a carta está na metade inferior da pilha)\n\nNeste caso, após um embaralhamento, ficarão acima dela as cartas precedentes da metade inferior, que são em número de $n-52-1=n-53$ e igual quantidade de cartas da metade superior. Logo, sua nova posição na pilha é $(n-53)+(n-53)+1=2 n-105$.\n\nEm particular, podemos agora completar a tabela, observando que $55=2 \\times 80-105$ e $5=2 \\times 55-105$.\n\n| número de embaralhamentos <br> a partir da situação inicial | 1 | 2 | 3 | 4 | 5 | 6 |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| posição da carta de número 5 <br> a partir do topo da pilha | $10^{\\mathrm{a}}$ | $20^{\\mathrm{a}}$ | $40^{\\mathrm{a}}$ | $80^{\\mathrm{a}}$ | $55^{\\mathrm{a}}$ | $5^{\\mathrm{a}}$ |\n\nb) Como visto acima, a carta que ocupa a posição $n$ passa a ocupar, após um embaralhamento, a posição $2 n$, se $n \\leq 52$ ou $2 n-105$, se $n>52$.\n\nc) Inicialmente, observamos que após um embaralhamento\n- as cartas da metade superior da pilha se movem para baixo, pois $2 n>n$ para todo $n$ positivo;\n- as cartas da metade inferior da pilha se movem para cima, pois $2 n-105<n$ para todo $n<105$. Logo, para que duas cartas troquem de posição entre si, uma delas deverá estar na metade superior da pilha e outra na metade inferior. Suponhamos que existam duas cartas com essa propriedade, e seja $n$ a posição da carta de metade superior. Após um embaralhamento ela se move para a posição $2 n$, e então a carta na posição $2 n$ deve passar para a posição $n$. Como a carta na posição $2 n$ está na metade inferior da pilha, devemos ter $2(2 n)-105=n$, donde $n=35$. E, de fato, as cartas nas posições $35$ e $70$ trocam de posição entre si a cada embaralhamento, pois $2 \\times 35=70$ e $2 \\times(2 \\times 35)-105=35$. Além disso, concluímos que não há outro par de posições com esta propriedade.\n\nd) Para simplificar a exposição, vamos escrever $x \\rightarrow y$ para indicar que a carta que está na posição $x$ vai para a posição $y$ após um embaralhamento. Suponhamos que exista um trio fixo, e seja $n$ a posição da primeira carta desse trio a contar do topo da pilha. O argumento do item c) mostra que as cartas não podem estar todas na metade superior ou todas na metade inferior da pilha; logo a posição $n$ está na metade superior da pilha. Após um embaralhamento temos $n \\rightarrow 2 n$; se $2 n$ está na parte superior da pilha então o trio fixo deve ser $n \\rightarrow 2 n \\rightarrow 4 n \\rightarrow n$; se $2 n$ está na metade inferior da pilha então o trio fixo deve ser $n \\rightarrow 2 n \\rightarrow 4 n-105 \\rightarrow n$. No primeiro caso, temos\n$$\nn=2(4 n)-105=8 n-105\n$$\ndonde $n=15$; no segundo temos\n$$\nn=2(4 n-105)-105=8 n-315\n$$\ndonde $n=45$. Agora basta verificar que $(15,30,60)$ e $(45,90,75)$ são efetivamente trios fixos.\n\nA título de curiosidade e/ou como exercício para o(a) leitor(a), listamos na tabela a seguir todas as $k$-uplas fixas, incluindo os casos $k=2$ e $k=3$ trabalhados nos itens c) e d) acima.\n\n| $\\mathbf{k}$ | k-uplas fixas |\n| :---: | :--- |\n| 2 | $(35,70)$ |\n| 3 | $(15,30,60), \\quad(45,90,75)$ |\n| 4 | $(7,14,28,56), \\quad(21,42,84,63), \\quad(49,98,91,77)$ |\n| 6 | $(5,10,20,40,80,55), \\quad(25,50,100,95,85,65)$ |\n| 12 | $(1,2,4,8,16,32,64,23,46,92,79,53),(3,6,12,24,48,96,87,69,33,66,27,54)$, <br> $(9,18,36,72,39,78,51,102,99,93,81,57),(11,22,44,88,71,37,74,43,86,67,29,58)$, <br> $(13,26,52,104,103,101,97,89,73,41,82,59), \\quad(17,34,68,31,62,19,38,76,47,94,83,61)$ |\n\nObservamos ainda que após $12$ embaralhamentos todas as cartas voltam à posição inicial.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26723,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOs seis triângulos da figura são retângulos e seus ângulos com vértice no ponto $A$ são iguais. Além disso, $AB=24~\\mathrm{cm}$ e $AC=54~\\mathrm{cm}$. Qual é o comprimento de $AD$?\n![](attached_image_1.png)\nA) $30~\\mathrm{cm}$\nB) $34~\\mathrm{cm}$\nC) $36~\\mathrm{cm}$\nD) $38~\\mathrm{cm}$\nE) $39~\\mathrm{cm}$",
    "options": [],
    "answer": "C",
    "solution": "Solution:\n\nVamos denotar as medidas, em centímetros, das hipotenusas dos triângulos retângulos que aparecem na figura por $a, b, x, d$ e $c$, como na figura abaixo. O nosso objetivo é achar $x=AD$.\n\nOs seis triângulos retângulos são semelhantes, pois têm em comum o ângulo de vértice $A$. Logo,\n$$\n\\frac{24}{a}=\\frac{a}{b}=\\frac{b}{x}=\\frac{x}{c}=\\frac{c}{d}=\\frac{d}{54}\n$$\nMultiplicando os três primeiros termos acima e, separadamente, os três últimos, obtemos $\\frac{24}{x}=\\frac{x}{54}$. Logo $x^{2}=24 \\times 54=2^{3} \\times 3 \\times 2 \\times 3^{3}=2^{4} \\times 3^{4}=4^{2} \\times 9^{2}=36^{2}$, donde $x=36$.\n\n![](attached_image_2.png)\n\nAlternativamente, seja $\\lambda=\\frac{24}{a}$. Multiplicando os seis termos da sequência de igualdades acima, obtemos $\\lambda^{6}=\\frac{24}{54}=\\frac{4}{9}=\\left(\\frac{2}{3}\\right)^{2}$, donde $\\lambda^{3}=\\frac{2}{3}$. Por outro lado, $\\lambda^{3}=\\frac{24}{a} \\times \\frac{a}{b} \\times \\frac{b}{x}=\\frac{24}{x}$ e obtemos $\\frac{24}{x}=\\frac{2}{3}$, donde $x=36$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26724,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA Figura I mostra um quadrado de $40~\\mathrm{cm}^2$ cortado em cinco triângulos retângulos isósceles, um quadrado e um paralelogramo, formando as sete peças do jogo Tangran. Com elas é possível formar a Figura II, que tem um buraco sombreado. Qual é a área do buraco?\n![](attached_image_1.png)\nFigura I\n![](attached_image_2.png)\nFigura II\nA) $5~\\mathrm{cm}^2$\nB) $10~\\mathrm{cm}^2$\nC) $15~\\mathrm{cm}^2$\nD) $20~\\mathrm{cm}^2$\nE) $25~\\mathrm{cm}^2$",
    "options": [],
    "answer": "C",
    "solution": "Solution:\nAbaixo vemos as figuras do enunciado da questão. A descrição das peças da Figura I implica que os pontos $M$ e $N$ são pontos médios dos lados $AB$ e $AC$. A Figura III, onde $P$ é o ponto médio de $BC$, mostra que a área do triângulo $AMN$ é igual à quarta parte da área do triângulo $ABC$, que por sua vez tem área igual à metade da área do quadrado. Logo, área $(AMN) = \\frac{1}{4} \\times \\frac{1}{2} \\times 40 = 5~\\mathrm{cm}^2$. A Figura II mostra que o buraco consiste de três triângulos iguais ao triângulo $AMN$; logo sua área é $15~\\mathrm{cm}^2$.\n![](attached_image_3.png)\nFigura I\n![](attached_image_4.png)\nFigura II\n![](attached_image_5.png)\nFigura III",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26725,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO múltiplo irado de um número natural é o menor múltiplo do número formado apenas pelos algarismos 0 e 1. Por exemplo, o múltiplo irado de 2, bem como de 5, é 10; já o múltiplo irado de 3 é 111 e o de 110 é ele mesmo.\n\na) Qual é o múltiplo irado de 20?\nb) Qual é o múltiplo irado de 9?\nc) Qual é o múltiplo irado de 45?\nd) Qual é o menor número natural cujo múltiplo irado é 1110?",
    "options": [],
    "answer": "a) 100; b) 111111111; c) 1111111110; d) 6",
    "solution": "Solution:\n\na) Os primeiros múltiplos de $20$ são $20, 40, 60, 80$ e $100$. Logo o múltiplo irado de $20$ é $100$.\n\nb) Se os algarismos de um número divisível por $9$ são apenas $0$ e $1$, nesse número devem aparecer pelo menos nove algarismos $1$. Para que esse múltiplo seja o menor possível, ele deve ter o menor número de algarismos possível; logo o múltiplo irado de $9$ é $111111111$.\n\nc) Um múltiplo de $45$ é múltiplo de $5$ e $9$; logo seu algarismo das unidades é $0$ ou $5$ e a soma de seus algarismos é divisível por $9$. Como múltiplos irados são formados apenas pelos algarismos $0$ e $1$, segue que o múltiplo irado de $45$ deve ter $0$ como algarismo das unidades; logo esse múltiplo é $1111111110$.\n\nd) O número $1110$ é o menor número que tem apenas os algarismos $0$ e $1$ e que é, ao mesmo tempo, múltiplo de $3$ (pois a soma de seus algarismos é $3$) e múltiplo de $2$ (pois seu último algarismo é $0$). Logo $1110$ é o múltiplo irado de $6$. Como os múltiplos irados de $1, 2, 3, 4$ e $5$ são, respectivamente, $1, 10, 111, 100$ e $10$, segue que o menor número cujo múltiplo irado é $1110$ é $6$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26726,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOs discos $A, B, C$ e $D$ representam polias de diâmetros $8, 4, 6$ e $2~\\mathrm{cm}$, respectivamente, unidas por correias que se movimentam sem deslizar. Quando o disco $A$ dá uma volta completa no sentido horário, o que acontece com o disco $D$?\n\n![](attached_image_1.png)\n\nA) Dá 4 voltas no sentido horário\nB) Dá 3 voltas no sentido horário\nC) Dá 6 voltas no sentido anti-horário\nD) Dá 4 voltas no sentido anti-horário\nE) Dá 3 voltas no sentido anti-horário",
    "options": [],
    "answer": "D",
    "solution": "Solution:\n\nA figura mostra que os discos $A$ e $B$ giram no mesmo sentido, os discos $B$ e $C$ em sentidos opostos e os discos $C$ e $D$ no mesmo sentido.\n\n![](attached_image_2.png)\n\nAssim, $D$ gira no sentido anti-horário. Lembramos que o perímetro $p$ de um círculo de raio $r$ é dado por $p = 2\\pi r$. Como o raio do disco $A$ é quatro vezes o de $D$, segue que o perímetro de $A$ também é quatro vezes o perímetro de $D$. Logo $D$ dá quatro voltas para cada volta de $A$.\n\nObservação: usamos, no argumento acima, o fato de que os raios dos discos $B$ e $C$ são irrelevantes para a resolução desta questão; é interessante mostrar isto rigorosamente. Denotando por $a, b, c$ e $d$ os raios de $A, B, C$ e $D$ e por $n_{a}, n_{b}, n_{c}$ e $n_{d}$ os números de voltas dados pelos discos $A, B, C$ e $D$, respectivamente, então:\n$$\nn_{a} 2\\pi a = n_{b} 2\\pi b, \\quad n_{b} 2\\pi b = n_{c} 2\\pi c, \\quad n_{c} 2\\pi c = n_{d} 2\\pi d\n$$\no que implica que:\n$$\nn_{a} 2\\pi a = n_{d} 2\\pi d,\n$$\nou seja,\n$$\n\\frac{n_{d}}{n_{a}} = \\frac{a}{d}\n$$\nAssim, se $n_{a} = 1$ então, usando que $a = 8$ e $d = 2$, obtemos que $n_{d} = \\frac{8}{2} = 4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26727,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUma folha retangular de $20~\\mathrm{cm}$ por $30~\\mathrm{cm}$ foi cortada ao longo das linhas tracejadas $AC$ e $BD$ em quatro pedaços: dois triângulos iguais e dois polígonos iguais de cinco lados cada um, como na Figura I. Os segmentos $AC$ e $BD$ têm o mesmo comprimento e se encontram no centro do retângulo formando ângulos retos.\n![](attached_image_1.png)\n\na) Qual é o comprimento do segmento $AB$?\n\nb) Qual é a área de um pedaço triangular? E de um pedaço de cinco lados?\n\nc) Com os quatro pedaços podemos montar um quadrado com um buraco retangular, como na Figura II. Qual é a área do buraco?",
    "options": [],
    "answer": "a) AB = 20 cm. b) Area of each triangular piece = 100 cm²; area of each five‑sided piece = 200 cm². c) Area of the rectangular hole = 200 cm².",
    "solution": "Solution:\n\na) Vamos representar a folha original pelo retângulo $PQRS$ na figura abaixo. Seja $M$ o ponto onde os segmentos $AC$ e $BD$ se encontram. Como o centro do retângulo é o centro de simetria da figura, concluímos que $AM = MC = \\frac{1}{2} AC$. Por outro lado, sabemos que $AC = BD$, donde $AM = BM = CM = DM$. Como os ângulos com vértice em $M$ são todos retos, os triângulos $AMB$, $BMC$, $CMD$ e $DAM$ são congruentes e, em particular, $AB = BC = CD = DA$ e os ângulos desses triângulos em $A$, $B$, $C$ e $D$ são iguais, donde $ABCD$ é um quadrado. Como $BPCQ$ é um retângulo, $BC = PQ = 20~\\mathrm{cm}$, donde $AB = 20~\\mathrm{cm}$.\n\n![](attached_image_2.png)\n\nb) A área de cada um dos triângulos $AMB$, $BMC$, $CMD$ e $DAM$ é igual a $\\frac{1}{4}$ da área do quadrado $ABCD$, que é $20~\\mathrm{cm} \\times 20~\\mathrm{cm} = 400~\\mathrm{cm}^2$; logo a área de um desses triângulos é $\\frac{400}{4} = 100~\\mathrm{cm}^2$. Como são os dois pedaços de cinco lados iguais, eles têm a mesma área. A folha original tem área igual a $20 \\times 30 = 600~\\mathrm{cm}^2$, e se subtrairmos dessa área as áreas dos dois pedaços triangulares $ABM$ e $DMC$, restará a área dos dois pedaços de cinco lados. Portanto, a área de cada pedaço de cinco lados, em centímetros quadrados, é igual a $\\frac{600 - 2 \\times 100}{2} = \\frac{600 - 200}{2} = \\frac{400}{2} = 200$.\n\nOutra solução para obtenção da área do triângulo: A base $AB$ do triângulo $ABM$ mede $20~\\mathrm{cm}$; a altura relativa a essa base é metade da altura da folha, ou seja, $\\frac{20~\\mathrm{cm}}{2} = 10~\\mathrm{cm}$. Portanto, a área de cada um dos dois triângulos é $\\frac{20~\\mathrm{cm} \\times 10~\\mathrm{cm}}{2} = 100~\\mathrm{cm}^2$.\n\nOutra solução para obtenção da área do polígono de cinco lados: Cada pedaço de cinco lados é formado por um dos quatro triângulos acima e por um retângulo de altura $20~\\mathrm{cm}$ e largura igual a $\\frac{20~\\mathrm{cm} - 10~\\mathrm{cm}}{2} = \\frac{10~\\mathrm{cm}}{2} = 5~\\mathrm{cm}$. Como a área de cada triângulo é de $100~\\mathrm{cm}^2$ e a área do retângulo é igual a $5 \\times 20~\\mathrm{cm}^2 = 100~\\mathrm{cm}^2$, concluímos que a área de cada pedaço de cinco lados é igual a $100~\\mathrm{cm}^2 + 100~\\mathrm{cm}^2 = 200~\\mathrm{cm}^2$.\n\nc) $1^{\\text{a}}$ solução: O quadrado formado pelos quatro pedaços e o buraco tem área igual a 8 vezes a área de cada pedaço triangular, conforme mostrado no desenho a seguir. Portanto, sua área é igual a $8 \\times 100~\\mathrm{cm}^2 = 800~\\mathrm{cm}^2$. Como a soma das áreas das quatro peças é igual à área da folha original, ou seja, $600~\\mathrm{cm}^2$, concluímos que a área do buraco é igual a $800~\\mathrm{cm}^2 - 600~\\mathrm{cm}^2 = 200~\\mathrm{cm}^2$.\n\n![](attached_image_3.png)\n\n$2^{a}$ solução: O buraco é um retângulo cuja altura é igual à altura da folha original, ou seja, $20~\\mathrm{cm}$. Seu comprimento é a diferença entre o comprimento da folha original e o segmento $AB$, ou seja, $30~\\mathrm{cm} - 20~\\mathrm{cm} = 10~\\mathrm{cm}$. Portanto, a área do buraco é $20~\\mathrm{cm} \\times 10~\\mathrm{cm} = 200~\\mathrm{cm}^2$.\n\n$3^{a}$ solução: Cada triângulo retângulo é isósceles com hipotenusa de medida $20~\\mathrm{cm}$. Se $a$ é a medida, em centímetros, de um dos catetos, temos $20^2 = a^2 + a^2 = 2a^2$, donde $a = \\sqrt{20^2 / 2} = \\sqrt{200} = 10\\sqrt{2}$. Assim, o quadrado grande tem lado igual a $10\\sqrt{2}~\\mathrm{cm} + 10\\sqrt{2}~\\mathrm{cm} = 20\\sqrt{2}~\\mathrm{cm}$ e sua área é $(20\\sqrt{2}~\\mathrm{cm})^2 = 800~\\mathrm{cm}^2$. Como a soma das áreas das quatro peças é igual à área da folha original, ou seja, $600~\\mathrm{cm}^2$, concluímos que a área do buraco é igual a $800~\\mathrm{cm}^2 - 600~\\mathrm{cm}^2 = 200~\\mathrm{cm}^2$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26728,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA figura mostra três circunferências de raios $1$, $2$ e $3$, tangentes duas a duas nos pontos destacados. Qual é o comprimento do segmento $AB$?\n![](attached_image_1.png)\nA) $1$\nB) $\\sqrt{2}$\nC) $\\frac{1+\\sqrt{5}}{2}$\nD) $\\frac{3}{2}$\nE) $\\sqrt{3}$",
    "options": [],
    "answer": "B",
    "solution": "Solution:\nLembramos primeiro que se duas circunferências são tangentes, então, a reta que passa por seus centros passa também pelo ponto de tangência. No nosso caso, chamando de $P$, $Q$ e $R$ os centros das circunferências (como na figura), isso mostra que $PR = 3$, $PQ = 4$ e $QR = 5$. Como $3^{2} + 4^{2} = 5^{2}$, segue que o triângulo $PQR$ é retângulo em $P$. Além disso, como $PA = PB = 1$, vemos que $AB$ é a diagonal de um quadrado de lado $1$, ou seja, $AB = \\sqrt{2}$.\n![](attached_image_2.png)",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26729,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUma tira de papel retangular, branca de um lado e cinza do outro, foi dobrada como na figura. Qual é a medida do ângulo $\\alpha$ ?\n![](attached_image_1.png)\nA) $110^\\circ$\nB) $115^\\circ$\nC) $120^\\circ$\nD) $125^\\circ$\nE) $130^\\circ$",
    "options": [],
    "answer": "C",
    "solution": "Solution:\n\nConsideremos o triângulo $ABC$ na figura ao lado. Ele é retângulo com $AB = 1\\ \\mathrm{cm}$ e $BC = 2\\ \\mathrm{cm}$, ou seja, um cateto é metade da hipotenusa. Segue que $D\\hat{C}B = A\\hat{C}B = 30^\\circ$ e, analogamente, $C\\hat{B}D = 30^\\circ$. Como a soma dos ângulos internos de um triângulo é $180^\\circ$, segue que $B\\hat{DC} = 180^\\circ - 30^\\circ - 30^\\circ = 120^\\circ$. Como $B\\hat{D}C$ e $\\alpha$ são opostos pelo vértice, concluímos que $\\alpha = 120^\\circ$.\n\n![](attached_image_2.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26730,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA figura mostra um dodecágono regular decomposto em seis triângulos equiláteros, seis quadrados e um hexágono regular, todos com lados de mesma medida.\n![](attached_image_1.png)\n\na) Se cada triângulo tem área igual a $1~\\mathrm{cm}^2$ qual é a área do hexágono?\n\nb) A figura abaixo foi obtida retirando doze triângulos equiláteros de um dodecágono regular cujo lado mede $1~\\mathrm{cm}$. Qua é a área dessa figura?\n![](attached_image_2.png)\n\nc) A figura abaixo foi obtida retirando dois hexágonos regulares de um dodecágono regular cujo lado mede $1~\\mathrm{cm}$. Qual é a área dessa figura?\n![](attached_image_3.png)\n",
    "options": [],
    "answer": "a) 6 cm^2; b) 6 cm^2; c) 6 cm^2",
    "solution": "Solution:\n\na) A figura abaixo mostra que o hexágono pode ser decomposto em seis triângulos iguais aos triângulos que fazem parte do dodecágono. Como cada um desses triângulos tem área $1~\\mathrm{cm}^2$, segue que o hexágono tem área $6~\\mathrm{cm}^2$.\n![](attached_image_4.png)\n\nb) $1^{a}$ solução: A figura do item anterior mostra que o dodecágono pode ser decomposto em doze triângulos equiláteros iguais e seis quadrados. Desse modo, ao retirar doze triângulos do dodecágono, a estrela que sobra tem área igual à área de seis quadrados. Como o lado do dodecágono mede $1~\\mathrm{cm}$, cada quadrado tem área $1~\\mathrm{cm}^2$ e assim a área da estrela é $6~\\mathrm{cm}^2$.\n![](attached_image_5.png)\n\n$2^{a}$ solução: Podemos decompor o hexágono central da estrela em seis triângulos e \"encaixá-los\" como indicado na figura abaixo. A figura assim obtida tem a mesma área da estrela e consiste de seis quadrados de lado $1~\\mathrm{cm}$; sua área é então $6~\\mathrm{cm}^2$.\n![](attached_image_6.png)\n\nc) A figura abaixo mostra que os dois hexágonos retirados têm a mesma área que doze triângulos equiláteros; como no item $b$ ), a região cinza tem a mesma área que seis quadrados de lado $1~\\mathrm{cm}$; sua área é então $6~\\mathrm{cm}^2$.\n![](attached_image_7.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26731,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA Princesa Telassim cortou uma folha de papel retangular em 9 quadrados de lados $1, 4, 7, 8, 9, 10, 14, 15$ e $18$ centímetros.\n\na) Qual era a área da folha antes de ser cortada?\n\nb) Quais eram as medidas da folha antes de ser cortada?\n\nc) A Princesa Telassim precisa montar a folha de novo. Ajude-a mostrando, com um desenho, como fazer esta montagem.",
    "options": [],
    "answer": "Area: 1056 square centimeters. Dimensions: 32 cm by 33 cm. The tiling is unique up to rotations and reflections (one explicit arrangement exists).",
    "solution": "Solution:\n\na) A área da folha era igual à soma das áreas dos nove quadrados, que é (em centímetros quadrados):\n$$\n1^2 + 4^2 + 7^2 + 8^2 + 9^2 + 10^2 + 14^2 + 15^2 + 18^2 = 1056\n$$\n\nb) Sejam $a$ e $b$ as dimensões da folha, onde supomos $a \\leq b$. Como a área de um retângulo é o produto de suas dimensões, temos $ab = 1056$. Além disso, como as medidas dos lados dos quadrados em que a folha foi cortada são números inteiros, segue que $a$ e $b$ devem ser números inteiros. Observamos, finalmente, que $a$ e $b$ devem ser maiores ou iguais a $18$, pois um dos quadrados em que a folha foi cortada tem lado com esta medida. Como $a$ e $b$ são divisores de $1056$, a fatoração em fatores primos $1056 = 2^5 \\times 3 \\times 11$ nos mostra que $a$ e $b$ são da forma $2^x \\times 3^y \\times 11^z$, onde $x, y$ e $z$ são inteiros tais que $0 \\leq x \\leq 5$, $0 \\leq y \\leq 1$ e $0 \\leq z \\leq 1$. Lembrando que $ab = 1056$ e que $a$ e $b$ são maiores que $18$, obtemos as seguintes possibilidades:\n\n| $\\mathbf{a}$ | $\\mathbf{b}$ |\n| :---: | :---: |\n| $2 \\times 11 = 22$ | $2^4 \\times 3 = 48$ |\n| $2^3 \\times 3 = 24$ | $2^2 \\times 11 = 44$ |\n| $2^5 = 32$ | $3 \\times 11 = 33$ |\n\nTemos agora que decidir quais destas possibilidades podem ocorrer como medidas da folha. Como o maior quadrado tem lado $18$, que é menor que $22$, $24$ e $32$, vemos que nenhum quadrado pode encostar nos dois lados de comprimento $b$ da folha. Isto quer dizer que $b$ pode ser expresso de duas maneiras como uma soma, na qual as parcelas são medidas dos lados dos quadrados, sendo que:\n- não há parcelas repetidas em nenhuma das duas expressões e\n- não há parcelas comuns às duas expressões.\n\nEste argumento mostra que $2b \\leq 1 + 4 + 7 + 8 + 9 + 10 + 14 + 15 + 18$, ou seja, $2b \\leq 86$. Logo $b \\leq 43$ e a única possibilidade é $b = 33$. Segue que as dimensões da folha eram $a = 32$ e $b = 33$.\n\nExistem outras maneiras de eliminar os pares $(22, 48)$ e $(24, 44)$, usando o argumento acima e mostrando, por exemplo, que não existem duas maneiras de escrever $22$ e $24$ como soma dos lados dos quadrados de duas maneiras com parcelas distintas e sem parcelas comuns. Esta solução depende do fato de que, em qualquer decomposição de um retângulo em quadrados, os lados dos quadrados são necessariamente paralelos a um dos lados do retângulo. Um argumento intuitivo para demonstrar este fato consiste em selecionar um vértice do retângulo e observar que o quadrado ao qual este vértice pertence tem seus lados apoiados sobre os lados do retângulo. Qualquer quadrado que toca este primeiro quadrado (mesmo que em apenas um vértice) tem seus lados necessariamente paralelos aos lados do retângulo, pois, caso contrário, teríamos ângulos diferentes de $90^\\circ$ ou $180^\\circ$ na decomposição, e estes ângulos não podem ser preenchidos com quadrados.\n\nc) A única possibilidade (a menos de rotações e simetrias) é mostrada a seguir:\n\n|    |    |    |    |\n|----|----|----|----|\n| 14 | 18 |    |    |\n| 10 | 4  |    |    |\n|    | 7  | 15 |    |\n| 1  | 7  |    |    |\n| 9  |    |    |    |",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26732,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA figura mostra um polígono regular de dez lados com centro $O$. Qual é a medida do ângulo $a$ ?\n![](attached_image_1.png)\nA) $15^{\\circ}$\nB) $18^{\\circ}$\nC) $20^{\\circ}$\nD) $30^{\\circ}$\nE) $36^{\\circ}$",
    "options": [],
    "answer": "B",
    "solution": "Solution:\nO triângulo $AOB$ é isósceles pois os lados $OA$ e $OB$ são iguais. Logo, os ângulos $OAB$ e $OBA$ também são iguais, ou seja, ambos têm medida $a$.\n\nNotamos agora que o ângulo central $AOB$ mede $\\frac{4}{10} \\times 360^{\\circ} = 144^{\\circ}$.\n\nComo a soma dos ângulos internos de um triângulo vale $180^{\\circ}$, segue que $2a + 144^{\\circ} = 180^{\\circ}$.\n\nLogo $a = \\frac{180^{\\circ} - 144^{\\circ}}{2} = \\frac{36^{\\circ}}{2} = 18^{\\circ}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26733,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm polígono convexo é elegante quando ele pode ser decomposto em triângulos equiláteros, quadrados ou ambos, todos com lados de mesmo comprimento. Abaixo, mostramos alguns polígonos elegantes, indicando para cada um deles uma decomposição e o número de lados.\n![](attached_image_1.png)\n4 lados\n![](attached_image_2.png)\n5 lados\n![](attached_image_3.png)\n6 lados\n![](attached_image_4.png)\n7 lados\na) Desenhe um polígono elegante de 8 lados, indicando uma decomposição.\nb) Quais são as possíveis medidas dos ângulos internos de um polígono elegante?\nc) Mostre que um polígono elegante não pode ter mais que 12 lados.\nd) Desenhe um polígono elegante de 12 lados, indicando uma decomposição.",
    "options": [],
    "answer": "Possible interior angles are 60, 90, 120, and 150 degrees; the number of sides is at most 12.",
    "solution": "Solution:\n\na) Um exemplo de polígono elegante com oito lados aparece abaixo.\n![](attached_image_5.png)\n\nb) Como um polígono elegante é convexo e é formado colocando lado a lado quadrados e triângulos equiláteros, seus ângulos são somas de parcelas iguais a $60^\\circ$ ou $90^\\circ$ que não ultrapassem $180^\\circ$. Os valores possíveis são então $60^\\circ$, $90^\\circ$, $120^\\circ = 60^\\circ + 60^\\circ$ e $150^\\circ = 60^\\circ + 90^\\circ$.\n\nc) Sabemos que a soma dos ângulos internos de um polígono com $n$ lados é $(n-2) \\times 180^\\circ$. Por outro lado, vimos no item b) que o maior valor possível do ângulo interno de um polígono elegante é $150^\\circ$; logo, a soma dos ângulos internos de um polígono elegante de $n$ lados é no máximo $n \\times 150^\\circ$. Temos então $180(n-2) \\leq 150 n$, e segue que $30 n \\leq 360$, ou seja, $n \\leq 12$.\n\nd) A figura abaixo mostra um polígono elegante de 12 lados.\n![](attached_image_6.png)",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26734,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO polígono $A B C D E F G H I J K L$ é regular e tem doze lados.\n![](attached_image_1.png)\n\na) Qual é a medida dos ângulos internos do polígono?\n\nb) O ponto $M$ é a interseção dos segmentos $A E$ e $D K$. Quais são as medidas dos ângulos $M \\hat{D} E$ e $D \\hat{M} E$?\n\nc) Qual é a medida do ângulo $C \\hat{B} M$?\n\nd) Prove que os pontos $B, M$ e $F$ estão alinhados.",
    "options": [],
    "answer": "a) 150 degrees. b) Angle MDE = 90 degrees and angle DME = 45 degrees. c) Angle CBM = 45 degrees. d) Points B, M, and F are collinear.",
    "solution": "Solution:\n\na) $1^{a}$ solução: Como a soma dos ângulos internos de um polígono de $n$ lados é $(n-2) \\times 180^{\\circ}$, a soma dos ângulos internos do dodecágono é $(12-2) \\times 180^{\\circ}=1800^{\\circ}$. Logo, cada um de seus ângulos internos mede $\\frac{1800^{\\circ}}{12}=150^{\\circ}$.\n\n$2^{2}$ solução: Outra solução usa a circunferência de centro $O$ circunscrito ao polígono. O ângulo $A \\hat{O} B$ mede $\\frac{360^{\\circ}}{12}=30^{\\circ}$. O triângulo $O A B$ é isósceles, pois $O A$ e $O B$ são iguais, como raios da circunferência. Logo, $\\hat{O A B}=O \\hat{B} A=\\frac{180^{\\circ}-30^{\\circ}}{2}=75^{\\circ}$. Pela simetria da figura, temos também $O \\hat{A} L=75^{\\circ}$, e então $B \\hat{A} L=2 \\times 75^{\\circ}=150^{\\circ}$.\n\n![](attached_image_2.png)\n\nAntes de prosseguir, lembramos um resultado básico de geometria elementar. Dado uma circunferência de centro $O$ e um arco $\\hat{A B}$ nesta circunferência (marcado em traço mais forte na figura abaixo), temos o ângulo central $A \\hat{O} B$ associado a este arco. Seja $P$ um ponto qualquer na circunferência que não pertence a $\\hat{A B}$. Então a medida do ângulo inscrito $A \\hat{P} B$ é a metade da medida do ângulo $O \\hat{A} B$, independentemente da posição de $P$. A figura abaixo ilustra esta situação; nela temos:\n$$\n\\beta=A \\hat{P} B=A \\hat{Q} B=\\frac{1}{2} A \\hat{O} B=\\frac{1}{2} \\alpha\n$$\n![](attached_image_3.png)\n\nb) $1^{\\text {a }}$ solução: Consideremos outra vez a circunferência de centro $O$ circunscrita ao polígono. Como $E$ e $K$ são diametralmente opostos, o ângulo $E \\hat{D} K$ está inscrito na semicircunferência e segue que $E \\hat{D} K=90^{\\circ}$. Como o ângulo central correspondente a um lado do dodecágono regular é $\\frac{180^{\\circ}}{12}=30^{\\circ}$, o ângulo central $A \\hat{O} D$ mede $90^{\\circ}$, e segue que $A \\hat{E} D=45^{\\circ}$. Finalmente, o triângulo $E D M$ tem ângulos $E \\hat{D} M=90^{\\circ}$ e $M \\hat{E} D=45^{\\circ}$; como a soma dos ângulos internos de um triângulo é $180^{\\circ}$, segue que $D \\hat{M} E=180^{\\circ}-(90^{\\circ}+45^{\\circ})=45^{\\circ}$.\n\n![](attached_image_4.png)\n\n$2^{\\mathrm{a}}$ solução: $O$ triângulo $I A E$ é equilátero, pois seus vértices estão igualmente espaçados no polígono regular; em particular $A \\hat{E} I=60^{\\circ}$. Além disso, os ângulos $A \\hat{E} D$ e $F \\hat{E} I$ são iguais (pois correspondem aos arcos iguais $A \\hat{C} D$ e $F \\hat{H} I$ ), donde\n$$\n150^{\\circ}=F \\hat{E} D=F \\hat{E} I+I \\hat{E} A+A \\hat{E} D=60^{\\circ}+2 \\times A \\hat{E} D\n$$\ne obtemos $A \\hat{E} D=45^{\\circ}$. Agora basta argumentar como na primeira solução para obter $E \\hat{D} M=90^{\\circ}$ e $D \\hat{M} E=45^{\\circ}$.\n\n![](attached_image_5.png)\n\n$3^{\\mathrm{a}}$ solução: A medida do ângulo $A \\hat{E} D=E \\hat{A} B$ (por simetria) também pode ser obtida através da soma dos ângulos do polígono de cinco lados $A E D C B$; temos\n$$\nA \\hat{E} D+E \\hat{A} B+3 \\times 150^{\\circ}=(5-2) \\times 180^{\\circ}=540^{\\circ}\n$$\ne então $2 \\times A \\hat{E} D=90^{\\circ}$, donde $A \\hat{E} D=45^{\\circ}$. A partir daí a solução procede como nas anteriores.\n\nc) Como o triângulo $E D M$ tem dois ângulos de $45^{\\circ}$, ele é isósceles; logo $M D=D E$, ou seja, $M D$ tem a mesma medida que os lados do polígono. Como $E \\hat{D C}=150^{\\circ}$ e $E \\hat{D M}=90^{\\circ}$, temos $M \\hat{D C}=60^{\\circ}$; e como $M D=D C$ segue que o triângulo $M D C$ é equilátero. Em particular, temos $M C D=60^{\\circ}$ e segue que $M \\hat{C} B=90^{\\circ}$. Finalmente, como $M C=C B$, o triângulo $M C B$ é isósceles e então $M \\hat{B C}=B \\hat{M C}=\\frac{90^{\\circ}}{2}=45^{\\circ}$.\n\nd) $1^{\\text {a }}$ solução: Temos $F \\hat{B C}=45^{\\circ}=M \\hat{B} C$. Logo os segmentos $F B$ e $M B$ fazem o mesmo ângulo com o segmento $B C$, e segue que os pontos $B, M$ e $F$ estão alinhados.\n\n$2^{\\mathrm{a}}$ solução: No quadrilátero $B C D M$ temos $M \\hat{B C}=45^{\\circ}, B \\hat{C} M=150^{\\circ}$ e $M \\hat{D C}=60^{\\circ}$; como a soma dos ângulos internos de um quadrilátero é $360^{\\circ}$, segue que $B \\hat{M} D=360^{\\circ}-(45^{\\circ}+150^{\\circ}+60^{\\circ})=105^{\\circ}$. Analogamente, no quadrilátero $M D E F$ temos $F \\hat{M} D=360^{\\circ}-(90^{\\circ}+150^{\\circ}+45^{\\circ})=75^{\\circ}$. Logo $F \\hat{M} B=F \\hat{M} D+D \\hat{M} B=75^{\\circ}+105^{\\circ}=180^{\\circ}$, e segue que os pontos $B, M$ e $F$ estão alinhados.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26735,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPatrícia escreveu, em ordem crescente, os inteiros positivos formados apenas por algarismos ímpares: $1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 31, 33, \\ldots$ Qual foi o $157^{\\circ}$ número que ela escreveu?\n\nA) 997\nB) 999\nC) 1111\nD) 1113\nE) 1115",
    "options": [],
    "answer": "D",
    "solution": "Solution:\n\nHá cinco algarismos ímpares: $1, 3, 5, 7$ e $9$. Contando apenas números inteiros positivos, existem então $5$ números formados por apenas um algarismo ímpar, $5 \\times 5 = 25$ números formados por dois algarismos ímpares e $5 \\times 5 \\times 5 = 125$ números formados por três algarismos ímpares. Assim, existem $5 + 25 + 125 = 155$ números inteiros positivos menores que $1000$ formados por algarismos ímpares. O $156^{\\circ}$ é então $1111$ e o $157^{\\circ}$ é $1113$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26736,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm número inteiro positivo esconde outro número quando, apagando alguns de seus algarismos, aparece o outro. Por exemplo, o número $123$ esconde os números $1$, $2$, $3$, $12$, $13$ e $23$, mas não esconde $32$, $123$ e $213$.\n\na) Qual é o maior número de três algarismos escondido por $47239$?\n\nb) Qual é o menor número que esconde simultaneamente $2009$ e $9002$?\n\nc) Ache um múltiplo de $2009$ que esconde $2009$ e cujo algarismo das unidades é $3$.",
    "options": [],
    "answer": "a) 739; b) 290029; c) 200914063",
    "solution": "Solution:\n\na) Para obter o maior número possível de três algarismos escondido por $47239$, devemos primeiro fazer com que esse número tenha o maior algarismo possível na casa das centenas. Para isso, devemos apagar o $4$ e deixar o $7$ na casa das centenas. Após isso buscamos o maior algarismo possível na casa das dezenas; para isso apagamos o $2$ e obtemos $739$, que é o número procurado.\n\nb) Como o número procurado esconde $2009$, entre seus algarismos aparecem $2$, $0$, $0$ e $9$, nesta ordem. Analogamente, como ele esconde $9002$ então entre seus algarismos aparecem $9$, $0$, $0$ e $2$ nesta ordem. Logo este número possui no mínimo seis algarismos: um $2$ e um $9$ à esquerda de dois $0$'s e um $2$ e um $9$ à direita dos mesmos. Há exatamente quatro números de seis algarismos deste tipo, a saber, $290029$, $290092$, $920029$ e $920092$. O menor deles é $290029$, que é o número procurado. Notamos que não é necessário pesquisar números de sete ou mais algarismos, pois eles são todos maiores que $290029$.\n\nc) Uma primeira ideia é encontrar um múltiplo de $2009$ que termina em $3$, o que é imediato: $7 \\times 2009 = 14063$. Esta não é a resposta procurada, pois $14063$ não esconde $2009$. Mas $200900000$ é múltiplo de $2009$, e então\n$$\n200914063 = 200900000 + 14063 = 100000 \\times 2009 + 7 \\times 2009 = 100007 \\times 2009\n$$\né um múltiplo de $2009$ que esconde $2009$ e termina em $3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26737,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEscreva os algarismos de $0$ até $9$ em uma linha, na ordem que você escolher. Na linha debaixo junte os vizinhos, formando nove números novos, e some esses números como no exemplo:\n\n| 2 |   | 1 |   | 3 |   | 7 |   | 4 |   | 9 |   | 5 |   | 8 |   | 0 |   | 6 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n|   | 21 |   | 13 |   | 37 |   | 74 |   | 49 |   | 95 |   | 58 |   | 80 |   | 06 |   |\n| 1 |   |   |   |   |   |   |   |   |   |   |   |   |   |   |   |   |   |   |\n\nQual é a maior soma que é possível obter desse modo?\nA) 506\nB) 494\nC) 469\nD) 447\nE) 432",
    "options": [],
    "answer": "B",
    "solution": "Solution:\n\nPara qualquer disposição dos algarismos, a soma dos vizinhos \"juntados\" terá sempre nove parcelas, sem repetição de algarismos nas unidades ou nas dezenas. O único algarismo que não aparece nas unidades é o primeiro e o único que não aparece nas dezenas é o último. Para que a soma seja máxima, o algarismo $0$ não deve comparecer nas dezenas e, portanto, deve ser o último; além disso, o menor dos algarismos $1,2, \\ldots, 9$ não deve aparecer nas unidades e, portanto, o $1$ deve ser o primeiro. Concluímos que a soma é máxima para qualquer escolha onde $1$ é o primeiro algarismo e $0$ o último. Nesse caso, a soma das unidades será $0+2+3+4+5+6+7+8+9=44$ e a soma das dezenas será $10+20+30+40+50+60+70+80+90=450$; a soma máxima é então $450+44=494$.\n\nAlgebricamente, podemos escrever esse argumento como segue. Seja $a_{1}, a_{2}, \\ldots, a_{10}$ uma disposição qualquer dos algarismos de $0$ até $9$ na primeira linha. Na segunda linha da tabela do enunciado dessa questão, aparecerão os números $a_{1} a_{2}, a_{2} a_{3}, \\ldots, a_{9} a_{10}$. Usando a representação decimal, a soma desses números pode ser escrita na forma\n$$\n\\begin{aligned}\nS & =a_{1} a_{2}+a_{2} a_{3}+\\cdots+a_{9} a_{10} \\\\\n& =\\left(10 a_{1}+a_{2}\\right)+\\left(10 a_{2}+a_{3}\\right)+\\cdots+\\left(10 a_{9}+a_{10}\\right) \\\\\n& =10 \\times\\left(a_{1}+a_{2}+\\cdots+a_{9}\\right)+\\left(a_{2}+\\cdots a_{9}+a_{10}\\right)-10 a_{10}-a_{1} \\\\\n& =11 \\times\\left(a_{1}+a_{2}+\\cdots+a_{9}+a_{10}\\right)-10 a_{10}-a_{1} \\\\\n& =45 \\times 11-10 a_{10}-a_{1} \\\\\n& =495-10 a_{10}-a_{1}\n\\end{aligned}\n$$\nLogo, o valor máximo de $S$ é atingido quando $a_{10}=0$ e $a_{1}=1$, e, nesse caso, vale $495-10 \\times 0-1=494$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26738,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPara qualquer número positivo $x$, dizemos que os números $x+1$ e $\\frac{x}{x+1}$ são filhos de $x$ e que os dois são irmãos. Por exemplo, $\\frac{3}{2}$ e $\\frac{1}{3}$ são irmãos, pois são filhos de $\\frac{1}{2}$; de fato, $\\frac{3}{2}=\\frac{1}{2}+1$ e $\\frac{1}{3}=\\frac{\\frac{1}{2}}{\\frac{1}{2}+1}$.\n\na) Encontre um irmão de $\\frac{5}{7}$.\n\nb) Um número pode ser filho de dois números positivos diferentes? Por quê?\n\nc) Mostre que $\\frac{1}{2008}$ é descendente de 1, isto é, ele é filho de um filho de um filho... de um filho de 1.",
    "options": [],
    "answer": "a) 7/2. b) No; a child has a unique positive parent. c) Yes; repeatedly taking the smaller child from one yields one over two thousand eight.",
    "solution": "Solution:\n\nObservação: para facilitar a escrita da solução, vamos dizer que $x$ é pai de $y$ se $y$ é filho de $x$.\n\na) Suponhamos que $\\frac{5}{7}$ seja filho de um número positivo $x$. Então $\\frac{5}{7}=x+1$ ou $\\frac{5}{7}=\\frac{x}{x+1}$. A primeira equação leva a\n$$\nx=\\frac{5}{7}-1=-\\frac{2}{7}\n$$\no que não pode acontecer pois $x>0$. A segunda equação leva a\n$$\n7 x=5(x+1)=5 x+5\n$$\ndonde $2 x=5$ e segue que $x=\\frac{5}{2}$. Logo o irmão de $\\frac{5}{7}$ é $\\frac{5}{2}+1=\\frac{7}{2}$.\n\nb) Suponhamos que $y$ seja filho de $x$ e de $z$. Temos então as seguintes quatro possibilidades $x+1=z+1$, $\\frac{x}{x+1}=\\frac{z}{z+1}$, $x+1=\\frac{z}{z+1}$ ou $\\frac{x}{x+1}=z+1$. No primeiro e no segundo caso obtemos que $x=z$. No terceiro obtemos $x(z+1)=-1$, que não tem solução pois $x$ e $z$ devem ser positivos. De maneira similar, o quarto caso, nos fornece que $z(x+1)=-1$ que também não tem solução. Logo, se $y$ tem pai, ele é único.\n\nc) Vamos chamar $\\frac{x}{x+1}$ de filho menor de $x$. Quando $x=\\frac{1}{n}$ o filho menor de $x$ é\n$$\n\\frac{x}{x+1}=\\frac{\\frac{1}{n}}{\\frac{1}{n}+1}=\\frac{1}{n+1}\n$$\nLogo o filho menor de 1 é $\\frac{1}{2}$, o filho menor de $\\frac{1}{2}$ é $\\frac{1}{3}$, o filho menor de $\\frac{1}{3}$ é $\\frac{1}{4}$ e assim por diante até obtermos $\\frac{1}{2008}$ como o filho menor de $\\frac{1}{2007}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26739,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm algarismo é afilhado de um número natural se ele é o algarismo das unidades de algum divisor desse número. Por exemplo, os divisores de $56$ são $1$, $2$, $4$, $7$, $8$, $14$, $28$ e $56$, logo os afilhados de $56$ são $1$, $2$, $4$, $6$, $7$ e $8$.\n\na) Quais são os afilhados de $57$?\n\nb) Ache um número que tenha $7$ e $9$ como afilhados, mas não $3$. Quais são os afilhados desse número?\n\nc) Explique porque $2$ e $5$ são afilhados de qualquer número que tenha $0$ entre seus afilhados.\n\nd) Explique porque $8$ é afilhado de qualquer número que tenha $0$ e $9$ entre seus afilhados.",
    "options": [],
    "answer": "a) The affiliates of 57 are 1, 3, 7, 9. b) One example is 49, whose affiliates are 1, 7, 9. c) Any number that has 0 among its affiliates also has 2 and 5 as affiliates. d) Any number that has 0 and 9 among its affiliates also has 8 as an affiliate.",
    "solution": "Solution:\n\na) Os divisores de $57$ são $1$, $3$, $19$ e $57$, donde seus afilhados são $1$, $3$, $9$ e $7$.\n\nb) O exemplo mais simples é $49$, cujos afilhados são $1$, $7$ e $9$.\n\nc) Se um número tem um divisor terminado em $0$ então este número é múltiplo de $10$. Logo ele é múltiplo de $2$ e de $5$, e portanto $2$ e $5$ são seus afilhados.\n\nd) Seja $N$ um número que tem $0$ e $9$ como afilhados. Pelo item anterior, $2$ é afilhado de $N$, logo $N$ é par. Como $9$ é afilhado de $N$, algum número ímpar terminado em $9$ é divisor de $N$. Portanto, $N$ é divisível pelo produto de $2$ por esse número, ou seja, $N$ é divisível por um número terminado em $8$. Logo, $8$ é afilhado de $N$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26740,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\na) Severina escreveu um número inteiro positivo em cada lado de um quadrado. Em seguida, escreveu em cada vértice o produto dos números escritos nos lados que se encontram nesse vértice. A soma dos números escritos em dois lados opostos é 60 e a soma dos números escritos nos outros lados é 85. Qual é a soma dos números escritos nos vértices?\n\nb) Catarina, por sua vez, escreveu em cada face de um cubo um número inteiro positivo. Em seguida, escreveu em cada vértice o produto dos números escritos nas três faces que se encontram nesse vértice. Se a soma dos números escritos nos vértices é 105, qual é a soma dos números escritos nas faces?",
    "options": [],
    "answer": "a) 5100; b) 15",
    "solution": "Solution:\na) $1^{a}$ solução: Sejam $a, b, c$ e $d$ os números escritos nos lados do quadrado no sentido horário. Os números associados aos vértices são, portanto, $ab$, $bc$, $cd$ e $da$, e sua soma é\n$$\nab + bc + cd + da = b(a + c) + d(a + c) = (a + c)(b + d) = 85 \\times 60 = 5100\n$$\n![](attached_image_1.png)\n\n$2^{a}$ solução: Sejam $a$ e $b$ dois lados adjacentes do quadrado e $60 - a$ e $85 - b$ os outros dois. Então, a soma dos produtos dos comprimentos de lados adjacentes é\n$$\nab + b(60 - a) + (60 - a)(85 - b) + (85 - b)a = 5100\n$$\n\nb) Sejam $(a, b), (m, n)$ e $(x, y)$ os pares de números escritos em faces opostas do cubo. Os números associados aos vértices são, portanto, $amx$, $anx$, $amy$, $any$, $bmx$, $bnx$, $bmy$ e $bny$, e sua soma é\n$$\n\\begin{aligned}\n105 &= amx + anx + amy + any + bmx + bnx + bmy + bny \\\\\n&= a(mx + nx + my + ny) + b(mx + nx + my + ny) \\\\\n&= (a + b)(mx + nx + my + ny) \\\\\n&= (a + b)[(m + n)x + (m + n)y] \\\\\n&= (a + b)(m + n)(x + y)\n\\end{aligned}\n$$\nComo 105 se fatora em fatores primos como $105 = 3 \\times 5 \\times 7$ e os números $a + b$, $m + n$ e $x + y$ são inteiros maiores que 1, segue que $a + b$, $m + n$ e $x + y$ devem ser iguais a 3, 5 e 7 (em alguma ordem). Logo $a + b + m + n + x + y = 3 + 5 + 7 = 15$.\n![](attached_image_2.png)\nNa figura os números $a$ e $b$ estão escritos, respectivamente, na frente e atrás do cubo, os números $m$ e $n$ embaixo e em cima e os números $x$ e $y$ à esquerda e à direita.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26741,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm número inteiro $n$ é simpático quando existem inteiros positivos $a, b$ e $c$ tais que $a < b < c$ e $n = a^{2} + b^{2} - c^{2}$. Por exemplo, os números 1 e 2 são simpáticos, pois $1 = 4^{2} + 7^{2} - 8^{2}$ e $2 = 5^{2} + 11^{2} - 12^{2}$.\n\na) Verifique que $(3x+1)^{2} + (4x+2)^{2} - (5x+2)^{2}$ é igual a $2x+1$, qualquer que seja $x$.\n\nb) Encontre números inteiros $m$ e $n$ tais que $(3x-m)^{2} + (4x-n)^{2} - (5x-5)^{2} = 2x$, qualquer que seja $x$.\n\nc) Mostre que o número 4 é simpático.\n\nd) Mostre que todos os números inteiros positivos são simpáticos.",
    "options": [],
    "answer": "m = 4, n = 3",
    "solution": "Solution:\n\na) Lembrando que $(a+b)^{2} = a^{2} + 2ab + b^{2}$, podemos simplificar a expressão $(3x+1)^{2} + (4x+2)^{2} - (5x+2)^{2}$ como segue:\n$$\n(3x+1)^{2} + (4x+2)^{2} - (5x+2)^{2} = 9x^{2} + 6x + 1 + 16x^{2} + 16x + 4 - 25x^{2} - 20x - 4\n$$\n$$\n= (9 + 16 - 25)x^{2} + (6 + 16 - 20)x + (1 + 4 - 4) = 2x + 1\n$$\n\nb) Notando que:\n$$\n(3x-m)^{2} + (4x-n)^{2} - (5x-5)^{2} = -(6m + 8n - 50)x + (m^{2} + n^{2} - 25)\n$$\ndevemos encontrar inteiros $m$ e $n$ tais que:\n$$\n-(6m + 8n - 50)x + (m^{2} + n^{2} - 25) = 2x\n$$\npara todos os valores de $x$.\nIsso só é possível se $m^{2} + n^{2} - 25 = 0$ e $-(6m + 8n - 50) = 2$, simultaneamente. Isto é equivalente a $m^{2} + n^{2} = 25$ e $3m + 4n = 24$.\nAs soluções para a equação $m^{2} + n^{2} = 25$ estão dispostas na tabela abaixo:\n\n| $\\mathbf{m}$ | 0 | $\\pm 3$ | $\\pm 4$ | $\\pm 5$ |\n| :---: | :---: | :---: | :---: | :---: |\n| $\\mathbf{n}$ | $\\pm 5$ | $\\pm 4$ | $\\pm 3$ | 0 |\n\nUma verificação direta mostra que, dentre as escolhas contidas nessa tabela, apenas os valores $m = 4$ e $n = 3$ satisfazem a equação $3m + 4n = 24$.\n\nc) Do enunciado temos $4^{2} + 7^{2} - 8^{2} = 1$. Multiplicando esta expressão por $2^{2}$, obtemos $2^{2} \\cdot 4^{2} + 2^{2} \\cdot 7^{2} - 2^{2} \\cdot 8^{2} = 4$, ou seja, $8^{2} + 14^{2} - 16^{2} = 4$, o que mostra que 4 é simpático. Outras expressões são $4 = 5^{2} + 10^{2} - 11^{2} = 6^{2} + 7^{2} - 9^{2} = 7^{2} + 22^{2} - 23^{2}$ e, mais geralmente, $4 = (3k+4)^{2} + (4k+2)^{2} - (5k+4)^{2}$ para $k > 2$.\n\nd) Vamos dividir o argumento para números ímpares e pares.\n\n- Números ímpares:\nSeja $n = 2k + 1$ um número ímpar maior que 1, ou seja, com $k > 0$. O item a) mostra que fazendo $a = 3k + 1$, $b = 4k + 2$ e $c = 5k + 2$ temos $n = a^{2} + b^{2} - c^{2}$. Notamos que $a < b < c$ segue de $k > 0$. Como já sabemos que 1 é simpático, segue que todo número ímpar positivo é simpático.\n\n- Números pares:\nSeja $n = 2k$ um número par maior que 4, ou seja, com $k > 2$. Aqui o item b) mostra que fazendo $a = 3k - 4$, $b = 4k - 3$ e $c = 5k - 5$ temos $n = a^{2} + b^{2} - c^{2}$. Notamos que $a < b < c$; de fato, $a < b$ vem do fato de $k$ ser positivo e $b < c$ decorre de $k > 2$. Como já sabemos que 2 e 4 são simpáticos, segue que todo número par positivo é simpático. Concluímos, então, que todos os inteiros positivos são simpáticos.\n\nCuriosidade: A fórmula geral que apresentamos abaixo (entre outras) mostra que todo número positivo $n$ é simpático:\n$$\nn = (n + 3)^{2} + \\left(\\frac{n^{2} + 5n + 8}{2}\\right)^{2} - \\left(\\frac{n^{2} + 5n + 8}{2} + 1\\right)^{2}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26742,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm conjunto de inteiros consecutivos é equilibrado se ele pode ser dividido em dois subconjuntos com o mesmo número de elementos, de modo que:\n1) os dois subconjuntos não tenham elementos em comum;\n2) a soma dos elementos de um dos subconjuntos seja igual à soma dos elementos do outro;\n3) a soma dos quadrados dos elementos de um dos subconjuntos seja igual à soma dos quadrados dos elementos do outro.\nPor exemplo, o conjunto $\\{7,8,9,10,11,12,13,14\\}$ é equilibrado, pois podemos dividi-lo nos subconjuntos $\\{7,10,12,13\\}$ e $\\{8,9,11,14\\}$, e\n$$\n\\begin{aligned}\n& 7+10+12+13=8+9+11+14 \\\\\n& 7^{2}+10^{2}+12^{2}+13^{2}=8^{2}+9^{2}+11^{2}+14^{2}\n\\end{aligned}\n$$\na) Verifique que o conjunto $\\{1,2,3,4,5,6,7,8\\}$ é equilibrado.\nb) Mostre que qualquer conjunto de oito inteiros consecutivos é equilibrado.\nc) Mostre que nenhum conjunto de quatro inteiros consecutivos é equilibrado.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na) Dividimos o conjunto $\\{1,2,3,4,5,6,7,8\\}$ nos subconjuntos $\\{1,4,6,7\\}$ e $\\{2,3,5,8\\}$. Como\n$$\n1+4+6+7=18=2+3+5+8\n$$\n$$\n1^{2}+4^{2}+6^{2}+7^{2}=102=2^{2}+3^{2}+5^{2}+8^{2}\n$$\nvemos que $\\{1,2,3,4,5,6,7,8\\}$ é equilibrado.\n\nb) Seja $A=\\{a+1, a+2, a+3, \\ldots, a+8\\}$ um conjunto arbitrário de 8 números inteiros consecutivos. Do item a), sabemos que $1+4+6+7=2+3+5+8$ e $1^{2}+4^{2}+6^{2}+7^{2}=2^{2}+3^{2}+5^{2}+8^{2}$. Da primeira igualdade segue que\n$$\n(a+1)+(a+4)+(a+6)+(a+7)=(a+2)+(a+3)+(a+5)+(a+8)\n$$\nou seja, podemos dividir $A$ nos subconjuntos $\\{a+1, a+4, a+6, a+7\\}$ e $\\{a+2, a+3, a+5, a+8\\}$ que têm a mesma soma. Para ver que a condição na soma dos quadrados também vale, basta calcular\n$$\n(a+1)^{2}+(a+4)^{2}+(a+6)^{2}+(a+7)^{2}=4 a^{2}+2 a(1+4+6+7)+(1^{2}+4^{2}+6^{2}+7^{2})\n$$\n$$\n(a+2)^{2}+(a+3)^{2}+(a+5)^{2}+(a+8)^{2}=4 a^{2}+2 a(2+3+5+8)+(2^{2}+3^{2}+5^{2}+8^{2})\n$$\nUsando as igualdades acima, concluímos que $A$ é equilibrado.\n\nc) Suponhamos que exista um número inteiro $a$ tal que o conjunto $\\{a, a+1, a+2, a+3\\}$ seja equilibrado. A soma dos elementos desse conjunto é $4 a+6$. Assim, para que ele satisfaça a primeira condição de um conjunto equilibrado, devemos dividi-lo em dois subconjuntos de dois elementos cada um e de modo que a soma dos elementos de cada um deles seja $\\frac{1}{2}(4 a+6)=2 a+3$. Isto só é possível quando os subconjuntos são $\\{a, a+3\\}$ e $\\{a+1, a+2\\}$. Para que a segunda condição de um conjunto equilibrado seja satisfeita, devemos ter\n$$\na^{2}+(a+3)^{2}=(a+1)^{2}+(a+2)^{2}\n$$\nou seja\n$$\n2 a^{2}+6 a+9=2 a^{2}+6 a+5\n$$\nSimplificando essa última igualdade chegamos a $4=0$, um absurdo. Logo nenhum conjunto com quatro inteiros consecutivos é equilibrado.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26743,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $N = 2012$. Find all $N^2$-uples of real numbers $a_{i,j}$, $1 \\le i, j \\le N$, such that the following limit is convergent:\n$$\n\\lim_{x \\to +\\infty} \\sum_{1 \\le i,j \\le N} \\left(a_{i,j} \\sqrt[j]{x+i}\\right)\n$$",
    "options": [],
    "answer": "Exactly those arrays satisfying sum_{i=1}^N a_{i,j} = 0 for every j = 1, 2, ..., N. In this case the limit exists and equals sum_{i=1}^N a_{i,1} · i.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26744,
    "subject": "Mathematics (Multi-modal)",
    "question": "Elaine uses each of the digits $1$ to $8$ and writes down two $4$-digit numbers.\n\na. If the sum of these numbers is the largest possible, what is their sum?\n\nb. If the sum of these numbers is the least possible, what is their minimum value?",
    "options": [],
    "answer": "a) 16173; b) 3825",
    "solution": "a. The largest sum is obtained when the largest digits are assigned to the leftmost positions, so it is equal to $(8+7) \\cdot 1000 + (6+5) \\cdot 100 + (4+3) \\cdot 10 + 2 + 1 = 16373$.\n\nb. The smallest sum is obtained when the smallest digits are assigned to the leftmost positions. So $1$ and $2$ are in the thousands, $3$ and $4$ are in the hundreds, $5$ and $6$ are in the tens and $7$ and $8$ are in the units. The biggest of the two numbers is the one beginning with $2$. The minimum of its difference is obtained by making the digits in the smallest number bigger and the digits in the biggest number smaller, so it is $2357 - 1468 = 889$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26745,
    "subject": "Mathematics (Multi-modal)",
    "question": "When two red amoebas join, the result in one blue amoeba; when a red amoeba and a blue amoeba join, they turn into three red amoeba; and when two blue amoeba join, they become four red amoeba. Fernando observes a test tube with initially $201$ blue amoebas and $112$ red amoebas.\n\na. Is it possible that after some amoebas transform the test tube contains $100$ blue amoebas and $314$ red amoebas?\n\nb. Is it possible that after some amoebas transform the test tube contains $99$ blue amoebas and $314$ red amoebas?",
    "options": [],
    "answer": "a) Yes, it is possible. b) No, it is not possible.",
    "solution": "If the number of blue amoebas is $b$ and the number of red amoebas is $r$ then $2b + r$ is invariant: indeed, whenever one blue amoeba appears/disappears, two red amoebas disappear/appear. In the problem, such number is $2 \\cdot 201 + 112 = 514$.\n\na. Since $2 \\cdot 100 + 314 = 514$, it can be possible. Indeed, if $50$ pairs of blue amoebas turn into $200$ red amoebas and a pair of one amoeba of each color turn into $3$ red amoebas, we have $201 - 2 \\cdot 50 - 1 = 100$ blue amoebas and $112 + 200 - 1 + 3 = 314$ red amoebas.\n\nb. Since $2 \\cdot 99 + 314 = 512$, it is not possible.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26746,
    "subject": "Mathematics (Multi-modal)",
    "question": "When two red amoebas join, the result in one blue amoeba; when a red amoeba and a blue amoeba join, they turn into three red amoeba; and when two blue amoeba join, they become four red amoeba. Fernando observes a test tube with initially $b$ blue amoebas and $r$ red amoebas.\nDetermine, in terms of $b$ and $r$, all possible quantities of amoebas in the test tube, specifying the quantities of amoebas of each color.",
    "options": [],
    "answer": "All reachable configurations are exactly those of the form: k blue amoebas and (2b + r − 2k) red amoebas, where k is any integer with 0 ≤ k ≤ b + floor(r/2). Equivalently, the possible total counts are 2b + r − k for such k. Here 2b + r is invariant under the allowed operations.",
    "solution": "If the number of blue amoebas is $b$ and the number of red amoebas is $r$ then $2b + r$ is invariant: indeed, whenever one blue amoeba appears/disappears, two red amoebas disappear/appear.\n\nIf there is only one amoeba, then it will never change. Otherwise, change all blue amoebas into red amoebas: this is possible because one can fuse a blue amoeba with a red amoeba to get three red amoebas repeatedly. So at this point we have $2b + r$ amoebas. Then change back red amoebas into blue amoebas. If we change $2k$ red amoebas, $0 < k \\leq b + \\lfloor r/2 \\rfloor$, we will get $k$ blue amoebas and $2b + r - 2k$ red amoebas, in a total of $2b + r - k$ amoebas.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26747,
    "subject": "Mathematics (Multi-modal)",
    "question": "The *Somanacci* sequence is generated by three initial terms, each a positive integer smaller than $2012$. Each term of the sequence, beginning from the fourth term, is the sum of all preceding numbers. How many distinct Somanacci sequences have $2012$ as one of their terms?",
    "options": [],
    "answer": "630261",
    "solution": "Let $S_n$ be the $n$-th Somanacci number. Then, for $n \\ge 5$, $S_n = S_{n-1} + S_{n-2} + \\dots + S_1 = S_{n-1} + S_{n-1} = 2S_{n-1}$, because $S_{n-1} = S_{n-2} + \\dots + S_1$. So, since $S_4 = S_3 + S_2 + S_1$, $S_n = (S_1 + S_2 + S_3) \\cdot 2^{n-4}$. Since $2012 = 2 \\cdot 1006 = 2^2 \\cdot 503$, the answer is the number of ways of writing $1006$ as sum of three positive integer numbers plus the number of ways of writing $503$ as sum of three positive integer numbers, that is $\\binom{1005}{2} + \\binom{502}{2} = 630261$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26748,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider all matrices of order $4n$ with $4n$ entries equal to $1$, $4n$ entries equal to $-1$ and all remaining entries equal to $0$. Find, in terms of $n$, the maximum of its determinant.",
    "options": [],
    "answer": "4^n",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26749,
    "subject": "Mathematics (Multi-modal)",
    "question": "Ana drew two distinct hexagons, *ABCDEF* and *PQRSTU*, each with all internal angles measuring $120^\\circ$.\n\na. Given that $AB = CD = 5$, $BC = 8$ and $EF = 3$, find the perimeter of *ABCDEF*.\n\nb. Given that $PQ = 3$, $QR = 4$, $RS = 5$ and $TU = 1$, find $ST + PU$.",
    "options": [],
    "answer": "a) 41; b) 14",
    "solution": "A hexagon with all internal angles measuring $120^\\circ$ can be inscribed in an equilateral triangle by extending three of its sides:\n\n![](attached_image_1.png)\n\nIn both items, $x$, $y$, $z$ and $w$ are given. The equilateral triangle has sidelength $x + y + z$, so the remaining two sides of the hexagon have lengths $x + y + z - (w + z) = x + y - w$ and $x + y + z - (x + w) = y + z - w$. Thus the perimeter of the hexagon is $x + y + z + w + (x + y - w) + (y + z - w) = 2x + 3y + 2z - w$.\n\na.\nWe have $x = z = 5$, $y = 8$ and $w = 3$, so the answer is $2 \\cdot 5 + 3 \\cdot 8 + 2 \\cdot 5 - 3 = 41$.\n\nb.\nNow $x = 3$, $y = 4$, $z = 5$ and $w = 1$, so the answer is $x + y - w + y + z - w = x + 2y + z - 2w = 3 + 2 \\cdot 4 + 5 - 2 \\cdot 1 = 14$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26750,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a triangle *ABC*, its *A*-excircle is the point where the external bisectors of angles $\\angle B$ and $\\angle C$ meet. Let $I_A$, $I_B$ and $I_C$ be the *A*, *B*, *C*-excenters of scalene triangle *ABC*, respectively, and *X*, *Y* and *Z* be the midpoints of $I_B I_C$, $I_C I_A$ and $I_A I_B$, respectively. The incircle of triangle *ABC* touches sides *BC*, *CA* and *AB* at *D*, *E* and *F*, respectively. Prove that lines *DX*, *EY* and *FZ* meet at a single point on line *IO* that connects the incenter *I* and the circumcenter *O* of triangle *ABC*.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We start with the following useful lemma:\n**Lemma.** Let $ABC$ be a triangle and $I_A$, $I_B$ and $I_C$ be its excenters. Then the midpoints of the sides of the triangle $I_A I_B I_C$ lie on the circumcircle of $ABC$.\n**Proof.** First notice that $AI_A$ and $I_B I_C$ are the internal and external bisectors of $\\angle BAC$, respectively, so $AI_A$ is the altitude relative to $I_B I_C$ in triangle $I_A I_B I_C$. So the circle that goes through the feet of the altitudes $A$, $B$ and $C$ is the nine point circle of $I_A I_B I_C$, and thus the midpoints of the sides of that triangle lie on it.\n\n![](attached_image_1.png)\n\nSince $AE = AF$ and $IE = IF$, $AI$ is the perpendicular bisector of $EF$, so $AI \\perp EF$. We also have $AI \\perp I_B I_C$. So $EF \\parallel I_B I_C$. Since $YZ \\parallel I_B I_C$ as well, we conclude that $EF \\parallel YZ$. Analogously, $DE \\parallel XY$ and $DF \\parallel XZ$, so triangles $DEF$ and $XYZ$ are homothetic. Let $\\phi$ be the homothety that maps $XYZ$ to $DEF$. Since $X$, $Y$, and $Z$ lie on the circumcircle of $ABC$ (by the lemma) and $D$, $E$ and $F$ lie on the incircle of $ABC$, $\\phi$ maps the circumcircle to the incircle, and thus $O$ to $I$. Hence $DX$, $EY$, $FZ$ and $OI$ pass through the center of the homothety $\\phi$ and we are done.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26751,
    "subject": "Mathematics (Multi-modal)",
    "question": "In this problem, a *box* is a parallelepiped $P \\in \\mathbb{R}^3$. We define the *size* of a box $P$ as $a^s + b^s + c^s$, $a, b, c$ being its dimensions and $s$ a fixed integer.\nFind all values of $s$ such that the following statement is true: if a box $P_1$ is inside box $P_0$ then the size of $P_1$ does not exceed size of box $P_0$.\n*The boxes are allowed to be in any position; in particular, they need not to have its edges parallel to the axes.*",
    "options": [],
    "answer": "s = 0 and s = 1",
    "solution": "",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 26752,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle, $M$ be the midpoint of side $AC$ and $N$ be the midpoint of side $AB$. Let $r$ and $s$ be the reflections of lines $BM$ and $CN$ across line $BC$, respectively. Lines $r$ and $s$ meet line $MN$ at points $D$ and $E$, respectively. The circumcircles of $BDM$ and $CEN$ meet at points $X$ and $Y$, lines $BE$ and $CD$ meet at $Z$ and lines $r$ and $s$ meet at $W$. Prove that lines $XY$, $WZ$ and $BC$ meet at a single point.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\nFirst, notice that $MN \\parallel BC$, so $\\frac{WD}{DB} = \\frac{EW}{EC}$. Applying Ceva's theorem to triangle $BCW$ and cevians $WP$, $CD$ and $BE$ we have $\\frac{WD}{DB} \\cdot \\frac{BP}{PC} \\cdot \\frac{CE}{EW} = 1 \\iff BP = PC$, so $WZ$ meets $BC$ at its midpoint $P$.\n\nNow it remains to prove that $P$ lies on line $XY$, which is the radical axis of the two circles. But, because of the reflection and the fact that $MN \\parallel BC$,\n\n$\\angle MBC = \\angle CBW = \\angle MDB$, and thus $BC$ is tangent to the circumcircle of $BDM$. The power of $P$ with respect to this circle is $PB^2 = BC^2/4$. Analogously, the power of $P$ with respect to the circumcircle of $CEN$ is $BC^2/4$, and thus $P$ lies on the radical axis of these circles, and we are done.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26753,
    "subject": "Mathematics (Multi-modal)",
    "question": "Christine has a deck of cards numbered from $1$ to $25$. She asked her friend Dorothy to choose six cards from the deck. Christine wrote down the chosen numbers and put the cards back in the deck. She then asked Dorothy to choose six cards again and, again, she wrote down the chosen numbers.\n\na. The first six numbers that Dorothy had chosen have the property that the difference between any two of the chosen numbers is a multiple of $4$ and only one of them is not a prime. Find the six numbers.\n\nb. In the second time around, Dorothy chose the numbers in a way that for each pair of the numbers except one, one of the numbers divide the other. Find the largest of the six numbers.",
    "options": [],
    "answer": "a) 3, 7, 11, 15, 19, 23; b) 24",
    "solution": "a. The prime numbers between $1$ and $25$ are $2$, $3$, $5$, $7$, $11$, $13$, $17$, $19$, $23$. The ones of the form $4k + 1$, $k \\in \\mathbb{Z}$, are $5$, $13$ and $17$; the ones of the form $4k + 3$ are $3$, $7$, $11$, $19$ and $23$. Since there are supposed to be five prime numbers, the numbers are $3$, $7$, $11$, $15$, $19$ and $23$.\n\nb. The largest number must be divisible by at least other four of the numbers, so it has at least five divisors (these four and itself). The only numbers between $1$ and $25$ with at least five divisors are $16$ and $24$. So the largest number is $16$ or $24$. If it is $16$, then $1$, $2$, $4$ and $8$ are four of the other numbers. The other lies between two consecutive powers of $2$; but it cannot divide the largest of them and it is not a multiple of the smallest, which is a contradiction.\n\nOn the other hand, there are a few examples with $24$ as last number as, say, $1$, $2$, $4$, $8$, $16$, $24$ and $1$, $2$, $4$, $6$, $12$, $24$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26754,
    "subject": "Mathematics (Multi-modal)",
    "question": "What is the smallest positive integer $n$ such that there exists a positive integer $k$ such that the last 2012 digits of $n^k$ are all equal to 1?",
    "options": [],
    "answer": "71",
    "solution": "*Answer: 71.*\n\nSince $\\underbrace{11\\dots1}_{2012 \\text{ ones}} = \\frac{10^{2012}-1}{9}$, the problem is equivalent to finding the smallest $n$ such that, for some $k$,\n$$\n\\begin{align*}\nn^k &\\equiv \\frac{10^{2012} - 1}{9} \\pmod{10^{2012}} \\\\\n9n^k &\\equiv -1 \\pmod{10^{2012}} \\\\\n&\\iff n^{-k} \\equiv -9 \\pmod{10^{2012}} \\\\\n&\\iff \\left\\{ \\begin{array}{l} n^{-k} \\equiv -9 \\pmod{2^{2012}} \\\\ n^{-k} \\equiv -9 \\pmod{5^{2012}} \\end{array} \\right.\n\\end{align*}\n$$\n\nLet's look at $n^{-k} \\bmod 8$ first. It's immediate that $n$ is odd. Since $1^2 \\equiv 3^2 \\equiv 5^2 \\equiv 7^2 \\equiv 1 \\pmod 8$, we must have $n \\equiv 7 \\pmod 8$ and $-k$ odd. Now look at it mod 16: let $k = -m$ and $n = 8u - 1$. Then\n$$\n(8u - 1)^m \\equiv -1 + 8um \\pmod{16}\n$$\n\nand\n$$\n\\begin{align*}\nn^m &\\equiv -9 \\pmod{16} \\\\\n8um - 1 &\\equiv -9 \\pmod{16} \\\\\n&\\iff um \\equiv 1 \\pmod{2},\n\\end{align*}\n$$\n\nthat is, $u$ is odd. Therefore $n = 8(2t + 1) - 1 \\equiv 7 \\pmod{16}$.\n\nNow look at $n^{-k} \\bmod 5$: since $-k$ is odd, and $1^3 \\equiv 1 \\pmod 5$, $2^3 \\equiv 3 \\pmod 5$, $3^3 \\equiv 2 \\pmod 5$ and $4^3 \\equiv 4 \\pmod 5$, we must have $n \\equiv 1 \\pmod 5$. Putting everything together, we have $n \\equiv 7 \\pmod{16}$ and $n \\equiv 1 \\pmod 5 \\implies n \\equiv 71 \\pmod{80}$, and thus $n \\ge 71$.\n\nWe claim that $n = 71$ does the job. First notice that, by the Lifting the Exponent lemma, $\\nu_2(71^N - 1) = \\nu_2(71^2 - 1) + \\nu_2(N) - 1 = \\nu_2(N) + 3$ and $\\nu_5(71^N - 1) = \\nu_5(71 - 1) + \\nu_5(N) = \\nu_5(N) + 1$.\n\nNow, notice that this implies $71^v \\equiv 71^w \\pmod{2^t}$ $\\iff 71^{v-w} \\equiv 1 \\pmod{2^t}$ $\\iff \\nu_2(v-w) + 3 \\ge t \\iff 2^{t-3} \\mid v-w \\iff v \\equiv w \\pmod{2^{t-3}}$. We analyze now the possible residues of $71^N \\pmod{2^t}$: first, we know that $71 \\equiv 7 \\pmod{16}$, and $71^2 \\equiv 1 \\pmod{16}$, so all residues $71^N \\pmod{2^t}$ are congruent to either 1 or $7 \\pmod{16}$, so one eighth of the residues appear, in a total of $2^{t-3}$ residues. But $71^v \\equiv 71^w \\pmod{2^t}$ $\\iff v \\equiv w \\pmod{2^{t-3}}$, so there are at most $2^{t-3}$ residues. This means that the residues are all numbers congruent to 1 or $7 \\pmod{16}$. So, for all $t \\ge 4$ there is $N$ such that $71^N \\equiv -9 \\pmod{2^t}$. Notice that $N$ is unique mod $2^{t-3}$.\n\nThe same argument applies mod $5^t$: all residues of $71^N \\pmod{5^t}$ are congruent to $1 \\pmod 5$ and $71^v \\equiv 71^w \\pmod{5^t} \\iff 71^{v-w} \\equiv 1 \\pmod{5^t} \\iff \\nu_5(v-w) + 1 \\ge t \\iff 5^{t-1} \\mid v-w \\iff v \\equiv w \\pmod{5^{t-1}}$. One fifth of the residues appear, all of them congruent to 1 mod 5. So the residues are all numbers congruent to 1 mod 5, and for all $t \\ge 2$ there is $N$ such that $71^N \\equiv -9 \\pmod{5^t}$. Notice that $N$ is unique mod $5^{t-1}$.\n\nWe can finish the problem now: we only need to find $N$ that is congruent to a number mod $2^t$ and to a number mod $5^t$, but it exists due to the Chinese remainder theorem. Now plug $t = 2012$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26755,
    "subject": "Mathematics (Multi-modal)",
    "question": "Each member of a population has two genes (possibly repeated) among the genes $G_1, G_2, \\dots, G_n$. Suppose that, in generation 0 of this population, $p_{ij}(0)$ ($1 \\le i, j \\le n$) is the proportion of members of this population with genotype $G_i G_j$ (or $G_j G_i$, which is biologically identical to $G_i G_j$).\nTo generate each member of generation $k+1$, two members of generation $k$ are chosen independently and at random, and one gene of each member is chosen independently and at random: these two genes will be the genotype of their breed. Let $p_{ij}(k)$ be the probability that a member of generation $k$ has genotype $G_i G_j$ (or $G_j G_i$). Suppose the population is big enough in a way that at least two members are generated.\nProve that $p_{ij}(2012) = p_{ij}(2013)$ for all $i, j$ ($1 \\le i, j \\le n$).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Discrete Mathematics",
    "subtopic": "Logic"
  },
  {
    "id": 26756,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine if there exist positive integers $n, a_1, a_2, \\dots, a_{2012}$ all greater than 1, such that\n$$\nn^2 = a_1^2 + a_2^3 + a_3^5 + \\dots + a_i^{p_i} + \\dots + a_{2012}^{p_{2012}},\n$$\nwhere $p_i$ is the *i*-th prime.",
    "options": [],
    "answer": "Yes",
    "solution": "Just pick any odd $a_2$, any even $a_3, \\dots, a_{2012}$. Then $a_2^3 + \\dots + a_{2012}^{p_{2012}} = 2N+1$ is odd. Then pick $a_1 = N$ and $n = N+1$, and $n^2 - a_1^2 = (N+1)^2 - N^2 = 2N+1$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26757,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the following diagram, *ABCDE* is a regular pentagon and *MNP* is an equilateral triangle. Find $\\angle CMD$.\n![](attached_image_1.png)",
    "options": [],
    "answer": "24°",
    "solution": "First notice that $BE \\parallel CD$, so $BEM$ is an equilateral triangle. So $BE = EM$ and $\\angle BEM = 60^\\circ$.\n\nSince $ABCDE$ is a regular pentagon, $\\angle BEC = 36^\\circ$, so $\\angle MEC = \\angle BEM + \\angle BEC = 60^\\circ + 36^\\circ = 96^\\circ$.\n\nWe also have $CE = BE = EM$, so triangle $CEM$ is isosceles and $\\angle CME = \\frac{180^\\circ - \\angle MEC}{2} = 42^\\circ$.\n\nHence $\\angle BMC = 60^\\circ - 42^\\circ = 18^\\circ$ and, by symmetry, $\\angle EMD = 18^\\circ$.\n\nFinally, $\\angle CMD = 60^\\circ - 2 \\cdot 18^\\circ = 24^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26758,
    "subject": "Mathematics (Multi-modal)",
    "question": "The real numbers $a$ and $b$ satisfy $(a+b)(a+1)(b+1) = 2$ and $a^3 + b^3 = 1$. Find $a+b$.",
    "options": [],
    "answer": "1",
    "solution": "Let $S = a+b$ and $P = ab$. Then $(a+1)(b+1) = ab + a + b + 1 = P + S + 1$ and $a^3 + b^3 = (a+b)(a^2 - ab + b^2) = S(S^2 - 3P)$. So\n$$\n\\left| \n\\begin{array}{l} \nS(P + S + 1) = 2 \\\\ \nS(S^2 - 3P) = 1 \n\\end{array} \n\\right. \n\\iff \n\\begin{array}{l} \n3SP + 3S^2 + 3S + S^3 - 3PS = 3 \\cdot 2 + 1 \\\\ \nS^3 + 3S^2 + 3S = 7 \\\\ \n(S + 1)^3 = 8 \\iff S = 1. \n\\end{array}",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26759,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f: (0, +\\infty) \\to (0, +\\infty)$ be a twice differentiable function such that $f'(x) < 0$ for all $x > 0$. For each $x > 0$ consider the triangle determined by the line tangent to the graph of $f$ at $(x, f(x))$ and the two coordinate axes. It is known that the area of this triangle is constant and equal to $C$.\nFind, in terms of $C$, all possible values of $f(1)$.",
    "options": [],
    "answer": "C/2",
    "solution": "",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 26760,
    "subject": "Mathematics (Multi-modal)",
    "question": "In how many ways one can color the cells of a $n \\times n$ table, each with one of four colors, such that no cells that share a side have the same color and all four colors appear in every $2 \\times 2$ square formed by neighboring cells?",
    "options": [],
    "answer": "3 * 2^(n+2) - 24",
    "solution": "Answer: $3 \\cdot 2^{n+2} - 24$.\nSuppose there are at least three different colors $A, B, C$ in the first row. Then they occur consecutively, say in the order $ABC$. Then the cell below the $B$ has the fourth color $D$, and all other cells are determined by the first row, because three cells in a $2 \\times 2$ square determine the other one, and we can fill the board from right to left and left to right, beginning at the $D$. Notice that the cells below $ABC$ are $CDA$, so the next row always has three different colors and can be filled. Finally, notice that the colors alternate in the columns in between $A, C$ and $B, D$, so all columns have two different colors. There are $\\binom{4}{2} = 6$ ways to choose the two colors for the top two cells in the first column and $2^n$ ways to choose the two colors for the top two cells in each next column: if we chose $\\{A, B\\}$ for the first column, then we can choose $(A, B)$ or $(B, A)$ for the first column, $(C, D)$ or $(D, C)$ for the next column, $(A, B)$ or $(B, A)$ for the next column, and so on. We only need to exclude the cases where there are only two colors in the first row: in this case, we still have 6 choices for the first column, and 2 choices for each of the first two columns orders. The other orders are determined to be the same as the preceding ones, so to repeat the pattern. So we must exclude $6 \\cdot 2 \\cdot 2 = 24$ cases.\nNow we deal with the case in which there are only two colors in the first row. There are $4 \\cdot 3$ ways to choose the colors in the first row. Then each following row has the other two colors, alternated in one of two ways. So in this case we have $12 \\cdot 2^{n-1}$ colorings.\nSo the grand total is $6 \\cdot 2^n - 24 + 12 \\cdot 2^{n-1} = 3 \\cdot 2^{n+2} - 24$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 26761,
    "subject": "Mathematics (Multi-modal)",
    "question": "When two red amoebas join, the result in one blue amoeba; when a red amoeba and a blue amoeba join, they turn into three red amoeba; and when two blue amoeba join, they become four red amoeba. Fernando observes a test tube with initially $19$ blue amoebas and $95$ red amoebas.\n\na. He observes that all the amoebas join in pairs, giving birth to a new generation of amoebas. Find the maximum number of amoebas in this generation.\n\nb. There is, at some point, $100$ amoebas. How many of them are blue?",
    "options": [],
    "answer": "a: 95; b: 33",
    "solution": "a. Suppose there are $x$ pairs of red amoebas, $y$ pairs of blue amoebas and $z$ pairs of one amoeba of each color. Then $2x + z = 95 \\iff x = (95 - z)/2$ and $2y + z = 19 \\iff y = (19 - z)/2$. The number of amoebas in the next generation is $x + 4y + 3z = (95 - z)/2 + 4 \\cdot (19 - z)/2 + 3z = (171 + z)/2$. So we need to maximize $z$, that is, take $z = 19$, and the maximum number of amoebas in the next generation is $(171 + 19)/2 = 95$.\n\nb. If the number of blue amoebas is $b$ and the number of red amoebas is $r$ then $2b + r$ is invariant: indeed, whenever one blue amoeba appears/disappears, two red amoebas disappear/appear. So if $b + r = 100$ then $2b + r = 2 \\cdot 19 + 95 \\iff b + 100 = 133 \\iff b = 33$. So there were $33$ blue amoebas.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26762,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDecide, with proof, if each of the following statement is true.\n\na. If $(a_n)_{n \\ge 1}$ is a decreasing sequence with positive terms such that $\\sum a_n = +\\infty$ then there exists a decreasing sequence $(b_n)_{n \\ge 1}$ with positive terms with $b_n \\le a_n$ for all $n$, $\\sum b_n = +\\infty$ and $\\lim(n b_n) = 0$.\n\nb. If $(a_n)_{n \\ge 1}$ is a decreasing sequence with positive terms such that $\\sum a_n = +\\infty$ then there exists a decreasing sequence $(b_n)_{n \\ge 1}$ with positive terms with $b_n \\le a_n$ for all $n$, $\\sum b_n = +\\infty$ and $(n b_n)_{n \\ge 1}$ decreasing.",
    "options": [],
    "answer": "a is false; b is false",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26763,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all surjective functions $f: (0, +\\infty) \\to (0, +\\infty)$ such that\n$$\n2x \\cdot f(f(x)) = (f(f(x)) + x) \\cdot f(x)\n$$\nfor all $x \\in (0, +\\infty)$.\n\nA function $f: A \\to B$ is said to be surjective iff the range of $f$ is $B$, that is, for all $y \\in B$ there exists $x \\in A$ such that $f(x) = y$.",
    "options": [],
    "answer": "f(x) = x",
    "solution": "First let's prove that $f$ is injective: suppose that $f(a) = f(b)$. Then\n$$\n\\begin{aligned}\n\\frac{2a f(f(a))}{2b f(f(b))} &= \\frac{f(a)(a+f(f(a)))}{f(b)(b+f(f(b)))} \\\\\n&\\iff \\frac{a}{b} = \\frac{a+f(f(a))}{b+f(f(b))} = \\frac{f(f(a))}{f(f(b))} = 1 \\\\\n&\\iff a = b.\n\\end{aligned}\n$$\n\nThen $f$ is invertible. Since the given equation is equivalent to\n$$\n\\frac{2}{f(x)} = \\frac{1}{x} + \\frac{1}{f(f(x))},\n$$\ndefining $\\frac{1}{f^{(n)}(x)} = a_n,\\ n \\in \\mathbb{Z}$, we have, plugging $x \\to f^{(n)}(x)$,\n$$\n2a_{n+1} = a_n + a_{n+2} \\iff a_{n+2} - a_{n+1} = a_{n+1} - a_n,\n$$\nthat is, $a_n$ is an arithmetic progression. Since $a_n > 0$ for all $n, x$, it has to be constant, and $a_1 = a_0 \\iff \\frac{1}{f(x)} = \\frac{1}{x} \\iff f(x) = x$, which can be easily verified to satisfy the equation.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26764,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUma tartaruga corredora anda em linha reta da seguinte maneira. No primeiro trecho do caminho, que mede $\\frac{1}{2}~\\mathrm{m}$, ela corre à velocidade de $3~\\mathrm{m}/\\mathrm{s}$. No segundo trecho, que mede $\\frac{1}{3}~\\mathrm{m}$, ela corre à velocidade de $4~\\mathrm{m}/\\mathrm{s}$. No terceiro trecho, que mede $\\frac{1}{4}~\\mathrm{m}$, ela corre à velocidade de $5~\\mathrm{m}/\\mathrm{s}$ e assim por diante.\n\na) Qual o tempo que a tartaruga leva para percorrer o primeiro trecho? Escreva o resultado como diferença de duas frações unitárias, ou seja, frações com numerador igual a 1.\n\nb) Faça o mesmo com respeito ao segundo trecho.\n\nc) Calcule o tempo que a tartaruga leva para percorrer os 2013 primeiros trechos.",
    "options": [],
    "answer": "a) 1/6 s = 1/2 − 1/3; b) 1/12 s = 1/3 − 1/4; c) 1/2 − 1/2015 s",
    "solution": "Solution:\n\na) Como velocidade é a razão entre a distância percorrida e o tempo gasto, temos que, no primeiro trecho,\n$$\n3 = \\frac{\\frac{1}{2}}{t}\n$$\ne daí obtemos $t = \\frac{1}{6}~\\mathrm{s}$. Para escrever $\\frac{1}{6}$ como diferença de duas frações unitárias, basta notar que\n$$\n\\frac{1}{6} = \\frac{1}{2} - \\frac{1}{3}\n$$\n\nb) Fazendo o mesmo do item anterior, obtemos\n$$\n4 = \\frac{\\frac{1}{3}}{t}\n$$\ne daí obtemos $t = \\frac{1}{4 \\cdot 3}~\\mathrm{s} = \\frac{1}{12}~\\mathrm{s}$. Para escrever $\\frac{1}{12}$ como diferença de duas frações unitárias, basta notar que\n$$\n\\frac{1}{12} = \\frac{1}{3 \\cdot 4} = \\frac{1}{3} - \\frac{1}{4}\n$$\n\nc) Os dois itens anteriores sugerem que os tempos percorridos em cada trecho são, de maneira ordenada, $\\frac{1}{2} - \\frac{1}{3}$, $\\frac{1}{3} - \\frac{1}{4}$, $\\frac{1}{4} - \\frac{1}{5}$, $\\ldots$ Para ver isso, basta notar que\n$$\n\\frac{1}{n(n+1)} = \\frac{1}{n} - \\frac{1}{n+1}\n$$\nLogo, somando os tempos gastos do primeiro até o $2013^{\\text{-}}$ trecho, obtemos\n$$\n\\left(\\frac{1}{2} - \\frac{1}{3}\\right) + \\left(\\frac{1}{3} - \\frac{1}{4}\\right) + \\left(\\frac{1}{4} - \\frac{1}{5}\\right) + \\ldots + \\left(\\frac{1}{2014} - \\frac{1}{2015}\\right)\n$$\nNessa soma, muitos termos se cancelam (o segundo de cada parênteses com o primeiro do próximo parênteses). Logo, obtemos como soma do tempo gasto nos 2013 primeiros termos o valor $\\left(\\frac{1}{2} - \\frac{1}{2015}\\right)~\\mathrm{s}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26765,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm quadrado mágico é uma tabela quadrada na qual a soma dos números em qualquer linha ou coluna é constante. Por exemplo,\n\n| 1 | 5 | 9 |\n| :--- | :--- | :--- |\n| 8 | 3 | 4 |\n| 6 | 7 | 2 |\n\né um quadrado mágico, o qual usa os números de 1 a 9. Como o leitor pode verificar, a soma em qualquer linha ou coluna é sempre igual a 15.\n\na) O quadrado abaixo é parte de um quadrado mágico que usa os números ímpares entre 1 e 17. Descubra qual número $X$ deve ser.\n\n|   | 1 |   |\n| :---: | :---: | :---: |\n| 5 |   | 13 |\n| $X$ |   | 3 |\n\nb) Um quadrado mágico é dito hipermágico quando a soma em qualquer linha, coluna, ou diagonal, é constante. Escreva os números de 1 a 9 no quadrado abaixo de modo que ele se torne hipermágico.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "a) X = 7; b) One valid hypermagic square: rows 8 1 6; 3 5 7; 4 9 2.",
    "solution": "Solution:\n\na) A soma de todos os ímpares de 1 a 17 é 81. Como são três colunas no quadrado, e todas as colunas (e linhas) têm a mesma soma, a soma em cada coluna deve ser $81 / 3 = 27$. Daí, deduzimos que o número que falta na terceira coluna (a dos números 13 e 3) é o número 11, pois aí teremos $11 + 13 + 3 = 27$. Da mesma forma, deduzimos que o número central é o 9. Na primeira linha já temos o 1 e o 11. Logo, o número que falta no canto esquerdo superior é o 15. Seguindo o argumento com as casas que faltam, chegamos à tabela\n\n| 15 | 1 | 11 |\n| :---: | :---: | :---: |\n| 5 | 9 | 13 |\n| 7 | 17 | 3 |\n\nde onde concluímos que $X = 7$.\n\nb) Existem várias soluções possíveis. Uma, por exemplo, seria\n\n| 8 | 1 | 6 |\n| :--- | :--- | :--- |\n| 3 | 5 | 7 |\n| 4 | 9 | 2 |",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26766,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nJúlio Daniel tem um quadrado de papel com vértices $A$, $B$, $C$ e $D$. Ele primeiro dobra este quadrado de papel $ABCD$ levando os vértices $B$ e $D$ até a diagonal, como mostra a figura a seguir:\n\n![](attached_image_1.png)\n\nEm seguida, Júlio Daniel leva o vértice $C$ até o vértice $A$, obtendo assim um pentágono, como é mostrado a seguir:\n\n![](attached_image_2.png)\n\na) Mostre que o ângulo $a$ mede $90^\\circ$.\n\nb) Determine a medida do ângulo $b$.",
    "options": [],
    "answer": "a = 90°, b = 112.5°",
    "solution": "Solution:\n\na) Abrindo o quadrado de papel dobrado, pode-se notar que o ângulo $a$ é o mesmo ângulo do vértice $A$ do quadrado $ABCD$. Logo, como todos os ângulos internos de um quadrado são iguais a $90^\\circ$, concluímos que $a = 90^\\circ$.\n\nb) Desdobrando o papel (apenas a última dobra), obtemos:\n\n![](attached_image_3.png)\n\nO ângulo $\\widehat{ACN}$ é metade da metade do ângulo original do vértice $C$ do quadrado. Como o quadrado tem todos os seus ângulos iguais a $90^\\circ$, concluímos que $\\widehat{ACN} = \\frac{90^\\circ}{4} = 22,5^\\circ$. Por simetria, o ângulo $\\widehat{CLN}$ é igual a $90^\\circ$. Como a soma dos ângulos internos de um triângulo é igual a $180^\\circ$, observando os ângulos do triângulo $CLN$, temos que\n$$\n90^\\circ + 22,5^\\circ + \\widehat{LNC} = 180^\\circ\n$$\nPortanto, $\\widehat{LNC} = 67,5^\\circ$. Como $\\widehat{LNC} + b = 180^\\circ$, obtemos a partir daí que $b = 112,5^\\circ$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26767,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPatrícia quer escrever algarismos nas faces de dois cubos de madeira de tal modo que qualquer dia do mês possa ser representado juntando um algarismo de uma face de um dos cubos e outro algarismo de uma face do outro cubo. Por exemplo, para representar o dia primeiro, Patrícia junta os cubos de modo a mostrar as faces:\n![](attached_image_1.png)\nPara representar o dia 26, Patrícia junta os cubos de maneira adequada mostrando as faces:\n$$2 \\mathbf{2} \\mathbf{6}$$\nO algarismo 6 pode ser usado para representar o 9, bastando para isso girar a face. Lembre-se também que os dias de um mês vão de 01 até 31.\n\na) Quais algarismos devem ser escritos em ambos os cubos?\n\nb) Encontre quais algarismos devem ser escritos em cada cubo.",
    "options": [],
    "answer": "a) The digits 0, 1, and 2 must appear on both cubes.\nb) One valid assignment is: Cube A: 0, 1, 2, 3, 4, 5; Cube B: 0, 1, 2, 6, 7, 8.",
    "solution": "Solution:\n\na) Os dias do mês que são escritos com dois algarismos iguais são 11 e 22. Logo, o 1 e o 2 devem aparecer em ambos os cubos. Além disso, embora não exista um dia 00 no mês, o zero deve aparecer em ambos os cubos, vejamos o porquê disso. Se o zero estivesse em apenas um dos cubos, para formar os dias de 01 até 09, no outro cubo deveriam estar os algarismos $\\{1,2,3,4,5,6,7,8\\}$ (note que o 9 não foi incluído, porque 06 pode ser usado para representá-lo). Como são apenas seis faces no cubo, e são oito elementos no conjunto $\\{1,2,3,4,5,6,7,8\\}$, isso é impossível. Logo, o zero deve aparecer em ambos os cubos. Falta mostrar que nenhum outro algarismo é escrito em ambos os cubos. Como já sabemos que os algarismos $0,1,2$ aparecem em ambos os cubos, restam seis faces a serem preenchidas com algarismos. Como ainda temos os seis algarismos $3,4,5,6,7,8$ para distribuir (o 9 é representado usando o 6 invertido) nas faces, fica claro que não podemos repetir mais nenhum algarismo em dois cubos diferentes.\n\nb) Há muitas soluções possíveis. Sabendo que 0, 1 e 2 aparecem em ambos os cubos, podemos distribuir os algarismos $3,4,5,6,7$ e 8 de qualquer maneira nas faces restantes. Por exemplo, um cubo poderia ter em suas faces os algarismos $0,1,2,3,4,5$ e o outro cubo os algarismos $0,1,2,6,7,8$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26768,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO professor Guilherme criou três estranhas máquinas. A máquina $A$ transforma um gato em um cachorro com probabilidade $\\frac{1}{3}$. A máquina $B$ transforma um gato em um cachorro com probabilidade $\\frac{2}{5}$. A máquina $C$ transforma um gato em um cachorro com probabilidade $\\frac{1}{4}$. E se o animal é um cachorro, nenhuma das máquinas faz transformação alguma.\nO professor Guilherme colocou um gato na máquina $A$, depois colocou o animal resultante da máquina $A$ na máquina $B$ e, por fim, colocou o animal resultante da máquina $B$ na máquina $C$. Qual a probabilidade de ter saído um cachorro da máquina $C$ ?",
    "options": [],
    "answer": "7/10",
    "solution": "Solution:\n\nPrimeiro, vamos calcular a probabilidade de um gato sair gato de cada máquina.\nComo a probabilidade de sair um cachorro da máquina $A$ é $\\frac{1}{3}$, a probabilidade de sair um gato desta máquina é $1-\\frac{1}{3}=\\frac{2}{3}$.\nComo a probabilidade de sair um cachorro da máquina $B$ é $\\frac{2}{5}$, a probabilidade de sair um gato desta máquina é $1-\\frac{2}{5}=\\frac{3}{5}$.\nComo a probabilidade de sair um cachorro da máquina $C$ é $\\frac{1}{4}$, a probabilidade de sair um gato desta máquina é $1-\\frac{1}{4}=\\frac{3}{4}$.\nPara que um gato saia gato depois de passar pelas três máquinas, é necessário que ele saia gato de cada uma delas, pois uma vez cachorro, nenhuma máquina o transforma de volta em gato. Logo, a probabilidade do gato sair gato depois de passar pelas três máquinas é o produto das três probabilidades:\n$$\n\\frac{2}{3} \\times \\frac{3}{5} \\times \\frac{3}{4}=\\frac{3}{10}\n$$\nA probabilidade que este gato saia cachorro depois de passar pelas três máquinas é um menos a probabilidade de que ele saia gato depois de passar pelas três máquinas. Como\n$$\n1-\\frac{3}{10}=\\frac{7}{10}\n$$\na resposta final é $\\frac{7}{10}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26769,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nConsidere um retângulo $ABCD$ onde os comprimentos dos lados são $\\overline{AB}=4$ e $\\overline{BC}=8$. Sobre os lados $BC$ e $AD$ se fixam os pontos $M$ e $N$, respectivamente, de modo que o quadrilátero $BMDN$ seja um losango. Calcule a área deste losango.",
    "options": [],
    "answer": "20",
    "solution": "Solution:\nObserve o seguinte desenho:\n\n![](attached_image_1.png)\n\nComo $BMDN$ é um losango, então os comprimentos dos segmentos $BM$, $MD$, $DN$ e $NB$ são iguais a um mesmo valor, digamos $x$. Pelo dado do problema, $\\overline{AD}=8$, logo $\\overline{AN}=8-x$.\n\n![](attached_image_2.png)\n\nPodemos agora aplicar o Teorema de Pitágoras no triângulo $BAN$:\n$$\n4^{2} + (8-x)^{2} = x^{2}\n$$\nResolvendo a equação obtemos $x=5$. Finalmente, a área do losango resulta $4 \\times x = 4 \\times 5 = 20$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26770,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSeja $a$ um número inteiro positivo tal que há exatamente 10 quadrados perfeitos maiores que $a$ e menores que $2a$.\na) Encontre o menor valor possível de $a$.\nb) Encontre o maior valor possível de $a$.",
    "options": [],
    "answer": "min a = 481, max a = 684",
    "solution": "Solution:\n\na) Como existem exatamente dez quadrados perfeitos maiores do que $a$ e menores do que $2a$, então esses dez quadrados perfeitos têm que ser consecutivos. Logo deve existir um número inteiro positivo $x$ tal que\n$$\n(x-1)^2 \\leq a < x^2 < (x+1)^2 < \\cdots < (x+9)^2 < 2a \\leq (x+10)^2\n$$\nEm particular, vale que:\n$$\na < x^2 \\quad \\text{e} \\quad (x+9)^2 < 2a\n$$\nDessas duas desigualdades temos que\n$$\n(x+9)^2 < 2a < 2x^2\n$$\nE, em particular, vale que\n$$\n(x+9)^2 < 2x^2\n$$\nDesenvolvendo essa desigualdade, obtemos que\n$$\n81 < x(x-18)\n$$\no que só pode ser satisfeito se\n$$\nx \\geq 22\n$$\nNote que, se $x \\geq 22$ então, vale que $(x+9)^2 \\geq 31^2 = 961$. Segue-se daí, junto com a segunda desigualdade em (.1), que\n$$\n961 = 31^2 \\leq (x+9)^2 < 2a\n$$\nlogo\n$$\na > \\frac{961}{2} > 480\n$$\no que nos fornece que\n$$\na \\geq 481\n$$\nAgora, observamos que, entre 481 e $2(481) = 962$, há exatamente dez quadrados perfeitos. De fato, como $22^2 = 484$ e $31^2 = 961$, temos que\n$$\n481 < 22^2 < 23^2 < 24^2 < 25^2 < 26^2 < 27^2 < 28^2 < 29^2 < 30^2 < 31^2 < 962\n$$\nPela equação (.2) não podemos escolher valores para $a$ menores do que 481. Concluímos assim que o menor valor possível de $a$ é 481.\n\nb) Assim como no item anterior, para que existam dez quadrados perfeitos entre $a$ e $2a$, é necessário que exista algum inteiro positivo $x$ tal que:\n$$\n(x-1)^2 \\leq a < x^2 < (x+1)^2 < \\cdots < (x+9)^2 < 2a \\leq (x+10)^2\n$$\nEm particular, vale que\n$$\n(x-1)^2 < a \\quad \\text{e} \\quad 2a < (x+10)^2\n$$\nDessas duas desigualdades, decorre que\n$$\n2(x-1)^2 \\leq 2a \\leq (x+10)^2\n$$\ne, então,\n$$\n2(x-1)^2 \\leq (x+10)^2\n$$\nDesenvolvendo essa desigualdade, obtemos que\n$$\nx(x-24) \\leq 98\n$$\no que só pode ser satisfeito se\n$$\nx \\leq 27\n$$\nNote que, se $x \\leq 27$, então $(x+10)^2 \\leq 37^2 = 1369$. Segue-se daí, junto com a segunda desigualdade em (.3), que\n$$\n2a < (x+10)^2 \\leq 37^2 = 1369\n$$\nLogo\n$$\na \\leq \\frac{1369}{2}\n$$\ne como $a$ é um inteiro, vale que\n$$\na \\leq 684\n$$\nAgora observamos que, entre 684 e 2(684) = 1368, há exatamente dez quadrados perfeitos. De fato, temos que $26^2 = 676 < 684$, $27^2 = 729$, $36^2 = 1296$ e $37^2 = 1369 > 1368$, logo:\n$$\n684 < 27^2 < 28^2 < 29^2 < 30^2 < 31^2 < 32^2 < 33^2 < 34^2 < 35^2 < 36^2 < 1368\n$$\nPela equação (.4) deduzimos que não é possível escolher um valor maior do que 684 para $a$. Com isto concluímos que o maior valor possível para $a$ é 684.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26771,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA figura a seguir mostra um cubo de aresta 1.\n![](attached_image_1.png)\n\na) Qual o menor comprimento possível para um caminho formado por arestas do cubo que passa por todos os 8 vértices?\n\nA figura abaixo mostra um cubo de aresta 1 no qual todas as 12 diagonais da face foram desenhadas. Assim, criou-se uma rede com 14 vértices (os 8 vértices do cubo e os 6 centros da face) e 36 arestas (as 12 do cubo e mais 4 sobre cada uma das 6 faces).\n![](attached_image_2.png)\n\nb) Qual o menor comprimento possível para um caminho formado por arestas dessa rede que passa por todos os 14 vértices?",
    "options": [],
    "answer": "a) 7; b) 6*sqrt(2) + 1",
    "solution": "Solution:\n\na) A figura a seguir representa um pedaço de folha de papel, contendo oito quadrados de lado 1 posicionados lado a lado.\n![](attached_image_3.png)\n\nObserve que, ao dobrarmos a folha de papel nas arestas que são comuns a dois quadrados, obteremos um cubo de lado 1 como aquele do enunciado do exercício. Assim é equivalente pensarmos em caminhos nessa figura ou no cubo do enunciado do exercício. Para facilitar a descrição dos caminhos, pensaremos sempre nessa figura.\n\nOs vértices do cubo resultante foram numerados da maneira como aparece nessa figura. Observe que os números 8, 7, 2, 3, 4 e 5 aparecem duas vezes nessa numeração. Isso se deve ao fato de que, ao dobrarmos a folha de papel, os dois vértices marcados com o mesmo número irão coincidir.\n\nComo a distância entre cada par de vértices adjacentes na figura acima é igual a 1, qualquer caminho passando por 8 vértices distintos deve ter, no mínimo, comprimento igual a 7.\n\nConsidere o caminho\n$$\n1 \\rightarrow 2 \\rightarrow 3 \\rightarrow 4 \\rightarrow 5 \\rightarrow 6 \\rightarrow 7 \\rightarrow 8\n$$\ndesenhado na figura a seguir.\n![](attached_image_4.png)\n\nEsse caminho passa por todos os vértices do cubo e tem comprimento igual a 7. Assim, a resposta para o exercício é 7.\n\nb) Como na resolução do item a), consideramos a figura abaixo que representa um pedaço de folha de papel, contendo oito quadrados de lado 1 posicionados lado a lado.\n![](attached_image_5.png)\n\nDessa vez acrescentamos ainda os vértices localizados nos centros de cada um dos quadrados bem como as diagonais dos quadrados que passam por esses vértices.\n\nComo no item anterior, ao dobrarmos a folha de papel nas arestas que são comuns a dois quadrados, obteremos um cubo de lado 1 como aquele do enunciado desse item. Assim é equivalente pensarmos em caminhos nessa figura ou no cubo do enunciado desse item. Para facilitar a descrição dos caminhos, pensaremos sempre nessa figura.\n\nUtilizamos os números de 1 a 8 para marcar os vértices que estão sobre as arestas do cubo, e as letras de $a$ a $f$, para marcar aqueles vértices localizados nos centros das faces do cubo. Assim o número total de vértices é igual a 14. Observe que os números aparecem duas vezes nessa numeração, devido ao fato de que, ao dobrarmos a folha de papel, os dois vértices marcados com o mesmo número irão coincidir.\n\nConsideremos o caminho\n$$\n1 \\rightarrow a \\rightarrow 2 \\rightarrow b \\rightarrow 3 \\rightarrow c \\rightarrow 4 \\rightarrow 5 \\rightarrow d \\rightarrow 6 \\rightarrow e \\rightarrow 7 \\rightarrow f \\rightarrow 8\n$$\nrealçado na figura a seguir:\n![](attached_image_6.png)\n\nEsse caminho consiste em 12 trechos ligando um vértice marcado com um número a um vértice marcado com uma letra (por exemplo, o trecho $1 \\rightarrow a$) e apenas um segmento ligando dois vértices marcados ambos com números (a saber, o trecho $4 \\rightarrow 5$). Note que cada trecho ligando um vértice marcado com um número a um vértice marcado com uma letra tem comprimento igual a $\\sqrt{2}/2$. Por outro lado, o trecho ligando o vértice 4 ao vértice 5 tem comprimento igual a 1. Assim, o comprimento total do caminho descrito acima é igual a $12 \\times \\sqrt{2}/2 + 1 = 6\\sqrt{2} + 1$.\n\nVamos agora mostrar que nenhum outro caminho contendo os 14 vértices pode ter um comprimento menor do que o caminho que destacamos acima. De fato, qualquer caminho contendo os 14 vértices deve, obrigatoriamente, conter uma quantidade mínima de 13 trechos. Cada trecho poderia a princípio ser de três tipos diferentes:\n- conectando um vértice marcado com número a um vértice marcado com número;\n- conectando um vértice marcado com número a um vértice marcado com letra.\n\nNo primeiro caso, o trecho tem comprimento igual a 1, enquanto que no segundo ele tem comprimento igual a $\\sqrt{2}/2$. Assim, para minimizar o comprimento dos caminhos, devemos utilizar o máximo possível de trechos com comprimento igual a $\\sqrt{2}/2$.\n\nUm caminho que só contivesse trechos conectando um vértice marcado com número a um vértice marcado com letra deveria conter o mesmo número de vértices de cada tipo, já que cada trecho começa em um número e termina em uma letra. Como temos um total de oito vértices marcados com números, e apenas seis marcados com letras, qualquer caminho que passe por todos os 14 vértices deve conter um trecho ligando dois vértices marcados com números, ou seja, um trecho de comprimento 1.\n\nAssim, o caminho que realçamos acima é um exemplo de caminho contendo todos os vértices do cubo e que contém a quantidade mínima permitida de trechos com comprimento igual a 1, logo tem o menor comprimento possível. Portanto, a resposta para esse item é $6 \\times \\sqrt{2} + 1$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26772,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDaniel inventou uma brincadeira na qual é permitido apenas realizar as seguintes operações:\n- somar quatro unidades;\n- multiplicar por quatro;\n- elevar ao quadrado.\nComeçando de um certo número, Daniel desafia um amigo a obter um outro número realizando sucessivamente qualquer uma das operações permitidas.\nPor exemplo, Daniel desafiou Alan a obter o número 152 a partir do número 3. Alan então conseguiu vencer o desafio realizando as seguintes operações:\n![](attached_image_1.png)\n\na) Daniel desafiou Alan a obter o número 340 a partir do número 3. Alan conseguiu vencer o desafio da maneira ilustrada abaixo:\n$$\n3 \\longrightarrow 9 \\longrightarrow 81 \\longrightarrow 85 \\longrightarrow 340\n$$\nDescreva qual foi a operação utilizada por Alan em cada uma das etapas.\n\nb) Mostre que Alan poderia também obter o número 340 começando do número 5.\n\nc) Suponha que Alan começa um desafio a partir de um número cuja divisão por 4 deixa resto 1. Mostre que após qualquer etapa do desafio o número obtido pode ter apenas resto 1 ou 0 .\n\nd) Mostre que é possível vencer o desafio de obter o número 43 a partir do número 3. Mostre também que não é possível vencê-lo começando do número 5.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na) As operações realizadas em cada etapa aparecem acima de cada seta abaixo:\n![](attached_image_2.png)\n\nb) Pode-se obter o $9$ a partir do $5$ somando-se $4$. A partir daí pode-se usar as três últimas etapas do item anterior como ilustrado abaixo:\n![](attached_image_3.png)\n\nc) Observe que somar quatro unidades nunca muda o resto da divisão por $4$.\n\nPor outro lado, após realizar uma multiplicação por $4$ o resultado obtido é sempre um múltiplo de $4$, logo sua divisão por $4$ deixa resto $0$.\n\nLogo, começando de um número cuja divisão por $4$ deixa resto $1$, após realizar as duas operações citadas acima, podemos apenas obter um resultado cuja divisão por $4$ com resto $1$ ou resto $0$. A primeira opção acontece somente quando não realizamos nenhuma multiplicação por $4$. Já a segunda opção acontece somente se realizarmos pelo menos uma multiplicação por $4$.\n\nUm número $x$ cuja divisão por $4$ deixa resto $0$ é um múltiplo de $4$, logo podemos escrevê-lo como $x = n \\times 4$ para algum número natural $n$. Dessa forma $x^{2} = 16 \\times n^{2}$, é um número que também é múltiplo de $4$, e assim ao ser dividido por $4$, deixa resto $0$.\n\nPor outro lado, se $x$ é um número cuja divisão por $4$ deixa resto $1$ então podemos escrevê-lo como $x = n \\times 4 + 1$ para algum número natural $n$. Assim $x^{2} = (n \\times 4 + 1)^{2} = n^{2} \\times 16 + 2 \\times n \\times 4 + 1$ que também deixa resto $1$ ao ser dividido por $4$.\n\nAssim, em todo caso, ao realizarmos as operações permitidas no desafio, o resultado obtido vai sempre deixar resto $0$ ou $1$ ao ser dividido por $4$.\n\nd) Para obter o número $43$ a partir do número $3$, basta somar $4$ dez vezes repetidamente.\n\nComo a divisão do número $5$ por $4$ deixa resto $1$, pelo item anterior, qualquer operação realizada deixaria resto $0$ ou $1$. Como a divisão de $43$ por $4$ deixa resto $3$, não é possível obtê-lo a partir do número $5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26773,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nConsidere o diagrama ilustrado abaixo:\n![](attached_image_1.png)\nAugusto gosta de contar caminhos partindo de algum ponto, chegando no ponto $A$ e nunca passando por um mesmo vértice duas vezes. Para isso, ele representa um caminho pela sequência dos pontos que o caminho visita. Por exemplo, o caminho pontilhado na figura abaixo é representado pela sequência $D C B A$.\n![](attached_image_2.png)\nAugusto chama um caminho de inusitado se a sequência que representa esse caminho está ordenada de maneira alfabética decrescente. Em outras palavras, o caminho é inusitado se nunca anda para a esquerda, seja subindo ou descendo. Por exemplo, o caminho $D C B A$ é inusitado. Já o caminho $D B C A$ não é inusitado, já que a letra $C$ aparece antes da letra $B$.\na) Quantos caminhos inusitados existem começando em $D$ e terminando em $A$ ?\nb) Mostre que o número de caminhos inusitados começando em $E$ é a soma do número de caminhos inusitados começando em $D$ com o número de caminhos inusitados começando em $C$.\nc) Augusto calculou o número de caminhos inusitados saindo de $K$ e chegando em $A$. Qual é esse número?",
    "options": [],
    "answer": "a) 3; c) 89",
    "solution": "Solution:\na) Os caminhos inusitados saindo de $D$ e chegando em $A$ são $D C B A$, $D B A$ e $D C A$.\n\nb) Um caminho inusitado saindo de $E$, tem que visitar logo em seguida um dos pontos $D$ ou $C$. No caso em que ele visita o ponto $D$, a sua continuação até o ponto $A$ pode ser qualquer um dos caminhos inusitados saindo de $D$ e chegando em $A$. No caso em que ele visita o ponto $C$, a sua continuação pode ser qualquer um dos caminhos inusitados saindo de $C$. Dessa forma, o número total de caminhos inusitados saindo de $E$ é a soma do número de caminhos inusitados saindo de $D$ com o número de caminhos inusitados saindo de $C$.\n\nc) Para solucionar essa questão vamos renomear os pontos do diagrama. Façamos $A_{0}=A$, $A_{1}=B$, $A_{2}=C$ e assim sucessivamente até $A_{10}=K$.\nSeja $N(i)$ o número de caminhos inusitados partindo do ponto $A_{i}$ e chegando no ponto $A_{0}=A$. Então temos que $N(i)=N(i-1)+N(i-2)$ para todo $i=2,3, \\ldots, 10$. Além disso, só existe um caminho inusitado partindo de $A_{1}=B$ e chegando em $A$, e existem dois caminhos inusitados partindo de $C$ e chegando em $A$, a saber $C B A$ e $C A$. Assim, $N(1)=1$ e $N(2)=2$. Então $N(3)=N(1)+N(2)=2+1=3$\n(conferir com a resposta do item $a$ ). De forma análoga, temos que $N(4)=3+2=5$, $N(5)=3+5=8$ e assim, sucessivamente.\nSeguindo esse procedimento podemos gerar a sequência abaixo fazendo cada número ser a soma dos seus dois antecessores:\n$$\n1,2,3,5,8,13,21,34,55,89,144, \\ldots\n$$\nO $n$-ésimo termo dessa sequência corresponde então ao número de caminhos inusitados saindo de $A_{n}$ e chegando em $A$. Assim, o número de caminhos inusitados saindo de $K=A_{10}$ e chegando em $A$ é o décimo termo dessa sequência, ou seja, $89$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26774,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nVanessa deseja escolher 50 números inteiros positivos distintos menores do que 100 e tais que a soma de quaisquer dois números escolhidos por ela seja sempre distinta de 99 e de 100.\n\na) Mostre como Vanessa pode atingir o seu objetivo.\n\nb) Mostre que há somente uma maneira pela qual Vanessa pode escolher esses 50 números para atingir o seu objetivo.",
    "options": [],
    "answer": "The unique selection is the set of integers 50, 51, 52, ..., 99.",
    "solution": "Solution:\n\na) Basta que ela escolha os números $50, 51, 52, \\ldots, 99$. Com efeito, se $a < b$ são dois desses números, então $a \\geq 50$ e $b \\geq 51$. Somando essas duas desigualdades obtemos $a + b \\geq 101$ e, em particular, a soma $a + b$ não pode ser igual a $99$ ou a $100$.\n\nb) Consideremos os seguintes 49 pares de números:\n$$\n\\{1,99\\},\\{2,98\\},\\{3,97\\}, \\ldots,\\{48,52\\},\\{49,51\\}\n$$\nVanessa pode escolher no máximo um número de cada um desses pares (já que em caso contrário ela escolheria dois números com soma $100$). Então ela pode escolher no máximo $49$ dentre os números\n$$\n1,2, \\ldots, 48,49,51,52, \\ldots, 100\n$$\nComo ela tem que escolher $50$ números no total, então ela é obrigada a escolher exatamente um número de cada um desses pares e também o único número restante, ou seja, o $50$.\n\nAgora, observe que como ela tem que escolher o número $50$, então não pode escolher o $49$ (já que nesse caso teria escolhido dois números com soma $99$). Mas como ela tem que escolher um número do par $\\{49,51\\}$, então ela deve escolher o número $51$.\n\nDa mesma maneira, como ela tem que escolher o $51$, então não pode escolher o $48$, e logo deve escolher o número $52$. Continuando com esse procedimento, chegaremos à conclusão de que ela deve escolher necessariamente os números $50,51,52, \\ldots, 99$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26775,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm tetraedro regular é um sólido de quatro faces, sendo todas elas triângulos equiláteros de mesmo tamanho. A figura abaixo mostra um tetraedro regular.\n![](attached_image_1.png)\nO comprimento de qualquer aresta de um tetraedro regular é o mesmo. Por exemplo, no tetraedro acima, $\\overline{AB} = \\overline{AC} = \\overline{CD} = \\overline{BC} = \\overline{AD} = \\overline{BD}$. Mostre como colocar um tetraedro de lado $\\sqrt{2}$ inteiramente dentro de um cubo de lado 1.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nComeçamos desenhando um cubo de lado 1:\n![](attached_image_2.png)\n\nVamos traçar duas arestas do tetraedro que queremos inscrever neste cubo: uma será $\\overline{AC}$ e outra será $\\overline{EG}$:\n![](attached_image_3.png)\n\nEm seguida, ligamos o ponto $A$ aos pontos $E$ e $G$, e ligamos o ponto $C$ aos pontos $E$ e $G$:\n![](attached_image_4.png)\n\nObserve pelo desenho que todas as arestas do tetraedro $ACEG$ são diagonais de alguma das faces do quadrado. Por exemplo, olhando a face $ABCD$ do cubo, temos\n![](attached_image_5.png)\n\nPelo Teorema de Pitágoras, $x^2 = 1^2 + 1^2$; portanto, $x = \\sqrt{2}$ é o comprimento de qualquer aresta do tetraedro $ACEG$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26776,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\na) Encontre todos os números inteiros positivos de dois algarismos $\\overline{a b}$ tais que:\n$$\n(a+1)(b+1)=\\overline{a b}+1\n$$\nb) Encontre todos os números inteiros positivos de três algarismos $\\overline{a b c}$ tais que:\n$$\n(a+1)(b+1)(c+1)=\\overline{a b c}+1\n$$",
    "options": [],
    "answer": "a) 19, 29, 39, 49, 59, 69, 79, 89, 99; b) 199, 299, 399, 499, 599, 699, 799, 899, 999",
    "solution": "Solution:\na) Devemos achar todos os números da forma $\\overline{a b}$ tais que\n$$\n(a+1)(b+1)=\\overline{a b}+1\n$$\nNotamos agora que $\\overline{a b}=10a+b$. Substituindo na expressão acima obtemos que:\n$$\n(a+1)(b+1)=10a+b+1\n$$\nSimplificando essa equação obtemos\n$$\nab=9a\n$$\nComo o número $\\overline{a b}$ tem dois algarismos, temos que $a \\neq 0$. Assim, podemos dividir os dois lados da equação acima por $a$ para obter que $b=9$. Portanto, os valores possíveis para $\\overline{a b}$ são:\n$$\n19, \\quad 29, \\quad 39, \\quad 49, \\quad 59, \\quad 69, \\quad 79, \\quad 89 \\quad \\text{e} \\quad 99\n$$\nÉ simples verificar que todos esses números cumprem com a condição do problema. Por exemplo, para o número 19, temos que $a=1$ e $b=9$, logo $(a+1)(b+1)=20=19+1=\\overline{a b}+1$. Para os demais números, deixamos a verificação a cargo do leitor!\n\nb) Devemos agora achar todos os números da forma $\\overline{a b c}$ tais que\n$$\n(a+1)(b+1)(c+1)=\\overline{a b c}+1\n$$\nNote que $\\overline{a b c}=100a+10b+c$. Assim a expressão acima se reescreve como:\n$$\n(a+1)(b+1)(c+1)=100a+10b+c+1\n$$\nO lado esquerdo pode ser escrito como:\n$(a+1)(b+1)(c+1)=a(b+1)(c+1)+(b+1)(c+1)=a(b+1)(c+1)+b(c+1)+c+1$.\nSubstituindo na expressão anterior e simplificando obtemos então que\n$$\na(b+1)(c+1)+b(c+1)=100a+10b\n$$\nAgora, arranjamos os termos para obter a seguinte equação equivalente:\n$$\n0=a\\{100-(b+1)(c+1)\\}+b\\{10-(c+1)\\}\n$$\nNo entanto $b$ e $c$ denotam algarismos, logo como números eles assumem valores menores ou iguais a 9. Logo, temos que $(b+1)(c+1) \\leq 100$. Decorre daí que $\\{100-(b+1)(c+1)\\} \\geq 0$. Do mesmo modo, observamos que $\\{10-(c+1)\\} \\geq 0$. Então, para que o termo da direita na equação (.6) seja igual a zero, deve-se ter que\n$$\na\\{100-(b+1)(c+1)\\}=0 \\quad \\text{e} \\quad b\\{10-(c+1)\\}=0\n$$\nComo $a \\neq 0$ então $\\{100-(b+1)(c+1)\\}=0$. Isso somente é possível se $b=c=9$. Concluímos que os únicos números possíveis são\n$$\n199, \\quad 299, \\quad 399, \\quad 499, \\quad 599, \\quad 699, \\quad 799, \\quad 899 \\text{e} \\quad 999\n$$\nAgora, é simples verificar que todos esses números cumprem com a condição do problema.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26777,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nJuarez utilizou os algarismos $1$, $2$, $3$, $4$ e $5$ para escrever o número $\\overline{abcde}$ de cinco algarismos distintos. Sem revelar qual é esse número, ele disse a Luciana que:\n- o número $\\overline{abc}$ é divisível por $4$;\n- o número $\\overline{bcd}$ é divisível por $5$;\n- o número $\\overline{cde}$ é divisível por $3$.\nEm seguida, Luciana disse a Juarez que é possível descobrir qual é o número $\\overline{abcde}$. Mostre que Luciana está correta, isto é, encontre o número $\\overline{abcde}$.",
    "options": [],
    "answer": "12453",
    "solution": "Solution:\nTodo número divisível por $5$ tem o algarismo das unidades igual a $5$ ou igual a $0$. Como o algarismo $0$ não foi usado por Juarez, então para que $\\overline{bcd}$ seja divisível por $5$, devemos ter, necessariamente, $d=5$.\n\nTodo número divisível por $3$ é tal que a soma dos seus algarismos também é divisível por $3$. Assim, como $\\overline{cde}$ é divisível por $3$, temos que $c+d+e$ é também divisível por $3$. Por outro lado, como $d=5$, então $c$ e $e$ são menores do que $5$. Logo, $c+d+e$ é menor do que $15$. Também vale que $c$ e $e$ são maiores ou igual do que $1$, logo $c+d+e$ também é maior do que $6$. Os únicos múltiplos de três que são maiores ou iguais do que $6$ e menores do que $15$ são o $9$ e o $12$. Logo $c+d+e$ pode assumir apenas os valores $9$ ou $12$. Usando novamente o fato de que $d=5$, temos que $c+e$ pode assumir apenas os valores $4$ ou $7$.\n\nTodo número divisível por $4$ que contenha três ou mais algarismos é tal que o número formado pelos seus algarismos da dezena e da unidade (nessa ordem) é também divisível por $4$. Assim, como $\\overline{abc}$ é divisível por $4$, temos que o número $\\overline{bc}$ é também divisível por $4$. Em particular, $c$ deve ser par, já que todo número divisível por quatro é também par.\n\nComo $c$ é par, então vale que $c=2$ ou $c=4$. Vamos mostrar agora que isso implica que não pode ser o caso que $c+e$ seja igual a $4$. De fato, caso seja $c=2$, então para que $c+e$ fosse igual a $4$, deveríamos ter que $e=2$, o que seria impossível visto que $e \\neq c$. Caso seja $c=4$, então para que $c+e$ fosse igual a $4$, deveríamos ter que $e=0$, o que seria também impossível, já que o algarismo $0$ não foi usado por Juarez. Assim, obrigatoriamente temos que $c+e=7$.\n\nAté aqui já descobrimos que:\n(1) $c$ é par;\n(2) $c+e=7$;\n(3) $\\overline{bc}$ é divisível por $4$.\n\nComo $d=5$, então $e \\neq 5$. Então, pelo item (2) acima, temos que $c \\neq 2$. Portanto, o único valor par possível para $c$ é $4$. Concluímos assim que $c=4$ e $e=3$.\n\nOs valores possíveis que restam para $b$ são $1$ e $2$. Mas, pelo item (3), concluímos que $b=2$, pois, se fosse $b=1$, então $\\overline{bc}$ seria igual a $14$, que não é um múltiplo de $4$. Finalmente, temos que $a=1$. Logo, o número escrito por Juarez é o $\\overline{abcde}=12453$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26778,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nArnaldo, Bernaldo, Cernaldo, Dernaldo e Ernaldo são estudantes de distintas partes do Brasil que foram escolhidos para representar o seu país nas olimpíadas internacionais. Depois de várias semanas de treino, algumas amizades foram formadas. Perguntamos, então, a cada um deles quantos amigos tinham feito no grupo. Arnaldo, Bernaldo, Cernaldo e Dernaldo responderam, respectivamente, que tinham feito 1, 2, 3 e 4 amigos dentro do grupo. Quantos dos integrantes do grupo são amigos de Ernaldo?",
    "options": [],
    "answer": "2",
    "solution": "Solution:\n\nDigamos que Ernaldo tem $x$ amigos dentro do grupo. Como Dernaldo tem 4 amigos, e o grupo tem 5 integrantes, então todos são amigos de Dernaldo. Tiremos Dernaldo do grupo. Assim, Arnaldo, Bernaldo, Cernaldo e Ernaldo ficam agora, respectivamente, com $0,1,2$ e $x-1$ amigos dentro do subgrupo. Como Arnaldo não tem mais amigos dentro do subgrupo podemos esquecê-lo. Na seguinte tabela mostramos o número de amigos que cada integrante tem dentro do subgrupo.\n\n| Integrante | № de amigos |\n| :--- | :---: |\n| Bernaldo | 1 |\n| Cernaldo | 2 |\n| Ernaldo | $x-1$ |\n\nObviamente os dois amigos de Cernaldo dentro do subgrupo são Bernaldo e Ernaldo. Aliás, como Bernaldo só tem um amigo dentro do grupo, Cernaldo deve ser esse único amigo. Em particular, Ernaldo e Bernaldo não são amigos. Portanto, Cernaldo é também o único amigo de Ernaldo dentro do subgrupo, ou seja $x-1=1$. Concluímos assim que Cernaldo e Dernaldo são os únicos amigos de Ernaldo. A resposta é 2 .",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26779,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nOs ciclistas têm aversão ao número zero (porque é oval) e ao número oito (porque assim ficam as rodas após os acidentes). Quantos sócios podem se inscrever num clube de ciclistas se cada um deve possuir uma identificação de três dígitos, sem usar o dígito zero nem o dígito oito?",
    "options": [],
    "answer": "512",
    "solution": "Solution:\nJá que os ciclistas não usam o dígito $0$ e nem o $8$, restam os dígitos $1, 2, 3, 4, 5, 6, 7$ e $9$. Assim, há $8$ possibilidades para a escolha de cada dígito. Temos que escolher números de três dígitos. Logo, temos $8$ opções para o primeiro dígito, $8$ opções para o segundo dígito e $8$ opções para o terceiro dígito. Daí, como\n$$\n8 \\times 8 \\times 8 = 512\n$$\nconcluímos que no máximo $512$ sócios podem se inscrever.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26780,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa cidade de Trocalândia, $20\\%$ dos gatos pensam que são cachorros e $25\\%$ dos cachorros pensam que são gatos. Certo dia, um psicólogo veterinário resolve testar todos os gatos e cachorros de Trocalândia, verificando que $30\\%$ do total pensava ser gato. Que proporção dos animais testados era de cães?",
    "options": [],
    "answer": "10/11",
    "solution": "Solution:\n\nSejam $C$ e $G$, respectivamente, o número de cães e gatos de Trocalândia. O número de gatos que pensam que são gatos é\n$$\n\\frac{80 G}{100}\n$$\nO número de cachorros que pensam que são gatos é\n$$\n\\frac{25 C}{100}\n$$\nLogo, o número total de animais que pensam que são gatos é\n$$\n\\frac{80 G+25 C}{100}\n$$\nConforme diz o psicólogo veterinário,\n$$\n\\begin{aligned}\n\\frac{80 G+25 C}{100}=\\frac{30}{100}(G+C) & \\Longrightarrow 80 G+25 C=30 G+30 C \\\\\n& \\Longrightarrow 80 G-30 G=30 C-25 C \\\\\n& \\Longrightarrow 10 G=C\n\\end{aligned}\n$$\nPortanto, a proporção de cães é\n$$\n\\frac{C}{C+G}=\\frac{10 G}{10 G+G}=\\frac{10}{11}\n$$\nsendo a resposta final 10/11.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26781,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOs cinco amigos Arnaldo, Bernaldo, Cernaldo, Dernaldo e Ernaldo disputaram entre si cinco corridas. Em cada corrida, o ganhador recebeu cinco pontos, o segundo colocado quatro pontos e assim sucessivamente até o último colocado, que recebeu apenas um ponto. Para obter a pontuação final de cada corredor, foram somadas as pontuações obtidas nas cinco corridas. Na pontuação final, Arnaldo ficou em primeiro lugar com 19 pontos, seguido de Bernaldo com 16 pontos. O terceiro, quarto e quinto lugares foram ocupados por Cernaldo, Dernaldo e Ernaldo, respectivamente. Não houve empate em nenhuma corrida e nem na pontuação final. Mas um fato curioso é que nem Arnaldo e tampouco Bernaldo ganharam sequer uma das cinco corridas.\n\na) Mostre que, em cada corrida tanto Arnaldo quanto Bernaldo obtiveram sempre o segundo e o terceiro lugar (em alguma ordem).\n\nb) Diga quantos pontos conseguiu Cernaldo.\n\nc) Quantas corridas ganhou cada um dos corredores?",
    "options": [],
    "answer": "a) In every race, Arnaldo and Bernaldo finished second and third in some order. b) Cernaldo scored 15 points. c) Cernaldo won 2 races, Dernaldo won 2 races, Ernaldo won 1 race, and Arnaldo and Bernaldo won none.",
    "solution": "Solution:\n\na) Como Arnaldo e Bernaldo obtiveram, respectivamente, 19 pontos e 16 pontos como pontuação final, o total de pontos por eles obtidos foi igual a 35. Como nenhum deles ganhou sequer uma das corridas, então, em cada uma das corridas, a soma dos pontos por eles obtidos foi, no máximo, igual a $4+3=7$. Como $7 \\times 5=35$ então, para que eles tenham obtido 35 pontos em cinco corridas, é necessário que, juntos, eles tenham obtido invariavelmente 7 pontos por corrida. Assim é necessário que eles tenham ocupado sempre os segundo e terceiro lugares.\n\nb) Em cada uma das cinco corridas são distribuídos $1+2+3+4+5=15$ pontos. Assim, o total de pontos distribuídos nas cinco corridas é igual a $(1+2+3+4+5) \\times 5=75$ pontos. Já que o total de pontos obtidos por Arnaldo e Bernaldo é igual a 35, então o total de pontos obtidos por Cernaldo, Dernaldo e Ernaldo é igual a $75-35=40$.\nSe o total de pontos obtidos pelo terceiro colocado fosse igual a 14, seria impossível que a soma dos pontos obtidos por Cernaldo, Dernaldo e Ernaldo fosse igual a 40. De fato, o total de pontos obtidos por Dernaldo e Ernaldo seria menor ou igual a 13 e 12 respectivamente. No entanto, $14+13+12=39<40$. Assim, o total de pontos obtidos por Dernaldo deve ser maior do que 14. No entanto, Bernaldo obteve 16 pontos e, como Cernaldo obteve menos pontos que ele, a única possibilidade é que Cernaldo tenha obtido 15 pontos.\n\nc) Do item a), temos que as segunda e terceira posições foram ocupadas por Arnaldo e Bernaldo em todas as corridas. Assim, caso Cernaldo tivesse ganhado apenas uma corrida, o máximo de pontos que ele poderia ter obtido seria igual a $5+2+2+2+2=13$ (que corresponde a cinco pontos obtidos na sua vitória somados com, no máximo dois pontos nas demais corridas).\nEntão, para que Cernaldo tenha conseguido 15 pontos, ele deveria ter vencido exatamente duas corridas. De fato, o único modo pelo qual Cernaldo pôde ter conseguido somar 15 pontos é o seguinte:\nCernaldo chegou primeiro em duas corridas, quarto em duas e quinto em uma.\nComo Cernaldo obteve 15 pontos, a soma dos pontos obtidos por Dernaldo e Ernaldo é igual a $40-15=25$. Assim, para que Dernaldo seja o quarto colocado na classificação final, a quantidade de pontos acumulados por ele deve ser ao menos igual a 13. Se Dernaldo tivesse vencido somente uma das corridas, o máximo de pontos que ele poderia ter obtido seria igual a $5+2+2+2+1=12$ (lembre-se de que Cernaldo chegou duas vezes em quarto lugar, logo Dernaldo não pode ter chegado mais do que três vezes em quarto lugar). Portanto, Dernaldo deve ter vencido duas corridas. Concluímos assim que Cernaldo e Dernaldo venceram duas corridas cada um. Por fim, Ernaldo venceu uma das cinco corridas.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26782,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa figura, $PQRS$ é um quadrado, $\\overline{QE} = 17$, e $\\overline{PH} = 12$.\n\n![](attached_image_1.png)\n\nCalcule $\\overline{SE}$.",
    "options": [],
    "answer": "7",
    "solution": "Solution:\n\nPodemos completar o desenho de modo que, na figura seguinte, $HAQE$ seja um retângulo.\n\n![](attached_image_2.png)\n\nSe chamamos $\\measuredangle HPS = a^\\circ$, então $\\measuredangle PQ = (90 - a)^\\circ$. Logo, $\\measuredangle PQA = a^\\circ$. Observe que, deste modo, os triângulos $PHS$ e $QAP$ possuem os mesmos ângulos internos (isto é, são semelhantes), como ilustrado na figura a seguir:\n\n![](attached_image_3.png)\n\nComo $SPQR$ é um quadrado, vemos que o comprimento da hipotenusa do triângulo $PHS$ coincide com o comprimento da hipotenusa do triângulo $QAP$. Concluímos que $PHS$ e $QAP$ são triângulos congruentes (isto é, os seus lados têm os mesmos comprimentos). Logo,\n$$\n\\overline{AQ} = \\overline{PH} = 12\n$$\nComo $HAQE$ é um retângulo, temos que $\\overline{AQ} = \\overline{HE}$, logo, temos que\n$$\n\\overline{HE} = 12\n$$\nPelo mesmo motivo, $\\overline{AH} = \\overline{QE} = 17$ e então $\\overline{AP} = \\overline{AH} - \\overline{PH} = 17 - 12 = 5$, como mostra a figura abaixo:\n\n![](attached_image_4.png)\n\nLembrando que os triângulos $PHS$ e $QAP$ são congruentes, concluímos que\n$$\n\\overline{HS} = \\overline{AP} = 5\n$$\nDe (.9) e (.10) segue-se que $\\overline{SE} = 12 - 5 = 7$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26783,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm número natural $N$ maior que 10 é chamado \"superquadrado\" se o número formado por cada dois algarismos consecutivos do número $N$ (considerados na mesma ordem) é sempre um quadrado perfeito. Por exemplo, $8164$ é \"superquadrado\" porque os números $81$, $16$ e $64$ são quadrados perfeitos. Outros exemplos de superquadrados são $25$ e $649$.\n\na) Quantos números \"superquadrados\" existem?\n\nb) Qual é o maior número \"superquadrado\"?",
    "options": [],
    "answer": "a) 14; b) 81649",
    "solution": "Solution:\n\na) Será útil adiante que observemos a lista de todos os quadrados perfeitos de dois dígitos:\n$$\n16, \\quad 25, \\quad 36, \\quad 49, \\quad 64, \\quad 81\n$$\nO número formado pelos primeiros dois algarismos de um superquadrado deve ser algum número da lista (.1). Assim, começemos contando os superquadrados a partir do número $16$.\n\n- O número $16$ é um superquadrado. Se queremos acrescentar um algarismo à direita de $16$ para formar um novo superquadrado, é necessário que tal algarismo seja um $4$, porque $64$ é o único número da lista (.1) começando com $6$. Assim, vemos que o único superquadrado de $3$ algarismos começando com $16$ é $164$.\nSendo agora $49$ o único número da lista (.1) começando com $4$, então o dígito $9$ é o único que pode ser acrescentado à direita de $164$ para formar um novo superquadrado.\nAgora, nenhum número da lista (.1) começa com $9$, portanto, não é possível criar um novo superquadrado adicionando um dígito à direita de $1649$. Obtemos assim que a lista de superquadrados que começam com $16$ é\n$$\n16, \\quad 164, \\quad 1649 \\text{.}\n$$\n\n- Como não existem números na lista (.1) começando com o dígito $5$, temos que o único superquadrado que começa com $25$ é $25$.\n\n- Seguindo o mesmo raciocínio utilizado acima para encontrar os superquadrados que começam com $16$, vemos que os superquadrados começando com $36$ são\n$$\n36, \\quad 364, \\quad 3649\n$$\n\n- Similar ao caso do $25$, o único superquadrado que tem $49$ como seus primeiros dois algarismos é o próprio $49$.\n\n- Os superquadrados começando com $64$ são\n$$\n64, \\quad 649\n$$\n\n- E finalmente, os superquadrados começando com $81$ são\n$$\n81, \\quad 816, \\quad 8164, \\quad 81649\n$$\n\nContando os superquadrados obtidos acima em cada um dos casos, concluímos que existem $14$ superquadrados:\n\n$16$, $164$, $1649$, $25$, $36$, $364$, $3649$, $49$, $64$, $649$, $81$, $816$, $8164$, $81649$.\n\nb) Pelo item anterior temos que o maior de todos superquadrados é o número $81649$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26784,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOs números $x$, $y$, $z$ e $w$ na figura são números inteiros todos diferentes entre si, maiores do que $1$, e foram colocados nas casas abaixo de modo que cada número (a partir de $y$) é divisor do número na casa da esquerda.\n\n| $x$ | $y$ | $z$ | $w$ |\n| :--- | :--- | :--- | :--- |\n\nDescubra todas as soluções possíveis para $x$, $y$, $z$ e $w$ sabendo que a soma deles é $329$.",
    "options": [],
    "answer": "x = 280, y = 28, z = 14, w = 7",
    "solution": "Solution:\n\nPelo enunciado do problema sabemos que\n$$\n1 < w < z < y < x < 329\n$$\nComo $w$ é divisor de $z$ e $z$ é divisor de $y$, então $w$ é divisor de $y$. Mas $y$ é divisor de $x$ então, $w$ é também divisor de $x$ e, portanto, de cada número colocado nas casas.\n\nO número $w$ é portanto divisor da soma, $329 = 7 \\times 47$, e concluímos daí que $w = 7$ ou $w = 47$. Vejamos cada caso.\n\n- Suponhamos primeiro que $w = 47$. Usando para $z$ o mesmo argumento que para $w$ concluímos que $z$ é divisor da soma\n$$\nx + y + z = 47 \\times 2 \\times 3\n$$\nLembre que $w$ é divisor de $z$ e $w < z < 329$. Então $z = 47 \\times 2$ ou $z = 47 \\times 3$. Mas $z < y < x$ e portanto para conseguir (.11) é necessário que $z < 47 \\times 2$. Concluímos dessa forma que não existe solução com $w = 47$.\n\n- Suponhamos agora que $w = 7$. Como no caso anterior, $z$ deve ser divisor da soma\n$$\nz + y + x = 7 \\times 2 \\times 23\n$$\nAs únicas alternativas para $z$ são $z = 7 \\times 23$ e $z = 7 \\times 2$. Mas para ter $z < y < x$ e conseguir (.12) é necessário que $z < 7 \\times 23$. Logo $z = 7 \\times 2$.\n\nAgora, $y$ é divisor da soma\n$$\ny + x = 7 \\times 2 \\times 2 \\times 11\n$$\nComo $z$ é divisor de $y$ e $z < y$ então $y = 7 \\times 2 \\times 2$ ou $y = 7 \\times 2 \\times 11$. Mas $y < x$ e para conseguir (.13) é necessário que $y < 7 \\times 2 \\times 11$.\n\nConcluímos assim que a única solução possível é\n$$\nw = 7, \\quad z = 7 \\times 2, \\quad y = 7 \\times 2 \\times 2, \\quad x = 7 \\times 2 \\times 2 \\times 10\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26785,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO pequeno Abel ganhou de presente um tabuleiro $2 \\times n$ e $n$ fichas de tamanho $2 \\times 1$. Por exemplo, a figura a seguir mostra o caso em que $n=10$, isto é, quando Abel tem um tabuleiro $2 \\times 10$ e 10 fichas de tamanho $2 \\times 1$.\n![](attached_image_1.png)\nEle brinca de preencher o tabuleiro usando as $n$ fichas. Por exemplo, para $n=10$ Abel poderia preenchê-lo dos modos ilustrados a seguir:\n![](attached_image_2.png)\nObserve, no entanto, que existem muitas outras maneiras pelas quais Abel pode preencher o seu tabuleiro.\na) Calcule o número total de maneiras pelas quais Abel pode preencher o seu tabuleiro nos casos em que $n=1, n=2$ e $n=3$, isto é, no caso em que os tabuleiros têm dimensões $2 \\times 1,2 \\times 2$ e $2 \\times 3$.\nb) Seja $a_{n}$ a quantidade de maneiras pelas quais Abel pode preencher um tabuleiro $2 \\times n$ utilizando $n$ fichas $2 \\times 1$. Mostre que $a_{10}=a_{9}+a_{8}$.\nc) Calcule o número total de maneiras pelas quais Abel pode preencher o seu tabuleiro quando $n=10$.",
    "options": [],
    "answer": "a1=1, a2=2, a3=3; a10=89",
    "solution": "Solution:\n\na) Podemos contar facilmente os primeiros casos e observar que $a_{1}=1, a_{2}=2$ e $a_{3}=3$, como mostra a figura abaixo.\n![](attached_image_3.png)\n\nb) Ao começar a preencher o seu tabuleiro $2 \\times 10$, Abel pode proceder de duas maneiras distintas:\n- Abel pode começar colocando uma ficha verticalmente na primeira coluna do tabuleiro:\n![](attached_image_4.png)\n- Caso contrário, ele deve começar colocando duas fichas horizontais nas duas primeiras casas das duas primeiras linhas do tabuleiro:\n![](attached_image_5.png)\nNo primeiro caso, o Abel deverá usar as nove fichas restantes para preencher o resto do tabuleiro, que coincide com um tabuleiro de tamanho $2 \\times 9$. Ele pode fazê-lo de $a_{9}$ maneiras.\nNo segundo caso, Abel deverá utilizar as oito fichas que restaram para preencher o resto do tabuleiro que coincide com um tabuleiro de tamanho $2 \\times 8$. Ele pode fazê-lo de $a_{8}$ maneiras.\nAssim, concluímos que a quantidade total de maneiras que existem para que Abel finalize o preenchimento do seu tabuleiro é igual a $a_{9}+a_{8}$. Concluímos então que:\n$$\na_{10}=a_{9}+a_{8}\n$$\n\nc) Repetindo o mesmo argumento que usamos no item anterior, temos que, de modo mais geral, as maneiras pelas quais podemos preencher um tabuleiro $2 \\times n$ dividem-se em dois grupos. O primeiro grupo, no qual inicia-se posicionando uma ficha vertical no lado esquerdo do tabuleiro:\n![](attached_image_6.png)\nE o outro grupo no qual inicia-se posicionando duas fichas horizontais no lado esquerdo do tabuleiro:\n![](attached_image_7.png)\nPara o primeiro grupo, o número de maneiras de continuar o preenchimento coincide com $a_{n-1}$ enquanto que, no segundo grupo, esse número de maneiras coincide com $a_{n-2}$. Concluímos que\n$$\na_{n}=a_{n-1}+a_{n-2}, \\quad \\text{ para todo } n \\geq 3\n$$\nDe fato, pelo item a), temos que $a_{1}=1$ e $a_{2}=2$. Agora podemos usar (.14) para conseguir os próximos valores de $a_{n}$. De fato, temos que $a_{3}=a_{2}+a_{1}=3$ (como já havíamos determinado no item a)). De maneira análoga temos que $a_{4}=a_{3}+a_{2}=3+2=5$. Continuando com o mesmo raciocínio obtemos que:\n$$\na_{5}=8, \\quad a_{6}=13, \\quad a_{7}=21, \\quad a_{8}=34, \\quad a_{9}=55 \\quad \\text{ e } \\quad a_{10}=89\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26786,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA figura abaixo é composta por um quadrado e um pentágono regular.\n![](attached_image_1.png)\nCalcule a soma dos ângulos $a^{\\circ}$ e $b^{\\circ}$.\nFatos que ajudam. (Você pode usá-los!). A soma dos ângulos internos de um triângulo é sempre igual a $180^{\\circ}$. Além disso, a soma dos ângulos internos de um quadrilátero é sempre igual a $360^{\\circ}$. Para ver isso, basta dividir o quadrilátero em dois triângulos, ligando dois vértices opostos.\n![](attached_image_2.png)\nCada triângulo tem $180^{\\circ}$ como soma dos ângulos internos, daí obtemos $180^{\\circ} \\times 2=360^{\\circ}$ como soma dos ângulos internos do quadrilátero. E a soma dos ângulos internos de um pentágono é igual a $180^{\\circ} \\times 3=540^{\\circ}$, pois podemos dividir um pentágono qualquer em três triângulos como mostra a figura a seguir.\n![](attached_image_3.png)",
    "options": [],
    "answer": "324°",
    "solution": "Solution:\nObserve o triângulo $APM$ na seguinte figura:\n\n![](attached_image_4.png)\n\nComo $A\\hat{M}P = (180 - a)^{\\circ}$, e a soma dos ângulos internos de um triângulo vale $180^{\\circ}$, então $A\\hat{P}M = (a - 90)^{\\circ}$. Aplicando o mesmo argumento ao triângulo $BQN$, obtemos que $B\\hat{Q}N = (b - 90)^{\\circ}$. Note que $S\\hat{P}Q = A\\hat{P}M = (a - 90)^{\\circ}$ e que $P\\hat{Q}R = B\\hat{Q}N = (b - 90)^{\\circ}$, já que $S\\hat{P}Q$ e $A\\hat{P}M$ são ângulos opostos pelo vértice $P$ e, analogamente, $P\\hat{Q}R$ e $B\\hat{Q}N$ são opostos pelo vértice $Q$.\n\nVamos nos concentrar no quadrilátero $SPQR$ ilustrado abaixo:\n\n![](attached_image_5.png)\n\nSendo o pentágono regular, todos os seus cinco ângulos internos são iguais, assim a soma dos seus ângulos internos é igual a $5 \\times P\\hat{S}R$. Como a soma dos ângulos internos do pentágono é igual a $540^{\\circ}$, logo temos que $5 \\times P\\hat{S}R = 540^{\\circ}$, de onde obtemos que $P\\hat{S}R = 108^{\\circ}$. Como a medida de $Q\\hat{R}S$ é igual à de $P\\hat{S}R$, temos que $Q\\hat{R}S = 108^{\\circ}$.\n\nA soma dos ângulos internos do quadrilátero $PQRS$ é igual a $360^{\\circ}$. Portanto,\n$$\n(a - 90)^{\\circ} + (b - 90)^{\\circ} + 108^{\\circ} + 108^{\\circ} = 360^{\\circ}\n$$\nLogo, temos que $(a + b)^{\\circ} - 180^{\\circ} + 216^{\\circ} = 360^{\\circ}$, de onde concluímos que $a^{\\circ} + b^{\\circ} = 324^{\\circ}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26787,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNos lados $AB$ e $BC$ de um triângulo equilátero $ABC$, fixam-se dois pontos $D$ e $E$, respectivamente, de modo que $\\overline{AD} = \\overline{BE}$.\n\n![](attached_image_1.png)\n\nSe os segmentos $AE$ e $CD$ se cortam no ponto $P$, determine $\\measuredangle APC$.",
    "options": [],
    "answer": "120°",
    "solution": "Solution:\n\nObserve os triângulos $DAC$ e $EBA$. Sabe-se que $\\overline{DA} = \\overline{EB}$. Além disso, como o triângulo $ABC$ é equilátero, então $\\overline{AC} = \\overline{BA}$. Mais ainda, em um triângulo equilátero todos os ângulos internos medem $60^\\circ$. Logo $\\measuredangle DAC = \\measuredangle EBA = 60^\\circ$. Isso implica que os triângulos $DAC$ e $EBA$ são congruentes e portanto $\\measuredangle DCA = \\measuredangle BAE$. Na figura abaixo representamos $DCA = \\measuredangle BAE = x$.\n\n![](attached_image_2.png)\n\nAgora note que $\\measuredangle PAC = 60^\\circ - \\measuredangle DAP = 60^\\circ - \\measuredangle PCA$. Isto é, $\\angle PAC + \\measuredangle PCA = 60^\\circ$. Como a soma dos ângulos interiores no triângulo $APC$ deve ser $180^\\circ$, temos que $\\measuredangle PAC + \\measuredangle PCA + \\measuredangle APC = 180^\\circ$. Concluímos então que $\\measuredangle APC = 120^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26788,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nConsidere a lista de números $a_{1}, a_{2}, \\ldots$, onde\n$$\na_{n}=\\underbrace{111111 \\ldots 1}_{3^{n} \\text{ algarismos }}\n$$\nou seja, $a_{1}=\\underbrace{111}_{\\text{três uns }}$, $a_{2}=\\underbrace{111111111}_{\\text{nove uns }}$, $a_{3}=\\underbrace{111 \\ldots 1}_{\\text{vinte e sete uns }}$, e assim por diante.\na) Mostre que $a_{1}$ é múltiplo de 3 mas não de 9.\nb) Mostre que $a_{2}$ é múltiplo de 9 mas não de 27.\nc) Mostre que $a_{3}$ é múltiplo de 27 mas não de 81.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\na) Lembremos que um número é múltiplo de $3$ se, e somente se, a soma dos seus algarismos é um múltiplo de $3$. De modo similar, um número é múltiplo de $9$ se, e somente se, a soma de seus algarismos é um múltiplo de $9$. Como $a_{1}=111$, vemos que a soma dos seus algarismos é igual a $3$. Usando então os critérios antes mencionados, $a_{1}$ resulta ser um múltiplo de $3$ mas não múltiplo de $9$.\n\nb) A soma dos algarismos de $a_{2}=111111111$ é $9$. Portanto, $a_{2}$ é múltiplo de $9$. Agora, observemos que $a_{2}=1001001 \\times a_{1}$:\n\n![](attached_image_1.png)\n\nA soma dos algarismos de $1001001$ é $3$. Então $1001001$ é múltiplo de $3$ mas não de $9$. Isto é, $1001001=3 \\times k$ onde $k$ é um número inteiro que não é divisível por $3$. Já sabíamos, pela parte a), que $a_{1}=3 \\times \\ell$, onde $\\ell$ é um número inteiro que não é divisível por $3$. Assim, podemos concluir que o produto\n$$\na_{2}=1001001 \\times a_{1}=9 \\times k \\times \\ell\n$$\né múltiplo de $9$ mas não de $27$, porque $k \\times \\ell$ não é divisível por $3$.\n\nc) Do mesmo modo que na parte b), podemos verificar facilmente que\n$$\na_{3}=1 \\underbrace{00 \\ldots 00}_{8 \\text{ vezes }} 1 \\underbrace{00 \\ldots 00}_{8 \\text{ vezes }} 1 \\times a_{2}\n$$\nAgora, a soma dos algarismos do número $1000000001000000001$ é $3$. Então, esse número é múltiplo de $3$ mas não de $9$. Isto é,\n$$\n1 \\underbrace{00 \\ldots 00}_{8 \\text{ vezes }} 1 \\underbrace{00 \\ldots 00}_{8 \\text{ vezes }} 1=3 \\times q\n$$\nonde $q$ é um inteiro que não é divisível por $3$. Como já sabíamos que $a_{2}$ é da forma $a_{2}=9 \\times m$ com $m$ um inteiro que não é divisível por $3$, então podemos concluir que o produto em (.2) é da forma\n$$\n1 \\underbrace{00 \\ldots 00}_{8 \\text{ vezes }} 1 \\underbrace{00 \\ldots 00}_{8 \\text{ vezes }} 1 \\times a_{2}=27 \\times q \\times m\n$$\nIsso mostra que $a_{3}$ é múltiplo de $27$ mas não de $27 \\times 3=81$.\n\nObservação: Continuando com o mesmo procedimento podemos mostrar, de forma geral, que $a_{n}$ é múltiplo de $3^{n}$, mas não é múltiplo de $3^{n+1}$ para qualquer valor de $n \\geq 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26789,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm comerciante recebeu quatro sacos de batatas e deseja medir o peso de cada um deles. Ele sabe que os pesos desses sacos em quilogramas são quantidades inteiras e distintas. Suponha que os pesos dos sacos (em quilogramas) sejam $a, b, c$ e $d$, com $a<b<c<d$.\n\na) Mostre que, ao pesar os sacos de dois em dois, o maior resultado é $c+d$, e que o segundo maior é $b+d$. Mostre também que o menor resultado é $a+b$ e o segundo menor é $a+c$.\n\nb) A balança do comerciante quebrou. Assim ela só consegue indicar pesos maiores ou iguais do que 100 quilogramas. Ele decide então pesar os sacos de dois em dois realizando assim seis pesagens. Em quatro das pesagens ele obteve como medidas (em quilogramas): 101, 112, 116 e 127. Nas outras duas, ele só conseguiu descobrir que as somas eram menores ou iguais a 100 quilogramas. Encontre os pesos dos quatro sacos.",
    "options": [],
    "answer": "a=41, b=45, c=56, d=71",
    "solution": "Solution:\n\na) Somando $a$ em cada lado da desigualdade $b<c$, obtemos a desigualdade $a+b < a+c$. Com argumentos similares, podemos obter as desigualdades:\n$$\na+c < a+d, \\quad a+c < b+c, \\quad a+d < b+d, \\quad b+c < b+d \\quad \\text{e} \\quad b+d < c+d\n$$\nUsando essas desigualdades, podemos concluir que:\n$$\na+b < a+c < a+d < b+d < c+d \\quad \\text{e} \\quad a+b < a+c < b+c < b+d < c+d\n$$\nDe (.15), vemos que as 4 somas $a+b, a+c, a+d$ e $b+c$ são menores do que as somas $b+d$ e $c+d$. Como $b+d < c+d$, então a maior soma é $c+d$ e a segunda maior é $b+d$. Também de (.15), vemos que $a+b$ e $a+c$ são menores do que cada uma das outras 4 somas $a+d, b+c, b+d$ e $c+d$. Como $a+b < a+c$, então $a+b$ é a menor soma e $a+c$ é a segunda menor.\n\nb) Da parte a) sabemos que a maior soma é $c+d$ e que a segunda maior é $b+d$. Temos assim que\n$$\nc+d = 127 \\quad \\text{e} \\quad b+d = 116\n$$\nLogo, as somas $a+d$ e $b+c$ são iguais aos números 101 e 112 em alguma ordem. Subtraindo as equações em (.16), obtemos $c = b + 11$. Somando $b$ a cada lado vemos que $b+c = 2b + 11$. Como $2b + 11$ é um número ímpar, descobrimos assim que $b+c$ é ímpar. Logo,\n$$\nb+c = 101 \\quad \\text{e} \\quad a+d = 112\n$$\nUsando que $c = b + 11$ em (.17) obtemos que $b = 45$. Usando que $b = 45$ em (.16) e (.17), vemos que $c = 56$ e $d = 71$. Finalmente, obtemos que $a = 41$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26790,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA figura abaixo contém um quadrado e dois triângulos retângulos congruentes.\n![](attached_image_1.png)\nCom esses polígonos formamos um retângulo e um trapézio como mostra a figura seguinte:\n![](attached_image_2.png)\nSabendo que o perímetro do retângulo é 58, e que o perímetro do trapézio é 60, calcule o lado do quadrado.\nObservação: Um triângulo é dito retângulo se um dos seus ângulos mede $90^{\\circ}$. Dois triângulos são ditos congruentes quando os dois possuem lados com os mesmos comprimentos.",
    "options": [],
    "answer": "12",
    "solution": "Solution:\nFica claro pela figura no enunciado desse exercício que o comprimento do lado do quadrado deve coincidir com o comprimento de algum dos catetos dos triângulos. Chamemos de $a$ tal comprimento. Além disso, chamemos de $b$ o comprimento da hipotenusa e de $x$ o comprimento do outro cateto. Pela figura abaixo, vemos que o perímetro do trapézio é dado por $2a + 2b + 2x$.\n\n![](attached_image_3.png)\n\nEntão, para que o perímetro do trapézio seja igual a 60, é necessário que a seguinte equação seja satisfeita:\n$$\n2a + 2b + 2x = 60\n$$\nDaí, concluímos que\n$$\nx = \\frac{60 - 2a - 2b}{2} = 30 - a - b\n$$\n\nNote que o perímetro do retângulo é dado por $4a + 2x$, como podemos observar abaixo.\n![](attached_image_4.png)\n\nLogo, utilizando a expressão para $x$ obtida acima, obtemos que esse perímetro é igual a:\n$$\n2(30 - a - b) + 4a\n$$\nquantidade que deve coincidir com 58 pelo dado do problema. Logo, a seguinte equação deve ser satisfeita:\n$$\n2(30 - a - b) + 4a = 58\n$$\nSimplificando a equação acima obtemos que $b = a + 1$. Substituindo essa expressão para $b$ na expressão $x = 30 - a - b$, obtemos que $x = 29 - 2a$.\nAssim, os comprimentos dos catetos de cada triângulo resultam ser $a$ e $29 - 2a$, enquanto o comprimento da hipotenusa resulta ser $b = a + 1$, como ilustramos na figura a seguir:\n\n![](attached_image_5.png)\n\nUsando o Teorema de Pitágoras, temos a equação\n$$\na^{2} + (29 - 2a)^{2} = (a + 1)^{2}\n$$\nUma simplificação nos fornece que $a$ satisfaz\n$$\n4a^{2} - 118a + 840 = 0\n$$\nFatorando o polinômio em $a$ que aparece do lado esquerdo, podemos reescrever a equação acima como\n$$\n2(a - 12)(2a - 35) = 0\n$$\nLogo, obtemos como soluções da equação de segundo grau, os números 12 e $35/2$. Mas se $a$ fosse igual a $35/2$, o comprimento $x = 29 - 2a$ seria negativo. Portanto, a única solução válida é $a = 12$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26791,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSobre um tabuleiro de $5 \\times 5$ casas foram distribuídos os números $1,2,3, \\ldots, 25$ de modo tal que cada casa seja ocupada por um único número e que dois números consecutivos sempre estejam colocados em casas vizinhas. A figura a seguir mostra um exemplo de como distribuir esses números.\n\n| 3 | 4 | 5 | 24 | 23 |\n| :---: | :---: | :---: | :---: | :---: |\n| 2 | 1 | 6 | 25 | 22 |\n| 9 | 8 | 7 | 20 | 21 |\n| 10 | 13 | 14 | 19 | 18 |\n| 11 | 12 | 15 | 16 | 17 |\n\na) Mostre que, em qualquer distribuição cumprindo tal condição, todos os números colocados nas casas das duas diagonais são ímpares.\n\nVamos chamar de diagonal principal a diagonal do tabuleiro pintada em branco na figura seguinte.\n\n![](attached_image_1.png)\n\nNote que a diagonal principal divide o tabuleiro em duas zonas, uma superior e outra inferior.\n\nb) Sejam $x$ e $y$ dois números em $\\{1,2, \\ldots, 25\\}$ tais que $x<y$. Mostre que se $x$ e $y$ pertencem a zonas distintas, então existe um número $z$ situado na diagonal principal tal que $x \\leq z \\leq y$.\n\nc) Determine qual é o menor valor que pode ser assumido pela soma dos números que são colocados sobre a diagonal principal quando distribuem-se números sobre o tabuleiro, respeitando a condição do problema.",
    "options": [],
    "answer": "33",
    "solution": "Solution:\n\na) Vamos colorir as casas do tabuleiro de branco e preto como indica a figura seguinte. Desse modo, as casas correspondentes a dois números consecutivos estarão sempre pintadas de cores distintas.\n\n![](attached_image_2.png)\n\nEm particular, se a casa com o número $1$ está pintada com a cor $A$ (branca ou preta), então a casa do número $2$ estará pintada com a outra cor, digamos $B$. Agora, $2$ e $3$ são também consecutivos. Então a cor da casa com o número $3$ deverá ser distinta de $B$ e será, assim, a cor $A$. Continuando com o mesmo raciocínio, podemos deduzir a cor da casa de qualquer número até o $25$. Concluímos assim que, ou as casas com os números ímpares $\\{1,3,5, \\ldots, 25\\}$ são todas pretas, e aquelas com os números pares $\\{2,4,6, \\ldots, 24\\}$ são todas brancas, ou vice-versa.\n\nMas, no conjunto $\\{1,2,3, \\ldots, 24,25\\}$, existem $13$ números ímpares e $12$ pares, enquanto que no tabuleiro existem $13$ casas pretas e $12$ casas brancas. Então, a única possibilidade é que as $13$ casas pretas sejam ocupadas pelos $13$ números ímpares. Em particular, como as casas nas diagonais do tabuleiro são todas pretas, então os números situados nessas casas são todos ímpares.\n\nb) Sejam $x, y \\in\\{1,2, \\ldots, 25\\}$, com $x<y$, os números do enunciado. Chamemos de zona $1$ aquela contendo $x$ e de zona $2$ aquela contendo $y$. Seja $A$ o conjunto dos números em $\\{x, x+1, \\ldots, y\\}$ que pertencem à zona $2$. Então $A$ é não vazio, já que $y \\in A$. Por isso, podemos escolher o menor número em $A$ e chamá-lo de $w$. Note que $x<w$, já que $x$ pertence à zona $1$. Então $(w-1) \\in\\{x, x+1, \\ldots, y\\}$. Como $w$ é o menor número em $A$, necessariamente temos que $(w-1) \\in\\{x, x+1, \\ldots, y\\}$ mas $(w-1) \\notin A$. Isso significa que\n$$\n(w-1) \\text{ não pode pertencer à zona } 2.\n$$\nObserve agora que duas casas em zonas distintas nunca são vizinhas. No entanto, $(w-1)$ e $w$ devem pertencer a casas vizinhas. Como $w$ pertence à zona $2$, concluímos por (.18), que $(w-1)$ tem que pertencer à diagonal. Portanto, $z=(w-1) \\in\\{x, x+1, \\ldots, y\\}$ é o número que procurávamos.\n\nc) Sejam $a<b<c<d<e$ os números sobre a diagonal principal. Observemos primeiro que os números maiores do que $e$ estarão situados em uma mesma zona. Caso contrário, pelo mostrado na parte b), existiria um número maior do que $e$ na diagonal principal e isso não é possível por ser $e$ o maior número nessa diagonal.\n\nEm particular, como em cada zona há $10$ casas, a quantidade de números maiores do que $e$ não pode ser superior a $10$. Logo, $e$ não pode ser menor do que $15$. Se fosse $e=15$, os $10$ números $\\{16,17, \\ldots, 25\\}$ preencheriam as $10$ casas de alguma zona. Foi observado no item a) que os números colocados em casas da mesma cor (preta ou branca) devem ter a mesma paridade. Mas, será impossível que os $5$ números pares e os $5$ números ímpares em $\\{16,17, \\ldots, 25\\}$ estejam situados nas $6$ casas brancas e $4$ pretas de uma zona. Isto mostra que $e$ não pode ser $15$. Concluímos assim que $15<e \\leq 25$.\n\nPelo item a), os números $\\{a, b, c, d, e\\}$ sobre a diagonal principal devem ser ímpares. Temos então as seguintes desigualdades:\n$$\na \\geq 1, \\quad b \\geq 3, \\quad c \\geq 5, \\quad d \\geq 7, \\quad e \\geq 17\n$$\nSomando essas desigualdades, vemos que $a+b+c+d+e \\geq 33$. Assim, se conseguirmos um exemplo com soma $33$ na diagonal principal, teremos provado que $33$ é de fato a soma mínima.\n\nOs números na diagonal devem ser necessariamente $\\{1,3,5,7,17\\}$. Além disso, os números $1$ e $3$ devem estar em casas que compartilham um vértice, porque, caso contrário, não seria possível colocar o $2$. O mesmo argumento funciona para os pares $\\{3,5\\}$ e $\\{5,7\\}$. Decorre daí que $17$ pode estar somente em uma das duas esquinas da diagonal principal. Temos essencialmente duas possibilidades:\n\n![](attached_image_3.png)\n\nAs outras duas são análogas. Aliás, sabemos que os números maiores que $17$ pertencem a uma mesma zona. Usando um argumento similar, podemos concluir que os $9$ números maiores que $7$ e menores que $17$ devem estar também em uma mesma zona. Claramente, essa zona deve ser distinta à zona dos números maiores do que $17$. Guiados por esses critérios, conseguimos formar o seguinte exemplo:\n\n| 17 | 16 | 15 | 14 | 13 |\n| :---: | :---: | :---: | :---: | :---: |\n| 18 | 1 | 2 | 11 | 12 |\n| 19 | 20 | 3 | 10 | 9 |\n| 22 | 21 | 4 | 5 | 8 |\n| 23 | 24 | 25 | 6 | 7 |\n\nA soma dos números na diagonal principal para esse exemplo é $17+1+3+5+7=33$. Como já foi provado acima que a soma deve ser maior ou igual do que $33$, concluímos que a soma mínima é $33$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26792,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSobre um tabuleiro de $n \\times n$, Chris planeja desenhar três O's. Ele deseja fazê-lo de modo que cada $\\mathbf{O}$ fique dentro de um quadradinho e de modo que os três quadradinhos utilizados estejam dispostos no formato de um L. A seguinte figura mostra alguns exemplos com $n=4$.\n\n![](attached_image_1.png)\n\nNosso objetivo será sabotar o plano de Chris! Chegaremos antes que ele e marcaremos alguns dos quadradinhos com um $\\mathbf{X}$, não permitindo assim que ele desenhe os seus O's dentro desses quadradinhos. Por exemplo, em um tabuleiro de $3 \\times 3$ podemos marcar cinco quadradinhos como mostra a figura seguinte.\n\n![](attached_image_2.png)\n\nVemos que assim não há uma maneira pela qual Chris possa cumprir com seu plano.\n\na) Para $n=2$, qual é a quantidade mínima de quadradinhos que devemos marcar com um $\\mathbf{X}$ para sabotar o plano de Chris?\n\nb) Para $n=3$, encontre um modo de sabotar o plano de Chris marcando exatamente quatro quadradinhos com um $\\mathbf{X}$.\n\nc) Para $n=4$, encontre um modo de sabotar o plano de Chris marcando exatamente oito quadradinhos.\n\nd) Para $n=4$, mostre que é impossível sabotar o plano de Chris marcando apenas sete quadradinhos com um X. (Sugestão: use o item a)).",
    "options": [],
    "answer": "a) 2. b) Possible with exactly 4 marked squares (example exists). c) Possible with exactly 8 marked squares (example exists). d) Impossible with only 7 marked squares; at least 8 are needed.",
    "solution": "Solution:\n\na) Se marcamos somente um dos quadradinhos do tabuleiro com um $\\mathbf{X}$, então os três quadradinhos que restarão formarão sempre um L. Chris poderia então escrever um $\\mathbf{O}$ em cada um dos quadradinhos atingindo assim o seu objetivo. Assim, marcando um quadradinho apenas, não seremos capazes de sabotar o plano de Chris. No entanto, marcando dois quadradinhos é possível sabotar os planos de Chris. De fato, marcando dois quadradinhos quaisquer, restariam apenas dois quadradinhos para que Chris desenhe, logo será impossível que ele desenhe um L.\n\nb) Sobre o tabuleiro $3 \\times 3$, podemos marcar com um $\\mathbf{X}$ quatro quadradinhos como na figura abaixo:\n\n![](attached_image_3.png)\n\nNão deixaremos assim nenhuma configuração de quadradinhos em forma de L para que Chris escreva!\n\nc) Sobre o tabuleiro de $4 \\times 4$, podemos marcar oito quadradinhos como mostrado na figura abaixo:\n\n![](attached_image_4.png)\n\nConseguimos assim cumprir o nosso propósito de sabotar os planos de Chris.\n\nd) Dividamos o tabuleiro em quatro tabuleiros $2 \\times 2$ como mostra a figura seguinte:\n\n![](attached_image_5.png)\n\nPelo que foi observado no item a), em cada um dos tabuleiros $2 \\times 2$ é necessário marcar, no mínimo, dois quadradinhos para não deixar a Chris três quadradinhos dispostos em L. Como há quatro tabuleiros $2 \\times 2$, será necessário marcar pelo menos $4 \\times 2$ quadradinhos sobre o tabuleiro. Logo, será impossível sabotar os planos de Chris marcando apenas sete quadradinhos.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26793,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSobre um quadro negro, Kiara escreveu 20 números inteiros, todos eles diferentes de zero. Em seguida, para cada par de números escritos por Kiara, Yndira escreveu sobre o mesmo quadro o respectivo produto entre eles (inclusive, se o resultado de algum produto já estava escrito, Yndira o repetiu).\n\nPor exemplo, caso os números $2$, $3$, $4$ e $6$ estivessem entre aqueles escritos por Kiara, então Yndira teria escrito os números $6, 8, 12, 12, 18$ e $24$, pois temos que $6=2 \\times 3$, $8=2 \\times 4$, $12=2 \\times 6=3 \\times 4$, $18=3 \\times 6$ e $24=4 \\times 6$. Note que o número $6$ teria sido escrito novamente mesmo já tendo sido escrito por Kiara, enquanto que o número $12$ teria sido escrito duas vezes por Yndira.\n\na) No total, quantos números foram escritos por Kiara e Yndira sobre o quadro negro?\n\nb) Suponhamos que, do total de números escritos sobre o quadro, exatamente $120$ são positivos. Se Kiara escreveu mais números positivos do que negativos, diga quantos dos números escritos por Kiara eram positivos.",
    "options": [],
    "answer": "a) 210; b) 14",
    "solution": "Solution:\n\na) A quantidade de números escritos por Yndira é igual ao número de pares que podem ser formados com os $20$ números escritos por Kiara.\n\nDe quantas maneiras podemos escolher um par de elementos em um conjunto contendo $20$ elementos? Para a escolha do primeiro elemento temos $20$ possibilidades. Para cada escolha fixa do primeiro elemento, nos restam $19$ possibilidades para a escolha do segundo elemento. Assim temos até o instante $20 \\times 19$ escolhas. Como não interessa a ordem (escolher o elemento $A$ e depois o elemento $B$ é o mesmo que escolher o elemento $B$ e depois o elemento $A$), na contagem $20 \\times 19$, contamos cada par duas vezes. Logo, devemos dividir $20 \\times 19$ por $2$ para obter o número de pares. Resumindo, o número de pares que podem ser escolhidos dentre $20$ elementos é\n$$\n\\frac{20 \\times 19}{2} = 190\n$$\nAssim, a quantidade de números escritos por Yndira foi $190$. Como Kiara já havia escrito $20$ números no quadro, temos então um total $20 + 190 = 210$ números escritos no quadro-negro.\n\nb) Se no total de números escritos no quadro, há $120$ números positivos, então, pela parte a), há $210 - 120 = 90$ números negativos escritos. Seja $p$ a quantidade de números positivos escritos por Kiara. Então a quantidade de números negativos escritos por ela é igual a $20 - p$. Para Yndira obter um produto negativo, ela precisa multiplicar um dos $20 - p$ números negativos por um dos $p$ números positivos. Então, a quantidade de números negativos escritos por Yndira é igual a $(20 - p)p$. Assim, a quantidade total de números negativos escritos no quadro-negro é igual a $(20 - p) + (20 - p)p = (20 - p)(p + 1)$. Logo, vale que $(20 - p)(p + 1) = 90$, o que pode ser reescrito como:\n$$\np^2 - 19p + 70 = 0\n$$\nResolvendo essa equação do segundo grau, encontramos que os únicos valores de $p$ que satisfazem essa última equação são $p = 5$ e $p = 14$. Mas como Kiara escreveu mais números positivos que negativos, então devemos ter que $p > 10$. Logo vale que $p = 14$, isto é, Kiara escreveu $14$ números positivos sobre o quadro.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26794,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm número inteiro positivo é chamado ziguezague, se satisfaz as seguintes três condições:\n- Seus algarismos são não nulos e distintos.\n- Não possui três algarismos consecutivos em ordem crescente.\n- Não possui três algarismos consecutivos em ordem decrescente.\nPor exemplo, $14385$ e $2917$ são ziguezague, mas $2564$ e $71544$ não.\n\na) Encontre o maior número ziguezague.\n\nb) Quantos números ziguezague de quatro algarismos existem?",
    "options": [],
    "answer": "a) 978563412; b) 1260",
    "solution": "Solution:\n\na) O número $978563412$ é, claramente, um número ziguezague. Vamos mostrar que, se $N$ é um número ziguezague, então\n$$\nN \\leq 978563412\n$$\nA maior quantidade de algarismos que um número ziguezague pode possuir é igual a $9$. Portanto, podemos supor que $N$ é da forma\n$$\nN=\\overline{a b c d e f g h i}\n$$\nja que, de outro modo, a desigualdade já estaria verificada. Aliás, $a=9$ porque, de outro modo, também já teríamos a desigualdade satisfeita. Se $b$ fosse igual a $8$, o próximo algarismo $c$ seria necessariamente menor que $8$ e $N$ não seria ziguezague. Portanto, temos que $b<8$. Se $b$ não fosse igual a $7$, já teríamos a desigualdade verificada. Podemos então supor que $b=7$. Logo, $c$ só pode valer $8$. Podemos então supor que $N$ é da forma\n$$\nN=\\overline{978 d e f g h i}\n$$\nO número $d$ não pode ser $6$, porque isso implicaria que $e<6$ e, assim, $N$ não seria um número ziguezague. Portanto, $d<6$ e podemos supor que $d=5$, caso contrário a desigualdade estaria verificada. Necessariamente $e=6$. Podemos continuar o argumento e chegar à conclusão que se $N$ não fosse igual a $978563412$ então seria, necessariamente, menor.\n\nb) Para produzir um número ziguezague de $4$ algarismos podemos usar o seguinte procedimento:\n\n(1) Escolhemos de $\\{1,2,3, \\ldots, 9\\}$ um subconjunto $\\{a, b, c, d\\}$ de $4$ números distintos. Digamos que $a<b<c<d$.\n\n(2) Podemos usar os algarismos $\\{a, b, c, d\\}$ para formar $10$ números ziguezague distintos:\n$$\n\\begin{aligned}\n& \\overline{a c b d}, \\overline{a d b c}, \\overline{b d a c}, \\overline{b c a d}, \\overline{c d a b} \\\\\n& \\overline{d b c a}, \\overline{d a c b}, \\overline{c a d b}, \\overline{c b d a}, \\text{ e } \\overline{b a d c}\n\\end{aligned}\n$$\nTodo número ziguezague de $4$ algarismos pode ser produzido por esse procedimento e escolhas distintas em cada passo do procedimento produzem números ziguezague distintos. Agora podemos contá-los. Para o primeiro passo do procedimento temos que escolher uma entre $\\binom{9}{4}$ escolhas possíveis, e para o último passo temos $10$ escolhas. Portanto, existem\n$$\n10 \\times \\binom{9}{4} = 10 \\times \\frac{9!}{5!4!} = 1260\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26795,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA figura mostra um retângulo $ABCD$ de lado $\\overline{AD}=4$ e onde o ponto $M$ é o ponto médio do segmento $AB$.\n![](attached_image_1.png)\nSabendo que os segmentos $AC$ e $DM$ são ortogonais, calcule a área do retângulo $ABCD$.",
    "options": [],
    "answer": "16√2",
    "solution": "Solution:\nChamemos de $P$ o ponto de interseção dos segmentos $AC$ e $DM$. Como a soma dos ângulos internos do triângulo $APM$ é $180^\\circ$ então\n$$\n\\measuredangle MAP = 90^\\circ - \\measuredangle AMP\n$$\nMas $\\measuredangle MAP + \\measuredangle PAD = 90^\\circ$. Então, concluímos que $\\measuredangle PAD = \\measuredangle AMP$. Acabamos de mostrar assim que os triângulos retângulos $MAD$ e $ADC$ possuem os mesmos ângulos internos. Eles são portanto semelhantes.\n![](attached_image_2.png)\nAplicando a semelhança entre esses triângulos temos:\n$$\n\\frac{\\overline{AD}}{\\overline{DC}} = \\frac{\\overline{MA}}{\\overline{AD}}\n$$\nSabemos que $\\overline{AD} = 4$ e que $\\overline{DC} = 2 \\times \\overline{MA}$. Usando esses fatos em (.21) obtemos $\\overline{MA} = 2\\sqrt{2}$ e portanto $\\overline{DC} = 4\\sqrt{2}$. Finalmente, a área do retângulo $ABCD$ resulta $\\overline{AD} \\times \\overline{DC} = 16\\sqrt{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26796,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOs 50 primeiros números naturais atravessarão um corredor que contém 50 portas numeradas de 1 a 50, todas elas inicialmente trancadas. O primeiro a atravessar será o número $1$, o segundo será o número $2$, em seguida o número $3$ e assim por diante, até o número $50$ que será o último a atravessar. Ao atravessar o corredor, o número $n$ carregará consigo as chaves das portas numeradas com múltiplo de $n$. Assim, por exemplo, o número $1$ carregará as chaves de todas as portas, enquanto que o número $2$ carregará somente as chaves das portas com numeração par e o número $25$ carregará somente as chaves das portas numeradas com $25$ e $50$. Durante o seu percurso, cada número usa as chaves que possui para trancar as portas que estiverem abertas e destrancar aquelas que estiverem fechadas.\n\na) Quais serão as portas destrancadas pelo número $15$?\n\nb) Mostre que, depois do número $50$ ter percorrido o corredor, a porta de número $10$ ficará destrancada enquanto que a porta de número $9$ ficará trancada.\n\nc) Depois do número $50$ ter percorrido o corredor, quais serão as portas destrancadas?",
    "options": [],
    "answer": "a) 30 and 45. b) Door 10 ends locked and door 9 ends unlocked. c) Unlocked doors: 1, 4, 9, 16, 25, 36, 49.",
    "solution": "Solution:\n\na) Como os múltiplos de $15$ que são menores que $50$ são os números $15$, $30$ e $45$, temos que o número $15$ carrega apenas as chaves para as portas numeradas com $15$, $30$ e $45$. Vamos analisar cada uma dessas três portas separadamente.\n- Porta $15$: Os divisores de $15$ são os números $1$, $3$, $5$ e $15$. Assim, esses são os números que modificarão o estado dessa porta. O número $1$ vai destrancá-la, enquanto o número $3$ vai trancá-la novamente. Em seguida o número $5$ vai destrancá-la e finalmente o número $15$ vai trancá-la. Concluímos que essa porta não será destrancada pelo número $15$.\n- Porta $30$: Os divisores de $30$ são os números $1$, $2$, $3$, $5$, $6$, $10$, $15$ e $30$. Assim, o número $15$ será o sétimo número a modificar o estado dessa porta. Como a porta começa trancada e sete é um número ímpar, temos que o número $15$ irá destrancar essa porta.\n- Porta $45$: Os divisores de $45$ são os números $1$, $3$, $5$, $9$, $15$ e $45$. Assim, o número $15$ será o quinto número a modificar o estado dessa porta, portanto ele irá destrancá-la.\nConcluímos assim que o número $15$ irá, de fato, destrancar as portas numeradas com o $30$ e o $45$.\n\nb) Os divisores de $10$ são os números $1$, $2$, $5$ e $10$ assim, ao atravessarem o corredor, esses são os únicos números que podem trancar ou destrancar a porta de número $10$. Segue daí que essa porta terá o seu estado modificado exatamente quatro vezes. Como ela começa trancada, ela acabará também trancada.\nOs divisores de $9$ são os números $1$, $3$ e $9$. Assim a porta de número $9$ terá o seu estado modificado exatamente três vezes. Como ela começa trancada, ela acabará destrancada.\nObservação: Seja $k$ a numeração de uma das portas. Da solução do item acima, podemos concluir que, se $k$ possui uma quantidade par de divisores, então essa porta acabará trancada. Por outro lado, se $k$ possui uma quantidade ímpar de divisores, então essa porta acabará destrancada.\n\nc) Para cada $k \\in\\{1,2, \\ldots, 50\\}$ denotemos por $D(k)$ o número de divisores de $k$. Pelo enunciado do problema sabemos que existem exatamente $D(k)$ números naturais que carregam a chave da porta número $k$. Decorre daí que a porta número $k$ terá o seu estado alterado exatamente $D(k)$ vezes. Portanto as portas que acabarão abertas serão as portas de número $k$ tais que $D(k)$ é ímpar.\nFixemos um $k$ número natural e escrevamos sua decomposição em fatores primos:\n$$\nk=p_{1}^{\\alpha_{1}} \\times p_{2}^{\\alpha_{2}} \\times \\cdots \\times p_{m}^{\\alpha_{m}}\n$$\nQualquer divisor de $k$ deve ser um elemento da forma:\n$$\np_{1}^{\\beta_{1}} \\times p_{2}^{\\beta_{2}} \\times \\cdots \\times p_{m}^{\\beta_{m}}\n$$\nonde cada $\\beta_{i}$ é um número inteiro escolhido em $\\left\\{0,1, \\ldots, \\alpha_{i}\\right\\}$. Então $D(k)$ é constituído pelos números dessa forma.\nNote que, a cada escolha dos expoentes $\\beta_{1}, \\beta_{2}, \\ldots, \\beta_{m}$, corresponde um único elemento de $D(K)$, ou seja, um único divisor de $k$. Podemos contar então o número de maneiras de se escolherem expoentes $\\beta_{1}, \\beta_{2}, \\ldots, \\beta_{m}$. Mas como cada $\\beta_{i}$ deve pertencer ao conjunto $\\left\\{0,1, \\ldots, \\alpha_{i}\\right\\}$, temos que a quantidade total de escolhas para o valor de $\\beta_{i}$ é igual a $\\alpha_{i}+1$. Logo, existem\n$$\n\\left(1+\\alpha_{1}\\right)\\left(1+\\alpha_{2}\\right) \\cdots\\left(1+\\alpha_{m}\\right)\n$$\nescolhas para os valores dos expoentes $\\beta_{1}, \\beta_{2}, \\ldots, \\beta_{m}$ e, como consequência,\n$$\n\\left(1+\\alpha_{1}\\right)\\left(1+\\alpha_{2}\\right) \\cdots\\left(1+\\alpha_{m}\\right)\n$$\ndivisores do número $k$.\nPara que $k$ tenha uma quantidade ímpar de divisores, é suficiente e necessário que cada um dos fatores $\\alpha_{i}+1$ seja um número ímpar, ou seja, que $\\alpha_{i}$ seja par.\nIsso mostra que $k$ possui um número ímpar de divisores se, e somente se, $k$ é um quadrado perfeito. Portanto as portas que ficam abertas são precisamente aquelas com números:\n$$\n1, \\quad 4, \\quad 9, \\quad 16, \\quad 25,36 \\text { e } 49\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26797,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUma folha de papel é retangular, com base igual a $20~\\mathrm{cm}$ e altura $10~\\mathrm{cm}$. Esta folha é dobrada nas linhas pontilhadas conforme a figura abaixo, e no final recortada por uma tesoura na linha indicada, a qual é paralela à base e está na metade da altura do triângulo.\n![](attached_image_1.png)\n\na) Depois de cortar no local indicado, em quantas partes a folha ficou dividida?\n\nb) Qual a área da maior parte?",
    "options": [],
    "answer": "a) 3 parts; b) 150 cm^2",
    "solution": "Solution:\n\na) Vamos marcar a linha cortada pela tesoura em cinza, e fazer o processo inverso, que corresponde a abrir a folha depois de cortada:\n![](attached_image_2.png)\nLogo, a folha foi dividida em três pedaços.\n\nb) Como se pode observar, os dois quadrados recortados nos cantos superior esquerdo e inferior direito têm lado igual a $5~\\mathrm{cm}$. Como a área de um quadrado é o lado ao quadrado, a área de cada quadrado é igual $5 \\times 5 = 25~\\mathrm{cm}^2$. A folha é um retângulo de base $20~\\mathrm{cm}$ e altura $10~\\mathrm{cm}$. Logo, como a área de um retângulo é base vezes altura, a área da folha é de $20 \\times 10 = 200~\\mathrm{cm}^2$. Subtraindo a área total pela área dos dois quadrados nos cantos, concluímos que a área do pedaço maior da folha após o corte pela tesoura é\n$$\n200 - 2 \\times 25 = 200 - 50 = 150~\\mathrm{cm}^2\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26798,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm hospital tem os seguintes funcionários:\nSara Dores da Costa: reumatologista\nIná Lemos: pneumologista\nEster Elisa: enfermeira\nEma Thomas: traumatologista\nAna Lisa: psicanalista\nInácio Filho: obstetra\n\na) De quantas maneiras os funcionários podem fazer uma fila?\n\nb) De quantas maneiras os mesmos funcionários podem sentar numa mesa redonda? Lembre-se que, numa mesa redonda, se todos se mudam para a cadeira da esquerda, a mesa continua igual!\n\nc) E de quantas maneiras os funcionários podem compor uma comissão formada por presidente, vice-presidente e suplente?",
    "options": [],
    "answer": "a) 720; b) 120; c) 120",
    "solution": "Solution:\n\na) Para ser o primeiro da fila, podemos escolher qualquer um dos seis funcionários. Logo, há 6 possibilidades. Escolhido o primeiro da fila, restam cinco funcionários a serem escolhidos para ser o segundo da fila (porque um já foi escolhido). Para o terceiro lugar temos 4 possibilidades, e assim por diante. Logo, há\n$$\n6 \\times 5 \\times 4 \\times 3 \\times 2 \\times 1 = 720\n$$\nmaneiras distintas de organizar a fila.\n\nb) Numa mesa redonda, só importa a posição relativa, como foi dito no enunciado. Se movermos cada funcionário para a cadeira à sua esquerda, a mesa continua igual. Quantas vezes podemos fazer isso? De 6, já que são seis funcionários. Se numerarmos as cadeiras, teríamos a mesma resposta do item anterior, 720. Como podemos girar as pessoas para a cadeira ao lado 6 vezes, concluímos que o número de maneiras de colocar os funcionários na mesa é $720 / 6 = 120$.\n\nc) Podemos escolher o presidente de 6 maneiras, já que são 6 funcionários. Para cada uma dessas escolhas, teremos 5 possibilidades para escolher o vice. E para uma dessas, 4 para o suplente. Logo, são $6 \\times 5 \\times 4 = 120$ maneiras de escolher a comissão.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26799,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSergio pediu para Ivan pensar em um número inteiro positivo. Depois, pediu para Ivan calcular a soma de seus algarismos e, finalmente, elevar ao quadrado o resultado. Sem falar o número em que pensou inicialmente, Ivan contou que obteve como resultado final $x$. Mostre a Sergio como chegar às seguintes conclusões:\n\na) Se Ivan tivesse pensado em um número com 3 ou menos algarismos, então $x$ seria menor do que 730.\n\nb) Se Ivan tivesse pensado em um número com 4 algarismos, então $x$ seria menor do que o número no qual Ivan pensou.\n\nc) Se Ivan tivesse pensado em um número com 5 ou mais algarismos, então $x$ seria menor do que o número que Ivan pensou.\n\nSergio fez depois o seguinte: Considerou o número $x$ que Ivan disse, calculou a soma dos seus algarismos e elevou ao quadrado o resultado. Quando Sergio falou para Ivan o número que obteve, Ivan disse com surpresa que esse foi o número que havia pensado.\n\nd) Determine todos os possíveis valores para o número que Ivan pensou.",
    "options": [],
    "answer": "1, 81, 169, 256",
    "solution": "Solution:\n\na) Se Ivan tivesse pensado em um número com 3 ou menos algarismos, a soma de seus algarismos seria no máximo $9+9+9=27$. Então o número final de Ivan $x$ seria no máximo $27^{2}=729$.\n\nb) Se Ivan tivesse pensado em um número com 4 algarismos, digamos $\\overline{a b c d}$, então $x=(a+b+c+d)^{2}$. Distinguimos duas possibilidades. Primeiro, se $a=1$, então\n$$\nx \\leq (1+9+9+9)^{2}=784<\\overline{a b c d}\n$$\nAgora, se $a \\geq 2$, então\n$$\nx \\leq (9+9+9+9)^{2}=1296<\\overline{a b c d}\n$$\nEm qualquer um dos casos, $x<\\overline{a b c d}$.\n\nc) Suponhamos que Ivan pensou em um número com $n \\geq 5$ algarismos, digamos $\\overline{a_{1} a_{2} \\ldots a_{n}}$. Então, $x=\\left(a_{1}+a_{2}+\\cdots+a_{n}\\right)^{2}$. Logo,\n$$\nx \\leq (9 n)^{2}=81 n^{2}\n$$\nObserve que\n$$\n10^{n-1} \\leq \\overline{a_{1} a_{2} \\ldots a_{n}}\n$$\nEntão, para mostrar que $x<\\overline{a_{1} a_{2} \\ldots a_{n}}$ bastaria mostrar que\n$$\n81 n^{2}<10^{n-1}, \\quad \\text{ para todo inteiro } n \\geq 5\n$$\ne usar (.22) junto com (.23). Para mostrar (.24), procederemos por indução. Para $n=5$, temos\n$$\n81 \\cdot 5^{2}=2025<10000=10^{5-1}\n$$\nSuponhamos então que essa desigualdade é válida para algum inteiro $k \\geq 5$, ou seja,\n$$\n81 k^{2}<10^{k-1}\n$$\nVamos mostrar que ela é válida para $k+1$. Observemos que, por ser $k \\geq 5$, então\n$$\n81(k+1)^{2}<81(2 k)^{2}=4\\left(81 k^{2}\\right)\n$$\nUsando a hipótese de indução (.25) vemos que\n$$\n4\\left(81 k^{2}\\right)<4\\left(10^{k-1}\\right)\n$$\nComo $4<10$ vemos que a última expressão em (.27) é menor do que $10 \\times 10^{k-1}=10^{k}$. De (.26), (.27) e a última observação obtemos que\n$$\n81(k+1)^{2}<10^{(k+1)-1}\n$$\nIsto conclui a prova por indução e (.24) é, portanto, verdadeiro.\n\nd) Seja $I$ o número pensado por Ivan. Vamos mostrar que $I<730$. Suponhamos que não seja assim, isto é que $I \\geq 730$. Como $I$ é o número final obtido por Sergio a partir de $x$, podemos aplicar a parte a) (para Sergio em lugar de Ivan) e concluir que $x$ tem 4 ou mais algarismos. Logo, usando as partes b) e c) para Sergio concluímos que $x>I$. Em particular, $x \\geq 730$. Pela parte a), sabemos que então $I$ deve ter 4 ou mais algarismos. Logo, usando as partes b) e c) concluímos que $I>x$. Chegamos assim a uma contradição. Fica provado então que $I<730$.\n\nAlém disso, como $I$ é o número final de Sergio, então $I$ é um quadrado perfeito. Finalmente, basta verificar quais dos valores em\n$$\n1^{2}, \\quad 2^{2}, \\quad 3^{2}, \\quad \\ldots, \\quad 26^{2} \\text{ e } \\quad 27^{2}=729\n$$\npode $I$ tomar. Depois dessa verificação vemos que os possíveis valores para o número que Ivan pensou são 1, 81, 169 e 256.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26800,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa cidade de Cordisburgo, foi realizada uma corrida de bicicleta num circuito circular, da qual participaram três ciclistas, Guimarães, Rosa e João. Na primeira hora da corrida, Guimarães fez exatamente 230 voltas completas, João fez exatamente 111 voltas completas, porém não se sabe quantas voltas Rosa realizou, sabe-se apenas que foi um número inteiro e que Rosa deu mais voltas que João e menos do que Guimarães. Além disso, cada um deles andou com velocidade constante, e todos partiram juntos do mesmo ponto. Considerando também as ultrapassagens feitas no tempo inicial, quantas ultrapassagens no total foram feitas nessa primeira hora de corrida?",
    "options": [],
    "answer": "238",
    "solution": "Solution:\n\nSeja $x$ o número de voltas que Rosa realizou nessa primeira hora de corrida. Como Guimarães deu mais voltas que Rosa, e ambos andaram com velocidade constante, o número de vezes que Guimarães ultrapassou Rosa foi\n$$\n230 - x\n$$\nComo Rosa deu mais voltas que João, e ambos andaram com velocidade constante, o número de vezes que Rosa ultrapassou João foi\n$$\nx - 111\n$$\nComo Guimarães deu mais voltas que João, e ambos andaram com velocidade constante, o número de vezes que Guimarães ultrapassou João foi\n$$\n230 - 111\n$$\nLogo, o número total de ultrapassagens nessa primeira hora de corrida foi\n$$\n\\begin{aligned}\n(230 - x) + (x - 111) + (230 - 111) & = 230 - 111 + 230 - 111 \\\\\n& = 238\n\\end{aligned}\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26801,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nPedro escreveu a lista de todos os números inteiros positivos menores que $10000$ nos quais cada um dos algarismos $1$ e $2$ aparecem uma única vez. Por exemplo, $1234$, $231$, $102$ foram escritos na lista, mas $1102$ e $235$ não estão na lista. Quantos números há na lista escrita por Pedro?",
    "options": [],
    "answer": "768",
    "solution": "Solution:\nNotemos que um número natural menor do que $10000$ pode ser representado por exatamente quatro algarismos escolhidos em $\\{0,1,2,3,4,5,6,7,8,9\\}$, possivelmente com repetições. Assim, temos quatro posições para serem preenchidas com esses algarismos. Por exemplo, o número $12$ seria representado por $0012$, isto é, o algarismo $0$ foi escolhido para preencher a primeira e a segunda posição, o algarismo $1$ foi escolhido para a terceira e o algarismo $2$ foi escolhido para a quarta.\n\nOs números da lista de Pedro devem conter, obrigatoriamente, os dígitos $1$ e $2$. Assim, para formar um número da lista de Pedro podemos seguir o seguinte procedimento:\n\n1. Escolhemos a posição do algarismo $1$ dentre as quatro possíveis.\n2. Escolhemos a posição do algarismo $2$ dentre as três que restam.\n3. Preenchemos cada uma das duas posições restantes com um dos oito algarismos escolhidos no conjunto $\\{0,3,4,5,6,7,8,9\\}$, podendo haver repetição.\n\nNote que qualquer número da lista de Pedro é obtido desse modo e, para que dois procedimentos resultem no mesmo número, é necessário que as escolhas em cada passo coincidam. Logo, para contar a quantidade de números presentes na lista, basta contar a quantidade de escolhas possíveis nesse procedimento.\n\nPara o primeiro passo do procedimento temos quatro escolhas. Fixada uma escolha para o primeiro passo, temos três escolhas para o segundo passo. Fixadas as escolhas para os primeiro e segundo passos, para o último passo teremos $8 \\times 8$ alternativas, já que temos oito algarismos para escolher para cada uma das posições e pode haver repetição. No total teremos $4 \\times 3 \\times 8 \\times 8 = 768$ formas de realizar o procedimento e, portanto, a lista de Pedro tem $768$ números.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26802,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEscreve-se, em ordem crescente, cada um dos múltiplos de $3$ cuja soma com $1$ é um quadrado perfeito:\n$$\n3, 15, 24, 48, \\ldots\n$$\na) Qual é o próximo número que aparecerá, nesta sequência, depois do $48$?\nb) Qual é o oitavo número desta sequência?\nc) Qual é o número que aparecerá, nesta sequência, na $2013^{a}$ posição?",
    "options": [],
    "answer": "a) 63; b) 168; c) 9120399",
    "solution": "Solution:\n\na) Note que $48$ é um termo da sequência que é um múltiplo de $3$ que satisfaz que $48+1=49=7^{2}$. Assim, para encontrar o seu sucessor, tentaremos obter um múltiplo de três que, acrescido de uma unidade, seja igual a $8^{2}=64$. Subtraindo uma unidade de $64$, obtemos $63$, que é realmente um múltiplo de $3$.\n\nEm outras palavras, temos que $63+1=64=8^{2}$, logo $63$ é o próximo termo da sequência.\n\nObservação: Note que o mesmo raciocínio não funciona para encontrar o termo da sequência que aparece após o termo $63$. De fato, se calcularmos $9^{2}-1=81-1=80$ acabamos encontrando um número que não é múltiplo de $3$. O procedimento desenvolvido nos itens b) e c) abaixo nos mostra como contornar esse tipo de dificuldade.\n\nb) Começamos escrevendo os primeiros quadrados perfeitos em ordem crescente:\n$$\n2^{2}, 3^{2}, 4^{2}, 5^{2}, 6^{2}, 7^{2}, 8^{2}, 9^{2}, 10^{2}, 11^{2}, 12^{2}, 13^{2}, 14^{2}, 15^{2}, \\ldots\n$$\nque é igual à sequência\n$$\n4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, \\ldots\n$$\nSubtraindo uma unidade de cada um dos termos, obtemos a nova sequência\n$$\n3, 8, 15, 24, 35, 48, 63, 80, 99, 120, 143, 168, 195, 224, \\ldots\n$$\nExcluímos dessa lista aqueles números que não são múltiplos de $3$, obtendo a lista:\n$$\n3, 15, 24, 48, 63, 99, 120, 168, 195, \\ldots\n$$\nEssa é então uma lista crescente dos múltiplos de $3$ cuja soma com $1$ resulta em um quadrado perfeito. Note que o seu oitavo termo é igual a $168$. Logo, esse é o número que estamos procurando.\n\nObservação: Notamos na solução do item acima que, ao subtrairmos uma unidade do quadrado de qualquer número que não seja um múltiplo de $3$, obtemos como resultado um múltiplo de $3$.\n\nPor exemplo, o número $4$ não é um múltiplo de $3$, no entanto $4^{2}-1=16-1=15$ que é um múltiplo de $3$. Também $11$ não é um múltiplo de $3$, mas $11^{2}-1=121-1=120$ que é um múltiplo de $3$.\n\nNote que os múltiplos de $3$ são os números $x$ do tipo $x=3n$ para algum número natural $n$. Os números que não são múltiplos de $3$ são aqueles que podem ser escritos na forma $x=3n+1$ ou $x=3n+2$. Esses dois tipos de números correspondem respectivamente àqueles cujo resto da divisão por $3$ é igual a $1$ ou $2$.\n\nConsideremos primeiramente um número $x$ do primeiro tipo, isto é, do tipo $x=3n+1$. Note que $x^{2}=(3n+1)^{2}=9n^{2}+6n+1$. Assim, $x^{2}-1=9n^{2}+6n$ que é um múltiplo de $3$.\n\nDe modo análogo, para números do segundo tipo $x=3n+2$, temos que $x^{2}=9n^{2}+12n+4$, logo $x^{2}-1=9n^{2}+12n+3$ que também é um múltiplo de $3$.\n\nPor último notamos que para um múltiplo de $3$, $x=3n$, vale que $x^{2}-1=9n^{2}-1$ que não pode ser múltiplo de $3$, já que $9n^{2}$ é um múltiplo de $3$ por si só.\n\nc) Estamos interessados em calcular o $2013^{\\circ}$ termo da lista:\n$$\n3, 15, 24, 48, 63, 99, 120, 168, 195, \\ldots\n$$\nque aparece no item anterior. Sejam os números do tipo $1$ e do tipo $2$, aqueles cuja divisão por $3$ deixam resto igual a $2$ e $1$, respectivamente. Pela observação acima, concluímos que, para gerar essa lista, podemos formar primeiramente uma lista crescente alternando os números do tipo $1$ e do tipo $2$. Em seguida devemos elevar ao quadrado cada termo e, por último, subtrair uma unidade de cada termo.\n\nComo estamos interessados em encontrar o $2013^{\\circ}$ número, podemos nos concentrar em encontrar o $1007^{\\circ}$ termo da sequência de números do tipo $1$. O primeiro termo dessa lista é igual a $3 \\times 0+2=2$, o segundo é igual a $3 \\times 1+2=5$, o terceiro é igual a $3 \\times 2+2=8$ e assim por diante. Logo o $1007^{\\circ}$ termo é igual a $3 \\times 1006+2=3020$. Assim, para gerar o $2013^{\\circ}$ termo da lista\n$$\n3, 15, 24, 48, 63, 99, 120, 168, 195, \\ldots\n$$\nprecisamos então elevar o número $3020$ ao quadrado e subtrair uma unidade, obtendo então o número $9120399$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26803,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm prédio tem três escadas diferentes, todas começando na base do prédio e terminando no topo. Uma escada tem 104 degraus, outra tem 117 degraus, e a outra tem 156 degraus. Sempre que os degraus das três escadas estão na mesma altura, há um andar. Quantos andares tem o prédio?",
    "options": [],
    "answer": "13",
    "solution": "Solution:\n\nVamos chamar de $A$, $B$ e $C$ as três escadas, que têm 104, 117 e 156 degraus, respectivamente. Seja $a$ o número de degraus da escada $A$ entre cada dois andares, $b$ o número de degraus da escada $B$ entre cada dois andares, e $c$ o número de degraus da escada $C$ entre cada dois andares. Dividindo o número total de degraus de uma escada pelo número de degraus que esta escada tem entre cada dois andares, obtém-se o número de andares! Logo,\n$$\n\\frac{104}{a} = \\frac{117}{b} = \\frac{156}{c} = d\n$$\nonde $d$ é o número de andares do prédio. Ou seja, $d$ é um divisor comum de 104, 117 e 156. Além disso, $d$ deve ser o maior possível, pois $a$, $b$ e $c$ são os menores possíveis! Logo, $d$ é o maior divisor comum $(mdc)$ de 104, 117 e 156. Calculando o $mdc$ de 104, 117 e 156, obtemos o número 13 como resposta. Logo, o número de andares deste prédio é 13.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26804,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUma pulga, que está no ponto $A$ de uma reta, pula exatamente $1~\\mathrm{m}$ de cada vez, sem nunca sair dessa reta.\n\na) Se a pulga quer chegar no ponto $B$ localizado sobre a reta, a uma distância de $5~\\mathrm{m}$ à direita de $A$, com exatamente 7 pulos, de quantas maneiras ela pode fazer isso?\n\nb) Se a pulga quer chegar no ponto $C$ localizado sobre a reta, a uma distância $5~\\mathrm{m}$ à direita de $A$, com exatamente 9 pulos, de quantas maneiras ela pode fazer isso?\n\nc) É possível que a pulga chegue no ponto $D$ localizado sobre a reta a uma distância de $2013~\\mathrm{m}$ de $A$, com exatamente 2028 pulos? Justifique.",
    "options": [],
    "answer": "a) 7; b) 36; c) No",
    "solution": "Solution:\n\na) Para chegar em $B$, a pulga deve dar exatamente um passo para a esquerda, e seis para a direita, em qualquer ordem. Como esse passo para a esquerda pode ser dado em qualquer momento, há 7 momentos possíveis para dá-lo! Logo, são 7 maneiras distintas da pulga chegar em $B$ com 7 passos.\n\nb) Para chegar em $C$, a pulga deve dar 7 passos para a direita e 2 para a esquerda, em qualquer ordem. De quantas maneiras a pulga pode fazer isso? Uma maneira é listar todas, são 36.\nOutra maneira, mais interessante, é pensar da forma seguinte: de quantas formas podemos ordenar 9 objetos distintos? A resposta é $9 \\times 8 \\times 7 \\times 6 \\times 5 \\times 4 \\times 3 \\times 2 \\times 1$. E se há 7 objetos iguais de um tipo e 2 de outro tipo? Então, da maneira anterior, estaríamos contando cada configuração muitas vezes. De fato, estaríamos contando $7 \\times 6 \\times 5 \\times 4 \\times 3 \\times 2 \\times 1$ vezes por causa de um tipo de objeto repetido e estaríamos contando $2 \\times 1$ vezes por causa do outro. Daí, basta fazer\n$$\n\\frac{9 \\times 8 \\times 7 \\times 6 \\times 5 \\times 4 \\times 3 \\times 2 \\times 1}{(7 \\times 6 \\times 5 \\times 4 \\times 3 \\times 2 \\times 1) \\times (2 \\times 1)} = \\frac{9 \\times 8}{2 \\times 1} = 36\n$$\ne, portanto, são 36 maneiras. Podemos agora pensar em um objeto de um tipo como sendo um salto para a direita, e o objeto do outro tipo como sendo um salto para a esquerda. Assim, cada maneira de ordenar os objetos, corresponde a uma maneira de ordenar os saltos da pulga. Concluímos que a pulga tem 36 maneiras distintas de chegar em $C$.\n\nc) A resposta é não! Porque $2028-2013=15$, que é ímpar.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26805,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO triângulo $ABC$ abaixo é equilátero, ou seja, tem seus três lados de mesmo comprimento e todos seus ângulos iguais a $60^{\\circ}$. O senhor Simas marca um ponto $H$ qualquer no lado $BC$ do triângulo. Em seguida, ele traça um segmento paralelo ao lado $AC$, começando em $H$ e terminando no ponto $I$ sobre o lado $AB$. Em seguida, traça um segmento paralelo ao lado $AB$, começando em $H$ e terminando no ponto $J$ sobre o lado $AC$, conforme a figura abaixo:\n![](attached_image_1.png)\n\na) Sabendo que o lado $AB$ tem comprimento igual a 1, calcule o perímetro do quadrilátero $AIHJ$.\n\nb) O senhor Simas segue desenhando, como mostra a figura a seguir, e traça os segmentos $LO$ e $MP$ de maneira perpendicular ao lado $BC$.\n![](attached_image_2.png)\n\nSeja $d$ o comprimento do segmento $IL$, seja $f$ o comprimento do segmento $JM$ e seja $x$ o comprimento do segmento $OP$. Mostre que\n$$\nx=\\frac{1+d+f}{2}\n$$",
    "options": [],
    "answer": "Perimeter of AIHJ = 2; and x = (1 + d + f) / 2.",
    "solution": "Solution:\n\na) Como o segmento $IH$ é paralelo ao segmento $AJ$, e o segmento $AI$ é paralelo ao segmento $JH$, temos que $\\overline{IH}=\\overline{AJ}$ e $\\overline{AI}=\\overline{JH}$. Observação: a barra em cima do segmento denota o comprimento do segmento.\n\nComo o lado do triângulo $ABC$ mede 1, temos que\n$$\n1=\\overline{AB}=\\overline{AI}+\\overline{IB}=\\overline{AI}+\\overline{IH}\n$$\nLogo, o perímetro do quadrilátero é dado por\n$$\n\\overline{AI}+\\overline{IH}+\\overline{HJ}+\\overline{JA}=2(\\overline{AI}+\\overline{IH})=2\n$$\n\nb) Vamos começar chamando $\\overline{BO}=m_{2}$ e $\\overline{CP}=m_{1}$, conforme a figura a seguir:\n![](attached_image_3.png)\n\nOs ângulos internos de um triângulo equilátero são todos iguais a $60^{\\circ}$. Logo, $\\measuredangle ACB=\\measuredangle ABC=60^{\\circ}$. Vamos olhar para o triângulo $PCM$. Como $MP$ é perpendicular a $PC$, este triângulo é um triângulo retângulo. Como o cosseno de um ângulo (cos) é definido como a razão entre cateto adjacente a esse ângulo e hipotenusa, temos que\n$$\n\\cos 60^{\\circ}=\\frac{1}{2}=\\frac{\\overline{PC}}{\\overline{MC}}=\\frac{m_{1}}{\\overline{MC}}\n$$\nDisso, concluímos que $\\overline{MC}=2m_{1}$. Por um argumento parecido, concluímos também que $\\overline{LB}=2m_{2}$.\n![](attached_image_4.png)\n\nComo o triângulo $HJC$ é equilátero, temos que $\\overline{HJ}=f+2m_{1}$. Como $\\overline{AI}=\\overline{HJ}$, concluímos que $\\overline{AI}=f+2m_{1}$. Como $\\overline{AB}=1$, temos que $2m_{1}+d+f+2m_{2}=1$, e portanto, $m_{1}+m_{2}=\\frac{1-d-f}{2}$. Por outro lado, como $\\overline{BC}=1$, temos que\n$$\nm_{1}+x+m_{2}=1\n$$\nSubstituindo uma equação na outra, obtemos\n$$\nx=\\frac{1+d+f}{2}.\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26806,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAbaixo, veem-se círculos grandes e pequenos. Os círculos grandes têm raio $2$, e os círculos pequenos têm raio $1$. Qual a área da região pintada de cinza?\n\n![](attached_image_1.png)\n\nObservação: A área de um círculo de raio $r$ é igual a $\\pi r^{2}$.",
    "options": [],
    "answer": "4π",
    "solution": "Solution:\n\nVamos numerar as regiões:\n\n![](attached_image_2.png)\n\nAs regiões $2$ e $4$ têm mesma área, pois uma pode ser obtida refletindo a outra. Além disso, as regiões $1$ e $6$ também possuem mesma área. Logo, a área da região em cinza acima é a mesma da região pintada em cinza abaixo:\n\n![](attached_image_3.png)\n\nA área da região pintada em cinza acima é a área do círculo de raio $2$, logo, igual a $\\pi 2^{2} = 4\\pi$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26807,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa cidade de Autolândia, a numeração de placas de carros é feita através de números de três dígitos, portanto indo da placa 000 até a placa 999. Para diminuir a poluição, o prefeito Pietro decidiu implementar um rodízio de carros, estabelecendo os dias nos quais as pessoas podem usar seus carros. As regras do rodízio são:\n- Segunda-feira: somente carros com placa ímpar;\n- Terça-feira: somente carros com placa cuja soma dos três dígitos é maior ou igual a 11;\n- Quarta-feira: somente carros com placa cujo número é múltiplo de 3;\n- Quinta-feira: somente carros com placa cuja soma dos três dígitos é menor ou igual a 14;\n- Sexta-feira: somente carros com placa contendo pelo menos dois dígitos iguais;\n- Sábado: somente carros cujo número na placa for estritamente menor do que 500;\n- Domingo: somente carros cuja placa tenha os três dígitos menores ou iguais a 5.\n\na) Em quais dias o carro com a placa 729 pode circular?\n\nb) Maria, a esposa do prefeito, quer um carro que possa circular todos os dias exceto aos domingos. Qual placa ela deve ter?\n\nc) O prefeito Pietro precisa de uma placa que o permita circular todos os dias. Que placa ele deve ter?\n\nd) Por que todos os habitantes de Autolândia podem circular pelo menos uma vez por semana?",
    "options": [],
    "answer": "a) Monday, Tuesday, Wednesday.\nb) 363.\nc) 255.\nd) Any plate either meets the Tuesday sum condition or, if not, its digit sum is small enough to meet the Thursday condition, so everyone can drive at least Tuesday or Thursday.",
    "solution": "Solution:\n\na) Como o número 729 é ímpar, o carro com a placa de número 729 pode circular às segundas-feiras. Como $7+2+9=18$, o carro também pode circular às terças-feiras, mas não às quintas-feiras. Como 729 é múltiplo de 3, este carro também pode circular às quartas-feiras. Como 729 não possui dígitos iguais, é maior do que 500, e tem um dígito maior do que 6, ele não pode circular às sextas-feiras, sábados e domingos.\n\nb) O leitor pode verificar que a resposta para essa questão é o número 363. Vamos em seguida mostrar uma maneira de se chegar a essa resposta.\n\nSeja $ABC$ o número da placa do carro de Maria. Como o carro deve circular aos sábados, devemos ter\n$$\nA \\in \\{0,1,2,3,4\\}\n$$\nAlém disso, o carro deve circular às terças-feiras e quintas-feiras. Logo,\n$$\nA+B+C \\in \\{11,12,13,14\\}\n$$\nMais ainda, para circular às segundas-feiras, devemos ter que\n$$\nC \\in \\{1,3,5,7,9\\}\n$$\nVamos dividir as possibilidades em quatro casos:\n\nPrimeiro caso: $A=0$.\nNesse caso, temos que $B+C=A+B+C \\in \\{11,12,13,14\\}$. É impossível obter essa igualdade se $B=0$ ou $C=0$. Logo, devemos supor que $B \\neq 0$ e que $C \\neq 0$. Em particular, vale que $B \\neq A$ e $C \\neq A$. Como o carro deve circular às sextas-feiras, então temos que $B=C$. Assim $2C=B+C \\in \\{11,12,13,14\\}$. Como devemos ter $C \\in \\{1,3,5,7,9\\}$, a única possibilidade é $ABC=077$. Mas esse número não é um múltiplo de 3. Assim o carro com placa de número 077 está impedido de circular às quartas-feiras. Logo, esse caso deve ser descartado.\n\nSegundo caso: $A=1$.\nNesse caso, temos que $B+C=A+B+C-1 \\in \\{10,11,12,13\\}$. Assim, se fosse $B=1$, então deveríamos ter obrigatoriamente $C=9$. Logo, teríamos $ABC=119$ que não é um múltiplo de 3, logo não pode circular às quartas-feiras. Por outro lado, se fosse $C=1$, então deveríamos ter obrigatoriamente $B=9$ obtendo que $ABC=191$ que também não é um múltiplo de 3. Em particular, vale que $B \\neq A$ e $C \\neq A$. Como o carro deve circular às sextas-feiras, então temos que $B=C$. Logo, $2C=B+C \\in \\{10,11,12,13\\}$. Como devemos ter $C \\in \\{1,3,5,7,9\\}$, a única escolha possível é $C=5$. Assim, temos que $ABC=155$ que não é um múltiplo de 3. Deste modo, esse caso deve ser descartado.\n\nTerceiro caso: $A=2$.\nNesse caso, temos que $B+C=A+B+C-2 \\in \\{9,10,11,12\\}$. Assim, se fosse $B=2$ então as únicas escolhas possíveis para $C$ seriam $C=7$ ou $C=9$. Logo, deveríamos ter $ABC=227$ ou $ABC=229$. Como nenhum deles é um múltiplo de 3, temos que descartar a opção $B=2$. Também $C \\neq 2$, já que $C$ deve ser ímpar. Em particular, temos que $B \\neq A$ e $C \\neq A$. Como o carro deve circular às sextas-feiras, então temos que $B=C$. Logo, $2C=B+C \\in \\{9,10,11,12\\}$. Como devemos ter $C \\in \\{1,3,5,7,9\\}$, a única escolha possível é $C=5$. Assim, temos que $ABC=255$. Porém um carro contendo a placa com esse número não estaria impedido de circular aos domingos, pois satisfaria a exigência de que todos os dígitos são menores do que 6. Assim, esse caso deve ser descartado.\n\nQuarto caso: $A=3$.\nNesse caso temos que $B+C=A+B+C-3 \\in \\{8,9,10,11\\}$. Assim, se fosse $B=3$ então as únicas escolhas possíveis para $C$ seriam $C=5$ ou $C=7$. Logo, deveríamos ter $ABC=335$ ou $ABC=337$. Como nenhum deles é um múltiplo de 3, temos que descartar a opção $B=3$. Se fosse também $C \\neq 3$, então em particular teríamos que $B \\neq A$ e $C \\neq A$. Para que o carro possa circular às sextas-feiras deveríamos ter $B=C$. Logo $2C=B+C \\in \\{8,9,10,11\\}$. Como deveríamos ter $C \\in \\{1,3,5,7,9\\}$, a única escolha possível é $C=5$. Logo, teríamos $ABC=355$, que não é um múltiplo de 3. Assim, temos também que descartar a opção $C \\neq 3$. Como $B+C \\in \\{8,9,10,11\\}$, vale que $B \\in \\{5,6,7,8\\}$. Logo, as escolhas possíveis para o valor de $ABC$ são 353, 363, 373 e 383. Dessas, a única que é um múltiplo de 3 é o 363. Logo, todas as outras opções devem ser descartadas. Como 363 possui um dígito que não é menor do que 6, o carro que contém esse número em sua placa não pode circular aos domingos. Assim, esse é o número procurado. Para mostrar que essa é a única escolha de placa possível, vamos descartar também o próximo caso.\n\nQuinto caso: $A=4$.\nNesse caso, temos que $B+C=A+B+C-4 \\in \\{7,8,9,10\\}$. Assim, se fosse $B=4$ então as únicas escolhas possíveis para $C$ seriam $C=3$ ou $C=5$. Então as opções possíveis são $ABC=443$ ou $ABC=445$ que não são múltiplos de 3. Como $C$ deve ser ímpar temos que $C \\neq 4$, logo devemos ter $C=B$. Dessa forma $2C=B+C \\in \\{7,8,9,10\\}$. A única opção possível é $C=5$. Nesse caso, $ABC=455$ que não é um múltiplo de 3. Deste modo, não pode ser $A=4$.\n\nc) O leitor pode verificar que a resposta para essa questão é o número 255. Vamos em seguida mostrar uma maneira de se chegar a essa resposta, que se assemelha muito com o procedimento no item anterior.\n\nSeja $ABC$ o número da placa do carro do prefeito Pietro. Como o carro deve circular aos sábados, devemos ter\n$$\nA \\in \\{0,1,2,3,4\\}\n$$\nAlém disso o carro deve circular às terças-feiras e segundas-feiras, logo\n$$\nA+B+C \\in \\{11,12,13,14\\}\n$$\nMais ainda, para circular às segundas-feiras, devemos ter que\n$$\nC \\in \\{1,3,5,7,9\\}\n$$\nVamos dividir as possibilidades em quatro casos:\n\nPrimeiro caso: $A=0$ (igual ao do item $c$)).\nNesse caso, temos que $B+C=A+B+C \\in \\{11,12,13,14\\}$. É impossível obter essa igualdade se $B=0$ ou $C=0$. Portanto, devemos supor que $B \\neq 0$ e que $C \\neq 0$. Em particular, vale que $B \\neq A$ e $C \\neq A$. Como o carro deve circular às sextas-feiras, temos que $B=C$. Assim, $2C=B+C \\in \\{11,12,13,14\\}$. Como devemos ter $C \\in \\{1,3,5,7,9\\}$, a única possibilidade é $ABC=077$. Mas esse número não é um múltiplo de 3. Assim, o carro com placa de número 077 está impedido de circular às quartas-feiras. Logo, esse caso deve ser descartado.\n\nSegundo caso: $A=1$ (igual ao do item $c$)).\nNesse caso, temos que $B+C=A+B+C-1 \\in \\{10,11,12,13\\}$. Assim, se fosse $B=1$, então deveríamos ter obrigatoriamente $C=9$. Logo teríamos $ABC=119$ que não é um múltiplo de 3, logo não podendo circular às quartas-feiras. Por outro lado, se fosse $C=1$, então deveríamos ter obrigatoriamente $B=9$, obtendo que $ABC=191$ que também não é um múltiplo de 3. Em particular, vale que $B \\neq A$ e $C \\neq A$. Como o carro deve circular às sextas-feiras, então temos que $B=C$. Logo, $2C=B+C \\in \\{10,11,12,13\\}$. Como devemos ter $C \\in \\{1,3,5,7,9\\}$, a única escolha possível é $C=5$. Assim, temos que $ABC=155$ que não é um múltiplo de 3. Logo, esse caso deve ser descartado.\n\nTerceiro caso: $A=2$.\nNesse caso, temos que $B+C=A+B+C-2 \\in \\{9,10,11,12\\}$. Assim, se fosse $B=2$ então as únicas escolhas possíveis para $C$ seriam $C=7$ ou $C=9$. Logo, deveríamos ter $ABC=227$ ou $ABC=229$. Como nenhum deles é um múltiplo de 3, temos que descartar a opção $B=2$. Também $C \\neq 2$, já que $C$ deve ser ímpar. Em particular, temos que $B \\neq A$ e $C \\neq A$. Como o carro deve circular às sextas-feiras, então temos que $B=C$. Logo $2C=B+C \\in \\{9,10,11,12\\}$. Como devemos ter $C \\in \\{1,3,5,7,9\\}$, a única escolha possível é $C=5$. Assim, temos que $ABC=255$. Esse número é um múltiplo de 3, logo pode circular também às quartas-feiras. Um carro contendo a placa com esse número pode ainda circular aos domingos, pois satisfaz a exigência de que todos os dígitos são menores do que 6. Assim, essa é a placa procurada. Para mostrar que essa é a única escolha de placa possível vamos descartar também os próximos dois casos.\n\nQuarto caso: $A=3$.\nNesse caso, temos que $B+C=A+B+C-3 \\in \\{8,9,10,11\\}$. Assim, se fosse $B=3$ então as únicas escolhas possíveis para $C$ seriam $C=5$ ou $C=7$. Logo deveríamos ter $ABC=335$ ou $ABC=337$. Como nenhum deles é um múltiplo de 3, temos que descartar a opção $B=3$. Se fosse também $C \\neq 3$, então em particular teríamos que $B \\neq A$ e $C \\neq A$. Para que o carro possa circular às sextas-feiras deveríamos ter $B=C$. Logo, $2C=B+C \\in \\{8,9,10,11\\}$. Como deveríamos ter $C \\in \\{1,3,5,7,9\\}$, a única escolha possível é $C=5$. Logo teríamos $ABC=355$, que não é um múltiplo de 3. Assim, temos também que descartar a opção $C \\neq 3$. Como $B+C \\in \\{8,9,10,11\\}$ vale que $B \\in \\{5,6,7,8\\}$. Logo as escolhas possíveis para o valor de $ABC$ são 353, 363, 373 e 383. Dessas, a única que é um múltiplo de 3 é o 363. Logo, todas as outras opções devem ser descartadas. Como 363 possui um dígito que não é menor do que 6 o carro que contém esse número em sua placa não pode circular aos domingos. Assim esse caso deve ser também descartado.\n\nQuinto caso: $A=4$ (igual ao do item $c$)).\nNesse caso, temos que $B+C=A+B+C-4 \\in \\{7,8,9,10\\}$. Assim, se fosse $B=4$ então as únicas escolhas possíveis para $C$ seriam $C=3$ ou $C=5$. Então as opções possíveis são $ABC=443$ ou $ABC=445$, que não são múltiplos de 3. Como $C$ deve ser ímpar temos que $C \\neq 4$, logo devemos ter $C=B$. Dessa forma, $2C=B+C \\in \\{7,8,9,10\\}$. A única opção possível é $C=5$. Nesse caso, $ABC=455$ que não é um múltiplo de 3. Assim, não pode ser $A=4$.\n\nd) Carros com placas cuja soma dos números é maior do que 10 podem circular às terças-feiras. Assim, caso um carro não possa circular às terças-feiras a sua placa contém um número cuja soma dos algarismos é menor do que $10$ e, em consequência, menor também do que 15. Assim, esse carro pode circular às quintas-feiras. Deste modo, qualquer carro que seja impedido de circular às terças-feiras pode circular às quintas-feiras.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26808,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nO personagem histórico mexicano Benito Juárez nasceu na primeira metade do século XIX (o século XIX vai do ano 1801 ao ano 1900). Sabendo que Benito Juárez completou $x$ anos no ano $x^{2}$, qual foi o ano do seu nascimento?",
    "options": [],
    "answer": "1806",
    "solution": "Solution:\nOs quadrados perfeitos que estão mais próximos de 1801-1900 são:\n$$\n\\begin{aligned}\n& 42 \\times 42=1764 \\\\\n& 43 \\times 43=1849 \\\\\n& 44 \\times 44=1936\n\\end{aligned}\n$$\nSeja $x$ a idade de Benito Juárez no ano $x^{2}$. O número $x$ não pode ser 42, pois neste caso Benito não teria nascido no século XIX (o ano 1764 não está no século XIX, que vai do ano 1801 ao ano 1900). Vamos testar agora o ano 1849. Se a idade de Benito em 1849 é 43, então Benito nasceu em $1849-43 = 1806$. Como 1806 pertence ao século XIX, esta é a resposta correta. Observe que é a única possível, pois do número 44 em diante, o quadrado do número menos ele é maior do que 1901 e portanto não pertenceria ao século XIX.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26809,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA partir do meio-dia, João faz, a cada 80 minutos, uma marca na posição do ponteiro das horas do seu relógio.\na) Depois de quanto tempo não será mais necessário fazer novas marcas no relógio?\nb) Qual a soma dos ângulos internos do polígono formado pelas marcas?",
    "options": [],
    "answer": "a) 720 minutes; b) 1260 degrees",
    "solution": "Solution:\na) O ponteiro das horas concluirá uma volta completa após $12 \\cdot 60 = 720$ minutos e ao longo dela nenhuma marca será repetida. Como $720$ é múltiplo de $80$, durante esse período são feitas exatamente $\\frac{12 \\cdot 60}{80} = 9$ marcas no relógio e, além disso, os dois ponteiros voltam às suas posições iniciais. Daí, como as próximas marcas serão repetidas, o tempo desejado é $720$ minutos.\n\nb) A soma dos ângulos internos de um polígono de $9$ lados é $180^{\\circ} \\cdot (9-2) = 1260^{\\circ}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26810,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nEncontre dois inteiros positivos $x$ e $y$ tais que:\n$$\n\\frac{1-\\sqrt{2}+\\sqrt{3}}{1+\\sqrt{2}-\\sqrt{3}}=\\frac{\\sqrt{x}+\\sqrt{y}}{2}\n$$",
    "options": [],
    "answer": "x = 2, y = 6",
    "solution": "Solution:\n$$\n\\begin{aligned}\n\\frac{1-\\sqrt{2}+\\sqrt{3}}{1+\\sqrt{2}-\\sqrt{3}} \\cdot \\frac{\\sqrt{2}}{\\sqrt{2}} & =\\frac{-2+\\sqrt{2}+\\sqrt{6}}{2+\\sqrt{2}-\\sqrt{6}} \\\\\n& =\\frac{-2+\\sqrt{2}+\\sqrt{6}}{2+\\sqrt{2}-\\sqrt{6}} \\cdot \\frac{2 \\sqrt{2}+2+2 \\sqrt{3}}{2 \\sqrt{2}+2+2 \\sqrt{3}} \\\\\n& =\\frac{-\\sqrt{2}+(1+\\sqrt{3})}{(\\sqrt{2}+1)-\\sqrt{3}} \\cdot \\frac{\\sqrt{2}+(1+\\sqrt{3})}{(\\sqrt{2}+1)+\\sqrt{3}} \\\\\n& =\\frac{(1+\\sqrt{3})^{2}-2}{(\\sqrt{2}+1)^{2}-3} \\\\\n& =\\frac{\\sqrt{2}+\\sqrt{6}}{2}\n\\end{aligned}\n$$\nPortanto, $x=2$ e $y=6$ satisfazem ao enunciado.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26811,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPedro decidiu levar todos os seus filhos, meninos e meninas, para tomar sorvete na sorveteria Sorvete Matemático. Na sorveteria, há 12 sabores diferentes de sorvete e cada criança pediu um combo com 3 bolas de sorvete. Depois de sair da sorveteria, Pedro percebeu que, no total, foram pedidas exatamente duas bolas de cada sabor disponível na sorveteria.\n\na) Sabendo que Pedro não tomou sorvete, qual o número total de seus filhos (meninas e meninos)?\n\nb) Pedro olhou com mais atenção os sabores que cada um pediu e notou que nenhum sabor foi pedido por um menino e por uma menina, ou seja, se um menino escolheu um sabor, nenhuma menina escolheu aquele mesmo sabor. Sabendo que pelo menos um de seus filhos é menino e que ele possui mais filhas do que filhos, determine o número de suas filhas.",
    "options": [],
    "answer": "a) 8; b) 6",
    "solution": "Solution:\n\na) Seja $n$ o número de filhos de Pedro. No total, foram pedidas $3 n$ bolas de sorvete. Como cada um dos 12 sabores foi pedido duas vezes, temos $3 n = 2 \\cdot 12$, ou seja, $n = 8$. Portanto, Pedro possui 8 filhos.\n\nb) Sejam $x$ o número de meninos e $y$ o número de meninas. Pelo item anterior, sabemos que $x + y = 8$. Como existe pelo menos um filho e há mais filhas do que filhos, sabemos que: $0 < x < y$. Dado que nenhum sabor foi pedido simultaneamente por meninos e por meninas, eles podem ser separados em sabores dos meninos e sabores das meninas. Então, $3x$ é igual ao dobro do número de sabores dos meninos, pois foram pedidas exatamente duas bolas de cada sabor e nenhuma menina pode pedir um dos sabores dos meninos. Consequentemente, $x$ é par. Do mesmo modo, $y$ também é par. Como $0 < x < y$ e $x + y = 8$, podemos concluir que $x = 2$ e $y = 6$. Daí, Pedro possui 6 filhas.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26812,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUma técnica muito usada para calcular somatórios é a Soma Telescópica. Ela consiste em \"decompor\" as parcelas de uma soma em partes que se cancelem. Por exemplo,\n$$\n\\begin{aligned}\n& \\frac{1}{1 \\cdot 2}+\\frac{1}{2 \\cdot 3}+\\frac{1}{3 \\cdot 4}+\\frac{1}{4 \\cdot 5}= \\\\\n& \\left(\\frac{1}{1}-\\frac{1}{2}\\right)+\\left(\\frac{1}{2}-\\frac{1}{3}\\right)+\\left(\\frac{1}{3}-\\frac{1}{4}\\right)+\\left(\\frac{1}{4}-\\frac{1}{5}\\right)= \\\\\n& \\frac{1}{1}-\\frac{1}{5}= \\\\\n& \\frac{4}{5}\n\\end{aligned}\n$$\nCom esta técnica, podemos achar uma forma de somar números ímpares consecutivos. Vejamos:\na) Contando os números ímpares de um por um e começando pelo 1, verifique que o número na posição $m$ é igual a $m^{2}-(m-1)^{2}$.\nb) Calcule a soma de todos os números ímpares entre 1000 e 2014.",
    "options": [],
    "answer": "764049",
    "solution": "Solution:\n\na) Veja que o primeiro número ímpar é $2 \\cdot 1-1$ e, sabendo que os números ímpares crescem de 2 em 2, podemos concluir que o número ímpar que estará na posição $m$ em nossa contagem é\n$$\n\\begin{aligned}\n2 \\cdot 1-1+\\underbrace{2+2+\\ldots+2}_{m-1 \\text{ vezes }} & =2 \\cdot 1-1+2(m-1) \\\\\n& =2 m-1\n\\end{aligned}\n$$\nPara verificar que ele coincide com o número do item $a$ ), basta calcularmos\n$$\nm^{2}-(m-1)^{2}=m^{2}-\\left(m^{2}-2 m-1\\right)=2 m-1\n$$\n\nb) Queremos somar os números ímpares desde $1001=2 \\cdot 501-1$ até $2013=2 \\cdot 1007-1$. Usando a expressão do item $a$ ), temos\n$$\n\\begin{aligned}\n1001 & =501^{2}-500^{2} \\\\\n1003 & =502^{2}-501^{2} \\\\\n1005 & =503^{2}-502^{2} \\\\\n& \\cdots \\\\\n2011 & =1006^{2}-1005^{2} \\\\\n2013 & =1007^{2}-1006^{2}\n\\end{aligned}\n$$\nSomando tudo, vemos que todos os números de $501^{2}$ até $1006^{2}$ são cancelados. Assim, o resultado é:\n$$\n\\begin{aligned}\n1001+1003+\\ldots+2013 & =1007^{2}-500^{2} \\\\\n& =(1007-500)(1007+500) \\\\\n& =507 \\cdot 1507 \\\\\n& =764049\n\\end{aligned}\n$$\nEntão a soma dos ímpares entre 1000 e 2014 é 763048.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26813,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNo desenho abaixo, $ABCD$ é um retângulo e os pontos $P$ e $Q$ pertencem à diagonal $AC$ de modo que $AQ = PQ = PC = 1$ e $\\angle AQD = \\angle BPC = 90^\\circ$. Encontre a área do retângulo $ABCD$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "3*sqrt(2)",
    "solution": "Solution:\n\nPelas relações métricas no triângulo retângulo, temos $DQ^{2} = AQ \\cdot QC = 2$. Pelo Teorema de Pitágoras nos triângulos $\\triangle DAQ$ e $\\triangle DQC$, temos:\n$$\n\\begin{aligned}\nAD^{2} & = DQ^{2} + AQ^{2} \\\\\n& = 2 + 1 \\\\\n& = 3 \\\\\nDC^{2} & = DQ^{2} + QC^{2} \\\\\n& = 2 + 4 \\\\\n& = 6\n\\end{aligned}\n$$\nA área do retângulo $ABCD$ é $AD \\cdot DC = \\sqrt{3} \\cdot \\sqrt{6} = 3\\sqrt{2}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26814,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO mágico Magimático diz para uma pessoa da plateia escolher uma peça qualquer de um dominó comum. Tal peça é formada por um par de números de 0 a 6. Em seguida, ele diz para a pessoa escolher um dos números da peça e realizar a seguinte sequência de operações:\n1. multiplicá-lo por 5;\n2. somar o resultado anterior com 15;\n3. multiplicar o último resultado por $2$ e, finalmente,\n4. somar o último resultado com o outro número da peça.\nRealizadas tais operações, o resultado é divulgado e Magimático impressiona a plateia dizendo exatamente os números escritos no dominó escolhido.\n\na) Sabendo que o resultado foi 62, como o mágico descobriu o número escolhido pelo membro da plateia?\n\nb) Se o resultado tivesse sido $n$, como descobrir os números da peça escolhida?",
    "options": [],
    "answer": "a) The tile is (3, 2).\nb) In general, if the announced result is n, then y is the units digit of n and x = (n − 30 − y)/10 (equivalently, x is the tens digit of n minus 3).",
    "solution": "Solution:\n\na) Vamos revelar o segredo do mágico. Suponha que o par de números escritos no dominó é $(x, y)$ e que o número escolhido para a sequência de operações foi o $x$. Assim, as operações realizadas pelo membro da plateia foram:\n$$\nx \\rightarrow 5x \\rightarrow 5x+15 \\rightarrow 2(5x+15) \\rightarrow 2(5x+15)+y\n$$\nO resultado divulgado é o número $2(5x+15)+y = 10x + y + 30$. Como $30 + 10x$ termina em 0, o dígito das unidades do resultado e o de $y$ são iguais. Dado que $y < 7$ e o resultado foi 62, então $y = 2$ e $10x + 30 = 60$. Ou seja, $x = 3$ e a peça escolhida foi a peça com os números $(3, 2)$.\n\nb) Se o resultado divulgado foi $n$, teremos $10x + 30 + y = n$. Então, como $y < 7$ e $10x + 30$ termina em $0$, $y$ é o dígito das unidades de $n$ e $x$ é o dígito das dezenas subtraído de 3.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26815,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nConsidere dois tambores de capacidade suficientemente grande, um deles vazio e o outro cheio de líquido.\na) Determine se é possível colocar exatamente um litro do líquido do tambor cheio, no vazio, usando dois baldes, um com capacidade de 5 litros e o outro com capacidade de 7 litros.\nb) Determine se é possível colocar exatamente um litro do líquido de um dos tambores no outro usando dois baldes, um com capacidade de $2-\\sqrt{2}$ litros e o outro com capacidade de $\\sqrt{2}$ litros.",
    "options": [],
    "answer": "a) Yes, it is possible. b) No, it is not possible.",
    "solution": "Solution:\na) Basta encher o tambor vazio com 15 litros ($3 \\times 5$ litros) usando três vezes o balde de 5 litros e, em seguida, retirar 14 litros ($2 \\times 7$ litros) usando o balde de 7 litros duas vezes. Dessa forma, transportamos $3 \\times 5 - 2 \\times 7 = 1$ litro.\n\nb) A quantidade $a$ que podemos transportar do tambor cheio para o vazio é da forma $k(2-\\sqrt{2}) + l(\\sqrt{2})$ litros, onde $k$ e $l$ são inteiros que indicam quantas vezes tiramos ou colocamos líquidos usando cada um dos baldes. Se $l-k \\neq 0$, podemos escrever:\n$$\n\\begin{aligned}\na & = k(2-\\sqrt{2}) + l \\sqrt{2} \\\\\na - 2k & = \\sqrt{2}(l-k) \\\\\n\\frac{a-2k}{l-k} & = \\sqrt{2}\n\\end{aligned}\n$$\nAssim, o número $\\sqrt{2}$ seria o quociente de dois inteiros o que resultaria em um número racional. Sabemos que isso não pode acontecer porque $\\sqrt{2}$ é irracional. Falta analisarmos o que acontece quando $l = k$. A equação se transforma em:\n$$\n\\begin{aligned}\na & = k(2-\\sqrt{2}) + l \\sqrt{2} \\\\\n& = k(2-\\sqrt{2}) + k \\sqrt{2} \\\\\n& = 2k\n\\end{aligned}\n$$\nVeja que $2k$ é par e assim não podemos levar um valor ímpar como $a = 1$. Em qualquer caso, não é possível colocar exatamente 1 litro usando os baldes com as capacidades dadas neste item.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26816,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nQuantos dígitos possui o número $3^{100}$? Bom, podemos dar uma resposta aproximada para esta pergunta, sem usar a calculadora, simplesmente comparando-o com potências de $10$. Veja que $3^{2}<10$ nos permite concluir que $(3^{2})^{50}=3^{100}<10^{50}$. Então, $3^{100}$ tem no máximo $50$ dígitos pois, $10^{50}$ é o primeiro número com $51$ dígitos. O número $3^{100}$ tem de fato $48$ dígitos! Agora é a sua vez. Seja $N$ a quantidade de dígitos do número $2^{100}$, determine um inteiro positivo $k$ tal que $k \\leq N \\leq k+5$.",
    "options": [],
    "answer": "29",
    "solution": "Solution:\n\nVeja que $2^{3}=8$ é menor que $10$, então\n$$\n2^{100}<2^{102}=(2^{3})^{34}<10^{34}\n$$\nComo $10^{34}$ é o primeiro número com $35$ dígitos, $2^{100}$ possui no máximo $34$ dígitos. Além disso, veja que $2^{7}=128$ é maior que $100$. Daí\n$$\n2^{100}>2^{98}=(2^{7})^{14}>(10^{2})^{14}=10^{28}\n$$\nComo $10^{28}$ é o menor número com $29$ dígitos, $2^{100}$ possui pelo menos $29$ dígitos. Fica então demonstrado que $k=29$ satisfaz a condição dado que $29 \\leq N \\leq 34$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26817,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nO herói de um desenho animado enfrenta mais uma vez seu arqui-inimigo e precisa desferir seu famoso golpe do Raio Reflexivo. No quadrado da figura abaixo, o raio deverá, partindo de $F$ ricochetear, exatamente uma vez nos lados $C D, A D$ e $A B$, nesta ordem, antes de atingir o inimigo na posição $E$. Sempre que o raio ricocheteia em um dos lados do quadrado, o ângulo de incidência é igual ao ângulo de saída como mostra a figura da direita. Sabendo que $B E=E F=F C=2~\\mathrm{m}$ e que o raio viaja a $1~\\mathrm{m}/\\mathrm{s}$, determine o tempo decorrido entre o disparo do raio em $F$ e sua chegada ao ponto $E$.\n![](attached_image_1.png)",
    "options": [],
    "answer": "2 sqrt(61) seconds",
    "solution": "Solution:\nSe a parede onde o raio reflete fosse um espelho, a sua trajetória também \"apareceria\" no outro lado do espelho como uma continuação em linha reta da trajetória inicial. Assim, após refletirmos o quadrado três vezes ao longo dos lados onde o raio incide, conseguiremos traçar uma trajetória imaginária com o mesmo comprimento da trajetória real. No desenho abaixo, o quadrado $A B C D$ foi refletido incialmente com respeito ao lado $D C$ e depois seguido das reflexões nos lados $D J$ e $J O$. A soma dos quatro segmentos que compõem a trajetória real coincide com o comprimento do segmento $F W$. Como $O W=B E=E F$, segue que $O E=F W$.\n![](attached_image_2.png)\nPelo Teorema de Pitágoras no triângulo $O M E$, temos:\n$$\nO E^{2}=O M^{2}+E M^{2}=12^{2}+10^{2}=244\n$$\nPortanto, $O E=2 \\sqrt{61}~\\mathrm{m}$. Consequentemente, o raio levará $2 \\sqrt{61}$ segundos para atingir o alvo.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26818,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSejam $a$ e $b$ números reais positivos quaisquer. Determine o valor da expressão\n$$\n\\frac{\\sqrt{\\frac{a b}{2}}+\\sqrt{8}}{\\sqrt{\\frac{a b+16}{8}+\\sqrt{a b}}}\n$$",
    "options": [],
    "answer": "2",
    "solution": "Solution:\nSeja $x=\\sqrt{\\frac{a b}{2}}+\\sqrt{8}$. Então:\n$$\n\\begin{aligned}\nx^{2} & =\\frac{a b}{2}+4 \\sqrt{a b}+8 \\\\\n& =4\\left(\\frac{a b+16}{8}+\\sqrt{a b}\\right) \\\\\n& =4\\left(\\sqrt{\\frac{a b+16}{8}+\\sqrt{a b}}\\right)^{2}\n\\end{aligned}\n$$\nAssim, o valor da expressão procurada é\n$$\n\\begin{aligned}\n\\frac{\\sqrt{\\frac{a b}{2}}+\\sqrt{8}}{\\sqrt{\\frac{a b+16}{8}+\\sqrt{a b}}} & =\\frac{x}{x / 2} \\\\\n& =2\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26819,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa figura abaixo, $ABCD$ é um retângulo e $E$ é o ponto médio de $AD$. O segmento $FG$ passa pelo ponto médio $M$ de $CE$. Determine a razão entre os comprimentos de $GM$ e $MF$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "1/3",
    "solution": "Solution:\n\nPelo ponto $M$, trace o segmento de reta $PQ$ perpendicular aos lados $AB$ e $CD$ do retângulo $ABCD$ como mostra a figura abaixo. Como $M$ é o ponto médio de $CE$, podemos concluir que $PM$ é base média relativa ao lado $DE$ do triângulo $ECD$. Assim, se $DE = EA = x$, $PM = DE / 2 = x / 2$. Como $E$ é ponto médio de $DA$, temos $PQ = DA = 2x$. Consequentemente, $MQ = 2x - PM = 3x / 2$. Os triângulos $\\triangle PMG$ e $\\triangle MFQ$ são semelhantes, pois possuem os mesmos ângulos. Portanto,\n$$\n\\frac{GM}{MF} = \\frac{PM}{MQ} = \\frac{x / 2}{3x / 2} = \\frac{1}{3}\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26820,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nObserve a equação:\n$$\n\\begin{aligned}\n(1+2+3+4)^{2} & =(1+2+3+4)(1+2+3+4) \\\\\n& =1 \\cdot 1+1 \\cdot 2+1 \\cdot 3+1 \\cdot 4+2 \\cdot 1+2 \\cdot 2+2 \\cdot 3+2 \\cdot 4+ \\\\\n& +3 \\cdot 1+3 \\cdot 2+3 \\cdot 3+3 \\cdot 4+4 \\cdot 1+4 \\cdot 2+4 \\cdot 3+4 \\cdot 4\n\\end{aligned}\n$$\nNote que são formados $4 \\times 4=16$ produtos ao calcularmos $(1+2+3+4)^{2}$ usando a propriedade distributiva.\na) Quantos produtos serão formados ao calcularmos $(1+2+3+4)^{3}$ também usando a propriedade distributiva?\nb) Qual a quantidade de números de dois algarismos que usam apenas os dígitos 1, 2, 3 e 4?\nc) Qual a soma dos produtos dos dígitos de todos os números com quatro algarismos formados apenas pelos dígitos $1,2,3$ e 4 ?",
    "options": [],
    "answer": "a) 64; b) 16; c) 10000",
    "solution": "Solution:\na) Cada produto que aparece na soma final é uma expressão do tipo $x \\cdot y \\cdot z$ onde $x$ é um número vindo do primeiro parênteses, $y$ é um número vindo do segundo e $z$ um número vindo do terceiro. Como existem 4 opções possíveis para cada um desses números, pelo princípio multiplicativo temos $4 \\cdot 4 \\cdot 4=64$ tais produtos.\n\nb) Tendo apenas 4 opções de dígitos, existem 4 opções de escolha para o dígito das dezenas e quatro opções de escolha para o dígito das unidades. Assim, temos $4 \\cdot 4=16$ números de dois algarismos usando apenas os dígitos 1, 2, 3 e 4 .\n\nc) Calculemos inicialmente a soma dos produtos dos dígitos de todos os números com dois algarismos que usam apenas os quatro dígitos dados. Pelo item anterior, existem 16 tais números, a saber: $11,12,13,14,21,22,23,24,31,32,33,34,41,42,43,44$. A soma dos 16 produtos dos dígitos é:\n$$\n\\begin{aligned}\n& 1 \\cdot 1+1 \\cdot 2+1 \\cdot 3+1 \\cdot 4+2 \\cdot 1+2 \\cdot 2+2 \\cdot 3+2 \\cdot 4= \\\\\n& 3 \\cdot 1+3 \\cdot 2+3 \\cdot 3+3 \\cdot 4+4 \\cdot 1+4 \\cdot 2+4 \\cdot 3+4 \\cdot 4=(1+2+3+4)^{2}\n\\end{aligned}\n$$\nIsso nos motiva a analisar o número $(1+2+3+4)^{4}=10000$. Veja que se aplicarmos a propriedade de distributividade irão aparecer $4 \\cdot 4 \\cdot 4 \\cdot 4=256$ produtos e cada um deles coincide com o produto dos dígitos dos números de quatro algarismos formados apenas pelos dígitos 1, 2, 3 e 4. Assim, o valor procurado é 10000 .",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26821,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nJoão trabalha vendendo pacotes de previsão astrológica. Para incrementar as vendas de suas previsões, ele oferece descontos caso pessoas de um mesmo signo queiram contratar seus serviços. No Horóscopo Grego, como existem exatamente 12 signos, portanto, em um grupo de 13 pessoas, sempre duas delas terão o mesmo signo e poderão se interessar pelo pacote promocional.\n\na) Qual o número mínimo de pessoas que um grupo deve possuir para ele ter certeza de que existirão pelo menos 3 pessoas de um mesmo signo do Horóscopo Grego?\n\nb) No Horóscopo Chinês, também existem exatamente 12 signos. Se João quiser ter certeza de que, em determinado grupo de pessoas existirão duas possuindo exatamente os mesmos signos, tanto no Horóscopo Grego quanto no Horóscopo Chinês, qual o número mínimo de pessoas que tal grupo deve ter?",
    "options": [],
    "answer": "a) 25; b) 145",
    "solution": "Solution:\n\na) O mínimo é 25. Se em um grupo de 24 pessoas cada signo aparecer no máximo duas vezes, teremos no máximo $2 \\cdot 12=24$ pessoas. Como $24<25$, isso mostra que pelo menos um dos signos deverá aparecer três vezes. De fato, esse é o mínimo onde tal propriedade ocorre pois se considerarmos 24 pessoas divididas em 12 pares com o mesmo signo, a propriedade do enunciado não será encontrada.\n\nb) O número mínimo é $12 \\cdot 12+1=145$. Veja que existem no máximo $12 \\cdot 12=144$ pares de combinações possíveis entre signos Gregos e Chineses. Se escolhermos 145 pessoas e as dividirmos de acordo com esses pares, pelo menos um deles deverá ser usado duas vezes. Não é possível concluirmos isso com menos de 145, pois é possível 144 pessoas apresentarem todos os pares possíveis de combinações sem repetições.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26822,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm número natural é bacana se a soma de todos os seus divisores positivos (incluindo $1$ e $n$) é maior ou igual ao dobro do número. Por exemplo, $12$ é bacana pois $1+2+3+4+6+12=28 \\geq 24=2 \\cdot 12$ enquanto que $4$ não é bacana pois $1+2+4<8=2 \\cdot 4$. Demonstre que existem infinitos números que são bacanas e infinitos números que não são bacanas.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nVeja que nenhum número primo é bacana, pois se $p$ é primo a soma dos seus divisores positivos é $p+1<2p$. Como existem infinitos números primos, segue que existem infinitos números que não são bacanas.\n\nSe $p$ é um primo maior que $3$, o número $n=12p$ possui pelo menos os seguintes divisores positivos distintos:\n$$\n\\begin{aligned}\np+2p+3p+4p+6p+12p &= 28p \\\\\n&> 24p \\\\\n&= 2n\n\\end{aligned}\n$$\nAssim, todo número da forma $12p$ com $p$ primo maior que $3$ é bacana. Novamente, como existem infinitos números primos, temos uma coleção infinita de números bacanas.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26823,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm decorador distribuirá flores em oito pontos ao redor de um arranjo quadrado de flores, como indicado na figura abaixo. Ele quer fazer isso de modo tal que, em cada lado do arranjo, as pessoas vejam sempre a mesma quantidade de flores. No exemplo abaixo, temos o total de 11 flores e em cada um dos 4 lados do quadrado são vistas exatamente 4 delas.\n\na) Qual o número máximo de flores que podem ser usadas, considerando que em cada lado do quadrado devem ser vistas exatamente 9 flores?\n\nb) Qual o número mínimo de flores que podem ser usadas, considerando que em cada lado do quadrado devem ser vistas exatamente 12 flores?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "a) 36; b) 24",
    "solution": "Solution:\n\na) A soma das flores vistas nos lados é $4 \\cdot 9=36$. Como as flores nos cantos são vistas por dois lados e as flores no meio dos lados são vistas apenas uma vez, podemos escrever:\n$$\n2C+M=36\n$$\nonde $C$ e $M$ indicam as quantidades de flores nos cantos e no meio. Consequentemente, $C+M=36-C \\leq 36$ e seu valor será no máximo 36, que ocorre quando $C=0$. Assim, devemos distribuir nas posições 2, 4, 6 e 8 exatamente 9 flores. Portanto, o número máximo de flores é 36.\n\nb) A soma das flores vistas nos lados agora é $4 \\cdot 12=48$ e a equação do item anterior se transforma em $2C+M=48$. Consequentemente, $2(C+M)=48+M \\geq 48$. Para atingir tal valor, como devemos ter $M=0$, basta distribuir 6 flores em cada um dos quatro cantos 1, 3, 5 e 7 do arranjo. Portanto, o número mínimo de flores é 24.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26824,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPedro acabou de se mudar para sua nova casa e ganhou um novo quarto. A figura a seguir mostra uma vista superior simplificada de seu novo quarto que possui $2~\\mathrm{m}$ de largura por $2{,}5~\\mathrm{m}$ de comprimento.\n\n![](attached_image_1.png)\n\nA porta indicada na figura tem $50~\\mathrm{cm}$ de comprimento e pode ser aberta até encontrar a parede lateral. A janela é dividida em duas portas de mesmo comprimento que quando abertas encostam nas paredes vizinhas. Os arcos da figura mostram as aberturas da porta e da janela. A mãe de Pedro disse que ele deve colocar seus móveis no quarto de modo que não fiquem nos caminhos de abertura da porta nem da janela. Quantos metros quadrados Pedro tem em seu quarto para colocar os seus móveis?",
    "options": [],
    "answer": "(80 - 5π)/16",
    "solution": "Solution:\n\nSeja $L$ o comprimento de cada porta da janela. Considerando que, quando as duas portas se abrem elas encostam nas paredes dos lados, temos então: $4 \\cdot L = 2$, ou seja, $L = 0,5~\\mathrm{m}$.\n\nChamemos de $A$ a área que Pedro tem para colocar seus móveis. Para determiná-la, basta considerar a área total e subtrair as áreas de abertura da porta e da janela. Sabendo que a porta abre $\\frac{1}{4}$ de circunferência e que as janelas abrem, cada uma, meia circunferência, temos:\n\n$$\n\\begin{aligned}\nA & = 2 \\cdot 2,5 - \\frac{1}{4} \\cdot \\pi \\cdot (0,5)^2 - 2 \\cdot \\frac{1}{2} \\cdot \\pi \\cdot (0,5)^2 \\\\\n& = 5 - \\frac{\\pi}{16} - \\frac{\\pi}{4} \\\\\n& = \\frac{80 - 5\\pi}{16}\n\\end{aligned}\n$$\n\nEntão, Pedro possui $\\frac{80 - 5\\pi}{16}$ metros quadrados para colocar seus móveis.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26825,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nArnaldo e Bernaldo decidem jogar um jogo que possui um número limitado de jogadas. Arnaldo escreve o número 1 no quadro em sua primeira jogada. Em seguida, Bernaldo escreve 2 ou 4 no quadro. Depois disso, Arnaldo escreve 3 ou 9 no quadro. Os dois continuam jogando alternadamente mantendo a regra de que na jogada $n$ o jogador escreve $n$ ou $n^{2}$ no quadro. Arnaldo vence o jogo se, após a última jogada, a soma dos números no quadro for divisível por 3. Se a soma não for divisível por 3, então Bernaldo vence.\n\na) Suponha que o jogo acabe na jogada de número 15. Mostre que Bernaldo pode garantir a vitória.\n\nb) Suponha que o jogo acabe na jogada de número 7. Nesse caso, qual dos dois jogadores poderá sempre garantir a vitória independentemente de como o seu adversário jogue? Como ele deverá jogar para vencer?",
    "options": [],
    "answer": "a) Bernaldo can force a win when the game ends on the fifteenth move. b) Arnaldo can force a win when the game ends on the seventh move; his winning strategy is: if Bernaldo writes two on his second move, Arnaldo writes twenty‑five on his fifth move; if Bernaldo writes four on his second move, Arnaldo writes five on his fifth move.",
    "solution": "Solution:\n\na) Veja que 15 é divisível por 3, então independente da última jogada, 15 ou $15^{2}$, o resto na divisão por 3 não será alterado. Vejamos a jogada de número 14. Veja que 14 deixa resto 2 na divisão por 3 enquanto que $14^{2}=196$ deixa resto 1 na divisão por 3. Como 14 é par, quem fará tal jogada é Bernaldo. Ele pode garantir sua vitória da seguinte forma:\n\ni) Se a soma dos 13 primeiros números deixar resto 0 ou 2 por 3, Bernaldo deve jogar 14, tornando o resto 2 ou 1 (que é o resto de $2+2=4$).\n\nii) Se a soma dos 13 primeiros deixar resto 1, então Bernaldo deve usar 196 tornando o resto total 2.\n\nLogo, Bernaldo pode garantir sua vitória em qualquer caso.\n\nb) Observe que 7 e $7^{2}=49$ deixam resto 1 na divisão por 3. Então a sétima jogada acrescentará 1 ao resto da soma dos números anteriores da escolha. Estendendo esse raciocínio, vemos que 1, 3, 4, 6 e 7 deixam os mesmos restos que seus quadrados na divisão por 3. Assim, não importa o que seja feito nessas jogadas correspondentes a esses números, elas contribuirão com os restos:\n$$\n1+0+1+0+1=3\n$$\nComo a soma anterior deixa resto 0 por 3, as jogadas relevantes são as de número 2 (que pertence Bernaldo) e a de número 5 (que pertence a Arnaldo). Para Arnaldo garantir sua vitória, basta que ele jogue da seguinte forma:\n\ni) Se Bernaldo jogar 2, Arnaldo deve jogar 25 totalizando $2+25=27$ que deixa resto 0 por 3.\n\nii) Se Bernaldo jogar 4, Arnaldo deve jogar 5 totalizando $4+5=9$ que deixa resto 0 por 3.\n\nDesse modo, Arnaldo pode garantir sua vitória em qualquer caso.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26826,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm tabuleiro de Xadrez tem suas linhas e colunas numeradas conforme a figura a seguir. Em cada casa é escrito o número que é a soma dos números da linha e da coluna dessa casa. Por exemplo, na casa que está na linha 4 e na coluna 5 é escrito o número $4+5=9$.\n![](attached_image_1.png)\n\na) Qual a soma dos números escritos em todas as casas do tabuleiro?\n\nb) Sejam $S_{\\text {pretas }}$ a soma de todos os números escritos nas casas pretas e $S_{\\text {brancas }}$ a soma de todos os números escritos em casas brancas. Quanto vale a diferença $S_{\\text {pretas }}-S_{\\text {brancas }}$ ?\n\nc) Quanto vale $S_{\\text {pretas }}$ ?",
    "options": [],
    "answer": "a) 576; b) 0; c) 288",
    "solution": "Solution:\n\na) Veja que na linha 1 o número 1 é somado em cada uma das 8 casinhas, na linha 2 o número 2 também é somado oito vezes, e assim por diante. Desse modo, podemos contabilizar a contribuição das linhas por:\n$$\n\\begin{aligned}\n8 \\cdot (1+2+3+4+5+6+7+8) &= 8 \\cdot 36 \\\\\n&= 288\n\\end{aligned}\n$$\nO mesmo se passa com a contribuição das colunas, também totalizando 288. Concluímos que a soma de todas as casas é $2 \\cdot 288 = 576$.\n\nb) Essa diferença também pode ser feita analisando-se a contribuição de cada linha e cada coluna, mas nesse caso para cada casa preta devemos somar o seu número, enquanto que para cada casa branca devemos subtraí-lo. Como cada linha e cada coluna possui exatamente quatro casas pretas e quatro brancas, o número escrito em uma linha ou coluna deve ser somado e subtraído 4 vezes, ou seja, contribui com $4-4=0$. Portanto, concluímos que $S_{\\text {pretas }}-S_{\\text {brancas }}=0$.\n\nc) Juntando as duas informações dos itens $a$) e $b$), temos\n$$\n\\left\\{\n\\begin{array}{l}\nS_{\\text {pretas }}+S_{\\text {brancas }}=576 \\\\\nS_{\\text {pretas }}-S_{\\text {brancas }}=0 .\n\\end{array}\n\\right.\n$$\nSomando as duas linhas, temos $2 S_{\\text {pretas }}=576$, ou seja, $S_{\\text {pretas }}=288$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26827,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDizemos que um número inteiro positivo $n$ é espertinho se existirem números inteiros positivos $a, b, c$ e $d$, não necessariamente distintos, tais que:\n$$\nn=\\frac{a^{2}-b^{2}}{c^{2}+d^{2}}\n$$\nPor exemplo, 12 é espertinho, pois:\n$$\n12=\\frac{16^{2}-4^{2}}{4^{2}+2^{2}}\n$$\nMostre que todos os números inteiros positivos são espertinhos.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nPodemos reescrever a equação da seguinte forma:\n$$\n\\begin{aligned}\nn & =\\frac{a^{2}-b^{2}}{c^{2}+d^{2}} \\\\\nn\\left(c^{2}+d^{2}\\right) & =(a+b)(a-b)\n\\end{aligned}\n$$\nSe conseguirmos inteiros positivos tais que $a+b=n$ e $a-b=c^{2}+d^{2}$, teremos uma solução para a equação. Se $n$ é par, basta fazermos $c=d=1$ e resolvermos o sistema\n$$\n\\left\\{\\begin{array}{l}\na+b=n \\\\\na-b=2\n\\end{array}\\right.\n$$\nSomando e subtraindo as equações, encontramos a solução $a=\\frac{n+2}{2}$ e $b=\\frac{n-2}{2}$. Se $n>2, a$ e $b$ são inteiros positivos. Para ver que $n=2$ também é espertinho, escrevemos\n$$\n2=\\frac{5^{2}-3^{2}}{2^{2}+2^{2}}\n$$\nSe $n$ é ímpar, podemos usar o mesmo raciocínio tomando $c=1$ e $d=2$. Daí, teremos $c^{2}+d^{2}=5$ e:\n$$\n\\left\\{\\begin{array}{l}\na+b=n \\\\\na-b=5\n\\end{array}\\right.\n$$\nNovamente, somando e subtraindo as equações, encontramos a solução $a=\\frac{n+5}{2}$ e $b=\\frac{n-5}{2}$. Se $n>5, a$ e $b$ são inteiros positivos. Para ver que 1, 3 e 5 também são espertinhos, escrevemos:\n$$\n\\begin{aligned}\n1 & =\\frac{7^{2}-6^{2}}{3^{2}+2^{2}} \\\\\n3 & =\\frac{8^{2}-5^{2}}{3^{2}+2^{2}} \\\\\n5 & =\\frac{9^{2}-4^{2}}{3^{2}+2^{2}}\n\\end{aligned}",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26828,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\na) \"Se um quadrilátero $ABCD$ é tal que $\\angle ABC = \\angle ADC = 90^\\circ$, então $AB^2 - CD^2 = AD^2 - BC^2$.\"\nb) \"Se um quadrilátero $ABCD$ é tal que $\\angle ACB = \\angle ADC = 90^\\circ$, então $AB^2 = BC^2 + CD^2 + AD^2$.\"\nProve esses resultados.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\na) Como $\\angle ABC = \\angle ADC = 90^\\circ$, sabemos que os triângulos $\\triangle ABC$ e $\\triangle ADC$ são retângulos. Pelo Teorema de Pitágoras, temos\n$$\n\\begin{aligned}\nAB^2 + BC^2 & = AC^2 \\\\\nAD^2 + CD^2 & = AC^2\n\\end{aligned}\n$$\nPortanto,\n$$\n\\left(AB^2 - CD^2\\right) - \\left(AD^2 - BC^2\\right) = \\left(AB^2 + BC^2\\right) - \\left(AD^2 + CD^2\\right) = AC^2 - AC^2 = 0\n$$\ne, consequentemente $AB^2 - CD^2 = AD^2 - BC^2$.\n\nb) De $\\angle ACB = \\angle ADC = 90^\\circ$, sabemos que os triângulos $\\triangle ACB$ e $\\triangle ADC$ são retângulos. Usando o Teorema de Pitágoras teremos\n$$\n\\begin{aligned}\nAC^2 & = AD^2 + CD^2 \\\\\nAB^2 & = BC^2 + AC^2 \\\\\n& = BC^2 + \\left(AD^2 + CD^2\\right) \\\\\n& = BC^2 + CD^2 + AD^2\n\\end{aligned}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26829,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDez crianças decidem correr ao redor de um lago circular com $200\\ \\mathrm{~m}$ de perímetro. No início da corrida, as dez crianças estão paradas ocupando posições distintas e cada uma delas correrá no sentido horário ou anti-horário, a depender de sua vontade, com velocidade de $\\frac{200}{k}\\ \\mathrm{~m} / \\mathrm{min}$, onde $k$ é um inteiro positivo. Mostre que depois de certo tempo, existirá um instante em que todas as crianças estarão exatamente sobre as suas mesmas posições iniciais.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSe uma criança tem velocidade $\\frac{200}{k}\\ \\mathrm{~m} / \\mathrm{min}$, então ela demorará $k$ minutos para dar uma volta completa no lago. Tendo isso em mente, se $k_{1}, k_{2}, \\ldots, k_{10}$ denotam os inteiros associados às velocidades das 10 crianças e $M$ é um múltiplo comum de todos eles, após $M$ minutos a primeira criança terá feito exatamente $\\frac{M}{k_{1}}$ voltas e estará sobre sua posição inicial. Do mesmo modo, cada uma das outras crianças terá feito uma quantidade inteira de voltas e estará sobre sua posição inicial. Concluímos então que após $M$ minutos todas as crianças estarão em suas posições iniciais.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26830,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSeja $n$ um número inteiro positivo. Se, para cada divisor primo $p$ de $n$, o número $p^{2}$ não divide $n$, dizemos então que $n$ é livre de quadrados. Mostre que todo número livre de quadrados tem uma quantidade de divisores que é igual a uma potência de 2.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSuponha que $n$ é um número livre de quadrados e considere sua fatoração em primos:\n$$\nn = p_{1}^{\\alpha_{1}} \\cdot p_{2}^{\\alpha_{2}} \\cdot \\ldots \\cdot p_{k}^{\\alpha_{k}}\n$$\nComo $n$ é livre de quadrados, os expoentes $\\alpha_{i}$ são todos iguais a $1$. Portanto,\n$$\nn = p_{1} \\cdot p_{2} \\cdot \\ldots \\cdot p_{k}\n$$\nPara contarmos os divisores de $n$, basta contarmos quantos números possuem ou não cada um desses primos $p_{i}$. Como temos duas possibilidades para cada um desses primos figurar em um divisor, a saber, estar ou não estar na fatoração dele, pelo princípio multiplicativo temos\n$$\n2 \\cdot 2 \\cdot \\ldots \\cdot 2 = 2^{k}\n$$\ndivisores.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26831,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCinco crianças sentam-se ao redor de uma mesa circular. Cada criança escolhe um número inteiro positivo e o relata para as outras. Em seguida, cada criança faz a seguinte conta: soma os números das duas crianças à sua esquerda, subtrai a soma dos números das outras duas crianças à sua direita e multiplica essa diferença pelo seu próprio número, chegando assim ao seu resultado final.\nProve que a soma dos resultados finais de todas as crianças é um valor fixo que não depende dos números que as crianças escolheram inicialmente e, em seguida, determine esse valor.",
    "options": [],
    "answer": "0",
    "solution": "Solution:\n\nVamos supor que os números em sentido horário são $a, b, c, d$ e $e$. Os valores obtidos como resultados finais são:\n$$\n\\begin{aligned}\na((e+d)-(b+c)) & = a e + a d - a b - a c \\\\\nb((a+e)-(c+d)) & = b a + b e - b c - b d \\\\\nc((b+a)-(d+e)) & = c b + c a - c d - c e \\\\\nd((c+b)-(e+a)) & = d c + d b - d e - d a \\\\\ne((d+c)-(a+b)) & = e d + e c - e a - e b\n\\end{aligned}\n$$\nVeja que cada produto de dois números escolhidos inicialmente aparece uma vez com o sinal $+$ e uma vez com o sinal $-$, por exemplo, $a e$ aparece positivo na primeira expressão e negativo na última. Isso acontece, pois se um número $x$ tem $y$ no lado esquerdo, aparecerá $+x y$ em seu resultado final associado, enquanto que $y$ tendo $x$ do lado direito produzirá $-x y$ também em seu resultado final associado. Sendo assim, a soma dos resultados é sempre igual a zero.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26832,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa figura abaixo, os trapézios retângulos $ABCD$ e $AEFG$, com $BC \\parallel EF$ e $CD \\parallel FG$, possuem a mesma área. Sabendo que $BC=4$, $AD=7$, $CT=1$ e $TD=2$, determine a medida do segmento $DG$.\n![](attached_image_1.png)",
    "options": [],
    "answer": "9/4",
    "solution": "Solution:\n\nTrace uma perpendicular por $B$ ao lado $DA$, intersectando $EF$ em $Y$ e $AD$ em $X$. Chamemos $DG$ de $x$ e $EY$ de $y$. Veja a figura abaixo:\n![](attached_image_2.png)\nOs triângulos $\\triangle EYB$ e $\\triangle AXB$ possuem os mesmos ângulos e, consequentemente, são semelhantes. Desse modo, $\\frac{y}{3} = \\frac{1}{1+2}$ e, por conseguinte, $y = 1$.\n\nCalculemos as áreas dos trapézios dados:\n$$\n\\begin{aligned}\n[ABCD] & = \\frac{CD \\cdot (BC + AD)}{2} \\\\\n& = \\frac{3 \\cdot (4 + 7)}{2} \\\\\n& = \\frac{33}{2}\n\\end{aligned}\n$$\ne\n$$\n\\begin{aligned}\n[AEFG] & = \\frac{FG \\cdot (EF + AG)}{2} \\\\\n& = \\frac{2 \\cdot ((x+5) + (x+7))}{2} \\\\\n& = 2x + 12\n\\end{aligned}\n$$\nComo $[ABCD] = [AEFG]$, temos $33 = 4x + 24$ e $DG = x = 9/4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26833,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSejam $a$ e $b$ dois dígitos diferentes de zero não necessariamente diferentes. O número de dois dígitos $\\overline{a b}$ é chamado de curioso, se ele for um divisor do número $\\overline{b a}$, que é formado pela troca da ordem dos dígitos de $\\overline{a b}$. Ache todos os números curiosos.\n\nObservação: O traço sobre os números serve para distinguir o produto $a \\cdot b$ do número de dois dígitos $\\overline{a b}$.",
    "options": [],
    "answer": "{11, 22, 33, 44, 55, 66, 77, 88, 99}",
    "solution": "Solution:\nO número de dois dígitos $\\overline{a b}$ pode ser escrito como $10 a + b$, assim como $\\overline{b a} = 10 b + a$. Se $10 a + b$ é divisor de $10 b + a$, temos $10 b + a = (10 a + b) k$, onde $k$ é um inteiro menor ou igual a 9 já que os dois números possuem dois dígitos. Segue que\n$$\n\\begin{aligned}\n10 b + a & = (10 a + b) k \\\\\n10 b + a + 10 a + b & = (10 a + b)(k + 1) \\\\\n11(a + b) & = (10 a + b)(k + 1)\n\\end{aligned}\n$$\nPela última equação, o número $(10 a + b)(k + 1)$ deve ser múltiplo de 11. Como $k \\leq 9$, temos $k + 1 \\leq 10$ e, consequentemente, $k + 1$ não possui fator 11, implicando que $10 a + b$ deve ser múltiplo de 11. Todos os números de dois dígitos múltiplos de 11 possuem dígitos iguais e isso nos permite concluir que necessariamente $a = b$. Quando $a = b$, $\\overline{a b} = \\overline{b a}$ e, certamente, um divide o outro. Portanto, o conjunto dos números curiosos é\n$\\{11, 22, 33, 44, 55, 66, 77, 88, 99\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26834,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\na)\nVerifique que $(1+\\operatorname{tg} k)\\left(1+\\operatorname{tg}\\left(45^\\circ-k\\right)\\right)=2$.\n\nb)\nDado que\n$$\n\\left(1+\\operatorname{tg} 1^\\circ\\right)\\left(1+\\operatorname{tg} 2^\\circ\\right) \\cdot \\ldots \\cdot\\left(1+\\operatorname{tg} 45^\\circ\\right)=2^n\n$$\nencontre $n$.",
    "options": [],
    "answer": "23",
    "solution": "Solution:\na)\n$$\n\\begin{aligned}\n\\operatorname{tg}\\left(45^\\circ-k\\right)+1 & =\\frac{\\operatorname{sen}\\left(45^\\circ-k\\right)}{\\cos \\left(45^\\circ-k\\right)}+1 \\\\\n& =\\frac{\\operatorname{sen} 45^\\circ \\cos k-\\cos 45^\\circ \\operatorname{sen} k}{\\cos 45^\\circ \\cos k+\\operatorname{sen} 45^\\circ \\operatorname{sen} k}+1 \\\\\n& =\\frac{\\sqrt{2} / 2 \\cos k-\\sqrt{2} / 2 \\operatorname{sen} k}{\\sqrt{2} / 2 \\cos k+\\sqrt{2} / 2 \\operatorname{sen} k}+1 \\\\\n& =\\frac{\\frac{\\cos k}{\\cos k}-\\frac{\\operatorname{sen} k}{\\cos k}}{\\frac{\\cos k}{\\cos k}+\\frac{\\operatorname{sen} k}{\\cos k}} \\\\\n& =\\frac{1-\\operatorname{tg} k}{1+\\operatorname{tg} k}+1 \\\\\n& =\\frac{2}{1+\\operatorname{tg} k}\n\\end{aligned}\n$$\nConsequentemente, $\\left(\\operatorname{tg}\\left(45^\\circ-k\\right)+1\\right)(\\operatorname{tg} k+1)=2$.\n\nb)\nO item anterior nos permite agrupar os primeiros 44 termos do produto dado, através de pares da forma $\\left(\\operatorname{tg}\\left(45^\\circ-k\\right)+1\\right)(\\operatorname{tg} k+1)$, em 22 produtos iguais a 2. Como $1+\\operatorname{tg} 45^\\circ=2$, segue que $2^n=2^{23}$ e $n=23$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26835,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNo quadrilátero $A B C D$, o lado $A D$ é tal que $A D = A B + C D$. Se $P$ é o ponto de encontro das bissetrizes de $\\angle B A D$ e $\\angle C D A$, mostre que $B P = P C$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nConsidere um ponto $E$ sobre o lado $A D$ de modo que $A E = A B$. Consequentemente, $E D = A D - A E = A D - A B = C D$. Além disso, como $A P$ e $B P$ são bissetrizes, temos $\\triangle B A P \\equiv \\triangle A P E$ e $\\triangle P E D \\equiv \\triangle P D C$ pelo caso de congruência (L.A.L.). Daí, de $B P = P E$ e $P E = P C$, segue que $B P = P C$.\n\n![](attached_image_2.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26836,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAlgumas lâmpadas de Natal são arranjadas usando fios mágicos. Cada lâmpada pode ser da cor verde ou amarela. Cada fio está ligado a duas lâmpadas e tem uma propriedade mágica: quando alguém toca em um fio unindo duas lâmpadas, cada uma delas troca de cor passando de verde para amarela ou de amarela para verde.\n\na) No arranjo a seguir, cada ponto representa uma lâmpada e os segmentos representam os fios mágicos. No começo todas elas são amarelas. Qual o menor número de fios que devemos tocar para que todas as lâmpadas se tornem verdes? Mostre um exemplo de como fazer essa mudança com esse número mínimo de fios.\n\n![](attached_image_1.png)\n\nb) Considere o arranjo da figura a seguir onde todas as lâmpadas estão com a cor amarela. Mostre que não é possível tocar em alguns fios mágicos e mudar a cor de todas as lâmpadas para o verde.\n\n![](attached_image_2.png)",
    "options": [],
    "answer": "a) 8; b) impossible",
    "solution": "Solution:\n\na) Cada fio que tocamos muda exatamente a cor de duas lâmpadas. Como existem $16$ lâmpadas amarelas, devemos encostar em pelo menos $8$ fios. A figura a seguir mostra um exemplo de escolhas de fios que torna isso possível:\n\n![](attached_image_3.png)\n\nb) Veja que a configuração dada no item b) possui exatamente $23$ lâmpadas. Note que ao encostar em um fio mágico temos três possibilidades de mudanças de lâmpadas:\n\ni) Duas verdes podem virar duas amarelas.\n\nii) Duas amarelas podem virar duas verdes.\n\niii) Uma verde e uma amarela viram uma amarela e uma verde.\n\nAssim, ou o número de lâmpadas verdes diminui $2$, aumenta $2$ ou permanece o mesmo. Desse modo, se no início começamos com $0$ verdes, sempre teremos uma quantidade par de lâmpadas verdes. Como $23$ é ímpar, não é possível chegar a tornar todas as $23$ lâmpadas verdes.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26837,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSejam $D$ um ponto no lado $AB$ do triângulo $\\triangle ABC$ e $F$ a interseção de $CD$ e da mediana $AM$. Se $AF = AD$, encontre a razão entre $BD$ e $FM$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "2",
    "solution": "Solution:\n\nSeja $E$ o ponto de interseção da reta paralela ao lado $AB$, que passa pelo ponto $M$, com o segmento $CD$. Como $ME \\parallel AB$, segue que $\\angle MED = \\angle EDA$. Além disso, como $\\angle AFD = \\angle EFM$ e o triângulo $\\triangle AFD$ é isósceles, podemos concluir que $\\triangle EFM$ também é isósceles com $EM = FM$. Dado que $M$ é ponto médio de $BC$ e $EM \\parallel BD$, o segmento $EM$ é uma base média do triângulo $\\triangle CDB$. Assim,\n$$\n\\frac{BD}{FM} = \\frac{BD}{EM} = 2\n$$\n\n![](attached_image_2.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26838,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nÉ dado um tabuleiro $8 \\times 8$.\na) Qual o número mínimo de casinhas que devemos marcar nesse tabuleiro, de modo que cada um de seus subtabuleiros $3 \\times 3$ possua pelo menos uma casinha marcada?\nb) Qual o número mínimo de casinhas que devemos marcar nesse tabuleiro, de modo que cada um de seus subtabuleiros $3 \\times 3$ possua pelo menos três casinhas marcadas?",
    "options": [],
    "answer": "a) 4; b) 16",
    "solution": "Solution:\n\na)\nConsidere a figura a seguir.\n\n![](attached_image_1.png)\n\nCada um dos quatro subtabuleiros $3 \\times 3$ assinalados na figura deve ter pelo menos uma casa marcada. Além disso, com as quatro casas marcadas na figura acima, temos a propriedade desejada. Portanto, o mínimo é $4$.\n\nb)\nConsidere a figura a seguir.\n\n![](attached_image_2.png)\n\nVeja que cada um dos seis pedaços $3 \\times 3$ deve ter pelo menos três casas marcadas. Veja também que existem duas casas que podem ser contadas para dois pedaços. Com isso, teremos no mínimo $6 \\times 3 - 2 = 16$ casas marcadas. Observe ainda que com as $16$ casas marcadas na figura temos a propriedade desejada.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26839,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUmberto e Doisberto jogam em um tabuleiro $3 \\times n$ colocando dominós sempre cobrindo duas casas adjacentes (com lado em comum) do tabuleiro. Umberto faz a primeira jogada, Doisberto faz a segunda e eles seguem jogando alternadamente. Perde o jogador que não conseguir jogar. Para cada um dos casos abaixo, diga quais dos jogadores pode bolar uma estratégia e sempre garantir a vitória independentemente de como o outro jogue.\n\na) $n=3$\n\nb) $n=4$",
    "options": [],
    "answer": "a) Doisberto\nb) Umberto",
    "solution": "Solution:\n\na) Doisberto pode sempre garantir a vitória. Basta ele realizar um movimento que complete um quadrado $2 \\times 2$ a partir do primeiro dominó de Umberto.\n\n![](attached_image_1.png)\n\nVeja na figura que sobram 5 casas. Independente da jogada de Umberto, na jogada seguinte de Doisberto, o jogo acaba com a sua vitória.\n\nb) Umberto pode sempre garantir a vitória. Basta ele jogar o primeiro dominó nas duas casas centrais do tabuleiro.\n\n![](attached_image_2.png)\n\nA partir daí, a cada jogada de Doisberto, Umberto deve jogar de forma simétrica em relação ao centro do tabuleiro, ou seja, como se ele imitasse a jogada de Doisberto. Por exemplo, se Doisberto colocar uma peça na horizontal começando no canto superior esquerdo, Umberto deve colocar outra peça também na horizontal começando no canto inferior direito. Desse modo, se Doisberto fizer uma jogada, certamente Umberto também poderá fazer a sua. Depois de algumas jogadas, Doisberto não poderá jogar e perderá o jogo.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26840,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa figura abaixo, $ABCD$ é um quadrado e os pontos $K$, $L$ e $M$ estão sobre os lados $AB$, $BC$ e $CD$ de modo que $\\triangle KLM$ é um triângulo isósceles retângulo em $L$. Prove que $AL$ e $DK$ são perpendiculares.\n\n![](attached_image_1.png)\n",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSejam $\\angle MLC = \\alpha$ e $\\angle BAL = \\beta$. Como $\\angle KLM = \\angle KBL = \\angle LCM = 90^{\\circ}$, segue que $\\angle KLB = \\angle LMC = 90^{\\circ} - \\alpha$. Além disso, como $KL = LM$, os triângulos retângulos $\\triangle LMC$ e $\\triangle BLK$ são congruentes. De $BL = CM$ e $BK = LC$, segue que\n$$\nAK = AB - BK = BC - LC = BL\n$$\nOs triângulos $\\triangle KAD$ e $\\triangle ABL$ são congruentes pois $AK = BL$, $AB = AD$ e $\\angle ABL = \\angle KAD$. Consequentemente, $\\angle ADK = \\beta$ e $\\angle LAD = 90^{\\circ} - \\angle BAL = 90^{\\circ} - \\beta$. Como os ângulos $\\angle LAD$ e $\\angle KDA$ são complementares, segue finalmente que $AL$ e $KD$ são perpendiculares.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26841,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\na) Mostre que não é possível separar os números do conjunto $A=\\{1,2,3, \\ldots, 10\\}$ em dois conjuntos em que o produto dos números em cada um deles é o mesmo.\n\nb) Qual o menor número de elementos que precisamos retirar do conjunto $A$ de modo que os elementos restantes possam ser divididos em dois conjuntos cujo produto de seus elementos sejam iguais? Mostre que números devem ser retirados e como separar os dois conjuntos.",
    "options": [],
    "answer": "a) Impossible to split {1,2,3,4,5,6,7,8,9,10} into two groups with equal product.\nb) Minimal removals: 1 (remove 7). One valid partition of the remaining set {1,2,3,4,5,6,8,9,10} is {1,10,3,6,4} and {5,9,2,8}, each with product 720.",
    "solution": "Solution:\na) Basta olharmos para o número $7$. Como ele é o único número de $A$ com fator $7$, não é possível dividi-los em dois com o mesmo produto de seus elementos, pois um desses produtos seria múltiplo de $7$ e o outro não.\n\nb) Retirando apenas o número $7$, mostraremos que é possível fazer tal divisão. Listemos as fatorações dos outros números em primos:\n\n| 1 | 2 | 3 | 4 | 5 | 6 | 8 | 9 | 10 |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| 1 | 2 | 3 | $2^{2}$ | 5 | $2 \\cdot 3$ | $2^{3}$ | $3^{2}$ | $2 \\cdot 5$ |\n\nExistem exatamente dois números com fatores $5$ e inevitavelmente $5$ e $10$ devem estar em conjuntos separados. Como existem quatro fatores $3$, distribuídos em três números, $9$ deve ficar em um conjunto enquanto $3$ e $6$ devem ir para o outro. Finalmente, basta dividir os oito fatores $2$ restantes. Um exemplo seria\n$$\n\\begin{aligned}\n& C_{1}=\\{1,10,3,6,4\\} \\mathrm{e} \\\\\n& C_{2}=\\{5,9,2,8\\}\n\\end{aligned}\n$$\nCada um dos conjuntos anteriores possui o produto dos elementos igual a $720$ .",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26842,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEm uma lousa são escritos os 2014 inteiros positivos de 1 até 2014. A operação permitida é escolher dois números $a$ e $b$, apagá-los e escrever em seus lugares os números $mdc(a, b)$ (Máximo Divisor Comum) e $mmc(a, b)$ (Mínimo Múltiplo Comum). Essa operação pode ser feita com quaisquer dois números que estão na lousa, incluindo os números que resultaram de operações anteriores. Determine qual a maior quantidade de números 1 que podemos deixar na lousa.",
    "options": [],
    "answer": "1007",
    "solution": "Solution:\n\nA maior quantidade de números 1 que podemos deixar é 1007. Primeiro vamos mostrar como obtê-los. Para isso, basta tomar os pares de números consecutivos, $(1,2),(3,4),(5,6), \\ldots, (2013,2014)$ e realizar a operação em cada par. Sabendo que números consecutivos não têm fator comum, cada um dos máximos divisores comuns será 1.\n\nNão é possível obter mais do que isso pois a quantidade de números pares não se altera no decorrer das operações. Isso ocorre pois, se operarmos com dois números pares, teremos como resultado dois números pares, se operarmos com dois números ímpares teremos como resultado dois números ímpares e se operarmos com um número par e um número ímpar obteremos também um número par e um número ímpar. Começamos com 1007 números pares e sempre teremos 1007 números pares.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26843,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nÉ dado um quadrado $3 \\times 3$ com números escritos em cada casinha $1 \\times 1$. As jogadas permitidas são escolher uma linha, uma coluna ou uma diagonal e somar ou subtrair 1 dos três números que estiverem nela. Prove que não é possível começar com os números na configuração da esquerda e chegar aos números na configuração da direita após algumas operações.\n\n| 0 | 1 | 0 |\n| :--- | :--- | :--- |\n| 1 | 0 | 1 |\n| 0 | 1 | 0 |\n\n| 1 | 0 | 1 |\n| :--- | :--- | :--- |\n| 0 | 1 | 0 |\n| 1 | 0 | 1 |",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSeja $S$ a soma de todos os números nas casinhas do quadrado. Ao somarmos 1 em três casinhas, trocamos $S$ por $S+3$ e, ao subtrairmos 1 em três casinhas, trocamos $S$ por $S-3$. Como estamos sempre somando ou subtraindo 3, o resto da soma de todos os números na divisão por 3 não se altera. Na primeira configuração temos soma 4 e na segunda configuração temos soma 5. Como 4 e 5 não deixam o mesmo resto na divisão por 3, não é possível ir de uma configuração para a outra.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26844,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNo desenho abaixo, $ABCD$ é um retângulo e $E$ é o pé da perpendicular traçada de $A$ até a diagonal $BD$. As distâncias do ponto $E$ aos lados $DC$, $BC$ e $AB$ são $n$, $1$ e $x$, respectivamente. Seja ainda $d$ o comprimento da diagonal $BD$.\n![](attached_image_1.png)\n\na) Verifique que $DE = x^{2} \\sqrt{1 + x^{2}}$.\n\nb) Verifique que $n = x^{3}$.\n\nc) Verifique que $d^{2/3} - x^{2/3} = 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na) Seja $y = DG$. Pelo Teorema de Pitágoras no triângulo $\\triangle BEF$, temos $BE^{2} = x^{2} + 1$. Pelas relações métricas do triângulo retângulo $\\triangle AEB$, temos\n$$\n\\begin{aligned}\nx^{2} & = y \\cdot 1 \\\\\n& = y \\\\\nAE^{2} & = y(y+1) \\\\\n& = x^{2}(x^{2}+1)\n\\end{aligned}\n$$\nAgora, pelas relações métricas no triângulo retângulo $\\triangle ABD$, temos\n$$\nx^{2}(1+x^{2}) = AE^{2} = BE \\cdot ED = \\sqrt{1+x^{2}} \\cdot ED\n$$\nou seja, $ED = x^{2} \\sqrt{1+x^{2}}$.\n\nb) Aplicando o Teorema de Pitágoras no triângulo $\\triangle EDG$, obtemos\n$$\nn^{2} = EG^{2} = ED^{2} - DG^{2} = x^{4}(1+x^{2}) - x^{4} = x^{6}\n$$\nAssim, $n = x^{3}$.\n\nc)\n$$\n\\begin{aligned}\nd & = BE + ED \\\\\n& = \\sqrt{1+x^{2}} + x^{2} \\sqrt{1+x^{2}} \\\\\n& = (1+x^{2})^{3/2}\n\\end{aligned}\n$$\nDaí, $d^{2/3} = 1 + x^{2} = 1 + n^{2/3}$ e o resultado segue.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26845,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFrancisco acaba de aprender em sua aula de geometria espacial a Relação de Euler para poliedros convexos:\n$$\nV+F=A+2\n$$\nNa equação acima, $V$, $A$ e $F$ representam o número de vértices, de arestas e de faces do poliedro, respectivamente. Podemos verificar que a Relação de Euler é válida no cubo abaixo, pois existem 6 faces, 12 arestas, 8 vértices e\n$$\nV+F=8+6=12+2=A+2\n$$\n![](attached_image_1.png)\nJoão decidiu verificar a Relação de Euler em outro poliedro obtido de um cubo de madeira. Ele marcou os pontos médios de cada aresta e, em cada face, os uniu formando quadrados, como mostra a figura abaixo. Em seguida, ele cortou as 8 pirâmides formadas em torno de cada vértice, obtendo um novo poliedro. Determine:\n![](attached_image_2.png)\na) o novo número de vértices;\nb) o novo número de arestas;\nc) o novo número de faces.",
    "options": [],
    "answer": "a) 12; b) 24; c) 14",
    "solution": "Solution:\n\na) Os vértices do novo poliedro são exatamente os pontos médios das arestas do cubo original. Como o cubo tem 12 arestas, o novo poliedro possui 12 vértices.\n\nb) Cada aresta do novo poliedro é um lado de um dos quadrados formados nas faces. Como o cubo possui 6 faces e cada uma delas possui os 4 lados de um dos quadrados, o total de arestas procurado é $4 \\cdot 6=24$.\n\nc) Existem 8 faces triangulares que são as bases das pirâmides removidas e 6 faces quadradas formadas nas faces do cubo original. Temos então $8+6=14$ faces.\nVeja que a Relação de Euler é válida também para esse novo poliedro, pois\n$$\nV+F=12+14=24+2=A+2\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26846,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm jogo comum de dominó é composto por 28 peças. Cada peça é formada por dois números inteiros que variam de $0$ a $6$, inclusive. Todas as possibilidades de combinações possíveis $(a, b)$, com $a \\leq b$, são listadas exatamente uma vez. Note que a peça $(4,2)$ é listada como a peça $(2,4)$, pois $2 \\leq 4$. Excluindo a peça $(0,0)$, para cada uma das outras 27 peças $(a, b)$, com $a \\leq b$, escrevemos num quadro a fração $\\frac{a}{b}$.\n\na) Quantos valores distintos estão escritos nas formas de frações no quadro? (Veja que as frações $\\frac{1}{2}$ e $\\frac{2}{4}$ têm o mesmo valor e devem ser contadas apenas uma vez.)\n\nb) Qual a soma dos valores distintos encontrados no item anterior?",
    "options": [],
    "answer": "a) 13; b) 13/2",
    "solution": "Solution:\n\na) Basta começar contando pelos maiores denominadores e não repetir quando aparecerem os menores.\n\ni) Para $b=6$, temos\n$$\n\\left(\\frac{0}{6}, \\frac{1}{6}, \\frac{2}{6}, \\frac{3}{6}, \\frac{4}{6}, \\frac{5}{6}, \\frac{6}{6}\\right)=\\left(0, \\frac{1}{6}, \\frac{1}{3}, \\frac{1}{2}, \\frac{2}{3}, \\frac{5}{6}, 1\\right)\n$$\nii) Para $b=5$, não devemos repetir $0=0 / 5$ e nem $1=5 / 5$, pois já foram contados, temos\n$$\n\\left(\\frac{1}{5}, \\frac{2}{5}, \\frac{3}{5}, \\frac{4}{5}\\right)\n$$\niii) Para $b=4$, só podemos adicionar frações irredutíveis de denominador 4, pois já contamos as de denominador 1 e 2 quando $b=6$, temos então\n$$\n\\left(\\frac{1}{4}, \\frac{3}{4}\\right)\n$$\niv) Quando $b$ for 1, 2 ou 3, teremos frações que já foram contadas no caso $b=6$. Verifique!\nLogo, o número de valores distintos é $7+4+2=13$.\n\nb) Um bom jeito de somarmos as 13 frações é considerarmos suas formas redutíveis vistas no item anterior, ou seja,\n$$\n\\begin{aligned}\n\\frac{0}{6}+\\frac{1}{6}+\\frac{2}{6}+\\frac{3}{6}+\\frac{4}{6}+\\frac{5}{6}+\\frac{6}{6} & =\\frac{21}{6} \\\\\n& =\\frac{7}{2} \\\\\n\\frac{1}{5}+\\frac{2}{5}+\\frac{3}{5}+\\frac{4}{5} & =\\frac{10}{5} \\\\\n& =2 \\\\\n\\frac{1}{4}+\\frac{3}{4} & =1\n\\end{aligned}\n$$\nEntão a soma total é $\\frac{7}{2}+2+1=\\frac{13}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26847,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nEncontre as soluções da equação\n$$\n\\sqrt{x+\\sqrt{4 x+\\sqrt{16 x+\\sqrt{\\ldots+\\sqrt{4^{n} x+3}}}}}=1+\\sqrt{x}\n$$",
    "options": [],
    "answer": "x = 2^{-2n}",
    "solution": "Solution:\nElevemos a equação dada ao quadrado $n$ vezes, obtendo\n$$\n\\begin{aligned}\nx+\\sqrt{4 x+\\sqrt{16 x+\\sqrt{\\ldots+\\sqrt{4^{n} x+3}}}} & =1+2 \\sqrt{x}+x \\\\\n\\sqrt{4 x+\\sqrt{16 x+\\sqrt{\\ldots+\\sqrt{4^{n} x+3}}}} & =1+2 \\sqrt{x} \\\\\n4 x+\\sqrt{16 x+\\sqrt{\\ldots+\\sqrt{4^{n} x+3}}} & =1+4 \\sqrt{x}+4 x \\\\\n\\sqrt{16 x+\\sqrt{\\ldots+\\sqrt{4^{n} x+3}}} & =1+4 \\sqrt{x} \\\\\n\\cdots & =\\cdots \\\\\n\\sqrt{4^{n-1} x+\\sqrt{4^{n} x+3}} & =1+2^{n-1} \\sqrt{x} \\\\\n4^{n-1} x+\\sqrt{4^{n} x+3} & =1+2^{n} \\sqrt{x}+2^{2 n-2} x \\\\\n\\sqrt{4^{n} x+3} & =1+2^{n} \\sqrt{x} \\\\\n4^{n} x+3 & =1+2^{n+1} \\sqrt{x}+2^{2 n} x \\\\\n3 & =1+2^{n+1} \\sqrt{x}\n\\end{aligned}\n$$\nPortanto, $\\sqrt{x}=2^{-n}$, ou seja, $x=2^{-2 n}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26848,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nJoão conseguiu pintar de preto e branco os quadrados de um tabuleiro $n \\times n$ de modo que as interseções de quaisquer duas linhas e de quaisquer duas colunas não eram constituídas por quadrados com a mesma cor. Qual o valor máximo de $n$ ?",
    "options": [],
    "answer": "4",
    "solution": "Solution:\n\nUm exemplo com $n=4$ é dado na figura abaixo:\n\n![](attached_image_1.png)\n\nQueremos mostrar agora que, se $n \\geq 5$, não é possível existir tal pintura. Considere então um tabuleiro $n \\times n$ com $n \\geq 5$.\n\nAnalisando os quadrados da primeira linha, pelo menos três deles serão de uma mesma cor. Digamos que esta cor seja preta (se fosse branca não faria a menor diferença para a nossa análise seguinte) e observemos agora as colunas $A$, $B$ e $C$ que contêm esses três quadrados pretos. A segunda linha deve intersectar essas três colunas em pelo menos dois quadrados brancos, pois, caso contrário, teríamos quatro interseções pretas entre as duas primeiras linhas e duas dessas três colunas.\n\nSuponha agora que as colunas que contêm dois quadrados pretos na primeira linha e dois brancos na segunda sejam as colunas $A$ e $B$ (se fossem $A$ e $C$ ou $B$ e $C$ a análise seguinte seria a mesma). A partir da terceira linha, como não podemos ter quadrados de mesma cor simultaneamente nas colunas $A$ e $B$, as distribuições de cores só podem ser as duas opções seguintes: preto e branco ou branco e preto. Daí, dentre as linhas 3, 4 ou 5, duas delas terão exatamente a mesma distribuição. Essas duas linhas com mesma distribuição de cores intersectam a coluna $C$ em dois quadrados, em que nenhum deles pode ser preto ou branco e isso impede a existência da pintura satisfazendo as condições do enunciado.\n\n![](attached_image_2.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26849,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\na) O Teorema do Bico diz que as distâncias de um ponto exterior a uma circunferência aos pontos onde suas tangentes tocam a circunferência são iguais. Na figura a seguir, $AP$ e $AQ$ são tangentes à circunferência. Mostre que $AP = AQ$.\n\n![](attached_image_1.png)\n\nb) Considere o hexágono da figura a seguir, no qual todos os lados tangenciam a circunferência. Determine o valor do lado desconhecido $x$.\n\n![](attached_image_2.png)\n",
    "options": [],
    "answer": "4",
    "solution": "Solution:\na) Trace $OA$. Observe que os triângulos $\\triangle OPA$ e $\\triangle OQA$ são congruentes pois são triângulos retângulos com a mesma hipotenusa e um dos catetos com a mesma medida. Desse modo, $AP = AQ$.\n\nb) Cada um dos lados é dividido pelo ponto de tangência em dois segmentos, conforme a figura. Pelo item anterior, dois desses segmentos, que compartilham um vértice do hexágono em comum, são iguais. Daí,\n$$\n\\begin{aligned}\n6 + 5 + 10 &= (x_6 + x_1) + (x_2 + x_3) + (x_5 + x_4) \\\\\n&= (x_1 + x_2) + (x_3 + x_4) + (x_5 + x_6) \\\\\n&= x + 9 + 8\n\\end{aligned}\n$$\nPortanto, $x = 21 - 17 = 4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26850,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa equação $x^{2}+p x+q=0$, os coeficientes $p$ e $q$ podem assumir quaisquer valores do intervalo $[-1,1]$. Quais são os possíveis valores das raízes de tal equação?",
    "options": [],
    "answer": "All real numbers in the interval [-(1+√5)/2, (1+√5)/2]",
    "solution": "Solution:\n\nAs raízes da equação são dadas por $r=\\frac{-p \\pm \\sqrt{p^{2}-4 q}}{2}$. O valor máximo $\\alpha$ de tal expressão deve ocorrer quando $-p$ e $\\pm \\sqrt{p^{2}-4 q}$ forem máximos. Como $p^{2}-4 q \\leq 1^{2}-4(-1)=5$ e $-p \\leq 1$, segue que\n$$\n\\begin{aligned}\n\\alpha & =\\frac{1+\\sqrt{1^{2}-4(-1)}}{2} \\\\\n& =\\frac{1+\\sqrt{5}}{2}\n\\end{aligned}\n$$\nDe modo similar, podemos verificar que o valor mínimo é $-\\alpha$. Se $y$ é uma raiz de tal equação e $m$ é tal que $|m| \\leq 1$, então $z=m y$ é uma raiz de $x^{2}+p m x+q m^{2}$ e os coeficientes ainda estão em $[-1,1]$. Consequentemente, todos os números do intervalo $[-\\alpha, \\alpha]$ podem ser raízes de tais equações e, como vimos no início, nenhum outro número fora deste intervalo pode sê-lo.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26851,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCerto matemático adora pensar em problemas e cozinhar bolos. Após cozinhar seus bolos, ele os corta em pedaços iguais. As três figuras a seguir mostram bolos circulares de mesmo raio em que os dois primeiros foram cortados em 3 e 4 pedaços iguais, respectivamente. Ele deseja cortar o terceiro bolo, mas a única marcação conhecida é o centro do bolo. Mostre que usando um compasso e uma faca, de tamanhos suficientemente grandes, e os dois primeiros bolos é possível cortar o terceiro em 12 pedaços iguais.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nO compasso servirá para transportar distâncias. O primeiro passo é marcar um ponto $A$ de referência na lateral do terceiro bolo. Usando os comprimentos de arcos do primeiro bolo e o compasso, pode-se marcar um ponto $C$ tal que o arco $AC$ meça $\\frac{1}{3}$ do perímetro de sua circunferência. Em seguida, usando o comprimento de arco do segundo bolo, deve-se marcar o ponto $B$ no terceiro bolo tal que o arco $AB$ meça $\\frac{1}{4}$ do perímetro de sua circunferência. Veja a figura abaixo.\n\n![](attached_image_2.png)\n\nA fração do comprimento da circunferência que representa o arco $BC$ é\n$$\n\\frac{1}{3}-\\frac{1}{4}=\\frac{4-3}{12}=\\frac{1}{12}\n$$\nUsando o compasso podemos transportar a distância $BC$ ao longo do perímetro do bolo 11 vezes dividindo-o assim em 12 arcos iguais. Basta agora usar a faca e efetuar cortes que comecem nos pontos marcados e terminem no centro do bolo.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26852,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOs números que são inteiros positivos elevados ao quadrado são chamados quadrados perfeitos, por exemplo, 16 é um quadrado perfeito pois é igual a $4^{2}$. Um fato curioso é que números que são quadrados perfeitos deixam apenas restos 0 ou 1 na divisão por 4. Com isso podemos provar, por exemplo, que 2014 não é um quadrado perfeito pois 2014 deixa resto 2 na divisão por 4.\n\na) Sabendo que todo número inteiro ímpar é da forma $2k+1$, mostre que os quadrados perfeitos ímpares deixam resto 1 na divisão por 8.\n\nb) É possível colocar 45 números inteiros ímpares em sequência de modo que a soma de quaisquer 5 consecutivos e de quaisquer 9 consecutivos sejam quadrados perfeitos?",
    "options": [],
    "answer": "a) Odd perfect squares leave remainder one upon division by eight. b) No, such a sequence of forty five odd integers does not exist.",
    "solution": "Solution:\n\na) Desenvolvamos o produto notável relativo ao quadrado de um número ímpar $(2k+1)^{2}$:\n$$\n\\begin{aligned}\n(2k+1)^{2} & = 4k^{2} + 4k + 1 \\\\\n& = 4k(k+1) + 1\n\\end{aligned}\n$$\nComo pelo menos um dentre $k$ e $k+1$ é par, segue que $k(k+1)$ é par e que $4k(k+1)$ é múltiplo de $4 \\cdot 2 = 8$. Consequentemente, $4k(k+1)+1$ deixa resto 1 na divisão por 8.\n\nb) Provaremos por contradição que não é possível. Suponha que seja possível encontrar tal sequência de números. Como a soma de 9 números ímpares é ímpar, separando todos os números da sequência em 5 grupos de 9 números consecutivos, podemos concluir que a soma total é igual à soma de 5 quadrados perfeitos de números ímpares. Assim, como cada um deles deixa resto 1 na divisão por 8, em virtude do item anterior, a soma total deixaria o mesmo resto que\n$$\n1+1+1+1+1=5\n$$\nna divisão por 8. A soma de 5 números ímpares também é um número ímpar. Consequentemente, separando a lista total em 9 grupos de cinco números consecutivos, podemos concluir que a soma total também é igual à soma de 9 quadrados perfeitos ímpares. Assim, como cada um deles deixa resto 1 na divisão por 8, novamente em virtude do item anterior, a soma total deixaria o mesmo resto que\n$$\n1+1+1+1+1+1+1+1+1=9\n$$\nna divisão por 8. Ou seja, a soma total deixaria resto 1 na divisão por 8.\n\nObtemos uma contradição, pois a soma dos 45 números está deixando dois restos diferentes, 1 e 5, na divisão por 8 e isso mostra que não é possível existir tal sequência.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26853,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSejam $p$, $q$ e $r$ três números primos maiores que $3$. Sabe-se que o número $p+q+r$ também é primo. Mostre que $p+q$, $p+r$ ou $q+r$ é um múltiplo de $6$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nPrimeiramente vejamos que, se um número $n$ é maior que $3$ e deixa resto $0$, $2$, $3$ ou $4$ na divisão por $6$, então esse número não pode ser primo. Para isso, basta mostrar que o número pode ser escrito como o produto de dois números maiores que $1$:\n$$\n\\begin{aligned}\n& n=6m+0=2 \\cdot 3 \\cdot m \\\\\n& n=6m+2=2 \\cdot (3m+1) \\\\\n& n=6m+3=3 \\cdot (2m+1) \\\\\n& n=6m+4=2 \\cdot (3m+2)\n\\end{aligned}\n$$\nLogo, números primos maiores que $3$ deixam resto $1$ ou $5$ na divisão por $6$. Apliquemos isso aos números do enunciado. Como $p+q+r$ também é primo, podemos concluir que os três não podem deixar todos restos iguais a $1$ ou todos restos iguais a $5$, pois isso faria com que $p+q+r$ fosse um múltiplo de $3$ maior que $3$ e naturalmente não poderia ser primo. Então, dentre os números $p$, $q$ e $r$, pelo menos um deixa resto $1$ e pelo menos outro deixa resto $5$ na divisão por $6$. Somando-os, chegamos a um número múltiplo de $6$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26854,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm erro que muitos alunos cometem é pensar que dois quadriláteros são congruentes se tiverem os seus respectivos lados iguais. Isso não é verdade. Nesse problema, veremos que quadriláteros podem ter lados correspondentes iguais, mas áreas distintas.\n\na) Mostre que a maior área possível para um quadrilátero que possui dois lados de comprimento $3$ e dois de comprimento $4$ é $12$.\n\nb) Mostre que, nos quadriláteros em que isso acontece, a soma dos ângulos opostos é $180^{\\circ}$.",
    "options": [],
    "answer": "12",
    "solution": "Solution:\n\na) Existem dois modos de montar o quadrilátero com pares de lados iguais: ou eles ficam juntos ou ficam separados. Nos dois casos, o quadrilátero pode ser dividido em dois triângulos que serão congruentes pelo caso (L.L.L.). Veja a figura abaixo.\n\n![](attached_image_1.png)\n\n![](attached_image_2.png)\n\nNa segunda figura logo abaixo, fixamos o lado de comprimento $4$ e fazemos variar o lado de comprimento $3$.\n\n![](attached_image_3.png)\n\nComo a base de comprimento $4$ está fixa, a maior área possível ocorrerá quando tivermos a maior altura possível a tal lado e isso ocorre quando o lado de comprimento $3$ for perpendicular à essa base. Qualquer altura diferente de $3$ seria cateto de um triângulo retângulo de hipotenusa $3$ e, consequentemente, menor que $3$.\n\nPortanto, a maior área para cada triângulo é $(3 \\cdot 4) / 2 = 6$. Dado que existem dois de tais triângulos em cada tipo de quadrilátero, a área máxima é $6 + 6 = 12$.\n\nb) Veja que a área máxima ocorre quando os triângulos formados são retângulos. Assim, a soma de ângulos opostos retos é $90^{\\circ} + 90^{\\circ} = 180^{\\circ}$. Como a soma dos ângulos internos do quadrilátero é $360^{\\circ}$, os outros dois ângulos também devem somar $180^{\\circ}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26855,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEm um torneio de xadrez, todos os jogadores enfrentaram todos os outros exatamente uma vez. Em cada partida, o jogador ganha 1 ponto se vencer, $1/2$ se empatar e 0 ponto se perder. Ao final do torneio, um repórter somou as pontuações de todos os jogadores e obteve 190 pontos. Nesse tipo de torneio, o vencedor é aquele que faz mais pontos.\n\na) Quantos jogadores participaram do torneio?\n\nb) André participou do torneio e fez 9 pontos. Mostre que, mesmo sem saber as outras pontuações, André não foi o vencedor do torneio.",
    "options": [],
    "answer": "a) 20 players. b) André did not win.",
    "solution": "Solution:\n\na) Seja $J$ o número de jogadores. Cada partida vale no total 1 ponto, seja $1+0=1$ ou $1/2+1/2=1$. Então a pontuação total é igual ao número de partidas. Como cada um dos $J$ jogadores enfrenta cada um dos outros $J-1$ jogadores, poderíamos pensar que o total de jogos seria $J(J-1)$ embates. Entretanto, cada partida acaba sendo contada duas vezes e portanto o total de partidas é $\\frac{J(J-1)}{2}$. Usando o número obtido pelo jornalista, temos\n$$\n\\begin{aligned}\n\\frac{J(J-1)}{2} & = 190 \\\\\nJ(J-1) & = 380 \\\\\n& = 20 \\cdot 19\n\\end{aligned}\n$$\nDaí $J=20$.\n\nb) Como no total foram 190 pontos para 20 competidores, a média de pontos é $\\frac{190}{20} = 9,5$ pontos. Como André está abaixo da média de pontos e sempre existe um jogador que fez pelo menos tantos pontos quanto a média, podemos concluir que ele não foi o vencedor do torneio.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26856,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm hexágono é chamado equiângulo quando possui os seis ângulos internos iguais. Considere o hexágono equiângulo $A B C D E F$ com lados $3, y, 5, 4, 1$ e $x$, da figura a seguir. Determine os comprimentos $x$ e $y$ desconhecidos.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "x = 6, y = 2",
    "solution": "Solution:\n\nComo um hexágono pode ser dividido em 4 triângulos por meio de suas diagonais, a soma de seus ângulos internos é $180^{\\circ}(6-2)=720^{\\circ}$. Dado que ele é equiângulo, cada um dos ângulos internos medirá $\\frac{720^{\\circ}}{6}=120^{\\circ}$. Sabendo disso, ao prolongarmos os lados formaremos, como indicado abaixo, triângulos equiláteros menores externos a três de seus lados e um triângulo equilátero maior $\\triangle X Y Z$ que o conterá.\n\n![](attached_image_2.png)\n\nComo os lados do triângulo $\\triangle X Y Z$ são iguais, temos\n$$\n3+y+5=5+4+1=1+x+3\n$$\nLogo, $x=6$ e $y=2$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26857,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nResolva em $\\mathbb{R}$ a equação $\\sqrt{x^{2}+9}+\\sqrt{x^{2}-6 x+10}=5$.",
    "options": [],
    "answer": "9/4",
    "solution": "Solution:\nPrimeira Solução: Podemos eliminar radicais elevando membros da equação abaixo ao quadrado:\n$$\n\\begin{aligned}\n\\sqrt{x^{2}+9}+\\sqrt{x^{2}-6 x+10} & =5 \\\\\n\\left(\\sqrt{x^{2}-6 x+10}\\right)^{2} & =\\left(5-\\sqrt{x^{2}+9}\\right)^{2} \\\\\nx^{2}-6 x+10 & =25-10 \\sqrt{x^{2}+9}+x^{2}+9 \\\\\n10 \\sqrt{x^{2}+9} & =6 x+24 \\\\\n25\\left(x^{2}+9\\right) & =(3 x+12)^{2} \\\\\n25 x^{2}+225 & =9 x^{2}+72 x+144 \\\\\n16 x^{2}-72 x+81 & =0 \\\\\n(4 x-9)^{2} & =0\n\\end{aligned}\n$$\nConsequentemente, $4 x-9=0$, ou seja $x=\\frac{9}{4}$. Para verificar que $x=\\frac{9}{4}$ é a solução, basta\nescrever\n$$\n\\begin{aligned}\n\\sqrt{x^{2}+9}+\\sqrt{x^{2}-6 x+10} & =\\sqrt{\\frac{81}{16}+9}+\\sqrt{\\frac{81}{16}-6 \\cdot \\frac{9}{4}+10} \\\\\n& =\\frac{15}{4}+\\frac{5}{4} \\\\\n& =5\n\\end{aligned}\n$$\n\n\nSegunda Solução: Considere no plano cartesiano os pontos $F, D$ e $E$ de coordenadas $(x, 0)$, $(0,3)$ e $(3,-1)$, respectivamente.\n![](attached_image_1.png)\nPodemos associar as distâncias entre alguns pontos aos radicais dados:\n$$\n\\begin{aligned}\nD F & =\\sqrt{x^{2}+9} \\\\\nF E & =\\sqrt{(x-3)^{2}+1^{2}} \\\\\n& =\\sqrt{x^{2}-6 x+10} \\\\\nD E & =\\sqrt{9+16} \\\\\n& =5\n\\end{aligned}\n$$\nPela desigualdade triangular,\n$$\n\\begin{aligned}\n\\sqrt{x^{2}+9}+\\sqrt{(x-3)^{2}+1^{2}} & =D F+F E \\\\\n& \\geq D E \\\\\n& =5\n\\end{aligned}\n$$\nCom igualdade apenas quando $D, F$ e $E$ são colineares. Portanto, o ponto $F$ deve coincidir com a interseção $G$ entre $D E$ e o eixo $O x$, ou seja, $x=9 / 4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26858,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUma propriedade interessante do número 2013 é que 3 é o último dígito da soma $2+0+1$. Repetindo-se esse processo, isto é, escrevendo-se à direita o último dígito da soma dos três dígitos anteriores, teremos uma sequência:\n$$\n2,0,1,3,4,8,5,7 \\ldots\n$$\na) Prove que começando com a sequência 2,0,1, nessa ordem, podemos também encontrar os três números consecutivos $1,2,2$, nessa ordem.\nb) Observe que se uma sequência de três números consecutivos aparecer novamente na mesma ordem, então toda a sequência se \"repetirá\" sucessivamente. Por exemplo, a sequência abaixo não é a sequência do enunciado, mas se repete a cada quatro números\n$$\n\\ldots \\underline\\{124312431243 \\ldots 12431243 \\ldots\\}\n$$\nVerifique que alguma sequência de três dígitos se repete na sequência do enunciado.\nc) Suponha que na primeira aparição de \" $a, b, c$ \" na sequência, o número imediatamente anterior seja $x$, e que na sua segunda aparição seja $y$, ou seja, na sequência iremos encontrar os números na seguinte ordem:\n$$\n\\ldots, x, a, b, c, \\ldots, y, a, b, c \\ldots\n$$\nMostre que $x=y$.\nd) Dado que 1,2,2 apareceu na sequência, nessa ordem, mostre que eventualmente aparecerá novamente a sequência de dígitos $2,0,1$, também nessa ordem.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na) Basta continuarmos a escrever mais termos da sequência seguindo a regra do enunciado:\n$$\n\\mathbf\\{2 , 0 , 1\\}, 3,4,8,5,7,0,2,9, \\mathbf\\{1\\}, \\mathbf\\{2\\}, \\mathbf\\{2\\}, \\ldots\n$$\n\nb) Veja que temos no máximo 10 valores possíveis para cada um dos três números $a, b$ e $c$. Portanto, existem no máximo $10 \\cdot 10 \\cdot 10=1000$ sequências distintas de três números. Então, se tomarmos os 1003 primeiros números, teremos 1001 sequências de três números consecutivos desde os três primeiros até os três últimos deles. Como só existem 1000 sequências possíveis, alguma delas se repetirá, ou seja, existem três números consecutivos $a, b, c$ que aparecem mais de uma vez na sequência.\n\nc) Em virtude do aparecimento do dígito $c, x+a+b$ e $y+a+b$ devem ser iguais a $c$, $10+c$ ou $20+c$. Efetuando-se a subtração $x-y$, podemos obter $-20,-10,0,10$ ou 20. Como $x$ e $y$ são dígitos, essa diferença só poderá assumir o valor 0 e, consequentemente $x=y$.\n\nd) Pelo item $b$ ), sabemos que os números se repetem em algum momento. Pelo item $c$ ), vemos que se três números se repetem o dígito imediatamente anterior a eles também se repete e, por conseguinte, todos os anteriores correspondentes também se repetem. Logo, a tripla de dígitos 2,0,1 irá se repetir. Como, pelo item a), partindo de 2,0,1 chegase a 1,2,2, então quando 2,0,1 se repetir, teremos \"partido\" de 1,2,2 e chegado a 2,0,1.\n$$\n\\mathbf\\{2 , 0 , 1\\}, 3,4,8,5,7,0,2,9, \\mathbf\\{1\\}, \\mathbf\\{2\\}, \\mathbf\\{2\\}, \\ldots 9, \\mathbf\\{2 , 0\\}, \\mathbf\\{1\\}, 3,4,8,5,7,0 \\ldots\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26859,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA folha do caderno de desenho de João é um enorme plano cartesiano quadriculado. Um dos seus desenhos preferidos é a criação de cobrinhas cobrindo os lados dos quadradinhos com sua caneta. Basicamente uma cobrinha é uma sequência de $2n$ pontos distintos $P_{1}, P_{2}, \\ldots, P_{2n}$ escolhidos nos vértices dos quadradinhos dos tabuleiros de modo que pontos com índices consecutivos estão no lado de um mesmo quadradinho do tabuleiro. Por exemplo, na figura abaixo, temos uma cobrinha unindo os seguintes pontos do plano cartesiano:\n$$\n\\begin{aligned}\n& P_{1}=(0,0), \\quad P_{2}=(1,0), \\quad P_{3}=(1,1), \\quad P_{4}=(1,2), \\quad P_{5}=(2,2), \\quad P_{6}=(2,3) \\\\\n& P_{7}=(3,3), \\quad P_{8}=(4,3), \\quad P_{9}=(4,4), \\quad P_{10}=(5,4), \\quad P_{11}=(6,4), \\quad P_{12}=(6,5)\n\\end{aligned}\n$$\n![](attached_image_1.png)\nDepois de desenhar cobrinhas no tabuleiro, João gosta de calcular a soma das coordenadas dos pontos de índices ímpares, isto é, dos pontos $P_{1}, P_{3}, \\ldots, P_{2n-1}$, e subtrair desse número o resultado da soma das coordenadas dos pontos de índices pares, isto é, dos pontos $P_{2}, P_{4}, \\ldots, P_{2n}$.\n\na) Para $n=3$, ou seja, com 6 pontos, desenhe \"cobrinhas\" em que o resultado obtido por João seja $-1, -3$, $1$ e $3$.\n\nb) Dependendo de $n$, quais os possíveis valores que João pode obter?\n\nObservação: A \"cobrinha\" pode também conter pontos com coordenadas negativas, basta que ela \"se mova\" para a esquerda do eixo $y$ ou para baixo do eixo $x$.",
    "options": [],
    "answer": "a) For six points, examples exist achieving totals −3, −1, 1, and 3 (as shown by suitable snakes). b) The possible values are exactly all integers between −n and n having the same parity as n, i.e., {−n, −(n−2), …, n−2, n}.",
    "solution": "Solution:\n\na) A figura a seguir mostra como obter os valores $-1, 1, 3$ e $-3$ para o caso $n=3$.\n![](attached_image_2.png)\n\nb) Sejam $S(P_{n})$ a soma das coordenadas do ponto $P_{n}$ e $S$ o número obtido por João. Queremos calcular os possíveis valores de $S$.\n$$\n\\begin{aligned}\nS & =\\left(S(P_{1})+S(P_{3})+\\ldots+S(P_{2n-1})\\right)-\\left(S(P_{2})+S(P_{4})+\\ldots+S(P_{2n})\\right) \\\\\n& =\\left(S(P_{1})-S(P_{2})\\right)+\\left(S(P_{3})-S(P_{4})\\right)+\\ldots+\\left(S(P_{2n-1})-S(P_{2n})\\right)\n\\end{aligned}\n$$\nVeja que cada diferença $\\left(S(P_{x})-S(P_{x+1})\\right)$ é igual a $1$ ou $-1$ pois eles são vértices de um mesmo quadradinho da folha de caderno. Desse modo, $S$ é adição de $n$ parcelas $\\pm 1$, ou seja,\n$$\nS= \\pm 1+ \\pm 1+\\ldots \\pm 1\n$$\nA soma máxima é $n$, quando todos os termos forem $+1$ e a mínima é $-n$, quando todos forem $-1$. Será que $S$ pode assumir qualquer valor entre $n$ e $-n$? Note que se $n$ é par, $S$ é a soma de uma quantidade par de parcelas ímpares e isso naturalmente produz um número par. Se $n$ é ímpar, $S$ é a soma de uma quantidade ímpar de parcelas ímpares e isso naturalmente produz um número ímpar. Portanto, $S$ só pode assumir os valores que diferem por um número par de $n$, ou seja, valores do conjunto\n$$\nA=\\{-n, -(n-2), \\ldots, n-2l, \\ldots, n-2, n\\}\n$$\nO exemplo anterior nos fornece uma ideia de como mostrar que qualquer número da forma $k=n-2l$ do conjunto $A$ pode ser obtido.\nDesenhe uma \"cobrinha\" com comprimento $n+k=2n-2l$ no eixo $y$ e com comprimento $n-k=2l$ no eixo $x$ como indicado no desenho abaixo\n![](attached_image_3.png)\nOs segmentos no eixo $y$ darão como contribuição o número $\\frac{n+k}{2}=n-l$ e os do eixo $x$ o número $-\\frac{n-k}{2}=l$. O resultado total é $\\frac{n+k}{2}-\\frac{n-k}{2}=k$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26860,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\na) Todo número real ao quadrado é maior ou igual a $0$, sendo $0$ apenas se o número elevado ao quadrado for o próprio $0$. Consequentemente, para quaisquer números reais $a$ e $b$ temos $(a-b)^2 \\geq 0$. Prove que\n$$\n\\frac{a^2+b^2}{2} \\geq a b\n$$\ncom igualdade ocorrendo somente quando $a=b$.\n\nb) A figura a seguir mostra um quadrado de lado $1$ com um vértice em comum com uma reta horizontal. Considerando todas as posições em que o quadrado \"encosta\" apenas um de seus vértices na reta, qual a maior área possível do pentágono $A B C E F$ onde $E$ e $F$ são as projeções ortogonais dos vértices $A$ e $C$ na reta horizontal?\n![](attached_image_1.png)",
    "options": [],
    "answer": "3/2",
    "solution": "Solution:\na) Desenvolvendo o produto notável, temos\n$$\n(a-b)^2 \\geq 0 \\Leftrightarrow a^2-2 a b+b^2 \\geq 0 \\Leftrightarrow a^2+b^2 \\geq 2 a b \\Leftrightarrow \\frac{a^2+b^2}{2} \\geq a b\n$$\nVeja que a igualdade da última expressão acontece apenas quando há igualdade na primeira, ou seja, quando $a=b$.\n\nb) Considere a figura abaixo.\n![](attached_image_2.png)\nSeja $\\angle F A D=\\alpha$. Como $\\angle A F D=\\angle A D C=90^\\circ$, temos:\n$$\n\\begin{aligned}\n\\angle F D A & =90^\\circ-\\angle F A D=90^\\circ-\\alpha \\\\\n\\angle C D E & =180^\\circ-\\angle A D F-90^\\circ=\\alpha \\\\\n\\angle D C E & =90^\\circ-\\angle C D E=90^\\circ-\\alpha\n\\end{aligned}\n$$\nComo os triângulos $\\triangle F A D$ e $\\triangle E D C$ possuem os mesmos ângulos e um lado correspondente de mesma medida, eles são congruentes pelo caso de congruência (A.L.A.). Sejam $a$ e $b$ os comprimentos de seus catetos como indicado na figura. Assim, a área do pentágono $A B C E F$ é\n$$\n[A B C E F]=[F A D]+[E D C]+[A B C D]=2 \\cdot \\frac{a \\cdot b}{2}+1 \\cdot 1=a b+1\n$$\nPelo Teorema de Pitágoras aplicado ao triângulo $\\triangle A F D$, sabemos que $a^2+b^2=1$. Finalmente, usando o resultado do item $a$ ), obtemos:\n$$\n\\begin{aligned}\n{[A B C E F] } & =a b+1 \\\\\n& \\leq \\frac{a^2+b^2}{2}+1 \\\\\n& =\\frac{3}{2}\n\\end{aligned}\n$$\ncom igualdade apenas quando $a=b$. Portanto, a maior área possível do pentágono $A B C E F$ é $3 / 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26861,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA bissetriz de um ângulo é uma semirreta com origem no vértice de um ângulo que o divide em dois outros ângulos congruentes. Por exemplo, no desenho abaixo, a semirreta $OC$ é bissetriz do ângulo $\\angle AOB$.\n![](attached_image_1.png)\n\na) A diferença entre dois ângulos consecutivos mas não adjacentes é $100^\\circ$. Determine o ângulo formado por suas bissetrizes.\n\nObservação: Lembre-se que dois ângulos são consecutivos se possuírem o mesmo vértice e pelo menos um lado em comum e que dois ângulos são adjacentes se não possuírem pontos interiores em comum.\n\nb) No desenho abaixo, $DA$ é bissetriz do ângulo $\\angle CAB$. Determine o valor do ângulo $\\angle DAE$ sabendo que $\\angle CAB+\\angle EAB=120^\\circ$ e $\\angle CAB-\\angle EAB=80^\\circ$.\n![](attached_image_2.png)",
    "options": [],
    "answer": "a) 50 degrees; b) 30 degrees",
    "solution": "Solution:\n\na) Sejam $\\angle BAD=2x$ e $\\angle BAC=2y$ os ângulos adjacentes.\n![](attached_image_3.png)\nO ângulo entre as bissetrizes é\n$$\n\\begin{aligned}\n\\angle EAF &= \\angle EAB - \\angle FAB \\\\\n&= x - y \\\\\n&= \\frac{2x}{2} - \\frac{2y}{2} \\\\\n&= \\frac{\\angle CAB}{2} - \\frac{\\angle DAB}{2} \\\\\n&= \\frac{100^\\circ}{2} \\\\\n&= 50^\\circ\n\\end{aligned}\n$$\n\nb) Sejam $x=\\angle CAD=\\angle DAB$ e $y=\\angle EAB$. Então $2x+y=120^\\circ$ e $2x-y=80^\\circ$. Somando as duas equações, obtemos $4x=200^\\circ$, ou seja, $x=50^\\circ$. Substituindo esse valor em $2x+y=120^\\circ$, temos $y=120^\\circ-2x=120^\\circ-100^\\circ=20^\\circ$. Portanto,\n$$\n\\angle DAE = x - y = 50^\\circ - 20^\\circ = 30^\\circ\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26862,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA calculadora científica de João possui uma tecla especial que transforma qualquer número $x$ escrito na tela e que seja diferente de $1$ no número $\\frac{1}{1-x}$.\n\na) O que acontece se o número $2$ estiver escrito na tela e apertarmos a tecla especial três vezes?\n\nb) O que acontece se o número $2$ estiver escrito na tela e apertarmos a tecla especial dez vezes?\n\nc) Finalmente, o que acontece se o número $2$ estiver escrito na tela e apertarmos a tecla especial $2015$ vezes?",
    "options": [],
    "answer": "a) 2; b) -1; c) 1/2",
    "solution": "Solution:\n\na) Após apertarmos a tecla três vezes, obtemos:\n$$\n2 \\xrightarrow{1^{a}} \\frac{1}{1-2} = -1 \\xrightarrow{2^{a}} \\frac{1}{1-(-1)} = \\frac{1}{2} \\xrightarrow{3^{a}} \\frac{1}{1-1/2} = 2\n$$\n\nb) Em virtude do item anterior, a cada três toques na tecla especial, tudo se passa como se o número $2$ não tivesse sido alterado. Assim, após o sexto e o nono uso da tecla especial, o número $2$ ainda estará na tela. Finalmente, com o décimo uso da tecla especial, o transformaremos em $\\frac{1}{1-2} = -1$. Esse padrão de repetição não é particular ao número $2$ como mostra a sequência:\n$$\nx \\xrightarrow{1^{a}} \\frac{1}{1-x} \\xrightarrow{2^{a}} \\frac{1}{1-\\frac{1}{1-x}} = -\\frac{1-x}{x} \\xrightarrow{3^{a}} \\frac{1}{1-\\left(-\\frac{1-x}{x}\\right)} = x\n$$\n\nc) Como a cada três usos da tecla especial o número $2$ continuará na tela, sempre após um número que é múltiplo de $3$ de usos de tal tecla ainda teremos o número $2$. Como $2013$ é múltiplo de $3$, basta analisarmos as duas últimas apertadas:\n$$\n\\ldots \\xrightarrow{2013^{a}} 2 \\xrightarrow{2014^{a}} \\frac{1}{1-2} = -1 \\xrightarrow{2015^{a}} \\frac{1}{1-(-1)} = \\frac{1}{2}\n$$\nPortanto, restará o número $\\frac{1}{2}$ na tela.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26863,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\na) José aprendeu um método para calcular produtos de dois números de uma forma mais rápida baseado na fatoração:\n$$\n(n-k)(n+k)=n^{2}-k^{2}\n$$\nPara calcular $23 \\cdot 17$, ele escolhe $n=20, k=3$ e calcula:\n$$\n23 \\cdot 17=20^{2}-3^{2}=400-9=391\n$$\nDetermine, sem usar a calculadora, o valor de $\\sqrt{1001 \\cdot 1003+1}$.\n\nb) Verifique que $(n(n+3)+1)^{2}=n(n+1)(n+2)(n+3)+1$.\n\nc) Determine, sem usar a calculadora, o valor de:\n$$\n\\sqrt{(2014)(2015)(2016)(2017)+1}\n$$",
    "options": [],
    "answer": "a) 1002; b) The identity holds: n(n+1)(n+2)(n+3)+1 equals (n(n+3)+1) squared; c) 4062239",
    "solution": "Solution:\n\na) Basta escolher $n=1002$ e $k=1$, pois\n$$\n\\begin{aligned}\n\\sqrt{1001 \\cdot 1003+1} & =\\sqrt{1002^{2}-1^{2}+1} \\\\\n& =\\sqrt{1002^{2}} \\\\\n& =1002\n\\end{aligned}\n$$\n\nb)\n$$\n\\begin{aligned}\n(n(n+3)+1)^{2} & =n^{2}(n+3)^{2}+2 n(n+3)+1 \\\\\n& =n(n+3)[n(n+3)+2]+1 \\\\\n& =n(n+3)\\left[n^{2}+3 n+2\\right]+1 \\\\\n& =n(n+3)[(n+1)(n+2)]+1 \\\\\n& =n(n+1)(n+2)(n+3)+1\n\\end{aligned}\n$$\n\nc) Usando o item anterior e escolhendo $n=2014$, temos\n$$\n\\begin{aligned}\n\\sqrt{(2014)(2015)(2016)(2017)+1} & =\\sqrt{(2014 \\cdot 2017+1)^{2}} \\\\\n& =2014 \\cdot 2017+1 \\\\\n& =4062239\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26864,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEm um grupo de 200 pessoas, apenas $1\\%$ é mulher. Determine o número de homens que devem abandonar o grupo para que $98\\%$ das pessoas restantes sejam do sexo masculino.",
    "options": [],
    "answer": "100",
    "solution": "Solution:\n\nO número de mulheres é $200 \\cdot \\frac{1}{100} = 2$. Para que tal número represente $2\\% = 100\\% - 98\\%$ da nova quantidade total de pessoas $x$, devemos ter $2 = x \\cdot \\frac{2}{100}$, ou seja, $x = 100$. Assim, devem sair $198 - 98 = 100$ pessoas do sexo masculino do grupo.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26865,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nNo desenho abaixo, os pontos $E$ e $F$ pertencem aos lados $AB$ e $BD$ do triângulo $\\triangle ABD$ de modo que $AE = AC$ e $CD = FD$. Se $\\angle ABD = 60^{\\circ}$, determine a medida do ângulo $\\angle ECF$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "60°",
    "solution": "Solution:\nSejam $2\\alpha = \\angle EAC$ e $2\\beta = \\angle FDC$. Como os triângulos $\\triangle EAC$ e $\\triangle FDC$ são isósceles, segue que $\\angle ACE = \\angle AEC = 90^{\\circ} - \\alpha$ e $\\angle DCF = \\angle CFD = 90^{\\circ} - \\beta$. Consequentemente, $\\angle ECF = \\alpha + \\beta$.\n\nAnalisando agora a soma dos ângulos do triângulo $\\triangle ABD$, temos $60^{\\circ} + 2\\alpha + 2\\beta = 180^{\\circ}$, ou seja, $60^{\\circ} = \\alpha + \\beta$. Como já sabemos que $\\angle ECF = \\alpha + \\beta$, então $\\angle ECF = 60^{\\circ}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26866,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUma progressão aritmética, costumeiramente chamada de P.A., é uma sequência em que cada termo, a partir do segundo, é igual à soma do termo anterior com um valor fixo $r$ chamado de diferença comum ou razão da progressão. Por exemplo, a sequência abaixo é uma progressão aritmética com termo inicial 3 e diferença comum 4.\n$$\na_{1}=3, a_{2}=7, a_{3}=11, a_{4}=15, a_{5}=19, a_{6}=23, a_{7}=27, a_{8}=31, a_{9}=35, \\ldots\n$$\nVeja que estamos denotando o número da posição $i$ pelo símbolo $a_{i}$.\n\na) Se o primeiro termo de uma progressão aritmética é 2 e sua diferença comum é 3 , qual é o valor do quarto termo?\n\nb) A professora de João pediu que ele calculasse o décimo primeiro termo de uma progressão aritmética. Infelizmente ele esqueceu qual era o termo inicial e a diferença comum. As únicas informações das quais ele lembrava eram:\n$$\n\\begin{aligned}\na_{4}+a_{7}+a_{10} & =207 \\\\\na_{5}+a_{6}+a_{7}+a_{8}+a_{9}+a_{10}+a_{11} & =553\n\\end{aligned}\n$$\nQuanto vale o décimo primeiro termo?",
    "options": [],
    "answer": "a) 11; b) 109",
    "solution": "Solution:\n\na) Se $a_{1}=2$ e $r=3$, temos\n$$\n\\begin{aligned}\n& a_{2}=a_{1}+3=5 \\\\\n& a_{3}=a_{2}+3=8 \\\\\n& a_{4}=a_{3}+3=11\n\\end{aligned}\n$$\n\nb) Sejam $a_{1}=d$ e $r$ a razão. Então, temos:\n$$\n\\begin{aligned}\n& a_{1}=d, \\quad a_{2}=d+r, \\quad a_{3}=d+2 r, \\quad a_{4}=d+3 r, \\quad a_{5}=d+4 r, \\quad a_{6}=d+5 r \\\\\n& a_{7}=d+6 r, \\quad a_{8}=d+7 r, \\quad a_{9}=d+8 r, \\quad a_{10}=d+9 r, \\quad a_{11}=d+10 r .\n\\end{aligned}\n$$\nPortanto,\n$$\n\\begin{aligned}\na_{4}+a_{7}+a_{10} & =(d+3 r)+(d+6 r)+(d+9 r) \\\\\n207 & =3(d+6 r) \\\\\na_{5}+a_{6}+a_{7}+a_{8}+a_{9}+a_{10}+a_{11} & =(d+4 r)+(d+5 r)+\\ldots+(d+10 r) \\\\\n553 & =7(d+7 r)\n\\end{aligned}\n$$\nConsequentemente,\n$$\n\\left\\{\\begin{aligned}\nd+6 r & =207 / 3=69 \\\\\nd+7 r & =553 / 7=79\n\\end{aligned}\\right.\n$$\nResolvendo o sistema anterior, obtemos $r=10$ e $d=9$. Assim, $a_{11}=d+10 r=109$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26867,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nEncontre todas as soluções, no conjunto dos números reais positivos, do sistema de equações:\n$$\n\\left\\{\\begin{array}{l}\nx(x+y+z)=26 \\\\\ny(x+y+z)=27 \\\\\nz(x+y+z)=28\n\\end{array}\\right.\n$$",
    "options": [],
    "answer": "(26/9, 3, 28/9)",
    "solution": "Solution:\nSomando as três equações, obtemos $(x+y+z)^2=81$, ou seja, $x+y+z=9$, pois queremos soluções positivas.\n\nSubstituindo tal valor em cada equação, temos:\n\n$x = 26/9$, $y = 27/9 = 3$ e $z = 28/9$.\n\nAssim, a única solução do sistema é $(x, y, z) = (26/9, 3, 28/9)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26868,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO mágico Magimático chama três pessoas da plateia: Ana, Beto e Caio, para ajudarem em sua matemágica. Ele diz para cada um pensar em um número de 1 a 50, sem revelá-lo ao mágico, e contá-lo para cada um dos outros dois participantes. Em seguida, cada um deles deve simultaneamente trocar o seu número pela soma dos números dos outros dois. Por exemplo, Ana passa a ter a soma dos números de Beto e Caio. Magimático pede então que eles repitam esse processo mais uma vez. Após concluir a segunda troca, ele pede que falem os seus números. Ana responde 104, Beto 123 e Caio 137. Para a surpresa de todos, Magimático acerta os números iniciais escolhidos pelos três. Quais foram os números escolhidos incialmente?",
    "options": [],
    "answer": "Ana 13, Beto 32, Caio 46",
    "solution": "Solution:\n\nVamos chamar o número de Ana de $A$, o de Beto de $B$ e o de Caio de $C$. Na primeira troca, Ana passou a ter $B+C$, Beto $A+C$ e Caio $A+B$. Após a segunda troca, Ana passou a ter $A+C+A+B=2A+B+C$, Beto passou a ter $A+2B+C$ e Caio passou a ter $A+B+2C$. A partir das respostas finais que eles deram, sabemos que:\n$$\n\\begin{aligned}\n2A+B+C &= 104 \\\\\nA+2B+C &= 123 \\\\\nA+B+2C &= 137\n\\end{aligned}\n$$\nSomando as três equações, obtemos $4A+4B+4C=104+123+137=364$, ou seja, $A+B+C=91$. Subtraindo esse valor de cada uma das três equações, obtém-se $A=13$, $B=32$ e $C=46$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26869,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDado um triângulo de lados $a \\leq b \\leq c$, pela lei dos cossenos temos:\n$$\n\\cos \\hat{C}=\\frac{a^{2}+b^{2}-c^{2}}{2 a b}\n$$\nSe o ângulo $\\hat{C}$ é obtuso, $\\cos \\hat{C}<0$. Como $2 a b$ é positivo, isso é o mesmo que $a^{2}+b^{2}-c^{2}<0$. Portanto, para um triângulo ser obtusângulo, o maior lado elevado ao quadrado é maior que a soma dos quadrados dos outros dois lados. Além disso, pela desigualdade triangular, sabemos que o maior lado é menor que a soma dos outros dois. Podemos resumir essas duas informações através das desigualdades\n$$\na^{2}+b^{2}<c^{2}<(a+b)^{2}\n$$\nQuantos triângulos obtusângulos podemos formar com lados inteiros positivos menores que 7 ?\nObservação: Considere que dois triângulos com os mesmos comprimentos de lado mas em ordens diferentes representam o mesmo triângulo.",
    "options": [],
    "answer": "8",
    "solution": "Solution:\n\nPrimeiro vamos assumir a mesma ordem de lados que o enunciado, ou seja, $a \\leq b \\leq c$. Dividamos o problema em casos a partir do valor de $a$. Para facilitar a análise, lembre-se que os sete primeiros quadrados perfeitos são:\n$$\n1^{2}=1, \\quad 2^{2}=4, \\quad 3^{2}=9, \\quad 4^{2}=16, \\quad 5^{2}=25, \\quad 6^{2}=36 \\quad \\text{ e } \\quad 7^{2}=49\n$$\nEm cada caso, substituiremos o valor de $a$ na desigualdade do enunciado.\ni) Se $a=1$, temos $1+b^{2}<c^{2}<(b+1)^{2}$. Nesse caso, não teremos soluções, pois a desigualdade anterior implica que $b<c<b+1$ e não existe um inteiro em tal intervalo.\nii) Se $a=2$, temos $4+b^{2}<c^{2}<(b+2)^{2}$. Nesse caso, teremos $b<c<b+2$, daí $c=b+1$. Então, como $c<7$, as soluções são: $(2,2,3),(2,3,4),(2,4,5)$ e $(2,5,6)$.\niii) Se $a=3$ temos $9+b^{2}<c^{2}<(b+3)^{2}$. Nesse caso, teremos $b<c<b+3$ e então $c=b+1$ ou $c=b+2$. Se $b=3$, teremos apenas $c=5$ como solução. Se $b=4$, só teremos $c=6$. Finalmente, se $b=5$, teremos apenas $c=6$. Temos então as soluções: $(3,3,5),(3,4,6) \\mathrm{e}$ $(3,5,6)$.\niv) Se $a=4$ e $b=4$ só podemos completar com $c=6$ e a solução é $(4,4,6)$ pois $5^{2}<4^{2}+4^{2}$. Se $a=4$ e $b=5$ teremos $a^{2}+b^{2}>c^{2}$. O mesmo ocorre se $a \\geq 5$ pois $a^{2}+b^{2} \\geq 5^{2}+5^{2}>7^{2}$.\nPortanto, podemos formar apenas 8 triângulos obtusângulos.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26870,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO professor Carlão decidiu fazer uma questão de matemática que vale no total 10 pontos e possui três itens: $a, b$ e $c$. Após elaborar os itens, ele ficou na dúvida sobre qual a melhor maneira de distribuir os 10 pontos entre os itens de modo que cada um valha um número inteiro positivo de pontos.\n\na) Joana, uma professora amiga de Carlão, sugeriu que o item $c$ deveria valer o mesmo tanto de pontos que a soma dos itens $a$ e $b$ pois, segundo ela, o item $c$ é mais difícil. Se Carlão seguir a sugestão de Joana, de quantos modos diferentes ele pode distribuir os pontos?\n\nb) Desconsiderando a sugestão de Joana, ou seja, considerando que Carlão vai distribuir os pontos de uma maneira qualquer, de quantos modos diferentes ele pode distribuir os 10 pontos da questão entre os três itens?",
    "options": [],
    "answer": "a) 4; b) 36",
    "solution": "Solution:\n\na) Se Carlão seguir a sugestão de Joana o item $c$ valerá 5 pontos e os itens $a$ e $b$ devem somar outros 5 pontos. Teremos então quatro divisões possíveis de itens $(a, b, c): (1,4,5), (2,3,5), (3,2,5)$ e $(4,1,5)$.\n\nb) Uma vez definidas as pontuações dos itens $a$ e $b$, o item $c$ valerá $10-a-b$ pontos e, portanto, bastará contarmos o número de maneiras de escolhermos $a$ e $b$. Se os itens $a$ e $b$ valem juntos $n$ pontos, então teremos $n-1$ possibilidades de pares de inteiros positivos $(a, b)$:\n$$\n(1, n-1), (2, n-2), (3, n-3), \\ldots, (n-1,1)\n$$\nComo a soma $a+b$ deve valer no máximo $9$, quando $c$ é mínimo e vale $1$, e no mínimo $1+1=2$ pontos, quando $a$ e $b$ são mínimos, o total de maneiras de distribuirmos esses pontos é\n$$\n\\begin{array}{r}\n(9-1)+(8-1)+(7-1)+(6-1)+(5-1)+(4-1)+(3-1)+(2-1)= \\\\\n8+7+6+5+4+3+2+1=\n\\end{array}\n$$\n$36$.\n\nAssim, Carlão pode distribuir os 10 pontos de 36 modos diferentes.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26871,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSabemos que\n$$\n\\frac{8^{x}}{2^{x+y}}=64 \\mathrm{e} \\frac{9^{x+y}}{3^{4 y}}=243\n$$\nDetermine o valor de $2 x y$.",
    "options": [],
    "answer": "7",
    "solution": "Solution:\nComo $8=2^{3}$ e $9=3^{2}$, temos\n$$\n\\begin{aligned}\n64 & =\\frac{8^{x}}{2^{x+y}} \\\\\n2^{6} & =2^{3 x-(x+y)} \\\\\n& =2^{2 x-y}\n\\end{aligned}\n$$\n$$\n\\begin{aligned}\n243 & =\\frac{9^{x+y}}{3^{4 y}} \\\\\n3^{5} & =3^{(2 x+2 y)-4 y} \\\\\n& =3^{2 x-2 y}\n\\end{aligned}\n$$\nAssim, temos o seguinte sistema:\n$$\n\\left\\{\\begin{array}{l}\n2 x-y=6 \\\\\n2 x-2 y=5\n\\end{array}\\right.\n$$\nSubtraindo a segunda equação da primeira, temos $y=1$. Substituindo este valor na primeira equação, temos $x=7 / 2$. Daí, $2 x y=2 \\cdot \\frac{7}{2} \\cdot 1=7$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26872,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nConsidere o seguinte tabuleiro quadriculado onde todos os números naturais foram escritos em diagonal.\n\n| $\\ddots$ |  |  |  |  |  |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| 10 | $\\ddots$ |  |  |  |  |\n| 6 | 9 | $\\ddots$ |  |  |  |\n| 3 | 5 | 8 | 12 | $\\ddots$ |  |\n| 1 | 2 | 4 | 7 | 11 | $\\ddots$ |\n\nCada quadradinho possui uma posição denotada por $(x, y)$, em que $x$ representa a coluna, contada da esquerda para a direita, e $y$ representa a linha, contada debaixo para cima. Por exemplo, 12 é o número escrito no quadradinho de posição $(4,2)$ :\n\na) Determine o número que está no quadradinho de posição $(4,4)$.\n\nb) Determine o número que está no quadradinho de posição $(1,2016)$.\n\nc) Determine o número que está no quadradinho de posição $(2013,2017)$.",
    "options": [],
    "answer": "a) 25; b) 2033136; c) 8116423",
    "solution": "Solution:\n\na. Podemos preencher mais casas do tabuleiro exibido para encontrar a casa $(4,4)$ :\n\n| 21 | 27 |  |  |  |  |  |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| 15 | 20 | 26 |  |  |  |  |\n| 10 | 14 | 19 | 25 |  |  |  |\n| 6 | 9 | 13 | 18 | 24 |  |  |\n| 3 | 5 | 8 | 12 | 17 | 23 |  |\n| 1 | 2 | 4 | 7 | 11 | 16 | 22 |\n\nPortanto, o número no quadradinho $(4,4)$ é o $25$.\n\nb. O número da casa $(1, n)$ está em uma diagonal contendo as casas $(x, y)$ tais que $x+y = n+1$. Ou seja, na diagonal que o contém, existem $n$ números, a saber, os números das casas $(1, n), (2, n-1), (3, n-2), \\ldots, (n, 1)$. Repetindo essa contagem para os números das casas $(1, n-1), (1, n-2), \\ldots, (1,1)$, podemos constatar que já foram escritos $1+2+\\ldots+(n-1)$ números nas outras diagonais. Portanto, o número escrito na casa $(1, n)$ corresponde ao inteiro\n$$\n1+2+3+\\ldots+n=\\frac{n(n+1)}{2}\n$$\nAssim, o número na casa $(1,2016)$ é $\\frac{2016 \\cdot 2017}{2}$.\n\nc. Como o quadradinho $(m, n)$ está na diagonal contendo os quadradinhos $(x, y)$ com $x+y=m+n$, inicialmente iremos descobrir o número escrito em $(m+n-1,1)$. Pelo item anterior, o quadradinho $(1, m+n-2)$ possui o número $\\frac{(m+n-2)(m+n-1)}{2}$. Portanto, o número escrito no quadradinho $(m+n-1,1)$ é $\\frac{(m+n-2)(m+n-1)}{2}+1$. Além disso, o quadradinho $(m, n)$ é o $(n-1)$-ésimo sucessor do número escrito no quadradinho $(m+n-1,1)$, ou seja, o número\n$$\n\\begin{aligned}\n\\frac{(m+n-2)(m+n-1)}{2}+1+(n-1) & = \\\\\n\\frac{m^{2}+mn-m+mn+n^{2}-n-2m-2n+2+2+2n-2}{2} & = \\\\\n\\frac{n(n+1)+m(m-1)+2(m-1)(n-1)}{2} & = \\\\\n\\frac{n(n+1)}{2}+\\frac{m(m-1)}{2}+(m-1)(n-1) &\n\\end{aligned}\n$$\nLogo, o número escrito no quadradinho $(2013,2017)$ é\n$$\n\\frac{2013 \\cdot 2012}{2}+\\frac{2017 \\cdot 2018}{2}+2012 \\cdot 2016\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26873,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nJoão estava estudando para as Olimpíadas de Matemática e se deparou com a seguinte equação\n$$\n\\frac{1}{x} + \\frac{1}{y} = \\frac{1}{2} + \\frac{1}{z}\n$$\nonde $x$, $y$ e $z$ são inteiros positivos. Após tentar encontrar todas as soluções sem sucesso, ele pediu ajuda para o professor Piraldo, que decidiu dar algumas dicas de como ele deveria proceder. Vamos ajudar João a interpretar as dicas.\n\na. Se $x=1$, determine todos os pares $(y, z)$ de inteiros positivos que satisfazem a equação.\n\nb. Se $x=2$ e $y \\geq 2$, determine todos os pares $(y, z)$ de inteiros positivos que satisfazem a equação.\n\nc. Se $x=3$ e $y \\geq 3$, determine todos os pares $(y, z)$ de inteiros positivos que satisfazem a equação.\n\nd. Se $x$ e $y$ são maiores que ou iguais a 4, verifique que a equação não possui solução.",
    "options": [],
    "answer": "a) With x = 1: (y, z) = (2, 1).\nb) With x = 2 and y ≥ 2: (y, z) = (n, n) for any integer n ≥ 2.\nc) With x = 3 and y ≥ 3: (y, z) ∈ {(3, 6), (4, 12), (5, 30)}.\nd) With x, y ≥ 4: no solutions.\nBy symmetry in x and y, swapping x and y in any listed triple yields another solution. Consequently, all solutions are obtained from the families above.",
    "solution": "Solution:\n\na. Substituindo $x=1$ na equação, temos\n$$\n\\begin{aligned}\n& \\frac{1}{1} + \\frac{1}{y} = \\frac{1}{2} + \\frac{1}{z} \\\\\n& \\frac{1}{2} + \\frac{1}{y} = \\frac{1}{z}\n\\end{aligned}\n$$\nSe $z \\geq 2$, então $\\frac{1}{2} + \\frac{1}{y} = \\frac{1}{z} \\leq \\frac{1}{2}$. Daí, $\\frac{1}{y} \\leq 0$, que é falso. Logo, $z$ tem que ser $1$. Desse modo, a única solução é $(y, z) = (2, 1)$.\n\nb. Substituindo $x=2$ na equação, temos\n$$\n\\begin{aligned}\n\\frac{1}{2} + \\frac{1}{y} & = \\frac{1}{2} + \\frac{1}{z} \\\\\n\\frac{1}{y} & = \\frac{1}{z} \\\\\ny & = z\n\\end{aligned}\n$$\nTemos soluções $(y, z) = (n, n)$ para qualquer inteiro positivo $n \\geq 2$. Note que se $x=2$ e $y<2$, a única solução possível é $(y, z) = (1, 1)$.\n\nc. Substituindo $x=3$ na equação, temos\n$$\n\\begin{aligned}\n\\frac{1}{3} + \\frac{1}{y} & = \\frac{1}{2} + \\frac{1}{z} \\\\\n\\frac{1}{y} & = \\frac{1}{6} + \\frac{1}{z}\n\\end{aligned}\n$$\nSe $y \\geq 6$, então $\\frac{1}{6} + \\frac{1}{z} = \\frac{1}{y} \\leq \\frac{1}{6}$. Daí, $\\frac{1}{z} \\leq 0$, que é falso. Logo, $y$ tem que ser $3$, $4$ ou $5$. Testando esses valores encontramos as soluções $(y, z) = (3, 6), (4, 12)$ ou $(5, 30)$. Note que se $x=3$ e $y<3$, temos apenas a solução $(y, z) = (2, 3)$.\n\nd. Se tivéssemos $x \\geq 4$ e $y \\geq 4$, então $\\frac{1}{x} \\leq \\frac{1}{4}$ e $\\frac{1}{y} \\leq \\frac{1}{4}$ e isso implicaria\n$$\n\\frac{1}{2} + \\frac{1}{z} = \\frac{1}{x} + \\frac{1}{y} \\leq \\frac{1}{4} + \\frac{1}{4} = \\frac{1}{2}\n$$\nDaí, $\\frac{1}{z} \\leq 0$. Como $z > 0$, concluímos que nesse caso a equação não possui solução.\n\nCom essas dicas, João pode encontrar todas as soluções. Veja que os papéis desempenhados por $x$ e $y$ na equação são simétricos. Portanto, se $(x, y, z) = (a, b, c)$ é solução, então $(b, a, c)$ também é solução. Então basta encontrarmos as soluções com $x \\leq y$. O último item mostra que pelo menos um dentre $x$ e $y$ é menor ou igual a $3$ e, aproveitando o estudo dos três itens iniciais, podemos listar todas as soluções.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26874,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm grupo de 10 estudantes participa de uma competição de matemática formada por equipes de 4 estudantes. Sabemos que quaisquer duas das equipes possuem exatamente um estudante em comum.\n\na) Qual o número máximo de equipes de que um estudante pode participar? Forneça um exemplo de distribuição de 10 alunos onde este número máximo possa ser verificado.\n\nb) A competição pode possuir 8 equipes?",
    "options": [],
    "answer": "a) 3; for example, one student appears in three teams while the other nine students are split into three disjoint groups of three. b) No.",
    "solution": "Solution:\n\na) Considere um estudante $A$ que participa do maior número de equipes e digamos que ele esteja em uma equipe com os três estudantes $B, C$ e $D$. Qualquer outra equipe que também tenha $A$ como um de seus membros, deverá conter outros três estudantes que não estão no conjunto $\\{B, C, D\\}$. Como existem apenas $10-1=9$ estudantes diferentes de $A$, o número máximo de equipes que podem conter $A$ é $9 / 3=3$. Um exemplo de distribuição de 10 estudantes, representados pelas letras do conjunto $\\{A, B, C, D, E, F, G, H, I, J\\}$ é\n\n| A | B | C | D |\n| :--- | :--- | :--- | :--- |\n| A | E | F | G |\n| A | H | I | J |\n\nCada linha indica uma equipe e todas elas possuem apenas o estudante $A$ em comum.\n\nb) Suponhamos, por absurdo, que possam existir 8 equipes. Como cada uma delas possui 4 estudantes, teremos ao todo pelo menos $8 \\cdot 4=32$ participações de alunos, contadas com repetições. Dado que existem apenas 10 estudantes e $32 / 10>3$, pelo menos um estudante deverá participar de 4 equipes. Isso contradiz o item anterior e esse absurdo mostra que não podemos ter 8 equipes.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26875,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSeja $ABCD$ um quadrilátero com $AD = BC$ e $\\angle DAB + \\angle ABC = 120^\\circ$. Um triângulo equilátero $DEC$ é construído no exterior do quadrilátero. Prove que o triângulo $AEB$ também é equilátero.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nSejam $\\angle ADC = x$ e $\\angle DCB = y$. Como a soma dos ângulos internos de um quadrilátero é $360^\\circ$, temos\n$$\n\\begin{aligned}\n\\angle ADC + \\angle DCB + \\angle CBA + \\angle BAD & = 360^\\circ \\\\\nx + y & = 360^\\circ - 120^\\circ \\\\\n& = 240^\\circ\n\\end{aligned}\n$$\n\nAnalisando agora os triângulos $ADE$ e $CBE$, temos $\\angle ECB = 60^\\circ + y$ e\n$$\n\\begin{aligned}\n\\angle ADE & = 360^\\circ - 60^\\circ - x \\\\\n& = 60^\\circ + y\n\\end{aligned}\n$$\n\nComo $AD = CB$ e $DE = CE$, segue pelo caso de congruência $LAL$, que os triângulos $ADE$ e $CBE$ são congruentes. Daí, $AE = EB$ e\n$$\n\\begin{aligned}\n\\angle AEB & = \\angle AED + \\angle DEB \\\\\n& = \\angle BEC + \\angle DEB \\\\\n& = 60^\\circ\n\\end{aligned}\n$$\n\nAssim, $AEB$ é um triângulo isósceles com ângulo do vértice igual a $60^\\circ$, e, consequentemente, é equilátero.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26876,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nNo desenho abaixo, $ABCD$ e $EFGC$ são quadrados. As retas $BG$ e $DE$ se encontram no ponto $H$.\na) Verifique que $\\angle BHD = 90^\\circ$ e conclua que o ponto $H$ está simultaneamente nas circunferências de diâmetros $BD$ e $EG$.\nb) Encontre o valor de $\\angle AHD + \\angle DHG + \\angle GHF$.\n![](attached_image_1.png)",
    "options": [],
    "answer": "180°",
    "solution": "Solution:\nAntes de resolvermos o problema, precisaremos fazer um comentário sobre quadriláteros cíclicos. Considere um triângulo $ABC$, seu circuncírculo $\\Gamma$ e um ponto $P$ no mesmo semiplano que $C$ determinado pela reta $AB$. Existem três possibilidades para o ponto $P$: ele pode estar no lado de fora de $\\Gamma$, sobre ele ou dentro dele. Cada uma dessas situações está representada no desenho abaixo.\n![](attached_image_2.png)\nNa primeira situação, quando $P$ está fora da circunferência circunscrita, considere o ponto $Q$ de interseção de um dos segmentos $PA$ ou $PB$ com $\\Gamma$. Como $\\angle ACB$ e $\\angle AQB$ são ângulos inscritos no mesmo arco $AB$, temos $\\angle ACB = \\angle AQB$. Pelo Teorema do Ângulo Externo, podemos escrever\n$$\n\\begin{aligned}\n\\angle ACB & = \\angle AQB \\\\\n& = \\angle APB + \\angle PAQ \\\\\n& > \\angle APB\n\\end{aligned}\n$$\nNa segunda situação, como $\\angle ACB$ e $\\angle APB$ são ângulos inscritos no mesmo arco $AB$, temos $\\angle ACB = \\angle APB$. Finalmente, na terceira situação, considere o ponto de interseção das retas $PA$ ou $PB$ com $\\Gamma$. Novamente pelo Teorema de Ângulo Externo, temos\n$$\n\\begin{aligned}\n\\angle ACB & = \\angle AQB \\\\\n& < \\angle AQB + \\angle PAQ \\\\\n& = \\angle APB\n\\end{aligned}\n$$\nPortanto, $\\angle ACB = \\angle APB$ se, e somente se, o ponto $P$ está no circuncírculo do triângulo $ABC$.\n\na. Como $EC = CG$, $DC = BC$ e $\\angle BCG = \\angle DCE = 90^\\circ$, segue que os triângulos $BCG$ e $ECD$ são congruentes. Daí, $\\angle CBG = \\angle CDE$. Pela observação anterior, isso nos garante que o ponto $B$ está no circuncírculo do triângulo $DCH$. Entretanto, sabemos que a circunferência que passa pelos pontos $B$, $C$ e $D$ é a circunferência de diâmetro $BD$. Consequentemente, $A$, $B$, $C$, $D$ e $H$ estão sobre uma mesma circunferência de diâmetro $BD$ e $\\angle BHD = \\angle BCD = 90^\\circ$. Além disso, de $\\angle EHG = 180^\\circ - \\angle BHE = 90^\\circ = \\angle EFG$, segue em virtude da observação inicial que $E$, $H$, $F$ e $G$ estão em uma mesma circunferência.\n![](attached_image_3.png)\n\nb. Considerando a circunferência circunscrita ao quadrilátero $ABCD$, como $\\angle AHD$ e $\\angle ACD$ estão inscritos no mesmo arco $AD$, segue que $\\angle AHD = \\angle ACD$. Analogamente, considerando a circunferência circunscrita ao quadrado $EFGC$, temos $\\angle GHF = \\angle FCG$. Daí,\n$$\n\\begin{aligned}\n\\angle AHD + \\angle DHG + \\angle GHF & = \\\\\n\\angle ACD + \\angle DHG + \\angle FCG & = \\\\\n45^\\circ + 90^\\circ + 45^\\circ & = 180^\\circ\n\\end{aligned}\n$$\nAssim, podemos concluir que os pontos $A$, $H$ e $F$ são colineares.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26877,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSeja $D$ um ponto no lado $AB$ do triângulo $ABC$ de modo que $CD = AC$, como indica a figura abaixo. O incírculo do triângulo $ABC$ é tangente aos lados $AC$ e $AB$ nos pontos $E$ e $F$, respectivamente. Sejam $I$ o incentro do triângulo $BCD$ e $P$ o ponto de encontro dos segmentos $AI$ e $EF$. Além disso, seja $G$ um ponto sobre o segmento $AB$ de modo que $IG$ e $EF$ sejam paralelos.\n![](attached_image_1.png)\n\na) Prove que $DI = IG$.\n\nb) Prove que $AP = PI$.\n\nObservação: O incentro de um triângulo é o ponto de encontro das bissetrizes dos ângulos internos do triângulo e o incírculo é a circunferência centrada no incentro e tangente aos três lados do triângulo.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na) Como $AF = AE$, segue que $\\angle AFE = \\angle AEF = \\frac{180^{\\circ} - \\angle CAB}{2} = 90^{\\circ} - \\frac{\\angle CAB}{2}$. Além disso, dado que $AC = CD$, temos $\\angle CAD = \\angle CDA$. Sabendo que $DI$ é bissetriz de $\\angle CDB$, temos\n$$\n\\begin{aligned}\n\\angle IDB & = \\frac{\\angle CDB}{2} \\\\\n& = \\frac{180^{\\circ} - \\angle CDA}{2} \\\\\n& = 90^{\\circ} - \\frac{\\angle CAB}{2}\n\\end{aligned}\n$$\nComo $EF \\parallel IG$, segue que $\\angle IGD = \\angle EFA = \\angle IDG$. Portanto, o triângulo $IDG$ é isósceles com $ID = IG$.\n\n\nb) Para provar que $AP = PI$, dado que $EF \\parallel IG$, basta mostrarmos que $AF = FG$. Sejam $w = AF = AE$, $k = EC$, $l = FH$, $x = DH$ e $r = HB$. Para o que segue, precisaremos de um lema simples a respeito do comprimento dos segmentos determinados nos lados de um triângulo pelos pontos de tangência de seu incírculo. Considere a figura abaixo:\n![](attached_image_2.png)\n$AE = AD$, $BE = BF$ e $CF = CD$, se $2p = AB + BC + AC$, temos $2AE = 2p - 2BC$ e, consequentemente, $AE = p - BC$. Analogamente, $BE = p - AC$ e $CF = p - AB$. Usando o lema anterior, e notando que $CD = w + k$ e $CB = k + l + r$, podemos concluir que\n$$\n\\begin{aligned}\nDH & = \\frac{BD + CD - CB}{2} \\\\\nx & = \\frac{(x + r) + (w + k) - (k + l + r)}{2} \\\\\n2x & = w + x - l \\\\\nx + l & = w\n\\end{aligned}\n$$\nAssim, $FH = x + l = w = AF$ e isso conclui a demonstração.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26878,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm triângulo $AB C$ tem lados de comprimentos $AB = 50~\\mathrm{cm}$, $BC = 20~\\mathrm{cm}$ e $AC = 40~\\mathrm{cm}$. Sejam $M$ e $N$ pontos no lado $AB$ tais que $CM$ é a bissetriz relativa ao ângulo $\\angle ACB$ e $CN$ é a altura relativa ao lado $AB$. Qual a medida, em centímetros, de $MN$?",
    "options": [],
    "answer": "11/3",
    "solution": "Solution:\n\nPela Lei dos Cossenos aplicado ao triângulo $\\triangle ABC$, temos\n$$\n\\begin{aligned}\nAC^2 & = AB^2 + BC^2 - 2 \\cdot AB \\cdot BC \\cdot \\cos \\angle ABC \\\\\n1600 & = 2500 + 400 - 2 \\cdot 50 \\cdot 20 \\cdot \\cos \\angle ABC \\\\\n\\cos \\angle ABC & = \\frac{13}{20}\n\\end{aligned}\n$$\nAlém disso, pelo Teorema da Bissetriz Interna, temos\n$$\n\\begin{aligned}\n\\frac{BM}{MA} & = \\frac{BC}{AC} \\\\\n\\frac{BM}{BM + MA} & = \\frac{BC}{BC + AC} \\\\\n\\frac{BM}{50} & = \\frac{20}{60}\n\\end{aligned}\n$$\nDaí, $BM = 50 / 3$. Finalmente,\n$$\n\\begin{aligned}\nMN & = BM - BN \\\\\n& = \\frac{50}{3} - BC \\cdot \\cos \\angle ABC \\\\\n& = \\frac{50}{3} - 20 \\cdot \\frac{13}{20} \\\\\n& = \\frac{11}{3}\n\\end{aligned}\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26879,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEm uma circunferência são escritos 99 números naturais. Se $a$ e $b$ são dois números vizinhos na circunferência, então $\\frac{a}{b}=2$, $a-b=1$ ou $a-b=2$. Prove que existe algum número na circunferência que é divisível por 3.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSejam $a_{1}, a_{2}, \\ldots, a_{99}$ os números na circunferência, escritos em ordem, e suponha, por absurdo, que nenhum deles é múltiplo de 3. Daí, se dois vizinhos $a$ e $b$ deixarem o mesmo resto na divisão por 3, como 3 não pode dividir as diferenças 1 e 2, segue que $a=2b$ ou $b=2a$. Sabendo que 3 divide $a-b$, em ambas as situações anteriores poderíamos concluir que um deles, e portanto os dois, deve ser múltiplo de 3. Isso contradiz nossa suposição inicial. Uma vez que $a$ e $b$ não são múltiplos de 3 e os seus restos são distintos, um deles deve deixar resto 1 e o outro deve deixar resto 2 na divisão por 3. Isso nos diz que $a_{i}+a_{i+1}$ é um múltiplo de 3 para todo $i \\leq 98$. Podemos escrever:\n$$\n\\begin{aligned}\n& S=\\left(a_{1}+a_{2}\\right)+\\ldots+\\left(a_{97}+a_{98}\\right)+a_{99} \\\\\n& S=a_{1}+\\left(a_{2}+a_{3}\\right)+\\ldots+\\left(a_{98}+a_{99}\\right)\n\\end{aligned}\n$$\nComo as somas em cada parênteses são múltiplas de 3, podemos concluir que $a_{1}$ e $a_{99}$ deixam o mesmo resto na divisão por 3. Entretanto, isso é uma contradição, pois números vizinhos não deixam o mesmo resto na divisão por 3. Isso nos diz que a suposição inicial é falsa e, consequentemente, pelo menos um deles deve ser múltiplo de 3.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26880,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n(a) Encontre todos os pares $(m, n)$ de inteiros não negativos que satisfazem a equação\n$$\n5 m + 8 n = 120\n$$\n\n(b) Sejam $a$, $b$ e $c$ números inteiros positivos com $c > 1$ tais que\n$$\n\\frac{1}{a} + \\frac{1}{b} = \\frac{1}{c}\n$$\nProve que pelo menos um dos números $a + c$ ou $b + c$ é um número composto, ou seja, possui algum divisor maior que 1 e menor do que ele mesmo.",
    "options": [],
    "answer": "Part (a): (m, n) = (0, 15), (8, 10), (16, 5), (24, 0).",
    "solution": "Solution:\n\na) Podemos reescrever a equação como\n$$\n\\begin{aligned}\n8 n & = 120 - 5 m \\\\\n& = 5(24 - m)\n\\end{aligned}\n$$\nTemos que $8$ divide o lado direito e não possui fator comum com $5$. Consequentemente $24 - m$ deve ser um múltiplo de $8$. Sabendo que $24$ é um múltiplo de $8$, temos que $m$ é um múltiplo de $8$ entre $0$ e $24$. Logo, as soluções da equação são $(m, n) = (0, 15), (8, 10), (16, 5)$ ou $(24, 0)$.\n\nb) Se $a$ e $c$ possuírem um fator $d > 1$ em comum, então $d$ é um divisor de $a + c$ maior que $1$ e menor que $a + c$. Podemos concluir que $a + c$ é composto. Do mesmo modo, se $b$ e $c$ possuírem algum fator em comum, então $b + c$ é um número composto. Suponha, por outro lado, que $c$ não possua fatores em comum nem com $a$ e nem com $b$. Podemos desenvolver a equação dada\n$$\n\\begin{aligned}\n\\frac{1}{a} + \\frac{1}{b} & = \\frac{1}{c} \\\\\n\\frac{a + b}{a b} & = \\frac{1}{c} \\\\\nc(a + b) & = a b\n\\end{aligned}\n$$\nIsso nos diz que $c$ é um divisor maior que $1$ do produto $a b$ que não possui fatores em comum com o produto $a b$. Isso não é possível. Logo, um dos dois casos acima deve acontecer e concluímos que pelo menos um dos números $a + c$ ou $b + c$ é composto.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26881,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nJoão possui uma folha de papel quadriculado com 10 quadrados de comprimento e 7 quadrados de largura. Ele escolheu 30 triângulos com vértices nas interseções das linhas desse papel quadriculado. Explique por que obrigatoriamente existem pelo menos dois triângulos escolhidos com vértices em comum.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nO papel quadriculado possui 11 linhas horizontais e 8 linhas verticais. Consequentemente, existem $11 \\cdot 8 = 88$ pontos de interseções entre elas. Se não existirem dois triângulos com vértices em comum, precisaremos de pelo menos $3 \\cdot 30 = 90$ vértices distintos. Como $90 > 88$, esse absurdo mostra que pelo menos dois triângulos devem compartilhar um vértice.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26882,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUma desigualdade simples, mas bastante útil é $x^{2} \\geq 0$, para todo $x$ real. Para prová-la, basta estudar separadamente as seguintes possibilidades: $x>0, x<0$ ou $x=0$. De fato, um número real positivo multiplicado por um número real positivo é positivo, um número real negativo multiplicado por outro número real negativo é também positivo e, finalmente, $0 \\cdot 0=0$. A partir dessa desigualdade, podemos derivar outras não tão elementares. Por exemplo, para quaisquer números reais $x$ e $y$ é verdade que $x^{2}+x y+y^{2} \\geq 0$, pois\n$$\n\\begin{aligned}\nx^{2}+x y+y^{2} & =\\left(x^{2}+\\frac{2 x y}{2}+\\frac{y^{2}}{4}\\right)+3\\left(\\frac{y^{2}}{4}\\right) \\\\\n& =\\left(x+\\frac{y}{2}\\right)^{2}+3\\left(\\frac{y}{2}\\right)^{2} \\\\\n& \\geq 0\n\\end{aligned}\n$$\nNa última desigualdade, usamos que $\\left(x+\\frac{y}{2}\\right)^{2} \\geq 0 \\mathrm{e}\\left(\\frac{y}{2}\\right)^{2} \\geq 0$.\nVeremos agora uma aplicação dessas desigualdades. Sejam $a$ e $b$ números reais tais que $a^{3}-b^{3}=2$ e $a^{5}-b^{5} \\geq 4$.\n\na. Sabendo que para quaisquer reais $x$ e $y$ vale $x^{3}-y^{3}=(x-y)\\left(x^{2}+x y+y^{2}\\right)$, verifique que $a>b$.\n\nb. Verifique que $a^{2}+b^{2} \\geq 2$.",
    "options": [],
    "answer": "a > b and a^2 + b^2 ≥ 2",
    "solution": "Solution:\n\na.\nPela fatoração sugerida,\n$$\n\\begin{aligned}\n0<2 & =a^{3}-b^{3} \\\\\n& =(a-b)\\left(a^{2}+a b+b^{2}\\right)\n\\end{aligned}\n$$\nComo $a^{2}+a b+b^{2} \\geq 0$, segue que $a-b>0$, ou seja, $a>b$.\n\n\nb.\nSabemos que $a^{2}, b^{2} \\geq 0$ e $a-b>0$. Daí,\n$$\n\\begin{aligned}\n\\left(a^{3}-b^{3}\\right)\\left(a^{2}+b^{2}\\right) & =a^{5}-a^{2} b^{3}+a^{3} b^{2}-b^{5} \\\\\n2\\left(a^{2}+b^{2}\\right) & =a^{5}-b^{5}+a^{2} b^{2}(a-b) \\\\\n& \\geq a^{5}-b^{5} \\\\\n& \\geq 4\n\\end{aligned}\n$$\nPortanto, $a^{2}+b^{2} \\geq 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26883,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSeja $n>1$ um inteiro positivo, chamamos de $d(n)$ a quantidade de divisores positivos de $n$. Para calcular $d(n)$, basta escrever a fatoração de $n$ em potências de primos distintos e multiplicar os sucessores dos expoentes. Por exemplo, para $2016$ temos a fatoração $2016=2^{5} \\cdot 3^{2} \\cdot 7^{1}$ e $d(2016)=(5+1)(2+1)(1+1)=36$.\n\na. Prove que se $n$ é um quadrado perfeito, então $d(n)$ é ímpar.\n\nb. Determine todos os $n$ menores que $400$ tais que $n$ e $d(n)$ sejam quadrados perfeitos.",
    "options": [],
    "answer": "[36, 100, 196, 225, 256]",
    "solution": "Solution:\n\na. Se $n$ é um quadrado perfeito, então existe um inteiro positivo $m$ tal que $n=m^{2}$. Na fatoração em primos de $n$, todos os expoentes são pares, pois cada fator primo de $m$ aparece o dobro de vezes na fatoração de $n$. Por exemplo, se $m=2^{2} \\cdot 3$, então $n=m^{2}=\\left(2^{2} \\cdot 3\\right)^{2}=2^{4} \\cdot 3^{2}$. Como o sucessor de um número par é ímpar e o produto de números ímpares também é ímpar, concluímos que $d(n)$ é ímpar.\n\nb. Escreva $n=m^{2}$. Como $m^{2}<400=20^{2}$ podemos concluir que $m \\leq 20$. Entre os inteiros positivos menores que $20$, temos apenas $3$ possíveis fatorações para $m$: $p^{k}$, $p \\cdot q$ ou $p^{2} \\cdot q$, com $p$ e $q$ primos distintos, pois qualquer produto de três primos distintos é maior que $20$. Além disso, $k \\leq 4$, pois $2^{5}>20$. No primeiro caso, $d(n)=d\\left(p^{2k}\\right)=2k+1$ deve ser quadrado perfeito e apenas $m=2^{4}$, ou seja, $n=16^{2}$ satisfaz as condições. No segundo caso, $d(n)=d\\left(p^{2} \\cdot q^{2}\\right)=(2+1)(2+1)=9$. Temos então os seguintes valores para $n$: $6^{2}$, $10^{2}$, $14^{2}$ e $15^{2}$. No terceiro caso, $d\\left(p^{4} \\cdot q^{2}\\right)=(4+1)(2+1)=15$, que não é quadrado perfeito. Portanto, os números $n$ menores que $400$ tais que $n$ e $d(n)$ são quadrados perfeitos são $6^{2}$, $10^{2}$, $14^{2}$, $15^{2}$ e $16^{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26884,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nQueremos cobrir um tabuleiro quadriculado com certas pecinhas sem sobreposição e de modo que nenhuma parte delas fique fora do tabuleiro. Usaremos pecinhas, formadas por quadradinhos, chamadas L-triminós e I-triminós e que podem ser rotacionadas nas posições descritas na figura a seguir.\n\n![](attached_image_1.png)\n\nPara provar que é possível realizar uma cobertura, basta mostrar uma maneira de posicionar as pecinhas. Por outro lado, para provar que não é possível realizar alguma cobertura, nem sempre é conveniente testar todas as configurações possíveis de peças e muitas vezes precisamos esboçar argumentos engenhosos. Por exemplo, provaremos que não é possível cobrir um tabuleiro $3 \\times 3$ usando apenas L-triminós.\n\n![](attached_image_2.png)\n\nObserve os quadradinhos pintados da figura. São 4 quadradinhos e não é possível cobrir dois deles usando um mesmo L-triminó. Assim, para cobrir os 4 quadradinhos teríamos que usar pelo menos $4 \\mathrm{~L}$-triminós, mas isso resultaria em $4 \\cdot 3=12$ quadradinhos cobertos, que claramente excede o total de 9 quadradinhos do tabuleiro inteiro. Portanto, não é possível cobrir o tabuleiro $3 \\times 3$ com L-triminós.\n\na) Mostre uma maneira de cobrir um tabuleiro $3 \\times 4$ usando apenas L-triminós.\n\nb) Prove que não é possível cobrir um tabuleiro $3 \\times 5$ usando apenas L-triminós.\n\nc) É possível cobrir o $3 \\times 5$ usando exatamente um I-triminó e alguns L-triminós. Determine as posições que o I-triminó pode ocupar de modo que o resto do tabuleiro possa ser coberto com L-triminós.",
    "options": [],
    "answer": "a) A covering of the three by four board using only L-triminos exists (one explicit arrangement is shown). b) It is impossible to cover the three by five board using only L-triminos. c) The straight trimino must cover two marked squares; it can occupy seven positions: four horizontal positions in the top and bottom rows, and three vertical positions in the first, third, and fifth columns.",
    "solution": "Solution:\n\na) A figura a seguir mostra uma maneira de cobrir o tabuleiro $3 \\times 4$ usando apenas L-triminós.\n\n![](attached_image_3.png)\n\nb) Considere o tabuleiro $3 \\times 5$ a seguir e os 6 quadradinhos pintados.\nComo um L-triminó não pode cobrir duas dessas casinhas pintadas, se fosse possível cobrir o tabuleiro inteiro teríamos que usar pelo menos $6 \\mathrm{~L}$-triminós. Porém, esses $6 \\mathrm{~L}$-triminós cobririam no total $6 \\cdot 3=18$ quadradinhos e isso ultrapassa o total de $3 \\cdot 5=15$ quadradinhos que o tabuleiro possui. Portanto, não é possível fazer essa cobertura.\n\nc) Observe que um I-triminó pode cobrir 0, 1 ou 2 dos quadradinhos pintados na figura anterior. Se cobrir 0 ou 1, podemos usar o argumento do item anterior para mostrar que não será possível concluir a cobertura, pois 5 L-triminós já cobrem 15 quadradinhos. Portanto, a única possibilidade é que o I-triminó cubra 2 quadradinhos. De fato, se o I-triminó cobrir dois quadradinhos marcados é sempre possível concluir a cobertura com os L-triminós como indicado na próxima figura:\n\n![](attached_image_4.png)\n\nSão 4 posições horizontais, duas na primeira linha e duas na terceira linha, e 3 verticais, cobrindo primeira, terceira e quinta colunas. No total o I-triminó pode ocupar 7 posições.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26885,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPara determinar a quantidade de divisores positivos de um número, basta fatorá-lo como potências de primos distintos e multiplicar os sucessores dos expoentes. Por exemplo, $2016=2^{5} \\cdot 3^{2} \\cdot 5^{1}$ possui $(5+1)(2+1)(1+1)=36$ divisores positivos. Considere o número $n=2^{7} \\cdot 3^{4}$.\n\na) Determine o número de divisores positivos de $n^{2}$.\n\nb) Quantos divisores de $n^{2}$ são maiores que $n$ ?\n\nc) Quantos divisores de $n^{2}$ são maiores que $n$ e não são múltiplos de $n$ ?",
    "options": [],
    "answer": "a) 135; b) 67; c) 28",
    "solution": "Solution:\n\na) A partir da fatoração de $n$, podemos determinar a fatoração de $n^{2}$ :\n$$\n\\begin{aligned}\nn^{2} & =\\left(2^{7} \\cdot 3^{4}\\right)^{2} \\\\\n& =2^{14} \\cdot 3^{8}\n\\end{aligned}\n$$\nEntão o número $n^{2}$ possui $(14+1)(8+1)=15 \\cdot 9=135$ divisores positivos.\n\nb) Note que todos os divisores positivos de $n^{2}$ podem ser organizados em pares de números distintos da forma $\\left(d, \\frac{n^{2}}{d}\\right)$ cujo produto é $n^{2}$, com exceção de $n$. Além disso, como os números em cada par são distintos e o produto é $n^{2}$, um deles é maior que $n$ e o outro é menor que $n$. Por exemplo, se considerarmos $n=6$, os divisores de $n^{2}=36$ diferentes de 6 podem ser organizados nos seguintes pares: $\\{1,36\\},\\{2,18\\},\\{3,12\\}$ e $\\{4,9\\}$. Dessa forma, podemos concluir que o número de divisores de $n^{2}$ maiores que $n$ é igual ao número de pares que podemos formar. Portanto, são $\\frac{135-1}{2}=67$ divisores de $n^{2}$ maiores que $n$.\n\nc) Vamos continuar o raciocínio do item anterior. Veja que em cada par temos $x \\cdot y= n^{2}$, em que $x<n<y$. Observando os fatores primos de $x$ e de $y, y$ é um múltiplo de $n$ nos casos em que $x$ é um divisor de $n$ menor que $n$. Como $n=2^{7} \\cdot 3^{4}$ possui $(7+1)(4+1)=40$ divisores positivos, existem 39 deles diferentes de $2^{7} \\cdot 3^{4}$. Portanto, excluindo-se os pares em que $x$ é um divisor de $n$ menor que $n$, podemos concluir que existem $67-39=28$ divisores de $n^{2}$ maiores que $n$ que não são múltiplos de $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26886,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSeja $n \\geq 3$ um inteiro positivo. Sobre uma reta, são marcados os $n$ pontos $P_{1}, P_{2}, P_{3}, \\ldots, P_{n}$, nessa ordem e igualmente espaçados entre si. Em seguida, cada um dos pontos deve ser pintado de azul ou de vermelho de modo que não existam três pontos $P_{x}, P_{\\frac{x+y}{2}}$ e $P_{y}$ pintados da mesma cor, sendo $x+y$ par.\n\na. Mostre que para $n=8$ existe uma maneira de colorir os pontos $P_{1}, P_{2}, \\ldots, P_{8}$ satisfazendo a condição dada.\n\nb. Mostre que qualquer pintura para $n=9$ não satisfaz a condição dada.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na. Marcaremos com $A$ e $V$ as cores dos pontos em ordem. Considere a seguinte forma de colorir os 8 pontos:\n$$\n\\begin{array}{cccccccc}\nP_{1} & P_{2} & P_{3} & P_{4} & P_{5} & P_{6} & P_{7} & P_{8} \\\\\nA & V & V & A & A & V & V & A .\n\\end{array}\n$$\nA condição dada requer que não existam pontos de mesma cor de modo que um deles seja o ponto médio do segmento formado pelos outros dois. É imediato verificar que essa condição é satisfeita no exemplo anterior.\n\n\nb. Considere uma pintura qualquer para $n=9$ satisfazendo as condições do enunciado. Assim, não podem existir trios da forma $\\left(P_{x}, P_{\\frac{x+y}{2}}, P_{y}\\right)$ com todos esses pontos com a mesma cor. Suponha inicialmente que $P_{3}$ e $P_{5}$ possuem a mesma cor, digamos azul. Então $P_{1}, P_{4}$ e $P_{7}$ deverão ter a cor oposta, ou seja, vermelho. Caso contrário, algum dos trios $\\left(P_{1}, P_{3}, P_{5}\\right),\\left(P_{3}, P_{4}, P_{5}\\right)$ ou $\\left(P_{3}, P_{5}, P_{7}\\right)$ teria três pontos pintados de azul. Entretanto, isso implicaria que ($P_{1}, P_{4}, P_{7}$) estão pintados de vermelho. Portanto, para a condição do enunciado ser satisfeita, concluímos que $P_{3}$ e $P_{5}$ não podem ter a mesma cor. O mesmo argumento anterior aplicado a $P_{5}$ e $P_{7}$ nos permite concluir que eles também possuem cores opostas. Suponha, sem perda de generalidade, que $P_{3}$ é azul e $P_{5}$ é vermelho. Logo, $P_{7}$ deve ter cor azul. Desse modo, temos a seguinte configuração:\n$$\n\\begin{array}{ccccccccc}\nP_{1} & P_{2} & P_{3} & P_{4} & P_{5} & P_{6} & P_{7} & P_{8} & P_{9} \\\\\n- & - & A & - & V & - & A & - & -.\n\\end{array}\n$$\nAnalisaremos agora as possíveis cores dos outros pontos. Tratemos inicialmente do caso em que $P_{1}$ é azul. Assim, teremos a seguinte configuração:\n$$\n\\begin{array}{ccccccccc}\nP_{1} & P_{2} & P_{3} & P_{4} & P_{5} & P_{6} & P_{7} & P_{8} & P_{9} \\\\\nA & - & A & - & V & - & A & - & -.\n\\end{array}\n$$\nPara que $\\left(P_{1}, P_{4}, P_{7}\\right)$ ou $\\left(P_{1}, P_{2}, P_{3}\\right)$ não sejam azuis, $P_{2}$ e $P_{4}$ devem ser vermelhos.\n$$\n\\begin{array}{ccccccccc}\nP_{1} & P_{2} & P_{3} & P_{4} & P_{5} & P_{6} & P_{7} & P_{8} & P_{9} \\\\\nA & V & A & V & V & - & A & - & -.\n\\end{array}\n$$\nPara que $\\left(P_{2}, P_{4}, P_{6}\\right)$ não sejam todos vermelhos, $P_{6}$ deve ser azul:\n$$\n\\begin{array}{ccccccccc}\nP_{1} & P_{2} & P_{3} & P_{4} & P_{5} & P_{6} & P_{7} & P_{8} & P_{9} \\\\\nA & V & A & V & V & A & A & - & -\n\\end{array}\n$$\nSe $P_{8}$ é vermelho, $\\left(P_{2}, P_{5}, P_{8}\\right)$ serão todos vermelhos e, se $P_{8}$ é azul, $\\left(P_{6}, P_{7}, P_{8}\\right)$ serão azuis. Em ambas as situações a condição do enunciado não é satisfeita. Resta analisarmos agora o que acontece quando $P_{1}$ é vermelho. Nesse caso, temos a seguinte configuração:\n$$\n\\begin{array}{ccccccccc}\nP_{1} & P_{2} & P_{3} & P_{4} & P_{5} & P_{6} & P_{7} & P_{8} & P_{9} \\\\\nV & & A & & V & & A & - & -\n\\end{array}\n$$\nSe $P_{9}$ é vermelho, $\\left(P_{1}, P_{5}, P_{9}\\right)$ serão todos vermelhos. Portanto, $P_{9}$ deve ser azul e a configuração deve ser:\n$$\n\\begin{array}{ccccccccc}\nP_{1} & P_{2} & P_{3} & P_{4} & P_{5} & P_{6} & P_{7} & P_{8} & P_{9} \\\\\nV & & A & & V & & A & - & A\n\\end{array}\n$$\nMas esse caso é o simétrico do caso $P_{1}$ pintado de azul. Repetindo os mesmos passos, veremos que $P_{2}$ não poderá ser colorido de vermelho nem de azul. Para qualquer coloração do $P_{1}$, não será possível pintar todos os 9 pontos de modo que não existam três pontos $P_{x}, P_{\\frac{x+y}{2}}$ e $P_{y}$ pintados da mesma cor.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26887,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nConsidere um triângulo $ABC$ e quadrados $ABDE$ e $ACFG$ construídos exteriormente sobre seus lados. Os segmentos $BG$ e $EC$ se intersectam em $P$.\n![](attached_image_1.png)\na) Prove que os triângulos $BAG$ e $EAC$ são congruentes.\nb) Como esses dois triângulos são congruentes e possuem o ponto $A$ em comum, podemos rotacionar um deles em torno do ponto $A$ para obter o outro. Determine o ângulo de rotação e calcule o ângulo $\\angle BPC$.",
    "options": [],
    "answer": "Rotation angle = 90°, and ∠BPC = 90°",
    "solution": "Solution:\nConsidere a figura a seguir em que marcamos os lados iguais dos dois quadrados.\n![](attached_image_2.png)\n\na) Podemos afirmar que os triângulos $BAG$ e $EAC$ são congruentes pelo caso $LAL$, já que $BA=EA$, $\\angle BAG=\\angle EAC=90^\\circ+\\angle BAC$ e $AG=AC$.\n\nb) Como o ângulo formado entre $AG$ e $AC$ é $90^\\circ$, podemos concluir que o ângulo de rotação para levar o triângulo $BAG$ no triângulo $EAC$ é $90^\\circ$ (podemos usar uma rotação no sentido horário em torno de $A$). Decorre da congruência anterior que $\\angle AEC=\\angle ABG$ e $\\angle AGB=\\angle ACE$. Sabendo que a soma dos ângulos internos de um quadrilátero é $360^\\circ$, pois podemos dividi-lo em dois triângulos com soma de seus ângulos internos igual a $180^\\circ$, podemos escrever para a soma dos ângulos internos do quadrilátero $AEPG$:\n$$\n\\begin{aligned}\n\\angle AEP+\\angle EPG+\\angle PGA+\\angle GAE & =360^\\circ \\\\\n\\angle AEP+\\angle EPG+\\angle ACP+\\left(\\angle EAC+90^\\circ\\right) & =360^\\circ \\\\\n(\\angle AEC+\\angle ECA+\\angle EAC)+\\angle EPG+90^\\circ & =360^\\circ \\\\\n\\angle EPG & =90^\\circ\n\\end{aligned}\n$$\nNote que $\\angle BPC=\\angle EPG$, pois são opostos pelo vértice $P$. Daí, $\\angle BPC=90^\\circ$.\n\nObservação: Perceba que o ângulo entre os segmentos correspondentes $EC$ e $BG$ coincide com o ângulo de rotação aplicado aos triângulos $AEC$ e $ABG$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26888,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSérgio escolhe dois números inteiros positivos $a$ e $b$. Ele escreve 4 números no seu caderno: $a$, $a+2$, $b$ e $b+2$. Em seguida, todos os 6 produtos de dois desses números são escritos na lousa. Seja $Q$ a quantidade de quadrados perfeitos escritos nela, determine o valor máximo de $Q$.",
    "options": [],
    "answer": "2",
    "solution": "Solution:\n\nInicialmente provaremos que o produto $a(a+2)$ não é um quadrado perfeito para qualquer escolha de $a$. Temos dois casos a considerar:\n\ni) Se $a$ é ímpar, então nenhum primo que divide $a$ poderá dividir $a+2$. Daí, $a$ e $a+2$ deverão ser, cada um, um quadrado perfeito. Isso claramente não tem solução para $a \\geq 1$, pois a diferença entre dois quadrados perfeitos é pelo menos 3.\n\nii) Se $a$ é par, temos $a=2k$ para algum inteiro positivo $k$. Daí, $a(a+2)=2k(2k+2)=(2k+1)^2-1$. Isso também não é possível, pois neste caso teríamos um quadrado perfeito entre dois quadrados perfeitos consecutivos: $(2k)^2 < a(a+2) < (2k+1)^2$.\n\nAssim, o produto $a(a+2)$ não pode ser um quadrado perfeito. Além disso, $ab$ e $b(a+2)$ não podem ser simultaneamente quadrados perfeitos, pois neste caso $ab \\cdot b(a+2) = b^2 \\cdot a(a+2)$ é um quadrado perfeito e isso implica $a(a+2)$ também quadrado perfeito, contradizendo o lema anterior. Analogamente, $a(b+2)$ e $(a+2)(b+2)$ também não podem ser simultaneamente quadrados perfeitos. Por simetria, $(a+2)b$ e $(a+2)(b+2)$ não podem ser ao mesmo tempo quadrados perfeitos.\n\nDessa forma, os possíveis quadrados seriam $ab$, $b(a+2)$, $a(b+2)$ e $(a+2)(b+2)$. Excluindo-se os produtos $a(a+2)$, $b(b+2)$ e notando que nos pares ($ab$, $b(a+2)$) e ($ab$, $a(b+2)$) apenas um de seus membros é um quadrado perfeito, podemos concluir que $Q \\leq 3$. Para termos $Q=3$, $b(a+2)$, $a(b+2)$, $(a+2)(b+2)$ são quadrados. Isso é um absurdo, pois já vimos que $b(a+2)$ e $(a+2)(b+2)$ não podem ser simultaneamente quadrados perfeitos. Veja que $Q=2$ pode ser obtido, por exemplo, com $a=b=1$, pois $ab=1^2$ e $(a+2)(b+2)=3^2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26889,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOs números inteiros do conjunto $\\{1,2, \\ldots, 20\\}$ serão pintados com duas cores, branco e preto, de modo que ambas as cores sejam usadas. Além disso, o produto dos números de uma cor não deve possuir fatores primos em comum com o produto dos números da outra cor. De quantos modos isso pode ser feito?",
    "options": [],
    "answer": "62",
    "solution": "Solution:\n\nIndependente das cores escolhidas para os outros números, temos duas opções de escolha para a cor do número $1$. Temos também duas opções para a cor do número $2$ e, uma vez que ela tenha sido escolhida, todos os números do conjunto $\\{4,6,8,10,12,14,16,18,20\\}$ devem possuir a mesma cor do número $2$. Como existem números desse conjunto que compartilham fatores primos em comum com os números do conjunto $\\{3,7,9,15\\}$, estes 4 números ímpares também devem possuir a cor do número $2$. Resta escolhermos a cor dos números do conjunto $\\{11,13,17,19\\}$ e cada uma delas pode ser feita de modo independente, pois não existe outro número no conjunto inicial com esses mesmos fatores primos. Assim, o total de pinturas é $2 \\cdot 2 \\cdot 2 \\cdot 2 \\cdot 2 \\cdot 2 = 2^{6}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26890,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm tabuleiro de xadrez é um quadrado $8 \\times 8$ em que as casinhas estão distribuídas em 8 linhas e 8 colunas.\n![](attached_image_1.png)\nUma torre em um tabuleiro de xadrez ataca todas as peças que estejam na sua linha ou na sua coluna. Sabendo disso, determine:\n\na) Qual o número máximo de torres que podemos colocar num tabuleiro de xadrez de modo que não haja duas se atacando?\n\nb) Qual o número máximo de torres que podemos colocar num tabuleiro de xadrez de modo que cada torre seja ameaçada de ataque por no máximo uma das outras torres?",
    "options": [],
    "answer": "a) 8; b) 10",
    "solution": "Solution:\n\na) A seguir, temos um exemplo com 8 torres sem que duas estejam se atacando.\n![](attached_image_2.png)\nSe colocarmos 9 ou mais torres, como existem apenas 8 linhas, haverá duas numa mesma linha e uma poderá atacar a outra. Portanto, o número máximo é 8.\n\n\nb) A seguir, temos um exemplo com 10 torres em que cada uma é atacada por no máximo uma outra torre.\n![](attached_image_3.png)\nAgora resta mostrar que se colocarmos 11 ou mais torres haverá uma torre ameaçada por pelo menos duas outras. Considere uma distribuição de torres em que cada uma é ameaçada por no máximo outra. Podemos classificá-las em dois grupos: aquelas que não são ameaçadas por outras torres, chamadas de solitárias, e as que são ameaçadas por exatamente uma outra torre, chamadas de torres com inimigas. Para cada torre solitária, podemos colocar uma etiqueta em sua linha e coluna. Gastaremos precisamente duas etiquetas. Para cada torre que é ameaçada por exatamente uma outra, podemos colocar uma etiqueta nas linhas e nas colunas que contêm ela e sua inimiga. Usaremos precisamente três etiquetas nesse caso. Se $x$ é a quantidade de torres solitárias e $2y$ a quantidade de torres inimigas, em que $y$ é a quantidade de pares de torres inimigas, então a quantidade de etiquetas usadas é $2x + 3y$. Além disso, como cada linha e coluna, que totalizam 16, recebeu apenas uma etiqueta, temos:\n$$\n\\begin{aligned}\n16 &\\geq 2x + 3y \\\\\n32 &\\geq 4x + 6y \\\\\n&\\geq 3x + 6y\n\\end{aligned}\n$$\nDaí, $32/3 \\geq x + 2y$. Como a quantidade de torres, que é dada por $x + 2y$, é um número inteiro e $32/3 < 11$, segue que podem existir no máximo 10 torres no tabuleiro satisfazendo as condições do segundo item.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26891,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA mediatriz de um segmento $X Y$ é a reta perpendicular ao segmento passando por seu ponto médio. A principal propriedade da mediatriz é que um ponto está sobre ela se, e somente se, a distância desse ponto até $X$ é igual à distância desse ponto até $Y$. Uma afirmação formada usando \"se, e somente se\" é equivalente a duas implicações. No presente caso, para verificar a afirmação feita, precisamos por um lado mostrar que se a distância até $X$ é igual à distância até $Y$, então o ponto está na mediatriz. Por outro lado, precisamos mostrar que se um ponto está sobre a mediatriz, então as distâncias aos pontos $X$ e $Y$ são iguais. Para provar a primeira parte, considere um ponto $P$ tal que $P X=P Y$ e seja $M$ o ponto médio de $X Y$.\n![](attached_image_1.png)\nOs triângulos $P M X$ e $P M Y$ são congruentes, pois possuem os três lados correspondentes de mesmo comprimento. Portanto, $\\angle P M X=\\angle P M Y$. Como $X, M$ e $Y$ são colineares, temos $\\angle P M X+\\angle P M Y=180^\\{\\circ\\}$. Consequentemente, $\\angle P M X=90^\\{\\circ\\}$ e $P$ está na reta perpendicular ao segmento que passa pelo seu ponto médio.\n\na) Prove a segunda parte da proposição, ou seja, prove que se $P$ é um ponto sobre a reta mediatriz de $X Y$, então $P X=P Y$.\n\nb) Prove que todo triângulo $A B C$ possui um ponto $O$ no mesmo plano tal que $O A=O B=O C$. Esse ponto é denominado circuncentro do triângulo, pois uma circunferência de centro $O$ e raio $A O$ passa pelos três vértices do triângulo $A B C$.\n\nc) Considere um quadrilátero $A B C D$. Suponha que as mediatrizes de $A B, A C$ e $A D$ são concorrentes, ou seja, as três passam por um mesmo ponto. Prove que existe uma circunferência que passa pelos 4 vértices do quadrilátero $A B C D$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na) Como os triângulos $P M X$ e $P M Y$ são retângulos e $M X=M Y$, pelo Teorema de Pitágoras, temos:\n$$\nP X^{2}=P M^{2}+M X^{2}=P M^{2}+M Y^{2}=P Y^{2}\n$$\nPortanto, $P X=P Y$.\n\nb) Veja que as mediatrizes de $A B$ e $B C$ não são paralelas, pois esses segmentos não são paralelos. Logo elas se encontram em um ponto, que denotaremos por $O$.\n![](attached_image_2.png)\nComo $O$ está na mediatriz de $A B, O A=O B$. Além disso, como $O$ está na mediatriz de $B C, O B=O C$. Logo $O A=O B=O C$.\n\nc) Seja $P$ o ponto de encontro das três mediatrizes. Como a mediatriz de um segmento é o lugar geométrico dos pontos que possuem mesma distância em relação aos extremos do segmento, podemos afirmar que $P A=P B=P C=P D$. Dessa forma, se traçarmos a circunferência de centro $P$ e raio $P A$, então ela passará pelos 4 vértices do quadrilátero $A B C D$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26892,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA figura a seguir representa um triângulo $ABC$, retângulo em $C$, com uma circunferência no seu interior tangenciando os três lados $AB$, $BC$ e $CA$ nos pontos $C_1$, $A_1$ e $B_1$, respectivamente. Seja $H$ o pé da altura relativa ao lado $A_1C_1$ do triângulo $A_1B_1C_1$.\n![](attached_image_1.png)\n\na) Calcule a medida do ângulo $\\angle A_1C_1B_1$.\n\nb) Mostre que o ponto $H$ está na bissetriz do ângulo $\\angle BAC$.",
    "options": [],
    "answer": "45°",
    "solution": "Solution:\n\nConsidere a figura a seguir.\n![](attached_image_2.png)\n\n(a) Como $\\angle ACB=90^\\circ$, então $\\angle CBA=90^\\circ-\\angle BAC=90^\\circ-\\angle A$. Dado que $AB_1$ e $AC_1$ são tangentes à circunferência, segue que $AB_1=AC_1$. De modo semelhante, $BA_1=BC_1$. Assim, como $A_1BC_1$ e $AB_1C_1$ são isósceles, segue que\n$$\n\\begin{aligned}\n& \\angle AB_1C_1=\\angle AC_1B_1=\\frac{180^\\circ-\\angle B_1AC_1}{2}=90^\\circ-\\frac{\\angle A}{2} \\\\\n& \\angle BC_1A_1=\\angle BA_1C_1=\\frac{180^\\circ-\\angle A_1BC_1}{2}=45^\\circ+\\frac{\\angle A}{2}\n\\end{aligned}\n$$\nDaí,\n$$\n\\begin{aligned}\n\\angle A_1C_1B_1 & =180^\\circ-\\angle AC_1B_1-\\angle A_1C_1B \\\\\n& =180^\\circ-\\left(90^\\circ-\\frac{A}{2}\\right)-\\left(45^\\circ+\\frac{A}{2}\\right) \\\\\n& =45^\\circ\n\\end{aligned}\n$$\n\n\n(b) Analisando agora o triângulo $B_1HC_1$, podemos obter $\\angle HB_1C_1=90^\\circ-\\angle A_1C_1B_1=45^\\circ$, ou seja, esse triângulo é isósceles com $B_1H=C_1H$. Assim, os triângulos $AB_1H$ e $AC_1H$ são congruentes, pois possuem os três lados de mesmo comprimento. Consequentemente, $\\angle B_1AH=\\angle C_1AH$ e $H$ está sobre a bissetriz do ângulo $\\angle BAC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26893,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUma sequência de números reais $x_{n}$ é uma lista ordenada de reais em que o primeiro número da lista é o termo $x_{1}$, o segundo é o termo $x_{2}$ e assim por diante. Por exemplo, a sequência usual dos números inteiros positivos pode ser descrita como $x_{n}=n$ para todo inteiro positivo $n$. Algumas sequências podem ser definidas por equações de recorrências, em que um termo é definido em função dos seus anteriores.\nPor exemplo, a sequência de inteiros positivos poderia ser definida por $x_{1}=1$ e $x_{n}=x_{n-1}+1$ para todo inteiro positivo $n \\geq 2$. Desse modo, poderíamos calcular $x_{2}=1+1=2$, $x_{3}=2+1=3$ e assim por diante.\nConsidere uma sequência de números reais definida por $x_{1}=1$ e $x_{n}=x_{n-1}+\\frac{1}{x_{n-1}^{2}}$ para todo inteiro positivo $n \\geq 2$.\n\na) Calcule $x_{2}, x_{3}$ e $x_{4}$.\n\nb) Verifique que a sequência é estritamente crescente, ou seja, que $x_{n}>x_{n-1}$ para todo inteiro positivo $n$.\n\nc) Perceba que a sequência parece crescer muito pouco. Após calcular alguns termos iniciais, poderíamos suspeitar que nenhum termo excede 2016, mas de fato vamos provar que existem termos maiores que 2016. Para isso, vamos usar a sequência auxiliar $y_{n}=x_{n}^{3}$. Prove que $y_{n}>y_{n-1}+3$ para todo $n \\geq 2$.\n\nd) Prove que existe um número $N$ tal que $x_{N}>2016$.",
    "options": [],
    "answer": "x2 = 2, x3 = 9/4, x4 = 793/324. Moreover, one can take N = 2016^3 to ensure x_N > 2016.",
    "solution": "Solution:\n\na) Basta usar a equação dada para $n=2, n=3$ e $n=4$.\n$$\n\\begin{gathered}\nx_{2}=x_{1}+\\frac{1}{x_{1}^{2}}=1+\\frac{1}{1^{2}}=2 \\\\\nx_{3}=x_{2}+\\frac{1}{x_{2}^{2}}=2+\\frac{1}{2^{2}}=\\frac{9}{4} \\\\\nx_{4}=x_{3}+\\frac{1}{x_{3}^{2}}=\\frac{9}{4}+\\frac{1}{\\left(\\frac{9}{4}\\right)^{2}}=\\frac{793}{324}\n\\end{gathered}\n$$\n\nb) Para qualquer $n$, se $x_{n-1} \\neq 0$, então $x_{n-1}^{2}>0$ e $\\frac{1}{x_{n-1}^{2}}>0$, pois todo quadrado de um número real não nulo é positivo. Assim,\n$$\n\\begin{aligned}\nx_{n} & =x_{n-1}+\\frac{1}{x_{n-1}^{2}} \\\\\n& >x_{n-1}\n\\end{aligned}\n$$\nComo $x_{1}=1>0$, pelo argumento anterior, temos $x_{2}>x_{1}>0$. Agora, usando $x_{2}$ no papel de $x_{1}$ no argumento anterior, temos $x_{3}>x_{2}>0$. Veja que podemos continuar repetindo o argumento, agora com $x_{3}$ no papel de $x_{2}$. Esse processo indutivo nos permite concluir que a sequência é estritamente crescente.\n\nc) Elevando a equação de recorrência ao cubo teremos\n$$\n\\begin{aligned}\nx_{n}^{3} & =\\left(x_{n-1}+\\frac{1}{x_{n-1}^{2}}\\right)^{3} \\\\\n& =x_{n-1}^{3}+3 x_{n-1}^{2} \\frac{1}{x_{n-1}^{2}}+3 x_{n-1} \\frac{1}{x_{n-1}^{4}}+\\frac{1}{x_{n-1}^{6}} \\\\\n& >x_{n-1}^{3}+3\n\\end{aligned}\n$$\nLogo,\n$$\ny_{n}=x_{n}^{3}>x_{n-1}^{3}+3=y_{n-1}+3\n$$\n\nd) A ideia agora é usar o crescimento de $y_{n}$ para chegar em alguma conclusão sobre o crescimento de $x_{n}$. Temos $x_{n}>2016$ se, e somente se, $y_{n}=x_{n}^{3}>2016^{3}$. No item anterior, provamos que $y_{n}>y_{n-1}+3$ para todo $n \\geq 2$. Aplicações sucessivas dessa desigualdade nos permitem concluir que:\n$$\n\\begin{aligned}\ny_{n} & >y_{n-1}+3 \\\\\n& >y_{n-2}+6 \\\\\n& >y_{n-3}+9 \\\\\n& \\cdots \\\\\n& >y_{1}+3(n-1) \\\\\n& =3 n-2\n\\end{aligned}\n$$\nBom, agora basta tomar $N$ que satisfaça $3 N-2>2016^{3}$, ou seja, $N>\\frac{2016^{3}+2}{3}$. Podemos tomar $N=2016^{3}$, por exemplo, já que esse número satisfaz a inequação anterior. Para esse valor de $N$, o termo $x_{N}$ da sequência será maior que 2016.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26894,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm número natural $n$ é interessante se a soma dos dígitos de $n$ é igual à soma dos dígitos de $3n+11$. Verifique que existem infinitos números interessantes.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nDado um número natural $n$, vamos denotar por $s(n)$ a soma dos dígitos de $n$. Podemos fazer uma tabela com os primeiros inteiros positivos para encontrar algum exemplo de número interessante.\n\n| $n$ | $3n+11$ | $s(n)$ | $s(3n+11)$ | $n$ | $3n+11$ | $s(n)$ | $s(3n+11)$ |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| 1 | 14 | 1 | 5 | 5 | 26 | 5 | 8 |\n| 2 | 17 | 2 | 8 | 6 | 29 | 6 | 11 |\n| 3 | 20 | 3 | 2 | 7 | 32 | 7 | 5 |\n| 4 | 23 | 4 | 5 | 8 | 35 | 8 | 8 |\n\nPodemos concluir que o número $8$ é interessante. Nossa estratégia será construir outros números interessantes a partir dele. Considere, por exemplo, um número obtido a partir de $8$ com a inserção de vários dígitos nulos à sua esquerda e do algarismo $1$, ou seja, da forma $n=10^{k}+8$. Veja que $3n+11=3 \\cdot 10^{k}+35$ e que $s(n)=1+8=9$ e $s(3n+11)=3+3+5=11$. Infelizmente ele não é interessante, mas a inserção de muitos dígitos nulos à esquerda do $8$ nos permite ainda identificar a soma dos dígitos $3+5=8$ no número $3n+11$. Para alcançar a propriedade que buscamos, basta trocarmos o dígito inicial $1$ de $n$ por algum outro dígito que ao ser multiplicado por $3$ ainda mantenha a mesma soma de seus dígitos. Considere agora $n=9 \\cdot 10^{k}+8$. Assim, $3n+11=27 \\cdot 10^{k}+35$ e $s(n)=17=2+7+3+5=s(3n+11)$. Como $k$ pode ser qualquer inteiro positivo, existem infinitos números naturais da forma $9 \\cdot 10^{k}+8$ que são interessantes.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26895,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nExistem 2017 cadeiras não ocupadas em uma fila. A cada minuto, uma pessoa chega e se senta em uma delas que esteja vazia e, no mesmo instante, caso esteja ocupada, uma pessoa em uma cadeira vizinha se levanta e vai embora. Qual o número máximo de pessoas que podem estar simultaneamente sentadas na fileira de cadeiras?",
    "options": [],
    "answer": "2016",
    "solution": "Solution:\n\nNão é possível todas as cadeiras estarem simultaneamente ocupadas, pois o último a sentar inevitavelmente sentaria ao lado de uma cadeira ocupada e forçaria, de acordo com a regra do enunciado, alguém a ir embora. Nosso objetivo agora é mostrar uma sequência de movimentos onde é possível 2016 pessoas estarem sentadas simultaneamente. Numere as cadeiras com os números de 1 até 2017. Inicialmente uma pessoa sentará na cadeira de número 1. Em seguida, outra pessoa sentará na cadeira de número 3. Nenhuma pessoa irá embora, pois as cadeiras 1 e 3 não são vizinhas. A próxima pessoa deverá sentar na cadeira de número 2 e assim a pessoa na cadeira de número 3 irá embora. Uma nova pessoa sentará na cadeira de número $4$ e em seguida, outra pessoa chegará e sentará na cadeira de número 3. Agora a cadeira de número 4 é desocupada, mas as três primeiras cadeiras estão com pessoas. Suponha que todas as cadeiras de números de 1 até $k$, com $k<2016$, estão com pessoas e as cadeiras restantes estão vazias. O próximo a sentar escolherá a cadeira $k+2$. Em seguida, outra pessoa sentará na cadeira $k+1$ e obrigará a pessoa na cadeira $k+2$ a ir embora. Assim, as cadeiras de 1 até $k+1$ estão ocupadas e as cadeiras restantes estão vazias. Podemos repetir esse procedimento enquanto $k<2016$ e quando ele for executado pela última vez teremos as cadeiras de 1 até 2016 ocupadas.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26896,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nConsidere um triângulo acutângulo $A B C$ e quadrados $A B Z X$ e $A C W Y$, construídos externamente sobre os seus lados. Os pontos $M$ e $P$ são os pontos médios dos segmentos $B C$ e $X Y$, respectivamente; os pontos $Q$ e $N$ são os centros dos quadrados $A B Z X$ e $A C W Y$, respectivamente; e os pontos $R$ e $S$ são pontos médios dos lados $A B$ e $A C$, respectivamente.\n![](attached_image_1.png)\nVamos provar que $M N P Q$ é um quadrado em alguns passos.\n\na) Verifique que os triângulos $Q R M$ e $M S N$ são congruentes.\n\nb) Verifique que $\\angle Q M N=90^\\{\\circ\\}$.\n\nc) Sabendo dos resultados anteriores, mostre que $M N P Q$ é um quadrado.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na) Uma base média de um triângulo é um segmento ligando dois pontos médios de seus lados. Esse segmento é paralelo ao lado oposto e possui metade do seu comprimento. Sabendo isso, temos $R M=\\frac{A C}{2}, S M=\\frac{A B}{2}$ e $\\angle B R M=\\angle B A C=\\angle M S C$. Além disso, como $Q$ e $N$ são centros dos quadrados, podemos afirmar que $Q R=\\frac{A B}{2}, N S=\\frac{A C}{2}$ e $\\angle Q R B=\\angle N S C=90^{\\circ}$. Dessa forma, os triângulos $Q R M$ e $M S N$ são congruentes pelo caso $L A L$, pois $Q R=M S, \\angle Q R M=\\angle M S N=90^{\\circ}+\\angle B A C$ e $R M=S N$.\n\n\nb) Como os triângulos $Q R M$ e $M S N$ são congruentes, $M S \\| A B$ e $\\angle Q R B=90^{\\circ}$, então\n$$\n\\begin{aligned}\n\\angle Q M N & =\\angle Q M R+\\angle R M S+\\angle S M N \\\\\n& =\\angle Q M R+\\angle B R M+\\angle M Q R \\\\\n& =\\angle Q M R+\\angle Q R M+\\angle M Q R-90^{\\circ} \\\\\n& =180^{\\circ}-90^{\\circ} \\\\\n& =90^{\\circ}\n\\end{aligned}\n$$\n\n\nc) Com as informações dos itens anteriores, sabemos que o triângulo $Q M N$ é um triângulo retângulo isósceles com $Q M=M N$. Daí, $\\angle M Q N=\\angle Q N M=45^{\\circ}$. De modo semelhante, podemos provar também que o triângulo $Q P N$ também é um triângulo retângulo isósceles com $\\angle Q P N=90^{\\circ} \\mathrm{e} Q P=P N$, pois basta repetir o argumento anterior trocando o papel do triângulo $A B C$ pelo do triângulo $X A Y$. Dado que os triângulos retângulos isósceles $Q N M$ e $P Q N$ possuem a mesma hipotenusa, podemos concluir que eles são congruentes e assim $P N M Q$ é um retângulo com todos os lados iguais, ou seja, um quadrado.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26897,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\na) Considere o tabuleiro $4 \\times 4$ da figura a seguir. As letras e os números foram colocados para ajudar a localizar os quadradinhos. Por exemplo, o quadradinho $A 1$ é o do canto superior esquerdo e o $D 4$ é o do canto inferior direito.\n![](attached_image_1.png)\nMostre uma maneira de escolhermos 12 quadradinhos de modo que em cada linha e em cada coluna sejam escolhidos exatamente 3 quadradinhos.\nb) Determine o número de maneiras diferentes de escolhermos 12 quadradinhos no tabuleiro $4 \\times 4$ de modo que em cada linha e em cada coluna sejam escolhidos exatamente 3 quadradinhos.\nc) Considere um tabuleiro $8 \\times 8$ com uma identificação de linhas e colunas similar ao tabuleiro anterior, mas colorido alternadamente de preto e branco.\n![](attached_image_2.png)\nDetermine o número de maneiras de escolhermos 56 quadradinhos do tabuleiro de modo que todos os quadradinhos pretos sejam escolhidos e exatamente 7 quadradinhos sejam escolhidos em cada linha e em cada coluna.",
    "options": [],
    "answer": "24 ways on the 4×4 board; 576 ways on the 8×8 board.",
    "solution": "Solution:\na) Existem várias maneiras de fazer isso (veremos exatamente quantas no próximo item). No exemplo da figura da esquerda abaixo, apenas os quadradinhos $A 1, B 2, C 3$ e $D 4$ não são escolhidos.\n![](attached_image_3.png)\n\nb) Vamos usar o princípio fundamental da contagem. Na primeira linha, temos 4 maneiras de selecionar o quadradinho que não será escolhido. Para cada uma dessas maneiras, teremos 3 maneiras de selecionar o quadradinho na segunda linha. Em seguida, 2 maneiras de selecionar na terceira e apenas uma maneira de selecionar na quarta linha. Com isso, teremos $4 \\cdot 3 \\cdot 2 \\cdot 1=4!=24$ maneiras de escolhermos os 12 quadradinhos.\n\nc) Se considerarmos que todos os quadradinhos pretos são escolhidos, resta escolher exatamente 3 quadradinhos brancos em cada fileira (linha ou coluna). Observe que o número de maneiras de fazer isto é igual ao número de maneiras de escolher 3 quadradinhos em cada fileira dos tabuleiros $4 \\times 4$, indicados à direita e ao centro na figura anterior, de modo que em cada fileira exatamente 3 quadradinhos sejam escolhidos. Como as escolhas em cada tabuleiro são independentes, para cada escolha de 12 quadradinhos do tabuleiro do centro e cada escolha de 12 quadradinhos do tabuleiro da direita, atendendo todas as condições, temos uma escolha de 24 quadradinhos brancos do tabuleiro $8 \\times 8$. Juntamente com os 32 pretos, eles formam os 56 que devem ser escolhidos. Dessa forma, o número de maneiras de escolher os 56 quadradinhos satisfazendo as condições do problema é $24 \\cdot 24=576$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26898,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nConsidere três pontos colineares $B$, $C$ e $D$ de modo que $C$ está entre $B$ e $D$. Seja $A$ um ponto que não pertence à reta $BD$ de modo que $AB = AC = CD$.\n\n![](attached_image_1.png)\n\na. Se $\\angle BAC = 36^{\\circ}$, então verifique que\n$$\n\\frac{1}{CD} - \\frac{1}{BD} = \\frac{1}{CD + BD}\n$$\n\nb. Suponha agora que vale\n$$\n\\frac{1}{CD} - \\frac{1}{BD} = \\frac{1}{CD + BD}\n$$\nVerifique que $\\angle BAC = 36^{\\circ}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n(a) Como $\\angle BAC = 36^{\\circ}$ e o triângulo $ABC$ é isósceles, temos $\\angle ABC = \\angle ACB = 72^{\\circ}$. Além disso, como $\\angle ACD = 108^{\\circ}$ e o triângulo $ACD$ também é isósceles, segue que $\\angle CAD = \\angle CDA = 36^{\\circ}$.\n\n![](attached_image_2.png)\n\nPortanto, os triângulos $ABC$ e $DBA$ são semelhantes, pois possuem todos os ângulos iguais. Consequentemente $\\frac{BC}{AB} = \\frac{AB}{BD}$ e\n$$\n\\begin{aligned}\n\\frac{BC}{BA} & = \\frac{AB}{BD} \\\\\n\\frac{BC}{CD} & = \\frac{CD}{BD} \\\\\nBD \\cdot BC & = CD^2 \\\\\nBD(BD - CD) & = CD^2 \\\\\nBD^2 - BD \\cdot CD & = CD^2 \\\\\nBD^2 - CD^2 & = BD \\cdot CD \\\\\n(BD - CD)(BD + CD) & = BD \\cdot CD \\\\\n\\frac{BD - CD}{BD \\cdot CD} & = \\frac{1}{BD + CD} \\\\\n\\frac{1}{CD} - \\frac{1}{BD} & = \\frac{1}{CD + BD}\n\\end{aligned}\n$$\n\n\n(b)\n![](attached_image_3.png)\n\nSuponha agora que vale a equação dada. A partir dela e repetindo os passos da última sequência de equações, na ordem inversa, podemos concluir que $\\frac{BC}{BA} = \\frac{BA}{DB}$. Como $\\angle ABC = \\angle DBA$, segue que $ABD$ e $ABC$ são semelhantes. Daí, se $\\angle BAC = x$, temos $\\angle BDA = x$. Lembrando que $AC = CD$ temos o triângulo $ACD$ isósceles implicando $\\angle CAD = x$. Somando os ângulos internos do triângulo $ACD$ temos $\\angle ACD = 180^{\\circ} - 2x$. De $AC = CD$, decorre que $\\angle CAD = \\angle CDA = x$ e, consequentemente, $\\angle BCA = 2x$. Finalmente, como $AB = AC$, temos $\\angle ABC = 2x$ e\n$$\n\\begin{aligned}\n\\angle ABC + \\angle BCA + \\angle CAB & = 180^{\\circ} \\\\\n2x + 2x + x & = 180^{\\circ} \\\\\nx & = 36^{\\circ}.\n\\end{aligned}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26899,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA figura a seguir representa um triângulo $A B C$, dois quadrados $A B D E$ e $A C F G$, ambos construídos sobre os lados do triângulo $A B C$, e outro triângulo $A E G$.\n![](attached_image_1.png)\nDizemos que os triângulos $A B C$ e $A E G$, posicionados dessa forma em relação a dois quadrados, são triângulos irmãos.\n\na. Verifique que $\\angle B A C+\\angle G A E=180^\\circ$.\n\nb. Após verificar o item anterior, podemos rotacionar o triângulo $A E G$, por $90^\\circ$ no sentido anti-horário e em torno de $A$, para chegar na posição $B A G'$. Vale ressaltar que $B A=A E$, pois são lados do quadrado $A B D E$. A partir disso, prove que os triângulos irmãos $A B C$ e $A E G$ possuem mesma área.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na. Como a soma de todos os ângulos ao redor de um ponto é $360^{\\circ}$, temos\n$$\n\\begin{aligned}\n\\angle B A C+\\angle C A G+\\angle G A E+\\angle E A B & =360^{\\circ} \\\\\n\\angle B A C+90^{\\circ}+\\angle G A E+90^{\\circ} & =360^{\\circ} \\\\\n\\angle B A C+\\angle G A E & =180^{\\circ}\n\\end{aligned}\n$$\n\nb. Pelo item anterior, $C$, $A$ e $G'$ estão sobre uma mesma reta, pois $\\angle C A G'=180^{\\circ}$. Seja $h$ a distância do ponto $B$ em relação a essa reta. Então podemos calcular a área do triângulo $A B C$ e do triângulo $B A G'$ usando essa distância como altura:\n$$\n[A B C]=\\frac{A C \\cdot h}{2}=\\frac{A G' \\cdot h}{2}=\\left[B A G'\\right]\n$$\nComo o triângulo $B A G'$ é congruente ao triângulo $A E G$, concluímos que os triângulos irmãos $A B C$ e $A E G$ possuem mesma área.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26900,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nConsidere um triângulo acutângulo $A B C$.\n![](attached_image_1.png)\nExplique como construir um triângulo $D B C$ com mesma área que $A B C$, satisfazendo $D B=A B$, mas que não seja congruente ao triângulo $A B C$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nPrimeiro, vamos criar uma construção em que o novo triângulo possua a mesma área. Para isso trace a reta paralela a $B C$ por $A$, pois para qualquer ponto $P$ dessa reta, a altura relativa a $B C$ é a mesma. Calculando a área em relação a essa base $B C$ teremos que a área de $P B C$ igual à área de $A B C$. Agora, devemos garantir que $D B=A B$. Para isso, trace a circunferência de centro $B$ passando por $A$. Os pontos dessa circunferência são exatamente os pontos cuja distância em relação a $B$ é igual à $A B$.\n![](attached_image_2.png)\nComo o triângulo é acutângulo, essa circunferência encontra a reta paralela a $B C$ por $A$ em um segundo ponto $D$. Veja que os triângulos $D B C$ e $A B C$ não são congruentes, pois $\\angle D B C>90^\\circ$ e $A B C$ é acutângulo.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26901,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAs peças a seguir são chamadas de $L$-triminós.\n![](attached_image_1.png)\nEssas peças são usadas para cobrir completamente um tabuleiro $6 \\times 6$. Nessa cobertura, cada $L$-triminó cobre exatamente 3 quadradinhos do tabuleiro $6 \\times 6$ e nenhum quadradinho é coberto por mais de um $L$-triminó.\n\na) Quantos $L$-triminós são usados para cobrir um tabuleiro $6 \\times 6$ ?\n\nb) Em uma cobertura de todo o tabuleiro, dizemos que uma fileira (linha ou coluna) corta um $L$-triminó quando a fileira possui pelo menos um dos quadradinhos cobertos por esse $L$-triminó. Caso fosse possível obter uma cobertura do tabuleiro $6 \\times 6$ na qual cada fileira cortasse exatamente a mesma quantidade de L-triminós, quanto seria essa quantidade?\n\nc) Prove que não existe uma cobertura do tabuleiro $6 \\times 6$ com $L$-triminós na qual cada fileira corte a mesma quantidade de $L$-triminós.",
    "options": [],
    "answer": "a) 12; b) 4; c) impossible",
    "solution": "Solution:\n\na) Seja $x$ o número de $L$-triminós usados para cobrir um tabuleiro $6 \\times 6$. Como esse tabuleiro possui exatamente $6 \\cdot 6=36$ quadradinhos e cada um deve ser coberto por exatamente um dos $L$-triminós, então $3x=36$, ou seja, $x=12$.\n\n\nb) Seja $y$ a quantidade de $L$-triminós que cada fileira corta. Considere todos os pares $(F, L)$, onde $F$ denota uma das 12 fileiras (linhas ou colunas) e $L$ um dos $12\\ L$-triminós que é cortado pela fileira $F$. Por um lado, como cada fileira corta $y$ triminós, temos $12y$ pares do tipo $(F, L)$. Por outro lado, cada um dos $12\\ L$-triminós é cortado por exatamente 4 fileiras (duas linhas e duas colunas) e isso nos dá o total de $12 \\cdot 4=48$ pares do tipo $(F, L)$. Essa contagem deve ser a mesma nas duas situações e daí $12y=48$, ou seja, $y=4$.\n\n\nc) Suponha, por absurdo, que exista uma cobertura em que cada fileira corta exatamente a mesma quantidade de $L$-triminós. Pelo item anterior, sabemos que cada fileira deve cortar exatamente $4\\ L$-triminós. Quando uma fileira corta um $L$-triminó, eles possuem 1 ou 2 quadradinhos em comum. Tendo isso em mente, considere agora uma fileira em que dos $4\\ L$-triminós cortados por ela, $a$ possuem 1 quadradinho na fileira e $(4-a)$ possuem 2 quadradinhos. Como a fileira possui 6 quadradinhos, $a+2(4-a)=6$, ou seja, $a=2$ e $4-a=2$. Consequentemente, podemos concluir que em qualquer fileira existem $2\\ L$-triminós cortados em 1 quadradinho e $2\\ L$-triminós cortados com 2 quadradinhos.\n\n![](attached_image_2.png)\n\nConsidere agora a primeira linha do tabuleiro da figura anterior. Existem 2 $L$-triminós que cobrem, cada um, exatamente 1 quadradinho da primeira linha e, consequentemente, eles mesmos cobrem 2 quadradinhos da segunda linha. Analogamente, os $2\\ L$-triminós que cobrem exatamente dois quadradinhos da primeira linha cobrem, cada um, exatamente 1 quadradinho da segunda linha. Note que isso já cobre totalmente a primeira e a segunda linhas, implicando que nenhum $L$-triminó poderia cruzar a separação entre a segunda e a terceira linhas. Podemos repetir o raciocínio para terceira e quarta linhas e quinta e sexta linhas. Também podemos fazer isso em colunas com a primeira e a segunda colunas, a terceira e a quarta colunas e a quinta e a sexta colunas. Dessa forma, o tabuleiro $6 \\times 6$ fica dividido em 9 subtabuleiros $2 \\times 2$ por faixas que não podem ser cruzadas por $L$-triminós. Isso implica que cada subtabuleiro tem que ser coberto por $L$-triminós. Isso é impossível, pois o número de quadradinhos em cada subtabuleiro não é um múltiplo de 3. Concluímos assim que não é possível cobrir o tabuleiro $6 \\times 6$ com $L$-triminós de modo que cada fileira corte a mesma quantidade de $L$-triminós.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26902,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nQuantos divisores de $88^{10}$ deixam resto 4 quando divididos por 6?",
    "options": [],
    "answer": "165",
    "solution": "Solution:\nComo $88=2^{3} \\cdot 11$, temos $88^{10}=\\left(2^{3} \\cdot 11\\right)^{10}=2^{30} \\cdot 11^{10}$. Se um inteiro $x$ deixa resto 4 por 6, então $x+2$ é múltiplo de 6. Consequentemente $x$ é par e deixa resto 1 na divisão por 3. Perceba agora que todos os divisores primos de $88^{10}$ deixam resto 2 na divisão por 3 e que o produto de qualquer quantidade par deles deixa resto 1 por 3 enquanto o produto de qualquer quantidade ímpar deixa resto 2 por 3. Portanto, os divisores que queremos contar são aqueles que possuem pelo menos um fator 2, mas uma quantidade par de fatores primos. Assim, eles podem ser escritos na forma $2^{m} \\cdot 11^{n}$ com $m \\geq 1$ e $m+n$ par. Vamos dividir a contagem deles em duas partes:\n\n1. Quando $m$ é par, podemos escolhê-lo de 15 maneiras, pois ele deve pertencer ao conjunto $\\{2,4, \\ldots, 30\\}$. Uma vez que ele tenha sido escolhido, temos 6 opções de escolha de outro número par para $n$, a saber, os elementos de $\\{0,2,4,6,8,10\\}$.\n\n2. Quando $m$ é ímpar, podemos escolhê-lo também de 15 maneiras, pois ele deve pertencer ao conjunto $\\{1,3,5, \\ldots, 29\\}$. Uma vez que ele tenha sido escolhido, temos 5 opções de escolhas, a saber, os elementos de $\\{1,3,5,7,9\\}$.\n\nPortanto, a quantidade de divisores é $15 \\cdot 6+15 \\cdot 5=165$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26903,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nExistem 2017 pessoas em uma festa. Em um determinado momento, cada uma delas dá um presente para um outro convidado (é possível que um convidado receba mais de um presente). Mostre que podemos encontrar um grupo de 673 pessoas na festa de modo que quaisquer duas delas não trocaram presentes entre si.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nNossa estratégia será classificar os convidados da festa em grupos de três cores de modo que pessoas associadas a uma mesma cor não trocaram presentes entre si. Pelo menos uma dessas cores terá 673 pessoas, pois caso contrário, o total de pessoas será no máximo $672 \\cdot 3=2016<2017$. Bastará então escolher o grupo formado pelas 673 pessoas de uma mesma cor para satisfazer a condição do enunciado.\n\nPara começar a classificação de cores, escolha uma pessoa qualquer e a coloque em uma mesa circular. Do seu lado esquerdo, sente a pessoa que recebeu um presente dela. Em seguida, caso ainda não esteja sentada, sente a pessoa que recebeu um presente da que acabou de sentar e repita esse processo. Como o número de pessoas é finito, eventualmente o processo de convite para que novas pessoas se sente será encerrado e teremos encontrado um ciclo, como ilustrado na figura abaixo:\n\n![](attached_image_1.png)\n\nUma vez que um ciclo de pessoas tenha sido construído, escolha uma delas e a associe a cor 1. Em seguida, numere alternadamente as pessoas no ciclo com cores 1 e 2. Caso o ciclo tenha tamanho ímpar, a última cor associada deve ser a 3. Dessa forma, pessoas associadas a uma mesma cor não trocam presentes entre si.\n\nAlém disso, para as pessoas que não estão no ciclo, associe a cor 1 se ela deu um presente para alguém da cor 2 e associe a cor 2 se ela deu presente para alguém da cor 1 ou 3. Depois repita o processo sucessivamente seguindo essa mesma regra. Não existirá ambiguidade na atribuição de cores, pois cada pessoa só entrega um presente. Caso existam pessoas que não estejam associadas a algumas das três cores, podemos concluir que elas trocaram presentes entre si e não com as pessoas que já estão associadas às cores anteriores. Nesse caso, podemos repetir o argumento para esse grupo de pessoas que trocaram presentes entre si, construir um ciclo e, em seguida, realizar a atribuição de cores seguindo o mesmo padrão. Como a cada vez que executamos esse processo o número\n\n![](attached_image_2.png)\n\nde pessoas que não estão associadas a cores diminui e o número de pessoas na festa é finito, em algum momento todas elas estarão com alguma cor. Pela forma como foi feita a atribuição, pessoas de uma mesma cor não trocaram presente entre si e isso conclui o argumento descrito no início da solução.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26904,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCada um dos números $1,2,3, \\ldots, 25$ é arranjado em uma das casas de um tabuleiro $5 \\times 5$. Em cada linha, eles aparecem em ordem crescente, da esquerda para a direita. Encontre os valores máximo e mínimo possíveis para as somas dos números da terceira coluna.",
    "options": [],
    "answer": "minimum 45, maximum 85",
    "solution": "Solution:\n\nPodemos numerar as linhas e colunas do tabuleiro com os números de 1 a 5, de cima para baixo e da esquerda para a direita. Além disso, podemos denotar o número escrito na linha de número $i$ e na coluna de número $j$ por $a_{ij}$, como indicado na figura abaixo:\n\n| $a_{11}$ | $a_{12}$ | $a_{13}$ | $a_{14}$ | $a_{15}$ |\n| :--- | :--- | :--- | :--- | :--- |\n| $a_{21}$ | $a_{22}$ | $a_{23}$ | $a_{24}$ | $a_{25}$ |\n| $a_{31}$ | $a_{32}$ | $a_{33}$ | $a_{34}$ | $a_{35}$ |\n| $a_{41}$ | $a_{42}$ | $a_{43}$ | $a_{44}$ | $a_{45}$ |\n| $a_{51}$ | $a_{52}$ | $a_{53}$ | $a_{54}$ | $a_{55}$ |\n\nEm cada linha $i$, sabemos que\n$$\na_{i1} < a_{i2} < a_{i3} < a_{i4} < a_{i5}\n$$\n\nPerceba que se trocarmos as posições de duas linhas entre si, a propriedade anterior continua sendo verdadeira e a soma dos elementos de cada coluna continua a mesma. Assim, realizando trocas entre as linhas até que os elementos da terceira coluna estejam ordenados de forma crescente, de cima para baixo, o problema se resume a calcular as somas máxima e mínima dos elementos da terceira coluna para essa configuração do tabuleiro.\n\nPara estimar a soma máxima, perceba que $a_{53}$ é no máximo 23, pois $a_{53} < a_{54} < a_{55}$. Existem pelo menos 5 inteiros no tabuleiro maiores que $a_{43}$, a saber, $a_{44}$, $a_{45}$, $a_{53}$, $a_{54}$ e $a_{55}$. Portanto, o valor máximo de $a_{43}$ é 20. Como $a_{33} < a_{34} < a_{35}$, $a_{33} < a_{43}$ e existem pelo menos 5 inteiros maiores que $a_{43}$, então existem pelo menos 8 inteiros no tabuleiro maiores que $a_{33}$ e daí o seu valor é no máximo 17.\n\nRepetiremos esse argumento mais duas vezes, usando que:\n$$\n\\begin{aligned}\n& a_{23} < a_{24} < a_{25} \\\\\n& a_{23} < a_{33}\n\\end{aligned}\n$$\ne\n$$\n\\begin{aligned}\n& a_{13} < a_{14} < a_{15} \\\\\n& a_{13} < a_{23}\n\\end{aligned}\n$$\npara concluir que existem pelo menos 11 inteiros maiores que $a_{23}$ e 14 inteiros maiores que $a_{13}$. Logo, $a_{23}$ é no máximo 14 e $a_{13}$ é no máximo 11. Isso mostra que a soma dos elementos da terceira coluna é no máximo\n$$\n23 + 20 + 17 + 14 + 11 = 85\n$$\n\nResta mostrarmos que realmente essa estimativa pode ser obtida. Basta exibirmos uma configuração do tabuleiro satisfazendo as regras mencionadas, como mostra o tabuleiro da esquerda da próxima figura.\n\n| 1 | 6 | 11 | 12 | 13 |\n| :---: | :---: | :---: | :---: | :---: |\n| 2 | 7 | 14 | 15 | 16 |\n| 3 | 8 | 17 | 18 | 19 |\n| 4 | 9 | 20 | 21 | 22 |\n| 5 | 10 | 23 | 24 | 25 |\n\nPara estimar a soma mínima, perceba que pelo menos dois elementos, a saber, $a_{11}$ e $a_{12}$ são menores que $a_{13}$. Portanto, $a_{13}$ é no mínimo 3. Em seguida, repetindo a estratégia anterior, temos pelo menos 5, 8, 11 e 14 elementos no tabuleiro que são menores que $a_{23}$, $a_{33}$, $a_{43}$ e $a_{53}$, respectivamente. Assim, a soma mínima dos elementos da terceira coluna é\n$$\n3 + 6 + 9 + 12 + 15 = 45\n$$\n\nPara mostrar que essa soma é realmente possível, basta ver o tabuleiro da direita da última figura.\n\n| 1 | 2 | 3 | 16 | 21 |\n| :---: | :---: | :---: | :---: | :---: |\n| 4 | 5 | 6 | 17 | 22 |\n| 7 | 8 | 9 | 18 | 23 |\n| 10 | 11 | 12 | 19 | 24 |\n| 13 | 14 | 15 | 20 | 25 |",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26905,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm Quadrado Latino é um tabuleiro $n \\times n$ preenchido com $n$ símbolos distintos de modo que em cada linha e em cada coluna não existam símbolos repetidos. Por exemplo, a figura abaixo mostra um exemplo de um Quadrado Latino de dimensões $3 \\times 3$. O nome foi inspirado em trabalhos do matemático Leonhard Euler, que usou caracteres latinos como símbolos.\n\n| 1 | 2 | 3 |\n| :--- | :--- | :--- |\n| 3 | 1 | 2 |\n| 2 | 3 | 1 |\n\nSabemos que existem 576 Quadrados Latinos distintos de dimensões $4 \\times 4$. De quantos modos podemos completar o quadrado abaixo, que já possui duas casas preenchidas, com os algarismos 1,2,3 e 4 de modo que em cada linha e coluna figurem os quatro algarismos?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "48",
    "solution": "Solution:\n\nObserve que dado um Quadrado Latino, quando trocamos todas as casas de um símbolo pelas casas de outro, ainda obtemos outro Quadrado Latino. No exemplo dado no enunciado, ao trocarmos as casas de número 1 e 2 de posição, obtemos:\n\n| 2 | 1 | 3 |\n| :--- | :--- | :--- |\n| 3 | 2 | 1 |\n| 1 | 3 | 2 |\n\nIsso nos permite construir uma correspondência biunívoca entre todos os Quadrados Latinos que possuem no canto superior esquerdo um símbolo em $\\{1,2,3,4\\}$ e todos os outros Quadrados Latinos com outro símbolo no mesmo conjunto. Daí, em $1/4$ do total de Quadrados Latinos $4 \\times 4$ deve possuir o algarismo 1 na casa do canto superior esquerdo. Dentre esses quadrados, qualquer permutação entre os símbolos de $\\{2,3,4\\}$ ainda irá gerar um quadrado onde a casa do canto superior esquerdo tem o símbolo 1.\n\n| 1 | $X$ | $Y$ | $Z$ |\n| :--- | :--- | :--- | :--- |\n|  |  |  |  |\n|  |  |  |  |\n|  |  |  |  |\n\nPortanto, em um terço dessas configurações o algarismo 2 se encontra na posição $X$, em outro terço na posição $Y$ e no último terço na posição $Z$ dos quadrados da primeira linha da figura anterior. Logo, o total de Quadrados Latinos procurados é\n\n$$\n\\frac{1}{4} \\cdot \\frac{1}{3} \\cdot 576 = 48\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26906,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEm uma festa, existem 25 crianças. Sabemos que quaisquer duas delas possuem pelo menos um de seus avós em comum (avô ou avó). Explique por que pelo menos 17 crianças possuem ou um mesmo avô ou uma mesma avó nessa família.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nPara cada criança, associe duas etiquetas, cada uma com o nome de um de seus avós. Se uma pessoa é o avô ou a avó de todas as crianças, então claramente a afirmação do enunciado é verdadeira. Tratemos então do caso em que nem todas as crianças são netas de uma mesma pessoa. Suponha que uma criança tenha as etiquetas das pessoas $A$ e $B$. Como nem todas as crianças são netas de uma mesma pessoa, deve existir uma criança que não tem a etiqueta $A$. Como essa criança deve possuir um avô ou avó em comum com a criança anteriormente mencionada, ele deve possuir etiquetas $B$ e $C$. Considere agora uma criança que não possua a etiqueta $B$ e a compare com as duas crianças anteriores. Para que ela tenha um avô ou avó em comum com as duas crianças, ela deve possuir as etiquetas $A$ e $C$. Uma criança que não possua apenas as etiquetas dos tipos $A, B$ ou $C$ não terá etiqueta em comum com alguma das três crianças anteriormente mencionadas e isso nos leva a concluir que todas as crianças são netas apenas das pessoas $A, B$ ou $C$. Foram usadas $2 \\cdot 25=50$ etiquetas. Se nenhuma dessas pessoas aparece mais que 16 vezes, teremos no máximo $3 \\cdot 16=48<50$ e isso é um absurdo. Portanto, alguma etiqueta é usada pelo menos 17 vezes e isso significa que uma pessoa é avô ou avó de pelo menos 17 crianças.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26907,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSeja $A$ um conjunto infinito de inteiros positivos. Sabe-se que se tomarmos qualquer subconjunto finito $B$ do conjunto $A$ existe um inteiro positivo $b$ maior que 1 tal que $b$ divide todos os elementos do conjunto $B$. Prove que existe um inteiro positivo $d$ maior que 1 que divide todos os elementos do conjunto $A$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nConsidere um elemento $a$ do conjunto $A$ e sejam $p_{1}, p_{2}, \\ldots, p_{k-1}, p_{k}$ os divisores primos de $a$. Se algum primo $p_{i}$ divide todos os elementos do conjunto $A$, então $d = p_{i}$ satisfaz o enunciado. Vamos provar que isso obrigatoriamente acontece. Suponha, por absurdo, que para cada primo $p_{i}$ podemos encontrar um elemento $a_{i}$ do conjunto $A$ que não é divisível por $p_{i}$. Veja que não necessariamente todos os elementos $a_{i}$ são distintos. Considere agora o conjunto $B$ formado por $a$ e pelos inteiros $a_{i}$, para todo $i \\in \\{1, 2, \\ldots, k\\}$. Pela condição do enunciado, existe $b > 1$ que divide todos os elementos de $B$. Como $b$ divide $a$, então ele possui algum dos fatores primos de $a$ em sua fatoração, digamos $p_{x}$. Dado que $a_{x} \\in B$, então $b$ divide $a_{x}$ e, consequentemente, $p_{x}$ deve ser um divisor de $a_{x}$. Esse absurdo mostra que a afirmação inicial feita sobre os primos $p_{i}$ é falsa. Logo, pelo menos um deles deve dividir todos os elementos do conjunto $A$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26908,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nJoão escreveu todas as potências de $2$, $3$ e $5$ maiores que $1$ e menores que $2017$ em uma folha de papel. Em seguida, ele realizou todos os produtos possíveis de dois números distintos dessa folha e os escreveu em outra folha de papel. Qual a quantidade de inteiros que João registrou na segunda folha?",
    "options": [],
    "answer": "155",
    "solution": "Solution:\n\nInicialmente, devemos encontrar as potências de $2$, $3$ e $5$ registradas na primeira folha. Como $2^{10}<2017<2^{11}$, $3^{6}<2017<3^{7}$ e $5^{4}<2017<5^{5}$, as potências escritas na primeira folha podem ser divididas em três conjuntos:\n$$\nP_{2} = \\{2^{1}, 2^{2}, \\ldots, 2^{10}\\}, \\quad P_{3} = \\{3^{1}, 3^{2}, \\ldots, 3^{6}\\} \\text{ e } P_{5} = \\{5^{1}, 5^{2}, 5^{3}, 5^{4}\\}\n$$\nEm virtude da fatoração única em números primos, os produtos obtidos pela multiplicação de potências de conjuntos distintos são também distintos. Assim, pelo Princípio Multiplicativo, existem $10 \\cdot 6 + 10 \\cdot 4 + 6 \\cdot 4 = 124$ produtos de potências distintas.\n\nResta agora contarmos quantos produtos existem entre potências de mesma base. Dado um número primo $q$ e o conjunto $P_{q} = \\{q^{1}, q^{2}, \\ldots, q^{k}\\}$, o menor produto de potências distintas é $q^{1} \\cdot q^{2} = q^{3}$ e o maior é $q^{k-1} \\cdot q^{k} = q^{2k-1}$. Verificaremos agora que todas as potências $q^{t}$ com expoente $t$ entre $3$ e $2k-1$ podem ser obtidas como produto de dois números desse conjunto.\n\nSe $t$ é par, podemos escrever $t = 2m$ e dado que $3 < t < 2k-1$, temos $1 < m < k-1$. Daí, basta multiplicar as potências $q^{m-1}$ e $q^{m+1}$, que fazem parte do conjunto $P_{q}$, para obtermos $q^{m-1} \\cdot q^{m+1} = q^{2m} = q^{t}$.\n\nSe $t$ é ímpar, podemos escrever $t = 2m+1$ e dado que $3 < t < 2k-1$, temos $0 < m < k-1$. Basta então multiplicar $q^{m}$ e $q^{m+1}$, que também fazem parte de $P_{q}$, para obtermos $q^{m} \\cdot q^{m+1} = q^{2m+1} = q^{t}$.\n\nIsso mostra que existem exatamente $2k-3$ produtos de potências distintas obtidas pela multiplicação de dois elementos de $P_{q}$.\n\nAplicando essa contagem com $q=2,3$ e $5$, podemos concluir que existem mais $17 + 9 + 5 = 31$ potências distintas de mesma base na segunda folha.\n\nLogo, o total de números da segunda folha é $124 + 31 = 155$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26909,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO conteúdo multiplicativo de um conjunto é o produto de seus elementos. Caso o conjunto possua um único elemento, seu conteúdo multiplicativo é este único elemento e, caso o conjunto seja vazio, seu conteúdo multiplicativo é $1$. Por exemplo, o conteúdo multiplicativo de $\\{1,2,3\\}$ é $1 \\cdot 2 \\cdot 3=6$.\n\na) Determine a soma dos conteúdos multiplicativos de todos os subconjuntos de $\\{1,2,3,4\\}$.\n\nb) Determine a soma dos conteúdos multiplicativos de todos os subconjuntos de\n$$\n\\left\\{1, \\frac{1}{2}, \\frac{1}{3}, \\ldots, \\frac{1}{2016}\\right\\}\n$$",
    "options": [],
    "answer": "a) 120; b) 2017",
    "solution": "Solution:\n\na) Observe que no produto $(1+a)(1+b)$, ao aplicarmos a propriedade distributiva da multiplicação dos números reais, aparecerão os produtos $1 \\cdot 1, 1 \\cdot b, a \\cdot 1$ e $a \\cdot b$. Esses produtos coincidem com os conteúdos multiplicativos dos conjuntos $\\varnothing, \\{b\\}, \\{a\\}$ e $\\{a, b\\}$, respectivamente. Esses são exatamente os subconjuntos de $\\{a, b\\}$. Na multiplicação $(1+a)(1+b)(1+c)$, novamente ao usarmos a propriedade distributiva da multiplicação, aparecerão $2^{3}=8$ termos. Cada um desses termos está associado a elementos dentro de cada parênteses que deverão ser escolhidos para figurar na parcela correspondente. Por exemplo, o termo $1 \\cdot b \\cdot c = b \\cdot c$ corresponde a não escolhermos o $a$ no primeiro parênteses e, em seguida, escolhermos tanto $b$ quanto $c$ nos próximos dois parênteses. Os subconjuntos de $\\{a, b, c\\}$ estão associados a todas essas 8 parcelas, até mesmo o conjunto vazio, que tem como conteúdo multiplicativo o número $1$ e é representado na soma pela parcela $1 \\cdot 1 \\cdot 1$. O mesmo se passa com qualquer quantidade de termos e assim a soma de todos os conteúdos multiplicativos do conjunto $\\{1,2,3,4\\}$ é o resultado $(1+1)(1+2)(1+3)(1+4)=120$.\n\nb) O argumento utilizado no item anterior também se aplica aqui e, consequentemente, o conteúdo multiplicativo procurado é\n$$\n\\begin{aligned}\n\\left(1+\\frac{1}{1}\\right)\\left(1+\\frac{1}{2}\\right)\\left(1+\\frac{1}{3}\\right) \\ldots\\left(1+\\frac{1}{2016}\\right) &= \\\\\n\\left(\\frac{2}{1}\\right)\\left(\\frac{3}{2}\\right)\\left(\\frac{4}{3}\\right) \\ldots\\left(\\frac{2017}{2016}\\right) &= 2017\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26910,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDetermine os números primos $p$ tais que a representação decimal da fração $\\frac{1}{p}$ tenha período de tamanho 5.\n\nObservação: Se a representação decimal de um número possui uma sequência de dígitos que se repete de forma periódica, o tamanho da menor sequência de dígitos que se repete é o tamanho da representação decimal. Por exemplo, $61 / 495=0,1232323 \\ldots$, apesar de 23, 2323 serem sequências de dígitos que se repetem na representação decimal, o tamanho da menor sequência é 2 e este é o tamanho do período.",
    "options": [],
    "answer": "41, 271",
    "solution": "Solution:\nSejam $q$ o tamanho da parte não periódica e $\\overline{abcde}$ o período da representação decimal de $1/p$. Assim, se representarmos os dígitos da parte não periódica por meio do símbolo $\\star$, temos\n$$\n\\begin{aligned}\n\\frac{1}{p} & = 0, \\star \\star \\ldots \\star \\text{ abcdeabcde } \\ldots \\\\\n\\frac{10^{q}}{p} & = \\star \\star \\ldots \\star, \\text{ abcdeabcde } \\ldots \\\\\n\\frac{10^{q}}{p} & = M + \\frac{\\text{abcde}}{10^{5}-1}\n\\end{aligned}\n$$\nNa última equação, representamos por $M$ o inteiro à esquerda da vírgula da representação decimal de $10^{q}/p$ ou, se preferir, a parte inteira do número em questão. Multiplicando a equação anterior por $\\left(10^{5}-1\\right)p$, obtemos\n$$\n\\begin{aligned}\n& 10^{q}\\left(10^{5}-1\\right) = \\left(10^{5}-1\\right) \\cdot p \\cdot M + \\text{abcde} \\cdot p \\\\\n& 10^{q}\\left(10^{5}-1\\right) = \\left[\\left(10^{5}-1\\right) \\cdot M + \\text{abcde}\\right] \\cdot p\n\\end{aligned}\n$$\nconsequentemente, $p \\mid 10^{q}\\left(10^{5}-1\\right)$. Os únicos divisores primos de $10^{q}$ são 2 e 5 e claramente nenhum deles produz uma dízima periódica de período 5. Resta analisarmos os fatores primos de $10^{5}-1 = 9 \\cdot 11111 = 3^{2} \\cdot 41 \\cdot 271$. Dentre os três números primos que aparecem na fatoração anterior, podemos checar que 41 e 271 são os únicos que produzem uma dízima periódica de período 5.\n\nObservação: Se $p$ é um primo diferente de 2 e 5 e $k$ é o menor expoente positivo tal que $p$ divide $10^{k}-1$, é possível mostrarmos que a expansão decimal de $1/p$ possui período de tamanho $k$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26911,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm quadrilátero é dito cíclico quando seus quatro vértices estão sobre uma mesma circunferência. Considere um quadrilátero $ABCD$ cíclico com diagonais $AC$ e $BD$ perpendiculares. Além disso, sejam $O$ o centro da circunferência que passa pelos vértices do quadrilátero e $P$ o ponto de encontro das diagonais.\n\n![](attached_image_1.png)\n\na) A partir do ponto $P$ traçamos uma reta $r$ perpendicular a $BC$. A reta $r$ corta $BC$ em $X$ e $AD$ em $M$. Verifique que $M$ é o ponto médio do lado $AD$.\n\nb) A partir do ponto $P$ trace a reta $s$ perpendicular a $AD$ cortando $AD$ no ponto $Y$ e $BC$ no ponto $N$. Verifique que o quadrilátero $OMPN$ é um paralelogramo.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na) Seja $\\angle BCA=\\theta$. Veja que $\\angle BCA=\\angle BDA=\\theta$, pois eles estão inscritos no mesmo arco $AB$. Considerando a soma dos ângulos do triângulo $CXP$, temos $\\angle CPX=90^{\\circ}-\\theta$ e, consequentemente, $\\angle BPX=\\angle BPC-\\angle CPX=\\theta$. Vale que $\\angle APM=\\angle XPC=90^{\\circ}-\\theta$, pois eles são ângulos opostos pelo vértice $P$, e $\\angle MPD=\\angle BPX=\\theta$ pelo mesmo motivo. Analisando agora a soma dos ângulos internos do triângulo $PAD$, temos $\\angle PAD+\\angle APD+\\angle PDA=180^{\\circ}$. Assim, como $\\angle APD=90^{\\circ}$, temos $\\angle PAD=90^{\\circ}-\\theta$. Daí, $\\angle MAP=\\angle APM$ e $\\angle MPD=\\angle MDP$, ou seja, $APM$ e $MDP$ são triângulos isósceles com $MA=MP=MD$. Concluímos então que $M$ é o ponto médio de $AD$.\n\nb) De modo semelhante ao item anterior, podemos concluir que $N$ é o ponto médio de $BC$. Como $O$ é equidistante dos extremos dos vértices $A, D, B$ e $C$, $OM$ e $ON$ são mediatrizes de $AD$ e $BC$, respectivamente. Decorre do item anterior que $OM \\parallel PN$ e $ON \\parallel PM$. Concluímos assim que $OMPN$ é um paralelogramo, pois seus lados opostos são paralelos.\n\n![](attached_image_2.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26912,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nExistem $2m$ miçangas de $m$ cores distintas, sendo duas de cada cor. Essas miçangas são distribuídas em $m$ caixas, com duas em cada caixa, de modo que é possível escolher uma miçanga em cada uma delas e obter um conjunto de $m$ miçangas de cores distintas. Prove que o número de maneiras de fazermos esse tipo de escolha é necessariamente uma potência de 2.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nAntes de descrevermos a solução para o caso geral, considere o caso particular com $m=5$, em que as cores das miçangas serão denotadas por $A, B, C, D$ e $E$, com as seguintes distribuições de caixas:\n$$\n\\begin{array}{ccccc}\nC_{1} & C_{2} & C_{3} & C_{4} & C_{5} \\\\\n\\hline A, B & B, C & C, A & D, E & D, E .\n\\end{array}\n$$\nContemos agora o número de escolhas de miçangas, uma de cada caixa, de modo que todas as escolhidas sejam de cores distintas. Analisando as duas últimas caixas, se escolhermos uma cor em $C_{4}$, somos forçados a escolher a outra cor em $C_{5}$. Essas escolhas não interferem nas escolhas das cores das primeiras caixas. Se escolhermos $A$ em $C_{1}$, somos obrigados a escolher $B$ em $C_{2}$ e $C$ em $C_{3}$. Por outro lado, se escolhermos $B$ em $C_{1}$, temos que escolher $C$ em $C_{2}$ e $A$ em $C_{3}$. Ou seja, existem 2 escolhas para as cores retiradas das três primeiras caixas e 2 para as das duas últimas. Essas $2 \\cdot 2=4$ escolhas totalizam o número de maneiras de escolhermos 5 miçangas distintas nesse exemplo particular.\n\nPara o caso geral, considere uma caixa genérica, que chamaremos de $C_{1}$ e denote por $m_{1}$ a cor de uma de suas miçangas. Em seguida, escolha uma outra caixa, que chamaremos de $C_{2}$, contendo uma miçanga de uma cor, que chamaremos de $m_{2}$, presente em $C_{1}$, mas distinta da cor $m_{1}$. Escolha agora um caixa $C_{3}$ contendo uma miçanga de cor $m_{3}$ presente em $C_{2}$, mas diferente de $m_{2}$. Continue esse processo até, eventualmente, obtermos para algum $l$ que $C_{l}=C_{1}$. Isso se traduz na seguinte distribuição:\n\n| $C_{1}$ | $C_{2}$ | $C_{3}$ | $\\ldots$ | $C_{l-1}$ |\n| :---: | :---: | :---: | :---: | :---: |\n| $m_{1}, m_{2}$ | $m_{2}, m_{3}$ | $m_{3}, m_{4}$ | $\\ldots$ | $m_{l-1}, m_{1}$. |\n\nPerceba que qualquer uma das duas escolhas possíveis das cores escolhidas da caixa $C_{1}$ determina unicamente quais miçangas devem ser escolhidas nas outras caixas do ciclo anterior para que as cores permaneçam distintas. Se ainda existirem caixas fora deste ciclo, podemos repetir o processo descrito anteriormente e agrupá-las em ciclos. Para cada um deles, existirão apenas duas maneiras de selecionarmos suas miçangas de modo que as cores sejam todas distintas. Como as escolhas nesses ciclos são independentes, se $k$ é o número deles, o total de escolhas de miçangas de cores diferentes é $2^{k}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26913,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUma partição do Conjunto dos Números Naturais é uma coleção de conjuntos $A_{1}, A_{2}, \\ldots, A_{k}$ de modo que cada número natural pertença a exatamente um deles. Veja que em qualquer partição do Conjunto dos Números Naturais pelo menos um desses conjuntos é infinito, pois caso contrário o Conjunto dos Números Naturais seria a união de uma quantidade finita de conjuntos finitos e seria, portanto, finito. Um exemplo de partição do Conjunto dos Números Naturais é considerar como $A_{1}$ o conjunto de todos os números naturais pares e como $A_{2}$ o conjunto de todos os números naturais ímpares. Existem várias partições possíveis e os próximos dois itens são fatos gerais que podem ser verificados em qualquer uma dessas partições.\n\na) Explique por que, para cada inteiro positivo $x$ fixado, existe sempre algum dos conjuntos $A_{i}$ com infinitos múltiplos de $x$.\n\nb) Pelo item anterior, dados dois inteiros positivos $p$ e $q$, existe um dos conjuntos da partição com infinitos múltiplos de $p$ e outro conjunto que contém infinitos múltiplos de $q$. Entretanto, esses dois conjuntos não precisam ser necessariamente iguais. Mostre agora que sempre algum desses conjuntos $A_{i}$ possui infinitos múltiplos de qualquer inteiro positivo.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na) Veja que $x$ possui infinitos múltiplos no conjunto dos números naturais que estarão divididos entre os conjuntos da partição. Se cada conjunto tivesse apenas uma quantidade finita de múltiplos de $x$, então o número total de múltiplos de $x$ entre os naturais, por ser uma união desses conjuntos, seria finito. Isso é uma contradição. Logo, podemos afirmar que algum dos conjuntos possui infinitos múltiplos de $x$.\n\n\nb) Novamente faremos uma demonstração por absurdo. Suponha que não existe nenhum conjunto que satisfaça a condição, ou seja, para cada conjunto $A_{i}$ existe pelo menos um inteiro positivo $n_{i}$ que não possui infinitos múltiplos em $A_{i}$. Considere o número $n=n_{1} n_{2} \\ldots n_{k}$, que é o produto de todos os números naturais $n_{i}$. Veja que $n$ é múltiplo de cada $n_{i}$ e isso implica que todo múltiplo de $n$ é múltiplo deles. Pelo item anterior, algum dos conjuntos, digamos $A_{j}$, deve possuir infinitos múltiplos de $n$. Daí, $A_{j}$ teria infinitos múltiplos de $n_{j}$ e isso contradiz nossa suposição inicial. Esse absurdo mostra que pelo menos um dos conjuntos possui infinitos múltiplos de qualquer inteiro positivo.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26914,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nQuantos são os pares ordenados $(a, b)$, com $a$ e $b$ inteiros positivos, tais que\n$$\na + b + mdc(a, b) = 33?\n$$",
    "options": [],
    "answer": "21",
    "solution": "Solution:\nSeja $d = mdc(a, b)$. Podemos reescrever a equação como:\n$$\n\\frac{a}{d} + \\frac{b}{d} + 1 = \\frac{33}{d}\n$$\nComo o lado esquerdo é uma soma de números inteiros, segue que $d$ divide $33$. Além disso,\n$$\n\\begin{aligned}\nmdc\\left(\\frac{a}{d}, \\frac{b}{d}\\right) & = mdc\\left(\\frac{a}{d}, \\frac{33}{d} - 1\\right) \\\\\n& = mdc\\left(\\frac{b}{d}, \\frac{33}{d} - 1\\right) \\\\\n& = 1\n\\end{aligned}\n$$\nFixado $d$, é suficiente encontrarmos pares de inteiros positivos $(x, y)$ com $mdc\\left(x, \\frac{33}{d} - 1\\right) = 1$ tais que $x + y = \\frac{33}{d} - 1$, pois daí obteremos também $mdc\\left(y, \\frac{33}{d} - 1\\right) = 1$ e que $(a, b) = (d x, d y)$ também é solução. Vejamos então as possibilidades para $d$:\n\ni) Para $d = 1$ e $x + y = 32$, temos 16 soluções, pois basta escolher $x$ ímpar.\n\nii) Para $d = 3$ e $x + y = 10$, temos 4 soluções, pois $x$ não pode ser par e nem múltiplo de 5.\n\niii) Para $d = 11$ e $x + y = 2$, temos 1 solução apenas.\n\niv) Não podemos ter $d = 33$, pois $a$ e $b$ são positivos.\n\nLogo, existem 21 pares de soluções.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26915,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nEncontre todas as soluções inteiras do sistema\n$$\n\\left\\{\\begin{array}{l}\nx z-2 y t=3 \\\\\nx t+y z=1\n\\end{array}\\right.\n$$",
    "options": [],
    "answer": "(3, 1, 1, 0), (-3, -1, -1, 0), (1, 0, 3, 1), (-1, 0, -3, -1)",
    "solution": "Solution:\nUma boa estratégia será aplicar alguma manipulação algébrica, como somar as equações, multiplicá-las, somar um fator de correção, entre outras para obtermos alguma fatoração envolvendo esses números. Elevando ambas as equações ao quadrado, temos:\n$$\n\\left\\{\\begin{aligned}\nx^{2} z^{2}-4 x y z t+4 y^{2} t^{2} & =9 \\\\\nx^{2} t^{2}+2 x y t z+y^{2} z^{2} & =1\n\\end{aligned}\\right.\n$$\nMultiplicando a segunda por dois e somando com a primeira, temos:\n$$\n\\begin{aligned}\nx^{2}\\left(z^{2}+2 t^{2}\\right)+2 y^{2}\\left(z^{2}+2 t^{2}\\right) & =11 \\\\\n\\left(x^{2}+2 y^{2}\\right)\\left(z^{2}+2 t^{2}\\right) & =11\n\\end{aligned}\n$$\nComo cada uma das parcelas anteriores é um inteiro não negativo, temos dois casos:\n$$\n\\begin{gathered}\n\\left\\{\\begin{array}{l}\nx^{2}+2 y^{2}=11 \\\\\nz^{2}+2 t^{2}=1\n\\end{array} \\Rightarrow(x, y, z, t)=( \\pm 3, \\pm 1, \\pm 1,0)\\right. \\\\\n\\left\\{\\begin{array}{l}\nx^{2}+2 y^{2}=1 \\\\\nz^{2}+2 t^{2}=11\n\\end{array} \\Rightarrow(x, y, z, t)=( \\pm 1,0, \\pm 3, \\pm 1)\\right.\n\\end{gathered}\n$$\nLogo, as únicas soluções possíveis são as quádruplas $( \\pm 1,0, \\pm 3, \\pm 1)$ e $( \\pm 3, \\pm 1, \\pm 1,0)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26916,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSeja $n$ um inteiro positivo. Se a equação $2x + 2y + z = n$ tem 28 soluções em inteiros positivos $x$, $y$ e $z$, determine os possíveis valores de $n$.",
    "options": [],
    "answer": "17, 18",
    "solution": "Solution:\n\nPerceba inicialmente que se $x$ e $y$ estão definidos, só existe uma possível escolha para $z$ e, além disso, $z$ e $n$ possuem a mesma paridade. Consideremos os seguintes casos:\n\ni) O número $n$ é par, ou seja, $n = 2i$. Assim, devemos ter $z = 2j$ e\n$$\n\\begin{aligned}\n2x + 2y + z &= n \\\\\nx + y &= i - j.\n\\end{aligned}\n$$\nTemos então as seguintes $i - j - 1$ possibilidades para o par $(x, y)$:\n$$\n(1, i-j-1), (2, i-j-2), \\ldots, (i-j-1, 1)\n$$\nComo devemos ter $1 \\leq i-j-1 \\leq \\frac{n-4}{2}$, fixado $n$ par, temos\n$$\n\\frac{n-4}{2} + \\frac{n-6}{2} + \\ldots + 1 = \\frac{(n-4)(n-2)}{8}\n$$\nsoluções.\n\nii) O número $n$ é ímpar, ou seja, $n = 2i + 1$. Assim, devemos ter $z = 2j + 1$ e\n$$\n\\begin{aligned}\n2x + 2y + z &= n \\\\\nx + y &= i - j.\n\\end{aligned}\n$$\nTemos então as seguintes $i - j - 1$ possibilidades para o par $(x, y)$:\n$$\n(1, i-j-1), (2, i-j-2), \\ldots, (i-j-1, 1)\n$$\nDe modo semelhante ao caso anterior, devemos ter $1 \\leq i-j-1 \\leq \\frac{n-3}{2}$ e, fixado $n$ ímpar, temos\n$$\n\\frac{n-3}{2} + \\frac{n-5}{2} + \\ldots + 1 = \\frac{(n-3)(n-1)}{8}\n$$\nsoluções.\n\nPortanto,\n$$\n\\begin{aligned}\n& \\frac{(n-4)(n-2)}{8} = 28 \\text{ ou } \\\\\n& \\frac{(n-3)(n-1)}{8} = 28\n\\end{aligned}\n$$\nAs únicas soluções positivas das equações anteriores são $n = 17$ e $n = 18$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26917,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm quadrado $9 \\times 9$ foi dividido em 81 quadrados $1 \\times 1$ e 8 deles foram pintados de preto e o restante de branco.\na) Mostre que independente de onde eles tenham sido pintados, sempre é possível encontrar um quadrado $3 \\times 3$, formado pela justaposição de quadradinhos e com lados paralelos aos lados do tabuleiro, de modo que todos os seus quadrados sejam brancos.\nb) Exiba um exemplo de pintura do tabuleiro na qual não seja possível encontrar qualquer retângulo com mais de 9 quadradinhos, também formado pela justaposição de quadradinhos e com lados paralelos aos lados do tabuleiro, sem conter um quadradinho pintado de preto.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na) Divida o tabuleiro $9 \\times 9$ em $3 \\times 3 = 9$ subtabuleiros de tamanho $3 \\times 3$ como indicado na figura abaixo:\n\n![](attached_image_1.png)\n\nComo foram pintados apenas 8 quadradinhos e $8 < 9$, pelo menos um dos quadrados $3 \\times 3$ terá todos os seus quadrados não pintados.\n\nb) A próxima figura exibe um exemplo.\n\n![](attached_image_2.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26918,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNeste problema, iremos estudar quantos fatores 2 aparecem na fatoração de números da forma $5^{2^{n}}-1$.\n\na. Sejam $x$ e $y$ dois números inteiros ímpares. Prove que $x^{2}+y^{2}$ possui exatamente um fator 2 em sua fatoração em primos.\n\nb. Usando a fatoração $a^{2}-b^{2}=(a-b)(a+b)$, determine quantos fatores 2 o número $5^{4}-1$ possui.\n\nc. O número $N=5^{2^{2017}}-1$ possui quantos fatores 2 ?\n\nd. Sabendo que o número $5^{20}$ possui 14 algarismos. Prove que $5^{2^{18}+20}$ possui 6 zeros consecutivos em sua representação decimal.",
    "options": [],
    "answer": "a) Exactly one factor of two. b) Four factors of two. c) Two thousand nineteen factors of two. d) The decimal representation has six consecutive zeros.",
    "solution": "Solution:\n\na. Como $x$ e $y$ são ímpares, então existem inteiros $x_{0}$ e $y_{0}$ tais que $x=2 x_{0}+1$ e $y=2 y_{0}+1$. Daí,\n$$\n\\begin{aligned}\nx^{2}+y^{2} & =\\left(2 x_{0}+1\\right)^{2}+\\left(2 y_{0}+1\\right)^{2} \\\\\n& =4 x_{0}^{2}+4 x_{0}+1+4 y_{0}^{2}+4 y_{0}+1 \\\\\n& =4 x_{0}^{2}+4 x_{0}+4 y_{0}^{2}+4 y_{0}+2 \\\\\n& =2\\left(2 x_{0}^{2}+2 x_{0}+2 y_{0}^{2}+2 y_{0}+1\\right) \\\\\n& =2\\left(2\\left(x_{0}^{2}+x_{0}+y_{0}^{2}+y_{0}\\right)+1\\right)\n\\end{aligned}\n$$\nO número $x^{2}+y^{2}$ é o produto de 2 por um número ímpar e, portanto, possui apenas um fator 2 .\n\nb. Como $5^{4}=\\left(5^{2}\\right)^{2}$, podemos usar a fatoração da diferença de quadrados duas vezes:\n$$\n\\begin{aligned}\n5^{4}-1 & =\\left(5^{2}\\right)^{2}-1^{2} \\\\\n& =\\left(5^{2}+1\\right)\\left(5^{2}-1\\right) \\\\\n& =\\left(5^{2}+1^{2}\\right)\\left(5^{2}-1^{2}\\right) \\\\\n& =\\left(5^{2}+1^{2}\\right)(5-1)(5+1)\n\\end{aligned}\n$$\nAgora basta contar os fatores $2$ em cada parcela. Pelo item anterior, $5^{2}+1^{2}$ possui um fator 2. Além disso, $5-1=4$ possui dois fatores 2 e $5+1$ possui apenas um fator 2 . Logo, $5^{4}-1$ possui $1+2+1=4$ fatores 2 .\n\nc. Com o expoente é $2^{2017}$ poderemos usar a fatoração da diferença de quadrados 2017 vezes:\n$$\n\\begin{aligned}\n5^{2^{2017}}-1 & =\\left(5^{2^{2016}}\\right)^{2}-1^{2} \\\\\n& =\\left(5^{2^{2016}}+1\\right)\\left(5^{2016}-1\\right) \\\\\n& =\\left(5^{2^{2016}}+1\\right)\\left(5^{2^{2015}}+1\\right)\\left(5^{2^{2015}}-1\\right) \\\\\n& =\\ldots \\\\\n& =\\left(5^{2^{2016}}+1\\right)\\left(5^{2^{2015}}+1\\right) \\ldots\\left(5^{2}+1\\right)(5+1)(5-1)\n\\end{aligned}\n$$\nTemos 2016 números da forma $x^{2}+y^{2}$, com $x$ e $y$ ímpares, e cada um contribui com apenas um fator 2. Além disso, $5+1=6$ tem apenas um fator 2 e $5-1=4$ tem dois fatores 2. Podemos concluir que o número $N=5^{2^{2017}}-1$ possui exatamente $2016 \\cdot 1+1+2=2019$ fatores 2 em sua fatoração.\n\nd. Pelo argumento do item anterior, existe um inteiro $k$ tal que $5^{2^{18}}-1=2^{20} k$. Consequentemente, $5^{2^{18}+20}=10^{20} k+5^{20}$. Como $10^{20}$ termina em 20 zeros e $5^{20}$ possui apenas 14 dígitos, segue que existem pelo menos $20-14=6$ dígitos iguais a zero consecutivos dentre os últimos dígitos da representação decimal de $5^{2^{18}+20}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26919,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nQual a maior quantidade de inteiros que podemos escolher no conjunto $\\{1,2,3, \\ldots, 2017\\}$ de modo que a diferença entre quaisquer dois deles não seja um número primo?",
    "options": [],
    "answer": "505",
    "solution": "Solution:\n\nA maior quantidade é $\\frac{2017-1}{4}+1=505$, que pode ser obtida escolhendo-se os elementos que deixam resto 1 na divisão por 4: $\\{1,5,9, \\ldots, 2017\\}$. A diferença entre quaisquer dois deles é sempre um múltiplo de 4 e consequentemente não é um número primo. Mostraremos agora que não é possível escolhermos mais que essa quantidade.\n\nConsidere os 8 inteiros positivos consecutivos $\\{n, n+1, n+2, n+3, n+4, n+5, n+6, n+7\\}$. Queremos estudar quantos inteiros podem ser escolhidos desse conjunto de modo que a diferença entre quaisquer dois deles seja um número primo.\n\ni) Se o inteiro $n$ for escolhido, não poderemos escolher $n+2, n+3, n+5$ e $n+7$, pois a diferença entre eles e $n$ é um número primo. Além disso, $n+1$ e $n+6$ não podem ser simultaneamente escolhidos, pois diferem por 5. Desse modo, caso escolhamos $n$, poderemos escolher apenas dois inteiros do conjunto.\n\nii) Suponha então que $n$ não é escolhido. Se escolhermos $n+1$, não poderemos escolher $n+3, n+4, n+6$. Além disso, do conjunto restante $\\{n+2, n+5, n+7\\}$ apenas um deles poderá ser escolhido. Novamente só poderemos escolher dois inteiros do conjunto.\n\niii) Suponha agora que $n$ e $n+1$ não são escolhidos. Se escolhermos $n+2$, não poderemos escolher $n+4, n+5$ e $n+7$. Além disso, do conjunto restante $\\{n+3, n+6\\}$ apenas um deles pode ser escolhido. Novamente o número máximo é dois.\n\niv) Se $n, n+1$ e $n+2$ não são escolhidos, mas $n+3$ é, não poderemos escolher $n+5$ e $n+6$. Do conjunto $\\{n+4, n+7\\}$ apenas um deles pode ser escolhido. O máximo de inteiros que podem ser escolhidos também é dois neste caso.\n\nv) Considerando os pares $(n+4, n+6)$ e $(n+5, n+7)$, apenas um inteiro de cada par pode ser escolhido. Portanto, se os quatro primeiros da lista não são escolhidos, só poderemos escolher no máximo dois.\n\nDividindo os números de 1 até 2016 em $2016/8 = 252$ grupos de 8 inteiros consecutivos, podemos concluir que em cada um deles poderemos escolher no máximo 2 inteiros. Assim, contando com o 2017, teremos no máximo $2 \\cdot 252+1=505$ inteiros que podem ser escolhidos satisfazendo as condições do enunciado.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26920,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDetermine se o número $\\underbrace{11 \\ldots 1}_{2016} 2 \\underbrace{11 \\ldots 1}_{2016}$ é um número primo ou um número composto.",
    "options": [],
    "answer": "composite",
    "solution": "Solution:\nSeja $x=\\underbrace{11 \\ldots 1}_{2017}$. Daí,\n$$\n\\begin{aligned}\n\\underbrace{11 \\ldots 1}_{2016}2\\underbrace{11 \\ldots 1}_{2016} & =10^{2016} \\cdot x + x \\\\\n& = x\\left(10^{2016}+1\\right)\n\\end{aligned}\n$$\nComo $x$ e $10^{2016}+1$ são divisores maiores que 1 do número dado, podemos concluir que ele é composto.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26921,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa figura abaixo, estão desenhados 16 triângulos equiláteros de lado $1$. Dizemos que dois deles são vizinhos se possuem um lado em comum. Determine se é possível escrevermos os números de $1$ até $16$ dentro desses triângulos de modo que todas as diferença entre os números colocados em dois triângulos vizinhos sejam $1$ ou $2$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Not possible",
    "solution": "Solution:\n\nObserve que entre quaisquer dois triângulos de lado $1$ da figura é possível construirmos um caminho formado por no máximo $5$ outros triângulos que são mutuamente vizinhos. Assim, se fosse possível fazer a distribuição dos $16$ números como indicado no enunciado, seria possível começar do triângulo com o número $1$ e realizar no máximo $6$ incrementos de $1$ ou $2$ unidades, através do caminho de triângulos vizinhos, e chegar no triângulo com o número $16$. Entretanto, $1+2+2+2+2+2+2=13<16$ e isso mostra que tal distribuição é impossível.\n\n![](attached_image_2.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26922,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOs pontos $X$, $Y$ e $Z$ estão marcados nos lados $AD$, $AB$ e $BC$ do retângulo $ABCD$, respectivamente. Dado que $AX = CZ$, mostre que $XY + YZ \\geq AC$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_2.png)\nConstrua o ponto $E$ sobre a reta $AD$ de modo que $AE = AX$, como indicado na figura acima. Como $AY$ é altura e mediana do triângulo $EYX$, podemos concluir que $YE = YX$. Além disso, como $AE = AX = CZ$ e $AE \\parallel CZ$, segue que $EZCA$ é um paralelogramo e, consequentemente, $EZ = AC$. Pela Desigualdade triangular, temos\n$$\n\\begin{aligned}\nXY + YZ &= EY + YZ \\\\\n&\\geq EZ \\\\\n&= AC\n\\end{aligned}\n$$\nVeja que a igualdade ocorre apenas quando $E$, $Y$ e $Z$ são colineares.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26923,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm ponto $D$ é escolhido no lado $BC$ do triângulo $ABC$. A reta tangente aos incírculos dos triângulos $ABD$ e $ADC$ e diferente de $BC$ e $AD$ intersecta o segmento $AD$ em $T$. Se $AB=40\\ \\mathrm{cm}$, $AC=50\\ \\mathrm{cm}$ e $BC=60\\ \\mathrm{cm}$, determine o valor do comprimento de $AT$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "30 cm",
    "solution": "Solution:\n\nSabemos que os comprimentos dos segmentos tangentes traçados de um ponto externo a um círculo são congruentes e que $XY=HI$. Assim,\n$$\n\\begin{aligned}\n2AT &= (AF-TF)+(AJ-TJ) \\\\\n&= AG+AK-(TX+TY) \\\\\n&= (AB-BG)+(AC-CK)-XY \\\\\n&= AB+AC-(BH+HI+IC) \\\\\n&= AB+AC-BC \\\\\n&= 30\n\\end{aligned}\n$$\nPortanto, $AT=30\\ \\mathrm{cm}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26924,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOs ângulos internos de um quadrilátero não convexo formado por uma linha poligonal que não se auto-intersecta são $45^{\\circ}$, $45^{\\circ}$, $45^{\\circ}$ e $225^{\\circ}$. Mostre que os pontos médios de seus lados formam um quadrado.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSejam $A B C D$ o quadrilátero não convexo, com $\\angle A D C=225^{\\circ}$ e $E, F, G$ e $H$ os pontos médios dos lados $A B, A D, C D$ e $B C$, respectivamente. Além disso, sejam $X$ e $Y$ os pontos de interseção de $C D$ e $B D$ com $A B$ e $A C$, respectivamente.\n\n![](attached_image_2.png)\n\nComo $\\angle A B C=\\angle B C D=45^{\\circ}$, segue que $\\angle B X C=90^{\\circ}$. Decorre daí, dado que $\\angle B A D=45^{\\circ}$, que $\\angle A D X=45^{\\circ}$. Assim, $A D X$ e $B C X$ são triângulos retângulos isósceles com $A X=D X$ e $B X=C X$. Pelo caso de congruência $L A L$, como $\\angle B X D=\\angle A X C$, podemos concluir que os triângulos $B D X$ e $A C X$ são congruentes. Isso nos produz $B D=A C$ e $\\angle A B D=\\angle A C D$. Como\n$$\n\\begin{aligned}\n& \\angle A B D+\\angle B A C= \\\\\n& \\angle A C D+\\angle B A C=90^{\\circ},\n\\end{aligned}\n$$\npodemos concluir que $\\angle B Y A=90^{\\circ}$. Veja que $E F$ e $G H$ são bases médias dos triângulos $A B D$ e $B C D$, respectivamente. Daí, $E F=H G=B D / 2$ e $E F\\|H G\\| B D$. De modo semelhante, $E H=F G=A C / 2$ e $E H\\|F G\\| A C$. Isso nos permite concluir que todos os lados do quadrilátero $E F G H$ são iguais e, como $B D$ é perpendicular a $A C$, os lados consecutivos são perpendiculares. Logo, $E F G H$ é um quadrado.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26925,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\na) Determine a quantidade de algarismos do produto $111111 \\cdot 1111111111$, em que o primeiro fator possui 6 algarismos e o segundo possui 10 algarismos.\n\nb) Os números $2^{2016}$ e $5^{2016}$ são escritos um ao lado do outro para formar um único número $N$ que possui uma quantidade de algarismos que é a soma das quantidades de algarismos dos dois números. Por exemplo, se fizéssemos isso com $2^{3}$ e $5^{3}$ iríamos obter o número 8125, que possui 4 algarismos. Determine a quantidade de algarismos de $N$.",
    "options": [],
    "answer": "a) 15; b) 2017",
    "solution": "Solution:\na) Sejam $A$ e $B$ os números com 6 e 10 algarismos e começados por 1, respectivamente. Temos $10^{5}<A<2 \\cdot 10^{5}$ e $10^{9}<B<2 \\cdot 10^{9}$. Com isso, multiplicando as duas desigualdades anteriores, podemos escrever:\n$$\n10^{14}<A \\cdot B<4 \\cdot 10^{14}<10^{15}\n$$\nConcluímos assim que o produto $A \\cdot B$ possui exatamente 15 algarismos.\n\nb) Chamaremos de $x$ e $y$ as quantidades de algarismos de $2^{2016}$ e $5^{2016}$, respectivamente. Desejamos encontrar $x+y$. Comparando com potências de 10, temos $10^{x-1}<2^{2016}<10^{x}$ e $10^{y-1}<5^{2016}<10^{y}$ (não pode ocorrer igualdade, pois $10^{k}$ não é uma potência de 2 e nem de 5 para $k \\geq 1$ ). Multiplicando as duas desigualdades anteriores, temos\n$$\n10^{x-1} \\cdot 10^{y-1}<2^{2016} \\cdot 5^{2016}<10^{x} \\cdot 10^{y}\n$$\nou seja,\n$$\n10^{x+y-2}<10^{2016}<10^{x+y} .\n$$\nDaí $x+y-2<2016<x+y$ e, consequentemente, $x+y=2017$. Então $N$ possui 2017 algarismos.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26926,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nConsidere a coleção de todos os números de 5 dígitos cuja soma dos dígitos é 43. Um desses números é escolhido ao acaso. Qual a probabilidade dele ser múltiplo de 11?",
    "options": [],
    "answer": "1/5",
    "solution": "Solution:\nA soma máxima dos dígitos de um número de cinco dígitos é 45, que corresponde à soma dos dígitos de $99999$. Para que um número possua soma de seus dígitos $43$ podem ocorrer dois casos: ou ele terá três dígitos iguais a $9$ e dois iguais a $8$ ou ele terá quatro dígitos iguais a $9$ e um igual a $7$.\n\nNo primeiro caso, podemos escolher a posição do primeiro dígito $8$ de $5$ maneiras e a posição do segundo de $4$ maneiras. Entretanto, como os dois dígitos são indistinguíveis, cada configuração foi contada duas vezes. Os dígitos $9$ devem ser colocados nas posições restantes e isso nos dá $\\frac{5 \\cdot 4}{2} = 10$ números.\n\nNo segundo caso, podemos escolher a posição do dígito $7$ de $5$ maneiras. Consequentemente, existem $10 + 5 = 15$ números de $5$ dígitos de modo que a soma de seus dígitos é $43$.\n\nPrecisamos agora recordar o critério de divisibilidade por $11$:\n\nUm número de $n$ dígitos $\\overline{a_n a_{n-1} a_{n-2} \\ldots a_1}$ é divisível por $11$ se, e somente se, a soma alternada $a_n - a_{n-1} + a_{n-2} - a_{n-3} + \\ldots + (-1)^n a_1$ é divisível por $11$.\n\nPelo critério anterior, o número de $5$ dígitos $\\overline{a_5 a_4 a_3 a_2 a_1}$ é divisível por $11$, se $11$ divide $a_5 - a_4 + a_3 - a_2 + a_1 = (a_5 + a_4 + a_3 + a_2 + a_1) - 2(a_3 + a_1)$. Supondo que a soma dos seus dígitos é $43$ e sabendo que $43 = 11 \\cdot 3 + 10$, segue que $11$ deve dividir $11 \\cdot 3 + 2(5 - a_3 - a_1)$. Portanto, $a_3 + a_1$ deixa resto $5$ na divisão por $11$ e as únicas possibilidades são $(a_3, a_1) = (7,9), (9,7)$ ou $(8,8)$.\n\nLogo, dos $15$ números da coleção, apenas $3$ deles são múltiplos de $11$ e, consequentemente, a probabilidade procurada é $\\frac{3}{15} = \\frac{1}{5}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26927,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAs somas das três colunas e das três linhas da tabela são iguais.\n\n| 4 | 9 | 2 |\n| :--- | :--- | :--- |\n| 8 | 1 | 6 |\n| 3 | 5 | 7 |\n\nQual é o menor número de casas da tabela, que devem ser alteradas, para que todas as novas seis somas sejam diferentes entre si?",
    "options": [],
    "answer": "4",
    "solution": "Solution:\n\nSe três ou menos casas são alteradas, ou existirão duas filas sem usarem casas alteradas ou uma casa é a única alterada em sua linha e coluna. No primeiro caso, essas duas filas sem casas alteradas possuem a mesma soma. No segundo caso, se apenas uma casa é a única alterada na sua linha e coluna, elas também possuem a mesma soma. A tabela a seguir representa um exemplo com apenas 4 alterações e todas as 6 somas de linhas e colunas distintas.\n\n| $\\mathbf{5}$ | $\\mathbf{9}$ | $\\mathbf{7}$ |\n| :--- | :--- | :--- |\n| 8 | $\\mathbf{3}$ | $\\mathbf{9}$ |\n| 3 | 5 | 7 |\n\nPortanto, o mínimo de alterações é 4.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26928,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nJuquinha marca pontos sobre uma circunferência e desenha triângulos ligando 3 destes pontos. Os comprimentos dos arcos de 2 pontos consecutivos são iguais.\n\na) Marcando 4 pontos na circunferência, quantos triângulos ele conseguirá desenhar?\n\nb) Marcando 5 pontos na circunferência, quantos triângulos equiláteros ele conseguirá desenhar?\n\nc) Quantos triângulos retângulos ele conseguirá desenhar se marcar 6 pontos?",
    "options": [],
    "answer": "a) 4; b) 0; c) 12",
    "solution": "Solution:\n\na) $4$. Temos que escolher $3$ pontos de um total de $4$. Em situações como esta, ao escolhermos um triângulo qualquer, por exemplo o triângulo $ABC$, todas as permutações dos pontos $A$, $B$ e $C$ continuam sendo o triângulo $ABC$. Dessa forma, ao utilizarmos o princípio fundamental da contagem, devemos dividir pela quantidade de permutações de $3$ pontos. Portanto, o total de triângulos que conseguimos desenhar, utilizando como vértices quatro pontos sobre uma circunferência, é $$\\frac{4 \\cdot 3 \\cdot 2}{3 \\cdot 2 \\cdot 1} = 4.$$\n\nb) Nenhum. Como os comprimentos dos arcos delimitados por $2$ pontos consecutivos são iguais, então cada um destes arcos mede $$\\frac{360^{\\circ}}{5} = 72^{\\circ}.$$ Ao desenharmos um triângulo equilátero inscrito a uma circunferência, cada arco mede $$\\frac{360^{\\circ}}{3} = 120^{\\circ}.$$ Dessa forma, deveríamos ter $120^{\\circ}$ como múltiplo de $72^{\\circ}$, o que é absurdo. Assim, não é possível desenhar triângulo equilátero com vértices em $3$ dos $5$ pontos desenhados simetricamente sobre uma circunferência.\n\nc) $12$. Um triângulo retângulo inscrito em uma circunferência deve ter um de seus lados (a hipotenusa) sobre o diâmetro. Como foram desenhados $6$ pontos simétricos sobre a circunferência, a medida de cada arco delimitado por $2$ pontos consecutivos é $$\\frac{360^{\\circ}}{6} = 60^{\\circ},$$ ou seja, devemos escolher $2$ pontos com $3$ arcos de cada lado. Isso é o mesmo que desenharmos um hexágono regular utilizando estes pontos e tomarmos a maior diagonal como sendo a hipotenusa. Dessa forma, existem $3$ destas diagonais (diâmetros) e para cada uma delas $4$ pontos para completar o triângulo. Sendo assim, o total de triângulos retângulos é $$3 \\cdot 4 = 12.$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26929,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm ladrilho, em forma de polígono regular, foi retirado do lugar que ocupava em um painel. Observou-se, então, que se esse ladrilho sofresse uma rotação de $40^{\\circ}$ ou de $60^{\\circ}$ em torno do seu centro, poderia ser encaixado perfeitamente no lugar que ficou vago no painel. Qual o menor número de lados que esse polígono pode ter?",
    "options": [],
    "answer": "18",
    "solution": "Solution:\n\nPara que seja possível efetuar tais rotações, com o encaixe do polígono, é necessário e suficiente que o ângulo central seja um divisor de $40^{\\circ}$ e $60^{\\circ}$. Se $n$ é o número de lados do ladrilho, o ângulo central é dado por $\\frac{360^{\\circ}}{n}$. Assim, as razões\n$$\n\\frac{40^{\\circ}}{\\frac{360^{\\circ}}{n}} = \\frac{n}{9} \\text{ e } \\frac{60^{\\circ}}{\\frac{360^{\\circ}}{n}} = \\frac{n}{6}\n$$\ndevem ser inteiras. O menor inteiro positivo múltiplo de $6$ e $9$ é $18$. Claramente um polígono regular de $18$ lados, por possuir ângulo central de $20^{\\circ}$, satisfaz a condição do enunciado.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26930,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nQuando Paulo fez 15 anos, convidou 43 amigos para uma festa. O bolo tinha a forma de um polígono regular de 15 lados e havia 15 velas sobre ele. As velas foram colocadas de tal maneira que não havia três velas em linha reta. Paulo dividiu o bolo em pedaços triangulares onde cada corte ligava duas velas ou ligava uma vela a um vértice. Além disso, nenhum corte cruzou outro já realizado. Explique por que, ao fazer isso, Paulo pôde dar um pedaço de bolo a cada um de seus convidados, mas ele próprio ficou sem comer.",
    "options": [],
    "answer": "43",
    "solution": "Solution:\n\nSeja $n$ o número de triângulos em que se pode dividir o bolo com as condições dadas. Somaremos os ângulos interiores destes triângulos de duas formas:\n\n- Por um lado, como cada triângulo possui soma dos ângulos internos igual a $180^{\\circ}$, a soma de todos os ângulos internos deles é $180^{\\circ} \\cdot n$.\n\n- Por outro lado, cada ângulo interno de um triângulo está associado a uma vela ou vértice do bolo. A soma dos ângulos internos associados às velas é $360^{\\circ} \\cdot 15$ e a soma dos ângulos internos associados aos vértices coincide com a soma dos ângulos internos do polígono, que é $180^{\\circ} \\cdot (15-2)$. Assim, a soma total também é igual a\n$$\n360^{\\circ} \\cdot 15 + 180^{\\circ} \\cdot (15-2)\n$$\nPortanto,\n$$\n180^{\\circ} \\cdot n = 360^{\\circ} \\cdot 15 + 180^{\\circ} \\cdot (15-2)\n$$\no que implica $n=43$. Como são 43 convidados, Paulo fica sem comer!",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26931,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA tabela abaixo mostra alguns dos resultados do último Festival de Pesca de Pirajuba, exibindo quantos competidores $q$ pescaram $n$ peixes para alguns valores de $n$.\n\n| $n$ | 0 | 1 | 2 | 3 | $\\ldots$ | 13 | 14 | 15 |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| $q$ | 9 | 5 | 7 | 23 | $\\ldots$ | 5 | 2 | 1 |\n\nA notícia publicada no jornal da cidade relatou que:\ni) o vencedor pescou 15 peixes;\nii) dentre aqueles que pescaram 3 ou mais peixes, a média foi de 6 peixes pescados; $\\mathrm{e}$\niii) dentre aqueles que pescaram 12 ou menos peixes a média de peixes pescados foi 5.\n\na) Qual foi o número total de peixes pescados durante o festival?\nb) Quantos competidores pescaram de 4 a 12 peixes?",
    "options": [],
    "answer": "a) 943; b) 123",
    "solution": "Solution:\nSejam $N$ e $P$, os números de competidores e peixes pescados no evento, respectivamente. Analisando os competidores que pescaram 3 ou mais peixes, temos:\n$$\nN-9-5-7=N-21 \\text{ pescadores e } P-0 \\cdot 9-1 \\cdot 5-2 \\cdot 7=P-19 \\text{ peixes, }\n$$\nsendo a média escrita como:\n$$\n\\begin{aligned}\n\\frac{P-19}{N-21} & =6 \\\\\nP-19 & =6 N-126 \\\\\nP & =6 N-126+19 \\\\\nP & =6 N-107\n\\end{aligned}\n$$\nObservando agora os competidores que pescaram 12 ou menos peixes, obtemos:\n$N-5-2-1=N-8$ pescadores e $P-13 \\cdot 5-14 \\cdot 2-5 \\cdot 1=P-98$ peixes\nresultando em\n$$\n\\begin{aligned}\n\\frac{P-98}{N-8} & =5 \\\\\nP-98 & =5 N-40 \\\\\nP & =5 N-40+98 \\\\\nP & =5 N+58 \\\\\n6 N-107 & =5 N+58 \\\\\n6 N-5 N & =58+107 \\\\\nN & =165 .\n\\end{aligned}\n$$\n\na) Substituindo o valor encontrado de $N$ na equação $P=6 N-107$, obtemos $P=6 \\cdot 165-107=883$ peixes no festival.\n\nb) Para o número de competidores que pescaram de 4 a 12 peixes basta fazermos a diferença entre o total de competidores e aqueles que pescaram 0 peixe (9 pescadores), 1 peixe (5), 2 peixes (7), 3 peixes (23), 13 peixes (5), 14 peixes (2) e, finalmente, 15 peixes (1):\n$$\n165-9-5-7-23-5-2-1=113\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26932,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\na) Helena e Gabriel pulam, simultaneamente, de uma jangada em um rio e nadam em direções opostas. Gabriel nada rio abaixo seguindo a corrente e Helena nada rio acima contra a corrente, possivelmente a uma velocidade diferente. Depois de 5 minutos, eles viram e voltam para a jangada, cada um mantendo uma velocidade constante durante todo o tempo. Quem chega primeiro? (Durante esse tempo, a jangada estava ancorada, ou seja, permaneceu parada, apesar da correnteza.)\nb) Em sua velocidade usual, um homem desce um rio de $15~\\mathrm{km}$ de comprimento em 5 horas a menos que o tempo que ele gasta nadando no mesmo rio percorrendo o caminho contrário. Se ele dobrar a sua velocidade usual, ele passa a descer o rio gastando apenas 1 hora a menos que o tempo gasto na volta. Considerando que a velocidade da correnteza do rio se mantém constante durante os trajetos, qual o seu valor em $\\mathrm{km} / \\mathrm{h}$ ?",
    "options": [],
    "answer": "a) Helena. b) 2 km/h",
    "solution": "Solution:\na) A distância percorrida por Gabriel no tempo de 5 minutos foi $D_{G} = (V_{G} + V_{C}) \\cdot 5$, em que $V_{G}$ e $V_{C}$ são as velocidades de Gabriel e da correnteza, respectivamente. Já Helena percorreu $D_{H} = (V_{H} - V_{C}) \\cdot 5$, com $V_{H}$ a velocidade dela. Na volta, o tempo $t_{G}$ de Gabriel será $D_{G} = (V_{G} - V_{C}) \\cdot t_{G}$ e $t_{H}$ de Helena será $D_{H} = (V_{H} + V_{C}) \\cdot t_{H}$. Perceba que\n$$\n\\begin{aligned}\n(V_{H} - V_{C}) \\cdot 5 &= (V_{H} + V_{C}) \\cdot t_{H} \\\\\nt_{H} &= 5 \\cdot \\frac{V_{H} - V_{C}}{V_{H} + V_{C}} \\\\\n&< 5\n\\end{aligned}\n$$\nAlém disso, temos\n$$\n\\begin{aligned}\n(V_{G} + V_{C}) \\cdot 5 &= (V_{G} - V_{C}) \\cdot t_{G} \\\\\nt_{G} &= 5 \\cdot \\frac{V_{G} + V_{C}}{V_{G} - V_{C}}\n\\end{aligned}\n$$\n$$\n> 5\n$$\nComo $t_{H} < t_{G}$, Helena chega antes.\n\nb) Sejam $c$ e $v$ as velocidades da correnteza do rio e do homem, respectivamente, em $\\mathrm{km} / \\mathrm{h}$. O homem leva $\\frac{15}{v + c}$ horas, nadando a favor da correnteza e $\\frac{15}{v - c}$ horas, nadando contra a correnteza. Assim, os dados do enunciado podem ser traduzidos no seguinte sistema:\n$$\n\\left\\{\n\\begin{array}{l}\n\\frac{15}{v + c} = \\frac{15}{v - c} - 5 \\\\\n\\frac{15}{2v + c} = \\frac{15}{2v - c} - 1\n\\end{array}\n\\right.\n$$\nMultiplicando a primeira equação por $(v^{2} - c^{2})$ e a segunda por $(4v^{2} - c^{2})$, obtemos\n$$\n\\left\\{\n\\begin{array}{l}\n15(v - c) = 15(v + c) - 5(v^{2} - c^{2}) \\\\\n15(2v - c) = 15(2v + c) - (4v^{2} - c^{2})\n\\end{array}\n\\right.\n$$\nPor comparação, segue que\n$$\n5(v^{2} - c^{2}) = 4v^{2} - c^{2}\n$$\nou seja,\n$$\nv^{2} = 4c^{2}\n$$\nUsando uma das equações, chegamos a\n$$\n\\begin{aligned}\n15(v - c) &= 15(v + c) - 5(v^{2} - c^{2}) \\\\\n-15c &= 15c - 5 \\cdot 3c^{2} \\\\\n15c^{2} - 30c &= 0 \\\\\nc^{2} - 2c &= 0\n\\end{aligned}\n$$\nPortanto, a velocidade da correnteza é $2~\\mathrm{km} / \\mathrm{h}$ (única solução positiva).",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26933,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNum concurso de tiros, 8 alvos são arrumados em duas colunas com 3 alvos e uma coluna com 2 alvos. As regras são:\n- O atirador escolhe livremente em qual coluna atirar.\n- Ele deve tentar o alvo mais baixo ainda não acertado.\n\n![](attached_image_1.png)\n\na) Se o atirador desconsiderar a segunda regra, de quantos modos ele poderá escolher apenas 3 posições dos 8 discos distintos para atirar?\n\nb) Se as regras forem cumpridas, então, de quantas maneiras os 8 alvos podem ser acertados?",
    "options": [],
    "answer": "a) 56; b) 560",
    "solution": "Solution:\n\na) Se $x$, $y$ e $z$ são as posições dos alvos, em princípio, o atirador possui 8 escolhas para $x$, $8-1=7$ para $y$, pois não podemos repetir a posição já escolhida, e $8-2=6$ escolhas para $z$, pois não podemos repetir nenhuma das duas posições já selecionadas. Isso dá $8 \\cdot 7 \\cdot 6$ escolhas. Entretanto, as escolhas\n$$\nx y z,\\ x z y,\\ y x z,\\ y z x,\\ z x y,\\ z y x\n$$\nproduzem o mesmo conjunto de posições escolhidas. Para evitar contagens repetidas, basta dividirmos a contagem anterior pelo número de vezes que cada conjunto de posições foi contado, ou seja, o número total de escolhas é\n$$\n\\frac{8 \\cdot 7 \\cdot 6}{6} = 56\n$$\nEm geral, o número de maneiras de escolhermos 3 objetos em um conjunto com $n$ objetos distintos é\n$$\n\\frac{n(n-1)(n-2)}{6}\n$$\n\nb) Suponha que as colunas estejam marcadas com as letras $A$, $B$, e $C$. Cada sequência de tiros pode ser codificada com uma disposição linear dessas letras seguindo a ordem dos tiros da esquerda para a direita, formando uma palavra com 3 letras $A$, 2 letras $B$ e 3 letras $C$. Por exemplo, $A A B C C C B A$, significa que iremos atirar duas vezes na primeira coluna, uma vez na segunda, três vezes na terceira, mais uma vez na segunda e, finalmente, uma vez na primeira. A ordem dos tiros em cada coluna está determinada pelas regras do concurso. Como existe uma correspondência entre essas palavras e a distribuição de tiros, basta contarmos o número de palavras distintas que podem ser criadas com essas letras. Para isso, vamos calcular o número de anagramas possíveis usando 3 $A$'s, 2 $B$'s e 3 $C$'s, numa sequência de 8 letras. Observando as 8 posições, podemos escolher as 3 de $A$'s de $\\frac{8 \\cdot 7 \\cdot 6}{6} = 56$ maneiras. Dentre os cinco espaços restantes, devemos escolher 2 para os $B$'s e isso pode ser feito de $\\frac{5 \\cdot 4}{2} = 10$ maneiras. As três posições restantes serão ocupadas pelos $C$'s. Portanto, há $56 \\cdot 10 = 560$ formas de acertar todos os alvos com as regras impostas.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26934,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm ciclo de três conferências teve sucesso constante, isto é, em cada sessão havia o mesmo número de participantes. No entanto, a metade dos que compareceram à primeira não voltou mais; um terço dos que compareceram à segunda conferência assistiu apenas a ela, e um quarto dos que compareceram à terceira não assistiu nem à primeira nem à segunda. Sabendo que havia 300 inscritos e que cada um assistiu a pelo menos uma conferência, determine:\n\na) Quantas pessoas compareceram a cada conferência?\n\nb) Quantas pessoas compareceram às três conferências?",
    "options": [],
    "answer": "a) 156 people attended each conference. b) 37 people attended all three conferences.",
    "solution": "Solution:\n\na) Chamemos de $P$ o número de presentes em cada conferência, $x, y, t$ e $z$ serão os números inteiros dos que foram às três conferências; para às primeira e segunda, apenas; primeira e terceira, apenas; e segunda e terceira, apenas, respectivamente.\n\n![](attached_image_1.png)\n\nPodemos então escrever que:\n- $y = 300 - \\left(P + \\frac{P}{2} + \\frac{P}{3}\\right) = 300 - \\frac{11P}{6}$;\n- $z = 300 - \\left(P + \\frac{P}{3} + \\frac{P}{4}\\right) = 300 - \\frac{19P}{12}$;\n- $t = 300 - \\left(P + \\frac{P}{2} + \\frac{P}{4}\\right) = 300 - \\frac{7P}{4}$;\n- $x = \\frac{P}{2} - (y + t) = \\frac{49P}{12} - 600$.\n\nComo $x$ é um inteiro, $P$ é múltiplo de $12$, então existe $k$ inteiro tal que $P = 12k$. Daí, ficamos com $y = 300 - 22k$, $z = 300 - 19k$, $t = 300 - 21k$ e $x = 49k - 600$, todos não negativos, logo:\n- $y = 300 - 22k \\geq 0 \\Rightarrow k \\leq 13,63$,\n- $z = 300 - 19k \\geq 0 \\Rightarrow k \\leq 15,78$,\n- $t = 300 - 21k \\geq 0 \\Rightarrow k \\leq 14,28$,\n- $x = 49k - 600 \\geq 0 \\Rightarrow k \\geq 12,24$.\n\nO único inteiro nesse intervalo é $k = 13$. Isso produz $P = 156$.\n\nb) Pelo item anterior, $x = 49k - 600 = 49 \\cdot 13 - 600 = 37$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26935,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNo triângulo $ABC$, retângulo em $A$, tem-se $AB = 8~\\mathrm{cm}$ e $AC = 6~\\mathrm{cm}$.\n\na. O ponto $P$, interior ao triângulo, dista $1~\\mathrm{cm}$ do lado $AB$ e $2~\\mathrm{cm}$ do lado $AC$. Qual é a distância de $P$ ao lado $BC$?\n\nb. Calcule o raio da circunferência que é tangente ao lado $AC$ e aos prolongamentos dos lados $AB$ e $BC$.",
    "options": [],
    "answer": "a) 2.8 cm; b) 4 cm",
    "solution": "Solution:\n\na.\nSe $x$ é a distância de $P$ ao lado $BC$ e usando a notação $(XYZ)$ para indicar a área do polígono $XYZ$, temos\n$$\n\\begin{aligned}\n(APB) + (BPC) + (CPA) &= (ABC) \\\\\n\\frac{8 \\cdot 1}{2} + \\frac{10 \\cdot x}{2} + \\frac{6 \\cdot 2}{2} &= \\frac{6 \\cdot 8}{2} \\\\\n8 + 10x + 12 &= 48\n\\end{aligned}\n$$\nPor fim, chegamos a $x = 2{,}8~\\mathrm{cm}$.\n\n![](attached_image_1.png)\n\nb.\nSendo $R$ o raio da circunferência, temos:\n$$\n\\begin{aligned}\n(APB) + (BPC) - (CPA) &= (ABC) \\\\\n\\frac{8 \\cdot R}{2} + \\frac{10 \\cdot R}{2} - \\frac{6 \\cdot R}{2} &= \\frac{6 \\cdot 8}{2}\n\\end{aligned}\n$$\nPor fim, obtemos $R = 4$.\n\n![](attached_image_2.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26936,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO pai de João possui entre 200 e 300 livros em sua biblioteca. Um quinto destes livros está em inglês, um sétimo em francês, um quarto em italiano e o resto são livros em espanhol. Qual o total de livros em espanhol nessa biblioteca?",
    "options": [],
    "answer": "114",
    "solution": "Solution:\n\nA quantidade de livros é um múltiplo de $5$, $7$ e $4$. Como esses números não possuem fatores primos em comum, necessariamente deve ser um múltiplo de $5 \\cdot 7 \\cdot 4 = 140$. No intervalo dado, existe um único múltiplo de $140$, a saber, o número $280$. Portanto, as quantidades de livros que não estão em espanhol é\n$$\n\\frac{280}{5} + \\frac{280}{7} + \\frac{280}{4} = 56 + 40 + 70 = 166\n$$\nAssim, existem $280 - 166 = 114$ livros de espanhol.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26937,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm hotel possui 5 quartos distintos, todos com camas individuais para até 2 pessoas. O hotel está sem outros hóspedes e 5 amigos querem passar a noite nele. De quantos modos os 5 amigos podem escolher seus quartos?",
    "options": [],
    "answer": "2220",
    "solution": "Solution:\n\nAnalisando apenas a quantidade de pessoas por quarto, sem levar em consideração a ordem, as possíveis distribuições podem ser associadas às listas:\n$$\n(1,1,1,1,1),\\ (1,1,1,2) \\text{ ou } (2,2,1)\n$$\nAnalisaremos agora lista por lista o número de maneiras de distribuir os amigos.\n\ni) Na lista $(1,1,1,1,1)$, todos os amigos ficarão em quartos distintos. Existem 5 escolhas de quarto para o primeiro amigo, 4 para o segundo, 3 para o terceiro e assim por diante. O total de escolhas é:\n$$\n5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1 = 120\n$$\n\nii) Na lista $(1,1,1,2)$, podemos escolher o quarto que terá dois amigos de 5 maneiras e o quarto que não terá ninguém de 4 maneiras. Assim, existem $5 \\cdot 4 = 20$ maneiras de distribuirmos os amigos nos 5 quartos. Para escolher os dois amigos que ficarão juntos, temos $\\frac{5 \\cdot 4}{2} = 10$ escolhas possíveis. Os outros três, a exemplo do item anterior, poderão ser distribuídos nos três quartos de $3 \\cdot 2 \\cdot 1 = 6$ modos. O total de distribuições nesse caso é $5 \\cdot 4 \\cdot 10 \\cdot 6 = 1200$.\n\niii) Na lista $(2,2,1)$, existem 5 modos de escolhermos o quarto que terá apenas um amigo. Dos quatro restantes, podemos escolher os que terão dois hóspedes de $\\frac{4 \\cdot 3}{2} = 6$ modos. Uma vez escolhida a quantidade de amigos por quarto, existem $\\frac{5 \\cdot 4}{2} = 10$ escolhas de dois amigos para o primeiro quarto duplo e $\\frac{3 \\cdot 2}{2} = 3$ escolhas de 2 amigos, dentre os restantes, para o outro quarto duplo. Finalmente, o amigo que sobrou ficará no quarto restante. O total de distribuições nesse caso é\n$$\n5 \\cdot 6 \\cdot 10 \\cdot 3 \\cdot 1 = 900\n$$\nPor fim, o total de distribuições é\n$$\n120 + 1200 + 900 = 2220\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26938,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO rei Luis estava desconfiado de alguns de seus cortesãos. Ele fez uma lista completa de cada um dos seus cortesãos e disse a cada um deles para espionar um outro cortesão. O primeiro da lista foi espionar o cortesão que estava espionando o segundo da lista, o segundo da lista foi espionar o cortesão que estava espionando o terceiro da lista, e assim sucessivamente, o penúltimo foi espionar o cortesão que estava espionando o último e o último foi espionar o cortesão que estava espionando o primeiro. Verifique que o rei Luis tinha um número ímpar de cortesãos.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSeja $n$ o número de cortesãos da lista e suponha que $n$ é par. Coloque-os sentados ao redor de uma mesa circular de modo que cada um esteja espionando o seu vizinho da esquerda.\n\n![](attached_image_1.png)\n\nO cortesão $1$ espia o cortesão $X$ que espia o cortesão $2$, o cortesão $2$ espia o cortesão $Z$ que espia o cortesão $3$, e assim sucessivamente até que o cortesão $\\frac{n}{2}$ espia o cortesão $Y$ que espia o cortesão $\\frac{n}{2}+1$. Como os números $1,2,3, \\ldots, n$ devem se alternar sobre o círculo, concluímos que o cortesão $\\frac{n}{2}+1$ é igual ao cortesão $1$, ou seja, $n=0$. Esse absurdo mostra que $n$ é ímpar.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26939,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nVinte e cinco garotos e vinte e cinco garotas estão sentados ao redor de uma mesa. Prove que é possível encontrar uma pessoa que tem garotas como vizinhas.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSuponha que não haja uma pessoa que possua duas garotas como vizinhas. Marquemos com $-1$ cada garoto e com $+1$ cada garota. Cada pessoa ou possui como vizinhos $(-1,-1)$ ou $(+1,-1)$. Somando todos os 50 pares de números associados aos vizinhos de cada pessoa, contaremos a contribuição de cada um duas vezes. Assim ficamos com $x \\cdot(-2)+y \\cdot(0)=50 \\cdot(+1)+50 \\cdot(-1)$, em que $x$ é o número de pessoas que têm 2 garotos como vizinhos e $y$ é o número de pessoas que têm um garoto e uma garota. Notemos ainda que $x+y=50$. A primeira equação produz $x=0$. Substituindo esse valor na segunda equação, podemos concluir que todas as pessoas têm um garoto e uma garota como vizinhos. Considerando duas pessoas vizinhas com $\\overline{+1}$ e $\\overline{-1}$, a observação anterior nos permite concluir que em alguma ordem os vizinhos delas em ambos os lados serão: $(\\ldots,-1,-1,+1, \\overline{+1},-1,-1,+1,+1, \\ldots)$. Daí, os garotos e as garotas terão que aparecer sempre aos pares e isso é um absurdo, pois obrigaria que a quantidade de cada um, que é $25$, fosse par.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26940,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEm torneios de xadrez, geralmente, cada vitória vale 1 ponto, cada empate, 0,5 ponto e cada derrota, zero ponto. No \"Campeonato Magistral\" de xadrez participaram apenas Mestres Internacionais - MI's e Grandes Mestres - GM's. O número de GM’s foi dez vezes o número de MI's. Cada enxadrista jogou apenas uma vez contra todos os adversários e assim, se $n$ foi o número de jogadores, então ocorreram $\\frac{n(n-1)}{2}$ jogos. A soma dos pontos de todos os GM's foi 4,5 vezes a soma de todos os MI's. Sendo assim, pergunta-se:\n\na) Quantos Mestres Internacionais participaram desta competição?\n\nb) Quantos Grandes Mestres participaram deste campeonato?\n\nc) Quantos jogos teve o torneio?",
    "options": [],
    "answer": "a) 1 b) 10 c) 55",
    "solution": "Solution:\n\na) Sejam $i$ e $g$, os números de MI's e GM's, respectivamente. Denote as somas de pontuações de cada categoria por $I$ e $G$ e o total de jogadores por $n$. Do enunciado, $g=10i$, $G=4,5I$ e $n=i+g=i+10i=11i$. Agora, o total de jogos (e de pontos disputados) foi\n$$\n\\begin{aligned}\n\\frac{n(n-1)}{2} &= \\frac{11i \\cdot (11i-1)}{2} \\\\\nI+G &= I+4,5I=5,5I\n\\end{aligned}\n$$\nAlém disso, sabemos que cada enxadrista jogará $(11i-1)$ jogos, que é também o máximo de pontos que um jogador pode alcançar. Portanto, podemos escrever que\n$$\n\\begin{aligned}\n\\frac{11i \\cdot (11i-1)}{2} &= 5,5I \\\\\n11i \\cdot (11i-1) &= 11I \\\\\ni \\cdot (11i-1) &= I\n\\end{aligned}\n$$\nA equação $I = i \\cdot (11i-1)$ pode ser interpretada como se todos os MI's tivessem feito pontuação máxima, pois $I$ é a soma total de pontos deles, $(11i-1)$ é o número de partidas (e a pontuação máxima) de cada MI e $i$, o número de MI's. No entanto, isto não poderia acontecer, porque com dois ou mais MI's no torneio, eles se enfrentariam resultando para um deles empate ou derrota. Daí, só havia 1 MI na competição.\n\nb) Como $g=10i$, então $g=10 \\cdot 1=10$. Dez GM's participaram do torneio.\n\nc) E houve $\\frac{11(11-1)}{2}=55$ jogos no campeonato.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26941,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEm um torneio com 5 times, não existem empates. De quantos modos podem ocorrer os $\\frac{5 \\cdot 4}{2}=10$ jogos do torneio de modo que, tanto não tenhamos um time que ganhou todas quanto um time que não perdeu todas as partidas?",
    "options": [],
    "answer": "544",
    "solution": "Solution:\n\nRepresente cada time por um vértice de um pentágono e o resultado de cada jogo por uma seta partindo do jogador que ganhou para o jogador que perdeu.\n\n![](attached_image_1.png)\n\nComo cada partida possui 2 resultados possíveis, existem $2^{10}$ possíveis resultados para o torneio. Suponha que o jogador $A$ ganhou todas as partidas, então os jogos restantes podem ser escolhidos de $2^{6}$ maneiras. No torneio, apenas um jogador pode ter ganho todas as partidas e assim, como existem 5 jogadores, existem $5 \\cdot 2^{6}$ distribuições das partidas em que um jogador ganhou todas as partidas. Do mesmo modo, existem $5 \\cdot 2^{6}$ distribuições das partidas em que um jogador perdeu todos os seus jogos. Entretanto, esses dois grupos de torneios não são disjuntos. Para contar quantas distribuições são comuns a essas duas contagens, perceba que podemos escolher o jogador que ganhou todas de 5 formas e o jogador que perdeu todas de $5-1=4$ formas. Feitas essas escolhas, já temos definidos os resultados de 7 jogos. Os 3 jogos restantes podem ser escolhidos de $2^{3}$ e não produzirão outros jogadores com o máximo de vitórias ou o mínimo de derrotas. Portanto, o número de torneios procurados é\n$$\n2^{10}-2 \\cdot 5 \\cdot 2^{6}+5 \\cdot 4 \\cdot 2^{3}=2^{5} \\cdot 17=544\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26942,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\na) Qual dos números é maior: $2^{100}+3^{100}$ ou $4^{100}$ ?\n\nb) Sejam $x$ e $y$ números naturais tais que\n$$\n2^{x} \\cdot 3^{y}=\\left(24^{\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4}+\\cdots+\\frac{1}{60}}\\right) \\cdot\\left(24^{\\frac{1}{3}+\\frac{1}{4}+\\frac{1}{5}+\\cdots+\\frac{1}{60}}\\right)^{2} \\cdot\\left(24^{\\frac{1}{4}+\\frac{1}{5}+\\frac{1}{6}+\\cdots+\\frac{1}{60}}\\right)^{3} \\cdot \\ldots \\cdot\\left(2^{\\frac{1}{60}}\\right)^{59}\n$$\nDeterminar o valor de $x+y$.",
    "options": [],
    "answer": "a) 4^100 is greater than 2^100 + 3^100. b) 3540",
    "solution": "Solution:\na) Perceba que\n$$\n4^{100}=4^{99}+4^{99}+4^{99}+4^{99}=3 \\cdot 4^{99}+4^{99}\n$$\nComparemos as parcelas separadamente:\n$$\n\\begin{aligned}\n4 & >3 \\\\\n4^{99} & >3^{99} \\\\\n3 \\cdot 4^{99} & >3 \\cdot 3^{99} \\\\\n3 \\cdot 4^{99} & >3^{100}\n\\end{aligned}\n$$\ne\n$$\n4^{99}=2^{198}>2^{100}\n$$\nSomando as desigualdades\n$$\n\\begin{aligned}\n3 \\cdot 4^{99} & >3^{100} \\\\\n4^{99} & >2^{100}\n\\end{aligned}\n$$\nobtemos\n$$\n4^{100}>3^{100}+2^{100}\n$$\n\nb) Observe que\n$$\n\\begin{gathered}\n\\left(24^{\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4}+\\cdots+\\frac{1}{60}}\\right) \\cdot\\left(24^{\\frac{1}{3}+\\frac{1}{4}+\\frac{1}{5}+\\cdots+\\frac{1}{60}}\\right)^{2} \\cdot\\left(24^{\\frac{1}{4}+\\frac{1}{5}+\\frac{1}{6}+\\cdots+\\frac{1}{60}}\\right)^{3} \\cdot \\ldots \\cdot\\left(24^{\\frac{1}{60}}\\right)^{59}= \\\\\n\\left(24^{\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4}+\\cdots+\\frac{1}{60}}\\right) \\cdot\\left(24^{\\frac{2}{3}+\\frac{2}{4}+\\frac{2}{5}+\\cdots+\\frac{2}{60}}\\right) \\cdot\\left(24^{\\frac{3}{4}+\\frac{3}{5}+\\frac{3}{6}+\\cdots+\\frac{3}{60}}\\right) \\cdot \\ldots \\cdot\\left(24^{\\frac{59}{60}}\\right)= \\\\\n\\left(24^{\\frac{1}{2}}\\right) \\cdot\\left(24^{\\frac{1}{3}+\\frac{2}{3}}\\right) \\cdot\\left(24^{\\frac{1}{4}+\\frac{2}{4}+\\frac{3}{4}}\\right) \\cdot\\left(24^{\\frac{1}{5}+\\frac{2}{5}+\\frac{3}{5}+\\frac{4}{5}}\\right) \\cdot \\ldots \\cdot\\left(24^{\\frac{1}{60}+\\frac{2}{60}+\\cdots+\\frac{59}{60}}\\right)\n\\end{gathered}\n$$\nComo $2^{x} \\cdot 3^{y}=24^{k}$, segue que\n$$\n\\begin{aligned}\nk & =\\frac{1}{2}+\\left(\\frac{1}{3}+\\frac{2}{3}\\right)+\\left(\\frac{1}{4}+\\frac{2}{4}+\\frac{3}{4}\\right)+\\left(\\frac{1}{5}+\\frac{2}{5}+\\frac{3}{5}+\\frac{4}{5}\\right)+\\cdots+\\left(\\frac{1}{60}+\\frac{2}{60}+\\cdots+\\frac{59}{60}\\right) \\\\\n& =\\frac{1}{2}+\\frac{2}{2}+\\frac{3}{2}+\\frac{4}{2}+\\cdots+\\frac{59}{2} \\\\\n& =\\frac{1}{2} \\cdot \\frac{(1+59) \\cdot 59}{2} \\\\\n& =15 \\cdot 59\n\\end{aligned}\n$$\nLogo, $2^{x} \\cdot 3^{y}=\\left(2^{3} \\cdot 3^{1}\\right)^{15 \\cdot 59}$, ou seja, $x=45 \\cdot 59$ e $y=15 \\cdot 59$. Portanto, $x+y=60 \\cdot 59=3540$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26943,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOs pontos $D$ e $E$ dividem o lado $AB$ do triângulo equilátero $ABC$ em três partes iguais, $D$ entre $A$ e $E$. O ponto $F$ está sobre o lado $BC$ de modo que $CF = AD$. Encontre a soma dos ângulos $\\angle CDF + \\angle CEF$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "30°",
    "solution": "Solution:\n\nSejam $\\angle CDF = \\alpha$ e $\\angle CEF = \\beta$. Como $BF = BD = \\frac{2}{3} \\cdot AB$ e $\\angle ABC = 60^\\circ$, segue que $\\triangle BDF$ é equilátero. Além disso, como $E$ é ponto médio de $BD$, temos que $EF$ é altura e bissetriz relativa ao vértice $F$. Assim, $\\angle BFE = 30^\\circ$. Pelo Teorema do Ângulo Externo, segue que $\\angle BCE = 30^\\circ - \\beta$. Novamente pelo Teorema do Ângulo Externo, como $\\angle BFD = 60^\\circ$, segue que $\\angle ECD = 30^\\circ + \\beta - \\alpha$. Finalmente, como $BE = AD$, $BC = AC$ e $\\angle CBE = \\angle CAD = 60^\\circ$, podemos concluir que $\\triangle CBE \\equiv \\triangle CAD$ e assim $\\angle ACD = \\angle BCE = 30^\\circ - \\beta$. Daí\n$$\n\\begin{aligned}\n\\angle ACB & = \\angle BCE + \\angle ECD + \\angle ACD \\\\\n& = (30^\\circ - \\beta) + (30^\\circ + \\beta - \\alpha) + (30^\\circ - \\beta) \\\\\n& = 90^\\circ - (\\alpha + \\beta)\n\\end{aligned}\n$$\nDaí,\n$$\n\\angle CDF + \\angle CEF = \\alpha + \\beta = 30^\\circ\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26944,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMaria e Pedro jogam em um tabuleiro $9 \\times 9$. Maria começa pintando de vermelho 46 quadradinhos do tabuleiro. Em seguida, Pedro deve escolher um quadrado $2 \\times 2$. Se o quadrado escolhido por Pedro tem 3 ou mais casinhas pintadas de vermelho, ele vence o jogo. Caso contrário, vence Maria. Qual dos dois pode sempre garantir a vitória independentemente da jogada do adversário?",
    "options": [],
    "answer": "Pedro",
    "solution": "Solution:\n\nPedro vence o jogo. Divida o tabuleiro em 20 tabuleiros $2 \\times 2$ como indicado na figura abaixo. Existem 5 quadradinhos do tabuleiro que não fazem parte desses tabuleiros e que estão indicados com a letra $X$. Com a pintura dos 46 quadradinhos, pelo menos $46-5=41=2 \\cdot 20+1$ serão colocados nesses 20 tabuleiros $2 \\times 2$. Pelo Princípio da Casa dos Pombos, independente de como eles estejam distribuídos, pelo menos 3 deles estarão no mesmo tabuleiro $2 \\times 2$.\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26945,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nConsidere um torneio de xadrez envolvendo terráqueos e alienígenas em que cada jogador joga contra todos os outros exatamente uma vez. Ao final do torneio, cada jogador obteve metade dos pontos que conquistou jogando contra terráqueos e metade jogando contra alienígenas. Sendo $t$ e $a$ os números de terráqueos e alienígenas, respectivamente, responda:\n\na) Qual o total de jogos desse torneio, em função de $t$ e $a$?\n\nb) Nesse torneio, cada vitória vale 1 ponto, cada empate vale 0,5 ponto e cada derrota vale zero ponto. Qual o total de pontos dos terráqueos em função de $t$ e $a$?\n\nc) Verifique que o número total de jogadores do torneio é um quadrado perfeito.",
    "options": [],
    "answer": "a) (t + a)(t + a − 1) / 2.\nb) t(t − 1).\nc) t + a = (a − t)^2, so the total number of players is a perfect square.",
    "solution": "Solution:\na) Sendo $n = t + a$ o total de enxadristas presentes, com cada dupla se enfrentando em partidas únicas, o torneio teve\n$$\n\\frac{n(n-1)}{2} = \\frac{(t+a)(t+a-1)}{2} \\text{ partidas. }\n$$\n\nb) Se somarmos a pontuação que cada atleta recebeu por jogo (independentemente do resultado) obteremos sempre a soma 1 por jogo. Assim, existe uma relação direta entre a quantidade de jogos e os pontos distribuídos. Como cada terráqueo ganhou metade dos seus pontos jogando contra adversário da mesma raça e outra metade contra alienígenas, o total de pontos que os terráqueos tiveram contra terráqueos é igual ao número de pontos que os terráqueos obtiveram contra os extraterrestres. Cada terráqueo jogou $\\frac{t(t-1)}{2}$ partidas contra terráqueos (total de pontos dos terráqueos contra terráqueos) e $\\frac{a(a-1)}{2}$ contra aliens (total de pontos dos terráqueos contra os ET's). O mesmo raciocínio vale para os alienígenas. Portanto, o total de pontos dos terráqueos foi de\n$$\n\\frac{t(t-1)}{2} + \\frac{a(a-1)}{2}\n$$\n\nc) Em virtude dos itens anteriores, a quantidade de pontos distribuídas coincide com o total de jogos $\\frac{n(n-1)}{2}$ e, além disso, é igual a soma dos pontos de todos os terráqueos contra seus idênticos e contra os de outra raça, mais a soma de todos os pontos obtidos pelos alienígenas entre si e contra os terráqueos. Sendo assim, podemos escrever que:\n$$\n\\begin{aligned}\n\\frac{(t+a)(t+a-1)}{2} & = 2 \\cdot \\frac{t(t-1)}{2} + 2 \\cdot \\frac{a(a-1)}{2} \\\\\n\\left(t^{2} + t a - t + a t + a^{2} - a\\right) & = 2\\left(t^{2} - t + a^{2} - a\\right) \\\\\na^{2} - 2 a t + t^{2} & = t + a \\\\\n(a-t)^{2} & = n\n\\end{aligned}\n$$\n\nAssim, conseguimos escrever $n$ como um quadrado perfeito.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26946,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEm uma classe com 35 estudantes, pesquisou-se sobre os gostos relativos a matemática e literatura e constatou-se que:\n- 7 homens gostam de matemática;\n- 6 homens gostam de literatura;\n- 5 homens e 8 mulheres disseram não gostar de ambos;\n- há 16 homens na classe;\n- 5 estudantes gostam de ambos; $\\mathrm{e}$\n- 11 estudantes gostam somente de matemática.\n\na) Quantos homens gostam de matemática e literatura?\nb) Quantas mulheres gostam apenas de literatura?",
    "options": [],
    "answer": "a) 2; b) 2",
    "solution": "Solution:\n\nSejam $H$ o conjunto dos homens e $U$ o conjunto total de pessoas, portanto $U-H$ é o conjunto das mulheres. Além deles, considere os conjuntos Mat e $L$ das pessoas que gostam de matemática e literatura, respectivamente. Se $x$ representa a quantidade de homens que gostam de matemática e literatura e $y$ as mulheres que gostam apenas de literatura, temos o seguinte diagrama:\n\n![](attached_image_1.png)\n\na) Como existem 16 homens na sala,\n$$\n\\begin{aligned}\n5+7-x+x+6-x & =16 \\\\\nx & =2\n\\end{aligned}\n$$\n\nb) Como a sala é composta por 35 estudantes\n$$\n\\begin{aligned}\n|H|+|U-H| & =35 \\\\\n16+4+x+5-x+y+8 & =35 \\\\\ny & =2\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26947,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDois quadrados de um tabuleiro $7 \\times 7$ são pintados de amarelo e o resto é pintado de verde. Dois esquemas de cores são equivalentes se um pode ser obtido do outro aplicando uma rotação no plano do tabuleiro. Quantos esquemas de cores não equivalentes podemos obter?",
    "options": [],
    "answer": "300",
    "solution": "Solution:\n\nComo o tabuleiro possui $7 \\cdot 7 = 49$ quadrados, existem $\\frac{49 \\cdot 48}{2} = 1176$ maneiras de escolhermos dois deles para receberem a cor amarela. Quando eles não são diametralmente opostos, podemos rotacioná-los em $90^{\\circ}$ três vezes e obter configurações que geram esquemas equivalentes. Para contarmos os pares de quadrados diametralmente opostos, escolha um quadrado distinto do centro de $49-1=48$ formas. O outro quadrado está completamente determinado por essa escolha. Entretanto, como a ordem de escolha entre eles é irrelevante, o total de pares é $\\frac{49-1}{2}=24$. Cada um desses 24 esquemas pode ser rotacionado uma vez por $180^{\\circ}$ gerando outro equivalente. Portanto, o número de esquemas é\n$$\n\\frac{1176-24}{4} + \\frac{24}{2} = 300\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26948,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nJoão possui um único pedaço de corrente com $n$ elos, cada um pesando $1~\\mathrm{g}$. Cada vez que um elo de algum pedaço de corrente é quebrado, obtemos 3 pedaços. Por exemplo, se um pedaço possui 9 elos e quebramos o quarto, passaremos a ter pedaços com as seguintes quantidades de elos: 3, 1 e 5. Um elo quebrado continua pesando $1~\\mathrm{g}$.\n\na) Se $n=8$, mostre um exemplo de como quebrarmos apenas 2 elos e obtermos pedaços que possam ser reunidos de modo a realizarmos qualquer peso de $1~\\mathrm{g}$ até $8~\\mathrm{g}$.\n\nb) Se $n=16$, mostre um exemplo de como quebrarmos apenas 2 elos e obtermos pedaços que possam ser reunidos de modo a realizarmos qualquer peso de $1~\\mathrm{g}$ até $16~\\mathrm{g}$.\n\nc) Se $n=63~\\mathrm{g}$, mostre um exemplo de como quebrarmos apenas 3 elos e obtermos pedaços que possam ser reunidos de modo a realizarmos qualquer peso de $1~\\mathrm{g}$ até $63~\\mathrm{g}$.\n\nd) Se $n=k \\cdot 2^{k}-1$, mostre um exemplo de como quebrarmos apenas $k-1$ elos e obtermos pedaços que possam ser reunidos de modo a realizarmos qualquer peso de $1~\\mathrm{g}$ até $ng$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\na) Basta quebrarmos o primeiro e o sexto elos obtendo os pedaços de comprimentos: 1, 4, 1 e 2. Para realizarmos qualquer peso de $1~\\mathrm{g}$ a $8~\\mathrm{g}$, considere as somas de pedaços:\n$$\n\\begin{aligned}\n& 1=1 \\\\\n& 2=2 \\\\\n& 3=1+2 \\\\\n& 4=4 \\\\\n& 5=1+4 \\\\\n& 6=2+4 \\\\\n& 7=1+2+4 \\\\\n& 8=1+2+4+1\n\\end{aligned}\n$$\n\nb) Basta quebrarmos o terceiro e o oitavo elos obtendo os pedaços de comprimentos: 2, 1, 4, 1 e 8. Usando as somas do item anterior, podemos obter os pesos de $1~\\mathrm{g}$ até $8~\\mathrm{g}$. Acrescentando a essas somas a parcela correspondente ao pedaço de 8 elos, poderemos obter as somas de $9~\\mathrm{g}$ até $16~\\mathrm{g}$.\n\nc) Basta quebrarmos o quinto, décimo quarto e trigésimo elos, obtendo os pedaços de comprimentos $4, 1, 8, 1, 16, 1$ e 32. Quando dividimos qualquer número $n$ do conjunto $\\{1,2,3, \\ldots, 63\\}$ por 4, obtemos um quociente $q$ e um resto $r \\in \\{0,1,2,3\\}$, ou seja, $n=4q+r$. Os pedaços de peso 1 são suficientes para a realização de qualquer valor de $r$. Os pedaços restantes são todos os quádruplos dos números do conjunto $\\{1,2,4,8\\}$. Como vimos nos primeiros itens, com pedaços de pesos 1, 2, 4 e 8 podemos formar qualquer inteiro de 1 até 15. Assim, com os seus quádruplos, podemos formar qualquer múltiplo de 4 até 60, que somados a $\\{0,1,2,3\\}$ (pesos para os pedaços de $1~\\mathrm{g}$), podemos obter qualquer peso de $1~\\mathrm{g}$ a $63~\\mathrm{g}$.\n\nd) Basta quebrar a corrente com $n$ elos para obter os pedaços indicados pelos parêntesis da soma abaixo:\n$$\n(k)+(1)+(2k)+(1)+(4k)+(1)+(8k)+(1)+\\ldots+(1)+\\left(k \\cdot 2^{k-1}\\right)=(k-1)+k\\left(2^{k}-1\\right)=k \\cdot 2^{k}-1\n$$\nObservação: Veja que não conseguiríamos com menos de $k-1$ pedaços, pois qualquer quantidade $r$ de quebras, com $r<k-1$, só poderia gerar $r$ parcelas de peso $1~\\mathrm{g}$ e no máximo $r+1$ parcelas de outros tamanhos. As parcelas não associadas às quebras geram no máximo $2^{r+1}$ somas. Combinando essas somas com o possível acréscimo de até $r$ elos de peso $1~\\mathrm{g}$, teremos no máximo $(r+1) \\cdot 2^{r+1} < (r+1) \\cdot 2^{k} < k \\cdot 2^{k}$. Daí o número máximo de somas é menor ou igual a $k \\cdot 2^{k}-2 < k \\cdot 2^{k}-1 = n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26949,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa figura a seguir, $ABC$ é um triângulo isósceles com $BA = BC$. O ponto $D$ está em seu interior de modo que $\\angle ABD = 13^{\\circ}$, $\\angle ADB = 150^{\\circ}$ e $\\angle ACD = 30^{\\circ}$. Além disso, $ADE$ é um triângulo equilátero. Determine o valor do ângulo $\\angle DBC$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "39°",
    "solution": "Solution:\n\nTrace os segmentos indicados na figura a seguir.\n\n![](attached_image_2.png)\n\nComo $\\triangle ADE$ é equilátero, segue que $\\angle ADE = 60^{\\circ}$. Daí, $\\angle BDE = 360^{\\circ} - 150^{\\circ} - 60^{\\circ} = 150^{\\circ}$. Além disso, de $AD = DE$, $BD = BD$ e $\\angle ADB = \\angle BDE$, segue pelo caso de congruência $LAL$ que $\\triangle ABD \\equiv \\triangle BDE$. Assim, $AB = BE$ e $\\angle ABE = 2 \\cdot \\angle ABD = 26^{\\circ}$.\n\nConsidere agora a circunferência $k$ que passa por $A$, $D$ e $C$. O ângulo $\\angle ACD$ está inscrito no arco correspondente à corda $AD$ e assim o centro dessa circunferência produz um ângulo central de $60^{\\circ}$ em relação a essa corda. Ou seja, $E$ é o centro de $k$. Portanto $AE = ED = EC$.\n\nFinalmente, de $BA = BC$, como os lados dos triângulos $ABE$ e $BCE$ são congruentes, o caso de congruência $LLL$ produz $\\triangle ABE \\equiv \\triangle BCE$ e isso nos dá $\\angle ABE = \\angle EBC$. Logo,\n\n$$\n\\angle DBC = \\angle DBE + \\angle EBC = 13^{\\circ} + 26^{\\circ} = 39^{\\circ}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26950,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO triângulo $ABC$ abaixo tem um ponto $D$ no seu lado $AB$, tal que $AB = CD$, $\\angle ABC = 100^{\\circ}$ e $\\angle DCB = 40^{\\circ}$.\n![](attached_image_1.png)\n\na) Qual a medida do ângulo $B\\hat{D}C$ ?\n\nb) Qual a medida do ângulo $A\\hat{C}D$ ?",
    "options": [],
    "answer": "a) 40°; b) 10°",
    "solution": "Solution:\n\na) Analisando a soma dos ângulos internos do triângulo $\\triangle BDC$, temos $100^{\\circ} + 40^{\\circ} + \\angle BDC = 180^{\\circ}$, ou seja, $\\angle BDC = 40^{\\circ}$.\n\nb) Em virtude do item anterior, $\\triangle BDC$ é isósceles de base $CD$ e então $BC = BD$. Construa o triângulo auxiliar $\\triangle BCE$, equilátero, e denote $\\angle ACD = \\alpha$.\n![](attached_image_2.png)\n\nPor construção, $\\angle DCE = 100^{\\circ}$ e $CE = BC$. De $AB = DC$ e $\\angle ABC = \\angle DCE$, podemos escrever $\\triangle ABC \\equiv \\triangle DCE$, pelo caso LAL. Daí, chegamos a\n$$\n\\angle ACB = \\angle DEC = 40^{\\circ} + \\alpha\n$$\nComo $\\triangle BCE$ é equilátero ($BC = CE = EB$); e $\\triangle BCD$ é isósceles ($BC = BD$), concluímos que $\\triangle BDE$ é isósceles com $BE = BD$ e base $DE$. Portanto, seus ângulos da base são iguais a $10^{\\circ}$. Por fim, como $\\angle BEC = 60^{\\circ} = 10^{\\circ} + 40^{\\circ} + \\alpha$, concluímos que $\\alpha = 10^{\\circ}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26951,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\na) Considere um primo $p$ que divide $10^{n}+1$ para algum $n$ inteiro positivo. Por exemplo, $p=7$ divide $10^{3}+1$. Analisando o período principal da representação decimal de $\\frac{1}{p}$, verifique que o número de vezes que o dígito $i$ aparece é igual ao número de vezes que o dígito $9-i$ aparece para cada $i \\in\\{0,1,2, \\ldots, 9\\}$.\n\nb) Considere um número primo $p$ que não divide 10 e suponha que o período da representação decimal de $\\frac{1}{p}$ seja $2k$. É sempre possível quebrarmos o período em dois números que somam $10^{k}-1$? Por exemplo, o período de $\\frac{1}{7}$ tem tamanho $6=2k$, pois é igual à 142857. Veja que $142+857=999=10^{3}-1=10^{k}-1$.\n\nc) Sendo\n$$\nx=\\frac{1}{1998}+\\frac{1}{19998}+\\frac{1}{199998}+\\ldots\n$$\nao escrevermos $2x$ como um número decimal, qual será o $59^{\\circ}$ algarismo após a vírgula?",
    "options": [],
    "answer": "2",
    "solution": "Solution:\na) Podemos escrever $10^{n}+1=p \\cdot a$ onde $a$ é um número com não mais que $n$ dígitos na base 10, digamos $a=a_{1} a_{2} \\ldots a_{n}$. Queremos dizer com isso que cada número $a_{i}$ é um dos dígitos de $a$. Mesmo que ele possua estritamente menos que $n$ dígitos, podemos colocar alguns $a_{i}$'s da esquerda como sendo 0. Temos\n$$\n\\begin{aligned}\n\\frac{1}{p} & =\\frac{a}{a \\cdot p} \\\\\n& =\\frac{a}{10^{n}+1} \\\\\n& =\\frac{a\\left(10^{n}-1\\right)}{10^{2 n}-1} \\\\\n& =\\frac{\\left[10^{n}(a-1)+\\left(10^{n}-1\\right)-(a-1)\\right]}{10^{2 n}-1}\n\\end{aligned}\n$$\nO número $10^{n}-1$ é constituído por $n$ números iguais a 9 e a diferença $\\left(10^{n}-1\\right)-(a-1)$ reduz cada um desses dígitos 9 por um dígito de $a$. Assim, a representação decimal do numerador é:\n$$\na_{1} a_{2} \\ldots a_{n-1}\\left(a_{n}-1\\right)\\left(9-a_{1}\\right)\\left(9-a_{2}\\right) \\ldots\\left(9-a_{n-1}\\right)\\left(10-a_{n}\\right)\n$$\nO número anterior representa o período da representação de $\\frac{1}{p}$ e cada dígito $i$ pode ser pareado com um outro dígito da forma $9-i$. Assim, as quantidades de aparições de tais dígitos são iguais. No exemplo do enunciado, o período de $1/7$ é 142857 e temos os seguintes pareamentos:\n$$\n\\begin{aligned}\n& 1 \\rightarrow 8 \\\\\n& 4 \\rightarrow 5 \\\\\n& 2 \\rightarrow 7\n\\end{aligned}\n$$\n\nb) Como $10^{2k}-1=\\left(10^{k}-1\\right)\\left(10^{k}+1\\right)$ e $p$ é primo, um dentre $10^{k}-1$ e $10^{k}+1$ é múltiplo de $p$. Não podemos ter $10^{k}-1$ múltiplo de $p$, pois caso contrário poderíamos escrever $\\frac{1}{p}=\\frac{\\left(10^{k}-1\\right) / p}{10^{k}-1}$ e obteríamos uma dízima periódica com período menor do que $2k$. Sendo assim, $p$ divide $10^{k}+1$ e podemos repetir a solução do item anterior para concluir que o período da representação decimal de $1/p$ é da forma:\n$$\na_{1} a_{2} \\ldots a_{k-1}\\left(a_{k}-1\\right)\\left(9-a_{1}\\right)\\left(9-a_{2}\\right) \\ldots\\left(9-a_{k-1}\\right)\\left(10-a_{k}\\right)\n$$\nSomando o número formado pelos $k$ primeiros dígitos com o número formado pelos $k$ últimos, obtemos $\\underbrace{99 \\ldots 9}_{k \\text{ vezes }}=10^{k}-1$.\n\nc) Perceba que\n$$\n\\begin{aligned}\n2x & =\\frac{1}{999}+\\frac{1}{9999}+\\frac{1}{99999}+\\ldots \\\\\n& =0, \\overline{001}+0, \\overline{0001}+0, \\overline{00001}+\\ldots\n\\end{aligned}\n$$\nNa primeira parcela, o 1 aparece nas ordens múltiplas de três, na segunda, nas múltiplas de quatro, na terceira, nas múltiplas de cinco, e assim sucessivamente. Ao somarmos as duas primeiras dízimas $0, \\overline{001}$ e $0, \\overline{0001}$, teremos a dízima periódica\n$$\n0,001101011002 \\ldots\n$$\nNas posições múltiplas de apenas um dos números do conjunto $\\{3,4\\}$, teremos o número 1. Nas que são simultaneamente múltiplas de 3 e 4, teremos 2 e nas demais o número 0. Além disso, a representação será periódica com período de tamanho 12. Para descobrirmos os algarismos em uma dada posição na soma precisamos estudar quantos divisores maiores que 2 a posição possui. Dentre os números naturais menores que 60, nenhum possui mais de 8 divisores inteiros positivos maiores que 2. Assim, na soma de dízimas:\n$$\n0, \\overline{001}+0, \\overline{0001}+0, \\overline{00001}+\\ldots+\\overline{0,00 \\ldots 01}\n$$\nnão ocorrerá a operação de \"vai um\" nas primeiras 59 casas e o algarismo na posição 59 será 1, pois 59 possui apenas um divisor maior que 2. As outras posições terão números menores ou iguais a 8, pois 48 é o número natural menor que 60 que mais possui divisores positivos e apenas 8 deles são maiores que 2. Como 60 possui 10 divisores positivos maiores que 2, a saber, $3,4,5,6,10,12,15,20,30$ e 60, ocorrerá a operação de \"vai um\" ao somarmos essa dízima com as anteriores e assim a casa na posição 59 terá o acréscimo de uma unidade e passará a ter o algarismo 2. Resta verificarmos qual a contribuição das demais dízimas nas posições de ordem maior que 60, levando em conta a operação de \"vai um\". A demais casas dessas dízimas e das restantes que ainda não foram contabilizadas, poderão contribuir através da operação de \"vai um\" no algoritmo da soma com um número estimado por\n$$\n\\begin{aligned}\n& \\frac{60}{10^{61}}+0, \\underbrace{\\overline{00 \\ldots 01}}_{61 \\text{ dígitos }}+0, \\underbrace{\\overline{00 \\ldots 01}}_{62 \\text{ dígitos }}+\\ldots< \\\\\n& \\frac{60}{10^{61}}+\\frac{1}{9 \\cdot 10^{60}}+\\frac{1}{9 \\cdot 10^{61}}+\\ldots= \\\\\n& \\frac{60}{10^{61}}+\\frac{1}{81 \\cdot 10^{59}}< \\\\\n& \\frac{1}{10^{59}}\n\\end{aligned}\n$$\nPortanto, não precisaremos contabilizar outras contribuições na posição 59 após a vírgula e o seu algarismo na soma é 2.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26952,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nO retângulo $ABCD$ de medidas $AB=240~\\mathrm{cm}$ e $BC=288~\\mathrm{cm}$ representa um papel que será dobrado pelo segmento $EF$, onde $E$ pertence a $AD$ e $F$ pertence a $BC$, de modo que o ponto $C$ ficará sobre o ponto médio de $AB$.\n![](attached_image_1.png)\n\na) Qual o comprimento de $CC^{\\prime}$ ?\n\nb) Qual o comprimento de $EF$ ?",
    "options": [],
    "answer": "CC' = 312 cm, EF = 260 cm",
    "solution": "Solution:\nA operação de dobrar a folha de papel equivale a produzir uma reflexão ao longo da dobra. Para ver isso, perceba que segmentos correspondentes possuem o mesmo comprimento, por exemplo, $C^{\\prime}F = CF$ e $C^{\\prime}E = CE$. Assim, $EF$ divide o segmento $CC^{\\prime}$ ao meio e é ortogonal a ele.\n![](attached_image_2.png)\n\na) Como o $\\triangle CBC^{\\prime}$ é retângulo em $B$, pelo Teorema de Pitágoras, temos $CC^{\\prime} = \\sqrt{120^{2} + 288^{2}} = 312~\\mathrm{m}$.\n\nb) Como $EF$ é ortogonal a $CC^{\\prime}$, segue que\n$$\n\\angle CC^{\\prime}B = 90^{\\circ} - \\angle C^{\\prime}CB = \\angle EFC\n$$\nTomando $G \\in BC$ tal que $EG \\perp BC$, podemos concluir que o $\\triangle EFG$ é semelhante ao $\\triangle CC^{\\prime}B$. Assim\n$$\n\\begin{aligned}\n\\frac{EF}{EG} & = \\frac{CC^{\\prime}}{BC} \\\\\n\\frac{EF}{240} & = \\frac{312}{288} \\\\\nEF & = 260~\\mathrm{m}\n\\end{aligned}\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26953,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAna multiplica dois números inteiros positivos cuja diferença é $202$, mas comete um erro e obtém um número $1000$ unidades menor que o correto. Ao dividir o resultado de Ana pelo menor dos números que deveria multiplicar, o quociente é $288$ e o resto é $67$. Quais os dois números que Ana multiplicou?",
    "options": [],
    "answer": "97 and 299",
    "solution": "Solution:\n\nSejam $x$ e $x+202$ os inteiros que Ana multiplica. A partir do resultado da divisão de Ana, podemos escrever:\n$$\nx(x+202)-1000=288x+67\n$$\nDaí,\n$$\nx^2-86x-1067=0\n$$\nResolvendo a equação anterior, encontramos como raízes $x=-11$ e $x=97$. Como $x$ é um inteiro positivo, devemos ter $x=97$. O outro número multiplicado por Ana é $97+202=299$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26954,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\na) Quantos números de quatro algarismos têm soma de seus algarismos par?\nb) Um número com dois dígitos distintos e não nulos é chamado de bonito se o dígito das dezenas é maior do que o dígito das unidades. Quantos números bonitos existem?\nc) Quantos números pares de quatro dígitos podemos formar utilizando os algarismos $0,1,2,3,4,5$ sem utilizar o mesmo algarismo duas vezes?\nd) Qual a média de todos os números de 5 algarismos que podem ser formados usando cada um dos dígitos $1,3,5,7$ e $8$ exatamente uma vez?",
    "options": [],
    "answer": "a) 4500, b) 45, c) 156, d) 53332.8",
    "solution": "Solution:\na) Para que a soma seja par, devemos utilizar quatro, dois ou nenhum algarismo par. Eles podem ocupar as seguintes ordens: unidades de milhar, centenas, dezenas ou unidades.\n\ni) No primeiro caso, o zero não pode ocupar a maior ordem e as outras podem ser ocupadas por quaisquer dos 5 números pares, resultando num total de\n$$\n4 \\times 5 \\times 5 \\times 5 = 500 \\text{ números. }\n$$\n\nii) No segundo caso, há duas formações, a saber:\n- Quando o primeiro dígito é ímpar, temos 5 opções para a maior ordem. Daí, ficamos com 3 opções para posicionar o outro algarismo ímpar. Separada a ordem desse algarismo, existem 5 opções de ímpares a serem colocados no local e, nas posições restantes, quaisquer um dos 5 pares podem ser postos. Isso gera\n$$\n5 \\times 3 \\times 5 \\times 5 \\times 5 = 1875 \\text{ números. }\n$$\n- Quando o primeiro dígito é par, mais uma vez, excluímos a possibilidade do zero ocupar a maior ordem. Existem três escolhas para a posição do outro par e as 5 opções para o seu valor. As duas ordens restantes podem ser ocupadas por qualquer um dos 5 ímpares, totalizando\n$$\n4 \\times 3 \\times 5 \\times 5 \\times 5 = 1500 \\text{ números. }\n$$\n\nNo último caso, quando nenhum algarismo é par, temos\n$$\n5 \\times 5 \\times 5 \\times 5 = 625 \\text{ possibilidades. }\n$$\n\nDaí, o total de número é $500 + 1875 + 1500 + 625 = 4500$.\n\nb) Começando pelo 9, temos o conjunto de números bonitos $\\{98,97, \\ldots, 90\\}$, que possui 9 números. Iniciando pelo 8, temos $\\{87,86, \\ldots, 80\\}$, que possui 8 números, e assim por diante até o conjunto dos que começam pelo 1, que é $\\{10\\}$. Totalizando assim $9 + 8 + 7 + \\cdots + 2 + 1 = 45$ números.\n\nc) Existem apenas três algarismos pares e a contagem pode ser dividida em dois casos: se o último algarismo for zero, temos $5 \\times 4 \\times 3 = 60$ possibilidades; se o último dígito não for zero, temos $4 \\times 4 \\times 3 \\times 2 = 96$. Portanto, o total de números é $60 + 96 = 156$.\n\nd) Fixado um algarismo da lista, ele irá aparecer em uma dada ordem em outros $4 \\cdot 3 \\cdot 2 \\cdot 1 = 24$ números. Cada vez que um número aparece em uma dada ordem, a sua contribuição na soma total é igual ao seu produto pela potência de 10 associada àquela ordem. Por exemplo, nos números 13578 e 13857, a contribuição do 5 é $5 \\cdot 10^{2}$ e $5 \\cdot 10^{1}$, respectivamente. Como todo número aparece igual número de vezes em cada ordem, a contribuição de cada algarismo na soma total é igual ao seu produto por $24 \\cdot \\left(10^{4} + 10^{3} + 10^{2} + 10^{1} + 10^{0}\\right) = 24 \\cdot 11111$. Além disso, a quantidade total de números é $5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1 = 120$. Portanto, a média é\n$$\n\\frac{(1+3+5+7+8) \\cdot 24 \\cdot 11111}{120} = 53332,8\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26955,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nJosé arrancou algumas folhas consecutivas de um livro com páginas numeradas com inteiros consecutivos e escritos em ambos os lados de cada folha. A soma dos números das páginas arrancadas é $344$.\n\na) Determine a fatoração em números primos do número $344$.\n\nb) Encontre a soma do primeiro e do último número dentre os que foram escritos nas páginas arrancadas.\n\nc) Qual a quantidade de páginas arrancadas?",
    "options": [],
    "answer": "a) 344 = 2^3 * 43; b) 43; c) 16",
    "solution": "Solution:\n\na) Como $44$ é múltiplo de $4$, então $344 = 4 \\cdot 86 = 2^{3} \\cdot 43$.\n\nb) Se $y$ é a quantidade de folhas arrancadas, sendo $2y$ a quantidade de páginas, e $x+1$ é o primeiro número que aparece nelas, temos\n$$\n\\begin{aligned}\n(x+1)+(x+2)+(x+3)+\\ldots+(x+2y) & = 344 \\\\\n2xy + (1+2+\\ldots+2y) & = 2^{3} \\cdot 43\n\\end{aligned}\n$$\nLembrando que\n$$\n1+2+3+\\ldots+n = \\frac{n(n+1)}{2}\n$$\nsegue que\n$$\n\\begin{aligned}\n2^{3} \\cdot 43 & = 2xy + \\frac{2y(2y+1)}{2} \\\\\n& = y(2x + 2y + 1)\n\\end{aligned}\n$$\nComo $2x + 2y + 1$, que é a soma dos números da primeira e última página arrancadas, é um divisor ímpar de $2^{3} \\cdot 43$ maior que $1$, segue que $2x + 2y + 1 = 43$.\n\nc) Como $2x + 2y + 1 = 43$, da equação do item anterior, chegamos a $y = 2^{3} = 8$. Portanto, o número de páginas arrancadas é $2y = 16$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26956,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\na) Encontre a soma dos dígitos de $99 \\cdot 33$.\nb) Usando $999999=10^{6}-1$, encontre a soma dos dígitos de $999999 \\cdot 333333$.\nc) Encontre a soma dos dígitos de $999\\ldots999 \\cdot 333\\ldots333$, em que cada dígito nos fatores anteriores aparece 2018 vezes.",
    "options": [],
    "answer": "a) 18; b) 54; c) 18162",
    "solution": "Solution:\na) Como $99 \\cdot 33 = 3267$, a soma procurada é $3+2+6+7=18$.\n\nb) Temos que:\n$$\n\\begin{aligned}\n999999 \\cdot 333333 &= (10^{6}-1) \\cdot 333333 \\\\\n&= 333333 \\cdot 10^{6} - 333333 \\\\\n&= 333332666667\n\\end{aligned}\n$$\nNote que os 12 algarismos podem ser agrupados em 6 pares de números com soma 9: $2+7=9$, $3+6=9$. Assim, a soma procurada é $9 \\cdot 6=54$.\n\nc) Repetindo o argumento do item anterior, temos\n$$\n\\begin{aligned}\n\\underbrace{99\\ldots99}_{2018 \\text{ vezes}} \\cdot \\underbrace{33\\ldots33}_{2018 \\text{ vezes}} &= (10^{2018}-1) \\cdot \\underbrace{33\\ldots33}_{2018 \\text{ vezes}} \\\\\n&= \\underbrace{33\\ldots33}_{2018 \\text{ vezes}} \\cdot 10^{2018} - \\underbrace{33\\ldots33}_{2018 \\text{ vezes}} \\\\\n&= \\underbrace{33\\ldots33}_{2017 \\text{ vezes}}2\\underbrace{66\\ldots667}_{2017 \\text{ vezes}}\n\\end{aligned}\n$$\nOs algarismos do número anterior podem ser agrupados em 2018 pares com soma 9. Assim, a soma dos dígitos é $2018 \\cdot 9=18162$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26957,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\na) Um campeonato terá 7 competidores. Cada um deles jogará exatamente um jogo contra todos os outros. Qual o total de jogos do campeonato?\n\nb) Um campeonato terá $n$ competidores e cada um deles jogará exatamente um jogo contra todos os outros. Verifique que o total de jogos é $\\frac{n \\cdot (n-1)}{2}$.\n\nc) Um competidor $A$ é chamado de mestre do campeonato se todo outro jogador ou perdeu para $A$ ou perdeu para alguém que perdeu para $A$. Verifique que todo campeonato em que cada jogador joga exatamente uma vez contra todos os outros e que não existem empates, sempre admite um mestre.\n\nd) João e Maria estavam num torneio em que cada atleta deveria jogar contra o outro exatamente uma vez. No entanto, Maria teve um problema de saúde e abandonou o torneio após disputar 10 jogos e João precisou sair após o seu primeiro jogo. Ao final do dia, foi contabilizado um total de 55 jogos. O jogo que João disputou foi contra Maria?",
    "options": [],
    "answer": "a) 21. b) n(n−1)/2. c) Every such tournament has a master; for example, the player with the most wins is a master. d) Yes, João’s only game was against Maria; in fact there were 10 other competitors.",
    "solution": "Solution:\na) Sejam $\\{a, b, c, d, e, f, g\\}$ os jogadores. O jogador $a$ participará dos jogos associados aos 6 pares: $(a, b), (a, c), (a, d), (a, e), (a, f)$ e $(a, g)$. A princípio, como já contamos o jogo associado a $(a, b)$, para contar as partidas do jogador $b$, basta contabilizarmos as partidas associadas aos 6 pares $(b, c), (b, d), (b, e), (b, f)$ e $(b, g)$. Ao descontarmos os jogos já contabilizados, repetindo essa análise para os próximos jogadores, obteremos as seguintes quantidades de partidas: 4, 3, 2 e 1. Portanto, o total de jogos é\n$$\n6+5+4+3+2+1=\\frac{6 \\cdot 7}{2}=21\n$$\n\nb) Repetindo a contagem do item anterior, se $\\{j_{1}, j_{2}, \\ldots, j_{n}\\}$ são os jogadores, as partidas que $j_{1}$ participará estão associadas aos $n-1$ pares:\n$$\n\\left(j_{1}, j_{2}\\right), \\left(j_{1}, j_{3}\\right), \\left(j_{1}, j_{4}\\right), \\ldots, \\left(j_{1}, j_{n}\\right)\n$$\nEm geral, após listados sem repetição os pares dos jogadores $j_{1}, j_{2}, \\ldots, j_{k-1}$, precisaremos contar para o jogador $j_{k}$ os $n-k-1$ jogos associados aos pares:\n$$\n\\left(j_{k}, j_{k+1}\\right), \\left(j_{k}, j_{k+2}\\right), \\left(j_{k}, j_{k+3}\\right), \\ldots, \\left(j_{k}, j_{n}\\right)\n$$\nPortanto, o total de jogos é\n$$\n1+2+3+\\ldots+n=\\frac{n(n-1)}{2}\n$$\nVeja que essa é exatamente a quantidade de pares $\\left(j_{l}, j_{m}\\right)$ com $l<m$ e poderia ser obtida com a seguinte contagem alternativa: Existem $n$ opções para o primeiro elemento do par e $n-1$ para o segundo elemento produzindo $n(n-1)$ pares da forma $\\left(j_{a}, j_{b}\\right)$ com $a \\neq b$. Entretanto, os dois pares $\\left(j_{a}, j_{b}\\right)$ e $\\left(j_{b}, j_{a}\\right)$ foram contados e apenas um deles possui o jogador de índice menor na primeira posição. Assim, basta dividir a contagem anterior por 2 obtendo $\\frac{n(n-1)}{2}$.\n\nc) Seja $A$ o jogador com mais vitórias do campeonato. Considere qualquer outro competidor $B$. podem ocorrer duas situações:\n1. $A$ ganhou de $B$.\n2. $A$ perdeu de $B$. Como $B$ não possui mais vitórias que $A$, em algum dos jogos de $B$ contra as pessoas que perderam de $A$, devemos encontrar uma partida em que $B$ foi o perdedor.\nDiante dessas condições, podemos afirmar que $A$ é um competidor mestre.\n\nd) Considere que havia $C_{1}, C_{2}, \\ldots, C_{n}$, João e Maria no torneio. Agora, perceba que tivemos $\\frac{n \\cdot (n-1)}{2}$ jogos entre os $C'$s. Se João não jogou contra Maria, a soma da quantidade de todos os jogos fica\n$$\n\\begin{aligned}\n\\frac{n \\cdot (n-1)}{2}+10+1 & =55 \\\\\n\\frac{n \\cdot (n-1)}{2} & =44 \\\\\nn \\cdot (n-1) & =88\n\\end{aligned}\n$$\nPodemos escrever 88 como produto de inteiros positivos das seguintes formas: $1 \\cdot 88$, $2 \\cdot 44, 4 \\cdot 22$ e $8 \\cdot 11$. Daí $n \\cdot (n-1)=88$, com $n$ inteiro positivo, não tem solução. Assim, conclui-se que João deve ter jogado contra Maria. Se João tiver jogado contra Maria, o total de jogos fica\n$$\n\\begin{aligned}\n\\frac{n \\cdot (n-1)}{2}+10 & =55 \\\\\n\\frac{n \\cdot (n-1)}{2} & =45 \\\\\nn \\cdot (n-1) & =90\n\\end{aligned}\n$$\nComo $90=10 \\cdot 9$, uma solução é $n=10$. A outra solução da equação do segundo grau $n^{2}-n-90=0$ é negativa e, portanto, não convém. Um exemplo de torneio satisfazendo às condições do enunciado com $10+2=12$ jogadores pode ser simbolizada pela seguinte lista de jogos:\n$\\left(j_{1}, j_{2}\\right), \\left(j_{2}, j_{3}\\right), \\left(j_{2}, j_{4}\\right), \\left(j_{2}, j_{5}\\right), \\left(j_{2}, j_{6}\\right), \\left(j_{2}, j_{7}\\right), \\left(j_{2}, j_{8}\\right), \\left(j_{2}, j_{9}\\right), \\left(j_{2}, j_{10}\\right), \\left(j_{2}, j_{11}\\right), \\left(j_{a}, j_{b}\\right)$, com $a<b$ e $a, b \\in \\{3,4,5, \\ldots, 12\\}$, em que $j_{1}$ representa João e $j_{2}$ Maria.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26958,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa figura a seguir, todos os retângulos são iguais e possuem o perímetro de $8~\\mathrm{cm}$. Qual o perímetro total da figura?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "32 cm",
    "solution": "Solution:\n\nPodemos deslizar os blocos e formar uma nova figura com o mesmo perímetro da anterior. Se o lado menor do bloco mede $a$ e o maior mede $b$ então $2a + 2b = 8~\\mathrm{cm}$. No perímetro da nova figura, temos 8 segmentos de tamanho $a$ e 8 de tamanho $b$. Assim, o seu perímetro é $8a + 8b = 4(2a + 2b) = 32~\\mathrm{cm}$.\n\n![](attached_image_2.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26959,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa figura a seguir, $ABC$ é um triângulo equilátero, $D$, $E$ e $F$ são seus pontos médios e $P$ é o seu centro. Qual a fração que a área sombreada representa do total do triângulo $ABC$?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "5/24",
    "solution": "Solution:\n\nOs pontos $I$, $G$ e $H$ são as interseções dos segmentos $DE$, $EF$ e $DF$ com os segmentos $AP$, $BP$ e $CP$. Pela simetria da figura, as áreas dos quadriláteros $EIPG$, $DIPH$ e $FGPH$ medem o mesmo valor $x\\ \\mathrm{cm}^2$. Além disso, pelo mesmo argumento, as áreas dos triângulos $AEI$ e $ADI$ também medem um mesmo valor $y\\ \\mathrm{cm}^2$. Os triângulos $ADE$, $DEF$, $DBF$ e $CEF$ possuem a mesma área $S\\ \\mathrm{cm}^2$ e assim $x = S/3$ e $y = S/2$. A fração procurada é\n$$\n\\begin{aligned}\n\\frac{A_{AEGP}}{A_{ABC}} & = \\frac{y + x}{4S} \\\\\n& = \\frac{S/2 + S/3}{4S} \\\\\n& = \\frac{5}{24}\n\\end{aligned}\n$$\n\n![](attached_image_2.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26960,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nConsidere o trapézio $ABCD$ de bases $BC$ e $AD$ de modo que $AB = BC = CD = 5$ e $AD = 10$. Seja $E$ o ponto de interseção das diagonais $AC$ e $BD$. A reta perpendicular a $AC$ traçada por $E$ intersecta o prolongamento de $AB$ em $F$ e a base $AD$ em $H$.\n![](attached_image_1.png)\na) Determine o comprimento de $AH$.\nb) Determine o comprimento de $AE$.\nc) Encontre a área do quadrilátero $A F C H$.",
    "options": [],
    "answer": "AH = 20/3, AE = 10√3/3, area(AFCH) = 50√3/3",
    "solution": "Solution:\na) Inicialmente, verificaremos que $ABCD$ é metade de um hexágono regular. Seja $M$ o ponto médio de $AD$. Como $BC$ e $AM$ são iguais e paralelos, $ABCM$ é um paralelogramo. Além disso, como $AM = AB = BC$, segue que $CM = AB = CD = DM$. Assim, $CDM$ é um triângulo equilátero. De modo semelhante, podemos obter $BM = CM = CD$. Daí os triângulos $ABM$, $BCM$ e $CDM$ são congruentes e a circunferência de centro $M$ e raio $CM$ passa por $A$, $B$, $C$ e $D$. Logo\n$$\n\\angle BAD = 60^\\circ \\text{ e } \\angle ABC = 180^\\circ - \\angle BAM = 120^\\circ\n$$\n![](attached_image_2.png)\nDe $AB = BC$ segue que $\\angle BAC = \\angle BCA = \\angle CAD$. Assim, $\\angle BAC = \\angle CAD = 30^\\circ$. Como $FH \\perp AE$, temos $\\triangle AFH$ equilátero. Além disso, $\\angle ACD = \\angle DBA = 90^\\circ$. Portanto, $EH \\parallel CD$. Como os triângulos $\\triangle BEC$ e $\\triangle AED$ são semelhantes, temos\n$$\n\\frac{AH}{HD} = \\frac{AE}{EC} = \\frac{10}{5} = 2\n$$\nPor conseguinte $HD = 10/3$ e $AH = 20/3$.\n\nb) No triângulo retângulo $AEH$, temos\n$$\nAE = AH \\cdot \\cos 30^\\circ = \\frac{20}{3} \\cdot \\frac{\\sqrt{3}}{2} = \\frac{10 \\sqrt{3}}{3}\n$$\n\nc) Como as diagonais de $AFCH$ são perpendiculares, temos $[AFCH] = \\frac{AC \\cdot FH}{2}$. Pelo Teorema de Pitágoras, segue que\n$$\nAC = \\sqrt{AD^2 - CD^2} = \\sqrt{100 - 25} = 5 \\sqrt{3}\n$$\nOutra forma de obter o comprimento desse segmento é calcular\n$$\nAC = AD \\cdot \\cos 30^\\circ = 5 \\sqrt{3}\n$$\nDaí,\n$$\n\\begin{aligned}\nA_{AFCH} & = \\frac{AC \\cdot FH}{2} \\\\\n& = \\frac{AC \\cdot AH}{2} \\\\\n& = \\frac{5 \\sqrt{3} \\cdot 20 / 3}{2} \\\\\n& = \\frac{50 \\sqrt{3}}{3}\n\\end{aligned}\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26961,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUma máquina de cortar grama mais eficiente está sendo desenvolvida. Para isso, em um vértice de um quadrado de grama, de lado $m$, prende-se a ponta de uma haste metálica de comprimento $p$ e na outra ponta da haste prende-se um triângulo equilátero, por um de seus vértices, de lado $l$, sendo $p+l<m$. O triângulo gira muito rápido ao redor do encaixe com a haste e a haste gira muito devagar ao redor do encaixe no vértice do quadrado, ou seja, toda grama abaixo do triângulo é cortada, formando parte de uma coroa circular. Qual a área cortada?",
    "options": [],
    "answer": "π p l",
    "solution": "Solution:\n\nQuando o triângulo gira ao redor de um de seus vértices, seu alcance é limitado por uma circunferência de raio $l$. Assim, o raio do maior arco da área de corte é $(p+l)$, enquanto que o raio do menor arco da área de corte é $(p-l)$. Portanto, a área de corte é\n$$\n\\frac{\\pi(p+l)^2-\\pi(p-l)^2}{4}=\\frac{\\pi}{4} \\cdot 4 p l=p l \\pi\n$$\n\n![](attached_image_1.png)",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26962,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO frasco de um perfume tem formato cilíndrico e uma embalagem em forma de flor deve ser construída para acondicioná-lo. Para a confecção desta embalagem será utilizada uma folha quadrada de lado $n$, quatro arcos de circunferência com centros nos vértices do quadrado e raio medindo $n$ e uma circunferência tangente a estes arcos.\n![](attached_image_1.png)\n\na) Qual a medida do raio da circunferência?\n\nb) Qual a medida da área mais clara da figura?",
    "options": [],
    "answer": "a) r = n*(2 - sqrt(2))/2. b) Area = n^2*(4 - 4*pi/3 - sqrt(3)).",
    "solution": "Solution:\n\na) O raio do arco mede $n$ e a diagonal do quadrado mede $n \\sqrt{2}$, então o raio da circunferência é $\\frac{n-(n \\sqrt{2}-n)}{2}=\\frac{2 n-n \\sqrt{2}}{2}$.\n\nb) Dividindo o quadrado como na figura, temos que a soma das áreas $A, B$ e $C$ é $\\frac{n^{2}}{2}$. Como $B=\\frac{\\pi \\cdot n^{2}}{6}$ (setor circular de $30^{\\circ}$), $C=\\frac{n^{2} \\sqrt{3}}{4} \\cdot \\frac{1}{2}=\\frac{n^{2} \\sqrt{3}}{8}$ (metade da área de um triângulo equilátero), então $A=\\frac{n^{2}}{2}-\\frac{\\pi n^{2}}{6}-\\frac{n^{2} \\sqrt{3}}{8}$. Assim, a área mais clara é $8 A= n^{2}\\left(4-\\frac{4 \\pi}{3}-\\sqrt{3}\\right)$.\n\n![](attached_image_2.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26963,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNo quadrado mágico a seguir, todo inteiro de $1$ a $25$ pode ser colocado nos quadradinhos de modo que as somas em toda linha e coluna, bem como nas diagonais, é a mesma. Dado que o número no centro do quadrado é $18$, qual é o valor da soma dos números escritos nos quadradinhos sombreados?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "119",
    "solution": "Solution:\n\nComo a soma de todos os números do quadrado mágico é $1+2+\\ldots+25=325$, a soma dos números em cada linha, coluna ou diagonal é $325 / 5=65$. Ao somar todos os números das duas diagonais, da linha do meio e da coluna do meio, estamos somando todos os números que não estão sombreados e assim obtemos $4 \\cdot 65=260$. Entretanto, o quadrado central foi somado quatro vezes e podemos concluir que a soma de todos os números não sombreados é $260-3 \\cdot 18=206$. Portanto, a soma dos números nos quadrados sombreados é $325-206=119$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26964,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSe $x$, $y$ e $z$ são números reais positivos e\n$$\n\\frac{x y}{x+y}=a, \\quad \\frac{x z}{x+z}=b, \\quad \\text{e} \\quad \\frac{y z}{y+z}=c\n$$\na) Verifique que\n$$\nx=\\frac{a y}{y-a}\n$$\nb) Verifique que\n$$\nx=\\frac{2 a b c}{a c+b c-a b}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\na) Da primeira equação do enunciado, temos\n$$\n\\begin{aligned}\n\\frac{x y}{x+y} & =a \\\\\nx y & =a x+a y \\\\\nx(y-a) & =a y \\\\\nx & =\\frac{a y}{y-a} .\n\\end{aligned}\n$$\nNote que a última divisão por $y-a$ é possível, pois se $y-a=0$ segue que $a y=0$ e isso é ímpossível, dado que $a$ e $x$ são positivos.\n\nb) De forma semelhante ao item anterior, podemos concluir que\n$$\nx=\\frac{z b}{z-b} \\quad \\text{e} \\quad y=\\frac{z c}{z-c}\n$$\nIsso nos permite eliminar $y$ na identidade daquele item:\n$$\n\\begin{aligned}\nx & =\\frac{a\\left(\\frac{z c}{z-c}\\right)}{\\frac{z c}{z-c}-a} \\\\\n& =\\frac{a z c}{z-c} \\cdot \\frac{z-c}{z c-a z+a c} \\\\\n& =\\frac{a z c}{z c-a z+a c}\n\\end{aligned}\n$$\nPodemos escrever $z$ em função de $b$ e $x$ a partir de\n$$\n\\begin{aligned}\nx & =\\frac{z b}{z-b} \\\\\nz x-b x & =z b \\\\\nz & =\\frac{b x}{x-b}\n\\end{aligned}\n$$\nFinalmente, podemos escrever\n$$\n\\begin{aligned}\nx & =\\frac{a z c}{z c-a z+a c} \\\\\n& =\\frac{\\frac{a b x c}{x-b}}{\\frac{b c x}{x-b}-\\frac{a b x}{x-b}+\\frac{a c(x-b)}{x-b}} \\\\\n& =\\frac{a b c x}{x-b} \\cdot \\frac{x-b}{b c x-a b x+a c(x-b)} \\\\\n& =\\frac{a b c x}{b c x-a b x+a c(x-b)}\n\\end{aligned}\n$$\nOu seja,\n$$\n\\begin{aligned}\nb c x-a b x+a c(x-b) & =a b c \\\\\nx(b c-a b+a c) & =2 a b c \\\\\nx & =\\frac{2 a b c}{a c+b c-a b}\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26965,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm quadrado $3 \\times 3$ está preenchido com os números $a, b, c, d, e, f, g, h$ e $i$ da seguinte forma:\n\n| $c$ | $f$ | $i$ |\n| :--- | :--- | :--- |\n| $b$ | $e$ | $h$ |\n| $a$ | $d$ | $g$ |\n\nSabemos que ele é um quadrado mágico, isto é, existe um valor $S$ que é igual as somas dos números em cada linha, coluna e cada uma das duas diagonais. Verifique que:\n\na) $2(a+c+g+i)=b+d+f+h+4 e$.\n\nb) $S=3 e$.\n\nc) $a c+c i+a g+g i=e(b+d+f+h)$.\n\nd) $2\\left(a^{2}+c^{2}+g^{2}+i^{2}\\right)=b^{2}+d^{2}+f^{2}+h^{2}+4 e^{2}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na) Somando os números das linhas e colunas que não contêm o quadradinho central, obtemos:\n$$\n(a+b+c)+(a+d+g)+(c+f+i)+(g+h+i)=4 S\n$$\nPor outro lado, somando a linha e a coluna que contêm o quadrado central, temos:\n$$\n2 S=(b+e+h)+(d+e+f)\n$$\nComparando as duas equações, podemos concluir que\n$$\n(a+b+c)+(a+d+g)+(c+f+i)+(g+h+i)=4 S=2(b+e+h)+2(d+e+f)\n$$\nFinalmente, subtraindo de cada membro o número $(b+d+f+h)$, ficamos com\n$$\n2(a+c+g+i)=b+d+f+h+4 e\n$$\n\nb) Considerando todas as linhas, colunas e diagonais que contêm o quadrado central, temos:\n$$\na+i=c+g=b+h=d+f=S-e\n$$\nUsando a equação anterior, podemos concluir que\n$$\n\\begin{aligned}\n2(a+c+g+i) & =b+d+f+h+4 e \\\\\n2((a+i)+(c+g)) & =(b+h)+(d+f)+4 e \\\\\n2(S-e+S-e) & =S-e+S-e+4 e\n\\end{aligned}\n$$\nDe $4 S-4 e=2 S+2 e$, obtemos $S=3 e$.\n\nc) Pelo item anterior, $S-e=2 e$ e daí\n$$\n\\begin{aligned}\na c+c i+a g+g i & =(a+i)(c+g) \\\\\n& =(S-e)(S-e) \\\\\n& =2 e(S-e) \\\\\n& =e(2 S-2 e)\n\\end{aligned}\n$$\nComo $S-e=b+h=d+f$, segue que\n$$\n\\begin{aligned}\nb+d+f+h & =(b+h)+(d+f) \\\\\n& =(S-e)+(S-e) \\\\\n& =2 S-2 e\n\\end{aligned}\n$$\nSubstituindo essa identidade na relação anterior, chegamos a\n$$\na c+c i+a g+g i=e(2 S-2 e)=e(b+d+f+h)\n$$\n\nd) Note agora que $a+c=S-b=h+e, c+i=S-f=d+e, g+i=S-h=b+e$ e $a+g=S-d=f+e$. Daí\n$$\n\\begin{aligned}\n(a+c)^{2}+(c+i)^{2}+(a+g)^{2}+(g+i)^{2} & =(h+e)^{2}+(d+e)^{2}+(f+e)^{2}+(b+e)^{2} \\\\\n2\\left(a^{2}+c^{2}+g^{2}+i^{2}\\right)+2(a c+c i+a g+g i) & =\\left(b^{2}+d^{2}+f^{2}+h^{2}\\right)+2 e(b+d+f+h)+4 e^{2}\n\\end{aligned}\n$$\nPelo item anterior, podemos cancelar na última equação os termos $2(a c+c i+a g+g i)$ e $2 e(b+d+f+h)$, obtendo\n$$\n2\\left(a^{2}+c^{2}+g^{2}+i^{2}\\right)=\\left(b^{2}+d^{2}+f^{2}+h^{2}\\right)+4 e^{2}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26966,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNo quadrilátero $ABCD$, temos:\n$$\n\\angle DAB = \\angle ABC = \\angle BCD = 30^\\circ, \\quad AB = 4\\ \\mathrm{cm}, \\quad BC = 2\\sqrt{3}\\ \\mathrm{cm}\n$$\na) Determine o valor do ângulo $\\angle DCA$.\nb) Determine o comprimento de $CD$.\nc) Encontre a área do quadrilátero $ABCD$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "∠DCA = 60°, CD = 1 cm, area = (3√3)/2 cm²",
    "solution": "Solution:\n\na) Como $\\frac{BC}{AB} = \\cos 30^\\circ$ e $\\angle ABC = 30^\\circ$, segue que $\\angle ACB = 90^\\circ$. Daí, $\\angle BAC = 60^\\circ$ e $\\frac{AC}{AB} = \\sen 30^\\circ = \\frac{1}{2}$, logo, $AC = \\frac{AB}{2} = 2\\ \\mathrm{cm}$. Consequentemente, $\\angle DAC = 30^\\circ$ e $\\angle DCA = 60^\\circ$.\n\n![](attached_image_2.png)\n\nb) Do item anterior e das relações trigonométricas no triângulo $ADC$, decorre que $\\frac{AD}{AC} = \\sen 60^\\circ$ e $\\frac{CD}{AC} = \\sen 30^\\circ$. Portanto,\n$$\nAD = \\sqrt{3}\\ \\mathrm{cm} \\text{ e } CD = 1\\ \\mathrm{cm}\n$$\n\nc) A área do quadrilátero $ABCD$ é\n$$\n\\frac{AC \\cdot BC}{2} - \\frac{CD \\cdot AD}{2} = \\frac{2 \\cdot 2\\sqrt{3}}{2} - \\frac{1 \\cdot \\sqrt{3}}{2} = \\frac{3\\sqrt{3}}{2}\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26967,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOs 2.019 armários dos 2.019 alunos de uma escola são numerados com os quadrados dos 2.019 primeiros naturais positivos, ou seja, o primeiro armário tem o número $1^{2}=1$, o segundo armário tem o número $2^{2}=4$, o terceiro armário tem o número $3^{2}=9$, e assim até o último armário que tem o número $2.019^{2}=4.076.361$.\n![](attached_image_1.png)\n\na) Quantos algarismos foram utilizados para pintar os cem primeiros armários?\n\nb) Somando todos os números dos armários, qual o algarismo das unidades deste resultado?",
    "options": [],
    "answer": "a) 358; b) 0",
    "solution": "Solution:\n\na) Vamos dividir em grupos pela quantidade de algarismos:\nI) 1 algarismo: 3 armários $\\left(1^{2}, 2^{2}, 3^{2}\\right)$;\nII) 2 algarismos: $9-3=6$ armários ( $4^{2}$ a $9^{2}$ );\nIII) 3 algarismos: $31-9=22$ armários $\\left(10^{2}\\right.$ a $\\left.31^{2}\\right)$;\nIV) 4 algarismos: $99-31=68$ armários $\\left(32^{2}\\right.$ a $\\left.99^{2}\\right)$;\nV) 5 algarismos: apenas o armário $100^{2}$.\nPortanto, foram pintados $3+2 \\cdot 6+3 \\cdot 22+4 \\cdot 68+5 \\cdot 1=358$ algarismos nos cem primeiros armários.\n\nb) O algarismo das unidades dos 10 primeiros quadrados desses armários são\n$$\n1,4,9,6,5,6,9,4,1,0\n$$\nEssa sequência de algarismos repete-se de 10 em 10. Como a soma de uma dessas sequências termina em 5 e são, até $2020^{2}$ (vamos considerar o $2020^{2}$ para termos uma quantidade inteira de sequências e como seu algarismo das unidades é zero, não altera o resultado) 202 somas que terminam em 5, que somadas, o resultado termina em 0.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26968,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nExistem 100 números reais distintos arranjados ao redor de um círculo. Verifique que existem quatro números consecutivos ao redor do círculo de modo que a soma dos dois números do meio é estritamente menor que a soma dos outros dois números.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSeja $a$ o menor número escrito no círculo e sejam $b$ e $c$ seus dois vizinhos, com $b < c$. Seja $d$ o outro vizinho de $b$. Assim, estarão escritos no círculo, em ordem, $d, b, a$ e $c$ ou $c, a, b$ e $d$. Em qualquer caso, como $a < d$ e $b < c$, temos $a + b < c + d$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26969,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCinco amigas são titulares de um time de vôlei. Suas camisas são numeradas nas costas com os 5 primeiros ímpares positivos. Ana é a número 1; Bia é a número 3; Cátia é a número 5; Dani é a número 7; e Esmeralda é a número 9. Durante os treinos as cinco amigas fazem filas para formar números com suas camisas, todos com 5 algarismos. Por exemplo, a fila com Esmeralda, Dani, Cátia, Bia e Ana, nesta ordem, formam o número 97.531.\n\na) Quantos números diferentes elas podem formar?\nb) Quantos números podem ser formados se Cátia não pode ser a primeira da fila?\nc) Quantos números podem ser formados se Esmeralda e Bia ficarem lado a lado?\nd) Fábia, camisa número 11, resolve participar da brincadeira. Quantos são os números formados agora?",
    "options": [],
    "answer": "a) 120; b) 96; c) 48; d) 600",
    "solution": "Solution:\n\na) Como são 5 algarismos para permutar, temos $5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1 = 120$ números diferentes.\n\nb) Se Cátia não pode ser a primeira, temos $4 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1 = 96$ números diferentes.\n\nc) Esmeralda e Bia lado a lado serão consideradas como uma só, podendo trocar de lugar uma com a outra apenas. Sendo assim, são $2 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1 = 48$ números.\n\nd) Como Fábia tem o número 11 e existe o número 1 (Ana), se as duas estiverem lado a lado e as trocarmos de lugar, continua o mesmo número. Então, vamos subtrair do total a metade da quantidade de situações nas quais elas aparecem lado a lado. Temos, portanto, $6 \\cdot 5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1 - 5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1 = 720 - 120 = 600$ números diferentes formados.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26970,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDois candidatos participaram de uma eleição com $p+q$ eleitores. O candidato $A$ recebeu $p$ votos e o candidato $B$ recebeu $q$ votos, com $p>q$. Durante a apuração, é registrado apenas um voto de cada vez em um quadro. Seja $r$ a probabilidade de que o número associado ao candidato $A$ no quadro seja sempre maior que o número associado ao candidato $B$ durante toda a apuração.\n\na) Determine o valor de $r$ se $p=3$ e $q=2$.\n\nb) Determine o valor de $r$ se $p=1010$ e $q=1009$.",
    "options": [],
    "answer": "a) 1/5; b) 1/2019",
    "solution": "Solution:\n\na) Podemos fazer listas com as letras $A$ e $B$ representando as possíveis ordens de votos apurados. Por exemplo, a lista $A A B A B$ indica que os dois primeiros e o quarto voto apurados foram para o candidato $A$, o terceiro e o quinto para o candidato $B$. Existem exatamente 10 listas com 3 letras $A$ e duas letras $B$:\n\n| A | A | A | B | B | A | A | B | A | B |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| A | B | A | A | B | B | A | A | A | B |\n| A | A | B | B | A | A | B | A | B | A . |\n| B | A | A | B | A | A | B | B | A | A |\n| B | A | B | A | A | B | B | A | A | A |\n\nDessas 10 listas, em apenas duas a quantidade de letras $A$ quando contabilizadas da esquerda para a direita é sempre superior a quantidade de letras $B$, a saber,\n\n$A A A B B$ e $A A B A B$\n\nPortanto,\n$$\nr=\\frac{2}{10}=\\frac{1}{5}\n$$\n\nb) Vamos chamar de empate um momento da apuração em que o número de votos de ambos os candidatos é o mesmo. Para que o candidato $A$ esteja sempre à frente, certamente não podemos ter empates. Qualquer apuração em que o primeiro voto foi para o candidato $B$ terá um empate, pois sabemos que no final o número de votos de $A$ é maior e se não tivermos empate em nenhum momento o candidato $B$ irá ganhar. Para qualquer sequência que começa em $A$ e atinja um empate, associe outra sequência trocando as letras $A$ 's por $B$ 's e vice-versa até a posição de primeiro empate. Por exemplo, na sequência\n\n$A A A B B A B B A B$.\n\nTemos empate nas apurações do oitavo e décimo votos. A posição de primeiro empate é a oitava e iremos trocar a sequência anterior por\n$$\nB B B A A B A A A B\n$$\nCom essa operação, perceba que as quantidades de letras $A$ 's e $B$ 's não se alteram e agora a sequência começa com a letra $B$. Com essa operação, para toda sequência que começa com $B$, que já sabemos possuir empates, podemos associar de modo único outra sequência começada por $A$ com empates, e vice-versa. Assim o número de sequências com empates começando com $A$ é igual ao número de sequências começadas por $B$. Com mais razão, podemos concluir que a probabilidade de uma sequência começar com $A$ e possuir empates é igual à probabilidade de uma sequência começar com $B$. Como existem $q$ letras $B$ em um universo de $p+q$ letras, a probabilidade de uma sequência começar em $B$ é $\\frac{q}{p+q}$. Assim, como toda sequência com empates começa com $A$ ou $B$, a probabilidade de escolhermos, dentre as sequências possíveis de $p$ letras $A$ e $q$ letras $B$ uma com empates é $\\frac{q}{p+q}+\\frac{q}{p+q}=\\frac{2q}{p+q}$. Finalmente, o valor de $r$ é o complementar dessa probabilidade:\n$$\n\\begin{aligned}\nr & =1-\\frac{2q}{p+q} \\\\\n& =\\frac{p-q}{p+q} \\\\\n& =\\frac{1}{2019}\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26971,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nOs denominadores de duas frações irredutíveis são $600$ e $700$. Qual é o menor valor possível do denominador de sua soma quando escrita como uma fração irredutível?\n\nObservação: Dizemos que a fração $p / q$ é irredutível se os inteiros $p$ e $q$ não possuem fatores primos em comum em suas fatorações. Por exemplo, $5/7$ é uma fração irredutível.",
    "options": [],
    "answer": "168",
    "solution": "Solution:\nSejam $a / 600$ e $b / 700$ as duas frações irredutíveis. Assim, $\\mathrm{mdc}(a, 600) = \\mathrm{mdc}(b, 700) = 1$. A soma das duas frações pode ser escrita como\n$$\n\\begin{aligned}\n\\frac{a}{600} + \\frac{b}{700} &= \\frac{7a + 6b}{6 \\cdot 7 \\cdot 100} \\\\\n&= \\frac{7a + 6b}{3 \\cdot 7 \\cdot 2^3 \\cdot 5^2}\n\\end{aligned}\n$$\nComo $a$ e $b$ são ímpares, $7a + 6b$ é ímpar e assim não possui o fator primo $2$ em sua fatoração. Como $7$ não divide $b$, segue que a soma $7a + 6b$ também não possui o fator primo $7$ em sua fatoração. De modo semelhante, como $3$ não divide $a$, podemos concluir que esse fator não está presente na fatoração de $7a + 6b$. Assim, apenas o fator primo $5$ pode ser comum ao numerador e ao denominador da soma e, por conseguinte, o denominador será pelo menos $3 \\cdot 7 \\cdot 2^3 = 168$.\n\nPara verificar que ele é admissível, basta encontrarmos $a$ e $b$ tais que $25$ seja um divisor de $7a + 6b$. Isso pode ser obtido com $a = 1$ e $b = 3$, por exemplo,\n$$\n\\frac{1}{600} + \\frac{3}{700} = \\frac{1}{168}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26972,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA logomarca de uma empresa deve ser criada sobrepondo-se um triângulo equilátero e um quadrado, conforme a figura. Se a medida do lado do triângulo é $12~\\mathrm{cm}$ e o lado do triângulo intercepta o lado do quadrado em seu ponto médio, qual a diferença entre a área sobreposta (escura) e a soma das áreas sem sobreposição (claras)?\n![](attached_image_1.png)",
    "options": [],
    "answer": "1026 - 576 sqrt(3) cm^2",
    "solution": "Solution:\nVamos chamar a medida do lado do quadrado de $2n$, então quatro triângulos retângulos que não foram sobrepostos (todos congruentes) têm catetos medindo $n$ e $(6-n)$ e o triângulo equilátero não sobreposto tem lado medindo $(4n-12)$.\n![](attached_image_2.png)\nComo os ângulos do triângulo equilátero medem $60^\\circ$, então:\n$$\n\\begin{aligned}\n\\frac{n}{6-n} & = \\sqrt{3} \\\\\nn & = 6\\sqrt{3} - \\sqrt{3}n \\\\\nn(\\sqrt{3} + 1) & = 6\\sqrt{3} \\\\\nn & = \\frac{6\\sqrt{3}}{\\sqrt{3} + 1} \\\\\nn & = 9 - 3\\sqrt{3}~\\mathrm{cm}\n\\end{aligned}\n$$\nVamos agora ao cálculo das áreas:\n$$\n\\begin{aligned}\nA_{\\text{escura}} & = (2n)^2 - n(6-n) \\\\\n& = 4n^2 + n^2 - 6n \\\\\n& = 5n^2 - 6n \\\\\n& = 5(9 - 3\\sqrt{3})^2 - 6(9 - 3\\sqrt{3}) \\\\\n& = 5(81 - 54\\sqrt{3} + 27) - 54 + 18\\sqrt{3} \\\\\n& = (486 - 252\\sqrt{3})~\\mathrm{cm}^2 \\\\\nA_{\\text{clara}} = & 2n(6-n) + \\frac{(4n-12)^2 \\sqrt{3}}{4} \\\\\n= & 12n - 2n^2 + 4\\sqrt{3}(n-3)^2 \\\\\n= & 12(9 - 3\\sqrt{3}) - 2(9 - 3\\sqrt{3})^2 + 4\\sqrt{3}(6 - 3\\sqrt{3})^2 \\\\\n= & 108 - 36\\sqrt{3} - 2(81 - 54\\sqrt{3} + 27) + 4\\sqrt{3}(36 - 36\\sqrt{3} + 27) \\\\\n= & 108 - 36\\sqrt{3} - 216 + 108\\sqrt{3} + 252\\sqrt{3} - 432 \\\\\n= & 324\\sqrt{3} - 540~\\mathrm{cm}^2\n\\end{aligned}\n$$\nPortanto, a diferença entre as áreas escura e clara é $(1.026 - 576\\sqrt{3})~\\mathrm{cm}^2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26973,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDe quantas maneiras podemos colocar 8 algarismos iguais a 1 e 8 algarismos iguais a $0$ em um tabuleiro $4 \\times 4$ de modo que as somas dos números escritos em cada linha e coluna sejam as mesmas?\n\n| 1 | 0 | 1 | 0 |\n| :--- | :--- | :--- | :--- |\n| 0 | 1 | 1 | 0 |\n| 1 | 0 | 0 | 1 |\n| 0 | 1 | 0 | 1 |",
    "options": [],
    "answer": "90",
    "solution": "Solution:\nComo a soma dos números de todas as casas do tabuleiro é 8, a soma dos números em cada linha e coluna é $8 / 4 = 2$. Ou seja, em cada linha e coluna temos exatamente dois algarismos iguais a 1 e dois algarismos iguais a 0. Podemos escolher a posição do primeiro 1 da primeira linha de 4 maneiras. Em seguida, podemos escolher a posição do segundo 1 de $4-1 = 3$ maneiras, pois não podemos colocá-lo em uma posição já escolhida. Entretanto, nessas $3 \\cdot 4$ escolhas, estamos contando cada maneira de colocá-los na primeira linha duas vezes, pois como eles são algarismos iguais, a inversão de posição entre eles gera a mesma escolha. Portanto, temos $3 \\cdot 4 / 2 = 6$ maneiras de dispormos os dois algarismos 1 na primeira fila.\n\nApós feita essa escolha, temos três casos a considerar para a segunda linha:\n\n(I) todos os seus algarismos são iguais aos das posições correspondentes na primeira linha,\n\n(II) todos os algarismos da segunda linha diferem dos seus correspondentes na primeira linha e\n\n(III) dois algarismos da segunda linha coincidem com os seus correspondentes na primeira linha.\n\nNão existem outros casos, porque se três algarismos da terceira linha coincidem com os correspondentes da primeira, como aparecem apenas dois algarismos de cada tipo nela, necessariamente o quarto algarismo também será igual. De modo semelhante, também podemos perceber que não é possível apenas um algarismo coincidir entre as duas primeiras linhas.\n\n![](attached_image_1.png)\n\nCaso I\n![](attached_image_2.png)\n\nCaso II\n![](attached_image_3.png)\n\nCaso III\nO total de tabuleiros do caso (I) é igual ao número de possíveis escolhas da primeira linha, que é 6.\n\nNo caso (II), as duas primeiras linhas diferem em todas as posições. Para a escolha da terceira linha, podemos determinar as posições dos dois algarismos iguais a 1 de 6 formas e as demais posições serão preenchidas com 0. Em seguida, a quarta linha só poderá ser escolhida de uma única forma, pois já terão sido definidos os três primeiros algarismos de cada coluna. Nesse caso, temos 6 escolhas possíveis para a primeira linha e outras 6 para a segunda linha. Isso dá um total de $6 \\cdot 6 = 36$ possibilidades.\n\n![](attached_image_4.png)\n\nNo caso (III), temos 6 escolhas possíveis para a primeira linha. Feita essa escolha, podemos escolher de 2 formas qual das posições da segunda linha repetirá o algarismo 1 da primeira linha e de outras 2 formas qual das posições da segunda linha repetirá o algarismo 0. Isso nos dá $6 \\cdot 2 \\cdot 2 = 24$ preenchimentos das duas primeiras linhas. Nas casas dessas duas primeiras linhas, em precisamente duas colunas, já teremos escrito 2 algarismos iguais a 1 e dois algarismos iguais a 0. Consequentemente, as demais casas dessas colunas estarão determinadas. Ainda analisando as possíveis escolhas de terceira linha, temos 2 escolhas possíveis para o quadradinho mais à esquerda ainda não preenchido. Uma vez que ele tenha sido escolhido, toda a terceira linha estará determinada e, finalmente, os algarismos da quarta linha também.\n\n| 1 | 1 | 0 | 0 |\n| :--- | :--- | :--- | :--- |\n| 0 | 1 | 1 | 0 |\n|   |   |   |   |\n|   |   |   |   |$\\rightarrow$| 1 | 1 | 0 | 0 |\n| :--- | :--- | :--- | :--- | :--- |\n| 0 | 1 | 1 | 0 |\n|   | 0 |   | 1 |\n|   | 0 |   | 1 |\n\nPortanto, nesse caso, temos $24 \\cdot 2 = 48$ possíveis tabuleiros.\n\nSomando as configurações encontradas nas três situações, temos $6 + 36 + 48 = 90$ possibilidades de dispormos os algarismos no tabuleiro.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26974,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNo Calendário Jupiteriano, os meses são Julius, Uranius, Plutônius, Ílius, Terrius, Eráclitus e Raley. Os meses que começam com consoantes possuem 17 dias e os meses que começam com vogais têm 19 dias. O ano começa em Július e segue a sequência mencionada, anteriormente, encerrando-se em Raley. Assim, como no nosso calendário, o Jupiteriano possui uma semana com 7 dias (domingo, segunda, terça, quarta, quinta, sexta e sábado). Diógenes, um menino Jupiteriano, nasceu em 11 de Plutônio de 1999, que foi em um domingo.\n\na) Que dia Diógenes completará 100 dias de vida?\n\nb) Que dia da semana Diógenes completará 20 anos?",
    "options": [],
    "answer": "a) 5 de Uranius de 2000. b) segunda.",
    "solution": "Solution:\n\na) Se ele nasceu em 11 de Plutônio de 1.999, viveu $17-11=6$ dias neste mês, 19 dias em Ílius, 17 dias em Terrius, 19 dias em Eraclitus, 17 dias em Raley, 17 dias em Julius e 5 em Uranius, ou seja, Diógenes completará 100 dias em 5 de Uranius de 2.000.\n\nb) Cada ano neste calendário possui $17 \\cdot 4+19 \\cdot 3=68+57=125$ dias. Sendo assim, 20 anos são $20 \\cdot 125=2.500$ dias. Quando dividimos 2.500 por 7, obtemos quociente 357 e resto 1, ou seja, são 357 semanas completas e mais um dia. Como Diógenes nasceu em um domingo, completará 20 anos em uma segunda (um dia após o domingo).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26975,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEm uma sequência de inteiros positivos, uma inversão é um par de posições em que o elemento da posição mais à esquerda é maior que o elemento da posição mais à direita. Por exemplo, a sequência $2,5,3,1,3$ tem $5$ inversões: entre a primeira e a quarta posição, entre a segunda e todas as demais para a direita e, finalmente, entre a terceira e a quarta. Dentre todas as sequências de inteiros positivos cuja soma de seus elementos é $n$, qual é o maior número possível de inversões se\n\na) $n=7$ ?\n\nb) $n=2019$ ?\n\nObservação: As sequências de inteiros positivos consideradas nesse problema podem ter mais de $5$ elementos.",
    "options": [],
    "answer": "a) 6; b) 509545",
    "solution": "Solution:\n\na) Primeiramente vamos mostrar que qualquer sequência maximizante do número de inversões precisa ser não-crescente. De fato, se existe um par de números consecutivos $a$ e $b$, com $a < b$, então a troca de posição desses elementos não altera a soma e aumenta o número de inversões em uma unidade. Para cada sequência não-crescente com soma $7$, indicaremos o seu número de inversões na coluna $I$ da tabela a seguir:\n\n|                | $I$ |                | $I$ |\n| :------------- | :-: | :------------- | :-: |\n| $(6,1)$        | 1   | $(5,2)$        | 1   |\n| $(5,1,1)$      | 2   | $(4,3)$        | 1   |\n| $(4,2,1)$      | 3   | $(4,1,1,1)$    | 3   |\n| $(3,3,1)$      | 2   | $(3,2,2)$      | 2   |\n| $(3,2,1,1)$    | 5   | $(3,1,1,1,1)$  | 4   |\n| $(2,2,2,1)$    | 3   | $(2,2,1,1,1)$  | 6   |\n| $(2,1,1,1,1,1)$| 5   | $(1,1,1,1,1,1,1)$ | 0 |\n\nO máximo de inversões é $6$ e pode ser obtido com a sequência $2,2,1,1,1$.\n\nb) Mostraremos que qualquer sequência não crescente que maximiza o número de inversões deve possuir apenas números iguais a $1$ e $2$. Suponha, por absurdo, que a sequência contém algum número $k > 2$. Troque o último $k$ por um par de elementos: $k-1$ na posição original e $1$ na posição final. Claramente essa operação não altera a soma. O $1$ final é parte de uma inversão com todo o elemento que era membro de uma inversão com o $k$ original, exceto pelos números $1$ à sua direita. O novo $k-1$ é parte de uma inversão com todo elemento que era menor que o $k$ original, incluindo as parcelas $1$ à sua direita. Assim, contabilizando a inversão criada entre o novo $k-1$ e o novo $1$, essa troca criada aumenta o número de inversões em pelo menos uma unidade.\n\nFinalmente, considerando uma sequência qualquer que maximiza o número de inversões e que possui a soma de seus elementos igual a $2019$, podemos supor que existem $a$ parcelas iguais a $2$ e $2019-2a$ parcelas iguais a $1$. O número de inversões é\n\n$$\n\\begin{aligned}\na(2019-2a) &= 2019a - 2a^2 \\\\\n&= \\frac{2019^2}{8} - 2\\left(a - \\frac{2019}{4}\\right)^2\n\\end{aligned}\n$$\n\nPara maximizar a expressão anterior, devemos minimizar $|a - 2019/4|$ e isso ocorre para $a = 505$. Portanto, o maior número de inversões é $505 \\cdot 1009$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26976,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nBronquinha consegue cortar a grama de seu quintal em 3 horas, mas se ele tomar suco de frutas Gummy, ele corta em 2 horas. Em determinado dia, Bronquinha começou a cortar a grama às 10 horas e, em certo momento, tomou o suco de frutas Gummy, terminando de cortar a grama às 12 horas e 30 minutos. Que horas Bronquinha tomou o suco de frutas Gummy?",
    "options": [],
    "answer": "11:30",
    "solution": "Solution:\n\nSeja $A$ a área de grama a ser cortada. Assim, sem tomar o suco, a velocidade com a qual Bronquinha corta a grama é $\\frac{A}{3}$, enquanto que tomando o suco a velocidade é $\\frac{A}{2}$. Seja $t$ o tempo que Bronquinha corta a grama sem tomar o suco, na situação proposta, temos:\n$$\n\\begin{aligned}\n\\frac{A}{3} \\cdot t + \\frac{A}{2} \\cdot (2,5 - t) & = A \\\\\n\\frac{t}{3} + \\frac{2,5 - t}{2} & = 1 \\\\\n2t + 7,5 - 3t & = 6 \\\\\nt & = 1,5\n\\end{aligned}\n$$\nPortanto, se Bronquinha começou a cortar a grama às $10\\ h$, ele tomou o suco às 11 horas e 30 minutos.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26977,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm número é dito TOP se possui 5 algarismos e quando o produto entre o $1^{\\circ}$ e o $5^{\\circ}$ é igual a soma do $2^{\\circ}$, $3^{\\circ}$ e $4^{\\circ}$. Por exemplo, $12.338$ é TOP, pois possui 5 algarismos e $1 \\cdot 8 = 2 + 3 + 3$.\n\na) Qual o valor de $a$ para que $23.4a8$ seja TOP?\n\nb) Quantos números TOP terminam com 2 e começam com 1?\n\nc) Quantos números TOP começam com 9?",
    "options": [],
    "answer": "a = 9; b = 6; c = 112",
    "solution": "Solution:\n\na) Temos que $2 \\cdot 8 = 3 + 4 + a$, segue que $a = 9$.\n\nb) Seja $1b.cd2$ um número TOP. Temos que $b + c + d = 2$, sendo que todas as possibilidades $(b, c, d)$ são $(0,0,2)$, $(0,1,1)$, $(0,2,0)$, $(1,0,1)$, $(1,1,0)$, $(2,0,0)$, ou seja, são 6 números TOP.\n\nc) Seja $9e.fgh$ um número TOP. Vamos analisar todos os casos iniciando pelos possíveis valores de $h$:\n\nI) Se $h = 0$, então $e + f + g = 0$, cuja única possibilidade é $(0,0,0)$.\n\nII) Se $h = 1$, então $e + f + g = 9$, ou seja, para $e = 0$, são 10 possibilidades para $f$ (de 0 a 9) e $g$ fica determinado; para $e = 1$, são 9 possibilidades para $f$ e $g$ fica determinado; para $e = 2$, são 8 possibilidades para $f$; e assim por diante até $e = 9$ e 1 possibilidade para $f$, ou seja, o total de possibilidades para $h = 1$ é $10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 55$.\n\nIII) Se $h = 2$, então $e + f + g = 18$, ou seja, para $e = 0$, existe apenas uma combinação para $f$ e $g$ que é $f = g = 9$; para $e = 1$ são 2 possibilidades para $f$ e $g$; para $e = 2$, são 3 possibilidades; para $e = 3$, são 4 possibilidades; e assim por diante até $e = 9$, que são 10 possibilidades. Neste caso, temos, portanto, $1 + 2 + 3 + 4 + \\ldots + 10 = 55$ possibilidades.\n\nIV) Se $h = 3$, então $e + f + g = 27$, que possui apenas a solução $e = f = g = 9$.\n\nV) Se $h > 3$ não existe número TOP.\n\nSendo assim, o total de números TOP que começam com 9 é $1 + 55 + 55 + 1 = 112$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26978,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\na) Encontre o valor da soma\n$$\n\\frac{1}{1+1/x}+\\frac{1}{1+x}\n$$\n\nb) Encontre o valor da soma\n$$\n\\frac{1}{2019^{-2019}+1}+\\ldots+\\frac{1}{2019^{-1}+1}+\\frac{1}{2019^{0}+1}+\\frac{1}{2019^{1}+1}+\\ldots+\\frac{1}{2019^{2019}+1}\n$$",
    "options": [],
    "answer": "a) 1; b) 4039/2",
    "solution": "Solution:\n\na) Temos\n$$\n\\begin{aligned}\n\\frac{1}{1+1/x}+\\frac{1}{1+x} & =\\frac{1}{(x+1)/x}+\\frac{1}{1+x} \\\\\n& =\\frac{x}{1+x}+\\frac{1}{1+x} \\\\\n& =1\n\\end{aligned}\n$$\n\nb) Em virtude do item anterior, considerando $x=a^{b}$, podemos agrupar as frações $\\frac{1}{a^{-b}+1}$ e $\\frac{1}{a^{b}+1}$ em pares que somam 1. Retirando o termo $\\frac{1}{2019^{0}+1}=1/2$, podemos reescrever a soma dos termos restantes como\n$$\n\\begin{aligned}\n& \\left(\\frac{1}{2019^{-2019}+1}+\\frac{1}{2019^{2019}+1}\\right)+\\left(\\frac{1}{2019^{-2018}+1}+\\frac{1}{2019^{2018}+1}\\right)+ \\\\\n& \\left(\\frac{1}{2019^{-2017}+1}+\\frac{1}{2019^{2017}+1}\\right)+\\left(\\frac{1}{2019^{-2016}+1}+\\frac{1}{2019^{2016}+1}\\right)+\\ldots\n\\end{aligned}\n$$\nA soma desses 2019 pares é 2019. Assim, a soma pedida vale $2019+1/2=\\frac{4039}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26979,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLuísa faz experiências com uma rolha e um copo d'água. Por conta de sua densidade, uma rolha fica com apenas $60\\%$ de seu volume imerso na água. A rolha da experiência de Luísa tem formato de um prisma hexagonal regular, ou seja, sua base é um hexágono regular, com $3~\\mathrm{cm}$ de altura e $2~\\mathrm{cm}$ de diâmetro da circunferência que circunscreve a base.\n\na) Se a rolha ficar \"em pé\", como na figura, qual a altura da parte não imersa?\n![](attached_image_1.png)\n\nb) Se a rolha ficar \"deitada\", como na figura, qual a altura da parte não imersa, se duas das faces laterais (a de cima e a de baixo) ficarem paralelas ao nível da água?\n![](attached_image_2.png)",
    "options": [],
    "answer": "a) 1.2 cm; b) (√255 − 5√3) / 10 cm",
    "solution": "Solution:\n\na) Como a altura da rolha é de $3~\\mathrm{cm}$, e a parte fora da água corresponde a $40\\%$, então esta altura corresponde a $0,4 \\cdot 3 = 1,2~\\mathrm{cm}$.\n\nb) Vamos analisar a figura abaixo, que representa uma seção transversal da rolha com sua parte inferior imersa. Vamos lembrar que um hexágono regular pode ser dividido em seis triângulos equiláteros e também que a altura de um triângulo equilátero de lado $R$ pode ser calculado como $\\frac{R \\sqrt{3}}{2}$.\n\n![](attached_image_3.png)\n\nPela figura, temos que $h$ é a altura de um triângulo equilátero de lado $2m$, ou seja, $h = m \\sqrt{3}$. Como a área não imersa é $40\\%$ da área do hexágono, temos:\n\n$$\n\\begin{aligned}\n\\frac{(R+R+2m) \\cdot h}{2} &= \\frac{4}{10} \\cdot \\frac{6 \\cdot R^{2} \\sqrt{3}}{4} \\\\\n\\left(R+\\frac{\\sqrt{3} h}{3}\\right) h &= \\frac{3}{5} \\cdot R^{2} \\sqrt{3} \\\\\n5 h^{2} + 5 R h \\sqrt{3} - 9 R^{2} &= 0 \\\\\nh &= \\frac{-5 R \\sqrt{3} \\pm \\sqrt{75 R^{2} + 180 R^{2}}}{10} \\\\\nh &= \\frac{-5 R \\sqrt{3} \\pm R \\sqrt{255}}{10} \\\\\nh &= \\frac{R}{10}( \\pm \\sqrt{255} - 5 \\sqrt{3}) .\n\\end{aligned}\n$$\n\nComo $R = 1~\\mathrm{cm}$, temos que a altura da parte da rolha não imersa é $h = \\frac{\\sqrt{255} - 5 \\sqrt{3}}{10}$ $\\mathrm{cm}$.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 26980,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO triângulo $ABC$ é isósceles com $AB = BC$. A bissetriz do ângulo $\\angle CAB$ encontra o lado $BC$ no ponto $D$. A diferença entre as medidas de dois ângulos internos do triângulo $ABD$ é $40^{\\circ}$. Encontre os possíveis valores do ângulo $\\angle ACB$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "4°, 20°, 40°, 68°",
    "solution": "Solution:\n\nSeja $x = \\angle BAC = \\angle ABC$. Assim, $\\angle ACB = 180^{\\circ} - 2x$. Consequentemente, os ângulos internos do triângulo $ABD$ são $\\angle BAD = x/2$, $\\angle DBA = x$ e $\\angle ADB = 180^{\\circ} - 3x/2$. Consideraremos todos os casos para os quais dois ângulos podem diferir por $40^{\\circ}$:\n\ni) Se $x - x/2 = 40^{\\circ}$, então $x = 80^{\\circ}$ e assim $\\angle ACB = 20^{\\circ}$.\n\nii) Como $x > 0$, não podemos ter $x/2 - x = 40^{\\circ}$.\n\niii) Se $180^{\\circ} - 3x/2 - x/2 = 40^{\\circ}$, então $x = 70^{\\circ}$ e $\\angle ACB = 40^{\\circ}$.\n\niv) Se $x/2 - (180^{\\circ} - 3x/2) = 40^{\\circ}$, temos $x = 110^{\\circ}$. Isso produz uma contradição, pois nesse caso $\\angle ACB = 180^{\\circ} - 2x < 0$.\n\nv) Se $(180^{\\circ} - 3x/2) - x = 40^{\\circ}$, então $x = 56^{\\circ}$ e $\\angle ACB = 68^{\\circ}$.\n\nvi) Finalmente, se $x - (180^{\\circ} - 3x/2) = 40^{\\circ}$, segue que $x = 88^{\\circ}$ e $\\angle ACB = 4^{\\circ}$.\n\nPortanto, os possíveis valores de $\\angle ACB$ são $4^{\\circ}$, $20^{\\circ}$, $40^{\\circ}$ e $68^{\\circ}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 26981,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEm um estacionamento existem motos, carros, ônibus e caminhões, em um total de 80 veículos e 540 rodas. Cada moto tem 2 rodas, cada carro tem 4, cada ônibus tem 6 e cada caminhão tem 8. O número de carros é a soma do número de motos com o número de ônibus. Quantos são os caminhões neste estacionamento, se este número é menor que 50 ?",
    "options": [],
    "answer": "48",
    "solution": "Solution:\n\nSejam as quantidades de motos, carros, ônibus e caminhões iguais a $m, k, o$ e $c$ respectivamente. Organizando as informações, temos:\n$$\n\\left\\{\\begin{array}{cccc}\nm+k+o+c & = & 80 & (I) \\\\\n2 m+4 k+6 o+8 c & = & 540 & (II) \\\\\nk & = & m+o & (III)\n\\end{array}\\right.\n$$\nSubstituindo (III) em (II) e (I), chegamos a:\n$$\n\\left\\{\\begin{array}{ccc}\n2 m+2 o+c & =80 & (IV) \\\\\n3 m+5 o+4 c & =270 & (V)\n\\end{array}\\right.\n$$\nSubstituindo (IV) em (V), temos:\n$$\n\\begin{aligned}\n3 m+5 o+4(80-2 m-2 o) & =270 \\\\\n3 m+5 o+320-8 m-8 o & =270 \\\\\n5 m+3 o & =50 \\\\\no & =\\frac{50-5 m}{3}\n\\end{aligned}\n$$\nVoltando a (IV), temos:\n$$\n\\begin{aligned}\n2 m+2\\left(\\frac{50-5 m}{3}\\right)+c & =80 \\\\\nc & =80-2 m+\\frac{10 m-100}{3} \\\\\nc & =\\frac{4 m+140}{3}\n\\end{aligned}\n$$\nSubstituindo agora os resultados encontrados em (I), chegamos a:\n$$\n\\begin{aligned}\nm+k+\\frac{50-5 m}{3}+\\frac{4 m+140}{3} & =80 \\\\\nk & =\\frac{50-2 m}{3}\n\\end{aligned}\n$$\nChegamos, assim, à solução do sistema\n$$\n\\{(m, k, o, c)\\}=\\left\\{\\left(m, \\frac{50-2 m}{3}, \\frac{50-5 m}{3}, \\frac{4 m+140}{3}\\right)\\right\\}\n$$\nDe $o=\\frac{50-5 m}{3} \\in \\mathbb{N}$, concluímos que $m \\in\\{1,4,7,10\\}$. Vamos analisar cada um dos casos:\n\n| caso | $m$ | $k$ | $o$ | $c$ |\n| :---: | :---: | :---: | :---: | :---: |\n| $1^{\\circ}$ | 1 | 16 | 15 | 48 |\n| $2^{\\circ}$ | 4 | 14 | 10 | 52 |\n| caso | $m$ | $k$ | $o$ | $c$ |\n| :---: | :---: | :---: | :---: | :---: |\n| $3^{\\circ}$ | 7 | 12 | 5 | 56 |\n| $4^{\\circ}$ | 10 | 10 | 0 | 60 |\n\nComo a quantidade de caminhões deve ser menor que 50, são 48 caminhões no estacionamento.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26982,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nConsidere as soluções do sistema\n$$\n\\left\\{\\begin{array}{l}\n2019=a+b-c \\\\\n2019=a^{2}+b^{2}-c^{2}\n\\end{array}\\right.\n$$\nem que $a$, $b$ e $c$ são inteiros.\na) Encontre pelo menos uma solução do sistema.\nb) Verifique que o número de soluções é finito.",
    "options": [],
    "answer": "One solution is (a, b, c) = (2020, 2039190, 2039191); the number of solutions is finite.",
    "solution": "Solution:\na) Da primeira equação, segue que $c=a+b-2019$. Substituindo na segunda equação, obtemos\n$$\n2019=a^{2}+b^{2}-(a+b-2019)^{2}=-2 a b+4038 a+4038 b-2019^{2}\n$$\nDaí,\n$$\n\\begin{aligned}\n2019-2019^{2} & =-2 a b+4038 a+4038 b-2 \\cdot 2019^{2} \\\\\n-2019 \\cdot 2018 & =-2(a-2019)(b-2019) \\\\\n2019 \\cdot 1009 & =(a-2019)(b-2019)\n\\end{aligned}\n$$\nPara obtermos uma solução, é suficiente que $a-2019=1$ e $b-2019=2019 \\cdot 1009$, ou seja,\n$$\n(a, b, c)=\\left(2020, \\frac{2019 \\cdot 2020}{2}, \\frac{2019 \\cdot 2020}{2}+1\\right)\n$$\n\nb) Em virtude do item anterior, $a-2019$ e $b-2019$ são divisores inteiros do número $2019 \\cdot 1009$, que possui um número finito de divisores inteiros.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26983,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nÀs margens de um lago circular, existem pedras numeradas de 1 a 10, no sentido horário. O sapo Frog parte da pedra 1 e salta no sentido horário apenas nestas 10 pedras.\na) Se Frog salta de 2 em 2 pedras, ou seja, ele vai da pedra 1 para a 3, da 3 para a 5 e assim por diante, após 100 saltos em que pedra estará?\nb) Se no primeiro salto, Frog vai para a pedra 2, no segundo para a pedra 4, no terceiro para a pedra 7, ou seja, em cada salto ele pula uma pedra a mais que no salto anterior. Em que pedra Frog estará após 100 saltos?",
    "options": [],
    "answer": "a) 1; b) 1",
    "solution": "Solution:\n\na) Depois de 5 saltos, Frog volta para a pedra 1 e inicia a mesma sequência. Como 100 é múltiplo de 5, no $100^{\\circ}$ salto ele vai para a pedra 1.\n\nb) No $1^{\\circ}$ salto ele se desloca 1 pedra; no $2^{\\circ}$, 2 pedras; no $3^{\\circ}$, 3 pedras e assim até o último salto quando se desloca 100 pedras. O total de deslocamentos foi de $\\frac{(1+100) \\cdot 100}{2} = 5.050$. Como a cada 10 deslocamentos, ele volta para a pedra 1 e 5.050 é múltiplo de 10, após 100 saltos, Frog volta para a pedra 1.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26984,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDizemos que uma tripla de inteiros $(x, y, z)$ é do tipo bacana se $x$, $y$ e $z$ são inteiros positivos, com $y \\geq 2$, e $x^{2}-3 y^{2}=z^{2}-3$.\n\na) Encontre uma tripla $(x, y, z)$ do tipo bacana com $x=5$ e $x=7$.\n\nb) Mostre que para todo $x \\geq 5$ e ímpar existem pelo menos duas triplas distintas $\\left(x, y_{1}, z_{1}\\right)$ e $\\left(x, y_{2}, z_{2}\\right)$ do tipo bacana.\n\nc) Encontre alguma tripla do tipo bacana com $x$ par.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na) Para $x=5$ e $x=7$, temos alguns exemplos de triplas do tipo bacana: $(x, y, z)=(5,2,4)$, $(5,3,1)$, $(7,3,5)$ e $(7,4,2)$.\n\nb) Os casos particulares do item anterior permitem conjecturar as seguintes triplas para $x$ ímpar:\n$$\n(x, y, z)=(2 n+1, n, n+2) \\text{ e } (x, y, z)=(2 n+1, n+1, n-1)\n$$\nPara verificar que elas satisfazem a equação, perceba que\n$$\n\\begin{aligned}\n(2 n+1)^{2}-3 n^{2} & = n^{2}+4 n+1 \\\\\n& = (n+2)^{2}-3\n\\end{aligned}\n$$\ne\n$$\n\\begin{aligned}\n(2 n+1)^{2}-3(n+1)^{2} & = n^{2}-2 n-2 \\\\\n& = (n-1)^{2}-3\n\\end{aligned}\n$$\n\nc) Considerando a fatoração $(x-z)(x+z)=3(y-1)(y+1)$, podemos concluir que $x-z$ e $x+z$ são divisores do membro direito da equação. Como $x$ é a média aritmética desses dois divisores, isso permite definir uma busca ordenada de possíveis soluções da equação com $x$ par. Escolhendo $y=4$, podemos analisar os possíveis pares de divisores positivos de $3 \\cdot 3 \\cdot 5$ :\n$$\n(3,3 \\cdot 5),\\ (3 \\cdot 3,5) \\text{ e } (1,3 \\cdot 3 \\cdot 5)\n$$\nComo $x-z<x+z$, temos os casos:\n$$\n\\left\\{\\begin{array}{l}\n x - z = 3 \\\\\n x + z = 15\n\\end{array}\\right. \\quad \\left\\{\\begin{array}{l}\n x - z = 5 \\\\\n x + z = 9\n\\end{array}\\right. \\quad \\left\\{\\begin{array}{l}\nx-z=1 \\\\\nx+z=45\n\\end{array}\\right.\n$$\nNão existem valores pares para $x$ nesse caso. A mesma análise para $y=7$ também mostra que não existem soluções nesse caso. Se $y=9$, a tripla $(x, y, z)=(16,9,4)$ é uma solução do tipo bacana com $x$ par.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26985,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSobre uma reta $r$, marcam-se os pontos $A$ e $B$, e sobre uma reta $s$, paralela à $r$, marcam-se os pontos $C$ e $D$, de maneira que $A B C D$ seja um quadrado. Marca-se também o ponto $E$ no segmento $C D$.\n![](attached_image_1.png)\n\na) Qual a razão entre as áreas dos triângulos $A B E$ e $B C D$, se $E$ for o ponto médio de $C D$ ?\n\nb) Qual a razão $\\frac{D E}{E C}$, para que a área do triângulo $B F E$ seja o dobro da área do $D F E$, sendo $F$ a intersecção dos segmentos $A E$ e $B D$ ?",
    "options": [],
    "answer": "a) 1; b) 1",
    "solution": "Solution:\n\na) Seja $2k$ a área do quadrado $A B C D$, então a área do triângulo $A B E$ é igual a $k$ e a área do triângulo $B C D$ também é igual a $k$, portanto, a razão entre as áreas é $1$.\n\nb) Se $\\frac{A_{B F E}}{A_{D F E}}=2$, então $\\frac{B F}{F D}=2$. Se $r \\parallel s$, então $\\triangle A B F \\sim \\triangle D E F$, de razão $2$ e, consequentemente, $\\frac{A B}{D E}=2$, segue que $E$ é o ponto médio do segmento $C D$, ou seja, $\\frac{D E}{E C}=1$.\n\n![](attached_image_2.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26986,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nConsidere o quadrado $ABCD$ com lados de comprimento $1$, como no desenho a seguir. Um segmento horizontal $EW$ e um segmento vertical $NS$, ambos de comprimento $1/2$, estão inteiramente dentro do quadrado e se intersectam no ponto $X$ formando um ângulo de $90^\\circ$.\n\na) Qual a área do quadrilátero $NESW$?\n\nb) Qual a soma das áreas dos triângulos $EDN$, $AES$, $BSW$ e $CNW$?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "a) 1/8; b) 3/8",
    "solution": "Solution:\n\na) A área do quadrilátero $NESW$ é dada por\n$$\n\\begin{aligned}\nA_{NESW} & = A_{ENW} + A_{ESW} \\\\\n& = \\frac{EW \\cdot NX}{2} + \\frac{EW \\cdot XS}{2} \\\\\n& = \\frac{EW \\cdot (NX + XS)}{2} \\\\\n& = \\frac{1/2 \\cdot 1/2}{2} \\\\\n& = \\frac{1}{8}\n\\end{aligned}\n$$\n\nb) A soma das áreas dos triângulos $CDN$ e $ABS$ é $1/4$, pois cada um tem base de comprimento $1$ e suas alturas somam $1/2$. Analogamente, a soma das áreas dos triângulos $AED$ e $BCW$ é também $1/4$. A soma das áreas dos quatro triângulos mencionados no enunciado é o complementar das áreas dos triângulos $CDN$, $ABS$, $ADE$, $BCW$ e do quadrilátero $NESW$, portanto, a área procurada é\n$$\n1 - \\frac{1}{8} - \\frac{1}{4} - \\frac{1}{4} = \\frac{3}{8}\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26987,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSe $n$ é um número inteiro positivo, qual o menor valor que a soma dos algarismos da representação decimal de $3 n^{2}+n+1$ pode assumir?",
    "options": [],
    "answer": "3",
    "solution": "Solution:\nSe $n=8$, temos que $3 n^{2}+n+1=201$ e soma de seus dígitos é $3$. Verificaremos agora que a soma dos dígitos de $3 n^{2}+n+1$ não pode ser $1$ ou $2$ e concluiremos que o menor valor possível é $3$.\n\nComo $n(n+1)$ é o produto de dois números consecutivos, ele é par e assim $3 n^{2}+n+1=2 n^{2}+n(n+1)+1$ é um número ímpar.\n\nPara que a soma dos algarismos seja $1$, devemos ter $3 n^{2}+n+1=10^{k}$. Isso é impossível, pois $3 n^{2}+n+1>1$ e $10^{k}$ é par para $k>1$.\n\nPara que a soma dos algarismos seja $2$, devemos ter $3 n^{2}+n+1=10^{i}+10^{j}$, com $i>j$, ou $3 n^{2}+n+1=2 \\cdot 10^{k}$. A segunda opção é inválida, pois $2 \\cdot 10^{k}$ é par. Na primeira opção, como $10^{i}+10^{j}$ precisa ser ímpar, devemos ter $j=0$, ou seja,\n$$\n\\begin{aligned}\n3 n^{2}+n+1 & =10^{i}+1 \\\\\nn(3 n+1) & =10^{i} \\\\\nn(3 n+1) & =2^{i} \\cdot 5^{i}\n\\end{aligned}\n$$\nSe algum número primo $p$ divide $n$, então ele divide $3 n$ e, consequentemente, não pode dividir o seu sucessor $3 n+1$. Assim, $n$ e $3 n+1$ não possuem fatores primos em comum. Como $n<3 n+1$, devemos ter $n=2^{i}$ e $3 n+1=5^{i}$. Isso é um absurdo, porque, nesse caso, para $i \\geq 2$,\n$$\n3 n+1=5^{i}>4^{i}>3 \\cdot 2^{i}+1=3 n+1\n$$\nQuando $i=1$, não há solução, pois $5 \\neq 3 \\cdot 2+1$. Isso termina nossa análise e mostra que a soma mínima dos algarismos é $3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26988,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSeis pontos são distribuídos dentro de um quadrado de lado $10~\\mathrm{cm}$ de tal modo que a distância entre quaisquer dois deles é um número inteiro em centímetros. Verifique que pelo menos duas dessas distâncias são iguais.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nA maior distância possível entre dois pontos do quadrado é $10 \\sqrt{2}~\\mathrm{cm}$, que ocorre quando dois pontos estão dispostos em extremos opostos de uma diagonal. Como $14 < 10 \\sqrt{2} < 15$, a maior distância inteira possível entre eles é de $14~\\mathrm{cm}$. Assim, existem $14$ distâncias inteiras possíveis entre os pontos: $1, 2, 3, \\ldots, 14$ centímetros.\n\nPara determinar o número de pares de pontos, perceba que podemos escolher qualquer um deles de $6$ formas possíveis e o outro de $5$ outras formas. Como a mudança de ordem entre eles gera o mesmo par, nessa contagem estaremos contando cada par duas vezes. Portanto, existem $6 \\cdot 5 / 2 = 15$ pares de distâncias possíveis entre os $6$ pontos.\n\nComo o número de segmentos é maior que o número de distâncias possíveis, pelo menos dois segmentos terão o mesmo comprimento.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26989,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLiu e Lia brincam no quadro da sala de aula. Um deles escreve dois números naturais positivos e o outro tem que fazer transformações MULTISSÔMICAS até transformar o menor no maior. Transformação MULTISSÔMICA é trocar um número $a = m + n$ por $m \\cdot n$, por exemplo, podemos trocar $10$ por $2 \\cdot 8 = 16$.\n\na) Liu escreve no quadro $6$ e $15$. Mostre como Lia pode transformar $6$ em $15$.\n\nb) Lia escreve $5$ e $2.019$. Mostre como Liu pode fazer a transformação.\n\nc) A professora gostou da brincadeira e resolveu participar, escrevendo $7$ e $x$ e perguntou quantos são os possíveis valores de $x$ para transformar $7$ em $x$ com exatamente duas transformações MULTISSÔMICAS. Qual deve ser a resposta de Liu e Lia?",
    "options": [],
    "answer": "13",
    "solution": "Solution:\n\na)\n$$\n6 = 2 + 4 \\rightarrow 2 \\cdot 4 = 8 \\rightarrow 8 = 3 + 5 \\rightarrow 3 \\cdot 5 = 15\n$$\n\nb) Qualquer natural positivo $n$ pode ser escrito como $n = (n-1) + 1$, consequentemente, pode ser transformado em $(n-1)$. Começando com $5$, temos:\n$$\n\\begin{aligned}\n5 = 2 + 3 &\\rightarrow 2 \\cdot 3 = 6 \\rightarrow 6 = 3 + 3 \\rightarrow 3 \\cdot 3 = 9 \\rightarrow 9 = 4 + 5 \\rightarrow 4 \\cdot 5 = 20 \\rightarrow 20 = 10 + 10 \\rightarrow \\\\\n&10 \\cdot 10 = 100 \\rightarrow 100 = 30 + 70 \\rightarrow 30 \\cdot 70 = 2.100.\n\\end{aligned}\n$$\nA partir de agora basta transformar cada número no seu antecessor até chegarmos a $2.019$.\n\nc) O $7$ pode ser transformado em $3$ números $(1 \\cdot 6 = 6, 2 \\cdot 5 = 10$ e $3 \\cdot 4 = 12)$. Vamos analisar cada um dos casos:\n\nI) $6$: são $3$ transformações $(1 \\cdot 5 = 5, 2 \\cdot 4 = 8$ e $3 \\cdot 3 = 9)$;\n\nII) $10$: são $5$ transformações $(1 \\cdot 9 = 9, 2 \\cdot 8 = 16, 3 \\cdot 7 = 21, 4 \\cdot 6 = 24, 5 \\cdot 5 = 25)$;\n\nIII) $12$: são $6$ transformações $(1 \\cdot 11 = 11, 2 \\cdot 10 = 20, 3 \\cdot 9 = 27, 4 \\cdot 8 = 32, 5 \\cdot 7 = 35, 6 \\cdot 6 = 36)$.\n\nEncontramos, portanto, $13$ valores diferentes para $x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26990,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAs cidades $A$, $B$ e $C$ estão posicionadas nos vértices de um triângulo equilátero com $60\\ \\mathrm{km}$ de lado. Entre elas, ligando duas a duas, existem três rodovias, $AB$, $AC$ e $BC$, todas em linha reta. Uma ferrovia será construída, também em linha reta, devendo interceptar $AB$ a $30\\ \\mathrm{km}$ de $A$; $AC$ a $20\\ \\mathrm{km}$ de $C$; e, por fim, a rodovia $BC$, depois de $C$.\n\na) Se $h_1$, $h_2$ e $h_3$ são as distâncias de $A$, $B$ e $C$ para a ferrovia, respectivamente, verifique que\n$$\n\\frac{AD}{BD} = \\frac{h_1}{h_2}, \\quad \\frac{BF}{CF} = \\frac{h_2}{h_3}, \\quad \\mathrm{e} \\quad \\frac{CE}{AE} = \\frac{h_3}{h_1}\n$$\n\nb) Qual a distância entre a cidade $C$ e a intersecção entre a ferrovia e a rodovia $BC$ ?",
    "options": [],
    "answer": "60 km",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\na) Na figura anterior, os triângulos retângulos $\\triangle ADI$ e $\\triangle BGD$ possuem os mesmos ângulos, pois $\\angle ADI = \\angle GDB$ e $\\angle AID = \\angle DGB = 90^\\circ$. Portanto, eles são semelhantes e daí\n$$\n\\frac{AD}{BD} = \\frac{h_1}{h_2}\n$$\nAs demais igualdades decorrem das semelhanças $\\triangle BGF \\sim \\triangle CHF$ e $\\triangle CEH \\sim \\triangle AEI$.\n\nb) Se denotarmos a distância $CF$ por $x$, pelo item anterior temos:\n$$\n\\begin{aligned}\n\\frac{AD}{DB} \\cdot \\frac{BF}{FC} \\cdot \\frac{CE}{EA} & = \\frac{h_1}{h_2} \\cdot \\frac{h_2}{h_3} \\cdot \\frac{h_3}{h_1} \\\\\n\\frac{30}{30} \\cdot \\frac{60 + x}{x} \\cdot \\frac{20}{40} & = 1 \\\\\n60 + x & = 2x \\\\\nx & = 60.\n\\end{aligned}\n$$\nPortanto, a distância do ponto de intersecção da rodovia $BC$ com a ferrovia à cidade $C$ é $60\\ \\mathrm{km}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26991,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nNo desenho a seguir, as cordas $DE$ e $BC$ são perpendiculares, sendo $BC$ um diâmetro do círculo com centro em $A$. Além disso, $\\angle CGF = 40^\\circ$ e $GH = 2\\ \\mathrm{cm}$.\n\na) Determine o valor do ângulo $\\angle CHF$.\n\nb) Encontre o comprimento de $HJ$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "(a) 40 degrees; (b) 2 cm",
    "solution": "Solution:\na) Como $BC$ é um diâmetro, segue que $\\angle BFC = 90^\\circ$. Assim, como também temos $\\angle CHG = 90^\\circ$, a circunferência $\\Gamma$ de diâmetro $CG$ passa por $F$ e $H$. Nessa circunferência, os ângulos $\\angle CGF$ e $\\angle CHF$ estão inscritos no mesmo arco $CF$ e assim $\\angle CHF = \\angle CGF = 40^\\circ$.\n\nb) Novamente observando o círculo $\\Gamma$, podemos concluir que $\\angle HCG = \\angle HFG$, pois ambos estão inscritos no arco $GH$. Considerando agora o círculo de diâmetro $BC$, temos $\\angle ICB = \\angle IFB$, porque ambos estão inscritos no arco $IB$. Assim, $\\angle ICH = \\angle BFH = \\angle HFG = \\angle HCG$. Daí como os triângulos retângulos $CHG$ e $CHJ$ possuem os mesmos ângulos e um cateto em comum, eles são congruentes, resultando em $HJ = GH = 2\\ \\mathrm{cm}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26992,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDois inteiros positivos $x$ e $y$ são tais que:\n$$\n\\frac{2010}{2011}<\\frac{x}{y}<\\frac{2011}{2012}\n$$\nEncontre o menor valor possível para a soma $x+y$.",
    "options": [],
    "answer": "8044",
    "solution": "Solution:\n\nComo $\\frac{2011}{2012}<1$, temos $x<y$ e assim $x=y-d$, com $d$ inteiro positivo. De\n$$\n\\frac{2011-1}{2011}<\\frac{y-d}{y}<\\frac{2012-1}{2012}\n$$\nsegue que\n$$\n\\frac{1}{2012}<\\frac{d}{y}<\\frac{1}{2011}\n$$\nAssim,\n$$\n2011 d<y<2012 d\n$$\nSe $d=1$, a desigualdade (1) não possui solução inteira. Se $d=2$, a única possibilidade é $y=4023$. Nesse caso, $x+y=8044$. Para $d \\geq 3$,\n$$\n\\begin{aligned}\nx+y & =2 y-d \\\\\n& >4021 d \\\\\n& \\geq 12063\n\\end{aligned}\n$$\nConsequentemente, o valor mínimo da soma é obtido com $d=2$ e nesse caso $x+y=8044$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26993,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSeja $m=999 \\ldots 99$ o número formado por 77 dígitos iguais a 9 e seja $n=777 \\ldots 77$ o número formado por 99 dígitos iguais a 7. Qual o número de dígitos de $m \\cdot n$ ?",
    "options": [],
    "answer": "176",
    "solution": "Solution:\n\nComo $m+1=10^{77}$, perceba que:\n$$\n\\begin{aligned}\nm \\cdot n & =(m+1) \\cdot n-n \\\\\n& =\\underbrace{777 \\ldots 77}_{99} \\underbrace{000 \\ldots 00}_{77}-\\underbrace{777 \\ldots 77}_{99} .\n\\end{aligned}\n$$\nComo $\\underbrace{777 \\ldots 77}_{99} \\underbrace{000 \\ldots 00}_{77}$ possui $99+77$ dígitos e $\\underbrace{777 \\ldots 77}_{99}$ é menor que $\\underbrace{777 \\ldots 77}_{175}$, o resultado da subtração anterior ainda terá 176 dígitos.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26994,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm triminó é um retângulo $3 \\times 1$ e um monominó é um único quadrado $1 \\times 1$. Quais são as possíveis posições de um monominó na cobertura de um tabuleiro $8 \\times 8$ usando 21 triminós e 1 monominó?\n![](attached_image_1.png)",
    "options": [],
    "answer": "(3,3), (3,6), (6,3), (6,6)",
    "solution": "Solution:\n\nPinte os quadradinhos do tabuleiro $8 \\times 8$ com as cores 1, 2 e 3 como indicado nos tabuleiros a seguir.\n\n| 1 | 2 | 3 | 1 | 2 | 3 | 1 | 2 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| 3 | 1 | 2 | 3 | 1 | 2 | 3 | 1 |\n| 2 | 3 | 1 | 2 | 3 | 1 | 2 | 3 |\n| 1 | 2 | 3 | 1 | 2 | 3 | 1 | 2 |\n| 3 | 1 | 2 | 3 | 1 | 2 | 3 | 1 |\n| 2 | 3 | 1 | 2 | 3 | 1 | 2 | 3 |\n| 1 | 2 | 3 | 1 | 2 | 3 | 1 | 2 |\n| 3 | 1 | 2 | 3 | 1 | 2 | 3 | 1 |\n\n| 2 | 1 | 3 | 2 | 1 | 3 | 2 | 1 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| 1 | 3 | 2 | 1 | 3 | 2 | 1 | 3 |\n| 3 | 2 | 1 | 3 | 2 | 1 | 3 | 2 |\n| 2 | 1 | 3 | 2 | 1 | 3 | 2 | 1 |\n| 1 | 3 | 2 | 1 | 3 | 2 | 1 | 3 |\n| 3 | 2 | 1 | 3 | 2 | 1 | 3 | 2 |\n| 2 | 1 | 3 | 2 | 1 | 3 | 2 | 1 |\n| 1 | 3 | 2 | 1 | 3 | 2 | 1 | 3 |\n\nComecemos a nossa análise pelo tabuleiro da esquerda. Nele pintamos 22 quadradinhos da cor 1, 21 da cor 2 e 21 da cor 3. Como todo triminó cobre exatamente um quadradinho de cada cor, a união deles cobrirá exatamente 21 quadradinhos de cada cor e assim o monominó deve ter a cor 1. Repetindo a mesma análise na pintura feita no tabuleiro da direita, também podemos concluir que o monominó deve ter a cor 1 naquele tabuleiro. Daí os únicos possíveis locais para os monominós são os que foram marcados com a cor 1 nos dois tabuleiros, ou seja, os quadradinhos pintados no desenho a seguir. Para verificar que para todos eles existe uma cobertura admissível, basta rotacionar o desenho do enunciado por $90^\\circ$, $180^\\circ$ e $270^\\circ$.\n\n![](attached_image_2.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26995,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nCada um dos números de $1$ a $13$ está escrito em um dos círculos do floco de neve da figura a seguir, de modo que as somas dos $5$ números em cada linha e a soma dos $7$ números no centro da figura sejam todas iguais. Encontre essa soma dado que ela é a menor possível dentre as que satisfazem essas condições.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "31",
    "solution": "Solution:\nSejam $s$ o valor da soma em cada linha e $a$ o valor escrito no círculo central. Então,\n$$\n3s = (1+2+\\ldots+13) + 2a = 91 + 2a\n$$\nconsequentemente $s = \\frac{91 + 2a}{3} \\geq \\frac{93}{3} = 31$. Para verificar que esse valor é possível, considere a figura a seguir.\n\n![](attached_image_2.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26996,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa figura abaixo, $D$ é o ponto médio do lado $AB$, $CE:DE=5:3$ e $BF:EF=1:3$. Se a área do triângulo $ABC$ é $192~\\mathrm{cm}^2$, determine a área do triângulo $BDF$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "9",
    "solution": "Solution:\n\nDenote os comprimentos de $BD$, $BF$ e $DE$ por $x$, $z$ e $3y$, respectivamente. Em virtude das proporções dadas, segue que $EF=3z$, $CE=5y$ e $AD=x$. Portanto,\n$$\n\\begin{aligned}\n& \\frac{A_{BDF}}{A_{BDE}}=\\frac{z}{3z+z} \\\\\n& \\frac{A_{BDE}}{A_{CDB}}=\\frac{3y}{3y+5y} \\\\\n& \\frac{A_{CDB}}{A_{ABC}}=\\frac{x}{x+x}\n\\end{aligned}\n$$\nMultiplicando essas equações, temos\n$$\n\\frac{A_{BDF}}{A_{ABC}}=\\frac{1}{4} \\cdot \\frac{3}{8} \\cdot \\frac{1}{2}\n$$\nDaí,\n$$\nA_{BDF}=\\frac{192 \\cdot 3}{64}=9\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26997,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSeja $A$ um subconjunto de $\\{1,2,3, \\ldots, 2019\\}$ possuindo a propriedade de que a diferença entre quaisquer dois de seus elementos não é um número primo. Qual é o maior número possível de elementos de $A$ ?",
    "options": [],
    "answer": "505",
    "solution": "Solution:\n\nSuponha que $a \\in A$. Então, nenhum elemento do conjunto $\\{a+2, a+3, a+5, a+7\\}$ pode pertencer a $A$ e entre os elementos de $\\{a+1, a+4, a+6\\}$, no máximo um deles pode pertencer a $A$. Assim, a cada 8 inteiros consecutivos, digamos os elementos do conjunto $\\{a, a+1, a+2, \\ldots, a+7\\}$, no máximo dois deles pertencem a $A$. Portanto, o número máximo de elementos de $A$ não é maior que o maior inteiro que não ultrapassa $2019/4$ mais $1$, ou seja, $505$. Essa quantidade pode ser obtida com o conjunto $\\{3,7,11, \\ldots, 2019\\}$. Note que a diferença entre quaisquer dois deles é um múltiplo de $4$ e, consequentemente, não pode ser um número primo. Portanto, o número máximo de elementos é $505$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 26998,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa figura a seguir, $ABC$ é um triângulo equilátero de papel com lado $1\\ \\mathrm{m}$ que foi dobrado ao longo do segmento $EF$ de modo que o vértice $A$ caísse sobre o lado $BC$, onde está o ponto $D$ na figura. Suponha que $DF$ é perpendicular a $BC$.\n![](attached_image_1.png)\n\na) Determine o ângulo $\\angle AED$.\nb) Determine o comprimento do segmento $CD$.\nc) Determine a razão entre as áreas dos triângulos $AEF$ e $ABC$.",
    "options": [],
    "answer": "a) 90 degrees; b) 2 − √3 m; c) (9√3 − 15)/2",
    "solution": "Solution:\n\na) Como $\\angle FDC = 90^{\\circ}$, segue que $\\angle DFC = 30^{\\circ}$ e $\\angle AFD = 180^{\\circ} - \\angle DFC = 150^{\\circ}$. A dobradura ao longo de $EF$ nos diz que os triângulos $AEF$ e $DEF$ são congruentes. Daí $\\angle AFE = \\angle EFD = \\frac{150^{\\circ}}{2} = 75^{\\circ}$ e $\\angle AEF = 180^{\\circ} - 75^{\\circ} - 60^{\\circ} = 45^{\\circ}$. Consequentemente, $\\angle AED = 2 \\cdot 45^{\\circ} = 90^{\\circ}$.\n\nb) Seja $x$ o comprimento de $CD$, então $\\frac{CD}{CF} = \\sen 30^{\\circ} = 1/2$ e assim $CF = 2x$. Além disso, $\\frac{DF}{CF} = \\cos 30^{\\circ} = \\frac{\\sqrt{3}}{2}$. Logo, $AF = DF = \\sqrt{3}x$. Finalmente, $1 = AF + FC = \\sqrt{3}x + 2x$ implica $x = \\frac{1}{2 + \\sqrt{3}} = (2 - \\sqrt{3})\\ \\mathrm{m}$.\n\nc) Como $\\angle EAF = \\angle EDF = 60^{\\circ}$, segue que $\\angle EDB = 30^{\\circ}$. Além disso, $\\angle BED = 180^{\\circ} - 90^{\\circ} = 90^{\\circ}$ e $BD = 1 - CD = \\sqrt{3} - 1$. Portanto, de $\\frac{BE}{BD} = \\sen 30^{\\circ} = \\frac{1}{2}$, temos $BE = \\frac{\\sqrt{3} - 1}{2}$ e $AE = 1 - BE = \\frac{3 - \\sqrt{3}}{2}$. Assim\n$$\n\\begin{aligned}\n\\frac{A_{AEF}}{A_{ABC}} & = \\frac{\\frac{AE \\cdot AF \\cdot \\sen \\angle EAF}{2}}{\\frac{AB \\cdot AC \\cdot \\sen \\angle EAF}{2}} \\\\\n& = \\frac{(3 - \\sqrt{3}) \\cdot (2\\sqrt{3} - 3)}{2} \\\\\n& = \\frac{9\\sqrt{3} - 15}{2}\n\\end{aligned}",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 26999,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nQual é o maior inteiro positivo $n$ para o qual existe um único inteiro $k$, tal que\n$$\n\\frac{8}{15}<\\frac{n}{n+k}<\\frac{7}{13} ?\n$$",
    "options": [],
    "answer": "112",
    "solution": "Solution:\nPodemos escrever a desigualdade como\n$$\n\\frac{13}{7}<\\frac{n+k}{n}<\\frac{15}{8}\n$$\nAssim, multiplicando os membros da desigualdade por $56 n$, obtemos a desigualdade equivalente\n$$\n\\begin{aligned}\n104 n & <56 n+56 k & <105 n \\\\\n48 n< & 56 k & <49 n\n\\end{aligned}\n$$\nPara que exista um único inteiro $k$ satisfazendo a desigualdade, o intervalo $(48 n, 49 n)$ deve possuir exatamente um múltiplo de 56. O comprimento de tal intervalo é $n$ e ele contém exatamente $n-1$ inteiros positivos. Se $n-1 \\geq 2 \\cdot 56=112$, o intervalo conterá pelo menos dois múltiplos de 56 e não servirá para a desigualdade. Portanto, $n \\leq 112$. Para verificar que $n=112$, é a solução máxima, basta notar que\n$$\n\\begin{aligned}\n48 \\cdot 112 & =56 \\cdot 96 \\\\\n& <56 \\cdot 97 \\\\\n& <56 \\cdot 98 \\\\\n& =49 \\cdot 112\n\\end{aligned}\n$$\nAssim, $k=97$ é o único inteiro possível para satisfazer a desigualdade quando $n=112$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27000,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\na) Dado que a representação decimal de $5^{2018}$ possui 1411 algarismos e começa com 3 (o dígito não nulo mais à esquerda é 3), para quantos inteiros $1 \\leq n \\leq 2017$ o número $5^{n}$ começa com 1?\nb) Os inteiros $4^{52}$ e $5^{52}$ ambos começam com o algarismo 2. Se as representações decimais das potências $4^{n}$ e $5^{n}$, com $n>0$ e inteiro, começam com o mesmo algarismo $d$, quais os possíveis valores desse algarismo?",
    "options": [],
    "answer": "a) 607; b) 2 and 4",
    "solution": "Solution:\na) Se $5^{k}$ começa com $a$ e possui $j$ algarismos, então\n$$\n10^{j} < 5^{k} < a \\cdot 10^{j+1}\n$$\ne assim\n$$\n\\begin{aligned}\n10^{j} & < 5 \\cdot 10^{j} \\\\\n& < 5 \\cdot 5^{k} \\\\\n& = 5^{k+1} \\\\\n& < 10 \\cdot 10^{j+1}\n\\end{aligned}\n$$\nIsso significa que a representação decimal de $5^{k+1}$ também possui $j$ algarismos. Por outro lado, se $5^{k+1}$ e $5^{k}$ possuem a mesma quantidade $j$ de algarismos, então o primeiro algarismo de $5^{k}$ é 1, pois caso contrário\n$$\n5^{k+1} = 5 \\cdot 5^{k} > 5 \\cdot 2 \\cdot 10^{j} = 10^{j+1}\n$$\npossuiria pelo menos $j+1$ algarismos. Então, o problema se resume a encontrarmos os valores de $k \\in \\{1,2, \\ldots, 2017\\}$ tais que $5^{k}$ e $5^{k+1}$ possuem a mesma quantidade de algarismos. Entre duas potências de 5 consecutivas, a quantidade de algarismos cresce em no máximo uma unidade. Como $5^{2018}$ possui 1411 algarismos, dentre as potências $5^{1}$, $5^{2}, \\ldots, 5^{2018}$, temos 1410 crescimentos nas quantidades de algarismos entre potências consecutivas e exatamente $2017-1410=607$ valores de $k$ em que $5^{k}$ e $5^{k+1}$ possuem a mesma quantidade de algarismos. Ou seja, o número procurado de potências que começam com 1 é 607.\n\nb) Como $4^{n}$ e $5^{n}$ começam com o mesmo algarismo $d$, existem inteiros não negativos $i$ e $j$, tais que\n$$\n\\begin{aligned}\n& d \\cdot 10^{i} \\leq 4^{n} < (d+1) \\cdot 10^{i} \\\\\n& d \\cdot 10^{j} \\leq 5^{n} < (d+1) \\cdot 10^{j}\n\\end{aligned}\n$$\nElevando ao quadrado a segunda dessas expressões e multiplicando o resultado pela primeira, obtemos\n$$\nd^{3} \\cdot 10^{i+2j} \\leq 10^{2n} < (d+1)^{3} \\cdot 10^{i+2j}.\n$$\nDaí,\n$$\n1 \\leq d^{3} \\leq 10^{2n-i-2j} < (d+1)^{3} \\leq (9+1)^{3} = 1000\n$$\nSe $2n-i-2j=0$, o único algarismo que satisfaz a desigualdade é $d=1$. Entretanto, nesse caso, como ocorre igualdade na última expressão, devemos ter igualdade nas duas iniciais. Assim, $1 \\cdot 10^{i} = 4^{n}$. Em virtude do Teorema Fundamental da Aritmética, isso só é possível se $i=0$ e $n=0$. Como $n$ é um inteiro positivo, esse caso não é válido. Então, $2n-i-2j>0$. Analisando os cubos\n$$\n1^{3}, 2^{3}, 3^{3}, \\ldots, 9^{3}, 10^{3}\n$$\npodemos concluir que os únicos algarismos $d$ para os quais o intervalo $\\left(d^{3}, (d+1)^{3}\\right)$ contém uma potência de 10 são $d=2$ e $d=4$. De fato, além do exemplo dado no enunciado garantindo que $d=2$ é possível, $4^{11}$ e $5^{11}$ começam com o algarismo 4.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27001,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDois estudantes precoces do Nível 3 participaram de um torneio de xadrez universitário. Cada participante joga contra todos os outros exatamente uma vez. Uma vitória vale 1 ponto, um empate vale 0,5 ponto e uma derrota vale 0 ponto. A soma das pontuações dos dois estudantes do Nível 3 é 6,5. Todos os estudantes universitários obtiveram a mesma pontuação. Quantos estudantes universitários participaram da competição?",
    "options": [],
    "answer": "11",
    "solution": "Solution:\n\nSeja $x$ a quantidade de estudantes universitários e $p$ a pontuação comum a todos eles. Como em cada jogo é disputado exatamente 1 ponto, segue que o total de pontos do torneio, que é $6,5 + p x$, coincide com o número de jogos, que é $\\frac{(x+2)(x+1)}{2}$. Além disso, a pontuação de cada participante é um múltiplo inteiro de $0,5$ e assim podemos escrever $p = k / 2$, para algum inteiro positivo $k$. Portanto:\n$$\n\\begin{aligned}\n\\frac{(x+2)(x+1)}{2} & = 6,5 + p x \\\\\n(x+2)(x+1) & = 13 + k x \\\\\nx^{2} + 3x + 2 & = 13 + k x \\\\\nx(x+3-k) & = 11\n\\end{aligned}\n$$\nComo $x$ e $x+3-k$ são inteiros, podemos concluir que $x$ é um divisor positivo de $11$, ou seja, $x=1$ ou $x=11$. Não podemos ter $x=1$, pois nesse caso o torneio teria apenas $3$ e não seria possível dois estudantes obterem $6,5$ pontos. Para mostrar que $x=11$ é solução, considere o torneio formado pelos universitários $U_{1}, U_{2}, \\ldots, U_{11}$ e pelos estudantes do Nível $3$ $E_{1}$ e $E_{2}$ com os seguintes resultados:\n\nI) Todos os jogos entre dois universitários terminaram em empate.\n\nII) $E_{1}$ perdeu para $U_{1}, U_{2}, \\ldots, U_{11}$ e $E_{2}$.\n\nIII) $E_{2}$ empatou com $U_{1}, U_{2}, \\ldots, U_{11}$.\n\nO torneio com esses resultados satisfaz o enunciado.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27002,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\na) Verifique que para qualquer inteiro positivo $a$, com $a>1$, a equação\n$$\n\\frac{1}{x}+\\frac{1}{y}=\\frac{1}{a}\n$$\npossui pelo menos três soluções da forma $(x, y)$, com $x$ e $y$ inteiros positivos. Por exemplo, para $a=3$, os pares $(6,6),(4,12)$ e $(12,4)$ são soluções.\n\nb) Encontre o número de pares de inteiros positivos $(x, y)$ que são soluções dessa equação quando $a=2019$.\nDica: Se a fatoração em primos do inteiro positivo $n$ é $p_{1}^{\\alpha_{1}} \\cdot p_{2}^{\\alpha_{2}} \\cdot \\ldots \\cdot p_{k}^{\\alpha_{k}}$, então ele possui $\\left(\\alpha_{1}+1\\right)\\left(\\alpha_{2}+1\\right) \\ldots\\left(\\alpha_{k}+1\\right)$ divisores positivos.",
    "options": [],
    "answer": "9",
    "solution": "Solution:\na) Podemos encontrar uma equação equivalente:\n$$\n\\begin{aligned}\n\\frac{1}{x}+\\frac{1}{y} & =\\frac{1}{a} \\Leftrightarrow \\\\\n(x-a)(y-a) & =a^{2}\n\\end{aligned}\n$$\nComo $1 / x$ e $1 / y$ são menores que $1 / a$, segue que $x-a$ e $y-a$ são positivos. Para encontrarmos soluções dessa última equação, considere os seguintes sistemas:\n$$\n\\left\\{\\begin{array} { l } \n{ x - a = 1 } \\\\\n{ y - a = a ^ { 2 } }\n\\end{array} \\quad \\left\\{\\begin{array} { l } \n{ x - a = a } \\\\\n{ y - a = a }\n\\end{array} \\quad \\left\\{\\begin{array}{l}\nx-a=a^{2} \\\\\ny-a=1\n\\end{array}\\right.\\right.\\right.\n$$\nAs soluções $(x, y)$ deles são, respectivamente, $\\left(a+1, a+a^{2}\\right),(2 a, 2 a)$ e $\\left(a+a^{2}, a+1\\right)$. Se $a>1$, essas soluções são distintas e satisfazem a equação dada.\n\nb) Em geral, se $d$ é um divisor qualquer de $a^{2}$, sempre existe uma solução em inteiros positivos para o sistema\n$$\n\\left\\{\\begin{array}{l}\nx-a=d \\\\\ny-a=\\frac{a^{2}}{d}\n\\end{array}\\right.\n$$\nque é dada por $(x, y)=\\left(a+d, a+\\frac{a^{2}}{d}\\right)$. Existe uma correspondência entre os pares $(x, y)$ que são soluções da equação original e os divisores positivos de $a^{2}$, pois para cada solução o inteiro $x-a$ corresponde a algum divisor positivo $d$ de $a^{2}$. Como $2019^{2}=3^{2} \\cdot 673^{2}$, o seu número de divisores positivos é $(2+1) \\cdot(2+1)=9$. Logo, o número de soluções é 9 .",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27003,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSeja $ABCD$ um trapézio, com $AD \\parallel BC$, tal que o lado $CD$ é tangente ao círculo com diâmetro $AB$. Se $G$ é o ponto médio de $CD$ e $CD = 8\\ \\mathrm{cm}$, determine a medida da altura $GF$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "4 cm",
    "solution": "Solution:\n\nSeja $O$ o centro do círculo de diâmetro $AB$. Como $O$ é ponto médio de $AB$ e $G$ é ponto médio de $CD$, segue que $GO$ é base média do trapézio $ABCD$. Daí, $GO$ é paralelo aos lados $AD$ e $BC$. Consequentemente, temos as seguintes igualdades de áreas: $A_{DGO} = A_{AGO}$ e $A_{COG} = A_{BGO}$.\n\nDaí,\n$$\n\\begin{aligned}\nA_{CDO} &= A_{DGO} + A_{CGO} \\\\\n&= A_{AGO} + A_{BGO} \\\\\n&= A_{ABG} \\\\\n&= \\frac{AB \\cdot h}{2}\n\\end{aligned}\n$$\n\n![](attached_image_2.png)\n\nPor outro lado, como $EO = AO = BO = \\frac{AB}{2}$, segue que\n$$\nA_{CDO} = \\frac{CD \\cdot EO}{2} = \\frac{CD \\cdot AB}{4}\n$$\n\nFinalmente, comparando as duas expressões para a área $A_{CDO}$, temos\n$$\n\\begin{aligned}\n\\frac{AB \\cdot h}{2} &= \\frac{CD \\cdot AB}{4} \\\\\nh &= \\frac{CD}{2} \\\\\n&= 4\\ \\mathrm{cm}\n\\end{aligned}",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27004,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nOs inteiros positivos $x$ e $y$ são tais que o número $x^{2019}+x+y^{2}$ é divisível por $x y$.\na) Dê um exemplo de tais inteiros $x$ e $y$, com $x>y$.\nb) Verifique que, necessariamente, $x$ é um quadrado perfeito.",
    "options": [],
    "answer": "(x, y) = (4, 2); necessarily, x is a perfect square.",
    "solution": "Solution:\na) Basta escolher $x=4$ e $y=2$, pois $4^{2019}+4+2^{2}=8 \\cdot (2^{4035}+1)$ é divisível por $4 \\cdot 2=8$.\n\nb) Seja $d=mdc(x, y)$. Assim, $x=d m$ e $y=d n$, com $mdc(m, n)=1$. Daí,\n$$\n\\frac{x^{2019}+x+y^{2}}{x y}=\\frac{d^{2018} m^{2019}+m+d n^{2}}{d m n}\n$$\né um inteiro. Como $d$ divide $d^{2018} m^{2019}+m+d n^{2}$, podemos concluir que $d$ divide $m$. Além disso, de $mdc(m, n)=1$ e $\\frac{d^{2018} m^{2019}+m+d n^{2}}{m} \\in \\mathbb{Z}$, podemos concluir que $m$ divide $d$. Como $m$ e $d$ são positivos, $m=d$ e $x=d m=d^{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27005,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\na) Verifique que se $a \\in \\{1,2,4\\}$, então $n(a+n)$ não é um quadrado perfeito para qualquer inteiro positivo $n$.\nb) Verifique que se $a=2^{k}$, com $k \\geq 3$, então existe um inteiro positivo $n$ tal que $n(a+n)$ é um quadrado perfeito.\nc) Verifique que se $a \\notin \\{1,2,4\\}$, então sempre existe um inteiro positivo $n$ tal que $n(a+n)$ é um quadrado perfeito.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\na) Para $a \\in \\{1,2,4\\}$ e $n$ inteiro positivo, em virtude das desigualdades\n$$\nn^{2}<n(n+1)<n(n+2)<(n+1)^{2}\n$$\ne\n$$\n(n+1)^{2}<n(n+4)<(n+2)^{2}\n$$\npodemos concluir que $n(n+a)$ está entre dois quadrados perfeitos consecutivos e, consequentemente, não pode ser um quadrado perfeito.\n\nb) Se $a=2^{k}$, com $k \\geq 3$, podemos escrever $a=8l$, com $l$ inteiro positivo. Basta escolher $n=l$, pois daí $n(a+n)=9l^{2}=(3l)^{2}$.\n\nc) Se $a$ não é uma potência de 2, podemos escrever $a=(2x+1)y$, $\\operatorname{com}~x \\geq 1$ e $y \\geq 1$. Basta escolher $n=x^{2}y$, pois daí\n$$\n\\begin{aligned}\nn(a+n) & =x^{2}y\\left((2x+1)y+x^{2}y\\right) \\\\\n& =x^{2}y^{2}\\left(x^{2}+2x+1\\right) \\\\\n& =(xy(x+1))^{2}\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27006,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nExistem $n$ números em um quadro negro. A seguinte operação é realizada sobre esses números: dois números $a$ e $b$ são apagados e, em seguida, é escrito o número $\\frac{a+b}{4}$. A operação é repetida $n-1$ vezes. Como resultado, um único número permanece no quadro. Prove que se todos os números originais são iguais a 1, então o número resultante não é menor que $1 / n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nComo todo quadrado é não negativo, temos\n$$\n\\begin{aligned}\n(a-b)^{2} & \\geq 0 \\\\\n(a+b)^{2} & \\geq 4 a b \\\\\n\\frac{1}{a}+\\frac{1}{b} & \\geq \\frac{4}{a+b}\n\\end{aligned}\n$$\nEssa desigualdade implica na soma $S$ dos inversos dos números escritos no quadro não aumentar após cada operação. Inicialmente, a soma é igual a $n$. Portanto, no final do processo, teremos $S \\leq n$, ou seja, o número restante $\\frac{1}{S}$ é pelo menos $\\frac{1}{n}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27007,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTodo lado de um triângulo equilátero é dividido em $n$ partes iguais. Linhas paralelas aos lados do triângulo são desenhadas através desses pontos dividindo o triângulo em $n^{2}$ triângulos menores. Dizemos que uma sequência de triângulos distintos é uma cadeia se dois triângulos sucessivos compartilham um lado em comum. Qual é o maior número possível de triângulos em uma cadeia?",
    "options": [],
    "answer": "n^2 - n + 1",
    "solution": "Solution:\n\nA resposta é $n^{2}-n+1$ e uma cadeia com essa quantidade de triângulos é mostrada a seguir. Para mostrar que esse número é o máximo, pinte os triângulos de forma alternada com duas cores como também indica a figura abaixo. Existem $n$ triângulos brancos a mais que cinzas. Como os triângulos da cadeia são de cores alternadas, pode existir no máximo um triângulo branco a mais que um cinza. Portanto, pelo menos $n-1$ triângulos brancos não fazem parte da cadeia.\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27008,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEm um certo país, existem exatamente $2019$ cidades e entre quaisquer duas delas existe exatamente um voo direto operado por alguma companhia aérea, isto é, dadas as cidades $A$ e $B$ ou existe um voo de $A$ para $B$ ou um voo de $B$ para $A$. Encontre o menor número de companhias aéreas que operam no país, sabendo que os voos diretos entre quaisquer três cidades distintas são operados por companhias diferentes.",
    "options": [],
    "answer": "2019",
    "solution": "Solution:\n\nA resposta é $2019$. Como existem $1009$ pares disjuntos de cidades, cada companhia aérea pode operar em no máximo $1009$ pares. Existem exatamente $2019 \\cdot 2018/2$ voos diretos e, portanto, o número de companhias aéreas é pelo menos $\\frac{2019 \\cdot 2018}{2 \\cdot 1009} = 2019$. Resta exibirmos um exemplo para verificar que essa quantidade é realmente possível. Denote as cidades por $c_{1}, c_{2}, \\ldots, c_{2019}$ e as companhias por $f_{1}, f_{2}, \\ldots, f_{2019}$. Conecte as cidades $c_{i}$ e $c_{j}$ por um voo da companhia $f_{l}$ se o resto na divisão de $i+j$ por $2019$ é $l$. Para verificar que essa escolha de companhias satisfaz o enunciado, considere três cidades quaisquer $c_{i}, c_{j}$ e $c_{k}$. Para que exista uma mesma companhia operando em dois pares delas, no conjunto\n$$\n\\{i+j, i+k, k+j\\}\n$$\ndevem existir dois números com o mesmo resto na divisão por $2019$. Digamos que esses dois números sejam $i+j$ e $i+k$. A diferença entre eles será então um múltiplo de $2019$ e podemos escrever, para algum $m$ inteiro, que\n$$\nj-k = (i+j)-(i+k) = 2019 \\cdot m\n$$\nComo $j, k \\in \\{1,2,3, \\ldots, 2019\\}$ são distintos, $|j-k| \\in \\{1,2,3, \\ldots, 2018\\}$. Esse conjunto não contém nenhum múltiplo de $2019$ e, consequentemente, a última equação não admite solução. Isso mostra que nessa distribuição as companhias que operam os voos diretos entre quaisquer três cidades são distintas.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27009,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUma competição de matemática consiste de três problemas, cada um dos quais recebe uma nota inteira de 0 a 7. Para quaisquer dois competidores, sabemos que existe no máximo um problema em que eles obtiveram a mesma pontuação. Encontre o maior número possível de competidores nessa competição.",
    "options": [],
    "answer": "64",
    "solution": "Solution:\n\nExistem 8 pontuações possíveis para cada problema e, consequentemente, $8 \\cdot 8 = 64$ pontuações distintas possíveis para os dois primeiros problemas. Como não podem existir dois competidores com exatamente as mesmas pontuações nos dois primeiros problemas, o total de competidores não pode ser maior que 64. Iremos mostrar agora que esse valor máximo é realizável. Para isso, basta exibirmos uma distribuição de pontuações entre 64 jogadores satisfazendo às condições do enunciado. Considere a tabela:\n\n|   | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| 0 | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |\n| 1 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 0 |\n| 2 | 2 | 3 | 4 | 5 | 6 | 7 | 0 | 1 |\n| 3 | 3 | 4 | 5 | 6 | 7 | 0 | 1 | 2 |\n| 4 | 4 | 5 | 6 | 7 | 0 | 1 | 2 | 3 |\n| 5 | 5 | 6 | 7 | 0 | 1 | 2 | 3 | 4 |\n| 6 | 6 | 7 | 0 | 1 | 2 | 3 | 4 | 5 |\n| 7 | 7 | 0 | 1 | 2 | 3 | 4 | 5 | 6 |\n\nAs linhas indicarão a pontuação no primeiro problema e as colunas as do segundo problema. Para cada uma das 64 combinações possíveis de linhas e colunas, que realizam todas as 64 pontuações possíveis nos dois primeiros problemas, escolha como pontuação do terceiro problema o número escrito na interseção delas. Por exemplo, a combinação da linha de número 2 com a coluna de número 5 gera a pontuação $(2,5,7)$. Como não existem números repetidos nas linhas e colunas, todas essas triplas geram pontuações satisfazendo as condições do enunciado.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27010,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nVinte jogadores participaram de um torneio de xadrez. Cada jogador enfrentou todos outro jogador exatamente uma vez e cada partida terminou com a vitória de um dos jogadores ou em empate. Nesse torneio, notou-se que para cada partida que terminou em empate, cada um dos demais 18 jogadores venceu pelo menos um dos dois jogadores envolvidos nela. Sabemos ainda que pelo menos dois jogos terminaram em empate. Mostre que é possível nomear os jogadores como $P_{1}, P_{2}, \\ldots, P_{20}$ de modo que o jogador $P_{k}$ ganhou do jogador $P_{k+1}$, para cada $k \\in\\{1,2,3, \\ldots, 19\\}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nInicialmente mostraremos que cada jogador participou de no máximo um empate. Suponha, por absurdo, que o jogador $A$ empatou com os jogadores $B$ e $C$. Como $A$ empatou com $B$, pelas regras mencionadas, $C$ ganhou de $A$ ou de $B$. Dado que $C$ empatou com $A$, a única possibilidade é $C$ ter ganho de $B$. Por simetria, também podemos concluir que $B$ ganhou de $C$. Isso é um absurdo.\n\nConsidere agora a maior cadeia de jogadores $P_{1}, P_{2}, \\ldots, P_{t}$ com $P_{k}$ perdendo para $P_{k+1}$, para $k=1,2, \\ldots, t-1$. Nosso objetivo é mostrar que $t=20$. Suponha, novamente por absurdo, que existe um jogador $A$ que não está na cadeia. Assim, $P_{1}$ não pode ter ganho de $A$, pois, caso contrário, ele poderia ser incluído na cadeia. Se $A$ venceu $P_{1}$, então $A$ não empatou com $P_{2}$. Se $A$ perdeu para $P_{2}$, podemos inserir $A$ entre $P_{1}$ e $P_{2}$ e aumentar a cadeia, obtendo assim uma contradição. Então, se $A$ venceu $P_{1}$, então também venceu $P_{2}$. Podemos repetir o argumento com os demais membros da cadeia e concluir que $A$ ganhou de todos eles, podendo assim ser incluído no final da cadeia. Isso também é uma contradição.\n\nResta analisarmos o caso em que $A$ empatou com $P_{1}$. Como $P_{1}$ empatou no máximo uma vez, podemos assumir $t=19$, pois, caso contrário, podemos trocar de jogador e recair na situação já tratada no último parágrafo. Se $A$ venceu $P_{2}$, também deve ter vencido $P_{3}$, do contrário poderíamos inserir $A$ entre $P_{3}$ e $P_{2}$. Repetindo esse argumento, podemos concluir que $A$ venceu todos os demais e assim pode ser incluído no final da cadeia, gerando um absurdo. Então $A$ perdeu para $P_{2}$ e, de forma semelhante, para todos os demais entre $P_{2}$ e $P_{19}$. Como existem pelo menos dois empates, devem existir $i$ e $j$, com $i, j \\geq 2$, tais que $P_{i}$ e $P_{j}$ empataram. Entretanto, $A$ não pode ter vencido nenhum desses dois jogadores e isso gera um novo absurdo. Logo, a cadeia maximal deve conter todos os jogadores.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27011,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nOs três inteiros positivos $a, b$ e $c$ satisfazem\n$$\n4^{a} \\cdot 5^{b} \\cdot 6^{c}=8^{8} \\cdot 9^{9} \\cdot 10^{10}\n$$\nDetermine o valor de $a+b+c$.",
    "options": [],
    "answer": "36",
    "solution": "Solution:\n$$\n\\begin{aligned}\n4^{a} \\cdot 5^{b} \\cdot 6^{c} & =8^{8} \\cdot 9^{9} \\cdot 10^{10} \\\\\n2^{2a} \\cdot 5^{b} \\cdot (2 \\cdot 3)^{c} & = (2^{3})^{8} \\cdot (3^{2})^{9} \\cdot (2 \\cdot 5)^{10} \\\\\n2^{2a} \\cdot 5^{b} \\cdot 2^{c} \\cdot 3^{c} & = 2^{24} \\cdot 3^{18} \\cdot 2^{10} \\cdot 5^{10} \\\\\n2^{2a+c} \\cdot 3^{c} \\cdot 5^{b} & = 2^{34} \\cdot 3^{18} \\cdot 5^{10} \\\\\n2^{2a+c-34} \\cdot 3^{c-18} \\cdot 5^{b-10} & = 1\n\\end{aligned}\n$$\nPelo Teorema Fundamental da Aritmética, todos os expoentes do membro esquerdo da última equação são nulos. Daí, $c=18$, $2a+c=34$ e $b=10$. Substituindo o valor de $c$ na segunda equação, encontramos $a=8$. Portanto, $a+b+c=18+8+10=36$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27012,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nInicialmente, o número $1$ e dois números positivos $x$ e $y$ estão escritos em um quadro negro. Em cada movimento, um jogador pode escolher dois números sobre o quadro, não necessariamente distintos, e escrever a sua soma ou a sua diferença no quadro. Também podemos escolher um número não nulo no quadro e escrever o seu inverso. Após um número finito de movimentos, descreva como podemos obter os seguintes números:\n\na) $x^{2}$.\n\nb) $x y$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na) Se $x=1$ não há o que fazer. Suponhamos então que $x \\neq 1$. Primeiramente, escreva $x+1$ e $x-1$. Usando o movimento do inverso, podemos escrever $\\frac{1}{x+1}$ e $\\frac{1}{x-1}$. Em seguida, podemos escrever a diferença desses dois números: $\\frac{2}{x^{2}-1}$. O inverso desse último número é $p=\\frac{x^{2}-1}{2}$. Calculando $p+p$ encontramos $x^{2}-1$. Finalmente, basta adicionar $1$ para obter $x^{2}-1+1=x^{2}$.\n\nb) Primeiramente, escreva $x+y$. Pelo item anterior, podemos escrever $(x+y)^{2}$, $x^{2}$ e $y^{2}$. Em seguida, com dois movimentos, podemos escrever $(x+y)^{2}-x^{2}$ e $(x+y)^{2}-x^{2}-y^{2}=2 x y$. Também podemos escrever o inverso desse último número: $q=\\frac{1}{2 x y}$. Finalmente, podemos escrever $q+q=\\frac{1}{x y}$ e o seu inverso $x y$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27013,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEm um determinado jogo, o número $1$ está escrito no quadro. Em qualquer momento, um movimento permitido consiste em trocar o número escrito no quadro pelo seu dobro ou por outro número que possui os mesmos dígitos que ele. Por exemplo, se estiver escrito no quadro o número $137$, um movimento permitido consiste em trocá-lo por $137 \\cdot 2 = 274$ ou por $173$, $317$, $371$, $713$ ou $731$. Determine se após um número finito de operações é possível obtermos os seguintes números:\n\na) $10^{3}$.\n\nb) $10^{9}$.\n\nc) $9876543210$.",
    "options": [],
    "answer": "a) yes; b) yes; c) no",
    "solution": "Solution:\n\na) Sim, é possível. Após realizar o movimento de multiplicação por $2$ nove vezes, podemos trocar o $1$ original por $2^{9} = 512$. Em seguida, podemos trocá-lo por $125$. Multiplicando-o por $2$ três vezes, podemos trocá-lo por $125 \\cdot 2^{3} = 1000$.\n\nb) Sim, também é possível. Note que $1000 = 1 \\cdot 10^{3}$. Podemos repetir as operações do item anterior trocando o fator $1$ por $10^{3}$:\n$$\n1 \\cdot 10^{3} \\rightarrow 512000 \\rightarrow 125000 \\rightarrow 1000000.\n$$\nEsse último número pode ser escrito como $1 \\cdot 10^{6}$. Novamente, repetindo as operações do primeiro item, podemos obter as seguintes trocas:\n$$\n1 \\cdot 10^{6} \\rightarrow 512 \\cdot 10^{6} \\rightarrow 125 \\cdot 10^{6} \\rightarrow 1000 \\cdot 10^{6}\n$$\nEsse último número é igual a $10^{9}$.\n\nc) Não é possível. Quando um número que não é múltiplo de $3$ é multiplicado por $2$, ele continua sendo um número que não é múltiplo de $3$. Em virtude do critério de divisibilidade por $3$, o mesmo acontece quando permutamos os seus dígitos. Como o número inicial não é divisível por $3$, não é possível após algumas das operações descritas trocá-lo por qualquer múltiplo de $3$. Como a soma dos dígitos de $9876543210$ é um múltiplo de $3$, é impossível obtê-lo.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27014,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm paralelepípedo deve ser construído com a sobreposição de cubinhos de $1~\\mathrm{cm}$ de medida de aresta, sendo seu comprimento composto por $n$ cubinhos, sua largura, por $p$ cubinhos e sua altura por $q$ cubinhos.\n\na) Qual o volume do paralelepípedo?\n\nb) Pintando as faces do paralelepípedo de vermelho, quantos cubinhos terão apenas uma de suas faces pintada de vermelho?\n\nc) Tomando um paralelepípedo, como o do enunciado, de forma que $n=p=q$. Se aumentarmos cada uma de suas dimensões em $a$, sendo $a$ um número natural, o novo cubo passa a ter 98 cubinhos a mais que o cubo inicial. Quais os valores de $n$ e $a$ ?",
    "options": [],
    "answer": "a) V = npq. b) 2[(p−2)(q−2) + (n−2)(p−2) + (n−2)(q−2)]. c) n = 3 and a = 2.",
    "solution": "Solution:\n\na) $V = n p q$.\n\nb) Nas faces $n \\times p$, apenas os cubinhos que não estão nas laterais terão apenas uma face pintada de vermelho, ou seja, são $2 \\cdot (n-2) \\cdot (p-2)$ cubinhos. De forma análoga, nas faces $n \\times q$, serão $2 \\cdot (n-2) \\cdot (q-2)$ e nas faces $p \\times q$ são $2 \\cdot (p-2) \\cdot (q-2)$. Portanto, são $2[(p-2)(q-2)+(n-2)(p-2)+(n-2)(q-2)]$ cubinhos com apenas uma face vermelha.\n\nc) Temos um cubo de aresta $n$, ou seja, formado de $n^{3}$ cubinhos. O novo cubo terá $(n+a)^{3}$ cubinhos, que equivale a 98 cubinhos a mais. Assim:\n$$\n\\begin{aligned}\n(n+a)^{3}-n^{3} & =98 \\\\\n3 n^{2} a+3 n a^{2}+a^{3} & =98 \\\\\na\\left(3 n^{2}+3 n a+a^{2}\\right) & =2 \\cdot 7^{2}\n\\end{aligned}\n$$\nComo $a$ é um número natural, os possíveis valores de $a$ pertencem ao conjunto $\\{1,2,7,14,49,98\\}$. Vamos analisar cada caso:\n\nI) Se $a=1$, então $3 n^{2}+3 n=97$, mas não teríamos $n \\in \\mathbb{N}$.\n\nII) Se $a=2$, então $3 n^{2}+6 n=45$, donde $n=3$.\n\nIII) Para $a=7$, $a=14$, $a=49$ ou $a=98$, não teríamos $n \\in \\mathbb{N}$.\n\nPortanto, $a=2$ e $n=3$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27015,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa figura a seguir, o quadrado maior possui área de $1\\ \\mathrm{cm}^2$ e o quadrado do meio área $M$. A área do quadrado menor, que possui um vértice sobre um lado do quadrado do meio, é $N$. Qual o valor de $N$ em função de $M$ ?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "N = ((1 - M)/2)^2",
    "solution": "Solution:\n\nSejam $AQ = x\\ \\mathrm{cm}$, $QR = a\\ \\mathrm{cm}$ e $AG = s\\ \\mathrm{cm}$. Como a área do quadrado maior é $1\\ \\mathrm{cm}^2$, segue que $AD = 1\\ \\mathrm{cm}$. Os triângulos retângulos $ARQ$ e $DQP$ possuem os mesmos ângulos, pois\n$$\n\\begin{aligned}\n\\angle AQR & = 180^\\circ - \\angle PQR - \\angle DQP \\\\\n& = 90^\\circ - \\angle DQP \\\\\n& = \\angle DPQ\n\\end{aligned}\n$$\nComo $QR = QP$, os triângulos $DPQ$ e $AQR$ são congruentes. Assim, $AR = DQ = 1 - x$. De $QF \\parallel AR$, segue que os triângulos $QGF$ e $QAR$ são semelhantes e assim\n$$\n\\begin{aligned}\n\\frac{s}{1-x} & = \\frac{x-s}{x} \\\\\nsx & = x - x^2 - s + sx \\\\\ns & = x - x^2\n\\end{aligned}\n$$\nPelo Teorema de Pitágoras aplicado ao triângulo $QAR$, segue que\n$$\n\\begin{aligned}\n(1-x)^2 + x^2 & = a^2 \\\\\n1 - a^2 & = 2x - 2x^2 \\\\\n1 - M & = 2s\n\\end{aligned}\n$$\nDaí,\n$$\nN = s^2 = \\left(\\frac{1 - M}{2}\\right)^2\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27016,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm conjunto contém 4 números. As seis somas de dois elementos desse conjunto são 189, 320, 287, 264, $x$ e $y$. Encontre o maior valor possível para $x+y$.",
    "options": [],
    "answer": "761",
    "solution": "Solution:\n\nSejam $a, b, c$ e $d$ os quatro números do conjunto. Temos dois casos a considerar:\n\nI) $x = a + b$ e $y = c + d$ (somas sem parcelas em comum). Então $a + c$, $a + d$, $b + c$ e $b + d$ são, em alguma ordem, os números 189, 320, 287 e 264. Adicionando essas quatro somas, obtemos $a + b + c + d = 530$. Assim, $x + y = 530$.\n\nII) $x = a + b$ e $y = a + c$ (somas com uma parcela em comum). Nesse caso, $a + d$, $b + c$, $b + d$ e $c + d$ são, em alguma ordem, os números 189, 320, 287 e 264. Temos a seguinte estimativa:\n$$\n\\begin{aligned}\nx + y & = 2(a + d) + 2(b + c) - (b + d) - (c + d) \\\\\n& \\leq 2(320 + 287) - (264 + 189) \\\\\n& = 761\n\\end{aligned}\n$$\nEsse valor pode ser obtido com o exemplo $(a, b, c, d) = (237, 181, 106, 83)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27017,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO segmento $AB$ de comprimento $16~\\mathrm{cm}$ é diâmetro de um círculo de centro $O$. Uma reta secante corta o círculo em $C$ e $D$ e a reta $AB$ em $P$, como indica a figura a seguir. Se $OD = DP$ e $\\angle APC = 18^\\circ$, qual o valor do ângulo $\\angle AOC$?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "54°",
    "solution": "Solution:\n\nComo $OD = DP$, segue que $\\angle DOP = \\angle OPD = 18^\\circ$. Pelo Teorema do Ângulo Externo, segue que $\\angle ODC = 2 \\cdot 18^\\circ = 36^\\circ$. O triângulo $COD$ é isósceles, pois $CO$ e $OD$ são raios do círculo. Assim, $\\angle OCD = \\angle CDO = 36^\\circ$. Novamente, pelo Teorema do Ângulo Externo, segue que $\\angle COA = \\angle OCP + \\angle CPO = 54^\\circ$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27018,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA figura abaixo representa uma estante com duas prateleiras com cinco pilhas de livros, sendo três delas com dois livros e duas delas com apenas um livro. Alice e Luiz inventaram um jogo no qual cada um deles, alternadamente, retira um ou dois livros de uma das pilhas de livros. Vence aquele que tirar o último livro. Alice começa o desafio. Qual deles tem uma estratégia vencedora, quaisquer que sejam as jogadas do adversário?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Alice",
    "solution": "Solution:\n\nAquele que deixar duas pilhas com 1 livro cada para o adversário jogar, vence o jogo. Para conseguir isso, basta Alice tirar, em seu primeiro lance, uma pilha com 2 livros. Após isso, Luiz terá 3 opções, que são:\n\nI) Tirar uma pilha com 2 livros: basta Alice tirar a outra pilha com 2 livros que chegará à posição vencedora citada no início da solução.\n\nII) Tirar uma pilha com apenas 1 livro: basta Alice tirar a outra pilha com 1 livro apenas, deixando duas pilhas com 2 livros cada, que ela vence no próximo lance ou chega à posição vencedora do início da solução.\n\nIII) Tirar 1 livro de uma pilha com 2 livros: basta Alice tirar 1 livro da pilha com 2 que restarão 4 pilhas com 1 livro e Alice chegará à posição vencedora do início da solução no próximo lance. Portanto, Alice tem a estratégia vencedora.\n\nOutra maneira é passar as quantidades das pilhas para notação binária: 10, 10, 10, 1, 1. Quando Alice for jogar, basta ela tirar uma quantidade que, somando os valores restantes, obtenha-se um resultado apenas com números pares, por exemplo, em seu primeiro lance, se ela tirar uma pilha com 2 livros, a soma dos valores restantes será $10+10+1+1=22$. Fazendo isso, em todos os seus lances ela conseguirá chegar à posição vencedora.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27019,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNo triângulo retângulo isósceles $A O B$, os pontos $P, Q$ e $S$ são escolhidos sobre os lados $O B, O A$ e $A B$, respectivamente, de modo que $P Q R S$ é um quadrado. Se os comprimentos de $O P$ e $O Q$ são $a$ e $b$, respectivamente, e a área do quadrado $P Q R S$ é $2/5$ da área do triângulo $A O B$, determine o valor de $a / b$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "2",
    "solution": "Solution:\n\nSeja $C$ o pé da perpendicular do ponto $S$ ao segmento $O B$. Os triângulos $S P C$ e $P Q O$ possuem os mesmos ângulos, pois\n$$\n\\begin{aligned}\n\\angle C P S & = \\angle 180^\\circ - \\angle S P Q - \\angle O P Q \\\\\n& = 90^\\circ - \\angle O P Q \\\\\n& = \\angle P Q O\n\\end{aligned}\n$$\nComo $P S = P Q$, esses triângulos são congruentes pelo caso A.L.A. Assim $P C = b$, $C S = a$. Uma vez que $B S C$ é um triângulo retângulo isósceles, obtemos $O B = 2a + b$. Consequentemente, a área do triângulo $A O B$ é $(2a + b)^2 / 2$. Pelo Teorema de Pitágoras, a área do quadrado $P Q R S$ é $P Q^2 = a^2 + b^2$. Daí\n$$\n\\begin{aligned}\n\\frac{A_{P Q R S}}{A_{O A B}} & = \\frac{2}{5} \\\\\n5\\left(a^2 + b^2\\right) & = (2a + b)^2 \\\\\na^2 + 4b^2 & = 4ab \\\\\n(a - 2b)^2 & = 0\n\\end{aligned}\n$$\nAssim, $a = 2b$ e a razão procurada é $a / b = 2$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27020,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAche todos os valores de $x$ satisfazendo\n$$\n\\frac{x+\\sqrt{x+1}}{x-\\sqrt{x+1}}=\\frac{11}{5}\n$$",
    "options": [],
    "answer": "8",
    "solution": "Solution:\n\nA equação pode ser reescrita como\n$$\n\\begin{aligned}\n\\frac{x+\\sqrt{x+1}}{x-\\sqrt{x+1}} &= \\frac{11}{5} \\\\\n5x + 5\\sqrt{x+1} &= 11x - 11\\sqrt{x+1} \\\\\n16\\sqrt{x+1} &= 6x \\\\\n8\\sqrt{x+1} &= 3x\n\\end{aligned}\n$$\nElevando os membros da última equação ao quadrado, temos\n$$\n\\begin{aligned}\n9x^{2} &= 64(x+1) \\\\\n9x^{2} - 64x - 64 &= 0\n\\end{aligned}\n$$\nAs raízes dessa equação são $x=8$ e $x=-\\frac{8}{9}$. Dessas soluções, apenas $x=8$ satisfaz a equação dada.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27021,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNo quadriculado da figura, dizemos que dois ou mais pontos nas intersecções das linhas são equilegais, em relação a um ponto fixo, quando suas distâncias são iguais a este. Por exemplo, $B$ e $C$ são equilegais em relação a $A$, mas $B$ e $D$ não são.\n\n![](attached_image_1.png)\n\na) Marque, no quadriculado a seguir, os pontos equilegais a $K$ em relação a $J$.\n\n![](attached_image_2.png)\n\nb) Se os pontos não precisarem ser nas intersecções das linhas, que figura será formada no plano do quadriculado dos pontos equilegais a $K$ em relação a $J$ ?\n\nc) Se os pontos não precisarem ser no plano do quadriculado, que figura será formada dos pontos equilegais a $K$ em relação a $J$ ?",
    "options": [],
    "answer": "a) All grid intersection points at distance sqrt(5) from J (i.e., those with displacements (±2, ±1) or (±1, ±2) from J). b) A circle of radius sqrt(5) centered at J. c) A sphere of radius sqrt(5) centered at J.",
    "solution": "Solution:\n\na) Supondo que a medida do lado de cada quadradinho seja $1~\\mathrm{cm}$, então a distância $d$ de $K$ a $J$ pode ser determinada pelo Teorema de Pitágoras: $d^{2} = 2^{2} + 1^{2}$, donde concluímos que $d = \\sqrt{5}~\\mathrm{cm}$. Todos os pontos das intersecções que estiverem à distância de $\\sqrt{5}~\\mathrm{cm}$ de $J$ serão equilegais a $K$. Vamos marcá-los:\n\n![](attached_image_3.png)\n\nb) O conjunto de pontos equidistantes de $J$ é uma circunferência cuja medida do raio é $\\sqrt{5}~\\mathrm{cm}$.\n\nc) No espaço, o conjunto de pontos equidistantes a um ponto qualquer é uma superfície esférica.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27022,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nJosé quer preencher as casas de um tabuleiro $2 \\times n$ com zeros e uns de modo que dois números vizinhos iguais, em uma mesma linha, impeçam que se preencha também com números iguais as casas correspondentes da outra linha. Por exemplo, no desenho abaixo, os valores de $A$ e $B$ não podem ser iguais.\n\n| 0 | 1 | 0 | $\\cdots$ | 1 | 1 | $\\cdots$ |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| 1 | 1 | 0 | $\\cdots$ | $A$ | $B$ | $\\cdots$ |\n\na) Encontre todos os preenchimentos possíveis do tabuleiro abaixo:\n\n| 0 | 0 | 1 |\n| :--- | :--- | :--- |\n| 1 |   |   |\n\nb) No tabuleiro, todas as colunas já estão preenchidas, exceto as duas últimas. De quantas maneiras os números das casas $A$ e $B$ podem ser escolhidos?\n\n| $\\cdots$ | 0 | $A$ |\n| :--- | :--- | :--- |\n| $\\cdots$ | 1 | $B$ |\n\nc) De quantas maneiras José pode preencher o tabuleiro se $n=2020$ ?",
    "options": [],
    "answer": "a) Bottom row can be 1 0 0 or 1 0 1. b) 3. c) 4*3^2019.",
    "solution": "Solution:\n\na) Temos os preenchimentos:\n\n| 0 | 0 | 1 |\n| :--- | :--- | :--- |\n| 1 | 0 | 0 |\n| 0 | 0 | 0 |\n| :--- | :--- | :--- |\n| 1 | 0 | 1 |\n| 0 | 0 | 1 |\n| :--- | :--- | :--- |\n| 1 | 0 | 1 |\n\nb) A princípio, existem 2 escolhas possíveis para $A$ e outras duas possíveis para $B$. Apenas a escolha $A=0$ e $B=1$ é proibida. Portanto, existem $2 \\cdot 2 - 1 = 3$ escolhas possíveis.\n\nc) Começando a preencher da esquerda para a direita, existem 4 tipos possíveis de colunas e as regras se resumem a não preenchermos uma certa coluna com a mesma configuração da coluna imediatamente anterior. Assim, uma vez que José escolheu os números de uma coluna, ele possui 3 opções de preenchimento para a próxima. No início, podemos escolher livremente como preencher a primeira coluna mais à esquerda e depois continuar a escolha das demais da esquerda para a direita. Pelo Princípio Multiplicativo, o total de preenchimentos é:\n$$\n4 \\cdot 3^{2019}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27023,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nJuca é um escoteiro que está explorando as proximidades do seu acampamento. Após coletar frutas e madeira, ele deve pegar água no rio e voltar para sua barraca. Vamos representar, na figura, Juca pela letra $J$, o rio pela letra $r$ e sua barraca pela letra $B$. A distância dos pés das perpendiculares $C$ e $E$, em $r$ dos pontos $J$ e $B$ é $180\\ \\mathrm{m}$.\n![](attached_image_1.png)\nQual a menor distância que Juca pode percorrer para voltar para sua barraca, passando pelo rio?",
    "options": [],
    "answer": "180√2 m",
    "solution": "Solution:\nVamos marcar o ponto $B'$, simétrico do ponto $B$, em relação à reta $r$. O ponto $B'$ também dista $80\\ \\mathrm{m}$ da reta $r$ e, com isso, temos os triângulos congruentes $\\triangle B C D$ e $\\triangle B' C D$ (lado-ângulo-lado), sendo $D$ o ponto que Juca pegará água no rio. Como a menor distância entre $J$ e $B'$ é o comprimento de um segmento de reta unindo esses dois pontos, $J$, $D$ e $B'$ devem estar alinhados. Seja $x$ a medida do segmento $\\overline{D E}$, vamos aplicar a razão de semelhança nos triângulos $\\triangle B C D$ e $\\triangle J E D$:\n$$\n\\begin{aligned}\n\\frac{180-x}{80} & = \\frac{x}{100} \\\\\n\\frac{180-x}{4} & = \\frac{x}{5} \\\\\n4x & = 900 - 5x \\\\\n9x & = 900 \\\\\nx & = 100\\ \\mathrm{m}\n\\end{aligned}\n$$\nAplicando agora o Teorema de Pitágoras nos triângulos $\\triangle B C D$ e $\\triangle J E D$, podemos encontrar a menor distância entre $J$ e $B$, passando por $r$:\n$$\n\\begin{aligned}\nJ D + D B & = \\sqrt{100^2 + 100^2} + \\sqrt{80^2 + 80^2} \\\\\n& = 100\\sqrt{2} + 80\\sqrt{2} \\\\\n& = 180\\sqrt{2}\\ \\mathrm{m}\n\\end{aligned}\n$$\n![](attached_image_2.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27024,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNo cubo $ABCD EFGH$, cuja aresta mede $6~\\mathrm{cm}$, o ponto $M$ é ponto médio de $\\overline{EF}$.\n![](attached_image_1.png)\n\na) Determine a área do triângulo $AMH$.\n\nb) Determine o volume da pirâmide $AMHE$. (O volume de uma pirâmide pode ser calculado pela terça parte do produto entre a área da base e a altura relativa a esta base).\n\nc) Calcule a medida da altura relativa à base $AMH$.",
    "options": [],
    "answer": "a) 9√6 cm²; b) 18 cm³; c) √6 cm",
    "solution": "Solution:\n\na) $\\overline{AH}$ é a diagonal de uma face, ou seja, $AH = 6 \\sqrt{2}~\\mathrm{cm}$. $\\overline{MH}$ e $\\overline{AM}$ são hipotenusas de triângulos cujos catetos medem $6~\\mathrm{cm}$ e $3~\\mathrm{cm}$, ou seja, $MH = AM = 3\\sqrt{5}~\\mathrm{cm}$. Traçando a altura $MM'$ do triângulo $AMH$, relativa ao lado $\\overline{AH}$, podemos encontrá-la aplicando o Teorema de Pitágoras, ou seja, $MM' = 3\\sqrt{3}~\\mathrm{cm}$. Por fim, a área do triângulo $AMH$ é $\\dfrac{6\\sqrt{2} \\cdot 3\\sqrt{3}}{2} = 9\\sqrt{6}~\\mathrm{cm}^2$.\n\n![](attached_image_2.png)\n\nb) Utilizando como base da pirâmide $AMHE$ o triângulo $EHM$, cuja área é $\\dfrac{6 \\cdot 3}{2} = 9~\\mathrm{cm}^2$, sua altura será $\\overline{AE}$ e seu volume será $\\dfrac{9 \\cdot 6}{3} = 18~\\mathrm{cm}^3$.\n\nc) Para o cálculo da medida da altura relativa à base $AMH$ da pirâmide $AMHE$, vamos calcular o volume desta pirâmide de duas maneiras: a primeira com base $AMH$ e altura $h$, que é o que procuramos, e a segunda com base $MHE$.\n\n$$\n\\begin{aligned}\n\\frac{9\\sqrt{6} \\cdot h}{3} &= 18 \\\\\nh &= \\frac{6}{\\sqrt{6}} \\\\\nh &= \\sqrt{6}~\\mathrm{cm}\n\\end{aligned}\n$$",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 27025,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOs triângulos $A B C$ e $A B D$ estão inscritos na mesma semicircunferência de diâmetro $A B$, que mede $15~\\mathrm{cm}$. Se traça por $D$ a perpendicular a $A B$ que intersecta $A B$ em $P$, o segmento $A C$ em $Q$ e o prolongamento do lado $B C$ em $R$. Além disso, $P R=\\frac{40}{3}~\\mathrm{cm}$ e $P Q=\\frac{15}{4}~\\mathrm{cm}$.\na) Verifique que $P Q \\cdot P R=B P \\cdot A P$.\nb) Encontre a medida do segmento $D P$.",
    "options": [],
    "answer": "5*sqrt(2)",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\na) Como $A B$ é um diâmetro, segue que $\\angle A C B=90^{\\circ}$. Daí\n$$\n\\angle P R B=90^{\\circ}-\\angle P B R=\\angle B A C\n$$\nOs triângulos $A P Q$ e $P R B$ são semelhantes, pois possuem ângulos de mesmas medidas. Portanto,\n$$\n\\begin{aligned}\n\\frac{P Q}{A P} & =\\frac{B P}{P R} \\\\\nP Q \\cdot P R & =B P \\cdot A P\n\\end{aligned}\n$$\n\nb) Os triângulos retângulos $A D P$ e $A B D$ são semelhantes, pois compartilham um dos ângulos agudos. O mesmo ocorre com os triângulos retângulos $A B D$ e $D B P$. Com mais razão, $A D P$ é semelhante a $D B P$ e assim\n$$\n\\begin{aligned}\n\\frac{D P}{A P} & =\\frac{B P}{D P} \\\\\nD P^{2} & =A P \\cdot B P\n\\end{aligned}\n$$\nPelo item anterior, segue que\n$$\n\\begin{aligned}\nD P^{2} & =P Q \\cdot P R \\\\\nD P & =\\sqrt{\\frac{40}{3} \\cdot \\frac{15}{4}} \\\\\n& =5 \\sqrt{2}\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27026,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEm um tabuleiro $4 \\times 4$, deve-se colocar os números de 1 a 16 nas casas, sem repetir, de forma que a soma dos números de cada linha, coluna e diagonal seja a mesma. Chamamos essa soma de Soma Mágica.\n\na) Qual a Soma Mágica deste tabuleiro?\n\nb) Se a soma das casas marcadas com $X$ no tabuleiro abaixo é 34, qual a soma das casas marcadas com $Y$?\n\n| $Y$ |   |   | $Y$ |\n| :--- | :--- | :--- | :--- |\n|   | $X$ | $X$ |   |\n|   | $X$ | $X$ |   |\n| $Y$ |   |   | $Y$ |\n\nc) Se preenchermos com naturais consecutivos de $k$ a $(k+15)$, de forma que a Soma Mágica seja 50, qual o valor de $k$ ?",
    "options": [],
    "answer": "a) 34; b) 34; c) 5",
    "solution": "Solution:\n\na) Como são 4 linhas (assim como 4 colunas), a Soma Mágica vale:\n$$\n\\frac{1+2+3+\\ldots+16}{4}=34\n$$\n\nb) Se somarmos as duas diagonais, teremos exatamente a soma das casas marcadas com $X$ e com $Y$. Assim, a soma das casas marcadas com $Y$ é $2 \\cdot 34 - 34 = 34$.\n\nc) Temos:\n$$\n\\begin{aligned}\n\\frac{k+(k+1)+(k+2)+\\ldots+(k+15)}{4} & = 50 \\\\\n16k + (1+2+3+\\ldots+15) & = 200 \\\\\n16k + \\frac{(1+15)\\,15}{2} & = 200 \\\\\n16k + 120 & = 200 \\\\\n16k & = 80 \\\\\nk & = 5\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27027,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEm uma escola, devem ser formados $n$ clubes, com $n \\geq 3$ e cada um com 3 integrantes, de modo que para cada par de clubes haja exatamente um estudante que integra ambos.\na) Dê um exemplo de uma distribuição de 7 clubes que satisfaçam as condições mencionadas.\nb) Verifique que se um estudante pertence a 4 clubes, então ele deve pertencer a todos os demais.\nc) Determine o valor mínimo de $n$ de modo que para qualquer conjunto de clubes que satisfaçam essas condições seja obrigatória a presença de um mesmo estudante em todos eles.",
    "options": [],
    "answer": "8",
    "solution": "Solution:\n\na) Nomeando os alunos por $A, B, C, D, E, F$ e $G$, os seguintes conjuntos representam 7 clubes que atendem às condições do enunciado:\n$$\n\\{A, B, C\\},\\{A, D, E\\},\\{A, F, G\\},\\{B, E, F\\},\\{B, D, G\\},\\{C, D, F\\},\\{C, E, G\\}\n$$\n\nb) Suponha que o estudante $A$ pertença aos seguintes 4 clubes:\n$$\n\\{A, B, C\\},\\{A, D, E\\},\\{A, F, G\\},\\{A, H, I\\}\n$$\nSuponha que existe um clube $K$ formado pelos alunos $\\{M, N, P\\}$ do qual $A$ não faz parte. Então, $K$ deve possuir pelo menos um elemento de cada um dos seguintes 4 conjuntos disjuntos:\n$$\n\\{B, C\\},\\{D, E\\},\\{F, G\\},\\{H, I\\}\n$$\nIsso é um absurdo, pois para que isso aconteça $K$ deveria ter pelo menos 4 elementos. Portanto, $A$ deve pertencer a todos os clubes.\n\nc) Afirmamos que o valor mínimo é $n=8$. Pelo item anterior, basta verificarmos que nesse caso algum aluno deve pertencer a pelo menos 4 clubes. Suponha que isso não ocorre, ou seja, que cada aluno pertence a no máximo 3 clubes e considere um clube genérico $w$ formado pelos alunos $\\{X, Y, Z\\}$. Cada um dos elementos de $w$ pode estar em no máximo mais outros 2 clubes. Como qualquer outro clube deve compartilhar um aluno em comum com $w$, contando os demais clubes que contêm $X, Y$ ou $Z$, concluímos que podem existir no máximo mais $2+2+2=6$ outros clubes distintos de $w$. Assim, o total de clubes seria no máximo $6+1=7$ e isso é um absurdo.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27028,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAnalisando os números naturais de 4 algarismos:\n\na) Quantos deles têm todos os algarismos diferentes?\n\nb) Quantos têm o algarismo 1 exatamente uma vez e todos os algarismos diferentes?\n\nc) Quantos têm o algarismo 1?",
    "options": [],
    "answer": "a) 4536; b) 1848; c) 3168",
    "solution": "Solution:\n\na) Para escolher a unidade do milhar, temos 9 possibilidades, já que não podemos utilizar o zero, pois não seria um número de quatro algarismos; para escolher o algarismo da centena, temos 9 possibilidades, pois já utilizamos um dos algarismos; para a dezena, são 8 possibilidades; e para a unidade são 7 possibilidades. Portanto são $9 \\cdot 9 \\cdot 8 \\cdot 7 = 4.536$ possibilidades ao todo.\n\nb) Vamos dividir em casos:\n\nI) Fixando o $1$ na casa da unidade do milhar, e não o utilizando mais, são $9 \\cdot 8 \\cdot 7 = 504$ possibilidades.\n\nII) Fixando agora o $1$ na casa da centena, da dezena ou da unidade, são $8 \\cdot 8 \\cdot 7 = 448$ possibilidades em cada um dos casos (lembrando que na casa da unidade do milhar deste caso não podemos utilizar $0$ nem $1$), ou seja, são $3 \\cdot 448 = 1.344$ possibilidades.\n\nPortanto, o total de possibilidades é $504 + 1.344 = 1.848$.\n\nc) A quantidade de números de 4 algarismos que NÃO possuem o algarismo $1$ é $8 \\cdot 9 \\cdot 9 \\cdot 9 = 5.832$. Como são $9.000$ números com 4 algarismos, destes, $9.000 - 5.832 = 3.168$ apresentam o algarismo $1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27029,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAs progressões geométricas $a_{1}, a_{2}, a_{3}, \\ldots$ e $b_{1}, b_{2}, b_{3}, \\ldots$ possuem a mesma razão, com $a_{1}=27$, $b_{1}=99$ e $a_{15}=b_{11}$. Encontre $o$ valor de $a_{9}$.",
    "options": [],
    "answer": "363",
    "solution": "Solution:\n\nSeja $r$ o valor da razão comum das duas progressões geométricas. Temos\n$$\n\\begin{aligned}\na_{15} & =a_{1} r^{14} \\\\\n& =27 r^{14} \\\\\nb_{11} & =b_{1} r^{10} \\\\\n& =99 r^{10}\n\\end{aligned}\n$$\nPortanto, $27 r^{14}=99 r^{10}$ e daí $3 r^{4}=11$. Finalmente\n$$\n\\begin{aligned}\na_{9} & =a_{1} r^{8} \\\\\n& =27\\left(r^{4}\\right)^{2} \\\\\n& =27\\left(\\frac{11}{3}\\right)^{2} \\\\\n& =\\frac{27 \\cdot 121}{9} \\\\\n& =363\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27030,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nExistem três cartões, cada um com um número do conjunto $\\{1,2, \\ldots, 10\\}$. Esses três cartões foram embaralhados e distribuídos a três pessoas, que registraram os números em seus respectivos cartões. Os cartões foram então coletados e o processo foi repetido novamente. Após algumas repetições, cada uma das três pessoas somou os seus registros. Sabendo que as somas obtidas foram 13, 15 e 23, quais eram os números nos cartões?",
    "options": [],
    "answer": "3, 5, 9",
    "solution": "Solution:\n\nSejam $x$, $y$ e $z$ os números escritos nos três cartões, com $x \\leq y \\leq z$. A cada etapa do processo de sorteio, a soma dos números dos três cartões é sempre $x+y+z$. Como $13+15+23=51=3 \\cdot 17$ e tanto $3$ quanto $17$ são números primos, segue que foram realizados $3$ sorteios e $x+y+z=17$.\n\nAnalisando a pessoa que obteve a soma $23$, devemos ter $z \\geq 8$, pois caso contrário a soma máxima seria $7 \\cdot 3=21$. Se $z=10$, nas somas $13$ e $15$ podem aparecer no máximo uma parcela $z$. Como $x+y=17-z=7$, segue que $(x, y, z)=(3,4,10)$, $(2,5,10)$ ou $(1,6,10)$. Nos dois primeiros casos, não é possível obter soma $13$ e no último não é possível obter soma $15$ com três parcelas.\n\nSe $z=8$, então novamente nas somas $13$ e $15$ podem aparecer no máximo uma parcela $z$. De $3 \\cdot 8>23>8+7+7$, podemos concluir que exatamente duas parcelas $8$ são usadas para obter a soma $23$ e assim um dos cartões deve possuir o número $23-8-8=7$. Sabendo que $x+y=17-z=9$, a única solução possível é $(x, y, z)=(2,7,8)$, mas essa tripla não pode gerar as somas $13$ e $15$.\n\nFinalmente, a única opção que resta a ser analisada é $z=9$. Nesse caso, nas somas $13$ e $15$ exatamente uma parcela $z$ é usada e, além disso, $x \\leq 3$, pois caso contrário a menor soma possível com a parcela $z$ seria $9+4+4=17>13$. Por outro lado, como $x+y=17-z=8$, as opções que nos restam são $(x, y, z)=(1,7,9)$, $(2,6,9)$ ou $(3,5,9)$. A primeira não pode gerar a soma $13$ e a segunda não pode gerar a soma $15$. A terceira opção é a única possível e um exemplo de sorteios está ilustrado na seguinte tabela:\n\n| Pessoa 1 | 3 | 5 | 5 | $=13$ |\n| :--- | :--- | :--- | :--- | :--- |\n| Pessoa 2 | 9 | 3 | 3 | $=15$ |\n| Pessoa 3 | 5 | 9 | 9 | $=23$ |",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27031,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO jogo \"Dado Geográfico\" consiste em cada participante jogar um dado uma quantidade $n$ de vezes e anotar a sequência, sendo que o primeiro lançamento indica a distância, em metros, que o participante andará para o Norte, o segundo para o Leste, o terceiro para o Sul, o quarto para o Oeste, o quinto para o Norte e assim por diante. Após isso mede-se a distância até a origem. Vence quem ficar mais próximo da posição inicial.\n\na) Márcia tirou 214365 (6 lançamentos). Qual a sua distância até a origem?\n\nb) Em 4 lançamentos, quantas possibilidades existem para que o participante volte para a origem após o último lançamento?\n\nc) Em 5 lançamentos, quantas possibilidades existem para que o participante volte para a origem após o último lançamento?",
    "options": [],
    "answer": "a) 5 m; b) 36; c) 90",
    "solution": "Solution:\n\na) Márcia andou $2+6=8$ m para o Norte, $1+5=6$ m para o Leste, $4$ m para o Sul e $3$ m para o Oeste. Na direção Norte-Sul, Márcia andou $8-4=4$ m para o Norte e na direção Leste-Oeste, Márcia andou $6-3=3$ m para o Leste. Como estas direções são ortogonais, basta aplicarmos o Teorema de Pitágoras para determinarmos a distância $d$ à origem:\n$$\nd^2 = 4^2 + 3^2\n$$\nsegue que $d = 5$ m.\n\nb) Para voltar à origem, o participante deve tirar o mesmo resultado para Norte e Sul, e também o mesmo para Leste e Oeste. Como existem $6$ possibilidades de valores iguais em dois lançamentos, o total de possibilidades de retorno à origem ao final do último lançamento é $6 \\cdot 6 = 36$.\n\nc) Vamos usar o mesmo raciocínio do item anterior, mas serão $2$ dados que determinarão deslocamento para o Norte (primeiro e quinto). Para a direção Leste-Oeste são $6$ possibilidades. Se no lançamento para o Sul, tirarmos $1$, não existe possibilidade de voltarmos para a origem com $2$ lançamentos para o Norte; se para o Sul tirarmos $2$, existe apenas $1$ possibilidade para o Norte $(1+1)$; se tirarmos $3$ para o Sul, são $2$ possibilidades para o Norte ($1+2$ ou $2+1$); se tirarmos $4$, serão $3$ possibilidades; se tirarmos $5$, serão $4$ possibilidades; se tirarmos $6$, serão $5$ possibilidades. Sendo assim, são $1+2+3+4+5=15$ possibilidades no sentido Norte-Sul. Portanto, serão $6 \\cdot 15 = 90$ possibilidades de voltarmos à origem após $5$ lançamentos.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27032,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nComeçando com um número inteiro positivo $n$, uma sequência é criada satisfazendo a seguinte regra: cada termo se obtém do anterior subtraindo-se o maior quadrado perfeito que é menor ou igual ao termo anterior, até chegar ao número zero. Por exemplo, se $n=142$, teremos a seguinte sequência de 5 termos:\n$$\na_{1}=142, a_{2}=21, a_{3}=5, a_{4}=1, a_{5}=0\n$$\npois $21=142-11^{2}$, $5=21-4^{2}$, $1=5-2^{2}$ e $0=1-1^{2}$.\na) Dê exemplo de uma sequência que tenha exatamente 6 termos.\nb) Encontre o menor valor de $n$ para que a sequência assim criada tenha exatamente 7 termos.",
    "options": [],
    "answer": "Example with 6 terms: 23 → 7 → 3 → 2 → 1 → 0. Smallest n for 7 terms: 167.",
    "solution": "Solution:\n\na) Um exemplo é a sequência\n$$\na_{1}=23, a_{2}=7=23-16, a_{3}=3=7-4, a_{4}=2=3-1, a_{5}=1=2-1, a_{6}=0=1-1\n$$\n\nb) Como $a_{n+1}=a_{n}-x^{2}$, $\\operatorname{com}~ x^{2} \\leq a_{n} < (x+1)^{2}$, segue que\n$$\n\\begin{aligned}\na_{n+1} & = a_{n} - x^{2} \\\\\n& < (x+1)^{2} - x^{2} \\\\\n& = 2x + 1\n\\end{aligned}\n$$\nDaí o inteiro $x$ satisfaz $x > \\frac{a_{n+1}-1}{2}$. Para obter o valor mínimo, a sequência será construída de trás para frente e em cada etapa será utilizado a estimativa mínima do incremento $x^{2}$ obtida anteriormente. Temos $a_{7}=0$ e $a_{6} \\geq 1^{2}$. Daí\n$$\n\\begin{aligned}\na_{5} & = a_{6} + x^{2} \\\\\n& \\geq 1 + 1^{2} \\\\\na_{4} & = a_{5} + x^{2} \\\\\n& \\geq 2 + 1^{2} \\\\\na_{3} & = a_{4} + x^{2} \\\\\n& \\geq 3 + 2^{2} \\\\\na_{2} & = a_{3} + x^{2} \\\\\n& \\geq 7 + 4^{2} \\\\\na_{1} & = a_{2} + x^{2} \\\\\n& \\geq 23 + 144\n\\end{aligned}\n$$\nAssim, o menor valor de $n$ é $23 + 144 = 167$ e a sequência de 7 termos que será criada é\n$$\na_{1}=167, a_{2}=23, a_{3}=7, a_{4}=3, a_{5}=2, a_{6}=1, a_{7}=0\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27033,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA professora Jacira propõe um jogo na lousa:\nI) Um dos alunos escreve uma sequência com $n$ algarismos não nulos;\nII) Um segundo aluno deve escrever outra sequência com $(n-1)$ algarismos na qual o primeiro é a diferença positiva dos dois primeiros da primeira sequência, o segundo é a diferença entre o segundo e o terceiro algarismo da primeira sequência e assim por diante;\nIII) Um terceiro aluno usa o mesmo processo, mas utilizando a segunda sequência.\nIV) Segue o processo até que o último aluno escreva apenas um algarismo na lousa.\nV) Se aparecer algum zero em uma das sequências, dizemos que é uma SEQUÊNCIA FURADA e não continuamos o processo.\nPor exemplo:\n\n| $1^{\\circ}$ aluno | 5 | 8 | 9 |   |\n| :---: | :---: | :---: | :---: | :---: |\n| $2^{\\circ}$ aluno | 3 |   |   | 6 |\n| $3^{\\circ}$ aluno |   | 2 | 5 |   |\n| $4^{\\circ}$ aluno |   |   |   |   |\n\nApós fazer a brincadeira algumas vezes, a professora propõe alguns desafios:\n\na) Continue as sequências abaixo, dada a primeira sequência escrita pelo primeiro aluno.\n\n| $1^{\\circ}$ aluno | 1 | 2 | 4 | 8 |\n| :--- | :--- | :--- | :--- |\n| $2^{\\circ}$ aluno |   |   |   |   |\n| $3^{\\circ}$ aluno |   |   |   |   |\n| $4^{\\circ}$ aluno |   |   |   |   |\n\nb) Se um jogo possui 5 sequências, sendo a última o algarismo 5, construa um exemplo para esta situação.\n\nc) Explique porque não é possível construir uma situação com 6 sequências, na qual a última é o algarismo 5.",
    "options": [],
    "answer": "a) Second sequence: 1 2 4; Third sequence: 1 2; Fourth sequence: 1.\nb) One valid construction:\n1st: 9 1 2 4 8\n2nd: 8 1 2 4\n3rd: 7 1 2\n4th: 6 1\n5th: 5\nc) Impossible: with six sequences ending in 5, the top row would require a digit at least 10, which is not allowed.",
    "solution": "Solution:\n\na) Utilizando o processo descrito no enunciado, temos:\n![](attached_image_1.png)\n\nb) Pensando do final para o começo, necessariamente, se o maior número do aluno da posição $k$ é $j$, então o aluno da posição $k-1$ (imediatamente anterior) tem seu maior número maior que $j$. Assim, se o $5^{\\circ}$ aluno escreveu $5$, o $4^{\\circ}$ deve ter escrito pelo menos $6$ sobre o $5$, o $3^{\\circ}$ deve ter escrito pelo menos $7$ sobre o $6$, o $2^{\\circ}$ pelo menos $8$ sobre o $7$ e o $1^{\\circ}$ deve ter escrito um $9$ sobre o $8$. Assim, já podemos construir um quadro com um algarismo por sequência.\n\n| $1^{\\circ}$ aluno | 9 |   |   |   |\n| :--- | :--- | :--- | :--- | :--- |\n| $2^{\\circ}$ aluno |   | 8 |   |   |\n| $3^{\\circ}$ aluno |   | 7 |   |   |\n| $4^{\\circ}$ aluno |   |   | 6 |   |\n| $5^{\\circ}$ aluno |   |   |   | 5 |\n\nBasta agora continuarmos o preenchimento das sequências.\n\n| $1^{\\circ}$ aluno | 9 |   | 1 |   | 2 |   | 4 |   | 8 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| $2^{\\circ}$ aluno |   | 8 |   | 1 |   | 2 |   | 4 |   |\n| $3^{\\circ}$ aluno |   |   | 7 |   | 1 |   | 2 |   |   |\n| $4^{\\circ}$ aluno |   |   |   | 6 |   | 1 |   |   |   |\n| $5^{\\circ}$ aluno |   |   |   | 5 |   |   |   |   |   |\n\nc) Como visto no item anterior, se finalizarmos com $5$ com o $5^{\\circ}$ aluno, precisamos de no mínimo um $9$ para o $1^{\\circ}$ aluno, mas se finalizarmos com $5$ com o $6^{\\circ}$ aluno, precisaríamos de $10$ para o $1^{\\circ}$ aluno, o que é absurdo, pois só podemos utilizar algarismos. Portanto, é impossível terminar com $5$ no $6^{\\circ}$ aluno.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27034,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSeja $ABC$ um triângulo retângulo com $\\angle BAC = 90^{\\circ}$ e $I$ o ponto de encontro de suas bissetrizes. Uma reta por $I$ corta os lados $AB$ e $AC$ em $P$ e $Q$, respectivamente. A distância de $I$ para o lado $BC$ é $1~\\mathrm{cm}$.\na) Encontre o valor de $PM \\cdot NQ$.\nb) Determine o valor mínimo possível para a área do triângulo $APQ$.\nDica: Se $x$ e $y$ são dois números reais não negativos, então $x+y \\geq 2 \\sqrt{x y}$.\n![](attached_image_1.png)",
    "options": [],
    "answer": "PM·NQ = 1 and the minimum area of triangle APQ is 2 (attained when AP = AQ = 2).",
    "solution": "Solution:\n\na. Se $\\angle APQ = \\alpha$, segue que $\\angle MIP = 90^{\\circ} - \\alpha$ e\n$$\n\\angle NIQ = 180^{\\circ} - \\angle MIN - \\angle MIP = \\alpha\n$$\nPortanto, os triângulos $IMP$ e $NIQ$ são semelhantes e daí\n$$\n\\frac{IM}{NQ} = \\frac{PM}{IN} \\Rightarrow PM \\cdot NQ = IM \\cdot IN = 1\n$$\n\nb. Sejam $x = MP$ e $y = NQ$. Como $AMIN$ é um quadrado de lado $1~\\mathrm{cm}$, segue que a área do triângulo $APQ$ é dada por\n$$\n\\begin{aligned}\n\\frac{AP \\cdot AQ}{2} &= \\frac{(1 + x)(1 + y)}{2} \\\\\n&= \\frac{1 + xy + x + y}{2}\n\\end{aligned}\n$$\nPelo item anterior, $xy = 1$. Além disso, pela desigualdade apresentada na dica, $x + y \\geq 2 \\sqrt{1} = 2$. Assim, a área mínima é $\\frac{1 + 1 + 2}{2} = 2$ e pode ser obtida fazendo $AP = AQ = 2$.\n\nObservação: A desigualdade apresentada como sugestão é um caso particular da desigualdade entre médias aritméticas e geométricas $(MA \\geq MG)$. Para verificar esse caso, de $(\\sqrt{x} - \\sqrt{y})^2 \\geq 0$, segue que\n$$\nx - 2\\sqrt{xy} + y \\geq 0 \\Rightarrow x + y \\geq 2\\sqrt{xy}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27035,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nJoana deve colocar três fichas em um tabuleiro $5 \\times 5$, no qual as casas são numeradas de 1 a 25, sendo uma em cada casa. De quantas maneiras ela pode fazer isso, se:\n\na) As 3 fichas são de cores diferentes?\n\nb) As 3 fichas são idênticas?\n\nc) As fichas são de cores diferentes e não podem estar duas a duas em uma mesma linha ou coluna?",
    "options": [],
    "answer": "a) 13800; b) 2300; c) 600",
    "solution": "Solution:\n\na) Como são 25 casas para a primeira ficha, temos 25 possibilidades, para a segunda ficha, temos 24 possibilidades e para a terceira ficha, temos 23 possibilidades. Portanto, são $25 \\cdot 24 \\cdot 23 = 13.800$ possibilidades.\n\nb) Para as peças da mesma cor, devemos descontar o número de situações repetidas que contaremos utilizando o raciocínio do item anterior. Vamos escolher as três primeiras casas do tabuleiro e as cores branca (B), amarela (A) e verde (V). Para estas casas, temos as possibilidades:\n\n| $\\mathrm{A}$ | $\\mathrm{B}$ | $\\mathrm{V}$ |\n| :---: | :---: | :---: |\n| $\\mathrm{A}$ | $\\mathrm{V}$ | $\\mathrm{B}$ |\n| $\\mathrm{B}$ | $\\mathrm{A}$ | $\\mathrm{V}$ |\n| $\\mathrm{B}$ | $\\mathrm{V}$ | $\\mathrm{A}$ |\n| $\\mathrm{V}$ | $\\mathrm{A}$ | $\\mathrm{B}$ |\n| $\\mathrm{V}$ | $\\mathrm{B}$ | $\\mathrm{A}$ |\n\nVemos que são 6 distribuições diferentes para cada conjunto de 3 casas. No caso das fichas com mesma cor, essas 6 distribuições são exatamente iguais, ou seja, apenas $\\frac{1}{6}$ das 13.800 distribuições anteriores. Portanto, o total de distribuições para fichas da mesma cor é $\\frac{13.800}{6} = 2.300$.\n\nc) Primeiro vamos escolher as 3 linhas que serão colocadas as fichas: usando o mesmo raciocínio do item anterior, temos $\\frac{5 \\cdot 4 \\cdot 3}{6} = 10$ combinações de 3 linhas de um total de 5. Agora vamos distribuir as fichas por uma das combinações de 3 linhas, por exemplo, as três primeiras linhas: na primeira linha são 5 possibilidades; na segunda 4, pois não podemos utilizar a coluna da ficha que está na primeira linha; e na última 3, pois não podemos utilizar as colunas das fichas já colocadas. Portanto, o total de maneiras de distribuir 3 fichas, uma de cada cor, em um tabuleiro sem repetir linha ou coluna é $10 \\cdot 5 \\cdot 4 \\cdot 3 = 600$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27036,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA bandeira da figura é formada por um retângulo $40~\\mathrm{cm} \\times 20~\\mathrm{cm}$ e possui duas cores (branca e cinza). A parte branca é composta por duas listras de mesma largura que se cruzam e são perpendiculares aos lados do retângulo. A parte cinza é a área do retângulo que não foi coberto pelas listras. Qual deve ser a medida da largura das listras para que a área branca seja igual a área cinza?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "30 − 10√5 cm",
    "solution": "Solution:\nSeja $x$ a medida da largura das faixas. Assim, uma das faixas mede $x$ por $40~\\mathrm{cm}$, enquanto que a outra mede $x$ por $20~\\mathrm{cm}$. Se somarmos as áreas das duas faixas, estaremos contando sua intersecção duas vezes e, portanto, precisaremos descontá-la. Como a área das faixas deve ser a metade da área da bandeira, temos:\n$$\n\\begin{aligned}\n40x + 20x - x^2 &= \\frac{40 \\cdot 20}{2} \\\\\n60x - x^2 &= 400 \\\\\nx^2 - 60x + 400 &= 0 \\\\\nx &= \\frac{60 \\pm \\sqrt{60^2 - 4 \\cdot 1 \\cdot 400}}{2} \\\\\n&= \\frac{60 \\pm \\sqrt{2000}}{2} \\\\\n&= 30 \\pm 10\\sqrt{5}\n\\end{aligned}\n$$\nComo $0 < x < 20$, temos que a largura das faixas é $x = (30 - 10\\sqrt{5})~\\mathrm{cm}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27037,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDizemos que um número é racional se ele pode ser escrito da forma $\\frac{p}{q}$, com $p$ e $q$ números inteiros. Se um número real não é racional, dizemos que ele é irracional. Por exemplo, $\\frac{1}{4}$ é um número racional e $\\sqrt{3}$ é irracional. Encontre todos os pares de números racionais $(a, b)$ tais que\n$$\n\\sqrt{a}+\\sqrt{b}=\\sqrt{2+\\sqrt{3}}\n$$",
    "options": [],
    "answer": "(1/2, 3/2) and (3/2, 1/2)",
    "solution": "Solution:\n\nElevando ao quadrado, obtemos as equações equivalentes:\n$$\n\\begin{aligned}\na+b+2 \\sqrt{a b} & =2+\\sqrt{3} \\\\\n2 \\sqrt{a b} & =2-a-b+\\sqrt{3}\n\\end{aligned}\n$$\nComo $a$ e $b$ são números racionais, $r=2-a-b$ também é um número racional. Eleve a última equação ao quadrado:\n$$\n\\begin{aligned}\n4 a b & =r^{2}+3+2 r \\sqrt{3} \\\\\n4 a b-r^{2}-3 & =2 r \\sqrt{3}\n\\end{aligned}\n$$\nSe $r \\neq 0$, temos\n$$\n\\sqrt{3}=\\frac{4 a b-r^{2}-3}{2 r}\n$$\nComo $r$ é um número racional, o número $\\frac{4 a b-r^{2}-3}{2 r}$ também é racional. Entretanto, isso gera uma contradição, pois $\\sqrt{3}$ é um número irracional. Logo, $r=0$ e assim das últimas equações temos\n$$\n\\begin{aligned}\na+b & =2 \\\\\na b & =3 / 4\n\\end{aligned}\n$$\nIsso mostra que $a$ e $b$ são raízes da equação $x^{2}-2 x+3 / 4=0$, ou seja, $(a, b)=(1 / 2,3 / 2)$ ou $(a, b)=(3 / 2,1 / 2)$. Esses números satisfazem a primeira equação e, consequentemente, são solução da equação do problema.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27038,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA professora Jane escreveu na lousa os números $1^{2}, 2^{2}, 3^{2}, \\ldots, 2020^{2}$. Ela propõe o seguinte jogo: Alice e Matias devem apagar números alternadamente, um número por vez, sendo que Matias começa, até que sobrem apenas dois números no quadro. Se a diferença entre estes dois números for múltiplo de 2021, Alice vence, caso contrário, Matias vence. Determine quem sempre pode garantir a vitória independentemente de como o outro jogador jogue.",
    "options": [],
    "answer": "Alice",
    "solution": "Solution:\nPerceba que $(2021-x)^{2}-x^{2}=2021(2021-2x)$, que é múltiplo de $2021$. Sendo assim, sempre que Matias apagar um número qualquer $k^{2}$, basta Alice apagar $(2021-k)^{2}$, que no final, os dois números restantes terão diferença múltipla de $2021$. Logo, usando essa estratégia desde o início, Alice pode garantir a vitória independentemente de como Matias jogue.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27039,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEm uma folha de papel estão desenhados 10 pontos, sendo 8 azuis e 2 vermelhos. Os dois pontos vermelhos são ligados a todos os pontos azuis, mas não são ligados um ao outro. Todos os pontos azuis são ligados um ao outro. Quando dois pontos são ligados, esta ligação é feita por um único segmento. De quantas maneiras diferentes podemos sair de um ponto vermelho e chegar ao outro passando no máximo uma única vez por cada ponto?",
    "options": [],
    "answer": "219200",
    "solution": "Solution:\n\nVamos chamar os dois pontos vermelhos de $A$ e $B$. Saindo de $A$, passando por exatamente um ponto azul, são 8 possibilidades; passando por dois pontos azuis, são $8 \\cdot 7 = 56$ possibilidades; passando por três pontos azuis, são $8 \\cdot 7 \\cdot 6 = 336$ possibilidades. Seguindo este raciocínio, vemos que o número de possibilidades de sair de $A$ e ir para $B$ é $8 + 8 \\cdot 7 + 8 \\cdot 7 \\cdot 6 + \\ldots + 8 \\cdot 7 \\cdot 6 \\cdot 5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1 = 109.600$. Claro que de $B$ para $A$ o resultado é o mesmo, ou seja, podemos ligar os dois pontos vermelhos de $2 \\cdot 109.600 = 219.200$ maneiras diferentes.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27040,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nConsidere os números reais $p$ e $q$ e a equação cúbica:\n$$\nx^{3}+p x+q=0\n$$\na) Se $x_{0}$ é uma raiz real da equação, então\n$$\nx^{3}+p x+q=\\left(x-x_{0}\\right)\\left(x^{2}+a x+b\\right)\n$$\nVerifique que $a=x_{0}$.\nb) Verifique que $p^{2} \\geq 4 x_{0} q$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na) Expandindo a equação dada, temos\n$$\n\\begin{aligned}\nx^{3}+p x+q & =\\left(x-x_{0}\\right)\\left(x^{2}+a x+b\\right) \\\\\n& =x^{3}+\\left(a-x_{0}\\right) x^{2}+\\left(b-a x_{0}\\right) x-b x_{0}\n\\end{aligned}\n$$\nComo o coeficiente de $x^{2}$ no termo da esquerda é nulo, devemos ter $a-x_{0}=0$, ou seja, $x_{0}=a$.\n\nb) Comparando os coeficientes de $x^{1}$ e $x^{0}$ em ambos os membros das equações anteriores, temos $p=b-a x_{0}=b-x_{0}^{2}$ e $q=-b x_{0}$. Daí, como todo quadrado de um número real é não negativo,\n$$\n\\begin{aligned}\np^{2}-4 x_{0} q & =\\left(b-x_{0}^{2}\\right)^{2}+4 b x_{0}^{2} \\\\\n& =b^{2}-2 b x_{0}^{2}+x_{0}^{4}+4 b x_{0}^{2} \\\\\n& =b^{2}+2 b x_{0}^{2}+x_{0}^{4} \\\\\n& =\\left(b+x_{0}^{2}\\right)^{2} \\\\\n& \\geq 0\n\\end{aligned}\n$$\nConsequentemente $p^{2} \\geq 4 x_{0} q$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27041,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nO professor M. A. Luco escreveu no quadro a expressão:\n$$\n\\frac{n^{2}-5 n+4}{n-4}\n$$\nEntão, ele diz aos alunos que $n$ pode ser qualquer número natural, com exceção de $4$.\n\na) Qual o valor da expressão para $n=1$?\n\nb) Marcos substituiu $n$ por um número natural e verificou que o valor da expressão é $5$. Marcos substituiu $n$ por qual número?\n\nc) Quais são os números naturais que não podem ser o valor numérico da expressão?",
    "options": [],
    "answer": "a) 0; b) 6; c) all natural numbers except 3",
    "solution": "Solution:\na) Para $n=1$, temos:\n$$\n\\begin{aligned}\n\\frac{n^{2}-5 n+4}{n-4} & =\\frac{1^{2}-5 \\cdot 1+4}{1-4} \\\\\n& =\\frac{1-5+4}{-3} \\\\\n& =0\n\\end{aligned}\n$$\n\nb) Igualando a expressão a $5$, temos:\n$$\n\\begin{aligned}\n\\frac{n^{2}-5 n+4}{n-4} & =5 \\\\\nn^{2}-5 n+4 & =5(n-4) \\\\\nn^{2}-10 n+24 & =0 \\\\\nn & =\\frac{10 \\pm \\sqrt{(-10)^{2}-4 \\cdot 1 \\cdot 24}}{2 \\cdot 1} \\\\\n& =\\frac{10 \\pm \\sqrt{100-96}}{2} \\\\\n& =\\frac{10 \\pm 2}{2} \\\\\n& =5 \\pm 1\n\\end{aligned}\n$$\nPortanto, Marcos substituiu $n$ por $6$, já que $n \\neq 4$.\n\nc) Fatorando o numerador e lembrando que $n \\neq 4$, temos:\n$$\n\\frac{n^{2}-5 n+4}{n-4}=\\frac{(n-4)(n-1)}{n-4}=n-1\n$$\nComo $n \\neq 4$, então $n-1 \\neq 3$. Portanto, a expressão pode assumir qualquer valor natural com exceção de $3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27042,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNo trapézio $ABCD$, os lados $AB$ e $CD$ são paralelos. Sejam $M$ o ponto médio da diagonal $AC$, $N$ o ponto médio da diagonal $BD$ e $P$ o ponto médio do lado $AB$. Sabemos que $AB = 15\\ \\mathrm{cm}$, $CD = 24\\ \\mathrm{cm}$ e a altura do trapézio é $h = 14\\ \\mathrm{cm}$.\n\na) Calcule a medida do comprimento do segmento $MN$.\n\nb) Calcule a área do triângulo $MNP$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "MN = 4.5 cm; Area of triangle MNP = 15.75 cm^2",
    "solution": "Solution:\n\n![](attached_image_2.png)\n\na) Se $P$ é o ponto médio de $AB$ e $L$ o ponto médio de $AD$, segue que $ML$ é base média do triângulo $ACD$ e assim $LM = 12\\ \\mathrm{cm}$. Como $LN$ é base média do triângulo $ABD$, segue que $LN = 7,5\\ \\mathrm{cm}$. Portanto, $MN = LM - LN = 4,5\\ \\mathrm{cm}$\n\nb) Como $MN$ está sobre a base média, a sua distância até a reta $CD$ é $h/2 = 7\\ \\mathrm{cm}$. Portanto, a altura do triângulo $MNP$ relativa ao lado $MN$ é $7\\ \\mathrm{cm}$ e assim\n$$\n\\text{Área}(MNP) = 1/2 \\cdot 7 \\cdot 4,5 = 15,75\\ \\mathrm{cm}^2\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27043,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUm fotógrafo deve tirar fotos de uma festa com 10 membros de uma mesma família. Cada um dos 45 possíveis pares de pessoas dessa família devem aparecer juntos em exatamente uma foto. Além disso, existem apenas dois tipo de fotos: as que possuem 2 ou 3 pessoas.\n\na) Verifique que cada pessoa da família deverá aparecer em pelo menos uma foto de apenas 2 pessoas.\n\nb) Verifique que o fotógrafo deverá tirar pelo menos 19 fotos.\n\nc) Dê um exemplo de distribuição de pessoas em 19 fotos satisfazendo as condições dadas.",
    "options": [],
    "answer": "19",
    "solution": "Solution:\n\na) Cada pessoa deve aparecer com outros 9 familiares exatamente uma vez em alguma foto. Nas fotos com 3 pessoas, uma pessoa aparece em exatamente 2 pares e, como 9 é ímpar, é necessário que cada pessoa apareça em pelo menos uma foto com exatamente 2 pessoas.\n\nb) Sejam $x$ o número de fotos com 3 pessoas e $y$ o número de fotos com 2 pessoas. Queremos estimar o total de fotos que é dado por $x+y$. Em cada foto com 3 pessoas, temos 3 pares de familiares e em cada foto com 2 pessoas apenas um par. Portanto,\n$$\n3x + y = 45\n$$\nPelo item anterior, como cada pessoa deve aparecer em pelo menos uma foto com 2 pessoas, devemos ter $y \\geq 5$. Consequentemente $x = (45 - y)/3 \\leq 13$. Temos, então:\n$$\n\\begin{aligned}\nx &\\leq 13 \\\\\n-x &\\geq -13 \\\\\n-2x &\\geq -26 \\\\\n45 - 2x &\\geq 19 \\\\\nx + y &\\geq 19\n\\end{aligned}\n$$\n\nc) Portanto, devemos ter pelo menos 19 fotos. Para construir um exemplo com exatamente 19 fotos, nomeie as 10 pessoas da família pelos números $0, 1, \\ldots, 9$. Vamos iniciar a construção do exemplo formando 4 fotos de 3 pessoas: $(1,2,3)$, $(3,4,5)$, $(5,6,7)$ e $(7,8,1)$. No diagrama abaixo, cada triângulo representa uma foto.\n\n![](attached_image_1.png)\n\nPara escolher as próximas fotos, vamos incluir no diagrama o parente de número $0$ e usar pares de parentes que ainda não apareceram em nenhuma foto. Por exemplo, escolha o seguinte grupo de 4 fotos de 3 pessoas: $(0,1,4)$, $(0,8,5)$, $(0,2,7)$ e $(0,3,6)$. No próximo diagrama essas fotos são representadas por triângulos com linhas tracejadas.\n\n![](attached_image_2.png)\n\nNote que neste diagrama ainda existem pares de familiares que não figuraram em nenhuma foto como, por exemplo, os familiares 1 e 6. Vamos incluir agora a pessoa de número 9 e formar os seguintes grupos de 4 fotos de 3 pessoas: $(9,1,6)$, $(9,2,5)$, $(9,3,8)$ e $(9,4,7)$. Veja que nessas fotos todas as pessoas de 1 a 8 aparecem com 0 e 9 em alguma foto. Dentre as pessoas de 1 a 8, alguns pares de familiares ainda não apareceram em fotos. Para terminar, considere a seguinte configuração com mais uma foto de 3 pessoas e 6 fotos de 2 pessoas:\n$$\n(2,4,6),\\ (2,8),\\ (6,8),\\ (0,9),\\ (1,5),\\ (3,7) \\text{ e } (4,8)\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27044,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNo quadrado $A B C D$, os pontos $M$ e $N$ são interiores aos lados $B C$ e $C D$ de modo que $\\angle M A N=45^{\\circ}$. Seja $O$ o ponto de interseção do círculo que passa por $C, M$ e $N$ com o segmento $A C$.\n![](attached_image_1.png)\n\na) Verifique que $O M=O N$.\nb) Verifique que $O$ é o centro da circunferência que passa por $A, M$ e $N$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na) Como $\\angle M C O=\\angle O C N=45^{\\circ}$, segue que os comprimentos das cordas $M O$ e $N O$ são iguais a um mesmo valor $L$.\n\nb) Seja $R$ o comprimento do raio da circunferência que passa por $A, M$ e $N$. Pela Lei dos Senos,\n$$\n\\frac{M N}{\\operatorname{sen}(\\angle M A N)}=2 R \\Rightarrow R=\\frac{M N}{\\sqrt{2}}\n$$\nComo $\\angle M C N=90^{\\circ}, M N$ é um diâmetro da circunferência que passa por $C, M$ e $N$. Consequentemente, $\\angle M O N=90^{\\circ}$. Pelo Teorema de Pitágoras,\n$$\n\\begin{aligned}\nM O^{2}+N O^{2} & =M N^{2} \\\\\n2 L^{2} & =M N^{2} \\\\\nL & =\\frac{M N}{\\sqrt{2}}\n\\end{aligned}\n$$\nDe $R=L$ e $M O=N O$, podemos concluir que $O$ é o centro da circunferência que passa por $A, M$ e $N$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27045,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSeja $ABCD$ um retângulo com $BC = 2 \\cdot AB$. Seja $E$ o ponto médio de $BC$ e $P$ um ponto arbitrário interno ao lado $AD$. Sejam $F$ e $G$ os pés das perpendiculares desenhadas de $A$ a $BP$ e de $D$ a $CP$. Sabemos que $\\angle BPC = 85^\\circ$.\n![](attached_image_1.png)\n\na) Verifique que os triângulos $BEF$ e $BEP$ são semelhantes.\n\nb) Determine o valor da soma dos ângulos $\\angle BEF + \\angle CEG$.",
    "options": [],
    "answer": "85°",
    "solution": "Solution:\n\na) Em virtude das relações métricas nos triângulos retângulos aplicadas ao triângulo $ABP$, temos\n$$\nBE^2 = AB^2 = BF \\cdot BP\n$$\nPortanto,\n$$\n\\frac{BE}{BP} = \\frac{BF}{BE}\n$$\nDada a relação de proporcionalidade da última equação e $\\angle EBF = \\angle EBP$, segue que os triângulos $BEF$ e $BEP$ são semelhantes.\n\nb) Da semelhança do item anterior, segue que $\\angle BEF = \\angle BPE$. De modo análogo, $\\angle CEG = \\angle EPC$. Assim,\n$$\n\\angle BEF + \\angle CEG = \\angle BPE + \\angle EPC = \\angle BPC = 85^\\circ.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27046,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSeja $x$ o menor número inteiro positivo que satisfaz simultaneamente as seguintes condições: $2x$ é o quadrado de um número inteiro, $3x$ é o cubo de um número inteiro e $5x$ é a quinta potência de um número inteiro. Encontre a fatoração em primos de $x$.",
    "options": [],
    "answer": "2^15 3^20 5^24",
    "solution": "Solution:\nDo enunciado, podemos escrever\n$$\n2x = a^{2}, \\quad 3x = b^{3}, \\quad 5x = c^{5}\n$$\nem que $a, b, c$ são inteiros. Daí podemos concluir que $x$ possui fatores primos $2, 3$ e $5$. Ou seja, a fatoração em primos pode ser escrita como $x = 2^{p} 3^{q} 5^{r} \\cdot m$, em que $m$ é o produto de primos maiores que $5$. Em virtude de (1), $m$ deve ser um quadrado, um cubo e uma quinta potência perfeita. Como queremos o menor $x$ possível, podemos admitir que $m = 1$.\n\nEm geral, a fatoração em primos de uma potência $n^{k}$ pode ser obtida a partir da fatoração em primos de $n$ multiplicando cada expoente de seus fatores primos por $k$. Daí os expoentes da fatoração em primos de um quadrado perfeito são todos pares, os de um cubo perfeito são todos múltiplos de $3$ e os de uma quinta potência são todos múltiplos de $5$. Portanto, a equação (1) nos diz que\n\ni) $p+1, q$ e $r$ são múltiplos de $2$;\nii) $p, q+1$ e $r$ são múltiplos de $3$;\niii) $p, q$ e $r+1$ são múltiplos de $5$.\n\nDe i) e ii), o menor valor possível para $p$ é $3 \\cdot 5 = 15$. Veja que este valor satisfaz i). De i) e iii), os valores possíveis para $q$ são $10, 20, 30, \\ldots$ O primeiro deles não satisfaz i) e assim o menor valor possível de $q$ é $20$. Finalmente, de i) e ii), os valores possíveis de $r$ são $6, 12, 18, 24, \\ldots$. Dessa lista, o menor que satisfaz a condição iii) é o valor $24$. Portanto, o menor valor de $x$ é $2^{15} 3^{20} 5^{24}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27047,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA Sequência de Fibonacci é definida recursivamente por $F_{n+2}=F_{n+1}+F_{n}$ para $n \\in \\mathbb{Z}$ e $F_{1}=F_{2}=1$. Determine o valor de:\n$$\n\\left(1-\\frac{F_{2}^{2}}{F_{3}^{2}}\\right)\\left(1-\\frac{F_{3}^{2}}{F_{4}^{2}}\\right) \\cdot \\ldots \\cdot\\left(1-\\frac{F_{2019}^{2}}{F_{2020}^{2}}\\right)\n$$",
    "options": [],
    "answer": "F_{2021} / (2 F_{2019} F_{2020})",
    "solution": "Solution:\n\nPodemos reescrever um termo genérico do produto dado como\n$$\n\\begin{aligned}\n1-\\frac{F_{i}^{2}}{F_{i+1}^{2}} & =\\frac{F_{i+1}^{2}-F_{i}^{2}}{F_{i+1}^{2}} \\\\\n& =\\frac{\\left(F_{i+1}-F_{i}\\right)\\left(F_{i+1}+F_{i}\\right)}{F_{i+1}^{2}} \\\\\n& =\\frac{F_{i-1} F_{i+2}}{F_{i+1}^{2}}\n\\end{aligned}\n$$\nPortanto,\n$$\n\\begin{aligned}\n\\left(1-\\frac{F_{2}^{2}}{F_{3}^{2}}\\right)\\left(1-\\frac{F_{3}^{2}}{F_{4}^{2}}\\right)\\left(1-\\frac{F_{4}^{2}}{F_{5}^{2}}\\right) \\ldots \\cdot\\left(1-\\frac{F_{2018}^{2}}{F_{2019}^{2}}\\right) \\cdot\\left(1-\\frac{F_{2019}^{2}}{F_{2020}^{2}}\\right) & = \\\\\n\\frac{F_{1} F_{4}}{F_{3}^{2}} \\cdot \\frac{F_{2} F_{5}}{F_{4}^{2}} \\cdot \\frac{F_{3} F_{6}}{F_{5}^{2}} \\cdot \\ldots \\cdot \\frac{F_{2017} F_{2020}}{F_{2019}^{2}} \\cdot \\frac{F_{2018} F_{2021}}{F_{2020}^{2}} & = \\\\\n\\frac{F_{1} F_{2} \\cdot\\left(F_{3} F_{4} \\ldots F_{2017} F_{2018}\\right) \\cdot F_{2021}}{F_{3} \\cdot\\left(F_{3} F_{4} \\ldots F_{2018} F_{2019}\\right) F_{2019} F_{2020}} & = \\\\\n\\frac{F_{2021}}{2 F_{2019} F_{2020}}\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27048,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA calculadora $MK$-2020 pode efetuar as seguintes três operações com números em sua memória:\n1) Determinar se dois números escolhidos são iguais.\n2) Adicionar dois números escolhidos.\n3) Para os números escolhidos $a$ e $b$, determinar as raízes reais da equação $x^{2} + a x + b = 0$ ou anunciar que tal equação não possui raízes reais.\nOs resultados de cada operação são acumulados em sua memória. Inicialmente, a memória contém apenas o número $z$ que é desconhecido por seus usuários.\n\na) Usando as operações, determine como descobrir se $z$ é diferente de $0$ ou não.\n\nb) Encontre as raízes da equação do segundo grau em $x$:\n$$\nx^{2} + 2z x + z = 0\n$$\n\nc) Como podemos determinar, usando a calculadora $MK$-2020, se $z$ é igual a $1$?",
    "options": [],
    "answer": "a) Compare z and 2z using the equality test; they are equal if and only if z = 0, so otherwise z ≠ 0. b) The roots are x = −z ± sqrt(z^2 − z). c) After confirming z ≠ 0, run the quadratic solver on x^2 + 2zx + z; it returns a single real root if and only if z = 1, otherwise z ≠ 1.",
    "solution": "Solution:\n\na) Podemos usar a segunda operação e gerar o número $2z$. Usando a primeira operação, podemos decidir se $z$ e $2z$ são iguais, ou seja, se $z$ é ou não igual a zero.\n\nb) Pela Fórmula de Bhaskara, as raízes de $x^{2} + 2z x + z = 0$ são dadas por:\n$$\nx = -z \\pm \\sqrt{z^{2} - z}\n$$\n\nc) Suponha que tenhamos descoberto que $z$ não é zero. Usando a terceira operação, podemos encontrar as raízes de $x^{2} + 2z x + z = 0$. Como já sabemos que $z \\neq 0$, o discriminante é nulo apenas quando $z = 1$. Assim, se a calculadora disser que existe apenas uma raiz real, saberemos que $z = 1$ e, no caso contrário, teremos $z \\neq 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27049,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa figura abaixo, os segmentos $A M$, $A K$, $B_{2} C_{2}$ e $B_{1} C_{1}$ são tangentes à circunferência de centro $O$. Se $\\angle B_{1} O B_{2} = 30^{\\circ}$, determine o valor do ângulo $\\angle B_{1} D B_{2}$.\n![](attached_image_1.png)",
    "options": [],
    "answer": "60°",
    "solution": "Solution:\n\nPelo Teorema do Ângulo Externo, temos\n$$\n\\begin{aligned}\n\\angle B_{1} O B_{2} &= \\angle A B_{1} O - \\angle A B_{2} O \\\\\n&= \\frac{\\angle A B_{1} C_{1} - \\angle A B_{2} C_{2}}{2} \\\\\n&= \\frac{\\angle B_{1} D B_{2}}{2}\n\\end{aligned}\n$$\nPortanto, $\\angle B_{1} D B_{2} = 2 \\cdot 30 = 60^{\\circ}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27050,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nCalcule o valor de\n$$\nA=\\frac{1001 \\cdot 1002 \\cdot 1003 \\cdot \\ldots \\cdot 2000}{1 \\cdot 3 \\cdot 5 \\cdot \\ldots \\cdot 1999}\n$$",
    "options": [],
    "answer": "2^{1000}",
    "solution": "Solution:\nSeja\n$$\n\\begin{aligned}\nB & =\\frac{2^{1000} \\cdot 1 \\cdot 2 \\cdot 3 \\cdot \\ldots \\cdot 1000}{2^{1000} \\cdot 1 \\cdot 2 \\cdot 3 \\cdot \\ldots \\cdot 1000} \\\\\n& =\\frac{2^{1000} \\cdot 1 \\cdot 2 \\cdot 3 \\cdot \\ldots \\cdot 1000}{2 \\cdot 4 \\cdot 6 \\cdot \\ldots \\cdot 2000}\n\\end{aligned}\n$$\nAssim\n$$\n\\begin{aligned}\nA \\cdot B & =\\frac{2^{1000} \\cdot(1 \\cdot 2 \\cdot 3 \\cdot \\ldots \\cdot 2000)}{(1 \\cdot 2 \\cdot 3 \\cdot \\ldots \\cdot 2000)} \\\\\n& =2^{1000}\n\\end{aligned}\n$$\nComo $B=1$, concluímos que $A=2^{1000}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27051,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSeja $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ uma função ímpar, isto é, uma função que satisfaz $-f(x)=f(-x)$ para todo $x \\in \\mathbb{R}$. Suponha que $f(x+5)=f(x)$ para todo $x \\in \\mathbb{R}$ e que $f(1 / 3)=1$. Determine o valor da soma:\n$$\nf(16 / 3)+f(29 / 3)+f(12)+f(-7)\n$$",
    "options": [],
    "answer": "0",
    "solution": "Solution:\n\nTemos\n$$\n\\begin{aligned}\nf\\left(\\frac{1}{3}\\right) & =1 \\\\\nf\\left(\\frac{1}{3}+5\\right) & =1 \\\\\nf\\left(\\frac{16}{3}\\right) & =1\n\\end{aligned}\n$$\nComo $f$ é uma função ímpar, i.e., $f(-x)=-f(x)$, segue que\n$$\n\\begin{aligned}\nf\\left(-\\frac{1}{3}\\right) & =-1 \\\\\nf\\left(-\\frac{1}{3}+5+5\\right) & =-1 \\\\\nf\\left(\\frac{29}{3}\\right) & =-1\n\\end{aligned}\n$$\ne\n$$\n\\begin{aligned}\nf(-7) & =f(-12+5) \\\\\n& =f(-12) \\\\\n& =-f(12)\n\\end{aligned}\n$$\nFinalmente,\n$$\n\\begin{aligned}\nf(16 / 3)+f(29 / 3)+f(12)+f(-7) & = \\\\\n1+(-1)+f(12)-f(12) & =0\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27052,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEm uma loja de chocolates, existem caixas com $8$, $9$ e $10$ chocolates. Observe que algumas quantidades de chocolates não podem ser compradas exatamente como, por exemplo, $12$ chocolates.\n\na) Encontre outra quantidade de chocolates que não pode ser comprada.\n\nb) Verifique que todo número maior que $56$ pode ser escrito na forma $8x + 9y$ com $x$ e $y$ inteiros não negativos.\n\nc) Qual é a maior quantidade de unidades de chocolates que não podemos comprar exatamente nessa loja?",
    "options": [],
    "answer": "31",
    "solution": "Solution:\n\na) Não é possível comprarmos $15$ chocolates, pois $15 > 10$ e a soma das quantidades de quaisquer duas caixas é maior que $15$.\n\nb) Inicialmente note que os números de $57$ a $64$ podem ser escritos na forma $8x + 9y$:\n\n| $x$ | $y$ | $8x + 9y$ |\n| :---: | :---: | :---: |\n| 9 | 1 | 57 |\n| 5 | 2 | 58 |\n| 4 | 3 | 59 |\n| 3 | 4 | 60 |\n| 2 | 5 | 61 |\n| 1 | 8 | 62 |\n| 0 | 7 | 63 |\n| 8 | 0 | 64 |\n\nSomando $8$ unidades a cada uma dessas representações, podemos escrever todos os números inteiros do intervalo $[65, 72]$ na forma $8x + 9y$. Por exemplo, como $60 = 8 \\cdot 3 + 9 \\cdot 4$, segue que $68 = 8 \\cdot 4 + 9 \\cdot 4$. Somando sucessivamente $8$, podemos concluir que todos os inteiros dos intervalos\n$$\n[73, 80], [81, 88], [89, 96], \\ldots\n$$\npodem ser escritos na forma $8x + 9y$, com $x$ e $y$ inteiros não negativos. Assim, todos os inteiros maiores que $56$ podem ser escritos na forma $8x + 9y$ com $x$ e $y$ inteiros não negativos.\n\nc) As quantidades de chocolates que podem ser compradas são os números da forma $8x + 9y + 10z$, com $x$, $y$ e $z$ inteiros não negativos representando as quantidades de cada tipo de caixa. Um número que pode ser escrito na forma $8x + 9y$ em particular também pode ser escrito na forma $8x + 9y + 10z$. Assim, em virtude do item anterior, basta analisarmos os números menores que $56$ para sabermos qual é o maior deles que não pode ser uma quantidade admissível de chocolates comprados na loja. A tabela a seguir indica como escrever todos os números de $32$ até $40$ na forma $8x + 9y + 10z$:\n\n| $x$ | $y$ | $z$ | $8x + 9y + 10z$ |\n| :---: | :---: | :---: | :---: |\n| 4 | 0 | 0 | 32 |\n| 3 | 1 | 0 | 33 |\n| 3 | 0 | 1 | 34 |\n| 2 | 1 | 1 | 35 |\n| 2 | 0 | 2 | 36 |\n| 1 | 1 | 2 | 37 |\n| 1 | 0 | 3 | 38 |\n| 0 | 1 | 3 | 39 |\n| 5 | 0 | 0 | 40 |\n\nSomando $8$ unidades a cada uma dessas representações, podemos escrever todos os números de $40$ a $48$. Repetindo esse processo, podemos escrever todos os números inteiros de $48$ a $56$ na forma $8x + 9y + 10z$, com $x$, $y$ e $z$ inteiros não negativos. Para concluir que $31$ é a maior quantidade de chocolate que não podemos comprar na loja, precisamos verificar que não existem $x$, $y$ e $z$ não negativos tais que\n$$\n8x + 9y + 10z = 31\n$$\nSe existissem tais inteiros, como $31$ é ímpar e $8$ e $10$ são pares, devemos ter $y \\neq 0$. Assim $y = 3$ ou $y = 1$. No primeiro caso, teríamos $8x + 10z = 4$, que claramente não possui solução em inteiros não negativos. No segundo caso, teríamos $8x + 10z = 22$, ou seja, $4x + 5z = 11$. Para $z = 0$, $z = 1$ e $z = 2$, deveríamos ter $4x = 11$, $4x = 6$ e $4x = 1$. Como nenhuma dessas equações possui soluções em inteiros, podemos concluir que a equação acima não possui solução em inteiros não negativos.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27053,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNós chamamos um número de telefone $d_{1} d_{2} d_{3}-d_{4} d_{5} d_{6} d_{7}$ de legal se o número $d_{1} d_{2} d_{3}$ for igual a $d_{4} d_{5} d_{6}$ ou a $d_{5} d_{6} d_{7}$. Por exemplo, $234-2347$ é um número de telefone legal. Assuma que cada $d_{i}$ pode ser qualquer dígito de 0 a 9. Quantos números de telefones legais existem?",
    "options": [],
    "answer": "19990",
    "solution": "Solution:\n\nSe $d_{1} d_{2} d_{3}$ é simultaneamente igual a $d_{4} d_{5} d_{6}$ e $d_{5} d_{6} d_{7}$, então todos os dígitos são iguais. Para contar o número de telefones com $d_{1} d_{2} d_{3}$ igual a $d_{4} d_{5} d_{6}$, basta escolhermos $d_{1} d_{2} d_{3}$ de $10 \\cdot 10 \\cdot 10 = 10^{3}$ maneiras e o valor de $d_{7}$ de 10 maneiras. Portanto, existem $10^{3} \\cdot 10 = 10^{4}$ telefones com $d_{1} d_{2} d_{3}$ igual a $d_{4} d_{5} d_{6}$. De modo semelhante, existem $10^{4}$ números de telefones com $d_{1} d_{2} d_{3}$ igual a $d_{5} d_{6} d_{7}$. Descontando os números de telefones que estão em ambos os grupos, temos $2 \\cdot 10^{4} - 10 = 19.990$ números de telefone legais.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27054,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa figura a seguir, $\\angle C A B = 2 \\cdot \\angle C B A$, $A D$ é uma altura e $M$ é o ponto médio de $A B$. Se $A C = 2~\\mathrm{cm}$, encontre o comprimento do segmento $D M$.\n![](attached_image_1.png)",
    "options": [],
    "answer": "1 cm",
    "solution": "Solution:\n\nSeja $K$ o ponto médio de $A C$. Como o círculo de centro $K$ e diâmetro $A C$ passa por $D$, segue que $C K = A K = D K = 1~\\mathrm{cm}$. Dessa última igualdade, decorre que o triângulo $A D K$ é isósceles e assim $\\angle K D A = \\angle K A D$. O segmento $K M$ é base média do triângulo $A B C$, implicando em $K M \\parallel B C$ e assim $\\angle K M D = \\angle C B A$. Pelo Teorema do Ângulo Externo,\n$$\n\\angle D K M = \\angle A D K - \\angle D M K = \\angle D M K\n$$\nFinalmente, como as últimas igualdades mostram que o triângulo $D K M$ é isósceles, segue que $D M = D K = 1~\\mathrm{cm}$.\n![](attached_image_2.png)",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27055,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\na) As letras $A$, $B$ e $C$ podem ser dispostas em linha de 6 formas distintas:\n$$\nA B C,\\ A C B,\\ B A C,\\ B C A,\\ C A B,\\ C B A\n$$\nNote que em 3 delas a letra $A$ aparece à esquerda da letra $B$:\n$$\nA B C,\\ A C B,\\ C A B\n$$\nDispondo as letras $A$, $B$, $C$ e $D$ em linha de todas as 24 formas distintas possíveis, em quantas delas a letra $A$ aparece à esquerda da letra $B$?\n\nb) Uma aranha tem uma meia e um sapato para cada um de seus oito pés. De quantas maneiras diferentes a aranha pode se calçar admitindo que as 8 meias e os 8 sapatos são distintos e que cada meia precisa ser colocada antes do seu respectivo sapato?",
    "options": [],
    "answer": "a) 12; b) 16!/2^8",
    "solution": "Solution:\na) Para cada palavra de 4 letras em que $A$ está à esquerda de $B$, trocando a posição dessas duas letras, obtemos uma palavra de 4 letras em que $A$ está à direita da letra $B$:\n$$\nC \\mathbf{A B} D \\leftrightarrow C \\mathbf{B A} D\n$$\nAgrupando as 24 palavras em 12 pares desse tipo, podemos concluir que em metade delas a letra $A$ está à esquerda de $B$ e na outra metade a letra $A$ está à direita de $B$.\n\nb) Se simbolizarmos as meias por $\\{m_{1}, m_{2}, \\ldots, m_{8}\\}$ e os sapatos por $\\{s_{1}, s_{2}, \\ldots, s_{8}\\}$, podemos interpretar cada palavra com essas 16 letras como uma ordem de uso desses objetos. Por exemplo, a palavra\n$$\nm_{1}\\ m_{2}\\ s_{1}\\ s_{2}\\ s_{3}\\ s_{4}\\ m_{5}\\ \\ldots\\ s_{8}\n$$\nindica que a aranha colocará inicialmente a meia $m_{1}$, depois a meia $m_{2}$, o sapato $s_{1}$, o sapato $s_{2}$ etc. Existem $16!$ permutações desses 16 símbolos. Por simetria, em metade deles, o símbolo $m_{1}$ vem antes do $s_{1}$ e na outra metade ocorre o contrário. Dessa metade em que $m_{1}$ vem antes de $s_{1}$, em metade delas o símbolo $m_{2}$ vem antes de $s_{2}$. Repetindo esse argumento para os demais pares de meia, podemos concluir que existem\n$$\n\\frac{16!}{2 \\cdot 2 \\cdot 2 \\cdot 2 \\cdot 2 \\cdot 2 \\cdot 2 \\cdot 2} = \\frac{16!}{2^{8}}\n$$\nsequências em que o símbolo $m_{i}$ vem antes de $s_{i}$ para todo $i$. Esse é o número procurado.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27056,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nO quadrado $A B C D$ de lado $1~\\mathrm{cm}$ está inscrito em uma circunferência de centro $O$. O ponto $M$ está sobre o arco $B C$, o segmento $A M$ encontra $B D$ no ponto $P$, o segmento $D M$ encontra $A C$ no ponto $Q$.\na) Verifique que $\\angle A Q D=\\angle P A D$.\n![](attached_image_1.png)\nb) Encontre a área do quadrilátero $A P Q D$.",
    "options": [],
    "answer": "1/2 cm^2",
    "solution": "Solution:\n\na) Como $\\angle A M D=45^{\\circ}=\\angle O A D=\\angle O D A$, temos\n$$\n\\begin{aligned}\n\\angle A Q D & =\\angle A M D+\\angle M A Q \\\\\n& =45^{\\circ}+\\angle M A Q \\\\\n& =\\angle O A D+\\angle M A Q \\\\\n& =\\angle P A D\n\\end{aligned}\n$$\n\nb) De modo semelhante ao item anterior, podemos mostrar que $\\angle A P D=\\angle A D Q$ e daí os triângulos $A P D$ e $Q D A$ são semelhantes. Daí\n$$\n\\frac{A Q}{A D}=\\frac{A D}{P D} \\Rightarrow A D^{2}=A Q \\cdot P D\n$$\nEm virtude do item anterior, a área do quadrilátero $A P Q D$ é dada por\n$$\n\\begin{aligned}\n\\frac{P D \\cdot A Q \\operatorname{sen}(\\angle P O Q)}{2} & =\\frac{A D^{2}}{2} \\\\\n& =\\frac{\\operatorname{Area}(A B C D)}{2} \\\\\n& =\\frac{1}{2}~\\mathrm{cm}^{2}\n\\end{aligned}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27057,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOs vértices do quadrilátero $ABCD$ estão em uma circunferência. Cada uma de suas diagonais bissecta um ângulo e trisecta o ângulo oposto. Determine as medidas dos ângulos do quadrilátero.\n\n![](attached_image_1.png)\n\nObservação: Dizemos que uma semirreta $OM$ trisecta um ângulo $\\angle AOB$ se a amplitude de um dos dois ângulos determinado por ela é um terço do valor da amplitude do ângulo $\\angle AOB$.",
    "options": [],
    "answer": "Either 72°, 72°, 108°, 108° or 720°/7, 720°/7, 540°/7, 540°/7.",
    "solution": "Solution:\n\nSuponha que a diagonal $AC$ bissecta o ângulo $\\angle DCB$. A outra diagonal bissecta o ângulo $\\angle ABC$ ou $\\angle ADC$. Podemos supor sem perda de generalidade que ela bissecta o ângulo $\\angle ADC$. Como $\\angle ADB = \\angle BDC$ e $\\angle ACB = \\angle ACD$, os arcos $AD$, $AB$ e $BC$ correspondem a um mesmo valor do ângulo central $x$. Analisando agora as trisecções dos ângulos, ou $\\angle DBC = 2 \\cdot \\angle ABD = x$ ou $\\angle DBC = \\frac{1}{2} \\cdot \\angle ABD = \\frac{x}{4}$. No primeiro caso, temos\n$$\n\\begin{aligned}\n\\angle DAB + \\angle ABC + \\angle BCD + \\angle CDA & = 360^\\circ \\\\\n\\frac{3x}{2} + \\frac{3x}{2} + x + x & = 360^\\circ \\\\\nx & = 72^\\circ\n\\end{aligned}\n$$\nAssim os ângulos do quadrilátero são $72^\\circ$, $72^\\circ$, $108^\\circ$ e $108^\\circ$. Procedendo de modo semelhante no segundo caso, temos $x = \\frac{720^\\circ}{7}$ e os ângulos do quadrilátero são $\\frac{720^\\circ}{7}$, $\\frac{720^\\circ}{7}$, $\\frac{540^\\circ}{7}$ e $\\frac{540^\\circ}{7}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27058,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSeja $f(x)=\\frac{9^{x}}{9^{x}+3}$.\na) Encontre o valor de $f(x)+f(1-x)$.\nb) Calcule o valor da soma\n$$\nf\\left(\\frac{1}{2020}\\right)+f\\left(\\frac{2}{2020}\\right)+f\\left(\\frac{3}{2020}\\right)+\\ldots+f\\left(\\frac{2019}{2020}\\right)\n$$",
    "options": [],
    "answer": "a) 1; b) 2019/2",
    "solution": "Solution:\n\na) Temos\n$$\n\\begin{aligned}\nf(x)+f(1-x) & =\\frac{9^{x}}{9^{x}+3}+\\frac{9^{1-x}}{9^{1-x}+3} \\\\\n& =\\frac{9^{x}}{9^{x}+3}+\\frac{9}{9+3 \\cdot 9^{x}} \\\\\n& =\\frac{9^{x}}{9^{x}+3}+\\frac{3}{3+9^{x}} \\\\\n& =1\n\\end{aligned}\n$$\n\nb) Em virtude do item anterior, podemos juntar os termos correspondendo às frações $\\frac{i}{2020}$ e $\\frac{2020-i}{2020}$ para obter o número 1 :\n$$\n\\begin{aligned}\n& f\\left(\\frac{1}{2020}\\right)+f\\left(\\frac{2}{2020}\\right)+f\\left(\\frac{3}{2020}\\right)+\\ldots+f\\left(\\frac{2019}{2020}\\right)= \\\\\n& \\left(f\\left(\\frac{1}{2020}\\right)+f\\left(\\frac{2019}{2020}\\right)\\right)+\\left(f\\left(\\frac{2}{2020}\\right)+f\\left(\\frac{2018}{2020}\\right)\\right)+ \\\\\n& \\left(f\\left(\\frac{3}{2020}\\right)+f\\left(\\frac{2017}{2020}\\right)\\right)+\\ldots+\\left(f\\left(\\frac{1009}{2020}\\right)+f\\left(\\frac{1011}{2020}\\right)\\right)+f\\left(\\frac{1010}{2020}\\right)= \\\\\n& 1009+\\frac{3}{3+3}= \\\\\n& \\frac{2019}{2} \\text{.}\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27059,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSeja $ABCD$ um retângulo tal que $AB = \\sqrt{2} BC$. Seja $E$ um ponto sobre o semicírculo com diâmetro $AB$, como indicado na figura a seguir. Sejam $K$ e $L$ as interseções de $AB$ com $ED$ e $EC$, respectivamente. Se $AK = 2\\ \\mathrm{cm}$ e $BL = 9\\ \\mathrm{cm}$, calcule, em $\\mathrm{cm}$, o comprimento do segmento $KL$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "6",
    "solution": "Solution:\n\nSejam $x$ e $y$ os comprimentos das projeções ortogonais dos segmentos $EK$ e $EL$ sobre o segmento $AB$ e $P$ a projeção ortogonal de $E$ sobre $AB$. Além disso, seja $h = EP$. Por semelhança de triângulos, temos\n$$\n\\frac{2}{x} = \\frac{BC}{h} \\text{ e } \\frac{9}{y} = \\frac{BC}{h}\n$$\nPortanto,\n$$\n\\begin{aligned}\nKL &= x + y \\\\\n   &= \\frac{11h}{BC}\n\\end{aligned}\n$$\nComo $EAB$ é um triângulo retângulo em $E$, segue das relações métricas dos triângulos retângulos que\n$$\n\\begin{aligned}\nh^2 &= AP \\cdot PB \\\\\n    &= (2 + x)(9 + y) \\\\\n    &= \\left(2 + \\frac{2h}{BC}\\right)\\left(9 + \\frac{9h}{BC}\\right) \\\\\n    &= 18\\left(1 + \\frac{h}{BC}\\right)^2\n\\end{aligned}\n$$\nPortanto,\n$$\n1 + \\frac{h}{BC} = \\frac{h}{3\\sqrt{2}}\n$$\nAlém disso, de\nsegue que\n$$\n\\sqrt{2} BC = AK + KL + LB = 11\\left(1 + \\frac{h}{BC}\\right)\n$$\n$$\n\\frac{h}{3\\sqrt{2}} = \\frac{\\sqrt{2} BC}{11}\n$$\nOu seja, $\\frac{h}{BC} = \\frac{6}{11}$ e $KL = \\frac{11h}{BC} = 6\\ \\mathrm{cm}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27060,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nEm uma turma existem 70 alunos, tais que:\nI) 14 meninos passaram em Matemática;\nII) 12 meninos passaram em Física;\nIII) 10 meninos e 16 meninas não passaram em Matemática nem em Física;\nIV) 32 são meninos;\nV) 10 passaram nas duas disciplinas;\nVI) 22 passaram apenas em Matemática.\nQuantas meninas passaram somente em Física?",
    "options": [],
    "answer": "4",
    "solution": "Solution:\nPara resolver o problema, vamos utilizar o diagrama abaixo, no qual o retângulo superior representa as quantidades de meninos em cada caso e o inferior as quantidades de meninas; na circunferência da esquerda, a quantidade de alunos que passou em matemática, enquanto que na da direita, a quantidade que passou em física, sendo que na intersecção, a quantidade que passou em ambos.\n![](attached_image_1.png)\nAgora, vamos preenchendo o diagrama, utilizando as informações, usando uma sequência conveniente. Por $III$, temos:\n![](attached_image_2.png)\nPor $IV$ e $I$, como 32 são meninos e 14 deles passaram em matemática, então $32-14=18$ não passaram em matemática, o que significa que $18-10=8$ deles passou apenas em física.\n![](attached_image_3.png)\nPor $II$, como 12 meninos passaram em física, então $12-8=4$ deles passaram também em matemática.\n![](attached_image_4.png)\nPor $I$, como 14 meninos passaram em matemática, então $14-4=10$ deles passaram apenas em matemática. Com isso, já determinamos todas as quantidades relacionadas aos meninos.\n|  | meninos |  |  |\n| :---: | :---: | :---: | :---: |\nPor $V$, como 10 alunos passaram nas duas disciplinas, então $10-4=6$ meninas passaram em ambas.\n![](attached_image_5.png)\nPor $VI$, como 22 passaram apenas em matemática, então $22-10=12$ meninas passaram apenas em matemática.\n![](attached_image_6.png)\nComo já temos $10+10+4+8+12+6+16=66$ alunos no diagrama, a quantidade de meninas que passou apenas em física é $70-66=4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27061,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSe $x$ e $y$ são números reais tais que $x+y=10$ e $x^{3}+y^{3}=400$, determine o valor de $x^{2}+y^{2}$.",
    "options": [],
    "answer": "60",
    "solution": "Solution:\n\nComo\n$$\n\\begin{aligned}\nx^{3}+y^{3} & =(x+y)\\left(x^{2}-x y+y^{2}\\right) \\\\\n& =(x+y)\\left((x+y)^{2}-3 x y\\right) \\\\\n& =10(100-3 x y)\n\\end{aligned}\n$$\nsegue que $400=10(100-3 x y)$, ou seja, $400=1000-30 x y$, logo $x y=20$. Daí\n$$\n\\begin{aligned}\nx^{2}+y^{2} & =(x+y)^{2}-2 x y \\\\\n& =100-40 \\\\\n& =60\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27062,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSeja $\\overline{a b}$ um número inteiro de dois dígitos. Um inteiro positivo é um parente de $\\overline{a b}$ se:\ni) o dígito das unidades de $n$ também é $b$.\nii) os outros dígitos de $n$ são distintos de zero e somam $a$.\nPor exemplo, os parentes de 31 são 31, 121, 211 e 1111. Encontre todos os números de dois dígitos que dividem todos os seus parentes.",
    "options": [],
    "answer": "30, 45, 90",
    "solution": "Solution:\n\nVamos dividir o problema em casos:\n\n1) Se $a=1$ e $n$ é parente de $\\overline{a b}$, então a única possibilidade é $n=\\overline{1 b}$. Esse número é sempre solução, pois $\\overline{1 b}$ divide $\\overline{1 b}$.\n\n2) Se $a=2$, como $\\overline{2 b}$ deve dividir $\\overline{11 b}$, então $\\overline{2 b}$ deve dividir $\\overline{11 b}-\\overline{2 b}=90$. Os divisores de $90=2 \\cdot 3^{2} \\cdot 5$ são\n$$\n1,2,3,5,6,9,10,15,18,30,45 \\text{ e } 90\n$$\nComo $20 \\leq \\overline{2 b}=20+b \\leq 29$, não temos soluções nesse caso.\n\n3) Se $a \\geq 3$, como $\\overline{a b}$ deve dividir a $\\overline{(a-3) 21 b}$ e $\\overline{(a-3) 12 b}$, então $\\overline{a b}$ divide a diferença $\\overline{(a-3) 21 b}-\\overline{(a-3) 12 b}=90$. Lembrando da análise de divisores do item anterior, as únicas possibilidades são 30, 45 e 90, dado que $\\overline{a b} \\geq 30$. Resta verificarmos que essas três opções dividem todos os seus parentes.\n\ni) Se $n$ é parente de 30, então $n$ termina em 0 e a soma de seus dígitos é um múltiplo de 3. Portanto, $n$ é múltiplo de $10 \\cdot 3=30$.\n\nii) Se $n$ é parente de 45, então $n=A \\cdot 10+5$, em que $A$ é um número cuja soma dos dígitos é 4. Como a soma dos dígitos de $n$ é um múltiplo de $9$, $n$ também é múltiplo de 9. Além disso, como $n$ termina em 5, então $n$ também é múltiplo de 5. Portanto, $n$ é múltiplo de $9 \\cdot 5=45$.\n\niii) Se $n$ é parente de 90, então $n=A \\cdot 10$, em que $A$ é um número cuja soma dos dígitos é 9. Como $n$ termina em 0 e é múltiplo de 9, segue que $n$ é múltiplo de $9 \\cdot 10=90$.\n\nLogo, os únicos inteiros positivos que dividem todos os seus parentes são\n$$\n30, 45 \\text{ e } 90\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27063,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPara cada número inteiro positivo $n$ se associa um inteiro não negativo $f(n)$ de modo que se cumpram as três regras seguintes:\ni) $f(a b)=f(a)+f(b)$.\nii) $f(n)=0$ se $n$ é um primo maior que 10$.$\niii) $f(1)<f(243)<f(2)<11$.\nSabendo que $f(2106)<11$, determine o valor de $f(96)$.",
    "options": [],
    "answer": "31",
    "solution": "Solution:\n\nPela propriedade $i)$, temos $f(243)=f\\left(3^{5}\\right)=5 f(3)$. Dado que\n$$\n0 \\leq f(1)<5 f(3)<f(2)<11\n$$\ne que $5 f(3)$ é um múltiplo de $5$, temos $5 f(3)=5$, ou seja, $f(3)=1$. Note que $2106=2 \\cdot 3^{4} \\cdot 13$. Assim, pela propriedade $i)$,\n$$\nf(2106)=f(2)+4 f(3)+f(13)=f(2)+4\n$$\nDe $f(2106)<11$, segue que $f(2)<7$. Usando $iii)$, temos $5<f(2)<7$ e assim $f(2)=6$. Logo,\n$$\nf(96)=f\\left(3 \\cdot 2^{5}\\right)=f(3)+5 f(2)=1+5 \\times 6=31\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27064,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUma escola tem 100 alunos e 100 armários numerados de 1 a 100. Inicialmente, todos os armários estão fechados. O primeiro aluno passa e abre todos os armários; o segundo passa e fecha todos os de números pares; o terceiro passa e muda a posição de todos os múltiplos de 3, ou seja, os que estão abertos ele fecha e os que estão fechados ele abre; o quarto aluno muda a posição de todos os armários de números múltiplos de 4; e assim por diante até o centésimo aluno, que muda a posição dos armários de números múltiplos de 100. Depois da passagem de todos os alunos, quantos armários ficam fechados?",
    "options": [],
    "answer": "90",
    "solution": "Solution:\n\nOs armários que ficam abertos são aqueles com números que possuem uma quantidade ímpar de divisores e isso só acontece com os números que são quadrados perfeitos. De 1 a 100, temos 1, 4, 9, 16, 25, 36, 49, 64, 81 e 100, que são quadrados perfeitos, ou seja, são 10 quadrados perfeitos. Portanto, o número de armários que ficam fechados é $100-10=90$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27065,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDizemos que um inteiro positivo é bacana se ao somarmos os quadrados de seus dígitos e repetirmos essa operação sucessivamente, obtivermos o número $1$. Por exemplo, $1900$ é bacana, pois\n$$\n1900 \\rightarrow 82 \\rightarrow 68 \\rightarrow 100 \\rightarrow 1\n$$\n\na) Encontre dois números de dois dígitos consecutivos bacanas.\n\nb) Verifique que para todo $n>2020$ existem dois números bacana de $n$ dígitos que são consecutivos.",
    "options": [],
    "answer": "Part a: 31 and 32. Part b: For each n > 2020, take A to be a number with thirty-one ones followed by enough zeros to have n digits, and let B be A plus one; both are consecutive n-digit numbers that eventually reach one under the process.",
    "solution": "Solution:\n\na) Os números $31$ e $32$ são bacanas, pois\n$$\n31 \\rightarrow 10 \\rightarrow 1\n$$\ne\n$$\n32 \\rightarrow 13 \\rightarrow 10 \\rightarrow 1\n$$\n\nb) Considere os números\n$$\nA=\\underbrace{111 \\ldots 11}_{31 \\text{ uns}}\\underbrace{000 \\ldots 00}_{n-31 \\text{ zeros}}\n$$\ne\n$$\nB=\\underbrace{111 \\ldots 11}_{31 \\text{ uns}}\\underbrace{000 \\ldots 00}_{n-32 \\text{ zeros}}1.\n$$\nNote que $A$ e $B$ são números bacanas consecutivos, pois\n$$\nA \\rightarrow 31 \\rightarrow 10 \\rightarrow 1\n$$\ne\n$$\nB \\rightarrow 32 \\rightarrow 13 \\rightarrow 10 \\rightarrow 1\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27066,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNo planeta $X$, existem 100 países alienígenas com conflitos entre si. Para evitar uma guerra mundial, esses países se organizam em grupos de alianças militares para proteção mútua. Sabemos que as alianças seguem as seguintes regras:\n1) Nenhuma aliança contém mais de 50 países.\n2) Quaisquer dois países pertencem a pelo menos uma aliança.\n\na) É possível que um país participe de menos de três alianças militares?\n\nb) Qual é o menor número possível de alianças para que essas duas condições sejam satisfeitas?",
    "options": [],
    "answer": "a) No; b) 6",
    "solution": "Solution:\n\na) Não é possível. Suponha que o país $A$ pertence a no máximo duas alianças. Nesse caso, como cada aliança tem no máximo 50 países, o país $A$ é membro de uma mesma aliança com no máximo $49+49=98$ países. Como existem 99 países distintos de $A$, pelo menos um deles não estará em uma aliança com $A$ e isso gera um conflito com a regra (2).\n\nb) Como cada país participa de pelo menos 3 alianças, se somarmos as quantidades de participações de todos os países em todas as alianças obteremos pelo menos o número $3 \\cdot 100=300$. Por outro lado, cada aliança contabiliza no máximo 50 participações. Assim, o número mínimo de alianças não pode ser menor do que $\\frac{300}{50}=6$. Para verificarmos que esse número é realmente o menor possível, basta exibirmos um exemplo de configuração em que ele é atingido. Divida os 100 países em 4 grupos de 25 países: $A$, $B$, $C$, $D$. Agora, forme as alianças de 50 países constituídas por todas as 6 combinações de dois desses grupos:\n$$\nA \\cup B,\\ A \\cup C,\\ A \\cup D,\\ B \\cup C,\\ B \\cup D,\\ C \\cup D\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27067,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNo desenho ao lado, o triângulo $A B C$ é equilátero e $B D = C E = A F = \\frac{A B}{3}$. A razão $\\frac{E G}{G D}$ pode ser escrita na forma $\\frac{m}{n}$, $\\operatorname{mdc}(m, n) = 1$. Quanto vale $m+n$ ?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "5",
    "solution": "Solution:\n\nVeja que $\\frac{E G}{G D} = \\frac{[E B G]}{[B G D]}$. Agora, $\\frac{[E B G]}{[B C F]} = \\frac{\\frac{1}{2} E B \\cdot B G \\cdot \\operatorname{sen}(\\angle E B G)}{\\frac{1}{2} B C \\cdot B F \\cdot \\operatorname{sen}(\\angle C B F)}$. Como $\\angle E B G = \\angle C B F$, temos\n$$\n\\frac{[E B G]}{[B C F]} = \\frac{E B \\cdot B G}{B C \\cdot B F}\n$$\nAnalogamente,\n$$\n\\frac{[B G D]}{[A B F]} = \\frac{B G \\cdot B D}{B A \\cdot B F}\n$$\nDividindo (1) por (2), obtemos $\\frac{[E B G]}{[B G D]} \\cdot \\frac{[A B F]}{[B C F]} = \\frac{E B \\cdot B A}{B C \\cdot B D}$. Finalmente, como $\\frac{[A B F]}{[B C F]} = \\frac{A F}{C F} = \\frac{1}{2}$, temos\n$$\n\\frac{[E B G]}{[B G D]} = \\frac{E B \\cdot B A \\cdot C F}{B C \\cdot B D \\cdot A F} = \\frac{2}{3} \\cdot 3 \\cdot 2 = 4\n$$\nLogo, $\\frac{E G}{G D} = \\frac{4}{1}$ e, portanto, $m+n=5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27068,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLembrando que\n$$\n\\begin{aligned}\n(a+b+c)^{3} = a^{3} + b^{3} + c^{3} + 3 a^{2} b + 3 a b^{2} + 3 a c^{2} + 3 a^{2} c + 3 b^{2} c + 3 b c^{2} + 6 a b c\n\\end{aligned}\n$$\nEncontre as soluções do sistema de equações\n$$\n\\begin{aligned}\na^{3} + 3 a b^{2} + 3 a c^{2} - 6 a b c & = 1 \\\\\nb^{3} + 3 b a^{2} + 3 b c^{2} - 6 a b c & = 1 \\\\\nc^{3} + 3 c a^{2} + 3 c b^{2} - 6 a b c & = 1\n\\end{aligned}\n$$",
    "options": [],
    "answer": "(1, 1, 1)",
    "solution": "Solution:\n\nSejam $A = a^{3} + 3 a b^{2} + 3 a c^{2} - 6 a b c$, $B = b^{3} + 3 b a^{2} + 3 b c^{2} - 6 a b c$ e $C = c^{3} + 3 c a^{2} + 3 c b^{2} - 6 a b c$. Usando a identidade algébrica mencionada no enunciado, temos\n$$\n\\begin{aligned}\n-A + B + C & = (-a + b + c)^{3} \\\\\nA - B + C & = (a - b + c)^{3} \\\\\nA + B - C & = (a + b - c)^{3}\n\\end{aligned}\n$$\nComo $A = B = C = 1$, segue que\n$$\n\\begin{aligned}\n(-a + b + c)^{3} & = 1 \\\\\n(a - b + c)^{3} & = 1 \\\\\n(a + b - c)^{3} & = 1\n\\end{aligned}\n$$\nPortanto\n$$\n-a + b + c = a - b + c = a + b - c = 1\n$$\nDe $2 = (-a + b + c) + (a - b + c) = 2c$, segue que $c = 1$. De forma semelhante, podemos concluir que $a = b = 1$ e que $(a, b, c) = (1, 1, 1)$ é a única solução do sistema.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27069,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEm um tabuleiro $7 \\times 7$, dizemos que 4 casas assinaladas com $X$ formam um castelo se elas são vértices de um retângulo com lados paralelos aos do tabuleiro, como indicado na figura a seguir:\n![](attached_image_1.png)\n\na) Marque 21 casas com $X$ no tabuleiro $7 \\times 7$ sem que exista qualquer castelo entre elas.\n\nb) Verifique que para qualquer escolha de 22 casas com $X$ sempre existirá um castelo.",
    "options": [],
    "answer": "21",
    "solution": "Solution:\n\na) Um exemplo é o que está na seguinte figura a seguir.\n\n| $X$ | $X$ |   | $X$ |   |   |   |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n|   | $X$ | $X$ |   | $X$ |   |   |\n|   |   | $X$ | $X$ |   | $X$ |   |\n|   |   |   | $X$ | $X$ |   | $X$ |\n| $X$ |   |   |   | $X$ | $X$ |   |\n|   | $X$ |   |   |   | $X$ | $X$ |\n| $X$ |   | $X$ |   |   |   | $X$ |\n\nb) Se existem 22 casas com um $X$, como há apenas 7 linhas, pelo menos uma delas terá 4 casas marcadas com $X$. Permutando linhas ou colunas do tabuleiro, nenhum castelo será criado ou destruído. Assim, podemos supor que a linha que contém as 4 casas marcadas é a primeira e que eles estão nas primeiras 4 colunas, como indicado na figura a seguir.\n![](attached_image_2.png)\n\nPodemos agora analisar duas situações distintas.\n\n1) Suponha que na primeira linha existem apenas 4 casas marcadas. Se em alguma linha restante das quatro colunas encontrarmos duas casas marcadas, teremos um castelo. Se isso não ocorre, nas primeiras 4 colunas teremos no máximo $4+6$ casas marcadas e, consequentemente, pelo menos $22-10=12$ casas com $X$ devem ser dispostas nas últimas 3 colunas. Isso garante que uma delas terá pelo menos 4 casas marcadas. Novamente após permutar linhas e colunas, podemos supor que essas 4 casas marcadas estão na quinta coluna como indicado na figura abaixo:\n![](attached_image_3.png)\n\nPara que não exista castelo, na região pontilhada pode existir no máximo mais uma casa marcada em cada uma das duas últimas colunas. Assim, essa região terá no máximo $4+2=6$ casas com $X$. Como nas últimas 3 colunas existem 12 casas marcadas, no retângulo tracejado devemos ter pelo menos $12-6=6$ casas marcadas e assim, inevitavelmente, teremos um castelo. Portanto, se na primeira linha existem exatamente 4 casas marcadas, sempre teremos um castelo.\n\n2) Suponha que na primeira linha existem mais de 4 casas marcadas e que essas casas estão nas primeiras 5 colunas. Como discutido anteriormente, se alguma linha restante das cinco primeiras colunas possuir mais de uma casa marcada, teremos um castelo. Se isso não ocorre, nas primeiras 5 colunas teremos no máximo $5+6$ casas marcadas e, consequentemente, pelo menos $22-11=11$ casas com $X$ devem ser dispostas nas últimas 2 colunas. Isso garante que uma delas terá pelo menos 6 casas marcadas. Novamente, após permutar linhas e colunas, podemos supor que 5 dessas casas marcadas estão na sexta coluna como indicado na figura abaixo:\n![](attached_image_4.png)\n\nPara que não exista castelo, na região pontilhada pode existir no máximo mais uma casa marcada na última coluna. Assim, essa região terá no máximo $5+1=6$ casas com $X$. Como nas últimas 2 colunas existem 11 casas marcadas, nos dois retângulos tracejados devemos ter pelo menos $11-6=5$ casas marcadas. Mas isso é um absurdo, pois nesses retângulos temos apenas 4 casas. Esse absurdo mostra que sempre teremos um castelo também nesse caso.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27070,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSeja $ABCD$ um quadrilátero tal que $AC = BC + CD$. Se $\\angle BCD = 120^{\\circ}$, $\\overline{CA}$ é bissetriz e $AB = x$, qual o valor de $BD$, em função de $x$?",
    "options": [],
    "answer": "x",
    "solution": "Solution:\n\nVamos analisar a figura:\n![](attached_image_1.png)\n\nAplicando a Lei dos Cossenos no triângulo $ABC$, temos:\n$$\n\\begin{aligned}\n& x^{2} = b^{2} + (b + c)^{2} - 2 \\cdot b \\cdot (b + c) \\cdot \\cos 60^{\\circ} \\\\\n& x^{2} = b^{2} + b^{2} + 2 b c + c^{2} - 2 b^{2} \\frac{1}{2} - 2 b c \\frac{1}{2} \\\\\n& x^{2} = b^{2} + b^{2} + 2 b c + c^{2} - b^{2} - b c \\\\\n& x^{2} = b^{2} + b c + c^{2}\n\\end{aligned}\n$$\n\nAplicando agora a Lei dos Cossenos no triângulo $BCD$, temos:\n$$\n\\begin{aligned}\nBD^{2} & = b^{2} + c^{2} - 2 \\cdot b \\cdot c \\cdot \\cos 120^{\\circ} \\\\\nBD^{2} & = b^{2} + c^{2} - 2 b c \\left(-\\cos 60^{\\circ}\\right) \\\\\nBD^{2} & = b^{2} + c^{2} + b c \\\\\nBD^{2} & = x^{2} \\\\\nBD & = x\n\\end{aligned}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27071,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDetermine o valor da soma\n$$\nS=\\operatorname{sen}^{2} 1^{\\circ}+\\operatorname{sen}^{2} 2^{\\circ}+\\operatorname{sen}^{2} 3^{\\circ}+\\ldots+\\operatorname{sen}^{2} 89^{\\circ}\n$$",
    "options": [],
    "answer": "89/2",
    "solution": "Solution:\n\nComo $\\sen x = \\cos (90^{\\circ} - x)$ e $\\sen^{2} x + \\cos^{2} x = 1$, segue que\n$$\n\\sen^{2} 1^{\\circ} + \\sen^{2} 89^{\\circ} = \\sen^{2} 2^{\\circ} + \\sen^{2} 88^{\\circ} = \\ldots = \\sen^{2} 44^{\\circ} + \\sen^{2} 46^{\\circ} = 1\n$$\nPortanto, pareando os termos da soma em pares do tipo $\\sen^{2} x + \\sen^{2}(90 - x)$, temos\n$$\nS = 44 + \\sen^{2} 45^{\\circ} = 44 + \\frac{1}{2} = \\frac{89}{2}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27072,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOs 2020 números\n$$\n1 \\cdot 2, 2 \\cdot 3, 3 \\cdot 4, \\ldots, 2020 \\cdot 2021\n$$\nsão escritos na lousa. Um movimento consiste em escolher três números $a, b, c$ escritos na lousa, apagá-los e escrever na lousa o número\n$$\n\\frac{a b c}{a b + b c + a c}\n$$\na) Verifique que\n$$\n\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} = \\left(\\frac{a b c}{a b + b c + a c}\\right)^{-1}\n$$\nb) Usando\n$$\n\\frac{1}{k(k+1)} = \\frac{1}{k} - \\frac{1}{k+1}\n$$\nencontre o valor da soma\n$$\n\\frac{1}{1 \\cdot 2} + \\frac{1}{2 \\cdot 3} + \\frac{1}{3 \\cdot 4} + \\ldots + \\frac{1}{2020 \\cdot 2021}\n$$\nc) Após 1009 movimentos, sobram dois números na lousa. Prove que se um deles é $4/3$, o outro é maior do que $4$.",
    "options": [],
    "answer": "The telescoping sum equals 2020/2021, and if one of the final two numbers is 4/3, the other is greater than 4.",
    "solution": "Solution:\n\na) Note que\n$$\n\\begin{aligned}\n\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} & = \\frac{b c + a c + a b}{a b c} \\\\\n& = \\left(\\frac{a b c}{a b + b c + a c}\\right)^{-1}\n\\end{aligned}\n$$\n\nb) Em virtude da identidade sugerida, temos\n$$\n\\begin{aligned}\n\\frac{1}{1 \\cdot 2} + \\frac{1}{2 \\cdot 3} + \\frac{1}{3 \\cdot 4} + \\ldots + \\frac{1}{2020 \\cdot 2021} & = \\\\\n\\left(\\frac{1}{1} - \\frac{1}{2}\\right) + \\left(\\frac{1}{2} - \\frac{1}{3}\\right) + \\left(\\frac{1}{3} - \\frac{1}{4}\\right) + \\ldots + \\left(\\frac{1}{2020} - \\frac{1}{2021}\\right) & = \\\\\n\\frac{1}{1} - \\frac{1}{2021} & = \\frac{2020}{2021}\n\\end{aligned}\n$$\n\nc) Pelo item anterior, a qualquer momento, a soma dos inversos dos números escritos na lousa é sempre a mesma. No final, se restam apenas os números $x$ e $4/3$, temos\n$$\n\\frac{1}{x} + \\frac{3}{4} = \\frac{2020}{2021} < 1\n$$\nDaí\n$$\n\\begin{aligned}\n\\frac{1}{x} & < 1 - \\frac{3}{4} \\\\\n& = 1/4\n\\end{aligned}\n$$\nOu seja, $x > 4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27073,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEm uma competição, os competidores ocupam todos os lugares de um salão retangular onde os assentos estão organizados em filas e colunas de tal modo que há mais de duas filas e em cada fila há mais de dois assentos. Em um dado momento, esses competidores recebem a ordem de cumprimentarem com um aperto de mão apenas os seus vizinhos diretos no salão, isto é, quem está na sua esquerda, direita, na frente, atrás e em diagonal.\n![](attached_image_1.png)\n\nAlguém percebeu que foram dados 1020 apertos de mãos. Seja $n$ o número de filas e $m$ a quantidade de lugares de cada fila.\n\na) Quais são as possíveis quantidades de apertos de mão que cada competidor do salão pode ter dado?\n\nb) Para cada uma das quantidades $x$ descritas no item anterior, calcule quantos competidores deram $x$ apertos de mão.\n\nc) Determine o número de competidores que estavam no salão.",
    "options": [],
    "answer": "a) Possible handshakes per person: 3, 5, or 8.\nb) Counts: 4 people have 3; 2(n−2)+2(m−2) people have 5; (n−2)(m−2) people have 8.\nc) Total competitors: 280, with (n, m) = (14, 20) or (20, 14).",
    "solution": "Solution:\n\na) Um competidor que sentou em um dos 4 cantos do salão deu 3 apertos de mão. Se um competidor sentou em algum dos 4 bordos, mas sem estar posicionado nos cantos, ele deu 5 apertos de mão. Finalmente, se um competidor sentou no interior do salão, ele deu 8 apertos de mão.\nA figura a seguir ilustra essas possibilidades.\n\n| 3 | 5 | 3 |\n| :---: | :---: | :---: |\n| 5 | 8 | 5 |\n| 3 | 5 | 3 |\n\nb) Existem 4 pessoas que apertaram a mão de 3 pessoas, $2(n-2)+2(m-2)$ pessoas que apertaram a mão de 5 pessoas e $(n-2)(m-2)$ pessoas que apertaram a mão de 8 pessoas.\n\nc) Multiplicando a quantidade de pessoas do item anterior pelos apertos de mão correspondentes, teremos contado cada um deles duas vezes, portanto\n$$\n\\begin{aligned}\n8 \\cdot (n-2)(m-2) + 5 \\cdot [2(n-2) + 2(m-2)] + 3 \\cdot 4 &= 2 \\cdot 1020 \\\\\n8 m n - 6 n - 6 m &= 2036\n\\end{aligned}\n$$\nDaí $m = \\frac{1018 + 3n}{4n - 3}$. Para que $m$ seja inteiro, $4n - 3$ deve dividir $1018 + 3n$ e consequentemente $4n - 3$ deve dividir $4(1018 + 3n) = 3(4n - 3) + 4081$. Como $4n - 3$ divide $3(4n - 3)$, segue que $4n - 3$ divide $4081 = 7 \\cdot 11 \\cdot 53$. Os divisores de 4081 são 1, 7, 11, 53, 77, 371, 583 e 4081. Note que $4n - 3$ deixa resto 1 na divisão por 4 e, dessa lista de divisores, os únicos que podem ser iguais a $4n - 3$ são: $1, 53, 77$ e $4081$. Se\n$$\n\\begin{aligned}\n& 4n - 3 = 1 \\Rightarrow n = 1 \\\\\n& 4n - 3 = 53 \\Rightarrow n = 14 \\\\\n& 4n - 3 = 77 \\Rightarrow n = 20 \\\\\n& 4n - 3 = 4081 \\Rightarrow n = 1021\n\\end{aligned}\n$$\nO primeiro e o quarto caso não são admissíveis, pois há mais de duas filas e em cada fila há mais de dois assentos. Assim $n = 14$ e $m = \\frac{1018 + 3 \\cdot 14}{4 \\cdot 14 - 3} = 20$ ou então $n = 20$ e $m = \\frac{1018 + 3 \\cdot 20}{4 \\cdot 20 - 3} = 14$. Portanto, estavam no salão $m \\cdot n = 280$ competidores.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27074,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCom 5 algarismos não nulos, podemos formar 120 números, sem repetir algarismo em um mesmo número. Seja $S$ a soma de todos esses números. Determine a soma dos algarismos de $S$, sendo:\n\na) 1, 3, 5, 7 e 9 os 5 algarismos;\n\nb) 0, 2, 4, 6 e 8 os 5 algarismos, lembrando que 02468 é um número com 4 algarismos e, portanto, não teremos 120 números neste caso.",
    "options": [],
    "answer": "a) 30; b) 12",
    "solution": "Solution:\n\na) São 120 números ao todo, com todas as combinações possíveis. Assim, em cada uma das posições (unidade, dezena, centena, unidade do milhar, dezena do milhar), cada um dos algarismos aparece a mesma quantidade de vezes, ou seja, $\\frac{120}{5}=24$. Por exemplo, nas unidades, o algarismo 1 aparece 24 vezes, assim como o 3, o 5, o 7 e o 9. Dessa forma, a soma de todas as unidades é:\n$$\n\\begin{aligned}\n24 \\cdot 1+24 \\cdot 3+24 \\cdot 5+24 \\cdot 7+24 \\cdot 9 & = \\\\\n24(1+3+5+7+9) & = \\\\\n24 \\cdot 25 & =600\n\\end{aligned}\n$$\nSendo assim, a soma $S$ é:\n$$\n\\begin{aligned}\nS & =600+600 \\cdot 10+600 \\cdot 100+600 \\cdot 1.000+600 \\cdot 10.000 \\\\\n& =600(1+10+100+1.000+10.000) \\\\\n& =600 \\cdot 11.111 \\\\\n& =6.666.600\n\\end{aligned}\n$$\nPor fim, a soma dos algarismos de $S$ é $6+6+6+6+6+0+0=30$.\n\nb) Vamos utilizar o mesmo raciocínio do item anterior, contando também os números que iniciam por 0, ou seja, que possuem apenas 4 algarismos. Chamando essa soma dos 120 números de $S'$, temos:\n$$\n\\begin{aligned}\nS' & =24 \\cdot(0+2+4+6+8) \\cdot 11.111 \\\\\n& =24 \\cdot 20 \\cdot 11.111 \\\\\n& =5.333.280\n\\end{aligned}\n$$\nPrecisamos descontar agora os números que começam com 0, que é a soma de todos os números de 4 algarismos que podemos formar com 2, 4, 6 e 8 (120-24 = 96 ao todo). Esta soma vale $24 \\cdot(2+4+6+8) \\cdot 1.111=533.280$. Portanto, $S=S'-533.280=4.800.000$, sendo a soma de seus algarismos igual a $4+8+0+0+0+0+0=12$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27075,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSeja $ABCD$ um quadrado de lado $28~\\mathrm{cm}$. Seja $P$ um ponto interior ao quadrado e $E$ um ponto no lado $CD$ tal que $PE$ é perpendicular a $CD$. Além disso, $AP = BP = PE$. Encontre o comprimento de $AP$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "17.5 cm",
    "solution": "Solution:\n\nSeja $M$ o ponto de interseção da reta $EP$ com o lado $AB$. Como $EP$ é perpendicular a $CD$, então $EM$ é perpendicular a $AB$. Além disso, como o triângulo $ABP$ é isósceles, $M$ é o ponto médio de $AB$. Se o comprimento de $AP$ é $x$, segue do Teorema de Pitágoras que\n$$\n\\begin{aligned}\nAP^2 & = AM^2 + PM^2 \\\\\nx^2 & = 14^2 + (28 - x)^2 \\\\\nx^2 & = 196 + 784 - 56x + x^2 \\\\\n56x & = 980 \\\\\nx & = 17,5~\\mathrm{cm}\n\\end{aligned}\n$$\n\n![](attached_image_2.png)",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27076,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNo paralelogramo $A B C D$, o ângulo $B A D$ é agudo e o lado $A D$ é menor que o lado $A B$. A bissetriz do ângulo $\\angle B A D$ corta o lado $C D$ em $E$. Por $D$ se traça uma perpendicular a $A E$ que corta $A E$ em $P$ e $A B$ em $F$. Traçamos por $E$ uma perpendicular a $A E$ que corta o lado $B C$ em $Q$. Além disso, o segmento $P Q$ é paralelo a $A B$ e o comprimento de $A B$ é $20 \\mathrm{~cm}$.\n![](attached_image_1.png)\n\na) Encontre o valor de $\\frac{C Q}{A D}$.\nb) Encontre a medida do comprimento do lado $A D$.",
    "options": [],
    "answer": "CQ/AD = 1/2; AD = 40/3 cm",
    "solution": "Solution:\n\na) De $\\angle D A P=\\angle P A F$ e $A P \\perp D F$, segue que o triângulo $A D F$ é isósceles e que $P$ é o ponto médio de $D F$. Como $E Q$ e $D F$ são perpendiculares a $A E$, temos que $D F \\| E Q$ e $\\angle C E Q=\\angle C D F=\\angle D F A$. No paralelogramo $A B C D$, $\\angle D A B=\\angle D C B$ e assim os triângulos $A D F$ e $E C Q$ são semelhantes. Portanto, $C E=C Q$ e\n$$\n\\frac{C Q}{A D}=\\frac{E Q}{D F}\n$$\nComo os lados opostos do quadrilátero $D E Q P$ são paralelos, segue que ele é um paralelogramo. Consequentemente $D P=E Q$ e\n$$\n\\frac{C Q}{A D}=\\frac{D P}{2 \\cdot D P}=\\frac{1}{2}\n$$\n\nb) Seja $x$ o comprimento do segmento $A D$. De $\\angle D A E=\\angle E A B=\\angle D E A$, podemos concluir que o triângulo $A D E$ é isósceles e assim $A D=D E=x$. Do item anterior, decorre que $C Q=x / 2$. Assim\n$$\n\\begin{aligned}\nC D & =D E+E C \\\\\n20 & =x+x / 2 \\\\\n& =3 x / 2\n\\end{aligned}\n$$\nFinalmente, $x=40 / 3 \\mathrm{~cm}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27077,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEm uma reunião de matemáticos, Carlos diz a Frederico: $O$ dobro do produto dos dois dígitos do número de matemáticos na reunião é exatamente a nossa quantidade. Qual a quantidade mínima de matemáticos que deve se juntar a nós para que nossa quantidade seja um número primo? Ajude Frederico a resolver o problema.",
    "options": [],
    "answer": "1",
    "solution": "Solution:\n\nVamos representar a quantidade de matemáticos por $\\overline{a b}$. Temos que:\n$$\n\\begin{aligned}\n2 a b & = \\overline{a b} \\\\\n2 a b & = 10 a + b \\\\\n2 a b - b & = 10 a \\\\\nb(2 a - 1) & = 10 a \\\\\nb & = \\frac{10 a}{2 a - 1}\n\\end{aligned}\n$$\nAnalisando os possíveis valores de $a$, lembrando que ambos são algarismos, com $a \\neq 0$, concluímos que $a = 3$ e $b = 6$. Portanto, o número de matemáticos é $36$ e com mais um matemático a quantidade será um número primo.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27078,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nNa figura, os triângulos $\\triangle A D E$ e $\\triangle A B C$ são isósceles. Se $\\angle D F C=150^{\\circ}$, qual a medida de $\\angle F D B$ ?\n![](attached_image_1.png)",
    "options": [],
    "answer": "40°",
    "solution": "Solution:\nComo $\\triangle A D E$ e $\\triangle A B C$ são isósceles, temos $\\angle D A E=\\angle D E A=\\angle A B C=\\alpha$. Além disso, $\\angle B F E=\\angle D F C=150^{\\circ}$ (OPV). Pela soma dos ângulos internos do quadrilátero $A B F E$, temos $3 \\alpha+150^{\\circ}=360^{\\circ}$, donde $\\alpha=70^{\\circ}$. Por fim, $\\angle F D B+70^{\\circ}+70^{\\circ}=180^{\\circ}$, ou seja, $\\angle F D B=40^{\\circ}$.\n![](attached_image_2.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27079,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNo triângulo $ABC$, temos os ângulos $\\angle ACB = 65^{\\circ}$ e $\\angle ABC = 70^{\\circ}$. Sobre os prolongamentos do lado $BC$, marcam-se o ponto $P$ de tal modo que $BP = AB$ e que $B$ esteja entre $P$ e $C$; e o ponto $Q$ de modo que $CQ = AC$ e que $C$ esteja entre $B$ e $Q$. Se $O$ é o centro da circunferência que passa por $A$, $P$ e $Q$, encontre os valores dos ângulos $\\angle OAQ$ e $\\angle OAP$.",
    "options": [],
    "answer": "∠OAP = 57°30′, ∠OAQ = 55°",
    "solution": "Solution:\n\nSejam $\\angle BAO = x$ e $\\angle OAC = y$. De $BP = AB$ e $AC = CQ$, podemos concluir que $\\angle APB = \\angle PAB = \\beta$ e $\\angle CAQ = \\angle AQC = \\gamma$. Como $O$ é o centro da circunferência que passa por $A$, $P$ e $Q$, temos $OP = OA = OQ$ e daí os triângulos $AOP$ e $AOQ$ são isósceles. Consequentemente,\n$$\n\\begin{aligned}\n\\angle APO & = \\angle PAO \\\\\n\\alpha + \\beta & = x + \\beta \\\\\nx & = \\alpha\n\\end{aligned}\n$$\nDe modo semelhante, podemos concluir que $y = \\alpha$. Pelo Teorema do Ângulo Externo, $\\angle ABC = 2\\beta$ e $\\angle ACB = 2\\gamma$. Assim, $\\beta = 70 / 2 = 35^{\\circ}$ e $\\gamma = 65 / 2^{\\circ} = 32^{\\circ} 30'$. Além disso, $2\\alpha = \\angle BAC = 45^{\\circ}$ e assim $\\alpha = 22^{\\circ} 30'$. Finalmente,\n$$\n\\begin{aligned}\n\\angle OAP & = \\beta + \\alpha \\\\\n& = 57^{\\circ} 30'\n\\end{aligned}\n$$\ne\n$$\n\\begin{aligned}\n\\angle OAQ & = \\gamma + \\alpha \\\\\n& = 55^{\\circ}\n\\end{aligned}\n$$\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27080,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEm um tabuleiro $5 \\times 5$, cada quadradinho possui uma peça em seu centro. O único movimento permitido para uma dessas peças é se deslocar para um quadradinho que compartilhe exatamente um vértice com o quadradinho em que ela está, como indicado na figura abaixo. Tanto é possível que várias peças ocupem um mesmo quadradinho quanto que um quadradinho fique vazio. Em um dado momento, todas as peças serão movidas simultaneamente.\na) Dê um exemplo de movimentos das 25 peças de modo que sobrem exatamente 5 quadradinhos vazios.\nb) Qual o número mínimo de quadradinhos vazios que poderão ser encontrados após esse momento?\n![](attached_image_1.png)",
    "options": [],
    "answer": "5",
    "solution": "Solution:\n\na) A figura a seguir mostra um conjunto de movimentos em que apenas 5 quadradinhos ficam vazios.\n![](attached_image_2.png)\n\nb) Pinte as colunas do tabuleiro de forma alternada de preto e branco. Assim existem 15 quadradinhos pretos e 10 brancos. Veja que após um movimento de uma peça, ela fica em um quadradinho de cor oposta ao que ela estava anteriormente. Assim, as peças dos 15 quadradinhos pretos ficarão nos 10 quadradinhos brancos e as 10 peças que estavam nos quadradinhos brancos poderão ocupar no máximo 10 quadradinhos pretos, ou seja, devem sobrar pelo menos 5 quadradinhos pretos vazios.\n![](attached_image_3.png)\n\nEm virtude do exemplo dado no item $a$, sabemos que esse mínimo é realizável.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27081,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSe $6 x y - \\sqrt{3} x^{2} = \\sqrt{3} y^{2}$, calcule\n$$\n\\left(\\frac{x}{y}\\right)^{4} + \\left(\\frac{y}{x}\\right)^{4}\n$$",
    "options": [],
    "answer": "98",
    "solution": "Solution:\nTemos:\n$$\n\\begin{aligned}\n6 x y - \\sqrt{3} x^{2} & = \\sqrt{3} y^{2} \\\\\n6 x y & = \\sqrt{3} (x^{2} + y^{2}) \\\\\n\\frac{x y}{x^{2} + y^{2}} & = \\frac{\\sqrt{3}}{6} \\\\\n\\frac{x^{2} + y^{2}}{x y} & = \\frac{6}{\\sqrt{3}} \\\\\n\\frac{x}{y} + \\frac{y}{x} & = 2 \\sqrt{3}\n\\end{aligned}\n$$\nElevando a última igualdade ao quadrado, temos:\n$$\n\\begin{aligned}\n\\left(\\frac{x}{y} + \\frac{y}{x}\\right)^{2} & = (2 \\sqrt{3})^{2} \\\\\n\\frac{x^{2}}{y^{2}} + 2 + \\frac{y^{2}}{x^{2}} & = 12 \\\\\n\\frac{x^{2}}{y^{2}} + \\frac{y^{2}}{x^{2}} & = 10\n\\end{aligned}\n$$\nFinalmente, elevando mais uma vez a expressão ao quadrado, temos:\n$$\n\\begin{aligned}\n\\left(\\frac{x^{2}}{y^{2}} + \\frac{y^{2}}{x^{2}}\\right)^{2} & = 10^{2} \\\\\n\\frac{x^{4}}{y^{4}} + 2 + \\frac{y^{4}}{x^{4}} & = 100 \\\\\n\\left(\\frac{x}{y}\\right)^{4} + \\left(\\frac{y}{x}\\right)^{4} & = 98\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27082,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nConsidere um quadrado $ABCD$ de centro $O$. Sejam $E, F, G$ e $H$ pontos no interior dos lados $AB, BC, CD$ e $DA$, respectivamente, tal que $AE = BF = CG = DH$. Sabe-se que $OA$ intersecta $HE$ no ponto $X$, $OB$ intersecta $EF$ no ponto $Y$, $OC$ intersecta $FG$ no ponto $Z$ e $OD$ intersecta $GH$ no ponto $W$. Sejam $x$ e $y$ as medidas dos comprimentos de $AE$ e $AH$, respectivamente.\na) Dado que Área $(EFGH) = 1\\ \\mathrm{cm}^2$, calcule o valor de $x^2 + y^2$.\nb) Verifique que $HX = \\frac{y}{x+y}$. Em seguida, conclua que $X, Y, Z$ e $W$ são vértices de um quadrado.\nc) Calcule\nÁrea $(ABCD) \\cdot$ Área $(XYZW)$.",
    "options": [],
    "answer": "x^2 + y^2 = 1 and Area(ABCD) × Area(XYZW) = 1",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\na) Sejam $x$ e $y$ as medidas dos comprimentos de $AE$ e $AH$, respectivamente. Dado que $AH = EB$, $AE = BF$ e $\\angle HAE = \\angle EBF$, segue que os triângulos $AEH$ e $EBF$ são congruentes. Daí\n$$\n\\angle HEF = 180^\\circ - \\angle HEA - \\angle BEF = 180^\\circ - \\angle EFB - \\angle BEF = 90^\\circ\n$$\nDe modo semelhante, podemos concluir que $\\angle EFG = \\angle FGH = \\angle GHE = 90^\\circ$. Pelo Teorema de Pitágoras, $HE^2 = x^2 + y^2$. O mesmo vale para os demais lados do retângulo $HEFG$, ou seja, $EH = EF = FG = GH = 1$. Portanto, a sua área é $1 = A_{HEFG} = x^2 + y^2$.\n\nb) Como $AC$ é bissetriz de $\\angle HAE$, decorre do Teorema da Bissetriz Interna que\n$$\n\\begin{aligned}\n\\frac{HX}{EX} & = \\frac{AH}{AE} \\\\\n\\frac{HX}{HX + EX} & = \\frac{AH}{AH + AE} \\\\\nHX & = \\frac{y}{x + y}\n\\end{aligned}\n$$\nDe modo semelhante, $EX = \\frac{x}{x + y}$. A diagonal $BD$ também é bissetriz de $\\angle EBF$ e $\\triangle EBF \\equiv AHE$. Daí $EY = HX = \\frac{y}{x + y}$ e podemos concluir por analogia ao argumento inicial, agora aplicado ao quadrado $HEFG$, que os pontos $X, Y, Z$ e $W$ são vértices de um quadrado.\n\nc) Aplicando o Teorema de Pitágoras no triângulo $EXY$, obtemos\n$$\n\\begin{aligned}\nXY^2 & = EX^2 + EY^2 \\\\\n& = \\frac{x^2}{(x + y)^2} + \\frac{y^2}{(x + y)^2}\n\\end{aligned}\n$$\nPortanto, a área do quadrilátero $XYZW$ é Área $(XYZW) = \\frac{x^2 + y^2}{(x + y)^2}$. Como Área $(EFGH) = 1$, segue que $x^2 + y^2 = 1$ e que\n$$\n\\text{Área}(ABCD) \\cdot \\text{Área}(XYZW) = (x + y)^2 \\cdot \\frac{x^2 + y^2}{(x + y)^2} = 1\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27083,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNa figura, $\\angle ABC = 100^\\circ$, $\\angle FAC = 3 \\angle ECB$ e $\\angle GCA = 3 \\angle DAB$. Determine a medida do ângulo agudo na interseção das bissetrizes internas dos triângulos $\\triangle ADB$ e $\\triangle CEB$ relativo aos ângulos $D$ e $E$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "35°",
    "solution": "Solution:\n\nVamos marcar na figura os pontos de intersecção e os ângulos que nos interessam, sendo $x$ o ângulo da intersecção das bissetrizes que procuramos.\n\n![](attached_image_2.png)\n\nPela soma dos ângulos internos do triângulo $\\triangle ABC$, temos:\n$$\n\\begin{aligned}\n100^\\circ + (180^\\circ - \\alpha - 3\\beta) + (180^\\circ - \\beta - 3\\alpha) &= 180^\\circ \\\\\n460^\\circ - 4\\alpha - 4\\beta &= 180^\\circ \\\\\n4\\alpha + 4\\beta &= 280^\\circ \\\\\n\\alpha + \\beta &= 70^\\circ\n\\end{aligned}\n$$\n\nPelos triângulos $\\triangle ABD$ e $\\triangle BCE$, temos $\\alpha + 2\\theta = 100^\\circ$ e $\\beta + 2\\lambda = 100^\\circ$. Somando estas duas últimas equações, chegamos a:\n$$\n\\begin{aligned}\n\\alpha + 2\\theta + \\beta + 2\\lambda &= 100^\\circ + 100^\\circ \\\\\n\\alpha + \\beta + 2\\theta + 2\\lambda &= 200^\\circ \\\\\n70^\\circ + 2(\\theta + \\lambda) &= 200^\\circ \\\\\n\\theta + \\lambda &= 65^\\circ\n\\end{aligned}\n$$\n\nAgora, vamos utilizar a soma dos ângulos internos do quadrilátero $KBLJ$.\n$$\n\\begin{aligned}\n\\angle KJL + \\angle BKJ + \\angle KBL + \\angle BLJ &= 360^\\circ \\\\\nx + (80^\\circ + \\theta) + 100^\\circ + (80^\\circ + \\lambda) &= 360^\\circ \\\\\nx + 260^\\circ + \\theta + \\lambda &= 360^\\circ \\\\\nx + \\theta + \\lambda &= 100^\\circ \\\\\nx + 65^\\circ &= 100^\\circ \\\\\nx &= 35^\\circ\n\\end{aligned}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27084,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nArătați că, pentru orice număr natural prim $p$, există numerele naturale $x, y, z$ și $t$, nu toate nule, astfel încât $t < p$ și\n$$\nx^{2} + y^{2} + z^{2} = t p\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27085,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $a \\geqslant b \\geqslant c \\geqslant d$ numere reale cu proprietatea că\n$$\n(a-b)(b-c)(c-d)(d-a) = -3\n$$\na) Dacă $a+b+c+d=6$, demonstrați că $d < 0,36$\n\nb) Dacă $a^{2}+b^{2}+c^{2}+d^{2}=14$, arătați că $(a+c)(b+d) \\leqslant 8$. Precizați cazurile de egalitate.",
    "options": [],
    "answer": "a) d < 0.36.\nb) (a + c)(b + d) ≤ 8, with equality when (a, b, c, d) = (3, 2, 1, 0) or (0, −1, −2, −3).",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27086,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $M, N$ și $P$ mijloacele laturilor $BC, CA$, respectiv $AB$, ale triunghiului ascuțitunghic $ABC$. Notăm cu $A'$, $B'$ și $C'$ punctele diametral opuse vârfurilor $A, B$, respectiv $C$ în cercul circumscris triunghiului $ABC$. Pe segmentele deschise $MA'$, $NB'$ și $PC'$ se consideră punctele $X, Y$, respectiv $Z$, astfel încât $\\frac{MX}{XA'}=\\frac{NY}{YB'}=\\frac{PZ}{ZC'}$.\n\na) Demonstrați că dreptele $AX, BY$ și $CZ$ sunt concurente într-un punct $S$.\n\nb) Arătați că $OS < OG$, unde $O$ este centrul cercului circumscris, iar $G$ este centrul de greutate al triunghiului $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na.\nFie $H$ ortocentrul triunghiului $ABC$. Patrulaterul $BHC A'$ este un paralelogram ($BH$ și $A'C$ sunt perpendiculare pe dreapta $AC$, iar $CH$ și $A'B$ sunt perpendiculare pe dreapta $AB$), prin urmare punctul $M$ este mijlocul segmentului $HA'$.\n\nDacă $\\frac{MX}{XA'}=k$, atunci\n$$\n\\frac{A'X}{XM}=\\frac{1}{k} \\Rightarrow \\frac{A'X}{A'M}=\\frac{1}{k+1} \\Rightarrow \\frac{A'X}{A'H}=\\frac{1}{2k+2} \\Rightarrow \\frac{A'X}{XH}=\\frac{1}{2k+1}\n$$\nFie $O$ centrul cercului circumscris triunghiului $ABC$ și $S$ punctul în care dreapta $AX$ intersectează dreapta Euler $OH$.\n\nFolosind teorema lui Menelaus în triunghiul $HOA'$ cu transversala $XSA$ obținem:\n$$\n\\frac{HS}{SO} \\cdot \\frac{OA}{AA'} \\cdot \\frac{A'X}{XH}=1 \\Rightarrow \\frac{HS}{SO} \\cdot \\frac{1}{2} \\cdot \\frac{1}{2k+1}=1 \\Rightarrow \\frac{HS}{SO}=4k+2\n$$\nConsiderând punctele de intersecție dintre dreapta Euler și dreptele $BY$, respectiv $CZ$, vom obține că ele împart segmentul $HO$ în același raport $4k+2$, deci coincid cu punctul $S$. Prin urmare, dreptele $AX, BY$ și $CZ$ sunt concurente (iar punctul lor comun este situat pe dreapta Euler).\n\nb.\nRaportul $k$ parcurge intervalul $(0, \\infty)$, deci raportul $\\frac{HS}{SO}=4k+2$ parcurge intervalul $(2, \\infty)$. Obținem $HS > 2 OS$, deci $OS < \\frac{OH}{3} = OG$.\n\n---\n\nSoluție alternativă pentru punctul a).\n\n$ABA'B'$ este dreptunghi, deci $A'B' = AB \\stackrel{\\text{not.}}{=} c$.\n$MN$ este linie mijlocie în triunghiul $ABC$, deci $MN \\parallel AB \\parallel A'B'$, așadar $MNB'A'$ este trapez. Deoarece $\\frac{MX}{XA'}=\\frac{NY}{YB'}$, rezultă că $XY \\parallel MN \\parallel A'B'$.\n\nFie $E \\in A'B'$, astfel încât $ME \\parallel NB'$, și $F = ME \\cap XY$.\nDeoarece $MNB'E$ este paralelogram și $FY \\parallel MN$, avem $EB' = FY = MN = \\frac{c}{2}$.\n\nDin asemănarea triunghiurilor $MXF$ și $MA'E$ obținem $\\frac{XF}{A'E} = \\frac{MX}{MA'} = t$, așadar $XF = t \\cdot A'E = t \\cdot \\frac{c}{2}$ și $XY = XF + FY = \\frac{(t+1) \\cdot c}{2}$.\n\nDeoarece $XY \\parallel A'B' \\parallel AB$, rezultă că $ABXY$ este trapez. Fie $S = AX \\cap BY$. Din asemănarea triunghiurilor $ASB$ și $XSY$ obținem $\\frac{SX}{SA} = \\frac{SY}{SB} = \\frac{XY}{AB} = \\frac{t+1}{2}$.\n\nAnalog rezultă că patrulaterul $BCYZ$ este trapez. Fie $S' = BY \\cap CZ$.\nCa mai înainte, deducem că $\\frac{S'Y}{S'B} = \\frac{t+1}{2}$, deci $S = S'$, așadar dreptele $AX, BY$ și $CZ$ sunt concurente.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27087,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDeterminați cel mai mare număr natural $n$ pentru care este adevărată afirmația:\nExistă $n$ numere naturale nenule distincte $x_{1}, x_{2}, \\ldots, x_{n}$ cu proprietatea că oricare ar fi numerele $a_{1}, a_{2}, \\ldots, a_{n} \\in\\{-1,0,1\\}$, nu toate nule, numărul $n^{3}$ nu divide numărul $a_{1} x_{1}+a_{2} x_{2}+\\ldots+a_{n} x_{n}$.",
    "options": [],
    "answer": "9",
    "solution": "Solution:\nPentru $n=9$ alegem $x_{1}=2^{0}, x_{2}=2^{1}, \\ldots, x_{9}=2^{8}$. Oricare ar fi numerele $a_{1}, a_{2}, \\ldots, a_{n} \\in\\{-1,0,1\\}$, avem:\n$$\n\\left|a_{1} x_{1}+a_{2} x_{2}+\\ldots+a_{n} x_{n}\\right| \\leq 1+2+\\ldots+2^{8}=2^{9}-1<9^{3}\n$$\nDacă $9^{3}$ divide $a_{1} x_{1}+a_{2} x_{2}+\\ldots+a_{n} x_{n}$, rezultă că $a_{1} \\cdot 1+a_{2} \\cdot 2+\\ldots+a_{n} \\cdot 2^{8}=0$. Din considerente de paritate obținem că $a_{1}=0$. Simplificând cu 2 și urmărind din nou paritatea, vom avea $a_{2}=0$ etc. Obținem că numerele $a_{1}, a_{2}, \\ldots, a_{n}$ sunt toate nule, contradicție. Ca urmare, $n=9$ are proprietatea din enunț.\n\nDacă $n \\geqslant 10$, atunci $2^{n}>n^{3}$.\nFie $A=\\left\\{x_{1}, x_{2}, \\ldots, x_{n}\\right\\}$ o mulțime de $n$ numere naturale nenule (distincte) și $\\mathcal{P}(A)$ mulțimea părților sale. Cum $|\\mathcal{P}(A)|=2^{n}>n^{3}$, folosind principiul lui Dirichlet obținem două submulțimi diferite $B$ și $C$ ale lui $A$ astfel încât\n$$\n\\sum_{x \\in B} x \\equiv \\sum_{x \\in C} x \\quad\\left(\\bmod n^{3}\\right)\n$$\nAlegând $a_{i}=1$ pentru elementele lui $B \\backslash C, a_{j}=-1$ pentru elementele lui $C \\backslash B$ și $a_{k}=0$ pentru celelalte elemente ale lui $A$, găsim o combinație $a_{1} x_{1}+a_{2} x_{2}+\\ldots+a_{n} x_{n}$ care se divide cu $n^{3}$. Așadar, numerele $n \\geqslant 10$ nu au proprietatea din enunț, deci numărul căutat este $n=9$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27088,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSe consideră două cercuri $\\mathcal{C}_1$ și $\\mathcal{C}_2$ tangente interior în punctul $P$ (cercul $\\mathcal{C}_2$ este interior cercului $\\mathcal{C}_1$).\nO coardă $AB$ din cercul $\\mathcal{C}_1$ este tangentă cercului $\\mathcal{C}_2$ în punctul $C$. Fie $D$ al doilea punct de intersectie dintre dreapta $CP$ și cercul $\\mathcal{C}_1$. O tangentă dusă din $D$ la $\\mathcal{C}_2$ intersectează a doua oară cercul $\\mathcal{C}_1$ in $E$ și cercul $\\mathcal{C}_2$ în $F$.\nArătați că punctul $F$ este centrul cercului înscris în triunghiul $ABE$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27089,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nÎn triunghiul scalen ascuțitunghic $A B C$ se notează cu $D$ piciorul bisectoarei din $A$ și cu $E$ piciorul înălțimii din $A$. Mediatoarea segmentului $A D$ intersectează semicercurile de diametre $A B$ și $A C$, construite în exteriorul triunghiului $A B C$, în $X$, respectiv $Y$. Demonstrați că punctele $X, Y, D$ și $E$ sunt conciclice.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27090,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $p_{i}\\left(i \\in \\mathbb{N}^{*}\\right)$ al $i$-ulea număr prim (în ordine crescătoare). Pentru fiecare număr natural nenul $k$, notăm cu $a_{k}$ numărul de numere naturale nenule $i$ cu proprietatea că produsul $p_{i} p_{i+1}$ divide numărul $k$.\nDacă $n$ este un număr natural nenul, arătați că\n$$\na_{1}+a_{2}+\\ldots+a_{n}<\\frac{n}{3}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27091,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAflați toate perechile de numere naturale $(a, b)$ pentru care numărul $\\frac{(a+b)^{2}}{4+4 a(a-b)^{2}}$ este întreg.",
    "options": [],
    "answer": "All pairs with a = b, and for any positive integer w, the pairs (a, b) = (4 w^4 − 2 w, 4 w^4) and (a, b) = (4 w^4 + 2 w, 4 w^4).",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27092,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nAflați câte numere naturale $k \\in \\{1,2,3, \\ldots, 2022\\}$ au proprietatea că, dacă pe un cerc se scriu 2022 numere reale astfel încât suma oricăror $k$ numere aflate pe poziții consecutive este egală cu 2022, atunci toate cele 2022 de numere sunt egale.",
    "options": [],
    "answer": "672",
    "solution": "Solution:\nFie $k \\in \\{1,2,3, \\ldots, 2022\\}$ un număr cu proprietatea că, date fiind numerele reale $x_{1}, x_{2}, \\ldots x_{2022}$ scrise pe un cerc, suma oricăror $k$ numere aflate pe poziții consecutive este egală cu 2022. Pentru orice $n \\in \\mathbb{N}$, definim $x_{n}=x_{r}$, unde $r \\in \\{1,2,3, \\ldots, 2022\\}$, astfel încât $n \\equiv r(\\bmod\\ 2022)$.\n\nA. Pentru orice $i \\in \\mathbb{N}$, avem\n$$\nx_{i}+x_{i+1}+\\ldots+x_{i+k-1}=x_{i+1}+x_{i+2}+\\ldots+x_{i+k}=2022\n$$\nde unde rezultă că $x_{i}=x_{i+k}$\n\nB. Dacă $(2022, k)=d \\geqslant 2$, notând $x=\\frac{2022 \\cdot d}{k}$ și scriind pe cerc, în ordine, numerele\n$$\nx, \\underbrace{0,0, \\ldots, 0}_{d-1\\ \\text{ori}} x, \\underbrace{0,0, \\ldots, 0, \\ldots x}_{\\text{de}\\ d-1\\ \\text{ori}}, \\underbrace{0,0, \\ldots, 0}_{\\text{de}\\ d-1\\ \\text{ori}}\n$$\nobservăm că suma oricăror $k$ numere aflate pe poziții consecutive este egală cu 2022, fără ca numerele să fie egale.\nCa urmare, numerele $k$ pentru care $(2022, k) \\neq 1$ nu sunt soluții.\n\nC. Arătam că toate numerele $k \\in \\{1,2,3, \\ldots, 2022\\}$ pentru care $(2022, k)=1$ sunt soluții. Cum $(k, 2022)=1$, există $u, v \\in \\mathbb{N}$, astfel încât $k \\cdot u=2022 v+1$. Atunci, pentru orice $m \\in \\mathbb{N}$, avem:\n$$\nx_{m}=x_{m+k \\cdot u}=x_{m+2022 v+1}=x_{m+1}\n$$\ndeci toate cele 2022 de numere sunt egale.\n\nD. În total sunt $\\varphi(2022)=672$ de soluții.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27093,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSe consideră o rețea formată din 49 de puncte, ce reprezintă vârfurile a 36 de pătrate de latură 1 în care este descompus un pătrat de latură 6.\nSpunem că un pătrat cu vârfurile în punctele rețelei este bun, dacă laturile și diagonalele sale nu sunt pe laturile pătratelor rețelei.\na) Aflați numărul de pătrate bune care se pot forma cu vârfurile rețelei.\nb) Arătați că există două pătrate bune disjuncte și necongruente, astfel încât cea mai mică distanță dintre punctele lor este $\\frac{\\sqrt{5}}{5}$.",
    "options": [],
    "answer": "a) 70; b) sqrt(5)/5",
    "solution": "Solution:\na) Spunem că un pătrat este normal, dacă are vârfurile în punctele rețelei și laturile sale se află pe drepte ale rețelei paralele cu laturile pătratului $6 \\times 6$, sau pe laturile pătratului $6 \\times 6$. Orice pătrat bun are vârfurile pe laturile unui pătrat normal, a cărui lungime a laturii poate fi egală cu $3,4,5$ sau $6$. Fie un pătrat normal de latură $x \\in \\{3,4,5,6\\}$. Latura unui pătrat bun înscris în acesta are lungimea de forma $\\sqrt{a^{2}+b^{2}}$, unde $a, b \\in \\{1,2, \\ldots, x-1\\}$, $a \\neq b$, astfel încât $a+b=x$.\n\nÎntr-adevăr, dacă $a=0$, sau $b=x$, atunci laturile pătratului bun sunt pe laturile pătratelor rețelei, iar dacă $a=b$, atunci laturile pătratului bun sunt paralele cu diagonalele pătratului mare, deci diagonalele sale sunt pe dreptele suport ale rețelei, fals.\n\nDacă $a, b \\in \\{1,2, \\ldots, x-1\\}$, cu $a \\neq b$ și $a+b=x$, fiecare pătrat normal de latură $x$ conține exact două pătrate bune generate de perechea $(a, b)$, având latura de lungime $\\sqrt{a^{2}+b^{2}}$ și ale căror vârfuri sunt pe laturile acestuia, ca în figura alăturată.\n\n![](attached_image_1.png)\n\nPentru $x=3=1+2$, în fiecare pătrat normal de latură $3$ sunt înscrise exact două pătrate bune, generate de $(a, b) \\in \\{(1,2),(2,1)\\}$. De aici, în fiecare bandă orizontală de forma $6 \\times 3$ există $2 \\cdot 4=8$ pătrate bune de latură $\\sqrt{5}$. Deoarece sunt $4$ benzi orizontale de lătime $3$, în total există $8 \\cdot 4=32$ pătrate bune de latură $\\sqrt{5}$.\n\nAnalog se arată că există:\n- $18$ pătrate bune de latură $\\sqrt{10}$, generate de $(a, b) \\in \\{(1,3),(3,1)\\}$,\n- $8$ pătrate bune de latură $\\sqrt{17}$, generate de $(a, b) \\in \\{(1,4),(4,1)\\}$,\n- $2$ pătrate bune de latură $\\sqrt{26}$, generate de $(a, b) \\in \\{(1,5),(5,1)\\}$,\n- $8$ pătrate bune de latură $\\sqrt{13}$, generate de $(a, b) \\in \\{(2,3),(3,2)\\}$,\n- $2$ pătrate bune de latură $\\sqrt{20}$, generated de $(a, b) \\in \\{(2,4),(4,2)\\}$.\n\nNumărul tuturor pătratelor bune este: $32+18+8+4+2+4+2=70$.\n\nb) Alegem pătratele bune $ABCD$ și $MNPQ$ ca în figura alăturată, cu $AB=\\sqrt{5}$ și $MN=\\sqrt{10}$, astfel încât $A, D, M$ și $N$ sunt pe laturile pătratului mare.\n\nFie dreapta $d$ care trece prin $Q$ și este paralelă cu $BC$. Deoarece $MNPQ$ este inclus în semiplanul $[dM$, rezultă că distanța căutată este cea de la $Q$ la $ABCD$. Alegem punctele $R$ și $F$ astfel încât $Q$ este mijlocul segmentului $BR$, triunghiul $BFQ$ este dreptunghic în $B$, cu $BF=2$, iar $F$ și $M$ sunt de o parte și de alta a dreptei $d$.\n\n![](attached_image_2.png)\n\nDeoarece triunghiurile $BQF$ și $RCB$ sunt congruente, rezultă că dreptele $BC$ și $FQ$ sunt perpendiculare, iar distanța căutată este egală cu lungimea segmentului $QE$, unde $Q$ este punctul de intersecție a dreptelor $BC$ și $FQ$. Cu teorema catetei, rezultă că $EQ=\\frac{\\sqrt{5}}{5}$.\n\nObservații.\n1) Dacă $a, b \\in \\{1,2, \\ldots, x-1\\}$, cu $a+b=x$, în fiecare bandă orizontală de forma $6 \\times (a+b)$ există $2 \\cdot (7-(a+b))$ pătrate bune de latură $\\sqrt{a^{2}+b^{2}}$. Deoarece sunt $7-(a+b)$ benzi orizontale de lătime $a+b$, în total există $2 \\cdot (7-a-b)^{2}$ pătrate bune de latură $\\sqrt{a^{2}+b^{2}}$, generate de perechile $(a, b)$ și $(b, a)$.\n2) Se poate arăta că singurele tipuri de pătrate bune necongruente care nu au vârfuri comune și au interioarele disjuncte sunt cele de mai înainte.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27094,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFie $n \\geqslant 2$ un număr natural și $M = \\mathbb{N} \\times \\mathbb{N} \\times \\ldots \\times \\mathbb{N}$ (de $n$ ori $\\mathbb{N}$), mulțimea $n$-uplelor de numere naturale. Pentru fiecare $a = (a_{1}, a_{2}, \\ldots, a_{n}) \\in M$ notăm cu $d_{a}$ numărul perechilor $(i, j)$, $i, j \\in \\{1, 2, \\ldots, n\\}$, pentru care $a_{i} - a_{j} = 1$.\nDeterminați valoarea maximă a numărului $d_{a}$ când $a$ parcurge mulțimea $M$.",
    "options": [],
    "answer": "floor(n^2/4)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27095,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFie $a$, $b$ și $c$ trei numere reale pozitive cu suma $3$. Arătați că:\n$$\n\\frac{a b}{a b+a+b}+\\frac{b c}{b c+b+c}+\\frac{c a}{c a+c+a}+\\frac{1}{9}\\left(\\frac{(a-b)^{2}}{a b+a+b}+\\frac{(b-c)^{2}}{b c+b+c}+\\frac{(c-a)^{2}}{c a+c+a}\\right) \\leqslant 1\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nSoluţia 1. Avem\n$$\n\\frac{a b}{a b+a+b}+\\frac{(a-b)^{2}}{9(a b+a+b)}=\\frac{a^{2}+7 a b+b^{2}}{9(a b+a+b)}\n$$\nVom arăta că\n$$\n\\frac{a^{2}+7 a b+b^{2}}{a b+a+b} \\leqslant a+b+1\n$$\nAceastă inegalitate se scrie echivalent\n$$\na^{2}+7 a b+b^{2} \\leqslant a^{2}+3 a b+b^{2}+a^{2} b+a b^{2}+a+b \\Leftrightarrow a^{2} b+a b^{2}+a+b \\geqslant 4 a b .\n$$\nDin inegalitatea mediilor avem $a^{2} b+b \\geqslant 2 a b$ și $a b^{2}+a \\geqslant 2 a b$, deci inegalitatea (1) este adevărată.\nScriind și inegalitățile analoage, deducem că:\n$$\n\\sum \\frac{a b}{a b+a+b}+\\frac{1}{9} \\sum \\frac{(a-b)^{2}}{a b+a+b} \\leqslant \\frac{1}{9} \\sum(a+b+1)=1\n$$\nSolution:\nSoluția 2. Avem $a b+a+b=a b+(a+b) \\cdot \\frac{a+b+c}{3}=\\frac{a^{2}+b^{2}+5 a b+a c+b c}{3}$\nPrin urmare,\n$$\n\\begin{aligned}\n\\frac{a b}{a b+a+b}+\\frac{(a-b)^{2}}{9(a b+a+b)}-\\frac{1}{3} & =\\frac{9 a b+\\left(a^{2}+b^{2}-2 a b\\right)-\\left(a^{2}+b^{2}+5 a b+a c+b c\\right)}{9(a b+a+b)} \\\\\n& =\\frac{b(a-c)}{9(a b+a+b)}+\\frac{a(b-c)}{9(a b+a+b)}\n\\end{aligned}\n$$\nAtunci\n$$\n\\begin{aligned}\n& \\sum \\frac{a b}{a b+a+b}+\\frac{1}{9} \\sum \\frac{(a-b)^{2}}{a b+a+b}-1=\\sum\\left(\\frac{b(a-c)}{9(a b+a+b)}+\\frac{a(b-c)}{9(a b+a+b)}\\right)= \\\\\n& =\\sum\\left(\\frac{b(a-c)}{9(a b+a+b)}+\\frac{b(c-a)}{9(b c+b+c)}\\right)=\\sum \\frac{-b(a-c)^{2}(b+c)}{9(a b+a+b)(b c+b+c)} \\leq 0\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27096,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSpunem despre un număr natural că este rotund, dacă numărul divizorilor săi este pătrat perfect. Pentru un număr natural rotund $n$ cu $d^{2}$ divizori construim un tablou $d \\times d$ și completăm celulele acestuia cu divizorii lui $n$. La fiecare pas, putem alege un rând al tabloului și muta divizorul de pe coloana 1 pe coloana 2, cel de pe coloana 2 pe coloana $3, \\ldots$, cel de pe coloana $d$ pe coloana 1. Oricare configurație a tabloului la care ajungem este numită fezabilă dacă există o coloană ce are elementele $a_{1}, a_{2}, \\ldots, a_{d}$ în această ordine și $a_{1}\\mid a_{2}\\mid \\ldots \\mid a_{d}$ sau $a_{d}\\mid a_{d-1}\\mid \\ldots \\mid a_{1}$.\nDeterminați numerele rotunde $n \\geqslant 2$ pentru care, oricum am completa initial tabloul divizorilor săi, există o configurație fezabilă la care putem ajunge după un număr finit de pași.",
    "options": [],
    "answer": "Exactly those n with τ(n) = 4, i.e., n = p^3 or n = pq with distinct primes p, q.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27097,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSpunem că o mulțime $A \\subset \\mathbb{R}$ cu cel puțin trei elemente este liberă de progresii aritmetice dacă pentru orice $a, b, c \\in A$ distincte, avem $a+b \\neq 2c$.\nArătați că mulțimea $\\{0,1,2, \\ldots, 3^{8}-1\\}$ conține o submulțime $A$ cu cel puțin 256 elemente, liberă de progresii aritmetice.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nAvem $6560 = 3^{8} - 1$.\nOrice număr din mulțimea $\\{0,1,2, \\ldots, 6560\\}$ se scrie în baza 3 cu 8 cifre (0, 1 sau 2). Definim mulțimea $A$ ca mulțimea numerelor care au doar cifrele 0 și 1, numărul lor fiind $2^{8} = 256$ și arătăm că ea este liberă de progresii aritmetice.\n\nDacă prin absurd ar exista $a, b, c \\in A$ astfel ca $a + b = 2c$, fie\n$$\n\\begin{aligned}\n& a = \\overline{a_{7} a_{6} a_{5} \\ldots a_{1} a_{0}} \\\\\n& b = \\overline{b_{7} b_{6} b_{5} \\ldots b_{1} b_{0}} \\\\\n& c = \\overline{c_{7} c_{6} c_{5} \\ldots c_{1} c_{0}}\n\\end{aligned}\n$$\ncu $a_{i}, b_{i}, c_{i} \\in \\{0,1\\},\\ i = \\overline{0,7}$. Cifrele lui $a + b$ sunt, în ordine, $a_{7} + b_{7}, a_{6} + b_{6}, \\ldots, a_{1} + b_{1}, a_{0} + b_{0}$, iar cele ale lui $2c$ sunt $2c_{7}, 2c_{6}, \\ldots, 2c_{1}, 2c_{0}$. Egalitatea are loc dacă și numai dacă:\n$$\na_{i} + b_{i} = 2c_{i} \\text{ (par), } i = \\overline{0,7}\n$$\nși atunci $a_{i} = b_{i} = 1$, sau $a_{i} = b_{i} = 0$, adică $a = b$, contradicție.\n\n\nSoluție alternativă.\n\nDemonstrăm prin inducție matematică după $n \\geq 2$ că: mulțimea $\\{0,1,2, \\ldots, 3^{n}-1\\}$ conține o submulțime cu $2^{n}$ elemente, liberă de progresii aritmetice.\n\nEtapa de verificare. Pentru $n=2$ submulțimea $A=\\{1,2,4,5\\}$ cu 4 elemente a lui $\\{0,1, \\ldots, 8\\}$ este liberă de progresii aritmetice.\n\nEtapa de demonstrație. Presupunem afirmația adevărată pentru $n$ și o demonstrăm pentru $n+1$.\nFie $X = \\{a_{1}, \\ldots, a_{2^{n}}\\} \\subset \\{0,1, \\ldots, 3^{n}-1\\}$ o submulțime liberă de progresii aritmetice, cu $a_{1} < \\cdots < a_{2^{n}}$.\n\nDemonstrăm că mulțimea $A = \\{a_{1}, \\ldots, a_{2^{n}}, a_{1} + 2 \\cdot 3^{n}, \\ldots, a_{2^{n}} + 2 \\cdot 3^{n}\\}$ este o submulțime liberă de progresii aritmetice a mulțimii $\\{0,1, \\ldots, 3^{n+1}-1\\}$, de cardinal $2^{n+1}$.\n\nFaptul că $A$ este submulțime rezultă din $a_{i} + 2 \\cdot 3^{n} < 3^{n} - 1 + 2 \\cdot 3^{n} = 3^{n+1} - 1$ și are $2^{n+1}$ elemente deoarece pentru orice $i < j$ avem $a_{i} < a_{j} < a_{i} + 2 \\cdot 3^{n} < a_{j} + 2 \\cdot 3^{n}$. În plus, se observă că $A \\cap \\{3^{n}, \\ldots, 2 \\cdot 3^{n} - 1\\} = \\varnothing\\ (^{*})$ și $A \\setminus X$ este liberă de progresii aritmetice.\n\nFaptul că $A$ este liberă de progresii aritmetice rezultă din:\n\n1) media aritmetică a lui $a_{i}$ și $a_{j}$ cu $i \\neq j$ nu este în $A$, deoarece $X$ este liberă de progresii aritmetice;\n\n2) pentru orice $i, j$ (pot fi și egale) avem\n$$\n\\frac{0 + 2 \\cdot 3^{n}}{2} = 3^{n} \\leq \\frac{a_{i} + (a_{j} + 2 \\cdot 3^{n})}{2} \\leq \\frac{3^{n} - 1 + 3^{n} - 1 + 2 \\cdot 3^{n}}{2} = 2 \\cdot 3^{n} - 1\n$$\ndeci, via $(^{*})$, media aritmetică a numerelor $a_{i}$ și $a_{j} + 2 \\cdot 3^{n}$ nu aparține lui $A$;\n\n3) pentru orice $i \\neq j$ media aritmetică a numerelor $a_{i} + 2 \\cdot 3^{n}$ și $a_{j} + 2 \\cdot 3^{n}$ nu este în $A$ observând că\n$$\n\\frac{2 \\cdot 3^{n} + 2 \\cdot 3^{n}}{2} = 2 \\cdot 3^{n} \\leq \\frac{(a_{i} + 2 \\cdot 3^{n}) + (a_{j} + 2 \\cdot 3^{n})}{2}\n$$\nadică media aritmetică nu se află în $X$, iar cum $A \\setminus X$ este liberă de progresii aritmetice obținem concluzia dorită. Inducția este completă și în particular, pentru $n=8$ se obține soluția problemei.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27098,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA right-angled trapezoid with area $10$ and altitude $4$ is divided into two circumscribed trapezoids by a line parallel to its bases. Find their inradii.",
    "options": [],
    "answer": "4/3, 2/3",
    "solution": "Solution:\nLet $ABCD$ be a right-angled trapezoid with area $10$ and altitude $AD = 4$. Let the line $MN \\parallel AB$, $M \\in AD$, $N \\in BC$, divide it into two circumscribed trapezoids. Set $AB = a$, $CD = b$ ($a > b$), $MN = c$, $AM = h_1$ and $DM = h_2$. Then $h_1^2 + (a - c)^2 = (a + c - h_1)^2$ and we get that $h_1 = \\frac{2ac}{a + c}$.\n\n![](attached_image_1.png)\n\nAnalogously $h_2 = \\frac{2bc}{b + c}$ which implies that $\\frac{h_1}{h_2} = \\frac{a(b + c)}{b(a + c)}$. On the other hand, $\\frac{h_1}{h_2} = \\frac{a - c}{c - b}$ and therefore $(a + b)(ab - c^2) = 0$. Hence $c^2 = ab$ and we get that\n$$\nh_1 = \\frac{2a\\sqrt{b}}{\\sqrt{a} + \\sqrt{b}}, \\quad h_2 = \\frac{2b\\sqrt{a}}{\\sqrt{a} + \\sqrt{b}}\n$$\nThen $\\sqrt{ab} = 2$ and $a + b = 5$, i.e., $a$ and $b$ are the roots of the equation $x^2 - 5x + 4 = 0$. Consequently $a = 4$, $b = 1$ and the radii equal $\\frac{h_1}{2} = \\frac{4}{3}$ and $\\frac{h_2}{2} = \\frac{2}{3}$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27099,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nCut $2003$ rectangles from an acute $\\triangle ABC$ such that any of them has a side parallel to $AB$ and the sum of their areas is maximal.",
    "options": [],
    "answer": "The maximum total area is 2003/2004 of the area of triangle ABC, achieved by stacking the rectangles so that the contact points on one side divide it into 2004 equal parts.",
    "solution": "Solution:\n1. It is not difficult to see that the rectangles must be placed one over another so that any of them has two vertices on the sides $AC$ and $BC$, and the first one has a base on the side $AB$.\nNow we shall prove by induction that the sum of the areas of $n$ such rectangles is maximal when the side $AC$ is divided into $n+1$ equal parts by the vertices of the rectangles lying on it. Then the sum equals $\\frac{n}{n+1} S$, where $S = S_{ABC}$.\nLet $MNPQ$ be a rectangle with $M, N \\in AB$, $P \\in BC$, $Q \\in AC$. Setting $\\frac{CQ}{AC} = x$, it easily follows that\n$$\nS_{MNPQ} = 2x(1-x) S\n$$\nHence $S_{MNPQ}$ is maximal if $x = \\frac{1}{2}$, i.e., when $Q$ is the midpoint of the side $AC$. This proves our statement for $n=1$.\nAssume that the statement holds true for some $k$ and consider $k+1$ rectangles $M_i N_i P_i Q_i$ with $M_i, N_i \\in P_{i-1} Q_{i-1}$ ($P_0 \\equiv A, Q_0 \\equiv B$) and $P_i \\in BC$ and $Q_i \\in AC$, $i=1,2, \\ldots, k+1$. Setting $\\frac{CQ_1}{AC} = x$, we get $S_{M_1 N_1 P_1 Q_1} = 2x(1-x) S$. The induction assumption implies that $\\sum_{i=2}^{k+1} S_{M_i N_i P_i Q_i}$ is maximal if $Q_1 Q_2 = Q_2 Q_3 = \\cdots = Q_k Q_{k+1} = Q_{k+1} C$. Therefore\n$$\n\\sum_{i=2}^{k+1} S_{M_i N_i P_i Q_i} \\leq \\frac{k S_{Q_1 P_1 C}}{k+1} = \\frac{k x^2 S}{k+1}\n$$\nThen\n$$\n\\begin{aligned}\n\\sum_{i=1}^{k+1} S_{M_i N_i P_i Q_i} & \\leq \\left(2x(1-x) + \\frac{k x^2}{k+1}\\right) S \\\\\n& = \\left[\\frac{k+1}{k+2} - \\frac{k+2}{k+1}\\left(x - \\frac{k+1}{k+2}\\right)^2\\right] S \\\\\n& \\leq \\frac{k+1}{k+2} S\n\\end{aligned}\n$$\nThe equality is attained if $x = \\frac{k+1}{k+2}$, i.e., when the points $Q_1, Q_2, \\ldots, Q_{k+1}$ divide the side $AC$ into equal parts.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27100,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $D$ be a point on the side $AC$ of $\\triangle ABC$ with $AC = BC$, and $E$ be a point on the segment $BD$. Prove that $\\Varangle EDC = 2 \\Varangle CED$ if $BD = 2AD = 4BE$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n1. Note that\n$$\n\\Varangle EDC = 2 \\Varangle CED \\Longleftrightarrow DI = EI,\n$$\nwhere $DI$ ($I \\in CE$) is the bisector of $\\Varangle CDE$. Setting $BD = 2AD = 4BE = 4x$ and $AC = BC = y$ we get\n$$\n\\begin{aligned}\nDI & = \\frac{\\sqrt{ED \\cdot CD \\left((ED + CD)^2 - CE^2\\right)}}{ED + CD} \\\\\n& = \\frac{\\sqrt{3x(y - 2x)\\left((x + y)^2 - CE^2\\right)}}{x + y} \\\\\n& EI = CE \\frac{ED}{ED + CD} = CE \\frac{3x}{x + y}\n\\end{aligned}\n$$\n\n![](attached_image_1.png)\n\nIt is easy to deduce from here that\n$$\nDI = EI \\Longleftrightarrow CE^2 = (x + y)(y - 2x)\n$$\nThe last identity follows by Stewart's theorem for $\\triangle BCD$:\n$$\n\\begin{aligned}\nCE^2 & = \\frac{BC^2 \\cdot DE + CD^2 \\cdot BE}{BD} - BE \\cdot DE \\\\\n& = \\frac{3y^2 + (y - 2x)^2}{4} - 3x^2 = (x + y)(y - 2x)\n\\end{aligned}\n$$\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27101,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all values of $a$, for which the equation\n$$\n\\sqrt{a x^{2}+a x+2}=a x+2\n$$\nhas a unique root.",
    "options": [],
    "answer": "a = -8 or a ≥ 1",
    "solution": "Solution:\nIf $a x+2<0$, then the equation has no real roots. If $a x+2 \\geq 0$, it is equivalent to $a x^{2}+a x+2=(a x+2)^{2}$, i.e., $(a^{2}-a) x^{2}+3 a x+2=0$. The last equation has a unique real root in the following three cases.\n\nCase 1. The coefficient of $x^{2}$ vanishes and the respective linear equation has a root $x$ such that $a x+2 \\geq 0$.\nIf $a=0$, then $2=0$ which is impossible. If $a=1$, then $x=-\\frac{2}{3}$ and $a x+2=-\\frac{2}{3}+2=\\frac{4}{3}>0$. Hence $a=1$ is a solution of the problem.\n\nCase 2. The coefficient of $x^{2}$ is non-zero, i.e., $a \\neq 0,1$, and the respective quadratic equation has a unique real root $x$ with $a x+2 \\geq 0$. Then $D=9 a^{2}-8(a^{2}-a)=a^{2}+8 a=0$ and hence $a=-8$. Then $x=\\frac{1}{6}$ and $a x+2= -8 \\cdot \\frac{1}{6}+2=\\frac{2}{3}>0$, i.e., $a=-8$ is a solution of the problem.\n\nCase 3. The coefficient of $x^{2}$ is non-zero, i.e., $a \\neq 0,1$, and the respective quadratic equation has two real roots $x_{1}<x_{2}$ such that $a x_{1}+2<0 \\leq a x_{2}+2$, i.e., $-\\frac{2}{a} \\in\\left(x_{1}, x_{2}\\right]$.\nIf $-\\frac{2}{a}=x_{2}$, then $(a^{2}-a)\\left(-\\frac{2}{a}\\right)^{2}+3 a\\left(-\\frac{2}{a}\\right)+2=0$. Hence $-\\frac{1}{a}=0$, contradiction. Therefore $-\\frac{2}{a} \\in\\left(x_{1}, x_{2}\\right)$ which is equivalent to\n$$\n(a^{2}-a)\\left((a^{2}-a)\\left(-\\frac{2}{a}\\right)^{2}+3 a\\left(-\\frac{2}{a}\\right)+2\\right)<0\n$$\nIt is easy to see that the solutions of the above inequality are $a>1$. So, the given equation has a unique real root for $a=-8$ and $a \\geq 1$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27102,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $k_{1}$ and $k_{2}$ be circles with centers $O_{1}$ and $O_{2}$, $O_{1}O_{2}=25$, and radii $R_{1}=4$ and $R_{2}=16$, respectively. Consider a circle $k$ such that $k_{1}$ is internally tangent to $k$ at a point $A$, and $k_{2}$ is externally tangent to $k$ at a point $B$.\n\na) Prove that the segment $AB$ passes through a constant point (i.e., independent on $k$).\n\nb) The line $O_{1}O_{2}$ intersects $k_{1}$ and $k_{2}$ at points $P$ and $Q$, respectively, such that $O_{1}$ lies on the segment $PQ$ and $O_{2}$ does not. Prove that the points $P, A, Q$ and $B$ are concyclic.\n\nc) Find the minimum possible length of the segment $AB$ (when $k$ varies).",
    "options": [],
    "answer": "12",
    "solution": "Solution:\n\na) We shall prove that the position of the point $S = O_{1}O_{2} \\cap AB$ does not depend on $k$. Let $O_{3}$ be the center of $k$. It follows by the Menelaus theorem for $\\triangle O_{1}O_{2}O_{3}$ and the line $AB$ that\n$$\n\\frac{O_{3}B}{BO_{2}} \\cdot \\frac{O_{2}S}{SO_{1}} \\cdot \\frac{O_{1}A}{AO_{3}} = 1\n$$\nSince $O_{3}B = AO_{3}$, we get $\\frac{O_{2}S}{SO_{1}} = \\frac{BO_{2}}{O_{1}A} = \\frac{R_{2}}{R_{1}} = \\frac{16}{4} = 4$.\nHence $S$ is a fixed point and the equalities $O_{1}O_{2} = 25 = O_{2}S + O_{1}S$ imply that $O_{2}S = 20$ and $O_{1}S = 5$.\n\nb) Setting $\\angle O_{1}O_{3}O_{2} = x$ and $\\angle O_{1}O_{2}O_{3} = y$, then $\\angle AO_{1}S = x + y$. Since $\\triangle O_{1}AP$ is isosceles, we have $\\angle APS = \\frac{x + y}{2}$. On the other hand, the triangles $AO_{3}B$ and $BO_{2}Q$ are also isosceles; hence $\\angle SBO_{3} = 90 - \\frac{x}{2}$ and $\\angle QBO_{2} = 90 - \\frac{y}{2}$, which implies that $\\angle SBQ = \\frac{x + y}{2}$. Therefore $\\angle APS = \\angle SBQ$, i.e. $PBQA$ is a cyclic quadrilateral.\n\nc) Note that $SP \\cdot SQ = SA \\cdot SB$ and $SP \\cdot SQ = (SO_{1} + R_{1})(SO_{2} - R_{2}) = 9 \\cdot 4 = 36$. The inequality\n$$\nAB = SA + SB \\geq 2 \\sqrt{SA \\cdot SB} = 2 \\sqrt{SP \\cdot SQ} = 12\n$$\nimplies that the minimum of $AB$ equals $12$ and it is attained if $SA = SB$.\nIt remains to show that there is a circle $k$ with $SA = SB$. Take a point $A \\in k_{1}$ such that $SA = 6$. Since the power of $S$ with respect to $k_{1}$ equals $SO_{1}^{2} - R_{1}^{2} = 5^{2} - 4^{2} = 9$ and $SA^{2} = 36 > 9$, it is easy to see that there is a circle $k$ passing through $A$ and satisfying the conditions of the problem. Then $SB = \\sqrt{SP \\cdot SQ} = 6$, i.e., $SA = SB$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27103,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThree nonintersecting circles $k_{i}(O_{i}, r_{i})$, $i=1,2,3$, where $r_{1}<r_{2}<r_{3}$, are tangent to the arms of an angle. One of the arms is tangent to $k_{1}$ and $k_{3}$ at points $A$ and $B$ and the other one is tangent to $k_{2}$ at point $C$. Let $K=AC \\cap k_{1}$, $L=AC \\cap k_{2}$, $M=BC \\cap k_{2}$ and $N=BC \\cap k_{3}$. The four lines through $C$ and $P=AM \\cap BK$, $Q=AM \\cap BL$, $R=AN \\cap BK$ and $S=AN \\cap BL$, meet $AB$ at the points $X, Y, Z$ and $T$, respectively. Prove that $XZ=YT$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nIf $E$ and $F$ are the second tangent points of $k_{1}$ and $k_{2}$ with the arms of the angle, then the equalities $AF^{2} = AL \\cdot AC$, $CE^{2} = CK \\cdot CA$ and $AF = CE$ imply that $AL = CK$. Hence $AK = CL$ and analogously $CM = BN$.\n\n![](attached_image_1.png)\n\nOn the other hand, Ceva's theorem gives\n$$\n\\frac{AX}{XB} = \\frac{AK \\cdot CM}{KC \\cdot MB} \\text{ and } \\frac{AT}{TB} = \\frac{AL \\cdot CN}{LC \\cdot NB}\n$$\nMultiplying these equalities gives $\\frac{AX}{XB} = \\frac{TB}{AT}$ and therefore\n$$\n\\frac{AX + XB}{XB} = \\frac{TB + AT}{AT} \\Longleftrightarrow AT = BX \\Longleftrightarrow AX = BT\n$$\nAnalogously $AZ = YB$ which implies that $XZ = YT$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27104,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $A$ be the set of all 4-tuples of $0$ and $1$. Two such 4-tuples are called neighbors if they coincide exactly at three positions. Let $M$ be a subset of $A$ with the following property: any two elements of $M$ are not neighbors and there exists an element of $M$ which is neighbor of exactly one of them. Find the minimum possible cardinality of $M$.",
    "options": [],
    "answer": "6",
    "solution": "Solution:\nConsider a table with rows corresponding to the elements of $A$ and columns corresponding to the elements of $M$. We write $\\times$ in a cell if the element of $A$ in the respective row is adjacent to the element of $M$ in the respective column. Let $|M| = k$. It follows from the given condition that there are no two equal rows, which means that $M$ has at least $16$ different subsets. Since a set with $n$ elements has $2^{n}$ subsets, we get $k \\geq 4$.\n\nAny row has exactly five adjacent and thus any column contains exactly five $\\times$, i.e., the total number of $\\times$ is $5k$. A minimal number of $\\times$ by rows is attained if one row contains no $\\times$, $k$ rows contain one $\\times$, $\\binom{k}{2}$ rows contain two $\\times$, etc.\n\nIf $k = 4$, then all the subsets of $M$ are $16$ and hence any subset of $M$ appears exactly once as an adjacent set to some element of $A$. Then we have one row with no $\\times$, $4$ rows with one $\\times$, $6$ rows with two $\\times$, $4$ rows with three $\\times$ and one row with four $\\times$. The total number of $\\times$ becomes $32 > 20$, a contradiction.\n\nFor $k = 5$ one has $25$ $\\times$. Their minimal number by rows is attained when one row contains no $\\times$, $5$ rows contain one $\\times$ and $10$ rows contain two $\\times$. Since $5 \\cdot 1 + 10 \\cdot 2 = 25$ the distribution of $\\times$ must be exactly the one described above. This means that any two elements of $M$ are simultaneously adjacent to some element of $A$.\n\nIt is easy to see that if two elements of $M$ do not coincide at at most two positions, then there is no element of $A$ that is adjacent to them. Hence any two elements of $M$ do not coincide at one or two positions. If there are two elements which do not coincide at one position, we may assume that they are $a = 0000$ and $b = 1000$. The adjacent to $a$ and $b$ in $M$ are $0000$ and $1000$. So the rows of $a$ and $b$ coincide, a contradiction. This shows that any two elements of $M$ are different at exactly two positions. We may assume that $0000 \\in M$. Then the remaining $4$ elements are among $0011$, $1100$, $0101$, $1010$, $1001$ and $0110$. But among any of the pairs $(0011, 1100)$, $(0101, 1010)$ and $(1001, 0110)$ at most one element can be chosen, a contradiction.\n\nFor $k = 6$ the set $M = \\{0000, 1111, 0111, 0100, 1001, 0101\\}$ separates any two elements of $A$.\n\nHence the desired minimal number is $6$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27105,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThree of $n$ equal balls are radioactive. A detector measures radioactivity. Any measurement of a set of balls gives as a result whether 0, 1 or more than 1 balls are radioactive. Denote by $L(n)$ the least number of measurements that one needs to find the three radioactive balls.\n\na) Find $L(6)$.\n\nb) Prove that $L(n) \\leq \\frac{n+5}{2}$.",
    "options": [],
    "answer": "L(6) = 4 and for all n, L(n) ≤ (n+5)/2",
    "solution": "Solution:\n\na. We shall prove that 4 measurements are enough. Denote the balls by $1,2,3,4,5,6$ and measure consecutively $\\{1,2\\}$, $\\{1,3\\}$, $\\{1,4\\}$ and $\\{1,5\\}$.\n\nCase 1. If all the measurements show radioactivity, then $1$ is a radioactive ball. If $\\{1, a\\}$, $a=2,3,4,5$, contains two radioactive balls, then $a$ is radioactive; otherwise, it is not. Hence we know which of the balls $1,2,3,4,5$ are radioactive and hence we also know whether $6$ is radioactive or not.\n\nCase 2. If some of the measurements shows no radioactivity, then $1$ is not a radioactive ball. Hence we again know which of the balls $1,2,3,4,6$ are radioactive and then whether $6$ is radioactive or not.\n\nAssume that $L(6) \\leq 3$, i.e., three measurements are enough. After two measurements we have $3^{2}=9$ possibilities. After the first measurement $1,2,3,4,5$ or $6$ balls can be chosen. It is clear that among any $5$ or $6$ balls there are at least $2$ radioactive; so a measurement with $5$ or $6$ gives no information.\n\nLet $x$, $x \\leq 4$, balls are chosen in the first measurement. If $x=1$ and the answer is \"one radioactive ball\", then the possibilities for the radioactive balls are $\\binom{5}{2}=10$. On the other hand, the number of the possibilities for the other two measurements are $3^{2}=9$. Analogously, for a first measurement of:\n\n- two balls and answer \"one radioactive ball\", the number of the possibilities is $\\binom{2}{1}\\binom{4}{2}=12>9$;\n- three balls and answer \"more than one radioactive ball\", the number of the possibilities is $\\binom{3}{2}\\binom{3}{1}+\\binom{3}{3}=10>9$;\n- four balls and answer \"more than one radioactive ball\", the number of the possibilities is $\\binom{4}{2}\\binom{2}{1}+\\binom{4}{3}=13>9$.\n\nThus three measurements are not enough and so $L(6)=4$.\n\nb. We shall show that $\\left[\\frac{n+5}{2}\\right]$ measurements are enough for finding the three radioactive balls which will imply the desired inequality. Let $n=2 t-\\varepsilon$, where $\\varepsilon \\in\\{0,1\\}$. Let us either pair the balls (if $\\varepsilon=0$) or take a ball and pair the remaining (if $\\varepsilon=1$). In both cases we check any pair and the taken ball (if $\\varepsilon=1$) for radioactivity. Two cases are possible.\n\n1. There is one set with two radioactive balls and one set with one radioactive ball.\n2. There are three sets with one radioactive ball.\n\nIn both cases three measurements are enough to find the radioactive balls. The total number of the measurements is $t-1+3=t+2$. Since\n$$\n\\left[\\frac{n+5}{2}\\right]=\\left[\\frac{2 t-\\varepsilon+5}{2}\\right]=t+2+\\left[\\frac{1-\\varepsilon}{2}\\right]=t+2\n$$\nwe get that $L(n) \\leq\\left[\\frac{n+5}{2}\\right]$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27106,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $a \\geq 2$ be a real number. Denote by $x_{1}$ and $x_{2}$ the roots of the equation $x^{2}-a x+1=0$ and set $S_{n}=x_{1}^{n}+x_{2}^{n}$, $n=1,2, \\ldots$\n\na) Prove that the sequence $\\left\\{\\frac{S_{n}}{S_{n+1}}\\right\\}_{n=1}^{\\infty}$ is decreasing.\n\nb) Find all $a$ such that\n$$\n\\frac{S_{1}}{S_{2}}+\\frac{S_{2}}{S_{3}}+\\cdots+\\frac{S_{n}}{S_{n+1}}>n-1\n$$\nfor any $n=1,2, \\ldots$.",
    "options": [],
    "answer": "a = 2",
    "solution": "Solution:\nIf $a \\geq 2$, then the roots $x_{1}$ and $x_{2}$ of the equation $x^{2}-a x+1=0$ are positive and $x_{1} x_{2}=1$. In particular, $S_{n}>0$ for $n=1,2, \\ldots$\n\na) We have\n$$\n\\begin{aligned}\n\\frac{S_{n-1}}{S_{n}} \\geq \\frac{S_{n}}{S_{n+1}} &\\Longleftrightarrow \\left(x_{1}^{n-1}+x_{2}^{n-1}\\right)\\left(x_{1}^{n+1}+x_{2}^{n+1}\\right) \\geq \\left(x_{1}^{n}+x_{2}^{n}\\right)^{2} \\\\\n&\\Longleftrightarrow x_{1}^{n-1} x_{2}^{n+1}+x_{2}^{n-1} x_{1}^{n+1} \\geq 2 x_{1}^{n} x_{2}^{n} \\\\\n&\\Longleftrightarrow \\left(x_{1} x_{2}\\right)^{n-1}\\left(x_{1}-x_{2}\\right)^{2} \\geq 0\n\\end{aligned}\n$$\nwhich obviously holds.\n\nb) Let $a \\geq 2$ have the desired property. Then a) implies that\n$$\nn \\frac{S_{1}}{S_{2}} \\geq \\frac{S_{1}}{S_{2}}+\\cdots+\\frac{S_{n}}{S_{n+1}}>n-1\n$$\ni.e., $\\frac{S_{1}}{S_{2}}>1-\\frac{1}{n}$. Since $\\lim _{n \\rightarrow \\infty} \\frac{1}{n}=0$, the last inequality gives $\\frac{S_{1}}{S_{2}} \\geq 1$. Using Vieta's formulas we get $S_{1}=a$, $S_{2}=a^{2}-2$ and therefore $\\frac{a}{a^{2}-2} \\geq 1 \\Longleftrightarrow \\frac{(a+1)(a-2)}{a^{2}-2} \\leq 0$. Since $a \\geq 2$ we get $a=2$.\n\nConversely, if $a=2$, then $x_{1}=x_{2}=1$ and $S_{n}=2$ for any $n=1,2, \\ldots$ Hence\n$$\n\\frac{S_{1}}{S_{2}}+\\frac{S_{2}}{S_{3}}+\\cdots+\\frac{S_{n}}{S_{n+1}}=n>n-1\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27107,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $M$ be an interior point of $\\triangle ABC$. The lines $AM$, $BM$ and $CM$ meet the lines $BC$, $CA$ and $AB$ at points $A_1$, $B_1$ and $C_1$, respectively, such that $S_{CB_1M} = 2 S_{AC_1M}$. Prove that $A_1$ is the midpoint of the segment $BC$ if and only if $S_{BA_1M} = 3 S_{AC_1M}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $A_1$ be the midpoint of the segment $BC$. Then Ceva's theorem implies that\n$$\n\\frac{AC_1}{C_1B} \\cdot \\frac{BA_1}{A_1C} \\cdot \\frac{CB_1}{B_1A} = 1\n$$\ni.e., $\\frac{AC_1}{C_1B} = \\frac{B_1A}{B_1C}$. Hence we have $B_1C_1 \\parallel BC$, i.e. $S_{BC_1M} = S_{CB_1M} = 2 S_{AC_1M}$ and we get that $S_{AB_1M} = S_{AC_1M}$. Then\n\n![](attached_image_1.png)\n\n$$\n\\frac{1}{3} = \\frac{S_{AC_1M}}{S_{AMC}} = \\frac{C_1M}{MC} = \\frac{S_{BC_1M}}{S_{BMC}} = \\frac{2 S_{AC_1M}}{2 S_{BA_1M}}\n$$\nand therefore $S_{BA_1M} = 3 S_{AC_1M}$.\n\nConversely, let $S_{AC_1M} = 1$, $S_{CB_1M} = 2$, $S_{BA_1M} = 3$, $S_{BC_1M} = x$, $S_{CA_1M} = 3y$ and $S_{AB_1M} = 2z$. We have to show that $y = 1$. Note that\n$$\n\\frac{1}{2(z+1)} = \\frac{S_{AC_1M}}{S_{AMC}} = \\frac{C_1M}{CM} = \\frac{S_{C_1MB}}{SCMB} = \\frac{x}{3(y+1)}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27108,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe incircle of $\\triangle ABC$ has radius $r$ and is tangent to the sides $AB$, $BC$ and $CA$ at points $C_1$, $A_1$ and $B_1$, respectively. If $N = BC \\cap B_1C_1$ and $AA_1 = 2A_1N = 2r\\sqrt{3}$, find $\\Varangle ANC$.",
    "options": [],
    "answer": "90°",
    "solution": "Solution:\n\nWe shall use the standard notation for the elements of $\\triangle ABC$. We may assume that $b > c$. Denote by $I$ the incenter of $\\triangle ABC$. Then the condition $A_1N = r\\sqrt{3}$ implies that $INA_1$ is a right-angled triangle with $\\Varangle NIA_1 = 60^\\circ$. We shall prove that $AA_1 \\perp IN$. If so, then $\\Varangle AA_1N = 60^\\circ$. Now, if $M$ is the midpoint of the segment $AA_1$, then $\\triangle MNA_1$ is equilateral and therefore $\\triangle ANM$ is isosceles with $\\Varangle ANM = \\Varangle MAN = 30^\\circ$. Thus $\\Varangle ANC = 90^\\circ$.\n\n![](attached_image_1.png)\n\nTo prove that $AA_1 \\perp IN$, note that this is equivalent to the equality $AI^2 - IA_1^2 = AN^2 - A_1N^2$. The Cosine theorem for $\\triangle ANB$ gives\n$$\nAN^2 = c^2 + BN^2 + 2c \\cdot BN \\cos \\beta = c^2 + BN^2 + 2c \\cdot BN \\frac{a^2 + c^2 - b^2}{2ac}\n$$\nOn the other hand, the Menelaus theorem for $\\triangle ABC$ and the line $B_1C_1$ implies that\n$$\n\\frac{CB_1}{B_1A} \\cdot \\frac{AC_1}{C_1B} \\cdot \\frac{BN}{NC} = 1\n$$\nHence $\\frac{p-c}{p-b} \\cdot \\frac{BN}{BN + a} = 1$, i.e., $BN = \\frac{a(p-b)}{b-c}$, where $p$ is the semiperimeter of $\\triangle ABC$. Then $A_1N = BN + p - b$ and therefore\n$$\n\\begin{aligned}\nAN^2 - A_1N^2 & = c^2 - (p-b)^2 + BN \\frac{a^2 + c^2 - b^2 - a^2 - ac + ab}{a} \\\\\n& = c^2 - (p-b)^2 - \\frac{a(p-b)}{b-c} \\cdot \\frac{2(b-c)(p-a)}{a} \\\\\n& = c^2 - (p-b)^2 - 2(p-a)(p-b) = (p-a)^2 \\\\\n& = AI^2 - IC_1^2 = AI^2 - IA_1^2\n\\end{aligned}\n$$\nwhich completes the proof.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27109,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAleksander writes a positive integer as a coefficient of a polynomial of degree four, then Elitza writes a positive integer as another coefficient of the same polynomial and so on till all the five coefficients of the polynomial are filled in. Aleksander wins if the polynomial obtained has an integer root; otherwise, Elitza wins. Who of them has a winning strategy?",
    "options": [],
    "answer": "Elitza",
    "solution": "Solution:\n\nWe shall prove that Elitza has a winning strategy. If the polynomial is $a_{0} x^{4} + a_{1} x^{3} + a_{2} x^{2} + a_{3} x + a_{4}$ and Aleksander writes $a_{0}$, $a_{1}$, $a_{2}$ or $a_{3}$, then Elitza writes respectively $a_{1} = a_{0}$, $a_{0} = a_{1}$, $a_{3} = a_{2}$ or $a_{2} = a_{3}$; if he writes $a_{4}$, she writes $a_{1} = 1$.\n\nIn a similar way Elitza is able to get $a_{1} \\leq a_{0}$ and $a_{3} \\leq a_{2}$ after her second move. Suppose that the polynomial obtained has an integer root $-y$. Then $y \\geq 1$ and hence $a_{4} = y^{3}(a_{1} - a_{0} y) + a_{3} - a_{2} y \\leq 0$, which is a contradiction.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 27110,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all positive integers $n$ for which there exists $n$ points in the plane such that any of them lies on exactly $\\frac{1}{3}$ of the lines determined by these $n$ points.",
    "options": [],
    "answer": "n = 6",
    "solution": "Solution:\nWe shall prove that $n=6$. If we take 6 points in general position (no three are collinear), then the lines are 15 and any point lies on 5 lines, i.e. $n=6$ is a solution of the problem.\n\nDenote by $k$ the number of the lines defined by the given $n$ points. Assume that there is a line $l$ containing 4 of the given points. Any of the points belongs to $\\frac{k}{3}-1$ lines different from $l$ which means that there are at least $4\\left(\\frac{k}{3}-1\\right)+1$ lines. Then\n$$\n4\\left(\\frac{k}{3}-1\\right)+1 \\leq k\n$$\ni.e., $k \\leq 9$. On the other hand, any point lying not on $l$ belongs to at least four lines (the lines through the point and the four points on $l$) and hence $k \\geq 12$, a contradiction. So any line contains at most 3 points. Let $a$ of the lines contain 2 points. Then each of the other $k-a$ lines contains 3 points.\n\nThe number of the points (any of them counted $\\frac{k}{3}$ times) is equal to $2a + 3(k-a)$ and then $2a + 3(k-a) = \\frac{nk}{3}$. On the other hand, since $n$ points define $\\frac{n(n-1)}{2}$ lines (some of them may coincide) and any line containing three points is counted three times, then $a + 3(k-a) = \\frac{n(n-1)}{2}$. Thus\n$$\na = \\frac{n(n-1)(9-n)}{2(2n-9)}, \\quad k = \\frac{3n(n-1)}{2(2n-9)}\n$$\nSince $k \\leq \\frac{n(n-1)}{2}$, then $2n-9 \\geq 3$, i.e., $n \\geq 6$. Now $a \\geq 0$ implies that $n \\leq 9$. For $n=7$ the values of $a$ and $k$ are not integers and hence $n=8$ or $n=9$. For $n=8$ one has that $k=12, a=4$ and for $n=9$ we get that $k=12, a=0$.\n\nDenote by $l$ the maximal number of points in general position among the given $n$ points. Then the remaining points belong to lines defined by these $l$ points.\n\nCase 1. Let $l=3$ and let the respective points be $A_{1}, A_{2}, A_{3}$. Any of the other points lies on one of the lines $A_{1}A_{2}, A_{1}A_{3}$ and $A_{2}A_{3}$. Since any line contains at most 3 points, then we have at most 6 points, a contradiction.\n\nCase 2. Let $l=4$ and let the respective points be $A_{1}, A_{2}, A_{3}, A_{4}$. Since the total number of the points is at least 8, we may find a point belonging to exactly one of the lines defined by $A_{1}, A_{2}, A_{3}, A_{4}$. We may assume that the point is $A_{5}$ and $A_{5} \\in A_{1}A_{2}$. Then the points $A_{3}, A_{4}, A_{1}, A_{5}$ as well as $A_{3}, A_{4}, A_{2}, A_{5}$ are in general position. Hence all the points must belong to the lines defined by $A_{1}, A_{2}, A_{3}, A_{4}; A_{3}, A_{4}, A_{1}, A_{5}$ and $A_{3}, A_{4}, A_{2}, A_{5}$. The only common lines are $A_{3}A_{4}$ and $A_{1}A_{2}A_{5}$, i.e., all the points lie on two lines. This is a contradiction to the fact that any line contains at most 3 points.\n\nCase 3. Let $l=5$ and let the respective points be $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}$. Any of the points $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}$ belongs to exactly 4 lines. This means that $A_{6}A_{i}, i=1,2, \\ldots, 5$, is one of these lines. We may assume that $A_{6} \\in A_{1}A_{2}$. Then $A_{6}A_{3}$ is one of the lines $A_{3}A_{4}, A_{3}A_{5}$ or $A_{4}A_{5}$. Let us have, for example, $A_{6} \\in A_{3}A_{4}$. Then $A_{6}A_{5}$ is a new line, a contradiction.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27111,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nConsider the polynomial $f(x) = 4x^{4} + 6x^{3} + 2x^{2} + 2003x - 2003^{2}$. Prove that:\n\na) the local extrema of $f'(x)$ are positive;\n\nb) the equation $f(x) = 0$ has exactly two real roots and find them.",
    "options": [],
    "answer": "(-1 + sqrt(4007))/2 and (-1 - sqrt(4007))/2",
    "solution": "Solution:\n\na) Since $\\lim_{x \\rightarrow +\\infty} f'(x) = +\\infty$ and $\\lim_{x \\rightarrow -\\infty} f'(x) = -\\infty$, it is enough to show that the local minimum $m$ of $f'(x)$ is positive. Since the equation $f''(x) = 0$ has two real roots $x_{1} > x_{2}$, it follows that $m = f'(x_{1}) > 0$. Now it is easy to check that $x_{1} \\in (-1 ; 0)$ and $m > 0$.\n\nb) It follows from a) that the equation $f'(x) = 0$ has a unique real root. Since $\\lim_{x \\rightarrow +\\infty} f(x) = \\lim_{x \\rightarrow -\\infty} f(x) = +\\infty$ and $f(0) < 0$, we conclude that the equation $f(x) = 0$ has exactly two real roots. To find them, set $y = 2003$ and consider $f(x) = 0$ as a quadratic equation with respect to $y$. We have\n$$\ny_{1,2} = \\frac{x \\pm x(4x + 3)}{2}\n$$\nand then either $2y = x - x(4x + 3)$ or $2y = x + x(4x + 3)$. For $y = 2003$ the first equation has no real roots and the second one has two real roots $x_{1,2} = \\frac{-1 \\pm \\sqrt{4007}}{2}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27112,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nConsider the functions\n$$\nf(x) = \\frac{\\cos^2 x}{1 + \\cos x + \\cos^2 x} \\text{ and } g(x) = k \\tan x + (1-k) \\sin x - x\n$$\nwhere $k$ is a real number.\n\na) Prove that $g'(x) = \\frac{(1-\\cos x)(k-f(x))}{f(x)}$.\n\nb) Find the image of $f(x)$ if $x \\in \\left[0 ; \\frac{\\pi}{2}\\right)$.\n\nc) Find all $k$ such that $g(x) \\geq 0$ for any $x \\in \\left[0 ; \\frac{\\pi}{2}\\right)$.",
    "options": [],
    "answer": "Range of the first function on the given interval: (0, 1/3]; Parameter values: k ∈ [1/3, ∞).",
    "solution": "Solution:\n\na) We have\n$$\n\\begin{aligned}\ng'(x) & = \\frac{(1-\\cos x)\\left[k\\left(1+\\cos x+\\cos^2 x\\right)-\\cos^2 x\\right]}{\\cos^2 x} \\\\\n& = \\frac{(1-\\cos x)\\left(1+\\cos x+\\cos^2 x\\right)}{\\cos^2 x}[k-f(x)] \\\\\n& = \\frac{(1-\\cos x)[k-f(x)]}{f(x)}\n\\end{aligned}\n$$\n\nb) Set $\\cos x = t$. Then $t \\in (0 ; 1]$ for $x \\in \\left[0 ; \\frac{\\pi}{2}\\right)$ and $f(x) = h(t) = \\frac{t^2}{1 + t + t^2}$. Since $h'(t) = \\frac{t^2 + 2t}{\\left(1 + t + t^2\\right)^2}$, we see that $h'(t) > 0$ for $t \\in (0 ; 1]$ and therefore $h(t)$ increases in this interval. Having in mind that $h(0) = 0$ and $h(1) = \\frac{1}{3}$, we conclude that the range of $h(t)$ for $t \\in (0 ; 1]$, i.e., the range of $f(x)$ for $x \\in \\left[0 ; \\frac{\\pi}{2}\\right)$, is the interval $\\left(0 ; \\frac{1}{3}\\right]$.\n\nc) If $k < 0$, then $\\lim_{x \\rightarrow \\frac{\\pi}{2},\\ x < \\frac{\\pi}{2}} g(x) = -\\infty$, i.e., any $k < 0$ is not a solution of the problem. It is also easy to see that $k = 0$ is not a solution of the problem.\n\nIt follows by b) that if $k \\geq \\frac{1}{3}$, then $g'(x) > 0$ for $x \\in \\left(0 ; \\frac{\\pi}{2}\\right)$, i.e., the function $g(x)$ increases in this interval. Now $g(0) = 0$ implies that $g(x) \\geq 0$ for $x \\in \\left[0 ; \\frac{\\pi}{2}\\right)$.\n\nIf $k \\in \\left(0 ; \\frac{1}{3}\\right)$, then the equation $g'(x) = 0$ has a unique root $x_0$ in the interval $\\left(0 ; \\frac{\\pi}{2}\\right)$. The considerations above show that $g'(x) < 0$ for $x \\in \\left(0 ; x_0\\right)$, i.e., $g(x)$ decreases in this interval. Then $g(0) = 0$ implies that the given inequality is not true.\n\nThus, the desired values of $k$ are $k \\in \\left[\\frac{1}{3} ; \\infty\\right)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27113,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $M$ be the centroid of $\\triangle ABC$ with $\\Varangle AMB = 2 \\Varangle ACB$. Prove that:\n\na) $AB^{4} = AC^{4} + BC^{4} - AC^{2} \\cdot BC^{2}$;\n\nb) $\\Varangle ACB \\geq 60^{\\circ}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\na) We shall use the standard notation for the elements of $\\triangle ABC$. The Cosine theorem for $\\triangle ABM$ gives\n$$\n\\cot \\Varangle AMB = \\frac{AM^{2} + BM^{2} - AB^{2}}{2 AM \\cdot MB \\sin \\Varangle AMB} = \\frac{a^{2} + b^{2} - 5c^{2}}{12 S_{ABC}}.\n$$\nSince $\\cot 2\\gamma = \\frac{2 \\cos^{2} \\gamma - 1}{2 \\sin \\gamma \\cos \\gamma}$, the condition $\\Varangle AMB = 2\\gamma$ becomes\n$$\n\\begin{aligned}\n& \\frac{a^{2} + b^{2} - 5c^{2}}{3ab} = 2 \\cos \\gamma - \\frac{1}{\\cos \\gamma} \\\\\n\\Longleftrightarrow & \\frac{a^{2} + b^{2} - 5c^{2}}{3ab} - \\frac{a^{2} + b^{2} - c^{2}}{ab} = -\\frac{2ab}{a^{2} + b^{2} - c^{2}} \\\\\n\\Longleftrightarrow & \\frac{a^{2} + b^{2} + c^{2}}{3ab} = \\frac{ab}{a^{2} + b^{2} - c^{2}} \\\\\n\\Longleftrightarrow & (a^{2} + b^{2})^{2} - c^{4} = 3a^{2}b^{2} \\\\\n\\Longleftrightarrow & c^{4} = a^{4} + b^{4} - a^{2}b^{2}\n\\end{aligned}\n$$\n\nb) It is easy to check that $a^{4} + b^{4} - a^{2}b^{2} \\geq (a^{2} + b^{2} - ab)^{2}$ for any $a, b > 0$. Then a) implies that $a^{2} + b^{2} - 2ab \\cos \\gamma = c^{2} \\geq a^{2} + b^{2} - ab$. Hence $\\cos \\gamma \\leq \\frac{1}{2}$, i.e., $\\gamma \\geq 60^{\\circ}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27114,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $M$, $N$ and $P$ be points on the sides $AB$, $BC$ and $CA$ of $\\triangle ABC$, respectively. The lines through $M$, $N$ and $P$, parallel to $BC$, $AC$ and $AB$, respectively, meet at a point $T$. Prove that:\n\na) if $\\frac{AM}{MB} = \\frac{BN}{NC} = \\frac{CP}{PA}$, then $T$ is the centroid of $\\triangle ABC$;\n\nb) $S_{MNP} \\leq \\frac{1}{3} S_{ABC}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nSet $PT \\cap BC = P_1$, $NT \\cap AB = N_1$ and $MT \\cap AC = M_1$. The triangles $N_1MT$, $PTM_1$ and $TP_1N$ are similar to $\\triangle ABC$. Set $k_1 = \\frac{N_1M}{AB}$, $k_2 = \\frac{PT}{AB}$ and $k_3 = \\frac{TP_1}{AB}$. Then\n$$\nk_1 + k_2 + k_3 = 1\n$$\nsince\n$$\nk_1 + k_2 + k_3 = \\frac{N_1M}{AB} + \\frac{PT}{AB} + \\frac{TP_1}{AB} = \\frac{N_1M}{AB} + \\frac{AN_1}{AB} + \\frac{MB}{AB} = 1\n$$\n\na) It is clear that $\\frac{AM}{MB} = \\frac{PT + N_1M}{AB} = \\frac{k_1 + k_2}{k_3}$. Analogously, $\\frac{BN}{NC} = \\frac{k_1 + k_3}{k_2}$, $\\frac{CP}{PA} = \\frac{k_2 + k_3}{k_1}$. It follows by $\\frac{AM}{MB} = \\frac{BN}{NC}$ that $\\frac{k_1 + k_2}{k_3} = \\frac{k_1 + k_3}{k_2}$, i.e., $(k_2 - k_3)(k_1 + k_2 + k_3) = 0$. Hence $k_2 = k_3$. We get in the same way that $k_1 = k_2$ and then $k_1 = k_2 = k_3$. Hence $PT = TP_1$ and since $PP_1 \\parallel AB$, it follows that the line $CT$ meets $AB$ at its midpoint. Analogously, the lines $BT$ and $AT$ meet $AC$ and $BC$ at their midpoints. Hence $T$ is the centroid of $\\triangle ABC$.\n\nb) We have\n$$\n\\begin{aligned}\nS_{MNP} & = S_{MNT} + S_{NPT} + S_{PMT} = S_{MBT} + S_{TNC} + S_{PAT} \\\\\n& = \\frac{1}{2}\\left(S_{MBP_1T} + S_{TNC M_1} + S_{PAN_1T}\\right) = \\frac{S_{ABC}}{2}\\left(1 - k_1^2 - k_2^2 - k_3^2\\right)\n\\end{aligned}\n$$\nIt follows by the inequality $k_1^2 + k_2^2 + k_3^2 \\geq \\frac{(k_1 + k_2 + k_3)^2}{3}$ and (1) that $k_1^2 + k_2^2 + k_3^2 \\geq \\frac{1}{3}$. Then\n$$\nS_{MNP} \\leq \\frac{S_{ABC}}{2}\\left(1 - \\frac{1}{3}\\right) = \\frac{1}{3} S_{ABC}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27115,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn a group of $n$ people there are three that are familiar to each other and any of them is familiar with more than half of the people in the group. Find the minimum possible triples of familiar people?",
    "options": [],
    "answer": "For n = 2k + 1, the minimum is k; for n = 2k, the minimum is k + 1.",
    "solution": "Solution:\n\nDenote by $A$, $B$ and $C$ the three familiar people in the group.\nLet $n=2k+1$ be an odd integer. Then any of $A$, $B$ and $C$ has at least $k+1$ familiar ( $k-1$ of them are not $A$, $B$ or $C$ ). Denote by $T$ the set of all people except $A$, $B$ and $C$ and let $a_{i}$, $i=0,1,2,3$, be the set of the people in $T$ who have exactly $i$ familiar among $A$, $B$ and $C$.\nThen $a_{0}+a_{1}+a_{2}+a_{3}$ is the number of all members of $T$, i.e. we have\n$$\na_{0}+a_{1}+a_{2}+a_{3}=2k-2\n$$\nOn the other hand, $a_{1}+2a_{2}+3a_{3}$ is the number of all familiar to $A$, $B$ and $C$, i.e. we have\n$$\na_{1}+2a_{2}+3a_{3} \\geq 3k-3\n$$\nHence\n$$\n\\begin{aligned}\n3k-3 &\\leq a_{1}+2a_{2}+3a_{3}=a_{0}+a_{1}+a_{2}+a_{3}+a_{2}+2a_{3} \\\\\n&=2k-2+a_{2}+2a_{3}\n\\end{aligned}\n$$\nand therefore $a_{2}+2a_{3} \\geq k-1$.\nSince any familiar to two of $A$, $B$ and $C$ is a member of a triple of familiar people and any familiar to $A$, $B$ and $C$ is member of three such triples, then the number of these triples is at least $1+a_{2}+3a_{3}$. Thus $1+a_{2}+3a_{3}>a_{2}+2a_{3} \\geq k-1$, which means that the number of the triples is not less than $k$.\nIt remains to construct an example with $k$ triples of familiar people. Let there be no familiar people in $T$. If $A$ is familiar to exactly $k-1$ people of $T$, and $B$ and $C$ to the remaining $k-1$, then the number of the triples is $k$.\n\nLet $n=2k$ be even. As in the previous case, we get that the number of the triples of familiar people is at least $k+1$. If $A$ and $B$ have exactly one common familiar person from $T$ (it is possible, since $|T|=2k-3$ and the familiar to $A$ and $B$ are at least $k-1$ ) who is not familiar to $C$, then the number of the triples is exactly $k+1$.\n\nSo the answer of the problem is $k$ for $n=2k+1$ and $k+1$ for $n=2k$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 27116,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $\\mathbb{R}$ be the set of real numbers. Find all $a>0$ such that there exists a function $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ with the following two properties:\n\na) $f(x)=a x+1-a$ for any $x \\in [2,3)$;\n\nb) $f(f(x))=3-2 x$ for any $x \\in \\mathbb{R}$.",
    "options": [],
    "answer": "a = 2 * 4^k for some integer k",
    "solution": "Solution:\nSetting $h(x)=f(x+1)-1$, it is easy to see that the conditions for $f(x)$ are equivalent to $h(x)=a x$ for any $x \\in [1,2)$ and $h(h(x))=-2 x$ for any $x \\in \\mathbb{R}$. Then $h(-2 x)=h(h(h(x)))=-2 h(x)$; in particular, $h(0)=0$.\n\nIt follows by induction that $h\\left(4^{n} x\\right)=4^{n} h(x)$ and hence $h(x)>0$ for $x \\in \\left[4^{n}, 2 \\cdot 4^{n}\\right)$, where $n$ is an arbitrary integer. Since $0>-2 x=h(h(x))=h(a x)$ for $x \\in [1,2)$, then $[a, 2 a) \\subset \\left[2 \\cdot 4^{k}, 4^{k+1}\\right)$ for some integer $k$. Therefore $a=2 \\cdot 4^{k}$.\n\nConversely, if $a$ has this form, then it is easy to check that the function\n$$\nh(x)=\\left\\{\\begin{aligned}\na x, & \\quad x \\in \\left[4^{n}, 2 \\cdot 4^{n}\\right) \\\\\n-\\frac{2 x}{a}, & \\quad x \\in \\left[2 \\cdot 4^{n}, 4^{n+1}\\right) \\\\\n0, & \\quad x=0 \\\\\na x, & \\quad x \\in \\left(-4^{n+1},-2 \\cdot 4^{n}\\right] \\\\\n-\\frac{2 x}{a}, & \\quad x \\in \\left(-2 \\cdot 4^{n},-4^{n}\\right]\n\\end{aligned}\\right.\n$$\nwhere $n$ runs over all integers, has the desired properties. One can easily show that this is the only function with the above properties.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27117,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $n$ be a positive integer. Ann writes down $n$ different positive integers. Then Ivo deletes some of them (possibly none, but not all), puts the signs $+$ or $-$ in front of each of the remaining numbers and sums them up. Ivo wins if $2003$ divides the result; otherwise, Ann wins. Who has a winning strategy?",
    "options": [],
    "answer": "Ann wins for n ≤ 10; Ivo wins for n ≥ 11.",
    "solution": "Solution:\n\nFor $n \\leq 10$ Ann wins by writing the numbers $1, 2, \\ldots, 2^{n-1}$. Indeed, the result Ivo can get is a non-zero integer between $-1023$ and $1023$, since it has the same sign as the largest remaining number $\\left(2^{j} > 2^{j}-1 = \\sum_{i=0}^{j-1} 2^{i}\\right)$.\n\nFor $n \\geq 11$ the set $C$ of Ann's numbers has $2^{n}-1 > 2003$ different nonempty subsets. Hence the sums of numbers of two of them, say $A$ and $B$, are congruent modulo $2003$. If Ivo puts $+$ in front of the numbers of $A \\setminus B$, $-$ in front of the numbers of $B \\setminus A$ and deletes the remaining numbers of $C$, he wins.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27118,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all functions $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ such that\n$$\nf\\left(x^{2}+y+f(y)\\right)=2 y+(f(x))^{2}\n$$\nfor any $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "f(x) = x",
    "solution": "Solution:\nIt follows by\n$$\nf\\left(x^{2}+y+f(y)\\right)=2 y+(f(x))^{2}\n$$\nthat the function $f$ is surjective. Note also that $(f(x))^{2}=(f(-x))^{2}$. In particular, we may choose $a$ such that $f(a)=f(-a)=0$. Setting $x=0, y= \\pm a$ in (1) gives $0=f( \\pm a)=(f(0))^{2} \\pm 2 a$, i.e., $a=0$. Plugging $y=-\\frac{(f(x))^{2}}{2}$ in (1), we get that $f\\left(x^{2}+y+f(y)\\right)=0$ and therefore $y+f(y)=-x^{2}$. Thus the function $y+f(y)$ takes any non-positive value. Since $f(0)=0$, it follows from (1) that\n$$\nf\\left(x^{2}\\right)=\\left((f(x))^{2} \\geq 0 \\text{ and } f(y+f(y))=2 y\\right.\n$$\nSetting $z=x^{2}, t=y+f(y)$, and using again (1) we deduce that $f(z+t)=f(z)+f(t)$ for any $z \\geq 0 \\geq t$. For $z=-t$ we get $f(-t)=-f(t)$ and then it is easy to check that $f(z+t)=f(z)+f(t)$ for any $z$ and $t$. Since $f(t) \\geq 0$ for $t \\geq 0$, it follows that $f$ is an increasing function. Suppose that $f(y)>y$ for some $y$. Then $f(f(y)) \\geq f(y)$ and we get\n$$\n2 y=f(y+f(y))=f(y)+f(f(y))>2 f(y)\n$$\na contradiction. Hence $f(y) \\leq y$. We see in the same way that $f(y) \\geq y$, so $f(x) \\equiv x$. This function obviously satisfies (1).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27119,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all real numbers $a$ such that $4[a n]=n+[a[a n]]$ for any positive integer $n$ ($[x]$ denotes the largest integer less than or equal to $x$).",
    "options": [],
    "answer": "2+\\sqrt{3}",
    "solution": "Solution:\nIt follows by the given condition that $4(a n-1)<n+a(a n)$ and $4 a n>n+a(a n-1)-1$, i.e.,\n$$\n1+a^2-\\frac{a+1}{n}<4 a<1+a^2+\\frac{4}{n}\n$$\nLetting $n \\rightarrow \\infty$ gives $1+a^2=4 a$, so $a=2-\\sqrt{3}$ or $a=2+\\sqrt{3}$.\nThe given equality for $n=1$ yields that the first case is impossible.\nIn the second case set $b=\\left[\\frac{n}{a}\\right]$ and $c=\\frac{n}{a}-b$. Since $a=4-\\frac{1}{a}$, then\n$$\n\\begin{aligned}\nn+[a[a n]] & =\\left[n+a\\left[4 n-\\frac{n}{a}\\right]\\right]=[n+a(4 n-b-1)] \\\\\n& =[a(4 n-1+c)]=\\left[\\left(4-\\frac{1}{a}\\right)(4 n-1+c)\\right] \\\\\n& =\\left[4(4 n-1)-4\\left(\\frac{n}{a}-c\\right)+\\frac{1-c}{a}\\right]=4(4 n-1-b) \\\\\n& =4\\left[4 n-\\frac{n}{a}\\right]=4[a n]\n\\end{aligned}\n$$\nTherefore $a=2+\\sqrt{3}$ is the only solution of the problem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27120,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nAt any lattice point in the plane a number from the interval $(0,1)$ is written. It is known that for any lattice point the number written there is equal to the arithmetic mean of the numbers written at the four closest lattice points. Prove that all written numbers are equal.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet $f(x, y)$ be the number written at the lattice point $(x, y)$. Then\n$$\nf(x, y) = \\frac{f(x+1, y) + f(x-1, y) + f(x, y+1) + f(x, y-1)}{4}\n$$\nAssume that not all the numbers are equal. Then there are two points at distance $1$ apart such that the numbers written there are different. Rotating the plane, if necessary, we may assume that $f(x_0+1, y_0) > f(x_0, y_0)$ for some $x_0, y_0 \\in \\mathbb{Z}$. Set\n$$\ng(x, y) = f(x+1, y) - f(x, y)\n$$\nThen $M = \\sup_{x, y \\in \\mathbb{Z}} g(x, y) \\in (0,1]$ and\n$$\ng(x, y) = \\frac{g(x+1, y) + g(x-1, y) + g(x, y+1) + g(x, y-1)}{4}\n$$\nIn particular, if $g(a, b) \\geq M - \\varepsilon$, where $\\varepsilon > 0$, then\n$$\n\\begin{aligned}\ng(a+1, b) &= 4g(a, b) - g(a-1, b) - g(a, b+1) - g(a, b-1) \\\\\n&\\geq 4(M-\\varepsilon) - 3M = M - 4\\varepsilon\n\\end{aligned}\n$$\nand we get by induction that $g(a+n, b) \\geq M - 4^n \\varepsilon$ for any $n \\in \\mathbb{N}$.\n\nChoose now $n \\geq \\frac{2}{M}$, $\\varepsilon \\in \\left(0, \\frac{M}{2 \\cdot 4^{n-1}}\\right]$ and integers $a$ and $b$ such that $g(a, b) \\geq M - \\varepsilon$. Then\n$$\n1 > f(a+n, b) > f(a+n, b) - f(a, b) = \\sum_{k=0}^{n-1} g(a+k, b) \\geq n \\frac{M}{2} \\geq 1\n$$\nwhich is a contradiction.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27121,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nConsider the sequence\n$$\ny_{1} = y_{2} = 1, \\quad y_{n+2} = (4k - 5) y_{n+1} - y_{n} + 4 - 2k, \\quad n \\geq 1\n$$\nFind all integers $k$ such that any term of the sequence is a perfect square.",
    "options": [],
    "answer": "k = 1 and k = 3",
    "solution": "Solution:\nLet $k$ have the given property. We have that $y_{3} = 2k - 2 = 4a^{2}$ ($a \\geq 0$), i.e., $k = 2a^{2} + 1$. Further, $y_{4} = 8k^{2} - 20k + 13$ and $y_{5} = 32k^{3} - 120k^{2} + 148k - 59 = 256a^{6} - 96a^{4} + 8a^{2} + 1$.\n\nIf $a = 0$ we get that $k = 1$ and the given sequence is $1, 1, 0, 1, 1, 0, \\ldots$ Hence $k = 1$ is a solution of the problem.\n\nLet $a > 0$. It is easy to check that\n$$\n\\left(16a^{3} - 3a\\right)^{2} \\geq y_{5} = 256a^{6} - 96a^{4} + 8a^{2} + 1 > \\left(16a^{3} - 3a - 1\\right)^{2}\n$$\nSince $y_{5}$ is a perfect square, the first inequality must be equality, i.e., $a = 1$ and then $k = 3$.\n\nWe shall prove that $k = 3$ is a solution of the problem. In this case the sequence is defined by $y_{1} = y_{2} = 1$ and $y_{n+2} = 7y_{n+1} - y_{n} - 2$ for $n \\geq 1$. Since $y_{3} = 2^{2}$, $y_{4} = 5^{2}$ and $y_{5} = 13^{2}$, it is natural to conjecture that $y_{n} = u_{2n-3}^{2}$ for $n \\geq 2$ where $\\{u_{n}\\}_{n=1}^{\\infty}$ is the Fibonacci sequence: $u_{1} = u_{2} = 1$ and $u_{n+2} = u_{n+1} + u_{n}$ for $n \\geq 1$. To prove this, we first note that $u_{n+2} = 3u_{n} - u_{n-2}$ and $u_{n+2}u_{n-2} - u_{n}^{2} = 1$ for any odd $n \\geq 3$. It follows that $\\left(u_{n+2} + u_{n-2}\\right)^{2} = 9u_{n}^{2}$ and thus\n$$\nu_{n+2}^{2} = 9u_{n}^{2} - u_{n-2}^{2} - 2u_{n-2}u_{n+2} = 7u_{n}^{2} - u_{n-2}^{2} - 2\n$$\nHence $y_{n} = u_{2n-3}^{2}$ for $n \\geq 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27122,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSome of the vertices of a convex $n$-gon are connected by segments such that any two of them have no a common interior point. Prove that for any $n$ points in general position (i.e., any three of them are not collinear) there is an one-to-one correspondence between the points and the vertices of the $n$-gon such that any two segments corresponding to the respective segments from the $n$-gon have no a common interior point.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $A_{1} A_{2} \\ldots A_{n}$ be a convex $n$-gon. Denote by $\\mathcal{B}$ the set of the connected vertices and let $B_{1} B_{2} \\ldots B_{k}$ be its convex hull. We shall prove by induction on $n$ that there is a map with the desired properties that in addition sends two given adjacent vertices of the $n$-gon to two given adjacent vertexes of $B_{1} B_{2} \\ldots B_{k}$.\n\nThe base of the induction $n=3$ is obvious. Suppose that our statement is true for any $k<n$. To prove it for $n$, it is enough to find a map for $n$ that sends $A_{1}$ and $A_{2}$ to $B_{1}$ and $B_{2}$, respectively. Note that there is a unique point $A_{i}$ that is connected with $A_{1}$ and $A_{2}$ (otherwise, some segments will have a common interior point). Consider the points $X_{1}, X_{2}, \\ldots, X_{s}$ from $\\mathcal{B}$ such that any of the triangles $B_{1} X_{i} B_{2}$ contains no points of $\\mathcal{B}$. It is easy to find a point among them, say $X_{l}$, such that the interiors of $\\Varangle B_{1} B_{2} X_{l}$ and $\\Varangle B_{2} B_{1} X_{l}$ contain at most $n-i$ and $i-3$ points of $\\mathcal{B}$, respectively. It is clear now that there are a line through $X_{l}$ and an interior point of the segment $B_{1} B_{2}$ that divides the set $\\mathcal{B}$ into two subsets $\\mathcal{B}_{1}$ and $\\mathcal{B}_{2}$, containing $n-i+1$ and $i-2$ points, respectively. Let $B_{1} X_{l}$ and $B_{2} X_{l}$ be sides of the convex hulls of these two sets. If $A_{i}$ is the corresponding point to $X_{l}$, then applying the induction assumption to the sets $A_{2} A_{3} \\ldots A_{i}$ and $\\mathcal{B}_{2} \\cup\\{X_{l}\\}$, and to $A_{i} A_{i+1} \\ldots A_{n} A_{1}$ and $\\mathcal{B}_{1} \\cup\\{X_{l}\\}$, we see that the statement is true for $n$ points. This completes the solution of the problem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27123,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $D$ be a point on the side $AC$ of $\\triangle ABC$ such that $BD = CD$. A line parallel to $BD$ intersects the sides $BC$ and $AB$ at points $E$ and $F$, respectively. Set $G = AE \\cap BD$. Prove that $\\Varangle BCG = \\Varangle BCF$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $H = AC \\cap EF$. Then $\\Varangle CDG = \\Varangle FHC$ and\n$$\n\\frac{CD}{DG} = \\frac{BD}{DG} = \\frac{FH}{HE} = \\frac{FH}{HC}\n$$\nIt follows that $\\triangle CDG \\sim \\triangle FHC$ which implies that $\\Varangle GCD = \\Varangle CFH$. Hence\n$$\n\\begin{aligned}\n\\Varangle BCG & = \\Varangle BCD - \\Varangle GCD \\\\\n& = \\Varangle CEH - \\Varangle CFH = \\Varangle BCF\n\\end{aligned}\n$$\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27124,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA set of at least three positive integers is called uniform if removing any of its elements the remaining set can be disjoint into two subsets with equal sums of elements. Find the minimal cardinality of a uniform set.",
    "options": [],
    "answer": "7",
    "solution": "Solution:\nLet $A=\\{a_{1}, a_{2}, \\ldots, a_{n}\\}$ be a uniform set. Set $S=a_{1}+a_{2}+\\cdots+a_{n}$. It follows from the given condition that $S-a_{i}$ is an even number for any $i=1,2, \\ldots, n$. Suppose that the number $S$ is even. Then all the numbers $a_{i}$ are even. Set $a_{i}=2 b_{i}, i=1,2, \\ldots, n$. Then it is easy to see that the set $B=\\{b_{1}, b_{2}, \\ldots, b_{n}\\}$ is uniform, too. So, we may assume that $S$ is an odd number and whence $a_{1}, a_{2}, \\ldots, a_{n}$ and $n$ are also odd numbers.\n\nWe shall prove that $n=7$. It is not difficult to check that $\\{1,3,5,7,9,11,13\\}$ is a uniform set. It remains to show that there are no uniform sets with 5 elements, since it is obvious that the sets with 3 elements are not uniform.\n\nSuppose that $A=\\{a_{1}, a_{2}, a_{3}, a_{4}, a_{5}\\}$ is a uniform set and let $a_{1}<a_{2}<a_{3}<a_{4}<a_{5}$. Considering the set $A \\setminus\\{a_{1}\\}$ we see that either $a_{2}+a_{5}=a_{3}+a_{4}$ or $a_{2}+a_{3}+a_{4}=a_{5}$. Considering the set $A \\setminus\\{a_{2}\\}$ we get that either $a_{1}+a_{5}=a_{3}+a_{4}$ or $a_{1}+a_{3}+a_{4}=a_{5}$.\n\n- If $a_{2}+a_{5}=a_{3}+a_{4}$ and $a_{1}+a_{5}=a_{3}+a_{4}$, then $a_{1}=a_{2}$.\n- If $a_{2}+a_{5}=a_{3}+a_{4}$ and $a_{1}+a_{3}+a_{4}=a_{5}$, then $a_{1}=-a_{2}$.\n- If $a_{2}+a_{3}+a_{4}=a_{5}$ and $a_{1}+a_{5}=a_{3}+a_{4}$, then $a_{1}=-a_{2}$.\n- If $a_{2}+a_{3}+a_{4}=a_{5}$ and $a_{1}+a_{3}+a_{4}=a_{5}$, then $a_{1}=a_{2}$.\n\nSince all the above possibilities lead to a contradiction, we conclude that there are no uniform sets with 5 elements.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27125,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFor any positive integer $n$ set\n$$\nA_{n} = \\{ j : 1 \\leq j \\leq n,\\ (j, n) = 1 \\}\n$$\nFind all $n$ such that the polynomial\n$$\nP_{n}(x) = \\sum_{j \\in A_{n}} x^{j-1}\n$$\nis irreducible over $\\mathbb{Z}[x]$.",
    "options": [],
    "answer": "n = 1, 2, 3, 4, 6",
    "solution": "Solution:\nWe have $P_{1}(x) = P_{2}(x) = 1$, $P_{3}(x) = x + 1$, $P_{4}(x) = x^{2} + 1$ and $P_{6}(x) = x^{4} + 1$. Hence the integers $n = 1, 2, 3, 4, 6$ have the desired property. We shall prove that these are the only solutions of the problem.\n\nIt is enough to show that for $n \\geq 3$ the polynomial $P_{n}(x)$ has a divisor of the form $1 + x^{r}$, $r \\geq 1$, which is a proper divisor for $n = 5$ and $n \\geq 7$.\n\nIf $n \\geq 3$ is a prime, this follows by the decomposition\n$$\nP_{n}(x) = (1 + x)\\left(1 + x^{2} + x^{4} + \\cdots + x^{n-3}\\right)\n$$\nNote also that $P_{4}(x) = x^{2} + 1$.\n\nNow we shall use induction on $n$. Suppose that the statement is true for all $m \\geq 3$ that are less than $n$. If $n \\geq 6$ is a composite integer, then $n = m p$, where $p$ is prime and $m \\geq 3$. Two cases are possible.\n\nCase 1. $p$ divides $m$. Then we have that $A_{n} = \\cup_{i=0}^{p-1} (A_{m} + i m)$ and hence $P_{n}(x) = P_{m}(x) \\sum_{i=0}^{p-1} x^{i m}$. It remains to use that $P_{m}(x)$ has a divisor of the form $1 + x^{r}$.\n\nCase 2. $p$ does not divide $m$. Using that\n$$\nA_{n} = \\left(\\cup_{i=0}^{p-1} (A_{m} + i m)\\right) \\setminus (p A_{m})\n$$\nand $x^{k p - 1} = x^{p - 1} (x^{p})^{k - 1}$, it follows that\n$$\nP_{n}(x) = P_{m}(x) \\sum_{i=0}^{p-1} x^{i m} - x^{p-1} P_{m}(x^{p})\n$$\nBy the induction assumption, $P_{m}(x)$ has a divisor of the form $1 + x^{r}$. Therefore $1 + x^{p r}$ divides $P_{m}(x^{p})$. We shall consider two subcases.\n\na) If $p \\geq 3$, then $p$ is odd and $1 + x^{r}$ divides $1 + x^{p r}$. Hence $1 + x^{r}$ divides $P_{n}(x)$.\n\nb) If $p = 2$, we may assume that $m$ is prime (otherwise, $m$ has an odd prime divisor and we may go either to a) or to the first case). Then\n$$\nP_{n}(x) = (1 + x^{m+1})\\left(1 + x^{2} + x^{4} + \\cdots + x^{m-3}\\right)\n$$\nIt remains to observe that $P_{n}(x) = 1 + x^{r}$ only for $n = 3, 4, 6$ which completes the solution.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 27126,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIs it true that for any permutation $a_{1}, a_{2}, \\ldots, a_{2002}$ of $1,2, \\ldots, 2002$ there are positive integers $m$ and $n$ of the same parity such that $1 \\leq m<n \\leq 2002$ and $a_{m}+a_{n}=2 a_{\\frac{m+n}{2}}$?",
    "options": [],
    "answer": "No",
    "solution": "Solution:\n\nThe answer is no. We shall prove by induction that for any $k \\geq 3$ there is a permutation $a_{1}, a_{2}, \\ldots, a_{k}$ of $1,2, \\ldots, k$ such that\n$$\na_{m}+a_{n} \\neq 2 a_{\\frac{m+n}{2}} \\text{ for any } 1 \\leq m<n \\leq k \\text{ of the same parity. }\n$$\nFor $k=3$ and $k=4$ take the permutations $1,3,2$ and $1,3,2,4$, respectively. Assume that our statement is true for any integer less than $k$. Start with the following permutation of $1,2, \\ldots, k$: the odd numbers are in the first block, the even numbers that are not divisible by $4$ are in the second block, etc. For example, if $k=12$, then we have $1,3,5,7,9,11 ; 2,6,10 ; 4,12 ; 8$.\nIf $a_{m}$ and $a_{n}$ are in different blocks, set $a_{m}=2^{s} b$ and $a_{n}=2^{t} c$, where $t>s>0$ and $b, c$ are odd integers. Then $\\frac{a_{m}+a_{n}}{2}=2^{s-1}\\left(b+2^{t-s} c\\right)$ is in the block before that of $a_{m}$, whence $a_{\\frac{m+n}{2}}$ is in a block between these of $a_{m}$ and $a_{n}$. So (1) holds.\nIt remains to reorder the integers in any block in such a way that (1) is satisfied for the numbers in this block. Consider the $(r+1)$-th block: $2^{r}, 3\\cdot2^{r}, \\ldots,(2 d-1) 2^{r}$, where $2 d-1 \\leq k$. By the induction assumption there is a permutation $b_{1}, b_{2}, \\ldots, b_{d}$ of $1,2, \\ldots, d$ satisfying (1). Set $c_{i}=2 b_{i}-1$ for $i=1,2, \\ldots, d$ and consider the permutation $2^{r} c_{1}, 2^{r} c_{2}, \\ldots, 2^{r} c_{2 d-1}$. Then\n$$\n\\frac{c_{m}+c_{n}}{2}=b_{m}+b_{n}-1 \\neq 2 b_{\\frac{m+n}{2}}-1=c_{\\frac{m+n}{2}}\n$$\nwhich completes the proof.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 27127,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $a$, $b$ and $c$ be rational numbers such that $a+b+c$ and $a^{2}+b^{2}+c^{2}$ are equal integers. Prove that the number $a b c$ can be written as a ratio of a perfect cube and a perfect square that are coprime.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet $a+b+c=a^{2}+b^{2}+c^{2}=t$. Then $t \\geq 0$. On the other hand, the Root mean square - Arithmetic mean inequality implies that\n$$\n\\frac{a^{2}+b^{2}+c^{2}}{3} \\geq \\frac{(a+b+c)^{2}}{9} \\Longleftrightarrow 3 t \\geq t^{2}\n$$\nHence $t \\in\\{0,1,2,3\\}$. If $t=0$ or $t=3$, then $a=b=c=0$ or $a=b=c=1$, respectively, and the statement is trivial.\n\nLet $t=1$. Denote by $d$ the product of the denominators of $|a|,|b|$ and $|c|$. Then $x=a d, y=b d$ and $z=c d$ are integers for which $x+y+z=d$ and $x^{2}+y^{2}+z^{2}=d^{2}$. We may assume that $z>0$. Note that\n$(x+y+z)^{2}=x^{2}+y^{2}+z^{2} \\Longleftrightarrow x y+y z+x z=0 \\Longleftrightarrow(x+z)(y+z)=z^{2}$.\nIt follows that $x+z=r p^{2}$, $y+z=r q^{2}$ and $z=|r| p q$, where $p$ and $q$ are coprime positive integers, and $r$ is a non-zero integer. Since $d=x+y+z= r\\left(p^{2}+q^{2}\\right)-|r| p q>0$ we get that $r>0$. Then\n$$\na=\\frac{x}{d}=\\frac{p(p-q)}{p^{2}+q^{2}-p q}, \\quad b=\\frac{y}{d}=\\frac{q(q-p)}{p^{2}+q^{2}-p q}, \\quad c=\\frac{z}{d}=\\frac{p q}{p^{2}+q^{2}-p q}\n$$\nand thus $a b c=\\frac{[p q(p-q)]^{2}}{\\left(p q-p^{2}-q^{2}\\right)^{3}}$.\nIt remains to show that $p q(p-q)$ and $p^{2}+q^{2}-p q$ are coprime integers. Suppose that $s$ is a prime divisor of $p q(p-q)$ and $p^{2}+q^{2}-p q$. Let $s \\mid p$. Since $s \\mid p^{2}+q^{2}-p q$, then $s \\mid q$, a contradiction. Analogously, $s \\nmid q$. Hence $s \\mid p-q$. Then $s \\mid(p-q)^{2}-\\left(p^{2}+q^{2}-p q\\right)=p q$, which is impossible.\n\nThe case $t=2$ can be reduced to the case $t=1$ by setting $a_{1}=1-a$, $b_{1}=1-b$ and $c_{1}=1-c$. Indeed, it is easy to check that $a_{1}+b_{1}+c_{1}= a_{1}^{2}+b_{1}^{2}+c_{1}^{2}=1$ and $a_{1} b_{1} c_{1}=-a b c$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27128,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $ABCD$ be a circumscribed quadrilateral and let $P$ be the orthogonal projection of its incenter on the diagonal $AC$. Prove that $\\Varangle APB = \\Varangle APD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nWe shall use the following fact: if $\\alpha, \\beta, \\gamma$ and $\\delta$ are angles such that $\\sin \\alpha \\sin \\delta = \\sin \\beta \\sin \\gamma$ and $\\alpha + \\beta = \\gamma + \\delta < 180^{\\circ}$, then $\\alpha = \\gamma$ and $\\beta = \\delta$.\n\nDenote by $M, N, R$ and $S$ the tangent points of the incircle of $ABCD$ centered at $O$ with the sides $AB, BC, CD$ and $DA$, respectively. Then the points $A, M, O, P$ and $S$ lie on the circle with diameter $AO$ and $\\Varangle APM = \\frac{\\widehat{AM}}{2} = \\frac{\\widehat{AS}}{2} = \\Varangle APS$.\n\n![](attached_image_1.png)\n\nAnalogously, $\\Varangle CPR = \\Varangle CPN$ and hence $\\Varangle SPR = \\Varangle MPN$.\n\nThe Sine theorem for $\\triangle BPM$ and $\\triangle BPN$ gives\n$$\n\\frac{\\sin \\Varangle MPB}{\\sin \\Varangle PMB} = \\frac{BM}{BP} = \\frac{BN}{BP} = \\frac{\\sin \\Varangle BPN}{\\sin \\Varangle BNP}\n$$\nand therefore $\\frac{\\sin \\Varangle MPB}{\\sin \\Varangle NPB} = \\frac{\\sin \\Varangle PMB}{\\sin \\Varangle PNB}$.\n\nSince\n$$\n\\begin{aligned}\n& \\Varangle PMB = 180^{\\circ} - \\Varangle AMP = 180^{\\circ} - \\Varangle AOP \\\\\n& \\Varangle PNB = 180^{\\circ} - \\Varangle CNP = 180^{\\circ} - \\Varangle COP\n\\end{aligned}\n$$\nthen $\\frac{\\sin \\Varangle MPB}{\\sin \\Varangle NPB} = \\frac{\\sin \\Varangle AOP}{\\sin \\Varangle COP}$.\n\nWe get in the same way that\n$$\n\\frac{\\sin \\Varangle SPD}{\\sin \\Varangle RPD} = \\frac{\\sin \\Varangle AOP}{\\sin \\Varangle COP}.\n$$\n\nApplying the fact mentioned above with $\\alpha = \\Varangle MPB$, $\\beta = \\Varangle NPB$, $\\gamma = \\Varangle SPD$ and $\\delta = \\Varangle RPD$ we conclude that $\\Varangle MPB = \\Varangle SPD$ and therefore $\\Varangle APB = \\Varangle APM + \\Varangle MPB = \\Varangle APS + \\Varangle SPD = \\Varangle APD$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27129,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the number of real solutions of the system\n$$\n\\left\\lvert\\,\n\\begin{aligned}\n& x + y + z = 3 x y \\\\\n& x^{2} + y^{2} + z^{2} = 3 x z \\\\\n& x^{3} + y^{3} + z^{3} = 3 y z\n\\end{aligned}\n\\right.\n$$",
    "options": [],
    "answer": "4",
    "solution": "Solution:\nNote first that the triple $(0, 0, 0)$ is a solution of the system.\n\nIf $y = 0$, then it follows from the first equation that $x = -z$ and the second one gives that $x = z = 0$.\n\nIf $y \\neq 0$, set $a = \\frac{x}{y}$ and $b = \\frac{z}{y}$. Then the system becomes\n$$\n\\left\\lvert\\,\n\\begin{aligned}\n& 1 + a + b = 3 a y \\\\\n& 1 + a^{2} + b^{2} = 3 a b \\\\\n& y\\left(1 + a^{3} + b^{3}\\right) = 3 b\n\\end{aligned}\n\\right.\n$$\nHence $y = \\frac{1 + a + b}{3 a}$ and therefore\n$$\n\\left\\lvert\\,\n\\begin{aligned}\n& (1 + a + b)\\left(1 + a^{3} + b^{3}\\right) = 9 a b \\\\\n& 1 + a^{2} + b^{2} = 3 a b\n\\end{aligned}\n\\right.\n$$\nSet $u = a + b$ and $v = a b$. Then\n$$\n\\left\\lvert\\,\n\\begin{aligned}\n& (1 + u)\\left(1 + u^{3} - 3 u v\\right) = 9 v \\\\\n& 1 + u^{2} - 2 v = 3 v\n\\end{aligned}\n\\right.\n$$\nHence $v = \\frac{u^{2} + 1}{5}$ and plugging it into the first equation gives\n$$\n0 = u^{4} + u^{3} - 6 u^{2} + u - 2 = (u - 2)\\left(u^{3} + 3 u^{2} + 1\\right)\n$$\nThe case $u = 2$ leads to $v = 1$, $a = b = 1$ and gives the solution $(x, y, z) = (1, 1, 1)$.\n\nThe function $f(u) = u^{3} + 3 u^{2} + 1$ has a local maximum at $u = -2$ and a local minimum at $u = 0$. Since $f(0) = 1 > 0$, the equation $f(u) = 0$ has only one real root $u_{0}$ and $u_{0} < -2$. Then $u_{0}^{2} - 4 \\frac{u_{0}^{2} + 1}{5} = \\frac{u_{0}^{2} - 4}{5} > 0$ which shows that the system\n$$\n\\left\\lvert\\,\n\\begin{aligned}\n& a + b = u_{0} \\\\\n& a b = \\frac{u_{0}^{2} + 1}{5}\n\\end{aligned}\n\\right.\n$$\nhas two solutions.\n\nThus the given system has four real solutions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27130,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA set $C$ of positive integers is called good if for any integer $k$ there exist $a, b \\in C, a \\neq b$, such that the numbers $a+k$ and $b+k$ are not coprime. Prove that if the sum of the elements of a good set $C$ equals $2003$, then there exists $c \\in C$ for which the set $C \\setminus \\{c\\}$ is good.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet $p_1, p_2, \\ldots, p_n$ be all prime divisors of all possible differences of two distinct numbers of $C$.\n\nSuppose that for any $p_i$ there exists an integer $\\alpha_i$ such that $c \\equiv \\alpha_i \\pmod{p_i}$ for at most one $c \\in C$. It follows by the Chinese remainder theorem that there is an integer $k$ such that $k \\equiv p_i - \\alpha_i \\pmod{p_i}$ for any $i$. Then the condition of the problem implies that $p_j$ divides $a+k$ and $b+k$ for some $j$ and some $a, b \\in C$. Then $a \\equiv b \\equiv \\alpha_j \\pmod{p_j}$, a contradiction.\n\nWe conclude that for some prime number $p$ each remainder modulo $p$ appears at least twice. Assuming that any remainder appears exactly twice we get that the sum of the elements of $C$ equals\n$$\np r + 2(0 + 1 + \\cdots + p-1) = p(r + p - 1), \\quad r \\geq 1\n$$\nThis is a contradiction, since $2003$ is a prime number. Hence some remainder appears at least three times. Removing an element of $C$ giving this remainder we obtain a new good set $C'$. (Indeed, for any $k$ we can find $a, b \\in C', a \\neq b$, such that $p$ divides $a+k$ and $b+k$.)",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27131,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all polynomials $P(x)$ with integer coefficients such that for any positive integer $n$ the equation $P(x) = 2^{n}$ has an integer solution.",
    "options": [],
    "answer": "All such polynomials are P(x) = a(x + b) with a ∈ {±1, ±2} and b any integer.",
    "solution": "Solution:\nDenote by $m$ and $a$ the degree and the leading coefficient of $P(x)$, respectively. Let $x_{n}$ be an integer solution of the equation $P(x) = 2^{n}$. Since $\\lim_{n \\rightarrow \\infty} |x_{n}| = +\\infty$, then\n$$\n\\lim_{n \\rightarrow \\infty} \\frac{a |x_{n}|^{m}}{2^{n}} = 1\n$$\nand hence\n$$\n\\lim_{n \\rightarrow \\infty} \\left| \\frac{x_{n+1}}{x_{n}} \\right| = \\sqrt[m]{2}.\n$$\nOn the other hand, $x_{n+1} - x_{n}$ divides $P(x_{n+1}) - P(x_{n})$ and thus $|x_{n+1} - x_{n}| = 2^{k_{n}}$ for some $k_{n} \\geq 0$. Then\n$$\n\\left| \\frac{x_{n+1}}{x_{n}} \\right| = \\frac{2^{k_{n}}}{|x_{n}|} + \\varepsilon_{n}\n$$\nwhere $\\varepsilon_{n} = \\pm 1$ and we get that\n$$\n\\sqrt[m]{2} = \\lim_{n \\rightarrow \\infty} \\left( \\frac{2^{k_{n}}}{|x_{n}|} + \\varepsilon_{n} \\right ) = \\lim_{n \\rightarrow \\infty} \\left( 2^{k_{n}} \\sqrt[m]{\\frac{a}{2^{n}}} + \\varepsilon_{n} \\right )\n$$\nNote that $\\varepsilon_{n}$ equals either $1$ or $-1$ for infinitely many $n$. Since the two cases are similar, we shall consider only the second one. Let $1 = \\varepsilon_{i_{1}} = \\varepsilon_{i_{2}} = \\cdots$. Then\n$$\n\\sqrt[m]{2} + 1 = \\sqrt[m]{a} \\lim_{j \\rightarrow \\infty} 2^{k_{i_{j}} - i_{j}}\n$$\nand hence the sequence of integers $k_{i_{j}} - i_{j}$ converges to some integer $\\ell$. It follows that $(\\sqrt[m]{2} + 1)^{m} = a 2^{m \\ell}$ is a rational number. According to the Eisenstein criteria, the polynomial $x^{m} - 2$ is irreducible. Hence $(x-1)^{m} - 2$ is the minimal polynomial of $\\sqrt[m]{2} + 1$. It follows that $(x-1)^{m} - 2 = x^{m} - a 2^{m \\ell}$ which is possible only for $m = 1$.\n\nLet $P(x) = a x + b$. Then $a(x_{2} - x_{1})$ divides $2$ and thus $a = \\pm 1, \\pm 2$. Now it follows easily that all polynomials with the desired property are of the form $P(x) = a(x + b)$, where $a = \\pm 1, \\pm 2$ and $b$ is an arbitrary integer.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27132,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nProve that there are no positive integers $m$ and $n$ such that\n$$\nm(m+1)(m+2)(m+3)=n(n+1)^2(n+2)^3(n+3)^4\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nWe shall use that if $\\left(\\sqrt{a^{2}-1}+1\\right)^{k}=x_{k} \\sqrt{a^{2}-1}+y_{k}$, then all the solutions of Pell's equation $\\left(a^{2}-1\\right) x^{2}+1=y^{2}$ are $\\left(x_{k}, y_{k}\\right)$. This implies that $\\left(a^{2}-1\\right) x^{2}+1$ is a perfect square if and only if $x$ is a term of the sequence defined by $x_{0}=0, x_{1}=1$ and $x_{k+2}=2 a x_{k+1}-x_{k}$ for $k \\geq 0$.\n\nSince $m(m+1)(m+2)(m+3)+1=\\left(m^{2}+3 m+1\\right)^{2}$, then $n(n+1)^{2}(n+2)^{3}(n+3)^{4}+1=\\left[(n+1)^{2}-1\\right]\\left[(n+1)(n+2)(n+3)^{2}\\right]^{2}+1$ is a perfect square. Applying the property mentioned above with $a=n+1$ gives that $(n+1)(n+2)(n+3)^{2}$ is a term of the sequence defined by $x_{0}=0, x_{1}=1$ and $x_{k+2}=(2 n+2) x_{k+1}-x_{k}$ for $k \\geq 0$.\n\nNow it is easy to see by induction on $k$ that the remainders of any $x_{k}$ modulo $2 n+1$ and $2 n+3$ are $0,1$ or $-1$. Hence $(n+1)(n+2)(n+3)^{2} \\equiv 0, \\pm 1 \\pmod{2 n+1}$ and then\n$$\n(2 n+2)(2 n+4)(2 n+6)^{2} \\equiv 0, \\pm 16 \\quad (\\bmod 2 n+1)\n$$\nUsing that $2 n+2 \\equiv 1 \\pmod{2 n+1}$, $2 n+4 \\equiv 3 \\pmod{2 n+1}$ and $2 n+6 \\equiv 5 \\pmod{2 n+1}$, it follows that $2 n+1$ divides $75, 59$ or $91$. Repeating the same arguments, we get that $2 n+3$ divides $7, 9$ or $25$. The only numbers satisfying both conditions are $n=1,2,3$. Now direct verifications complete the solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27133,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $A_{1}$, $B_{1}$ and $C_{1}$ be respectively the midpoints of the sides $BC$, $CA$ and $AB$ of $\\triangle ABC$ with centroid $M$. The line through $A_{1}$ and parallel to $BB_{1}$ meets the line $B_{1}C_{1}$ at a point $D$. Prove that if the points $A$, $B_{1}$, $M$ and $C_{1}$ are concyclic, then $\\Varangle ADA_{1} = \\Varangle CAB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nSince $A_{1}D \\parallel MB_{1}$ and $A$, $B_{1}$, $M$, $C_{1} \\in k$, it follows that\n$$\n\\Varangle AA_{1}D = \\Varangle AMB_{1} = \\Varangle AC_{1}B_{1} = \\Varangle ABC.\n$$\nUsing that $B_{1}C_{1} \\parallel BA_{1}$ and $A_{1}D \\parallel BB_{1}$ we conclude that $BA_{1}DB_{1}$ is a parallelogram. Hence $BA_{1} = B_{1}D = B_{1}C_{1}$. On the other hand, $AB_{1} = B_{1}C$ and therefore $AC_{1}CD$ is a parallelogram. In particular, $AD \\parallel CC_{1}$ and then\n$$\n\\begin{aligned}\n\\Varangle DAA_{1} & = \\Varangle CMA_{1} = \\Varangle AMC_{1} \\\\\n& = \\Varangle AB_{1}C_{1} = \\Varangle ACB.\n\\end{aligned}\n$$\n![](attached_image_1.png)\nTherefore\n$$\n\\Varangle ADA_{1} = 180^{\\circ} - \\Varangle DAA_{1} - \\Varangle DA_{1}A = 180^{\\circ} - \\Varangle ABC - \\Varangle ACB = \\Varangle CAB \\text{.}\n$$\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27134,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the least positive integer $m$ such that $2^{2000}$ divides $2003^{m} - 1$.",
    "options": [],
    "answer": "2^1998",
    "solution": "Solution:\nSet $m = 2^{k} p$, where $p > 1$ is an odd integer. Then\n$$\n2003^{m} - 1 = 2003^{2^{k} p} - 1 = \\left(2003^{2^{k}} - 1\\right) K\n$$\nwhere $K$ is a sum of $p$ even integers and $1$; in particular, $K$ is odd. Hence $m$ must be of the form $m = 2^{k}$. In this case we have that\n$$\n\\begin{aligned}\n2003^{2^{k}} - 1 & = \\left(2003^{2^{k-1}} - 1\\right)\\left(2003^{2^{k-1}} + 1\\right) \\\\\n& = \\left(2003^{2^{k-1}} + 1\\right)\\left(2003^{2^{k-2}} + 1\\right) \\ldots (2003 + 1)(2003 - 1)\n\\end{aligned}\n$$\nSince $2003^{2^{i}} + 1 \\equiv 2 \\pmod{4}$, then $2^{k+2}$ divides $2003^{2^{k}} - 1$ but $2^{k+3}$ does not (use that $2003 + 1 \\equiv 4 \\pmod{8}$ and $2003 - 1 \\equiv 2 \\pmod{4}$). Therefore $k + 2 = 2000$, i.e. $k = 1998$. Thus the desired number is equal to $2^{1998}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27135,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABC$ be an isosceles triangle with $AC = BC$ and let $k$ be a circle with center $C$ and radius less than the altitude $CH$, $H \\in AB$. Lines through $A$ and $B$ are tangent to $k$ at points $P$ and $Q$ lying on the same side of the line $CH$. Prove that the points $P$, $Q$ and $H$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nFirst solution. Since $CP = CQ$, $CA = CB$ and $\\angle APC = \\angle BQC = 90^\\circ$, then $\\triangle APC \\cong BQC$. Hence $\\angle CAP = \\angle CBQ$. Setting $AP \\cap BQ = T$, it follows that the quadrilateral $ABTC$ is cyclic. Then $\\angle BAC = \\angle QTC$ and now $\\angle TQC = \\angle AHC = 90^\\circ$ implies that $\\angle QCT = \\angle ACH$. The equalities $\\angle AHC = \\angle APC = \\angle CPT = \\angle CQT = 90^\\circ$ show that $AHPC$ and $CPTQ$ are cyclic quadrilaterals. Thus $\\angle APH = \\angle ACH$ and $\\angle QPT = \\angle QCT$ which means that $\\angle APH = \\angle QPT$. Hence the points $H$, $P$ and $Q$ are collinear.\n\n![](attached_image_1.png)\n\nSecond solution. Set $S = HQ \\cap k$. Since the quadrilateral $BHCQ$ is cyclic, and the triangles $ABC$ and $CQS$ are isosceles, it follows that $\\angle BAC = \\angle ABC = \\angle HQC = \\angle CSQ$. Then $AHSC$ is a cyclic quadrilateral and therefore $\\angle ASC = \\angle AHC = 90^\\circ$. Hence $S = P$, i.e., the points $H$, $P$ and $Q$ are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27136,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all real values of $a$ such that the system\n$$\n\\left\\lvert\\,\n\\begin{aligned}\n& \\frac{a x+y}{y+1}+\\frac{a y+x}{x+1}=a \\\\\n& a x^{2}+a y^{2}=(a-2) x y-x\n\\end{aligned}\n\\right.\n$$\nhas a unique solution.",
    "options": [],
    "answer": "a = 0 or a = -1/3",
    "solution": "Solution:\nIf $x \\neq -1$ and $y \\neq -1$, we easily get that $y = a$. Plugging it in the second equation gives\n$$\na x^{2} - (a^{2} - 2a - 1)x + a^{3} = 0\n$$\nIf $a = 0$ the system has a unique solution $(0 ; 0)$. If $a \\neq 0$, we consider the following two cases.\n\nCase 1. $-1$ is a root of $(*)$. Then $(*)$ gives that $a = 1$ or $-1$. In both cases the system has no a solution.\n\nCase 2. The equation $(*)$ has a double root. Then $a = -1$, $a = 1$ or $a = -\\frac{1}{3}$. In the first two cases we get that $y = -1$ or $x = -1$, i.e., the system has no a solutions. If $a = -\\frac{1}{3}$ the system has a unique solution $(x ; y) = \\left(\\frac{1}{3} ; -\\frac{1}{3}\\right)$.\n\nThus the desired values of $a$ are $0$ and $-\\frac{1}{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27137,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all values of $a$, for which the equation\n$$\n\\frac{2 a}{(x+1)^{2}}+\\frac{a+1}{x+1}-\\frac{2(a+1) x-(a+3)}{2 x^{2}-x-1}=0\n$$\nhas two real roots $x_{1}$ and $x_{2}$ satisfying the relation $x_{2}^{2}-a x_{1}=a^{2}-a-1$.",
    "options": [],
    "answer": "a = 3 and a = -1 - sqrt(2)",
    "solution": "Solution:\nThe given equation is equivalent to\n$$\na x^{2}+(1-2 a) x+(1-a)=0\n$$\nwhere $x \\neq -1, -\\frac{1}{2}, 1$. Hence this equation has two real roots $x_{1}$ and $x_{2}$ such that\n$$\nx_{2}^{2}-a x_{1}=a^{2}-a-1\n$$\nSince $x_{1}+x_{2}=\\frac{2 a-1}{a}$ we get that\n$$\nx_{2}^{2}+a x_{2}-a^{2}-a+2=0\n$$\nThis together with the identity\n$$\na x_{2}^{2}+(1-2 a) x_{2}+1-a=0\n$$\nimplies that\n$$\n\\left(a^{2}+2 a-1\\right) x_{2}=a^{3}+a^{2}-3 a+1=\\left(a^{2}+2 a-1\\right)(a-1)\n$$\nThe coefficient of $x_{2}$ vanishes if $a=-1 \\pm \\sqrt{2}$.\nIf $a=-1+\\sqrt{2}$, then\n$$\n(-1+\\sqrt{2}) x_{2}^{2}+(3-2 \\sqrt{2}) x_{2}+(2-\\sqrt{2})=0\n$$\nwhich is impossible, since the discriminant of this quadratic equation equals $33-24 \\sqrt{2}<0$, i.e. it has no real roots.\nIf $a=-1-\\sqrt{2}$ we get the equation\n$$\n(-1-\\sqrt{2}) x_{2}^{2}+(3+2 \\sqrt{2}) x_{2}+(2+\\sqrt{2})=0\n$$\nthat has two real roots, which are not equal to $\\pm 1$ and $-\\frac{1}{2}$.\nLet now $a \\neq -1 \\pm \\sqrt{2}$. Then $x_{2}=a-1$ and hence $a(a-1)(a-3)=0$. Since $a \\neq 0,1$ we get $a=3$. In this case the roots of the given equation are $-\\frac{1}{3}$ and $2$, and they satisfy the given condition.\nThus the desired values of $a$ are $-1-\\sqrt{2}$ and $3$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27138,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $ABCD$ be a parallelogram and let $\\angle BAD$ be acute. Denote by $E$ and $F$ the feet of the perpendiculars from the vertex $C$ to the lines $AB$ and $AD$, respectively. A circle through $D$ and $F$ is tangent to the diagonal $AC$ at a point $Q$ and a circle through $B$ and $E$ is tangent to the segment $QC$ at its midpoint $P$. Find the length of diagonal $AC$ if $AQ=1$.",
    "options": [],
    "answer": "5/3",
    "solution": "Solution:\nLet $DH \\perp AC$ ($H \\in AC$). Then $\\triangle AHD \\sim \\triangle AFC$ and $\\triangle CHD \\sim \\triangle AEC$. Hence\n$$\n\\begin{aligned}\nAC^{2} &= AH \\cdot AC + HC \\cdot AC \\\\\n&= AF \\cdot AD + AE \\cdot CD \\\\\n&= AQ^{2} + AE \\cdot AB = AQ^{2} + AP^{2}\n\\end{aligned}\n$$\nSetting $QP = PC = x$, we get the equation\n![](attached_image_1.png)\n$(1+2x)^{2} = 1 + (1+x)^{2} \\Longleftrightarrow 3x^{2} + 2x - 1 = 0$, which has a unique positive root $x = \\frac{1}{3}$. Therefore $AC = 1 + 2x = \\frac{5}{3}$.\n\n![](attached_image_1.png)",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 27139,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe dragon Spas has one head. His family tree consists of Spas, the Spas parents, their parents, etc. It is known that if a dragon has $n$ heads, then his mother has $3 n$ heads and his father has $3 n+1$ heads. A positive integer is called good if it can be written in a unique way as a sum of the numbers of the heads of two dragons from the Spas' family tree. Prove that $2003$ is a good number and find the number of the good numbers less than $2003$.",
    "options": [],
    "answer": "2003 is a good number; there are 404 good numbers less than 2003.",
    "solution": "Solution:\n\nWe shall say that the order of Spas is $1$, the order of his parents is $2$, the order of their parents is $3$, etc. Write the number of the heads of any dragon in ternary base: Spas has $1$ head, his mother $10$ heads, his father $11$ heads, etc. It follows by induction on the order of the dragons that the heads of the dragons of order $n$ are $n$-digit numbers whose ternary representation contains no digit $2$.\n\nIt follows that if two dragons have $a + b = k$ ($a > b$) heads in total and $k$ is written in ternary base, then $a$ and $b$ have $0$ at the positions where $k$ has $0$ and $1$ at the positions where $k$ has $2$. Moreover, if $k$ has $1$ at some position, then one of $a$ and $b$ has $1$ and the other one has $0$ at the same position.\n\nSo if the ternary representation of $k$ contains no $1$, then $a = b$ and hence $k$ is not a good number. If this representation contains at least two $1$'s ($k = \\ldots 1 \\ldots 1 \\ldots$), we have two possibilities: $(a = \\ldots 1 \\ldots 1 \\ldots, b = \\ldots 0 \\ldots 0 \\ldots)$ and $(a = \\ldots 1 \\ldots 0 \\ldots, b = \\ldots 0 \\ldots 1 \\ldots)$. Such a $k$ is not a good number except for the case when $k$ contains exactly two $1$'s and the other digits are $0$.\n\nIf $k$ contains exactly one $1$, then $a$ has $1$ and $b$ has $0$ at this position and the other digits of $a$ and $b$ are uniquely determined. The only exception is the case when the remaining digits of $k$ are $0$. All the numbers with one $1$ and at least one $2$ are good.\n\nSince $2003 = 2202012_{3}$, the number $2003$ is good.\n\nLet us count all good numbers with at most $7$ digits. One has $\\binom{7}{2} = 21$ numbers with exactly two $1$'s and $0$ at the other positions. There are $2^{6} - 1 = 63$ non-zero numbers with at most $6$ digits equal to $0$ or $2$. There exist $7$ possibilities to put $1$ in such numbers. So we get $7 \\cdot 63 = 441$ numbers and adding the above $21$ numbers, we obtain $462$ good numbers with at most $7$ digits.\n\nWe shall count the $7$-digit good numbers greater than $2202012_{3}$. They are $2202021_{3}, 2202100_{3}, 2202102_{3}, 2202120_{3}, 2202122_{3}, 2202201_{3}, 2202210_{3}$, $2202212_{3}, 2202221_{3}$, $16$ numbers of the form $\\overline{221mnpq}_{3}$, where $m, n, p, q \\in \\{0 ; 2\\}$ and $4 \\cdot 8 = 32$ numbers of the form $\\overline{222mnpq}_{3}$, where exactly one of the digits $m, n, p, q$ is equal to $1$. Thus, there are $1 + 9 + 16 + 32 = 58$ $7$-digit good numbers greater than $2202012_{3}$.\n\nHence we have $462 - 58 = 404$ good numbers smaller than $2003$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27140,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the number of positive integers $a$ less than $2003$, for which there exists a positive integer $n$ such that $3^{2003}$ divides $n^{3}+a$.",
    "options": [],
    "answer": "463",
    "solution": "Solution:\nWe shall prove that the desired numbers have one of the forms $9k \\pm 1$, $3^{3}(9k \\pm 1)$ or $3^{6}(9k \\pm 1)$.\n\nSuppose that $3$ does not divide $a$. Since $n^{3} \\equiv 0, \\pm 1 \\pmod{9}$, then $a \\equiv \\pm 1 \\pmod{9}$.\n\nConversely, let $a \\equiv \\pm 1 \\pmod{9}$. Since $9$ divides $1^{3}-1$ and $2^{3}+1$, then there is $n_{0}$ such that $n_{0}^{3}+a=3^{s} t$, where $s \\geq 2$ and $t$ is not divisible by $3$. We shall prove that if $n_{1}=n_{0}+2 \\cdot 3^{s-1} t$, then $3^{s+1}$ divides $n_{1}^{3}+a$. We have that\n$$\n\\left(n_{0}+2 \\cdot 3^{s-1} t\\right)^{3}+a=3^{s} t\\left(2 n_{0}^{2}+1\\right)+4 n_{0} 3^{2s-1} t^{2}+8 \\cdot 3^{3s-3} t^{3}\n$$\nSince $3$ does not divide $n_{0}$, then $2 n_{0}^{2}+1$ is divisible by $3$. Moreover, $2s-1 \\geq s+1$ and $3s-3 \\geq s+1$. Hence $n_{1}^{3}+a$ is divisible by $3^{s+1}$ but $3$ does not divide $n_{1}$. Repeating the same argument, we get a positive integer $n_{p}$ such that $3^{2003}$ divides $n_{p}^{3}+a$.\n\nLet now $3$ divides $a<2003$. Then $a=3^{s} b$, where $s \\leq 6$. Hence $n$ is divisible by $3$, i.e., $n=3^{p} n_{0}$, where $p \\geq 1$ and $3$ does not divide $n_{0}$. If $p \\geq 3$, then $3^{9}$ divides $n^{3}$ and does not divide $a$ which implies that $3^{2003}$ does not divide $n^{3}+a$. Hence $p=1$ or $p=2$ and it is easy to see that $s=3$ or $s=6$, respectively.\n\nIn the first case we get that $3^{2000}$ divides $n_{0}^{3}+b$, where $3$ does not divide $b$ and $27b<2003$. It follows as above that $b \\equiv \\pm 1 \\pmod{9}$.\n\nIn the second case we get similarly that $3^{1997}$ divides $n_{0}^{3}+b$, where $729b<2003$ and $b \\equiv \\pm 1 \\pmod{9}$.\n\nThe number of the positive integers $b \\equiv \\pm 1 \\pmod{9}$ such that $b<2003$, $27b<2003$ or $729b<2003$ equals $2 \\cdot 222+1=445$, $2 \\cdot 8+1=17$ or $1$, respectively. Hence the desired number is equal to $445+17+1=463$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27141,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $n$ be a positive integer. Find all positive integers $m$, for which there exists a polynomial $f(x) = a_{0} + a_{1} x + \\cdots + a_{n} x^{n} \\in \\mathbb{Z}[x],\\ a_{n} \\neq 0$, such that $\\gcd(a_{0}, a_{1}, \\ldots, a_{n}, m) = 1$ and $f(k)$ divides $m$ for any integer $k$.",
    "options": [],
    "answer": "All positive integers m that divide n!",
    "solution": "Solution:\n1. We shall use the following well-known facts.\n\nLEMMA 1. For any integer $x$ and any positive integer $t$ the number $t!$ divides $(x+1)(x+2) \\ldots (x+t)$.\n\nProof. The statement is obvious for $x \\in \\{0, -1, -2, \\ldots, -t\\}$. We have to prove it for $x > 0$. Let $p$ be a prime divisor of $t$. Then the power of $p$ in the prime factorization of $t!$ is $\\left[\\frac{t}{p}\\right] + \\left[\\frac{t}{p^{2}}\\right] + \\cdots$. This power does not exceed the power of $p$ in the prime factorization of $(x+1)(x+2) \\ldots (x+t) = \\frac{(x+t)!}{x!}$ since the last power equals $\\left[\\frac{x+t}{p}\\right] + \\left[\\frac{x+t}{p^{2}}\\right] + \\cdots - \\left[\\frac{x}{p}\\right] - \\left[\\frac{x}{p^{2}}\\right] - \\cdots = \\left[\\frac{x+t}{p}\\right] - \\left[\\frac{x}{p}\\right] + \\left[\\frac{x+t}{p^{2}}\\right] - \\left[\\frac{x}{p^{2}}\\right] - \\cdots$ and $[a+b] \\geq [a] + [b]$.\n\nLEMMA 2. For $g(x) \\in \\mathbb{R}[x]$ with $\\deg g = n$ one has that $g(\\mathbb{Z}) \\subset \\mathbb{Z}$ if and only if $g(x) = \\sum_{i=0}^{n} b_{i} \\binom{x}{i}$, where $b_{0}, b_{1}, \\ldots, b_{n} \\in \\mathbb{Z}$.\n\nProof. Obviously, if $g$ has the above form, then $g(\\mathbb{Z}) \\subset \\mathbb{Z}$.\nConversely, let $g(\\mathbb{Z}) \\subset \\mathbb{Z}$. Set $g(i) = \\alpha_{i} \\in \\mathbb{Z},\\ i = 0, 1, \\ldots, n$ and apply the Lagrange interpolation formula with knots $0, 1, \\ldots, n$. Then\n$$\n\\begin{aligned}\ng(x) & = \\sum_{i=1}^{n} \\frac{x(x-1) \\ldots (x-i+1)(x-i-1) \\ldots (x-n) \\alpha_{i}}{i(i-1) \\ldots (i-(i-1))(i-(i+1)) \\ldots (i-n)} \\\\\n& = \\sum_{i=1}^{n} (-1)^{n+i} \\frac{\\alpha_{i} (n+1) \\binom{x}{n+1}}{x-i}\n\\end{aligned}\n$$\nIt remains to use that any polynomial\n$$\n\\frac{(n+1) \\binom{x}{n+1}}{x-i} = \\frac{x(x-1) \\ldots (x-i+1)(x-i+1) \\ldots (x-n)}{n!}\n$$\ncan be written in the given form (compare the respective coefficients). The lemma is proved.\n\nLet now $m$ be a divisor of $n!$ and consider the polynomial $f(x) = (x+1)(x+2) \\ldots (x+n)$. It follows by Lemma 1 that $m$ divides $f(k)$ for any $k \\in \\mathbb{Z}$. Moreover, $f$ is a monic polynomial (i.e., the leading coefficient of $f$ equals $1$) and hence $\\gcd(a_{0}, a_{1}, \\ldots, a_{n}, m) = 1$. So all divisors of $n!$ are solutions of the problem.\n\nAssume that $m$ does not divide $n!$ and $m$ is a solution of the problem. Set $r = \\frac{m}{(m, n!)}$. It is clear that $r > 1$ is an integer and $(r, n!) = 1$. Let $f(x) = a_{0} + a_{1} x + \\cdots + a_{n} x^{n},\\ a_{n} \\neq 0$, be a polynomial with the desired properties. Then for $g(x) = \\frac{f(x)}{r}$ one has that $g(\\mathbb{Z}) \\subset \\mathbb{Z}$ and, by Lemma 2, $g(x) = \\sum_{i=0}^{n} b_{i} \\binom{x}{i}$, where $b_{0}, b_{1}, \\ldots, b_{n} \\in \\mathbb{Z}$.\n\nHence $f(x) = \\sum_{i=0}^{n} m b_{i} \\binom{x}{i}$ and using that $(r, i!) = 1$ for $i = 0, 1, \\ldots, n$, we get that $r$ divides all the coefficients of $f$, a contradiction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27142,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nIs there a set $A \\supset \\{1,2, \\ldots, 2004\\}$ of positive integers such that the product of its elements is equal to the sum of their squares?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nThere exists. Let us take $a_{0}=1$, $a_{i}=2004!a_{0} a_{1} \\ldots a_{i-1}-1$, $i \\geq 1$ and $A_{i}=\\{2,3, \\ldots, 2004, a_{0}, a_{1}, \\ldots, a_{i}\\}$, $i \\geq 0$. Then\n$$\n\\begin{gathered}\n\\left(\\prod_{a \\in A_{i-1}} a-\\sum_{a \\in A_{i-1}} a^{2}\\right)-\\left(\\prod_{a \\in A_{i}} a-\\sum_{a \\in A_{i}} a^{2}\\right)-1 \\\\\n\\quad=a_{i}^{2}-1-\\left(a_{i}-1\\right) \\prod_{a \\in A_{i-1}} a \\\\\n\\quad=\\left(a_{i}-1\\right)\\left(a_{i}+1-2004!a_{0} a_{1} \\ldots a_{i-1}\\right)=0\n\\end{gathered}\n$$\nHence\n$$\n\\prod_{a \\in A_{n}} a=\\sum_{a \\in A_{n}} a^{2} \\text{ for } n=\\prod_{a \\in A_{0}} a-\\sum_{a \\in A_{0}} a^{2}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27143,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all $k > 0$ such that there is a function $f : [0,1] \\times [0,1] \\to [0,1]$ satisfying the following conditions:\n\na) $f(f(x, y), z) = f(x, f(y, z))$;\nb) $f(x, y) = f(y, x)$;\nc) $f(x, 1) = x$;\nd) $f(z x, z y) = z^{k} f(x, y)$,\n\nfor any $x, y, z \\in [0,1]$.",
    "options": [],
    "answer": "k = 1 or k = 2",
    "solution": "Solution:\nNow let $x \\leq y \\leq z$, $x, y, z \\in (0,1)$. Then using (a), we get that $f\\left(x y^{k-1}, z\\right) = f\\left(x, y z^{k-1}\\right)$. Hence it follows from the above that\n$$\n\\{x y^{k-1} z^{k-1}, x^{k-1} y^{(k-1)^2} z\\} \\cap \\{x y^{k-1} z^{(k-1)^2}, x^{k-1} y z^{k-1}\\} \\neq \\varnothing\n$$\nThis easily implies that either $k=1$ and $f(x, y) = \\min\\{x, y\\}$ or $k=2$ and $f(x, y) = x y$. A direct verification shows that both functions are solutions for the respective $k$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27144,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSolve the inequality\n$$\n\\sqrt{x^{2}-1}+\\sqrt{2 x^{2}-3}+x \\sqrt{3}>0\n$$",
    "options": [],
    "answer": "(-∞, -\\frac{\\sqrt{5+\\sqrt{33}}}{2}) \\cup [\\frac{\\sqrt{6}}{2}, +∞)",
    "solution": "Solution:\nThe inequality makes sense for $x \\in \\left(-\\infty,-\\frac{\\sqrt{6}}{2}\\right] \\cup \\left[\\frac{\\sqrt{6}}{2},+\\infty\\right)$. All $x$ in the second interval are solutions of the inequality.\n\nLet $x \\in \\left(-\\infty,-\\frac{\\sqrt{6}}{2}\\right]$. Then the inequality is equivalent to\n$$\n\\sqrt{\\left(x^{2}-1\\right)\\left(2 x^{2}-3\\right)}>2 \\Longleftrightarrow 2 x^{4}-5 x^{2}-1>0\n$$\nSolving the last inequality gives $x \\in \\left(-\\infty,-\\frac{\\sqrt{5+\\sqrt{33}}}{2}\\right)$.\n\nSo, the solutions of the given inequality are\n$$\nx \\in \\left(-\\infty,-\\frac{\\sqrt{5+\\sqrt{33}}}{2}\\right) \\cup \\left[\\frac{\\sqrt{6}}{2},+\\infty\\right)\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27145,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $f(x) = x^{2} - a x + a^{2} - 4$, where $a$ is a real number. Find all $a$, for which:\n\na) the equation $f(x) = 0$ has two real roots $x_{1}$ and $x_{2}$ such that $|x_{1}^{3} - x_{2}^{3}| \\leq 4$;\n\nb) the inequality $f(x) \\geq 0$ holds for all integers $x$.",
    "options": [],
    "answer": "a) a ∈ [−4√3/3, −√5] ∪ [√5, 4√3/3].\nb) a ∈ (−∞, (−1 − √13)/2] ∪ [(1 + √13)/2, ∞).",
    "solution": "Solution:\n\na.\nUsing Vieta's formulas we get $x_{1}^{2} + x_{1} x_{2} + x_{2}^{2} = (x_{1} + x_{2})^{2} - x_{1} x_{2} = 4$. Hence $|x_{1}^{3} - x_{2}^{3}| \\leq 4 \\Longleftrightarrow |x_{1} - x_{2}| \\leq 1$. Let $D = 16 - 3 a^{2}$ be the discriminant of $f(x)$. Then $D \\geq 0$ and $|x_{1} - x_{2}| = \\sqrt{D}$. It follows that $0 \\leq 16 - 3 a^{2} \\leq 1$, and therefore $a \\in \\left[ -\\frac{4 \\sqrt{3}}{3}, -\\sqrt{5} \\right] \\cup \\left[ \\sqrt{5}, \\frac{4 \\sqrt{3}}{3} \\right]$.\n\nb.\nIf $D = 16 - 3 a^{2} \\leq 0$, i.e. $a \\in \\left( -\\infty, -\\frac{4 \\sqrt{3}}{3} \\right] \\cup \\left[ \\frac{4 \\sqrt{3}}{3}, \\infty \\right)$, then $f(x) \\geq 0$ for every $x$. If $D > 0$, then $|x_{1} - x_{2}| \\leq 1$ (since otherwise there is an integer $x \\in (x_{1}, x_{2})$, i.e., $f(x) < 0$). Hence $a \\in \\left[ -\\frac{4 \\sqrt{3}}{3}, -\\sqrt{5} \\right] \\cup \\left[ \\sqrt{5}, \\frac{4 \\sqrt{3}}{3} \\right]$ and we have to find all $a$ from these intervals such that $f(x) \\geq 0$ for every integer $x$.\nIf $f(x) < 0$ for some integer $x$, then the distance between $x$ and $\\frac{a}{2}$ is at most $\\frac{1}{2}$. Since $-\\frac{3}{2} < -\\frac{2 \\sqrt{3}}{3} \\leq \\frac{a}{2} \\leq -\\frac{\\sqrt{5}}{2} < -1$ or $\\frac{3}{2} > \\frac{2 \\sqrt{3}}{3} \\geq \\frac{a}{2} \\geq \\frac{\\sqrt{5}}{2} > 1$, we conclude that $x = \\pm 1$ and by the inequalities $f(-1) \\geq 0$ and $f(1) \\geq 0$ we get $a \\in \\left[ -\\frac{4 \\sqrt{3}}{3}, \\frac{-1 - \\sqrt{13}}{2} \\right] \\cup \\left[ \\frac{1 + \\sqrt{13}}{2}, \\frac{4 \\sqrt{3}}{3} \\right]$. In conclusion, the desired values of $a$ are\n$$\na \\in \\left( -\\infty, \\frac{-1 - \\sqrt{13}}{2} \\right] \\cup \\left[ \\frac{1 + \\sqrt{13}}{2}, +\\infty \\right)\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27146,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $f(x) = x^{4} - x^{3} + 8a x^{2} - a x + a^{2}$ and $g(y) = y^{2} - y + 6a$.\n\na) Prove that $f(x) = \\left(x^{2} - y_{1} x + a\\right)\\left(x^{2} - y_{2} x + a\\right)$, where $y_{1}$ and $y_{2}$ are the roots of the equation $g(y) = 0$.\n\nb) Find all values of $a$ such that the equation $f(x) = 0$ has four distinct positive roots.",
    "options": [],
    "answer": "(1/25, 1/24)",
    "solution": "Solution:\nSince $y_{1} + y_{2} = 1$ and $y_{1} y_{2} = 6a$, we have\n$$\n\\begin{gathered}\n\\left(x^{2} - y_{1} x + a\\right)\\left(x^{2} - y_{2} x + a\\right) \\\\\n= x^{4} - \\left(y_{1} + y_{2}\\right) x^{3} + \\left(2a + y_{1} y_{2}\\right) x^{2} - a\\left(y_{1} + y_{2}\\right) x + a^{2} = f(x)\n\\end{gathered}\n$$\n\nb) The roots of $f(x) = 0$ are the roots $x_{1}, x_{2}$ and $x_{3}, x_{4}$ of the equations\n$$\nx^{2} - y_{1} x + a = 0 \\text{ and } x^{2} - y_{2} x + a = 0\n$$\nrespectively. It is easy to see that $x_{1}, x_{2}, x_{3}$ and $x_{4}$ are real, distinct and positive exactly when the following three conditions are simultaneously satisfied:\n\n1) $a \\neq 0, y_{1} \\neq y_{2}$ are real $\\Longleftrightarrow a \\neq 0, D = 1 - 24a > 0 \\Longleftrightarrow a \\neq 0$, $a < \\frac{1}{24}$;\n\n2) $D_{1} = y_{1}^{2} - 4a > 0$ and $D_{2} = y_{2}^{2} - 4a > 0 \\Longleftrightarrow y_{1} > 10a$ and $y_{2} > 10a$;\n\n3) $x_{1} + x_{2} = y_{1} > 0, x_{1} x_{2} = a > 0, x_{3} + x_{4} = y_{2} > 0$ and $x_{3} x_{4} = a > 0$ $\\Longleftrightarrow y_{1} > 0, y_{2} > 0$ and $a > 0$.\n\nThese conditions are equivalent to $0 < a < \\frac{1}{24}, y_{1} > 10a$ and $y_{2} > 10a$, which is the same as $0 < a < \\frac{1}{24}, g(10a) > 0$ and $10a < \\frac{1}{2} \\Longleftrightarrow 0 < a < \\frac{1}{24}$, $4a(25a - 1) > 0$ and $a < \\frac{1}{20}$.\n\nHence the required values of $a$ are $a \\in \\left(\\frac{1}{25}, \\frac{1}{24}\\right)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27147,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABCDE$ be a cyclic pentagon with $AC \\parallel DE$. Denote by $M$ the midpoint of $BD$. If $\\Varangle AMB = \\Varangle BMC$, prove that $BE$ bisects $AC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $BE$ meet $AC$ at point $N$ and $P$ be the midpoint of $AB$. Set $\\Varangle BAC = \\Varangle BDC = \\alpha$, $\\Varangle ABE = \\Varangle CBD = \\beta$ and $\\Varangle ADB = \\Varangle ACB = \\gamma$. Then $\\triangle ABN \\sim \\triangle DBC$, and we conclude that $\\triangle BPN \\sim \\triangle BMC$.\n\nLet $\\Varangle AMB = \\Varangle BMC = \\varphi$. We have from the above that $\\Varangle BPN = \\varphi$.\n\n![](attached_image_1.png)\n\nWe shall use the following fact: if two chords of a circle bisect a third one and determine equal angles with it, then they are equal and their intersection point divides them into respectively equal parts (use congruent triangles or symmetry through a line). Let the ray $AM \\rightarrow$ meet the circle at point $F$. Then $CM = FM$ and therefore $\\triangle BMC \\cong \\triangle DMF$. Hence we have $BC = DF$ and $\\Varangle MAD = \\Varangle BDC = \\alpha$.\n\nIt follows from $\\triangle AMD$ that $\\varphi = \\alpha + \\gamma$ and using $\\triangle APN$ we get $\\Varangle ANP = \\varphi - \\alpha = \\gamma = \\Varangle ACB$. Hence $NP \\parallel BC$, which means that $N$ is the midpoint of $AC$. This completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27148,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $ABCD$ be a cyclic quadrilateral. Denote by $I$ and $J$ the incenters of $\\triangle ABD$ and $\\triangle BCD$. Prove that $ABCD$ is a circumscribed quadrilateral if and only if the points $A$, $I$, $J$ and $C$ are either collinear or concyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nIt is easy to see that if the points $A$, $I$, $J$ and $C$ are collinear, then $AB = AD$ and $BC = CD$. Hence $ABCD$ is a circumscribed quadrilateral.\n\nSuppose that the points $A$, $I$, $J$ and $C$ are concyclic. Since $\\angle AIC > \\angle AIB$ or $\\angle AIC > \\angle AID$, it follows that $\\angle AIC > 90^\\circ$. Analogously $\\angle AJC > 90^\\circ$ and therefore $\\angle AIC + \\angle AJC > 180^\\circ$. It follows that the points $I$ and $J$ are on the same side of the line $AC$.\n\n![](attached_image_1.png)\n\nLet $S = AI \\cap CJ$ and let the lines $AI$ and $CJ$ meet the circumcircle of $ABCD$ at points $P$ and $Q$, respectively. Since $P$ and $Q$ are the midpoints of the arcs $BCD$ and $BAD$, respectively, it follows that $PQ \\perp BD$.\n\nOn the other hand, $\\angle SJI = \\angle CAI = \\angle CQP$, which implies that $IJ \\parallel PQ$. Therefore $IJ \\perp BD$, which means that the incircles of $\\triangle ABD$ and $\\triangle BCD$ are tangent to each other at a point $T \\in BD$. Then\n$$\nDT = \\frac{AD + BD - AB}{2} = \\frac{BD + CD - BC}{2} \\Longleftrightarrow AB + CD = BC + AD\n$$\ni.e. $ABCD$ is a circumscribed quadrilateral.\n\nConversely, assume that $ABCD$ is a circumscribed quadrilateral. Note that if $I \\in AC$, then $J \\in AC$. Suppose that $I \\notin AC$. It follows from the equality $AB + CD = BC + AD$ that the incircles of $\\triangle ABD$ and $\\triangle BCD$ are tangent to each other at a point of $BD$. Hence $IJ \\perp BD \\Rightarrow IJ \\parallel PQ \\Rightarrow \\angle SJI = \\angle CQP = \\angle CAI$, and therefore $ACJI$ is a cyclic quadrilateral.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27149,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $M$ be the centroid of $\\triangle ABC$. Prove that:\n\na) $\\cot \\angle AMB = \\frac{BC^2 + CA^2 - 5 AB^2}{12 S_{ABC}}$.\n\nb) $\\cot \\angle AMB + \\cot \\angle BMC + \\cot \\angle CMA \\leq -\\sqrt{3}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nWe shall use the standard notation for the elements of $\\triangle ABC$.\n\na) The Cosine theorem for $\\triangle AMB$ gives\n$$\n\\cos \\angle AMB = \\frac{AM^2 + BM^2 - AB^2}{2 \\cdot AM \\cdot MB}\n$$\nThis together with the equality $\\sin \\angle AMB = \\frac{2S}{3 AM \\cdot MB}$ and the median formula implies that\n$$\n\\cot \\angle AMB = \\frac{3\\left(AM^2 + BM^2 - AB^2\\right)}{4S} = \\frac{a^2 + b^2 - 5c^2}{12S}\n$$\n\nb) It follows as above that\n$$\n\\cot \\angle BMC = \\frac{b^2 + c^2 - 5a^2}{12S}, \\quad \\cot \\angle CMA = \\frac{c^2 + a^2 - 5b^2}{12S}.\n$$\nHence\n$$\n\\cot \\angle AMB + \\cot \\angle BMC + \\cot \\angle CMA = -\\frac{a^2 + b^2 + c^2}{4S}.\n$$\nIt remains to show that $a^2 + b^2 + c^2 \\geq 4S \\sqrt{3}$. The Heron formula and the Arithmetic mean–Geometric mean inequality give\n$$\nS^2 = p(p-a)(p-b)(p-c) \\leq p\\left(\\frac{p-a + p-b + p-c}{3}\\right)^3 = \\frac{p^4}{27}\n$$\nTherefore\n$$\nS \\leq \\frac{p^2}{3\\sqrt{3}} = \\frac{(a+b+c)^2}{12\\sqrt{3}} \\leq \\frac{a^2 + b^2 + c^2}{4\\sqrt{3}}\n$$\nby the Root mean square inequality.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27150,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the largest positive integer $n$ for which there exists a set $\\{a_{1}, a_{2}, \\ldots, a_{n}\\}$ of composite positive integers with the following properties:\n(i) any two of them are coprime;\n(ii) $1 < a_{i} \\leq (3n+1)^{2}$ for $i = 1, \\ldots, n$.",
    "options": [],
    "answer": "14",
    "solution": "Solution:\nSuppose that $n$ has the required property. For every $j = 1, 2, \\ldots, n$ denote by $q_{j}$ the least prime divisor of $a_{j}$ and let $q = \\max_{1 \\leq i \\leq n} q_{i}$. Without loss of generality we may assume that $q = q_{1}$. Then\n$$\n(3n+1)^{2} \\geq a_{1} \\geq q_{1}^{2} \\geq p_{n}^{2}\n$$\nwhere $p_{n}$ is the $n$-th prime number. Therefore we have $p_{n} \\leq 3n+1$. It is easy to show (by induction) that $p_{n} > 3n+1$ for every $n \\geq 15$. Hence $n \\leq 14$. Since the set $\\{2^{2}, 3^{2}, 5^{2}, \\ldots, p_{14}^{2}\\}$ has the required properties, we conclude that $n = 14$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27151,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn a school there are $m$ boys and $j$ girls, $m \\geq 1, 1 \\leq j < 2004$.\nEvery student has sent a post card to every student. It is known that the number of the post cards sent by the boys is equal to the number of the post cards sent by girl to girl. Find all possible values of $j$.",
    "options": [],
    "answer": "j ∈ {2, 10, 65, 442}",
    "solution": "Solution:\n\nIt follows by the given condition that $m(m+j-1)=j(j-1)$ and hence $m^{2}=(j-m)(j-1)$. If $p$ is a prime divisor of $j-m$ and $j-1$, then it divides $m$. Therefore $p$ divides $j$ and $1$. This contradiction shows that $j-m$ and $j-1$ are coprime. Then $j-m=u^{2}$ and $j-1=v^{2}$, where $u$ and $v$ are non-negative integers. It follows that $uv=m$ and $u^{2}+uv=j=v^{2}+1$, and the condition $1 \\leq j < 2004$ gives $0 \\leq v \\leq 44$. So, we have to solve in non-negative integers the equation\n$$\nu^{2}+uv=v^{2}+1$$\nIf $v=0$, then $u=1$. Assume that the pair $(u_{0}; v_{0})$ is a solution of $(*)$ and $v_{0} \\geq 1$. Then $u_{0} \\geq 1$. Since $u_{0}v_{0} \\geq 1$, it follows that $u_{0} \\leq v_{0}$. Moreover, it is clear that $v_{0}<2u_{0}$. Set $v_{1}=v_{0}-u_{0}$, $0 \\leq v_{1}<v_{0}$. We have that $u_{0}^{2}=v_{0}(v_{0}-u_{0})+1=(u_{0}+v_{1})v_{1}+1=u_{0}v_{1}+v_{1}^{2}+1$. Set $u_{1}=u_{0}-v_{1}=2u_{0}-v_{0}>0$. Then\n$$\n(u_{1}+v_{1})^{2}=(u_{1}+v_{1})v_{1}+v_{1}^{2}+1 \\Longleftrightarrow u_{1}^{2}+u_{1}v_{1}=v_{1}^{2}+1\n$$\ni.e., we obtain a new solution of $(*)$. If $v_{1}=0$, then $u_{1}=1$. If $v_{1} \\geq 1$, then $u_{1} \\geq 1$. Setting $v_{2}=v_{1}-u_{1}<v_{1}$ and $u_{2}=u_{1}-v_{2}$, we get in the same way a new solution. So, we get a sequence of non-negative integers $v_{0}>v_{1}>\\cdots$. If $v_{k}=0$ for some $k$, then $u_{k}=1$ and writing $u_{k-1}, v_{k-1}, \\ldots, u_{0}, v_{0}$, we get the Fibonacci sequence.\n\nThus solutions of $(*)$ are $(u; v)=(1; 0), (1; 1), (2; 3), (5; 8), (13; 21)$, and for the other solutions we have that $v>44$. The first solution of $(*)$ gives $j=v^{2}+1=0^{2}+1=1$ and then $m=uv=0$, which does not satisfy the given condition. The other solutions give the following solutions of the problem: $j=1^{2}+1=2$, $j=3^{2}+1=10$, $j=8^{2}+1=65$ and $j=21^{2}+1=442$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27152,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nConsider the function\n$$\nf(x) = (a^{2} + 4a + 2)x^{3} + (a^{3} + 4a^{2} + a + 1)x^{2} + (2a - a^{2})x + a^{2}\n$$\nwhere $a$ is a real parameter.\n\na) Prove that $f(-a) = 0$.\n\nb) Find all values of $a$ such that the equation $f(x) = 0$ has three different positive roots.",
    "options": [],
    "answer": "a ∈ (-2 - √2, -3) ∪ (-3, (-11 - √137)/8) ∪ (-1, -2 + √2)",
    "solution": "Solution:\n\na) It follows by a direct verification.\n\nb) Writing the equation in the form\n$$\n(x + a)\\left((a^{2} + 4a + 2)x^{2} + (1 - a)x + a\\right) = 0\n$$\nwe get that $a < 0$. Moreover, the quadratic polynomial in (3) must have two distinct real zeros, i.e.\n$$\nD = (1 - a)^{2} - 4a(a^{2} + 4a + 2) > 0 \\Longleftrightarrow (a + 1)(-4a^{2} - 11a + 1) > 0.\n$$\nSolving this inequality and having in mind that $a < 0$, we get that\n$$\na \\in \\left(-\\infty, \\frac{-11 - \\sqrt{137}}{8}\\right) \\cup (-1, 0)\n$$\nThe roots of the quadratic polynomial in (1) are positive if and only if\n$$\n\\frac{a}{a^{2} + 4a + 2} > 0 \\quad \\text{and} \\quad -\\frac{1 - a}{a^{2} + 4a + 2} > 0.\n$$\nThen $a \\in (-2 - \\sqrt{2}, -2 + \\sqrt{2})$ and using (4), we obtain that\n$$\na \\in \\left(-2 - \\sqrt{2}, \\frac{-11 - \\sqrt{137}}{8}\\right) \\cup (-1, -2 + \\sqrt{2})\\text{.}\n$$\nIt remains to see when $-a$ is a zero of the quadratic polynomial in (3). We have that\n$$\n(a^{2} + 4a + 2)(-a)^{2} + (1 - a)(-a) + a = 0 \\Longleftrightarrow a^{2}(a^{2} + 4a + 3) = 0\n$$\nand hence $a = -3, -1, 0$. So, the answer of the problem is\n$$\na \\in (-2 - \\sqrt{2}, -3) \\cup \\left(-3, \\frac{-11 - \\sqrt{137}}{8}\\right) \\cup (-1, -2 + \\sqrt{2})\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27153,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $O$ and $G$ be respectively the circumcenter and the centroid of $\\triangle ABC$ and let $M$ be the midpoint of the side $AB$. If $OG \\perp CM$, prove that $\\triangle ABC$ is isosceles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nSet $\\vec{a}=\\overrightarrow{OA}$, $\\vec{b}=\\overrightarrow{OB}$, $\\vec{c}=\\overrightarrow{OC}$. We have that $\\overrightarrow{OM}=\\frac{1}{2}(\\vec{a}+\\vec{b})$ and hence\n$$\n\\overrightarrow{OG}=\\frac{1}{6}(3\\vec{a}+\\vec{b}+2\\vec{c})\n$$\nOn the other hand, $\\overrightarrow{CM}=\\frac{1}{2}(\\vec{a}+\\vec{b}-2\\vec{c})$. Then\n![](attached_image_1.png)\n$$\n\\begin{aligned}\n0 &=\\overrightarrow{OG} \\cdot \\overrightarrow{CM}=(3\\vec{a}+\\vec{b}+2\\vec{c})(\\vec{a}+\\vec{b}-2\\vec{c}) \\\\\n&=3R^{2}+3\\vec{a}\\vec{b}-6\\vec{a}\\vec{c}+\\vec{a}\\vec{b}+R^{2}-2\\vec{b}\\vec{c}+2\\vec{a}\\vec{c}+2\\vec{b}\\vec{c}-4R^{2}\n\\end{aligned}\n$$\nand therefore $0=4\\vec{a}(\\vec{b}-\\vec{c})$. Hence $OA \\perp BC$, i.e. $AB=AC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27154,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nProve that any graph with $10$ vertices and $26$ edges contains at least $4$ triangles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nDenote by $V$ and $E$ the sets of the vertices and the edges of $G$, respectively. For any vertex $x \\in V$, let $\\Gamma(x)$ be the set of the edges of $G$ which are adjacent to $x$ and let $d(x) = |\\Gamma(x)|$. Then for $x, y \\in V$ one has that\n$$\n|\\Gamma(x) \\cap \\Gamma(y)| = |\\Gamma(x)| + |\\Gamma(y)| - |\\Gamma(x) \\cup \\Gamma(y)| \\geq d(x) + d(y) - |V|.\n$$\nSumming up these inequalities for all the edges $(x, y) \\in E$, we get that\n$$\n\\begin{aligned}\n3 t(G) & = \\sum_{(x, y) \\in E} |\\Gamma(x) \\cap \\Gamma(y)| \\geq \\sum_{(x, y) \\in E} (d(x) + d(y)) - |V| \\cdot |E| \\\\\n& = \\sum_{x \\in V} d^{2}(x) - |V| \\cdot |E|\n\\end{aligned}\n$$\n(here $t(G)$ is the number of the triangles in $G$). Hence\n$$\n3 t(G) \\geq \\frac{1}{|V|} \\left( \\sum_{x \\in V} d(x) \\right)^{2} - |V| \\cdot |E| = \\frac{4|E|^{2}}{|V|} - |V| \\cdot |E|.\n$$\nIn our case we have that $|V| = 10$ and $|E| = 26$. Therefore $t(G) \\geq \\frac{52}{15}$ and hence $t(G) \\geq 4$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 27155,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nProve that if $a, b, c \\geq 1$ and $a+b+c=9$, then\n$$\n\\sqrt{a b+b c+c a} \\leq \\sqrt{a}+\\sqrt{b}+\\sqrt{c}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nSet $a=\\frac{9 x^{2}}{x^{2}+y^{2}+z^{2}}$, $b=\\frac{9 y^{2}}{x^{2}+y^{2}+z^{2}}$, $c=\\frac{9 z^{2}}{x^{2}+y^{2}+z^{2}}$, where $x, y, z>0$ and $x+y+z=1$. We have to prove that\n$$\nx^{2}+y^{2}+z^{2} \\geq 9\\left(x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}\\right)\n$$\nSince $a \\geq 1$, then\n$$\n9 x^{2}=a\\left(x^{2}+y^{2}+z^{2}\\right) \\geq x^{2}+y^{2}+z^{2} \\geq \\frac{(x+y+z)^{2}}{3}=\\frac{1}{3}\n$$\nthat is, $x \\geq \\frac{1}{3 \\sqrt{3}}$. Analogously $y \\geq \\frac{1}{3 \\sqrt{3}}$ and $z \\geq \\frac{1}{3 \\sqrt{3}}$.\nWe may assume that $x \\geq y \\geq z$. Then\n$$\n\\frac{1}{3 \\sqrt{3}} \\leq z \\leq \\frac{1}{3}, \\quad \\frac{1}{3} \\leq \\frac{x+y}{2}=\\frac{1-z}{2} \\leq \\frac{3 \\sqrt{3}-1}{6 \\sqrt{3}}\n$$\nLet $f(x, y, z)=x^{2}+y^{2}+z^{2}-9\\left(x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}\\right)$. Then $f(x, y, z)-f\\left(\\frac{x+y}{2}, \\frac{x+y}{2}, z\\right)=\\frac{(x-y)^{2}}{2}\\left[1-9 z^{2}+\\frac{9}{8}\\left((x+y)^{2}+4 x y\\right)\\right] \\geq 0$.\nIt is enough to prove that $f(t, t, 1-2 t) \\geq 0$, where $t=\\frac{x+y}{2}$. One has that\n$$\n\\begin{aligned}\nf(t, t, 1-2 t) & =2 t^{2}+(1-2 t)^{2}-9\\left(t^{4}+2 t^{2}(1-2 t)^{2}\\right) \\\\\n& =\\frac{(p-1)^{2}}{3}\\left(3+2 p-3 p^{2}\\right) \\geq 0\n\\end{aligned}\n$$\nwhere $p=3 t\\left(1 \\leq p \\leq \\frac{3 \\sqrt{3}-1}{2 \\sqrt{3}}\\right)$, i.e., $3+2 p-3 p^{2}>0 \\Longleftrightarrow \\frac{1-\\sqrt{10}}{3}<p<\\frac{1+\\sqrt{10}}{3}$. It remains to check that $\\frac{3 \\sqrt{3}-1}{2 \\sqrt{3}}<\\frac{1+\\sqrt{10}}{3}$.\nIn fact, we have proved the inequality under the weaker conditions $a+b+c=9$ and $a, b, c \\geq \\frac{89-28 \\sqrt{10}}{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27156,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all values of $a$ such that the equation\n$$\n4^{x} - (a^{2} + 3a - 2) 2^{x} + 3a^{3} - 2a^{2} = 0\n$$\nhas a unique solution.",
    "options": [],
    "answer": "(-∞, 0) ∪ (0, 2/3] ∪ {1, 2}",
    "solution": "Solution:\nSetting $y = 2^{x}$, we have to find all values of $a$ such that the equation\n$$\ny^{2} - (a^{2} + 3a - 2) y + 3a^{3} - 2a^{2} = 0 \\Longleftrightarrow (y - a^{2})(y - 3a + 2) = 0\n$$\nhas exactly one positive root.\n\nObviously $a = 0$ is not a solution. For $a \\neq 0$ the equation has a positive root $y_{1} = a^{2}$. It is unique if either $y_{2} = 3a - 2 \\leq 0$ or $y_{2} = y_{1}$.\n\nIn the first case we obtain $a \\leq \\frac{2}{3}$ and in the second one we have $a^{2} = 3a - 2$, whence $a_{1} = 1$ and $a_{2} = 2$.\n\nFinally, $a \\in (-\\infty, 0) \\cup (0, \\frac{2}{3}) \\cup \\{1\\} \\cup \\{2\\}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27157,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all real numbers $a$ such that the equation\n$$\n\\log_{4 a x}(x-3 a)+\\frac{1}{2} \\log_{x-3 a} 4 a x=\\frac{3}{2}\n$$\nhas exactly two solutions.",
    "options": [],
    "answer": "(-∞, -1/2) ∪ (-1/2, 0) ∪ (0, 1/6) ∪ (1/6, 1/4)",
    "solution": "Solution:\nThe equation is defined if\n$$\n4 a x > 0,\\ 4 a x \\neq 1,\\ x-3 a > 0,\\ x-3 a \\neq 1\n$$\nSetting $t = \\log_{4 a x}(x-3 a)$ we get the equation $t + \\frac{1}{2 t} = \\frac{3}{2}$ with roots $t_1 = 1$ and $t_2 = \\frac{1}{2}$.\nIf $\\log_{4 a x}(x-3 a) = 1$, then $x_1 = \\frac{3 a}{1-4 a}$, $a \\neq \\frac{1}{4}$.\nIf $\\log_{4 a x}(x-3 a) = \\frac{1}{2}$ we obtain the equation $x^2 - 6 a x + 9 a^2 = 4 a x$ with roots $x_2 = 9 a$ and $x_3 = a$.\nHence we have to find all $a$ for which exactly two of the numbers $x_1 = \\frac{3 a}{1-4 a}\\ (a \\neq \\frac{1}{4})$, $x_2 = 9 a$ and $x_3 = a$ are different solutions of the given equation. It is clear that $a \\neq 0$. We shall consider two cases.\n\n1. Let $a > 0$. Then $x_3 - 3 a = -2 a < 0$, which implies that $x_1$ and $x_2$ have to be the solutions. Since $x_1 = \\frac{3 a}{1-4 a}$ satisfies (*), it follows that $4 a x_1 = \\frac{12 a^2}{1-4 a} > 0$, $4 a x_1 = \\frac{12 a^2}{1-4 a} \\neq 1$, $x_1 - 3 a = \\frac{3 a}{1-4 a} - 3 a > 0$ and $x_1 - 3 a = \\frac{3 a}{1-4 a} - 3 a \\neq 1$. This implies that $a < \\frac{1}{4}$ and $a \\neq \\frac{1}{6}$. It is easy to check that if $a \\neq \\frac{1}{6}$, then $x_1 \\neq x_2$ and $x_2$ satisfies (*). So the desired numbers $a$ in this case are\n$$\na \\in \\left(0, \\frac{1}{6}\\right) \\cup \\left(\\frac{1}{6}, \\frac{1}{4}\\right)\n$$\n\n2. Let $a < 0$. Then $x_2 - 3 a = 6 a < 0$, which implies that $x_1$ and $x_3$ have to be the two solutions. Since $x_1 = \\frac{3 a}{1-4 a}$ again satisfies $(*)$, we find as in the first case that $a \\neq -\\frac{1}{2}$. Now it is easy to check that $x_1 \\neq x_3$ and $x_3$ satisfies $(*)$. Thus\n$$\na \\in \\left(-\\infty, -\\frac{1}{2}\\right) \\cup \\left(-\\frac{1}{2}, 0\\right)\n$$\n\nCombining both cases, it follows that the answer of the problem is\n$$\na \\in \\left(-\\infty, -\\frac{1}{2}\\right) \\cup \\left(-\\frac{1}{2}, 0\\right) \\cup \\left(0, \\frac{1}{6}\\right) \\cup \\left(\\frac{1}{6}, \\frac{1}{4}\\right)\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27158,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all values of $x \\in (-\\pi, \\pi)$ such that the numbers $2^{\\sin x}$, $2-2^{\\sin x+\\cos x}$ and $2^{\\cos x}$ are consecutive terms of a geometric progression.",
    "options": [],
    "answer": "x = -π/4 or x = 3π/4",
    "solution": "Solution:\nThe numbers form a geometric progression if and only if\n$$\n2^{\\sin x} \\cdot 2^{\\cos x} = \\left(2 - 2^{\\sin x+\\cos x}\\right)^2\n$$\nThis is equivalent to $4^{\\sin x+\\cos x} - 5 \\cdot 2^{\\sin x+\\cos x} + 4 = 0$. Setting $y = 2^{\\sin x+\\cos x}$ gives $y^2 - 5y + 4 = 0$, i.e., $y_1 = 4$ or $y_2 = 1$.\n\nIf $2^{\\sin x+\\cos x} = 4$, then $\\sin x + \\cos x = 2$, i.e., $\\sin \\left(x + \\frac{\\pi}{4}\\right) = \\sqrt{2}$, which is impossible.\n\nIf $2^{\\sin x+\\cos x} = 1$, then $\\sin x + \\cos x = 0$, i.e., $\\sin \\left(x + \\frac{\\pi}{4}\\right) = 0$. Hence $x = -\\frac{\\pi}{4} + k\\pi$, where $k$ is an integer. Since $x \\in (-\\pi, \\pi)$, we get finally that $x = -\\frac{\\pi}{4}$ and $x = \\frac{3\\pi}{4}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27159,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $AA_{1}$, $BB_{1}$ and $CC_{1}$ be the altitudes of an acute $\\triangle ABC$ ($A_{1} \\in BC$, $B_{1} \\in CA$ and $C_{1} \\in AB$). Denote by $O$ the circumcenter of $\\triangle ABC$, and by $H_{1}$ the orthocenter of $\\triangle A_{1}B_{1}C_{1}$. Prove that the midpoint of the segment $OH_{1}$ coincides with the incenter of the triangle with vertices at the midpoints of the sides of $\\triangle A_{1}B_{1}C_{1}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nDenote by $H$ the orthocenter of $\\triangle ABC$, and by $G_{1}$ the centroid of $\\triangle A_{1}B_{1}C_{1}$. Let $O_{1}$ be the midpoint of the segment $OH$. It is well-known that $O_{1}$ is the circumcenter of $\\triangle A_{1}B_{1}C_{1}$, and $H$ is its incenter. Then $\\overrightarrow{H_{1}G_{1}} = 2\\, \\overrightarrow{G_{1}O_{1}}$.\n\nNote that the dilation with center $G_{1}$ and ratio $-\\frac{1}{2}$ maps $\\triangle A_{1}B_{1}C_{1}$ into $\\triangle A_{2}B_{2}C_{2}$ formed by the midpoints of the segments $B_{1}C_{1}$, $A_{1}C_{1}$ and $A_{1}B_{1}$. Hence the image of $H$ under this dilation is the incenter $I_{2}$ of $\\triangle A_{2}B_{2}C_{2}$. Since\n\n![](attached_image_1.png)\n\n$G_{1}$ is the centroid of $\\triangle OHH_{1}$, it follows that $I_{2}$ is the midpoint of the segment $OH_{1}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27160,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nThe lines through the vertices $A$ and $B$ that are tangent to the circumcircle of an acute $\\triangle ABC$ meet at a point $D$. If $M$ is the midpoint of the side $AB$, prove that $\\Varangle ACM = \\Varangle BCD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nThe Sine theorem for $\\triangle AMC$ and $\\triangle BMC$ gives\n$$\n\\frac{AM}{CM} = \\frac{\\sin \\Varangle ACM}{\\sin \\alpha}, \\quad \\frac{BM}{CM} = \\frac{\\sin (\\gamma - \\Varangle ACM)}{\\sin \\beta}\n$$\nSince $AM = BM$, we get that $\\frac{\\sin \\Varangle ACM}{\\sin \\alpha} = \\frac{\\sin (\\gamma - \\Varangle ACM)}{\\sin \\beta}$ and therefore\n$$\n\\tan \\Varangle ACM = \\frac{\\sin \\alpha \\sin \\gamma}{\\sin \\beta + \\sin \\alpha \\cos \\gamma}\n$$\nThe Sine theorem for $\\triangle ADC$ and $\\triangle BDC$ implies that\n$$\n\\begin{aligned}\n\\frac{AD}{CD} & = \\frac{\\sin (\\gamma - \\Varangle BCD)}{\\sin (\\alpha + \\gamma)} \\\\\n\\frac{BD}{CD} & = \\frac{\\sin \\Varangle BCD}{\\sin (\\beta + \\gamma)}\n\\end{aligned}\n$$\nThen the equality $AD = BD$ shows that\n![](attached_image_1.png)\n$\\frac{\\sin (\\gamma - \\Varangle BCD)}{\\sin \\beta} = \\frac{\\sin \\Varangle BCD}{\\sin \\alpha}$, i.e.\n$$\n\\tan \\Varangle BCD = \\frac{\\sin \\alpha \\sin \\gamma}{\\sin \\beta + \\sin \\alpha \\cos \\gamma}.\n$$\nIt follows by (1) and (2) that $\\tan \\Varangle ACM = \\tan \\Varangle BCD$. Since these angles are acute, they are equal.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27161,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $m \\geq 3$ and $n \\geq 2$ be integers. Prove that in a group of $N = m n - n + 1$ people such that there are two familiar people among any $m$, there is a person who is familiar with $n$ people. Does the statement remain true if $N < m n - n + 1$?",
    "options": [],
    "answer": "No",
    "solution": "Solution:\n\nConsider a group with maximal number of people such that any two of them are not familiar. It is clear that if there are $l$ people in this group, then $l \\leq m-1$. Moreover, the maximality of $l$ implies that any of the other $N-l$ people is familiar to at least one of the $l$ people in the group. Hence some of these $l$ people is familiar to at least $\\frac{N-l}{l}$ people. Since\n$$\n\\frac{N-l}{l} = \\frac{N}{l} - 1 \\geq \\frac{N}{m-1} - 1 = n + \\frac{1}{m-1} - 1 > n-1\n$$\nthere is a person who is familiar to $n$ people.\n\nLet $N = m n - n = (m-1) n$ and consider $m-1$ groups by $n$ people such that any two people from one group are familiar and there are no familiar people from different groups. Then among any $m$ people there are two from one and the same group, i.e. they are familiar. On the other hand, any of the people is familiar to $n-1$ of the other and hence the statement is not true if $N < m n - n + 1$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 27162,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $k$ be an integer, $1 < k < 100$. For every permutation $a_{1}, a_{2}, \\ldots, a_{100}$ of the integers $1, 2, \\ldots, 100$, set $a_{101} = 0$ and choose the least integer $m > k$ such that $a_{m}$ is less than at least $k-1$ of the numbers $a_{1}, a_{2}, \\ldots, a_{k}$. Find all $k$ for which the number of permutations with $a_{m} = 1$, is equal to $\\frac{100!}{4}$.\nPeter Boyvalenkov, Emil Kolev, Nikolai Nikolov",
    "options": [],
    "answer": "45 and 55",
    "solution": "Solution:\nConsider the more general problem for $n+1$ numbers instead of $100$ and denote by $p_{k, n+1}$ the probability that $a_{m}$ be equal to $1$. If $1$ is not the last number of a permutation of the positive integers $1, 2, \\ldots, n+1$, then $a_{m} = 1$ with probability $p_{k, n}$ (we assume that $p_{n, n} = 0$). Otherwise, $a_{m} = 1$ only if $2$ and $3$ are among the first $k$ numbers of the permutation. Then\n$$\np_{k, n+1} = \\frac{n}{n+1} p_{k, n} + \\frac{1}{n+1} \\cdot \\frac{k}{n} \\cdot \\frac{k-1}{n-1}\n$$\nand hence\n$$\n\\begin{aligned}\n& (n+1) p_{k, n+1} = \\sum_{j=k}^{n} \\left((j+1) p_{k, j+1} - j p_{k, j}\\right) \\\\\n& = k(k-1) \\sum_{j=k}^{n} \\frac{1}{j(j-1)} = \\frac{k(n+1-k)}{n}\n\\end{aligned}\n$$\nIt follows that $p_{k, n} = \\frac{k(n-k)}{n(n-1)}$ and therefore $p_{k, 100} = \\frac{1}{4}$ for $k = 45$ and $k = 55$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 27163,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $a > 1$ be a positive integer. The sequence $a_{1}, a_{2}, \\ldots, a_{n}, \\ldots$ is defined by $a_{1} = 1$, $a_{2} = a$ and $a_{n+2} = a \\cdot a_{n+1} - a_{n}$ for $n \\geq 1$. Prove that the prime factors of its terms are infinitely many.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nWe shall prove by induction on $m$ that $a_{n+m} = a_{m} a_{n+1} - a_{m-1} a_{n}$ for every two positive integers $n, m \\geq 2$. For $m = 2$ this is the given recurrence relation. If for some $m \\geq 2$ the equality is satisfied for every $n$, then\n$$\n\\begin{aligned}\na_{m+1+n} & = a_{m+(n+1)} = a_{m} a_{n+2} - a_{m-1} a_{n+1} = \\\\\n& = a_{m} \\left(a a_{n+1} - a_{n}\\right) - a_{m-1} a_{n+1} = \\\\\n& = \\left(a a_{m} - a_{m-1}\\right) a_{n+1} - a_{m} a_{n} = a_{m+1} a_{n+1} - a_{m} a_{n}\n\\end{aligned}\n$$\nwhich completes the induction.\n\nThe recurrence relation shows that $\\gcd\\left(a_{n}, a_{n-1}\\right) = 1$ for every $n \\geq 2$. This and $a_{n+m} = a_{m} a_{n+1} - a_{m-1} a_{n}$ imply that $\\gcd\\left(a_{m+n}, a_{m}\\right) = \\gcd\\left(a_{m}, a_{n}\\right)$. Using induction again we conclude that for every two positive integers $m$ and $n$ we have $\\gcd\\left(a_{m}, a_{n}\\right) = a_{\\gcd(m, n)}$. Now the assertion follows immediately: if $1 < n_{1} < n_{2} < \\cdots < n_{k} < \\cdots$ is an infinite sequence of relatively prime integers then $\\gcd\\left(a_{n_{i}}, a_{n_{j}}\\right) = a_{\\gcd\\left(n_{i}, n_{j}\\right)} = a_{1} = 1$, i.e., $a_{n_{1}}, a_{n_{2}}, \\ldots, a_{n_{k}}, \\ldots$ are relatively prime. Therefore the set of their prime factors is infinite.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27164,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe points $D$ and $E$ lie respectively on the perpendicular bisectors of the sides $AB$ and $BC$ of $\\triangle ABC$. It is known that $D$ is an interior point for $\\triangle ABC$, $E$ does not and $\\Varangle ADB = \\Varangle CEB$. If the line $AE$ meets the segment $CD$ at a point $O$, prove that the areas of $\\triangle ACO$ and the quadrilateral $DBEO$ are equal.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSince the isosceles triangles $ABD$ and $CBE$ are similar, we have $\\frac{AB}{DB} = \\frac{CB}{BE}$. Using that $\\Varangle ABC = \\Varangle DBE$, we get that $\\triangle ABC \\sim \\triangle DBE$ and $\\frac{DB}{AB} = 2 \\cos \\varphi$, where $\\varphi = \\Varangle ABD$. Hence $\\Varangle ACB = \\Varangle DEB = \\gamma$.\n\nThen\n$$\n\\begin{aligned}\nS_{ACE} & = \\frac{1}{2} AC \\cdot CE \\cdot \\sin (\\varphi + \\gamma) \\\\\n& = \\frac{1}{2} b \\cdot \\frac{a}{2 \\cos \\varphi} \\cdot \\sin (\\varphi + \\gamma)\n\\end{aligned}\n$$\nand\n$$\n\\begin{aligned}\nS_{EBDC} & = \\frac{1}{2} CB \\cdot DE \\cdot \\sin (\\varphi + \\gamma) \\\\\n& = \\frac{1}{2} a \\cdot \\frac{b}{2 \\cos \\varphi} \\cdot \\sin (\\varphi + \\gamma)\n\\end{aligned}\n$$\n\n![](attached_image_1.png)\n\nHence $S_{ACE} = S_{EBDC}$, which implies the statement of the problem.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27165,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $a$, $b$ and $c$ be positive integers such that one of them is coprime with any of the other two. Prove that there are positive integers $x$, $y$ and $z$ such that $x^{a} = y^{b} + z^{c}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nWe consider two cases.\n\nCase 1. Let $(a, b) = (a, c) = 1$. Then $(a, b c) = 1$ and hence there are integers $u$ and $v$ such that $u a + v b c = 1$. This means that $a$ divides $-v b c + 1$. If $k \\geq 1$ is a positive integer such that $a$ divides $-v - k$, then $a$ divides $k b c + 1$, i.e. $k b c + 1 = a t$. Hence setting $x = 2^{t}$, $y = 2^{k c}$ and $z = 2^{k b}$ we have that\n$$\ny^{b} + z^{c} = 2^{k b c} + 2^{k b c} = 2^{k b c + 1} = \\left(2^{t}\\right)^{a} = x^{a}\n$$\n\nCase 2. Let $(c, a) = (c, b) = 1$. Then $(c, a b) = 1$ and as above we find a positive integer $k$ such that $c$ divides $k a b + 1$, i.e., $k a b + 1 = c t$. Hence setting $x = 2\\left(2^{a} - 1\\right)^{k b}$, $y = \\left(2^{a} - 1\\right)^{k a}$ and $z = \\left(2^{a} - 1\\right)^{t}$ one has that\n$$\nx^{a} - y^{b} = 2^{a}\\left(2^{a} - 1\\right)^{k a b} - \\left(2^{a} - 1\\right)^{k a b} = \\left(2^{a} - 1\\right)^{k a b + 1} = z^{c}\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27166,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOne chooses a point in the interior of $\\triangle ABC$ with area $1$ and connects it with the vertices of the triangle. Then one chooses a point in the interior of one of the three new triangles and connects it with its vertices, etc. At any step one chooses a point in the interior of one of the triangles obtained before and connects it with the vertices of this triangle. Prove that after the $n$-th step:\n\na) $\\triangle ABC$ is divided into $2n+1$ triangles;\n\nб) there are two triangles with common side whose combined area is not less than $\\frac{2}{2n+1}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na) At each step the triangle from which we choose a point is divided into three new triangles, i.e., the number of triangles increases by two. Hence at the $n$-th step we have $2n+1$ triangles.\n\nb) We shall prove by induction on $n$ that removing any triangle, there is a pairing of the remaining triangles such that the triangles in any pair have a common side.\n\nThe statement is trivial for $n=1$. Assume that it is true for $n=k$. We shall prove it for $n=k+1$. Let a point $O$ in $\\triangle MNP$ be added at the $(k+1)$-th step. Remove any $\\triangle XYZ$. If it is some of $\\triangle OMN$, $\\triangle OMP$ or $\\triangle ONP$ (we may assume $\\triangle OMN$), we consider the configuration obtained by removing $\\triangle MNP$ at the $k$-th step.\n\nBy the induction hypothesis, the remaining triangles can be paired in such a way that the triangles in any pair have a common side. Adding the pair $(\\triangle OMP, \\triangle ONP)$ we obtain the desired pairing.\n\nIf $\\triangle XYZ$ does not coincide with $\\triangle OMN$, $\\triangle OMP$ and $\\triangle ONP$, we consider the configuration obtained by removing $\\triangle XYZ$ at the $k$-th step. Assume that $\\triangle MNP$ is paired with $\\triangle QMN$. Then replacing the pair $(\\triangle MNP, \\triangle QMN)$ with the pairs $(\\triangle OMN, \\triangle QMN)$ and $(\\triangle OMP, \\triangle OPN)$ completes the induction.\n\nSince the area of $\\triangle ABC$ equals $1$, then the minimal area of a triangle does not exceed $\\frac{1}{2n+1}$. Remove a triangle of minimal area. Then as we proved above, the remaining triangles can be paired such that the triangles in any pair have a common side. Since the number of the pairs is equal to $n$, there is a pair of total area at least $\\frac{1-\\frac{1}{2n+1}}{n} = \\frac{2}{2n+1}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27167,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA table with $m$ rows and $n$ columns is given. At any move one chooses some empty cells such that any two of them lie in different rows and columns, puts a white piece in any of these cells and then puts a black piece in the cells whose lines and columns contain white pieces. The game is over if it is not possible to make a move. Find the maximum possible number of the white pieces that can be put on the table.",
    "options": [],
    "answer": "m + n - 1",
    "solution": "Solution:\nWe put $m+n-1$ white pieces by $m+n-1$ moves consecutively in the cells $(1,1), (2,1), \\ldots, (m,1), (1,2), (1,3), \\ldots, (1,n)$.\n\nWe shall call a closed chain of cells such that at least two numbers of rows and columns change alternatively a zig-zag cycle.\n\nNote that there is at least one black piece in the cells of a zig-zag cycle. Indeed, assume the contrary and consider the last white piece. Its neighbors (in the zig-zag cycle) are white pieces that have been put before it. Then the last piece must be black, a contradiction.\n\nAssume now that the game is over and there are more than $m+n-1$ pieces. Remove a row or a column with at most one white piece, repeat the same operation, etc. Since such a removing can be done at most $m+n-1$ times, any row and column in the new table will contain at least two white pieces. This new table, and hence the given table, has a zig-zag cycle, a contradiction to the proved above.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27168,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSolve in integers the equation\n$$\n2^{a} + 8 b^{2} - 3^{c} = 283\n$$",
    "options": [],
    "answer": "(a,b,c) = (2,6,2) and (2,-6,2)",
    "solution": "Solution:\nIt is easy to see that $a, c \\geq 0$. Since $3^{c}$ is congruent to $1$ or $3$ modulo $8$, then $0 \\leq a \\leq 2$. If $a=0$ or $a=1$, then $2$ divides $3^{c}$ or $8$ divides $3^{c}+1$, a contradiction.\n\nLet $a=2$, i.e., $8 b^{2} - 3^{c} = 279$. The cases $c=0,1$ are impossible and hence $c \\geq 2$. Then $3$ divides $b$ and setting $b=3 d$ gives $8 d^{2} - 3^{c-2} = 31$. If $c \\geq 3$, then $3$ divides $d^{2}+1$, a contradiction. Therefore $c=2$ and $d= \\pm 2$, i.e. $a=2$, $b= \\pm 6$ and $c=2$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27169,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all real numbers $a$ such that the graphs of the functions $x^{2}-2 a x$ and $-x^{2}-1$ have two common tangent lines and the perimeter of the quadrilateral with vertices at the tangent points is equal to $6$.",
    "options": [],
    "answer": "a = ±√2/2",
    "solution": "Solution:\nThe equation of a common tangent line to the graphs of $f(x)$ and $g(x)$ at points $(x_{1}, f(x_{1}))$ and $(x_{2}, g(x_{2}))$ has the form\n$$\ny = f(x_{1}) + f'(x_{1})(x - x_{1}) = g(x_{2}) + g'(x_{2})(x - x_{2})\n$$\nHence $f'(x_{1}) = g'(x_{2})$ and $f(x_{1}) - f'(x_{1}) x_{1} = g(x_{2}) - g'(x_{2}) x_{2}$. Since $f'(x) = 2x - 2a$ and $g'(x) = -2x$, we get that $x_{1} + x_{2} = a$ and $x_{1}^{2} + x_{2}^{2} = 1$. Then $x_{1} x_{2} = \\frac{a^{2} - 1}{2}$ and hence $x_{1}$ and $x_{2}$ are the roots of the quadratic equation $x^{2} - a x + \\frac{a^{2} - 1}{2} = 0$. Since the graphs of $f(x) = x^{2} - 2a x$ and $g(x) = -x^{2} - 1$ have two common tangent lines, it follows that they are disjoint, $x_{1} \\neq x_{2}$, $a^{2} < 2$, and the tangent points are $M(x_{1}, f(x_{1})), N(x_{2}, f(x_{2})), P(x_{2}, g(x_{2}))$ and $Q(x_{1}, g(x_{1}))$. Then\n$$\n\\begin{aligned}\nPQ^{2} & = (x_{1} - x_{2})^{2} + (g(x_{1}) - g(x_{2}))^{2} \\\\\n& = (x_{1}^{2} + x_{2}^{2} - 2 x_{1} x_{2})^{2} (1 + x_{1} + x_{2}) = (2 - a^{2})(1 + a^{2}) \\\\\nMQ & = |f(x_{1}) - g(x_{1})| = |x_{1}^{2} - 2a x_{1} + x_{1}^{2} + 1| = 2 - a^{2}\n\\end{aligned}\n$$\nand similarly\n$$\nMN^{2} = (2 - a^{2})(1 + a^{2}), \\quad NP = 2 - a^{2}\n$$\nHence $MNPQ$ is a parallelogram with perimeter\n$$\n2\\left(\\sqrt{(2 - a^{2})(1 + a^{2})} + 2 - a^{2}\\right) = 6\n$$\nThen $\\sqrt{(2 - a^{2})(1 + a^{2})} = 1 + a^{2}$ and we get that $a = \\pm \\frac{\\sqrt{2}}{2}$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 27170,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $a_{1}>0$ and $a_{n+1}=a_{n}+\\frac{n}{a_{n}}$ for $n \\geq 1$. Prove that:\n\na) $a_{n} \\geq n$ for $n \\geq 2$;\n\nb) the sequence $\\left\\{\\frac{a_{n}}{n}\\right\\}_{n \\geq 1}$ converges and find its limit.",
    "options": [],
    "answer": "1",
    "solution": "Solution:\n\na) We have $a_{2}=a_{1}+\\frac{1}{a_{1}} \\geq 2$. If $a_{n} \\geq n$, then\n$$\na_{n+1}-n-1=a_{n}+\\frac{n}{a_{n}}-n-1=\\frac{\\left(a_{n}-1\\right)\\left(a_{n}-n\\right)}{a_{n}} \\geq 0\n$$\nand the assertion follows by induction.\n\nb) Let $n \\geq 2$. It follows from a) that $a_{n+1} \\leq a_{n}+1$. Then $a_{n} \\leq a_{2}+n-2$, whence $1 \\leq \\frac{a_{n}}{n} \\leq 1+\\frac{a_{2}-2}{n}$. Therefore the sequence $\\left(\\frac{a_{n}}{n}\\right)_{n \\geq 1}$ is convergent and its limit equals $1$.\n\nRemark. One can prove the stronger statement that $\\lim _{n \\rightarrow \\infty}\\left(a_{n}-n\\right)=0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27171,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe incircle of $\\triangle ABC$ is tangent to the sides $AC$ and $BC$, $AC \\neq BC$, at points $P$ and $Q$, respectively. The excircles to the sides $AC$ and $BC$ are tangent to the line $AB$ at points $M$ and $N$. Find $\\Varangle ACB$ if the points $M, N, P$ and $Q$ are concyclic.",
    "options": [],
    "answer": "90°",
    "solution": "Solution:\n\nThe perpendicular bisector of the segment $AB$ and the bisector of $\\Varangle ACB$ meet at the midpoint $D$ of the arc $\\overparen{AB}$ of the circumcircle of $\\triangle ABC$ which does not contain $C$.\n\n![](attached_image_1.png)\n\nNext we use the standard notation for the elements of $\\triangle ABC$. Since $AM = BN = CP = CQ = p-c$, the condition of the problem is equivalent to the equality $DM = DP$. The Cosine theorem gives\n$$\n\\begin{aligned}\nDP^{2} & = DC^{2} + CP^{2} - 2 DC \\cdot CP \\cos \\frac{\\gamma}{2} \\\\\nDM^{2} & = DA^{2} + AM^{2} + 2 DA \\cdot AM \\cos \\frac{\\gamma}{2}\n\\end{aligned}\n$$\nSubtracting these equalities, we get\n$$\nDC - DA = 2(p-c) \\cos \\frac{\\gamma}{2}\n$$\nOn the other hand, we have $DC = \\frac{(a+b) DA}{c}$ by Ptolemy's theorem. Since $DA = \\frac{c}{2 \\cos \\frac{\\gamma}{2}}$, we get $DC = \\frac{a+b}{2 \\cos \\frac{\\gamma}{2}}$. Now (1) implies that $\\cos^{2} \\frac{\\gamma}{2} = \\frac{1}{2}$, i.e. $\\gamma = 90^{\\circ}$.\n\nRemark. The solution above shows that the points $M, N, P$ and $Q$ are concyclic if and only if $AC = BC$ or $\\Varangle ACB = 90^{\\circ}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27172,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn triangle $ABC$ with orthocenter $H$ one has that\n$$\nAH \\cdot BH \\cdot CH = 3 \\text{ and } AH^2 + BH^2 + CH^2 = 7\n$$\nFind:\na) the circumradius of $\\triangle ABC$;\nb) the sides of $\\triangle ABC$ with maximum possible area.",
    "options": [],
    "answer": "a) The circumradius is either 1 or 3/2. b) The maximum area occurs for an acute isosceles triangle with side lengths sqrt(6), sqrt(6), and sqrt(8).",
    "solution": "Solution:\n\na) If $\\triangle ABC$ is acute, then by the Law of cosines for $\\triangle AHB$ we get that\n$$\nAB^2 = AH^2 + BH^2 - 2 AH \\cdot BH \\cos(\\pi - \\gamma)\n$$\nSince $AB = 2R \\sin \\gamma$ and $CH = 2R \\cos \\gamma$ (by the Extended Law of sines), we obtain $AB^2 + CH^2 = 4R^2$. Therefore\n$$\n4R^2 = AH^2 + BH^2 + CH^2 + \\frac{AH \\cdot BH \\cdot CH}{R}\n$$\nThen $4R^3 = 7R + 3$, i.e. $(R+1)(2R+1)(2R-3) = 0$, whence $R = \\frac{3}{2}$.\nIf $\\triangle ABC$ is obtuse, then we get analogously that\n$$\n4R^2 = AH^2 + BH^2 + CH^2 - \\frac{AH \\cdot BH \\cdot CH}{R}\n$$\nand therefore $4R^3 = 7R - 3$, i.e. $(R-1)(2R-1)(2R+3) = 0$. Since $3 = AH \\cdot BH \\cdot CH < (2R)^3$, we conclude that $R = 1$.\nThe existence of $\\triangle ABC$ with $R = \\frac{3}{2}$ and $R = 1$ follows from b).\n\nb) Denote by $S$ the area of $\\triangle ABC$. Since $S = \\frac{AB \\cdot BC \\cdot CA}{4R}$, we have\n$$\nS^2 = \\frac{(4R^2 - AH^2)(4R^2 - BH^2)(4R^2 - CH^2)}{16R^2}\n$$\nSetting $x = AH^2$, $y = BH^2$, $z = CH^2$ and $t = 4R^2$, we get\n$$\nS^2 = \\frac{t^3 - 7t^2 + t(xy + yz + zx) - 9}{4t}\n$$\nWithout loss of generality we may assume that $x \\geq y \\geq z$. Then $x \\geq \\frac{7}{3}$ and therefore\n$$\nxy + yz + zx = \\frac{9}{x} + x(7 - x) = 15 - \\frac{(x-3)^2(x-1)}{x} \\leq 15\n$$\nwhere the equality is attained if $x = 3$. Hence\n$$\nS^2 \\leq \\frac{t^3 - 7t^2 + 15t - 9}{4t}\n$$\nSince $R = \\frac{3}{2}$ or $R = 1$, we conclude that $S_{\\max} = \\sqrt{8}$ and it is achieved for an acute $\\triangle ABC$ with $R = \\frac{3}{2}$, $AH = BH = \\sqrt{3}$ and $CH = 1$. The sides of this triangle are $\\sqrt{6}$, $\\sqrt{6}$ and $\\sqrt{8}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27173,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all values of $a$ such that the maximum of the function $f(x) = \\frac{a x - 1}{x^{4} - x^{2} + 1}$ is equal to $1$.",
    "options": [],
    "answer": "a = 2 or a = -2",
    "solution": "Solution:\nSince the denominator of the function is positive, the given condition means that $a x - 1 \\leq x^{4} - x^{2} + 1$ for any $x$, and that the equality is attained for some $x$.\n\nLet $a \\geq 0$. Then $a x \\leq 0$ for $x \\leq 0$ and hence $a$ is the minimum of the function $g(x) = \\frac{x^{4} - x^{2} + 2}{x}$ for $x > 0$. Since $g'(x) = \\frac{(3 x^{2} + 2)(x + 1)(x - 1)}{x^{2}}$, it follows that this minimum is equal to $g(1) = 2$.\n\nThe case $a \\leq 0$ can be reduced to the previous one replacing $a$ by $-a$ and $x$ by $-x$. So $a = \\pm 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27174,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nProve that for any integer $a \\geq 4$ there exist infinitely many squarefree positive integers $n$ that divide $a^{n}-1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nFirst we shall prove the following:\n\nLEMMA. Let $p \\geq 3$ be an odd divisor of $b$. Then there exists an odd prime $q$ that divides $(b+1)^{p}-1$ but does not divide $b$.\n\nProof of the lemma. If $b=pc$, then\n$$\n\\begin{aligned}\n(b+1)^{p}-1 & =b\\left((b+1)^{p-1}+\\cdots+b+1\\right) \\\\\n& =b\\left(B b^{2}+\\frac{p(p-1)}{2} b+p\\right)=b p\\left(b\\left(B c+\\frac{p-1}{2}\\right)+1\\right)=b p d\n\\end{aligned}\n$$\nand it remains to choose a prime divisor of $d$. Note that $d$ is odd (if $b$ is even, then $d=b K+1$ is odd; if $b$ is odd, then $(b+1)^{p}-1$ is odd, and hence $d$ is odd, too).\n\nWe shall prove now that if $a \\neq 2^{k}+1$, then there exists a sequence of odd primes $p_{1}, p_{2}, \\ldots$ such that $p_{1}$ divides $a-1$, and if $P_{n}=a^{p_{0} p_{1}} p_{n}-1$ (here $p_{0}=1$ ), then $p_{n+1}$ divides $P_{n}$, but does not divide $P_{n-1}$, $n \\geq 1$.\n\nLet $p_{1}$ be an odd prime divisor of $a-1$ and we have already chosen the primes $p_{1}, \\ldots, p_{k}$. Applying the lemma for $b=P_{k}$ and $p=p_{k}$, we find an odd prime $p_{k+1}$ that divides $P_{k}$ but does not divide $P_{k-1}$.\n\nSince $P_{k-1}$ is divisible by $p_{1}, p_{2}, \\ldots, p_{k}$, we conclude that $p_{k+1}$ differs from them. Therefore the numbers $p_{1}, p_{2}, \\ldots, p_{k}$ have the required property.\n\nIf $a=2^{l}+1, l \\geq 2$, then $a^{2} \\neq 2^{m}+1$ and it remains to multiply by 2 the numbers already found for $a^{2}$.\n\nRemark. It can be proved that if $n$ divides $2^{n}-1$, then $n=1$, and if $n$ divides $3^{n}-1$ then $n=1, n=2$, or $n$ is divisible by 4. The above solution shows that for $a=3^{4}$ there exist infinitely many positive square-free odd integers $n$ such that $4 n$ divides $3^{4 n}-1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27175,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA plane bisects the volume of the tetrahedron $A B C D$ and meets the edges $A B$ and $C D$ respectively at points $M$ and $N$ such that $\\frac{A M}{B M}= \\frac{C N}{D N} \\neq 1$. Prove that the plane passes through the midpoints of the edges $A C$ and $B D$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet a plane $\\pi$ bisect the volume of $A B C D$ and meet the edges $A B$, $B C$, $C D$ and $D A$ at points $M$, $Q$, $N$ and $P$, respectively. Set $x=\\frac{A M}{B M}$, $y=\\frac{C N}{D N}$, $z=\\frac{A P}{D P}$ and $t=\\frac{C Q}{B Q}$. If $T=\\pi \\cap A C$ (we assume $T=\\infty$ if $\\pi \\parallel A C$), then the Menelaus theorem implies that $\\frac{x}{t}=\\frac{A T}{C T}=\\frac{z}{y}$, i.e., $x y=z t$.\n\nOn the other hand,\n$$\n\\begin{aligned}\n\\frac{1}{2} & =\\frac{V_{A M Q C N P}}{V_{A B C D}}=\\frac{V_{A M Q C P}+V_{Q C N P}}{V_{A B C D}}=\\frac{S_{A M Q C}}{S_{A B C}} \\cdot \\frac{A P}{A D}+\\frac{S_{Q C N}}{S_{B C D}} \\cdot \\frac{D P}{A D} \\\\\n& =\\left(1-\\frac{B M}{A B} \\cdot \\frac{B Q}{B C}\\right) \\frac{A P}{A D}+\\frac{C N}{C D} \\cdot \\frac{C Q}{C B} \\cdot \\frac{D P}{A D} \\\\\n& =\\left(1-\\frac{1}{(1+x)(1+t)}\\right) \\frac{z}{1+z}+\\frac{y t}{(1+y)(1+t)(1+z)}\n\\end{aligned}\n$$\nand hence\n$$\n2 z(1+y)(x+t+x t)+2 y t(1+x)=(1+x)(1+y)(1+z)(1+t)\n$$\nLet now the plane from the condition of the problem meet the edge $B C$ in an interior point. Then it meets the edge $A D$. Since $x=y$ and $x y=z t$, then (1) easily implies that $(x-1)\\left(t^{2}+t(x+1)^{2}+x^{2}\\right)=0$. But $x \\neq 1$ and the second factor is positive. Hence this case is impossible.\n\nThus the given plane meets the edges $A C$ and $B D$. Replacing $C$ by $D$ and $D$ by $C$, we get $x y=1=z t$ and the inequality still holds. Then $(x-1)\\left(t^{2}-1\\right)=0$, i.e. $t=z=1$, which completes the solution.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 27176,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $ABCD$ be a circumscribed quadrilateral. Find $\\Varangle BCD$ if $AC = BC$, $AD = 5$, $E = AC \\cap BD$, $BE = 12$ and $DE = 3$.",
    "options": [],
    "answer": "90°",
    "solution": "Solution:\nIf the perpendicular bisector of $CD$ meets $BD$ at point $O$, then\n$$\n\\begin{aligned}\n\\Varangle COD & = 180^\\circ - 2 \\Varangle ODC \\\\\n& = 180^\\circ - 2 \\Varangle BAC \\\\\n& = \\Varangle ACB = \\Varangle ADO\n\\end{aligned}\n$$\nand therefore $AD \\parallel CO$. Hence $\\frac{OE}{3} = \\frac{CO}{AD} = \\frac{OE + 3}{5}$, which implies that $OE = \\frac{9}{2}$. This shows that $O$ is the midpoint of $BD$ and then\n\n![](attached_image_1.png)\n\n$\\Varangle BCD = 90^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27177,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\na) A set $A$ of positive integers less than $2000000$ is called good if $2000 \\in A$ and $a$ divides $b$ for any $a, b \\in A$, $a < b$. Find the maximum possible cardinality of a good set.\n\nb) Find the number of the good sets of maximal cardinality.",
    "options": [],
    "answer": "Maximum cardinality: 17; Number of good sets of maximal cardinality: 350",
    "solution": "Solution:\n\na) Let $a_{1} < \\cdots < a_{n-1} < a_{n} = 2000 < a_{n+1} < \\cdots < a_{m}$ be the elements of a good set. Since $a_{i+1} \\geq 2 a_{i}$, then $2000000 > a_{m} \\geq 2^{m-n} 2000$ and hence $m-n \\leq 9$.\n\nOn the other hand, the equality $2000 = 2^{4} 5^{3}$ shows that $a_{i} = 2^{k_{i}} 5^{l_{i}}$ for $i \\leq n-1$, where $0 \\leq k_{i} \\leq k_{i+1} \\leq 4$, $0 \\leq l_{i} \\leq l_{i+1} \\leq 3$ and $k_{i} + l_{i} \\leq 6$. Hence $n \\leq 8$ and so $A$ has at most $8 + 9 = 17$ elements. An example of a good set of $17$ elements is obtained by setting $a_{i} = 2^{i-1}$, $1 \\leq i \\leq 5$, $a_{i} = 2^{4} 5^{i-5}$, $6 \\leq i \\leq 8$, $a_{i} = 2^{i-4} 5^{3}$, $9 \\leq i \\leq 17$.\n\nb) For a good set of maximal cardinality one has that $m = 17$ and $n = 8$, i.e. $a_{8} = 2000$. Moreover, $k_{i} + l_{i} = i-1$ for $1 \\leq i \\leq 7$, which shows that $a_{1} = 1$ and that the subset $\\{a_{2}, \\ldots, a_{7}\\}$ is determined by the numbers $1 \\leq i_{1} < i_{2} < i_{3} \\leq 7$ such that $l_{i_{1}} = 0$, $l_{i_{1}+1} = 1$, $l_{i_{2}} = 1$, $l_{i_{2}+1} = 2$, $l_{i_{3}} = 2$ and $l_{i_{3}+1} = 3$.\n\nThere are $\\binom{7}{3} = 35$ possibilities for this subset. Since $2^{9} < 2^{8} \\cdot 3 < 1000 < 2^{10}$, it follows that either $a_{i} = 2^{i-4} 5^{3}$ for $9 \\leq i \\leq 17$, or there is an index $j$, $9 \\leq j \\leq 17$ such that $a_{i} = 2^{i-4} 5^{3}$ for $8 \\leq i < j$ and $a_{i} = 2^{i-5} 5^{3} 3$ for $j \\leq i \\leq 17$. Hence there are $10$ possibilities for the subset $\\{a_{9}, \\ldots, a_{17}\\}$. So, the number of the good sets of maximal cardinality equals $35 \\cdot 10 = 350$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27178,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFind all non-constant polynomials $P(x)$ and $Q(x)$ with real coefficients such that $P(x) Q(x+1) = P(x+2004) Q(x)$ for any $x$.",
    "options": [],
    "answer": "All solutions are given by choosing any nonconstant real polynomial P, and setting Q(x) = c · P(x) P(x+1) … P(x+2003) for some nonzero real constant c.",
    "solution": "Solution:\n\nSet $R(x) = P(x) P(x+1) \\ldots P(x+2003)$. It follows by the given condition that if $x$ is greater than the largest real root of $P(x)$, then\n$$\n\\frac{Q(x)}{R(x)} = \\frac{Q(x+1)}{R(x+1)}\n$$\nWe get by induction that $\\frac{Q(x)}{R(x)} = \\frac{Q(x+n)}{R(x+n)}$ for any positive integer $n$. Note that $\\lim_{n \\rightarrow \\infty} \\frac{Q(x+n)}{R(x+n)}$ is either a finite number, independent of $x$, or $\\infty$. On the other hand, this limit is equal to $\\frac{Q(x)}{R(x)}$. Hence $Q(x) = c R(x)$ for any $x$, where $c \\neq 0$ is a constant.\n\nConversely, it is clear that, if $Q(x) = c P(x) P(x+1) \\ldots P(x+2003)$, then the condition of the problem is satisfied.\n\nRemark. Getting equality (1), the solution can be completed by comparing the coefficient of the respective polynomials.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 27179,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nProve that if $a_{1}, a_{2}, \\ldots, a_{n}, b_{1}, b_{2}, \\ldots, b_{n} \\geq 0$ and $c_{k}=\\prod_{i=1}^{k} b_{i}^{\\frac{1}{k}}$, $1 \\leq k \\leq n$, then\n$$\nn c_{n}+\\sum_{k=1}^{n} k\\left(a_{k}-1\\right) c_{k} \\leq \\sum_{k=1}^{n} a_{k}^{k} b_{k}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nThe Arithmetic mean - Geometric mean inequality (for any $k=2,3, \\ldots, n$) implies that\n$$\nk a_{k} c_{k}=k a_{k} b_{k}^{\\frac{1}{k}} \\underbrace{c_{k-1}^{\\frac{1}{k}} \\ldots c_{k-1}^{\\frac{1}{k}}}_{k-1 \\text{ times }} \\leq a_{k}^{k} b_{k}+(k-1) c_{k-1}\n$$\nSumming up these inequalities and adding the equality $a_{1} c_{1}=a_{1} b_{1}$ gives the desired inequality.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27180,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFor any positive integer $n$ the sum $1+\\frac{1}{2}+\\cdots+\\frac{1}{n}$ is written in the form $\\frac{p_{n}}{q_{n}}$, where $p_{n}$ and $q_{n}$ are coprime numbers.\n\na) Prove that $3$ does not divide $p_{67}$.\n\nb) Find all $n$, for which $3$ divides $p_{n}$.",
    "options": [],
    "answer": "Part a: 3 does not divide p_67. Part b: n = 2, 7, 22.",
    "solution": "Solution:\n\na) Set $S_{n}=1+\\frac{1}{2}+\\cdots+\\frac{1}{n}$. We have that $S_{2}=\\frac{3}{2}$, $S_{7}=\\frac{3121}{140}$,\n$$\n\\begin{aligned}\nS_{22}-S_{7}= & \\frac{1}{8}+\\frac{1}{22}+\\frac{1}{10}+\\frac{1}{20}+\\frac{1}{11}+\\frac{1}{19}+\\frac{1}{14}+\\frac{1}{16}+\\frac{1}{13}+\\frac{1}{17}+ \\\\\n& \\frac{1}{9}+\\frac{1}{12}+\\frac{1}{15}+\\frac{1}{18}+\\frac{1}{21}=\\frac{30 a}{b}+\\frac{51}{140}=\\frac{3 c}{d}\n\\end{aligned}\n$$\nwhere $(a, b)=(c, d)=1$. It is easy to see that $a \\equiv b \\pmod{3}$. Hence $3 \\nmid c, d$, $c \\not \\equiv d \\pmod{3}$, $p_{22}=3 p_{22}'$ and $3 \\nmid p_{22}'$. Similarly, we have that $S_{67}-S_{22}= \\frac{90 e}{f}+\\frac{c}{d}$, where $3 \\nmid f$. It follows that $3 \\nmid p_{67}, q_{67}$.\n\nb) Set $S_{n}=\\frac{k_{n}}{3^{m_{n}} l_{n}}$, where $3 \\nmid k_{n}, l_{n}$. Then\n$$\n\\begin{aligned}\nS_{3 n} & =\\frac{S_{n}}{3}+1+\\frac{1}{2}+\\frac{1}{4}+\\frac{1}{5}+\\cdots+\\frac{1}{3 n-2}+\\frac{1}{3 n-1} \\\\\n& =\\frac{k_{n}}{3^{m_{n}+1} l_{n}}+3 \\cdot \\frac{a_{n}}{b_{n}}=\\frac{k_{n} b_{n}+3^{m_{n}+2} l_{n} a_{n}}{3^{m_{n}+1} l_{n} b_{n}}\n\\end{aligned}\n$$\nwhere $3 \\nmid b_{n}$. Therefore, if $m_{n} \\geq -1$, then $m_{3 n}=m_{n}+1$. Analogously, we have that $m_{3 n+2}=m_{n}+1$ for $m_{n} \\geq -1$ and $m_{3 n+1}=m_{n}+1$ for $m_{n} \\geq 0$. Since $m_{1}=0$, $m_{2}=m_{7}=m_{22}=-1$ and $m_{67}=0$, it is easy to see that the answer is $n=2,7,22$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27181,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn a group of $n$ tourists, among every three of them there are at least two that are not familiar. For any partition of the group into two groups, there are at least two familiar tourists in some of the groups. Prove that there is a tourist who is familiar with at most $\\frac{2n}{5}$ tourists.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nConsider a graph $G$ with $n$ vertices corresponding to the tourists and two vertices connected when they are familiar.\n\nThe first condition of the problem means that there is no triangle in $G$.\n\nThe second condition of the problem means that there is a cycle of odd length in the graph. Indeed, if all cycles are of even length, then the vertices can be partitioned into two groups such that there is no edge in any of them.\n\nLet $A_{1}, A_{2}, \\ldots, A_{k}$ be a cycle with minimal odd length in $G$. Since there is no triangle in $G$, and because of minimality, it follows that every vertex outside this cycle is connected with at most two vertices from the cycle. Hence the number of edges of the form $(X, A_{i}), X \\neq A_{j}, j=1,2, \\ldots, k$ does not exceed $2(n-k)$. Denote by $d(A_{i})$ the degree of the vertex $A_{i}$ and set $\\delta= \\min_{1 \\leq i \\leq k} d(A_{i})$. Obviously $\\sum_{i=1}^{k} d(A_{i})=|E^{*}|+2k$, where $E^{*}$ is the set of edges $X A_{i}$. We have that\n\n$$\n2(n-k) \\geq |E^{*}| = \\sum_{i=1}^{k} d(A_{i}) - 2k \\geq k\\delta - 2k\n$$\n\nHence $\\delta \\leq \\frac{2n}{k}$ and $k \\geq 5$ implies that $\\delta \\leq \\frac{2n}{5}$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 27182,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the maximum possible value of the inradius of a triangle with vertices in the interior or on the boundary of a unit square.",
    "options": [],
    "answer": "(sqrt(5)-1)/4",
    "solution": "Solution:\nIt is easy to see that if a triangle contains another triangle, then its inradius is greater than the inradius of the second one. So we may consider triangles with vertices on the boundary of the square. Moreover, we may assume that at least one vertex of the triangle is a vertex of the square and the other two vertices belong to the sides of the square containing not the first vertex. So we shall consider $\\triangle OAB$ such that $O=(0,0)$, $A=(a,1)$ and $B=(1,b)$, $0 \\leq a, b \\leq 1$.\n\nConsider also $\\triangle OCD$, where $C=(a+b, 1)$ and $D=(1,0)$. Denote by $S$ and $P$ the area and perimeter of $\\triangle OAB$, respectively. Set $x=OA=\\sqrt{1+a^{2}}$, $y=AB=\\sqrt{(1-a)^{2}+(1-b)^{2}}$, $z=OB=\\sqrt{1+b^{2}}$, $u=OC=\\sqrt{1+(a+b)^{2}}$ and $v=CD=\\sqrt{1+(1-a-b)^{2}}$. Note that $OD=1$, $u \\geq z \\geq 1$, $x \\geq 1$ and $v \\geq 1$.\n\nComparing the perimeters of $\\triangle OAB$ and $\\triangle OCD$ gives\n$$\n\\begin{aligned}\n(u+v+1)-(x+y+z) & =\\frac{u^{2}-x^{2}}{u+x}+\\frac{v^{2}-y^{2}}{v+y}+\\frac{1-z^{2}}{1+z} \\\\\n& =\\frac{2ab+b^{2}}{u+x}+\\frac{2ab}{v+y}-\\frac{b^{2}}{1+z} \\\\\n& \\leq \\frac{2ab+b^{2}}{1+z}+\\frac{2ab}{v+y}-\\frac{b^{2}}{1+z} \\\\\n& =2ab\\left(\\frac{1}{v+y}+\\frac{1}{1+z}\\right) \\leq 3ab \\leq (u+v+1)ab\n\\end{aligned}\n$$\nHence\n$$\n(u+v+1)(1-ab) \\leq x+y+z \\Longleftrightarrow \\frac{1}{u+v+1} \\geq \\frac{1-ab}{x+y+z}=\\frac{2S}{P}=r\n$$\nOn the other hand,\n$$\nu+v+1=\\sqrt{1+(a+b)^{2}}+\\sqrt{1+(1-a-b)^{2}}+1 \\geq \\min_{x \\geq 1} F(x)\n$$\nwhere $F(x)=\\sqrt{1+x^{2}}+\\sqrt{1+(1-x)^{2}}+1$. Since $\\min_{x \\geq 1} F(x)=\\sqrt{5}+1$, we get that $r \\leq \\frac{1}{\\sqrt{5}+1}=\\frac{\\sqrt{5}-1}{4}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27183,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $A=\\{1,2, \\ldots, n\\}, n \\geq 4$. For any function $f: A \\rightarrow A$ and any $a \\in A$ define $f_{1}(a)=f(a)$, $f_{i+1}(a)=f\\left(f_{i}(a)\\right)$, $i \\geq 1$. Find the number of the functions $f$ such that $f_{n-2}$ is a constant function but $f_{n-3}$ is not.",
    "options": [],
    "answer": "n!(2n-5)/2",
    "solution": "Solution:\nDefine an oriented graph $G$ with vertices the elements of $A$ and oriented edge $xy$ if $f(x)=y$. We have to count the graphs $G$ such that:\n- there are no cycles with length greater than $1$;\n- there is a chain $a_{2} \\ldots a_{n}$ with length $n-2$ and there is no chain with length $n-1$;\n- the only edge outside this chain has the form $a_{1} a_{j}$, where $3 \\leq j \\leq n$;\n- there is a unique loop $a_{n} a_{n}$.\nThe chain can be chosen in $n!$ ways, and the edge outside it - in $n-2$ ways. Note that the graphs for which this edge is $a_{1} a_{3}$, are counted two times and hence their number is equal to $\\binom{n}{2}(n-2)!$. So the answer of the problem is\n$$\nn!(n-2)-\\binom{n}{2}(n-2)!=\\frac{n!(2n-5)}{2}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27184,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the maximum possible value of the product of different positive integers with sum $2004$.",
    "options": [],
    "answer": "63!/11",
    "solution": "Solution:\nLet $x_{1} + x_{2} + \\cdots + x_{k} = 2004$, $x_{1}, x_{2}, \\ldots, x_{k} \\in \\mathbb{N}$, $x_{1} < x_{2} < \\cdots < x_{k}$ and the product $x_{1} x_{2} \\ldots x_{k}$ is maximal. Assume that for some $i, j$, $1 \\leq i < j \\leq k$ one has that $x_{i} \\leq x_{i+1} - 2$ and $x_{j} \\leq x_{j+1} - 2$. Then replacing $x_{i}$ and $x_{j}$ by $x_{i} + 1$ and $x_{j} - 1$, respectively (the sum is the same, i.e. $2004$), we get a larger product since\n$$\n(x_{i} + 1)(x_{j} - 1) = x_{i} x_{j} + x_{j} - x_{i} - 1 > x_{i} x_{j}\n$$\na contradiction. Hence $x_{1}, x_{2}, \\ldots, x_{k}$ are consecutive integers but at most one.\n\nLet the numbers be $\\{x_{1}, x_{2}, \\ldots, x_{k}\\} = \\{x, x+1, \\ldots, x+\\ell, x+\\ell+n, x+\\ell+n+1, \\ldots, x+k+n-2\\}$ and $k = n + \\ell - 2$. If $n \\geq 3$, we replace $x+\\ell$ and $x+\\ell+n$ by $x+\\ell+1$ and $x+\\ell+n-1$, respectively, and, as above, we get a larger product.\n\nLet $n = 1$ and the numbers be $x, x+1, \\ldots, x+k-1$. If $x \\geq 5$ we replace $x$ by the numbers $x-2$ and $2$. The sum remains $2004$ and the product increases, since $2(x-2) > x$. If $1 \\leq x \\leq 4$, a direct verification shows that either we have a larger product or the sum is not equal to $2004$ (for $x=2$ and $x=3$).\n\nIt remains to consider the case $n=2$. Let the numbers be\n$$\nx, x+1, \\ldots, x+\\ell, x+\\ell+2, x+\\ell+3, \\ldots, x+k, \\quad \\ell \\geq 0, k \\geq \\ell+2\n$$\nAs above, we get a larger product for $x=1$ and $x \\geq 4$.\n\nIf $x=2$, then $2+3+\\cdots+(\\ell+2)+(\\ell+4)+\\cdots+(k+2)=2004$ and hence $(k+2)(k+3)=2(2008+\\ell)$. Since $0 \\leq \\ell \\leq k-2$, it follows that $4016 \\leq (k+2)(k+3) \\leq 4012+2k$ and then $k=61, \\ell=8$. So the numbers are $2,3, \\ldots, 10,12,13, \\ldots, 63$ with product $\\frac{63!}{11}$.\n\nFor $x=3$ we get analogously $k=60, \\ell=5$ and the numbers are $3,4, \\ldots, 8,10,11, \\ldots, 63$ with product $\\frac{63!}{18}$ which is smaller than $\\frac{63!}{11}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27185,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn any word with letters $a$ and $b$ the following changes are allowed: $a b a \\rightarrow b$, $b \\rightarrow a b a$, $b b a \\rightarrow a$ and $a \\rightarrow b b a$. Is it possible to obtain the word $b \\underbrace{a a \\ldots a}_{2003}$ from the word $\\underbrace{a a \\ldots a}_{2003} b$?",
    "options": [],
    "answer": "No",
    "solution": "Solution:\n\nThe answer is no.\n\nWe shall prove that applying any of the given changes the number of $a$'s in odd (even) positions does not change its parity. Indeed, let the replacement $a b a \\rightarrow b$ be applied to the word $w_{1} a b a w_{2}$. In the new word $w_{1} b w_{2}$ all $a$'s in $w_{1}$ do not change their positions and all $a$'s in $w_{2}$ shift by two positions to the left. Hence such $a$'s keep the parity of their positions. Deleting both $a$'s in $a b a$ decreases the number of $a$'s in odd (even) positions by two.\n\nAnalogously, the same is true by applying the operation $b b a \\rightarrow a$ to the word $w_{1} b b a w_{2}$.\n\nSince the replacements $a \\rightarrow a b a$ and $a \\rightarrow b b a$ are converse to the above, they have the same property.\n\nIt remains to observe that the number of $a$'s in even positions in the words $b \\underbrace{a a \\ldots a}_{2003}$ and $\\underbrace{a a \\ldots a}_{2003} b$ is $1002$ and $1001$, respectively.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27186,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $a$, $b$, $c$ and $d$ be positive integers such that there are exactly $2004$ ordered pairs $(x, y)$, $x, y \\in (0,1)$, for which $a x + b y$ and $c x + d y$ are integers. If $(a, c) = 6$, find $(b, d)$.",
    "options": [],
    "answer": "{1, 7, 49}",
    "solution": "Solution:\nSuppose first that $a d \\neq b c$. The set of points $(a x + b y, c x + d y)$, $x, y \\in (0,1)$, coincides with the interior of the parallelogram with vertices $A = (0, 0)$, $B = (a, c)$, $C = (b, d)$ and $D = (a + b, c + d)$. Its area $S$ equals $|a d - b c|$. The Pick formula implies that $S = n + \\frac{m}{2} - 1$, where $n$ (respectively $m$) denotes the number of lattice points (i.e., with integer coordinates) in the interior (respectively on the boundary) of the parallelogram. Set $e = (a, c)$, $f = (b, d)$, $a = e a_1$, $c = e c_1$, $b = f b_1$ and $d = f d_1$. The interior points of the side $A B$ have coordinates $(a x, c x)$, $x \\in (0,1)$. Thus the number of such lattice points is $e - 1$. Analogously, the number of the interior lattice points on $B D$, $C D$ and $A C$ equals $f - 1$, $e - 1$ and $f - 1$, respectively. Consequently $\\frac{m}{2} = e + f$ and the first condition of the problem can be written as\n$$\ne f\\left|a_1 d_1 - b_1 c_1\\right| = 2003 + e + f\n$$\nSince $e = (a, c) = 6$, it follows that $f$ divides $2009 = 7^2 \\cdot 41$ and $6 f$ divides $2009 + f$. This is possible only for $f = 1, 7, 49$. For each of these values of $f$ the numbers $e = 6$, $a_1 = 1 + \\frac{2009 + f}{6 f}$, $b_1 = c_1 = d_1 = 1$ satisfy (1) and hence $(b, d) = 1, 7$ or $49$.\n\nSuppose now that $a d = b c$. It is easy to see that $a_1 = b_1$ and $c_1 = d_1$. For $x \\in \\left(0, \\frac{1}{e}\\right)$ set $y = \\frac{1 - x e}{f} < \\frac{1}{f}$. Then $a x + b y = a_1$ and $c x + d y = c_1$ are integers, and $e x + f y = 1$. It follows that in this case there are infinitely many pairs $(x, y)$ satisfying (1), a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27187,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $p(x)$ and $q(x)$ be polynomials with $m \\geq 2$ non-zero coefficients. If $\\frac{p(x)}{q(x)}$ is not a constant function, find the least possible number of the non-zero coefficients of the polynomial $f(u, v)=p(u) q(v)-p(v) q(u)$.",
    "options": [],
    "answer": "2m-2",
    "solution": "Solution:\nConsidering the polynomials $p(x)=x^{m-1}+x^{m-2}+\\cdots+x+1$ and $q(x)=x^{m-1}+x^{m-2}+\\cdots+x+a$, $a \\neq 1$, shows that the desired minimal number does not exceed $2m-2$ (one has that $f(u, v)=(a-1)\\left(u^{m-1}+u^{m-2}+\\cdots+u\\right)+(1-a)\\left(v^{m-1}+v^{m-2}+\\cdots+v\\right)$). We shall prove by induction on $m$ that the number of the non-zero coefficients is at least $2m-2$.\n\nIf $p(x)$ or $q(x)$ contains a monomial which does not appear in the other polynomial, then the non-zero coefficients in $f(u, v)$ are at least $2m$. So we may assume that $p(x)$ and $q(x)$ contain the same monomials. Note also that multiplying some of $p(x)$ and $q(x)$ by a non-zero number does not change the non-zero coefficients of $f(u, v)$.\n\nFor $m=2$ one has that $p(x)=a x^{n}+b x^{k}$, $q(x)=c x^{n}+d x^{k}$ and $ad-bc \\neq 0$. Then $f(u, v)=(ad-bc) u^{n} v^{k}+(bc-ad) u^{k} v^{n}$ has exactly two non-zero coefficients.\n\nLet $m=3$ and let $p(x)=x^{k}+a x^{n}+b x^{\\ell}$, $q(x)=x^{k}+c x^{n}+d x^{\\ell}$ and $ad-bc \\neq 0$. Then\n$$\n\\begin{aligned}\nf(u, v)= & (ad-bc) u^{\\ell} v^{k}+(bc-ad) u^{\\ell} v^{n}+(c-a) u^{k} v^{n}+(a-c) u^{n} v^{k} \\\\\n& +(d-b) u^{k} v^{\\ell}+(b-d) u^{\\ell} v^{k}\n\\end{aligned}\n$$\nThe first two coefficients are non-zero. Since the equalities $a=c$ and $b=d$ do not hold simultaneously, then at least two of the last four coefficients are also non-zero.\n\nLet now $m \\geq 4$ and let $p(x)=p_{1}(x)+a x^{n}+b x^{k}$, $q(x)=q_{1}(x)+c x^{n}+d x^{k}$, $ad-bc \\neq 0$ and any of the polynomials $p_{1}(x)$ and $q_{1}(x)$ has $m-2 \\geq 2$ non-zero coefficients. Then $f(u, v)=f_{1}(u, v)+f_{2}(u, v)+f_{3}(u, v)$, where\n$$\n\\begin{aligned}\nf_{1}(u, v)= & p_{1}(u) q_{1}(v)-p_{1}(v) q_{1}(u) \\\\\nf_{2}(u, v)= & \\left(a u^{n}+b u^{k}\\right) q_{1}(v)+\\left(c v^{n}+d v^{k}\\right) p_{1}(u) \\\\\n& -\\left(a v^{n}+b v^{k}\\right) q_{1}(u)-\\left(c u^{n}+d u^{k}\\right) p_{1}(v) \\\\\nf_{3}(u, v)= & (ad-bc) u^{n} v^{k}+(bc-ad) u^{k} v^{n}\n\\end{aligned}\n$$\nand the different polynomials have no similar monomials. If $p_{1}(x) \\neq \\alpha q_{1}(x)$, then, by the induction hypothesis, $f_{1}(u, v)$ has at least $2(m-2)-2=2m-6$\nnon-zero coefficients. Moreover, $f_{2}(u, v)$ has at least two non-zero coefficients and $f_{3}(u, v)$ has two non-zero coefficients.\n\nIf $p_{1}(x)=\\alpha q_{1}(x)$, $\\alpha \\neq 0$, then\n$f_{2}(u, v)=q_{1}(v)\\left[(a-c\\alpha) u^{n}+(b-d\\alpha) u^{k}\\right]+q_{1}(u)\\left[(c\\alpha-a) v^{n}+(d\\alpha-b) v^{k}\\right]$.\nSince the equalities $a-c\\alpha=0$ and $b-d\\alpha=0$ do not hold simultaneously, the polynomial $f_{2}(u, v)$ has at least $2m-2$ non-zero coefficients (two times more than those of $q_{1}(x)$). Counting the two non-zero coefficients of $f_{3}(u, v)$, we get the desired result.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 27188,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $H$ be the orthocenter of $\\triangle ABC$. The points $A_1 \\neq A$, $B_1 \\neq B$ and $C_1 \\neq C$ lie respectively on the circumcircles of $\\triangle BCH$, $\\triangle CAH$ and $\\triangle ABH$, and $A_1H = B_1H = C_1H$. Denote by $H_1$, $H_2$ and $H_3$ the orthocenters of $\\triangle A_1BC$, $\\triangle B_1CA$ and $\\triangle C_1AB$, respectively. Prove that $\\triangle A_1B_1C_1$ and $\\triangle H_1H_2H_3$ have the same orthocenter.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "$$\n\\sin \\angle C_1AH = \\frac{C_1H}{2R_1} = \\frac{A_1H}{2R_2} = \\sin \\angle A_1BH\n$$\nand analogously\n$$\n\\sin \\angle C_1AH = \\sin \\angle A_1BH = \\sin \\angle A_1CH = \\sin \\angle C_1BH = \\sin \\angle B_1AH.\n$$\nLet $\\angle C_1AH = \\angle A_1BH = \\angle B_1CH = \\varphi$. Since $\\angle AHB = \\alpha + \\beta = 180^\\circ - \\gamma$, then $\\angle AC_1B = \\gamma$, and hence $\\angle AH_3B = 180^\\circ - \\gamma$. Analogously, $\\angle BA_1C = \\alpha$, $\\angle BH_1C = 180^\\circ - \\alpha$, $\\angle CB_1A = \\beta$ and $\\angle CH_2A = 180^\\circ - \\beta$. It follows that the points $H_1$, $H_2$ and $H_3$ belong to the circumcircle of $\\triangle ABC$.\n\nNow we shall use now vectors. Denote by $[\\vec{a}]$ the image of the vector $\\vec{a}$ under rotation through $360^\\circ - 2\\varphi$. We get from the proved above that $\\angle C_1AH_3 = 90^\\circ - \\gamma = \\angle CAH$. Hence $\\angle CAH_3 = \\varphi$ and then $\\angle COH_3 = 2\\varphi$. Analogously, $\\angle AOH_1 = \\angle BOH_2 = 2\\varphi$.\n\nLet $S$ and $T$ be the orthocenters of $\\triangle A_1B_1C_1$ and $\\triangle H_1H_2H_3$, respectively. We shall prove that $\\overrightarrow{HS} = \\overrightarrow{HT}$.\n\nOne has that\n$$\n\\begin{aligned}\n& \\overrightarrow{HS} = \\overrightarrow{HA_1} + \\overrightarrow{HB_1} + \\overrightarrow{HC_1} = \\overrightarrow{HO} + \\overrightarrow{OA_1} + \\overrightarrow{HO} + \\overrightarrow{OB_1} + \\overrightarrow{HO} + \\overrightarrow{OC_1} \\\\\n& = 3\\overrightarrow{HO} + \\overrightarrow{OA_1} + \\overrightarrow{OB_1} + \\overrightarrow{OC_1} \\\\\n& = \\overrightarrow{HO} + [\\overrightarrow{OA_1} + \\overrightarrow{OB_1} + \\overrightarrow{OC_1} + 2\\overrightarrow{HO}] \\\\\n& = \\overrightarrow{HO} + [\\overrightarrow{OH_1} + \\overrightarrow{OH_2} + \\overrightarrow{OH_3}] \\\\\n& = \\overrightarrow{HO} + [\\overrightarrow{OH}]\n\\end{aligned}\n$$\n\nSimilarly,\n$$\n\\overrightarrow{HT} = \\overrightarrow{HO} + \\overrightarrow{OT} = \\overrightarrow{HO} + \\overrightarrow{OH_1} + \\overrightarrow{OH_2} + \\overrightarrow{OH_3} = \\overrightarrow{HO} + [\\overrightarrow{OH}]\n$$\nwhich means that $S \\equiv T$.\n\nThe case $\\angle C_1AH = \\angle A_1CH = \\angle B_1CH = \\varphi$ can be considered in a similar way and we omit the details.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27189,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $M$ be a point on a circle $k$. A circle $k_{1}$ with center $M$ meets $k$ at points $C$ and $D$. A chord $AB$ of $k$ is tangent to $k_{1}$ at point $H$. Prove that the line $CD$ bisects the segment $MH$ if and only if $AB$ is a diameter of $k$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet $AB$ be a diameter of $k$. Since $\\Varangle D H A = \\Varangle D C H = \\alpha$, $\\Varangle C H B = \\Varangle C D H = \\beta$, $\\Varangle D M H = 2\\alpha$, $\\Varangle C M H = 2\\beta$ and $\\Varangle M D C = 90^{\\circ} - \\alpha - \\beta$, the Sine theorem for $\\triangle D M O$ gives\n$$\nMO = \\frac{r \\cos (\\alpha + \\beta)}{\\cos (\\alpha - \\beta)}\n$$\nwhere $r$ is the radius of $k_{1}$ and $O = MH \\cap CD$. On the other hand, if $MH \\cap k = P$,\n![](attached_image_1.png)\nthen\n$$\n\\Varangle MAP = \\frac{1}{2} \\widehat{MP} = \\frac{1}{2}(\\widehat{DM} + \\widehat{CP}) = \\Varangle DOM = 90^{\\circ} - \\alpha + \\beta\n$$\nApplying the Sine theorem for $\\triangle APM$ we get that $2r = MP = 2R \\cos (\\alpha - \\beta)$, where $R$ is the radius of $k$. The Sine theorem for $\\triangle DMC$ implies that\n$$\nr = MC = 2R \\cos (\\alpha + \\beta)\n$$\nand hence $2 \\cos (\\alpha + \\beta) = \\cos (\\alpha - \\beta)$. Now (1) shows that $O$ is the midpoint of $MH$.\n\nConversely, it is easy to see that if $O$ is the midpoint of $MH$, then $H$ is the midpoint of $MP$, i.e. $AB$ is a diameter of $k$.\n\nAlternative solution.\nLet $AB$ be a diameter of $k$ and $MH$ meets $k$ for (the) second time at the point $P$. It is clear that $MH = HP = r$, where $r$ is the radius of $k_{1}$. Consider the inversion with respect to the circle with center $M$ and radius $r$. Then $P$ is the image of the midpoint $T$ of $MH$. The image of $k$ is the line $DC$. Hence the image of $P$ lies on $DC$, i.e. $T \\in DC$.\n\nConsidering the same inversion also implies easily the converse statement.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27190,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $p$ be a prime number and let $0 \\leq a_{1} < a_{2} < \\cdots < a_{m} < p$ and $0 \\leq b_{1} < b_{2} < \\cdots < b_{n} < p$ be arbitrary integers. Denote by $k$ the number of different remainders of the numbers $a_{i} + b_{j}$, $1 \\leq i \\leq m$, $1 \\leq j \\leq n$, modulo $p$. Prove that:\n\na) if $m + n > p$, then $k = p$;\n\nb) if $m + n \\leq p$, then $k \\geq m + n - 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na.\nLet $t \\in \\{0, 1, 2, \\ldots, p-1\\}$. Consider the remainders of $t - a_{i}$, $1 \\leq i \\leq m$ and $b_{j}$, $1 \\leq j \\leq n$, modulo $p$. Their number is $m + n > p$ and hence two of them are equal. Since the remainders of $t - a_{i}$ and $t - a_{j}$, $b_{i}$ and $b_{j}$, respectively, $i \\neq j$, are different, it follows that $t - a_{r} \\equiv b_{s} \\pmod{p}$, i.e., $a_{r} + b_{s} \\equiv t \\pmod{p}$ for some $r$ and $s$. Since $t$ is an arbitrary remainder modulo $p$, we conclude that $k = p$.\n\nb.\nLet $A = \\{a_{1}, a_{2}, \\ldots, a_{m}\\}$ and $B = \\{b_{1}, b_{2}, \\ldots, b_{n}\\}$. For any two sets $X$ and $Y$ denote $X + Y = \\{x + y \\pmod{p} \\mid x \\in X, y \\in Y\\}$. We have to prove that $k = |A + B| \\geq m + n - 1$. To do this, we may assume that $m \\leq n$ and we shall use induction on $m$.\n\nFor $m = 1$ and any $n$ the statement is true, since $a_{1} + b_{i} \\neq a_{1} + b_{j} \\pmod{p}$ if $i \\neq j$ and $|a_{1} + B| = |B| = n = 1 + n - 1$.\n\nSuppose that the statement is true for any two sets $X$ and $Y$ such that $|X| < m$, $|X| < |Y|$ and $|X| + |Y| \\leq p$. Let $|A| = m > 1$ and $|B| = n$, where $m \\leq n$ and $m + n \\leq p$. Then $n < p$ and hence there exists $c \\notin B$. Take different $a_{1}, a_{2} \\in A$. As the sequence $c + t(a_{2} - a_{1}) \\pmod{p}$, $t = 1, 2, \\ldots, p-1$, contains all remainders except $c$, then $b = c + t(a_{2} - a_{1}) \\in B$ for some $t$. Let $t$ be the minimal number with this property. The set $A' = \\{b - a_{2}\\} + A$ contains the elements $b - a_{2} + a_{1}$ and $b - a_{2} + a_{2} = b$. Note that $b - a_{2} + a_{1} = c + (t-1)(a_{2} - a_{1}) \\notin B$. Since $|A' + B| = |\\{b - a_{2}\\} + A + B|$, it is enough to prove that $|A' + B| \\geq m + n - 1$.\n\nSet $F = A' \\cap B$ and $G = A' \\cup B$. Since $b \\in F$, $b - a_{2} + a_{1} \\notin F$ and $b - a_{2} + a_{1} \\in A'$, then $F$ is a proper non-empty subset of $A'$. So $B$ is a proper subset of $G$. It follows that $0 < |F| < m \\leq n < |G|$. On the other hand, $m + n = |A'| + |B| = |A' \\cap B| + |A' \\cup B| = |F| + |G|$. Note also that $F + G \\subset A' + B$ (for $f \\in F$ and $g \\in G$, we may assume that $g \\in A'$ and then $f \\in F \\subset B$ implies that $f + g \\in A' + B$). Thus $|A'| + |B| \\geq |F| + |G|$. Then the inequalities $0 < |F| < m \\leq n < |G|$, $|F| + |G| \\leq p$ and the induction hypotheses imply that the statement is true for the sets $F$ and $G$. Hence\n\n$$\n|A + B| = |A' + B| \\geq |F + G| \\geq |F| + |G| - 1 = |A'| + |B| - 1 = m + n - 1\n$$\nwhich completes the induction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27191,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn any cell of an $n \\times n$ table a number is written such that all the rows are different. Prove that one can remove a column such that the rows in the new table are still different.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nConsider a graph with vertices the lines of the table and edges that join two vertices if the respective lines are different exactly in one position. Write on the edges the respective elements that are different.\n\nAssume that the given statement is not true. Then the graph has $n$ vertices and $n$ edges. It is not difficult to prove by induction on $n$ that there is a cycle in the graph. We may suppose that this cycle is $A_{1} A_{2} \\ldots A_{k}$.\n\nStarting from $A_{1}$, we remove $x_{1}$ to get $A_{2}$, then remove or add $x_{2}$ to get $A_{3}$, etc. When we turn back to $A_{1}$, we obtain a second copy of $A_{1}$ which does not contain $x_{1}$ (any edge contains no $x_{1}$), a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27192,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe points $P$ and $Q$ lie respectively on the diagonals $AC$ and $BD$ of a quadrilateral $ABCD$ and $\\frac{AP}{AC} + \\frac{BQ}{BD} = 1$. The line $PQ$ meets the sides $AD$ and $BC$ at points $M$ and $N$. Prove that the circumcircles of the triangles $AMP$, $BNQ$, $DMQ$ and $CNP$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $AC \\cap BD = O$ and $X$ be the second intersection point of the circumcircles of $\\triangle AOB$ and $\\triangle BOC$. Set $\\Varangle XBO = \\Varangle XCO = \\alpha$ and $\\Varangle XAO = \\Varangle XDO = \\beta$. Since $\\triangle AXC \\sim \\triangle DXB$, then $\\frac{XD}{XA} = \\frac{BD}{AC}$. This and the condition of the problem implies that $\\frac{AP}{AC} = 1 - \\frac{BQ}{BD} = \\frac{DQ}{BD}$ and hence $\\frac{AP}{DQ} = \\frac{AC}{BD} = \\frac{XA}{XD}$. Then $\\triangle APX \\sim \\triangle DQX$ and so $\\Varangle APX = \\Varangle DQX$. The last equality means that the points $X, Q, O$ and $P$ are concyclic and thus $\\Varangle XQP = \\Varangle XOP = \\Varangle XDA$.\n\n![](attached_image_1.png)\n\nThis implies that the points $X, Q, D$ and $M$ are concyclic, i.e. the circumcircle of $\\triangle DMQ$ passes through $X$. Then $\\Varangle XMN = \\Varangle \\beta$ and hence the points $X, A, P$ and $M$ are concyclic, i.e., the circumcircle of $\\triangle AMP$ passes through $X$.\n\nIt follows in the same way that the circumcircle of $\\triangle CNP$ passes through $X$ and then analogously the circumcircle of $\\triangle BNQ$ passes through $X$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27193,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $A_{1}, A_{2}, \\ldots, A_{n}$ be finite sets such that\n$$\n\\left|A_{i} \\cap A_{i+1}\\right|>\\frac{n-2}{n-1}\\left|A_{i+1}\\right|\n$$\nfor any $i=1,2, \\ldots, n$ ($A_{n+1} \\equiv A_{1}$). Prove that their intersection is a nonempty set.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nWe may assume the set $A_{1}$ has maximal cardinality. Denote $A_{i} \\cap A_{i+1} = B_{i}$, $i=1,2, \\ldots, n$. Since $A_{n} \\supset B_{n-1} \\cup B_{n}$, then\n$$\n\\begin{aligned}\n\\left|A_{n}\\right| & \\geq \\left|B_{n-1} \\cup B_{n}\\right| = \\left|B_{n-1}\\right| + \\left|B_{n}\\right| - \\left|B_{n-1} \\cap B_{n}\\right| \\\\\n& > \\frac{n-2}{n-1}\\left|A_{n}\\right| + \\frac{n-2}{n-1}\\left|A_{1}\\right| - \\left|B_{n-1} \\cap B_{n}\\right|\n\\end{aligned}\n$$\nHence\n$$\n\\left|B_{n-1} \\cap B_{n}\\right| > \\frac{n-2}{n-1}\\left|A_{1}\\right| - \\frac{1}{n-1}\\left|A_{n}\\right| \\geq \\frac{n-3}{n-1}\\left|A_{1}\\right|\n$$\ni.e., $\\left|A_{n-1} \\cap A_{n} \\cap A_{1}\\right| > \\frac{n-3}{n-1}\\left|A_{1}\\right|$. Further, if $C = A_{n-1} \\cap A_{n} \\cap A_{1}$, then $A_{n-1} \\supset C \\cup B_{n-2}$ and\n$$\n\\begin{aligned}\n\\left|A_{n-1}\\right| & \\geq \\left|B_{n-2} \\cup C\\right| = \\left|B_{n-2}\\right| + |C| - \\left|B_{n-2} \\cap C\\right| \\\\\n& > \\frac{n-2}{n-1}\\left|A_{n-1}\\right| + \\frac{n-3}{n-1}\\left|A_{1}\\right| - \\left|B_{n-2} \\cap C\\right|\n\\end{aligned}\n$$\nSo $\\left|B_{n-2} \\cap C\\right| > \\frac{n-3}{n-1}\\left|A_{1}\\right| - \\frac{1}{n-1}\\left|A_{n-1}\\right| \\geq \\frac{n-4}{n-1}\\left|A_{1}\\right|$, i.e.\n$$\n\\left|A_{n-2} \\cap A_{n-1} \\cap A_{n} \\cap A_{1}\\right| > \\frac{n-4}{n-1}\\left|A_{1}\\right|\n$$\nWe get by induction that\n$$\n\\left|A_{n-k} \\cap A_{n-k+1} \\cap \\cdots \\cap A_{n-1} \\cap A_{n} \\cap A_{1}\\right| > \\frac{n-k-2}{n-1}\\left|A_{1}\\right|\n$$\nfor $k=1,2, \\ldots, n-2$. In particular, $\\left|A_{2} \\cap A_{3} \\cap \\cdots \\cap A_{n-1} \\cap A_{n} \\cap A_{1}\\right| > 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27194,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $a$, $b$ and $n$ be positive integers. Denote by $K(n)$ the number of the representations of $1$ as a sum of $n$ numbers of the form $\\frac{1}{k}$, where $k$ is a positive integer. Let $L(a, b)$ be the least positive integer $m$ such that the equation $\\sum_{i=1}^{m} \\frac{1}{x_{i}}=\\frac{a}{b}$ has a solution in positive integers and set $L(b)=\\max \\{L(a, b), 1 \\leq a \\leq b\\}$. Prove that the number of the positive divisors of $b$ does not exceed $2 L(b)+K(L(b)+2)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nThe function $K(n)$ is increasing for $n \\geq 3$, since $\\sum_{i=1}^{n} \\frac{1}{x_{i}}=1$ implies that\n$$\n\\sum_{i=1}^{n-1} \\frac{1}{x_{i}}+\\frac{1}{x_{n}+1}+\\frac{1}{x_{n}(x_{n}+1)}=1\n$$\nThus it is enough to find $t \\leq L(b)$ such that $K(t+2)+2 L(b) \\geq d(b)$, where $d(b)$ is the number of the distinct positive integers that divide $b$. Let $t$ be the minimal positive integer, for which the equation $\\sum_{i=1}^{t} \\frac{1}{x_{i}}=1-\\frac{1}{b}$ has a solution. Then $t \\leq L(b)$. Fix now $t$, $b$, $x_{1}, \\ldots, x_{t}$.\n\nNote that the numbers of the solution of the equation $\\frac{1}{y_{1}}+\\frac{1}{y_{2}}=\\frac{1}{b}$ such that $b$ divides $y_{2}$ and $y_{1} \\leq y_{2}$, is equal to $d(b)$. Indeed, if $\\frac{1}{b}=\\frac{1}{y_{1}}+\\frac{1}{k b}$ for some $k \\geq 2$, then $y_{1}=b+\\frac{b}{k-1}$. So $k-1$ divides $b$ and there are exactly $d(b)$ possibilities for $k$.\n\nHence $K(t+2)$ is not less than $d(b)$ minus the number of the cases, when $y_{i}=x_{j}$. This cases are at most $2 L(b)$ which implies the desired inequality.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27195,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe edges of a graph with $2n$ vertices, $n \\geq 4$, are colored in blue and red such that there is no blue triangle and there is no red complete subgraph with $n$ vertices. Find the least possible number of the blue edges.",
    "options": [],
    "answer": "f(4) = 10; for n ≥ 5, f(n) = n + 5",
    "solution": "Solution:\n\nWe call a graph, satisfying the given condition, $n$-purple. Let $f(n)$ be the smallest possible number of blue edges in an $n$-purple graph.\n\nSuppose that $f(n) < n + 5$ for $n \\geq 5$. If any vertex of an $n$-purple graph $G$ with $f(n)$ blue edges is a head of at least two blue edges, then the total number of the blue edges is at least $2n$. Since $2n \\geq n + 5 > f(n)$ for $n \\geq 5$, we may find a vertex $a$ of $G$ that is the head of at most one blue edge. If $a$ is not head of a blue edge, then there is a vertex $b$ of the graph $G \\setminus \\{a\\}$ that is a head of a blue edge and hence $G \\setminus \\{a, b\\}$ is an $(n-1)$-purple graph. If $a$ and $b$ are joined by a blue edge, then $G \\setminus \\{a, b\\}$ is an $(n-1)$-purple graph. In both cases the obtained $(n-1)$-purple graph has at least one blue edge less than $G$ and so $f(n) \\geq f(n-1) + 1$. Then $f(n-1) < n + 4$, in particular, $f(4) < 9$.\n\nNow we shall compute $f(4)$. It is well-known that there are $3$-purple graphs. Then the above arguments show that any vertex of a $4$-purple graph is a head of at least two blue edges. Hence $f(4) \\geq 8$. If $f(4) = 8$, then any vertex is a head of exactly two blue edges. There are two such graphs containing no blue triangles but these graphs are not purple. If $f(4) = 9$, the two vertices are heads of three blue edges, and the remaining six vertices are heads of two blue edges. There are six such graphs containing no blue triangles but they are not purple. The following example shows that $f(4) = 10$ — a regular octagon with blue sides and two blue adjacent main diagonals, and the remaining diagonals are red.\n\nSo $f(n) \\geq n + 5$ for $n \\geq 5$. It is easy to see that the graph with $2n$ vertices such that its blue edges form three disjoint cycles with lengths $5, 5$ and $2n - 10$ is $n$-purple. Hence $f(n) = n + 5$ for $n \\geq 5$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 27196,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe bisectors of $\\angle A$, $\\angle B$ and $\\angle C$ of $\\triangle ABC$ meet its circumcircle at points $A_1$, $B_1$ and $C_1$, respectively. Set $AA_1 \\cap CC_1 = I$, $AA_1 \\cap BC = N$ and $BB_1 \\cap A_1C_1 = P$. Denote by $O$ the circumcenter of $\\triangle IPC_1$ and let $OP \\cap BC = M$. If $BM = MN$ and $\\angle BAC = 2 \\angle ABC$, find the angles of $\\triangle ABC$.",
    "options": [],
    "answer": "∠A = 72°, ∠B = 36°, ∠C = 72°",
    "solution": "Solution:\n\nSet $\\angle CAB = \\alpha$, $\\angle ABC = \\beta$ and $\\angle BCA = \\gamma$. Then\n$$\n\\begin{aligned}\n\\angle IPC_1 & = \\frac{1}{2}\\left(\\overparen{BA_1} + \\overparen{C_1B_1}\\right) \\\\\n& = \\frac{1}{2}\\left(\\overparen{BA_1} + \\overparen{AC_1} + \\overparen{AB_1}\\right) \\\\\n& = \\frac{1}{2}(\\alpha + \\beta + \\gamma) = 90^\\circ.\n\\end{aligned}\n$$\nHence $O$ is the midpoint of the segment $IC_1$ and it follows that\n\n![](attached_image_1.png)\n\n$$\n\\angle IOP = 2 \\angle IC_1P = \\overparen{CA_1} = \\alpha.\n$$\nWe also have $\\angle CC_1B = \\alpha$ and therefore $OP \\parallel C_1B$. Since $C_1O = OI$ and $BM = MN$, we conclude that $IN \\parallel C_1B$, i.e. $\\angle CIA_1 = \\alpha$. On the other hand,\n$$\n\\angle CIA_1 = \\frac{1}{2}\\left(\\overparen{CA_1} + \\overparen{AC_1}\\right) = \\frac{1}{2}(\\alpha + \\gamma) = 90^\\circ - \\frac{\\beta}{2}.\n$$\nThis implies that $\\alpha = 90^\\circ - \\frac{\\beta}{2} = \\frac{\\alpha + \\gamma}{2}$, i.e. $\\alpha = \\gamma$. Since $\\alpha = 2\\beta$, we obtain that $\\alpha = \\gamma = 72^\\circ$ and $\\beta = 36^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27197,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn a volleyball tournament for the Euro-African cup the European teams are 9 more than the African teams. Every two teams met exactly once and the European teams gained 9 times more points than the African teams (the winner takes 1 point and the loser takes 0 point). What are the maximum possible points gained by an African team?",
    "options": [],
    "answer": "11",
    "solution": "Solution:\n\nDenote by $x$ the number of African teams. Then the number of European teams equals $x+9$. The African teams played each other $\\frac{(x-1)x}{2}$ games and therefore the points won by them are $\\frac{(x-1)x}{2} + k$, where $k$ is the number of wins over European teams.\n\nFurther, the points won by the Europeans are $\\frac{(x+8)(x+9)}{2} + x(x+9) - k$. Thus,\n$$\n9\\left(\\frac{(x-1)x}{2} + k\\right) = \\frac{(x+8)(x+9)}{2} + x(x+9) - k\n$$\nand so $3x^2 - 22x + 10k - 36 = 0$. Since $x$ is a positive integer, we have that $121 - 3(10k - 36) = 229 - 30k$ is a perfect square. Then $k \\leq 7$ and a direct verification shows that we obtain perfect squares only for $k=2$ and $k=6$. For $k=2$ we have $x=8$ and therefore the best African team could have at most $7+2=9$ points.\n\nFor $k=6$ we get $x=6$ and therefore there are 6 African and 15 European teams. In this case the best African team has at most $5+6=11$ points, which happens if it wins over all other African teams and 6 European teams (the other African teams lost their games against all European teams). Finally, the answer is $11$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27198,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn every cell of an $n \\times n$ table one of the numbers $-1, 0$ and $1$ is written. Is it possible the sums of the numbers in every row and every column to be $2n$ mutually different numbers, if:\n\na) $n=4$;\n\nb) $n=5$?",
    "options": [],
    "answer": "a) Yes; b) No",
    "solution": "Solution:\n\na) Yes. Here is an example:\n$$\n\\left(\\begin{array}{rrrr}\n1 & 0 & 1 & 1 \\\\\n1 & -1 & -1 & -1 \\\\\n1 & -1 & 1 & 0 \\\\\n1 & -1 & 1 & -1\n\\end{array}\\right)\n$$\n\nb) No. We have 11 possibilities for these sums: $0, \\pm 1, \\pm 2, \\pm 3, \\pm 4$ and $\\pm 5$. Denote by $a_{i}$ the sum of the numbers in the $i$-th row and by $b_{j}$ the sum of the numbers in the $j$-th column. Obviously\n$$\na_{1}+a_{2}+a_{3}+a_{4}+a_{5}=b_{1}+b_{2}+b_{3}+b_{4}+b_{5}\n$$\nwhich shows that the number of odd sums $a_{i}$ and the number of odd sums $b_{j}$ are of the same parity. Therefore all odd sums must be achieved.\n\nWithout loss of generality we may assume that $b_{1}=5$ and then none of $a_{i}$ equals $-5$. Thus, we may suppose that $b_{2}=-5$. At least one of the sums equals $4$ or $-4$ and suppose it is $4$. This is possible only if there is a column with four $1$'s and one $0$. Let $b_{3}=4$ and the zero is in the last row. Therefore $a_{i} \\neq -3$ for every $i$ and we may assume that $b_{4}=-3$. Then in the fourth column we have at least three $-1$'s. If they are in the first four rows we may assume that they are in the first three rows. So $a_{1}, a_{2}, a_{3}$ is a permutation of $-1, 0, 1$. Therefore $b_{5} \\neq 3$ and since $a_{5} \\neq 3$, it follows that $a_{4}=3$, i.e., there are $1$'s in the last two cells of the fourth row. Since $b_{4}=-3$, the number in the last cell of the fourth column is $-1$. Now every possibility for the number in the 5th row and 5th column leads to a contradiction.\n\nIt remains to consider the case when there are at most two $-1$'s in the first four cells of the fourth column. We may assume that the numbers in the fourth column are consequently $-1, -1, 0, 0$ and $-1$. Since the sum of the first four numbers of the rows equal $0, 0, 1, 1$ and $-1$ we have that none of the rows equals $3$ and therefore $b_{5}=3$. We must use different numbers for rows 1 and 2 and for rows 3 and 4, so we may have at most three $1$'s. Hence we may assume that the numbers in the fifth column are $1, 0, 1, 0, 1$ and then $a_{1}=a_{4}=1$, a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27199,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nProve that among any $2n+1$ irrational numbers there are $n+1$ numbers such that the sum of any $2,3, \\ldots, n+1$ of them is an irrational number.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet the given numbers be $a_{1}, a_{2}, \\ldots, a_{2n+1}$. Choose first 1 and then choose at any step (if it is possible) a number that is not a linear combination with rational coefficients of the already chosen numbers. We may assume that the chosen numbers are $a_{0}=1, a_{1}, a_{2}, \\ldots, a_{k}$, $1 \\leq k \\leq 2n+1$. It is easy to see that any linear combination with rational coefficients of the given numbers can be uniquely presented as a linear combination of these numbers.\n\nLet $a_{i}=\\sum_{j=0}^{k} \\alpha_{ij} a_{j}$, where $\\alpha_{ij} \\in \\mathbb{Q}$, $1 \\leq i \\leq 2n+1$, $0 \\leq j \\leq k$. Then a sum of $a_{i}$'s is a rational number if and only if the sum of the corresponding numbers $b_{i}=a_{i}-\\alpha_{i0}$ vanishes. Since $b_{1}, b_{2}, \\ldots, b_{2n+1}$ are irrational numbers, they are non-zero. In particular, at least $n+1$ of them have the same sign and hence the corresponding $a_{i}$'s have the desired property.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27200,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all values of $a$ such that the equation\n$$\n(a^{2}-a-9) x^{2}-6 x-a=0\n$$\nhas two distinct positive roots.",
    "options": [],
    "answer": "a ∈ (-3, (1 − √37)/2)",
    "solution": "Solution:\nThe equation has two distinct positive roots if and only if\n$$\n\\left\\lvert\\,\n\\begin{aligned}\n& D=a^{3}-a^{2}-9 a+9>0 \\\\\n& x_{1}+x_{2}=\\frac{6}{a^{2}-a-9}>0 \\\\\n& x_{1} x_{2}=\\frac{-a}{a^{2}-a-9}>0\n\\end{aligned}\n\\right.\n$$\nThe first inequality is satisfied for $a \\in(-3,1) \\cup(3,+\\infty)$, the second one for $a \\in((1-\\sqrt{37}) / 2,(1+\\sqrt{37}) / 2)$, and the third - for every $a \\in(-\\infty, 0)$. Therefore the required values of $a$ are $a \\in(-3,(1-\\sqrt{37}) / 2)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27201,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all values of $a$ such that the equation\n$$\n\\sqrt{\\left(4 a^{2}-4 a-1\\right) x^{2}-2 a x+1}=1-a x-x^{2}\n$$\nhas exactly two solutions.",
    "options": [],
    "answer": "a = 1/3, a = 1/2, and a in (5/6, 3/2] excluding 1",
    "solution": "Solution:\nSquaring the equation\n$$\n\\sqrt{\\left(4 a^{2}-4 a-1\\right) x^{2}-2 a x+1}=1-a x-x^{2}\n$$\ngives the equation\n$$\nx^{2}\\left(x^{2}+2 a x-3 a^{2}+4 a-1\\right)=0\n$$\nwith roots $x_{1}=0$, $x_{2}=1-3 a$ and $x_{3}=a-1$. It is clear that $x_{1}=0$ is a root of (1) for any $a$. On the other hand, $x_{2}=1-3 a$ is a root of (1) if its right-hand side is non-negative, i.e., if\n$$\n1-a(1-3 a)-(1-3 a)^{2} \\geq 0 \\Longleftrightarrow 5 a-6 a^{2} \\geq 0 \\Longleftrightarrow a \\in\\left[0, \\frac{5}{6}\\right]\n$$\nAnalogously, $x_{3}=a-1$ is a root of (1) for $a \\in\\left[0, \\frac{3}{2}\\right]$. Two cases are possible.\n\nCase 1. Some of the numbers $x_{1}$, $x_{2}$ and $x_{3}$ are equal. This implies that $a=\\frac{1}{3}$, $\\frac{1}{2}$ or $1$. It follows from above that $a=\\frac{1}{3}$ and $a=\\frac{1}{2}$ are solutions of the problem.\n\nCase 2. The numbers $x_{1}$, $x_{2}$ and $x_{3}$ are pairwise different. Then it is easy to see that $a \\in\\left(\\frac{5}{6}, \\frac{3}{2}\\right] \\setminus\\{1\\}$.\n\nSo, the desired values of $a$ are $a=\\frac{1}{3}$, $a=\\frac{1}{2}$ and $a \\in\\left(\\frac{5}{6}, \\frac{3}{2}\\right] \\setminus\\{1\\}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27202,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nConsider the system\n$$\n\\begin{cases}\n  x^{2} + y^{2} = a^{2} + 2 \\\\\n  \\dfrac{1}{x} + \\dfrac{1}{y} = a\n\\end{cases}\n$$\nwhere $a$ is a real number.\n\na) Solve the system for $a=0$.\n\nb) Find all $a$, for which the system has exactly two solutions.",
    "options": [],
    "answer": "a) (1, −1) and (−1, 1). b) a ∈ (−√2, √2).",
    "solution": "Solution:\n\na) If $a=0$, then $x=-y$ and hence $2x^{2}=2$. It follows that $(x, y) = (1, -1)$ or $(x, y) = (-1, 1)$.\n\nb) We know from a) that $a=0$ is one of the desired numbers. Let $a \\neq 0$. Setting $x+y=p$, $xy=q$, we have $p=a q$ and $p^{2}-2q=a^{2}+2$. Then $a^{2} q^{2}-2q-a^{2}-2=0$ and hence $(p, q)=(-a, -1)$ or $(p, q)=\\left(\\dfrac{a^{2}+2}{a}, \\dfrac{a^{2}+2}{a^{2}}\\right)$. Note that these pairs of numbers are different. The first case $x+y=-a$, $xy=-1$ leads to the quadratic equation $z^{2}+a z-1=0$ which has two distinct real roots $z_{1}$ and $z_{2}$. It follows that $(x, y) = (z_{1}, z_{2})$ and $(x, y) = (z_{2}, z_{1})$ are solutions of the given system. Thus we have to find all $a$ for which the second case is impossible. This means that the discriminant of the quadratic equation $z^{2}-\\dfrac{a^{2}+2}{a} z+\\dfrac{a^{2}+2}{a^{2}}=0$ is negative, i.e., $a \\in (-\\sqrt{2}, \\sqrt{2}) \\setminus \\{0\\}$. So the answer to b) is $a \\in (-\\sqrt{2}, \\sqrt{2})$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27203,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $I$ be the incenter of $\\triangle ABC$ and $M$ be the midpoint of the side $AB$. Find the least possible value of $\\Varangle CIM$ if $CI = MI$.",
    "options": [],
    "answer": "150°",
    "solution": "Solution:\nWe may assume that $AC < BC$. Since $\\Varangle ACI$ and $\\Varangle AMI$ are acute, then $\\triangle ACI \\cong \\triangle AMI$.\nHence $AC = AM$ and $\\Varangle AIC = \\Varangle AIM$, i.e.\n$$\n\\Varangle CIM = 360^\\circ - 2 \\Varangle AIC = 180^\\circ - \\Varangle ABC.\n$$\nNote that $\\Varangle ABC$ is maximal if $BC$ is tangent to the circle with center $A$ and radius $AM$.\n\n![](attached_image_1.png)\n\nThen $\\Varangle ACB = 90^\\circ$, $\\Varangle ABC = 30^\\circ$, and hence the least possible value of $\\Varangle CIM$ is $150^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27204,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $A_1$ and $B_1$ be points on the sides $AC$ and $BC$ of $\\triangle ABC$ such that $4 AA_1 \\cdot BB_1 = AB^2$. If $AC = BC$, prove that the line $AB$ and the bisectors of $\\Varangle AA_1B_1$ and $\\Varangle BB_1A_1$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nDenote by $M$ the midpoint of $AB$. It follows that $\\frac{AM}{BB_1} = \\frac{AA_1}{BM}$. Then $\\triangle AMA_1 \\sim \\triangle BB_1M$ and hence $\\frac{AA_1}{BM} = \\frac{MA_1}{B_1M}$, i.e., $\\frac{AA_1}{AM} = \\frac{MA_1}{MB_1}$. Moreover, $\\Varangle AA_1M = \\Varangle BMB_1$ and therefore\n$$\n\\begin{aligned}\n\\Varangle A_1MB_1 & = 180^\\circ - \\Varangle AMA_1 - \\Varangle BMB_1 \\\\\n& = 180^\\circ - \\Varangle AMA_1 - \\Varangle AA_1M \\\\\n& = \\Varangle A_1AM.\n\\end{aligned}\n$$\n![](attached_image_1.png)\nThus $\\triangle AMA_1 \\sim \\triangle MB_1A_1$ which implies that $\\Varangle AA_1M = \\Varangle MA_1B_1$. Since $\\triangle BB_1M \\sim \\triangle AMA_1 \\sim \\triangle MB_1A_1$, it follows that $\\Varangle BB_1M = \\Varangle MB_1A_1$. Hence $M$ is the intersection point of the bisectors of $\\Varangle AA_1B_1$ and $\\Varangle BB_1A_1$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27205,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe diagonals $AC$ and $BD$ of a cyclic quadrilateral $ABCD$ with circumcenter $I$ intersect at a point $E$. If the midpoints of segments $AD$, $BC$ and $IE$ are collinear, prove that $AB = CD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nDenote the midpoints of $AD$ and $BC$ by $M$ and $N$, respectively. Without loss of generality we may assume that $I$ lies in the interior of $AMNB$ and $E$ lies in the interior of $MDCN$. Then we have\n$$\n\\begin{aligned}\nS_{MIN} & = S_{AMNB} - S_{ABI} - S_{AMI} - S_{BNI} \\\\\n& = S_{AMNB} - S_{ABI} - \\frac{1}{2}\\left(S_{ADI} + S_{BCI}\\right) \\\\\n& = S_{AMNB} - \\frac{1}{2} S_{ABI} - \\frac{1}{2}\\left(S_{ABCD} - S_{CDI}\\right)\n\\end{aligned}\n$$\n\n![](attached_image_1.png)\n\nAnalogously, $S_{MEN} = S_{MDCN} - \\frac{1}{2} S_{DCE} - \\frac{1}{2}\\left(S_{ABCD} - S_{ABE}\\right)$. Now using $S_{MIN} = S_{MEN}$ we get\n\n$S_{AMNB} - \\frac{1}{2} S_{ABI} - \\frac{1}{2}\\left(S_{ABCD} - S_{CDI}\\right) = S_{MDCN} - \\frac{1}{2} S_{DCE} - \\frac{1}{2}\\left(S_{ABCD} - S_{ABE}\\right)$,\n\n$$\n\\begin{aligned}\n& \\frac{1}{2} S_{ADN} + \\frac{1}{2} S_{ABC} - \\frac{1}{2} S_{ABI} - \\frac{1}{2}\\left(S_{ABCD} + \\frac{1}{2} S_{CDI}\\right) \\\\\n= & \\frac{1}{2} S_{ADN} + \\frac{1}{2} S_{CBD} - \\frac{1}{2} S_{DCE} - \\frac{1}{2}\\left(S_{ABCD} + \\frac{1}{2} S_{ABE}\\right) \\\\\n& S_{ABC} - S_{CBD} + S_{DCE} - S_{ABE} = S_{ABI} - S_{CDI} \\\\\n& S_{ABE} - S_{CDE} + S_{DCE} - S_{ABE} = S_{ABI} - S_{CDI}\n\\end{aligned}\n$$\n\nTherefore $S_{ABI} = S_{CDI}$, whence $AB = CD$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27206,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $a, b, c > 0$ and $a + b + c = 1$. Prove that\n$$\n\\frac{9}{10} \\leq \\frac{a}{1 + b c} + \\frac{b}{1 + c a} + \\frac{c}{1 + a b} < 1\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nTo prove the right inequality, it is enough to use that the denominators are greater than $1$. Hence\n$$\n\\frac{a}{1 + b c} + \\frac{b}{1 + c a} + \\frac{c}{1 + a b} < a + b + c = 1\n$$\n\nTo show the left inequality, we may assume that $a \\leq b \\leq c$. Then\n$$\n\\frac{1}{1 + b c} \\leq \\frac{1}{1 + c a} \\leq \\frac{1}{1 + a b}\n$$\nApplying consecutively the Chebyshev inequality, the Arithmetic mean - Harmonic mean inequality and the well-known inequality $(a + b + c)^2 \\geq 3(a b + b c + c a)$ we get that\n$$\n\\begin{aligned}\n& 3\\left(\\frac{a}{1 + b c} + \\frac{b}{1 + c a} + \\frac{c}{1 + a b}\\right) \\geq (a + b + c)\\left(\\frac{1}{1 + b c} + \\frac{1}{1 + c a} + \\frac{1}{1 + a b}\\right) \\\\\n& = \\frac{1}{1 + b c} + \\frac{1}{1 + c a} + \\frac{1}{1 + a b} \\geq \\frac{9}{3 + a b + b c + c a} \\geq \\frac{9}{3 + \\frac{(a + b + c)^2}{3}} = \\frac{27}{10}\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27207,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the least number of colors with the following property: the integers $1,2, \\ldots, 2004$ can be colored such that there are no integers $a < b < c$ of the same color for which $a$ divides $b$ and $b$ divides $c$.",
    "options": [],
    "answer": "6",
    "solution": "Solution:\nDenote by $f(n)$ the least number of colors such that the integers $1,2, \\ldots, n$ can be colored in the required way. We shall prove that $f(n) = \\lfloor (k+1)/2 \\rfloor$, where $2^{k-1} \\leq n < 2^{k}$.\n\nObserve that in the sequence $1, 2, 2^{2}, \\ldots, 2^{k-1}$ we have no three numbers of the same color. This means that $f(n) \\geq \\lfloor (k+1)/2 \\rfloor$.\n\nConsider the following coloring by $\\lfloor (k+1)/2 \\rfloor$ colors (each color is identified with an integer among $1,2, \\ldots, \\lfloor (k+1)/2 \\rfloor$). If $m = p_{1}^{\\alpha_{1}} p_{2}^{\\alpha_{2}} \\ldots p_{t}^{\\alpha_{t}} \\leq n$, where $p_{i}$ are primes, then we have $h(m) := \\alpha_{1} + \\cdots + \\alpha_{t} < k$ and we can correctly color $m$ by the color $\\lfloor (h(m)+1)/2 \\rfloor$.\n\nIf $a$ divides $b$ and $b$ divides $c$, then we have $h(a) < h(b) < h(c)$, i.e., $h(c) - h(a) \\geq 2$. This means that the numbers $a$ and $c$ have different colors. Hence $f(n) = \\lfloor (k+1)/2 \\rfloor$.\n\nNow applying the above formula for $n = 2004$ we get $f(2004) = 6$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27208,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all odd prime numbers $p$ which divide the number $1^{p-1}+2^{p-1}+\\cdots+2004^{p-1}$.",
    "options": [],
    "answer": "17, 2003",
    "solution": "Solution:\nNote that $k^{p-1} \\equiv 0 \\pmod{p}$ if $p$ divides $k$ and $k^{p-1} \\equiv 1 \\pmod{p}$ otherwise (by Fermat's little theorem). Then\n$$\n0 \\equiv 1^{p-1}+2^{p-1}+\\cdots+2004^{p-1} \\equiv 0 \\cdot\\left[\\frac{2004}{p}\\right]+1 \\cdot\\left(2004-\\left[\\frac{2004}{p}\\right]\\right) \\pmod{p}\n$$\nwhich implies\n$$\n2004 \\equiv \\left[\\frac{2004}{p}\\right] \\pmod{p}\n$$\n(in particular, $p<2004$). Let $2004=qp+r$, where $0 \\leq r \\leq p-1$. Then $\\left[\\frac{2004}{p}\\right]=\\left[q+\\frac{r}{p}\\right]=q$ and $(1)$ is equivalent to $r \\equiv q \\pmod{p}$.\n\nFor $q<p$, this congruence gives $r=q$. Then\n$$\n2004=(p+1)q \\leq p^{2}-1\n$$\nand therefore $p \\geq 47$. Since $p+1$ divides $2004=3 \\cdot 4 \\cdot 167$, we get $p=2003$ which is a solution of the problem.\n\nFor $q \\geq p$, we have that $2004 \\geq pq \\geq p^{2}$, i.e., $p \\leq 43$. A direct verification of (1) shows that $p=17$ is the only solution in this case.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27209,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSolve in integers the equation\n$$\nx^{3}+10 x-1=y^{3}+6 y^{2}\n$$",
    "options": [],
    "answer": "(x, y) = (6, 5) and (2, -3)",
    "solution": "Solution:\nIt is clear that $x$ and $y$ have different parity. Then $k = x - y$ is an odd number and\n$$\n(3k - 6) y^{2} + (3k^{2} + 10) y + k^{3} + 10k - 1 = 0\n$$\nThe discriminant of this equation is equal to\n$$\nD = -3k^{4} + 24k^{3} - 60k^{2} + 252k + 76\n$$\nand must be a perfect square. Since $D = -k^{2}(k^{2} - 24k + 60) + 252k + 76$, then $D < 0$ for $k \\leq -1$. On the other hand, $D = 3k^{3}(8 - k) + 2(38 - k^{2}) + 2k(126 - 29k)$ and therefore $D < 0$ for $k \\geq 8$. Since $D = -71 < 0$ for $k = 7$, it remains to check the cases $k = 1, 3, 5$. We have $D = 289 = 17^{2}$, $D = 697$ and $D = 961 = 31^{2}$, respectively, that give the solutions $x = 6$, $y = 5$ and $x = 2$, $y = -3$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27210,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA square $n \\times n$ ($n \\geq 2$) is divided into $n^{2}$ unit squares colored in black or white such that the squares at the four corners of any rectangle (containing at least four squares) have not the same color. Find the maximum possible value of $n$.",
    "options": [],
    "answer": "4",
    "solution": "Solution:\n\nWe shall show that the desired value is equal to $4$.\n\nLet $(i, j)$ be the unit square in the $i$-th row and the $j$-th column of a $4 \\times 4$ square. It is easy to check that if the squares $(1,1), (1,2), (2,1), (2,3), (3,2), (3,4), (4,3), (4,4)$ are white and the others are black, then the condition of the problem is satisfied.\n\nTo prove that for $n \\geq 5$ there is no coloring with the required properties, it is enough to deal only with the case $n=5$ (any $n \\times n$ square contains a $5 \\times 5$ square). Consider an arbitrary black-white coloring of a $5 \\times 5$ square. At least $13$ of the unit squares have the same coloring, for example, black. Three cases are possible.\n\nCase 1. One of the rows, say $l$, contains only black squares. Then one of the other rows contains at least $2$ black squares. Hence the corners of the rectangle with two vertices at these squares and the other two in $l$ are black.\n\nCase 2. One of the rows, say $l$, contains exactly $4$ black squares. Then one of the other rows contains at least $3$ black squares. Then at least two of them do not correspond to the white square in $l$. Hence the corners of the rectangle with two vertices at these squares and the other two in $l$ are black.\n\nCase 3. Any row contains at most $3$ black squares. Then at least $3$ rows contain exactly $3$ black squares, say the rows $1,2,3$ from above. We call a column black (white) if its upper square is black (white). There are exactly $3$ black columns. If a row contains $2$ black squares that lie in black columns, then obviously there is a rectangle with black corners. Otherwise, two of the black squares in row $2$ and two of the black squares in the two white columns are black corners of a rectangle.\n\nThe solution is completed.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27211,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nConsider the equations\n$$\n[x]^3 + x^2 = x^3 + [x]^2 \\text{ and } \\left[x^3\\right] + x^2 = x^3 + \\left[x^2\\right]\n$$\nwhere $[t]$ is the greatest integer that does not exceed $t$. Prove that:\na) any solution of the first equation is an integer;\nb) the second equation has a non-integer solution.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\na) Let $x$ satisfy the equality $[x]^3 + x^2 = x^3 + [x]^2$. Then setting $t = [x]$ and $\\alpha = x - t \\in [0,1)$ one has that\n$$\nt^3 - t^2 = (t + \\alpha)^3 - (t + \\alpha)^2 \\Longleftrightarrow \\alpha\\left(\\alpha^2 + (3t - 1)\\alpha + 3t^2 - 2t\\right) = 0\n$$\nHence either $\\alpha = 0$, or $\\alpha$ is a root of the equation in the brackets. In the second case the discriminant $(3t + 1)(1 - t)$ of this equation must be non-negative. Since $t$ is an integer, it follows that $t = 0$ or $t = 1$. Then either $\\alpha = 0$, $\\alpha = 1$ or $\\alpha = -1$. Now $\\alpha \\in [0,1)$ implies that $\\alpha = 0$, i.e., $x$ is an integer.\n\nb) The degree of the polynomial $y^3 - y^2 - 1$ is odd. Hence this polynomial has a real zero $\\alpha$. Obviously, $\\alpha$ is not an integer (in fact, $\\alpha$ is unique and $\\alpha \\in (1,2)$). Then $\\left[\\alpha^3\\right] = \\left[\\alpha^2 + 1\\right] = \\left[\\alpha^2\\right] + 1$ and hence $\\left[\\alpha^3\\right] - \\left[\\alpha^2\\right] = 1 = \\alpha^3 - \\alpha^2$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27212,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all triples $(x, y, z)$ of positive integers such that\n$$\n\\sqrt{\\frac{2005}{x+y}} + \\sqrt{\\frac{2005}{x+z}} + \\sqrt{\\frac{2005}{y+z}}\n$$\nis a positive integer.",
    "options": [],
    "answer": "All permutations of (28070, 4010, 4010).",
    "solution": "Solution:\n1. We first prove the following lemma.\n\nLEMMA. If $p, q, r$ and $\\sqrt{p} + \\sqrt{q} + \\sqrt{r}$ are rational numbers then $\\sqrt{p}$, $\\sqrt{q}$ and $\\sqrt{r}$ are also rational numbers.\n\nProof of the lemma. Let $\\sqrt{p} + \\sqrt{q} + \\sqrt{r} = s$, where $pqr \\neq 0$ and $s$ is a rational number. Then $\\sqrt{p} + \\sqrt{q} = s - \\sqrt{r}$ and we get by squaring that\n$$\np + q + 2\\sqrt{pq} = s^2 + r - 2s\\sqrt{r} \\Longleftrightarrow 2\\sqrt{pq} = s^2 + r - p - q - 2s\\sqrt{r}\n$$\nSquaring the last identity gives $4pq = M^2 + 4s^2 r - 4Ms\\sqrt{r}$, where $M = s^2 + r - p - q > 0$. Therefore $\\sqrt{r}$ is rational and we see in the same way that $\\sqrt{p}$ and $\\sqrt{q}$ are rational. This completes the proof of the lemma.\n\nLet the positive integers $x, y$ and $z$ have the required property. Then the lemma implies that $\\sqrt{\\frac{2005}{x+y}}$, $\\sqrt{\\frac{2005}{x+z}}$ and $\\sqrt{\\frac{2005}{y+z}}$ are rational numbers. Set $\\sqrt{\\frac{2005}{x+y}} = \\frac{a}{b}$, where $a$ and $b$ are coprime positive integers. Then $2005 b^2 = (x+y) a^2$ and it follows that $a^2$ divides $2005$. Hence $a = 1$ and therefore $x + y = 2005 b^2$. In the same way we obtain $x + z = 2005 c^2$ and $y + z = 2005 d^2$, where $c$ and $d$ are positive integers. Then\n$$\n\\sqrt{\\frac{2005}{x+y}} + \\sqrt{\\frac{2005}{x+z}} + \\sqrt{\\frac{2005}{y+z}} = \\frac{1}{b} + \\frac{1}{c} + \\frac{1}{d}\n$$\nis a positive integer. Since $b, c$ and $d$ are positive integers, we have\n$$\n1 \\leq \\frac{1}{b} + \\frac{1}{c} + \\frac{1}{d} \\leq 3\n$$\nIf $\\frac{1}{b} + \\frac{1}{c} + \\frac{1}{d} = 3$, then $b = c = d = 1$ and the system $x + y = x + z = y + z = 2005$ has no solutions in positive integers.\n\nIf $\\frac{1}{b} + \\frac{1}{c} + \\frac{1}{d} = 2$, then one of the numbers $b, c$ and $d$ is equal to $1$ and the other two are equal to $2$. Again, the system for $x, y$ and $z$ has no solutions in positive integers.\n\nIt remains to consider the case $\\frac{1}{b} + \\frac{1}{c} + \\frac{1}{d} = 1$. Suppose that $b \\geq c \\geq d > 1$. Then $\\frac{3}{d} \\geq 1$ and therefore $d = 2$ or $d = 3$. If $d = 3$ we get $b = c = 3$, and if $d = 2$ we have $\\frac{1}{b} + \\frac{1}{c} = \\frac{1}{2}$. This equation has two solutions: $b = 3, c = 6$ and $b = c = 4$.\n\nThe inspection shows that the system for $x, y$ and $z$ has a solution in positive integers only when $d = 2, b = c = 4$ and in this case $x = 14 \\cdot 2005$, $y = z = 2 \\cdot 2005$. Therefore the solutions of the problem are all triples $(x, y, z)$ in which two numbers are equal to $2 \\cdot 2005$ and the third one is equal to $14 \\cdot 2005$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27213,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $ABC$ be an acute triangle. Find the locus of the points $M$ in the interior of $\\triangle ABC$ such that\n$$\nAB - FG = \\frac{MF \\cdot AG + MG \\cdot BF}{CM}\n$$\nwhere $F$ and $G$ are the feet of the perpendiculars from $M$ to the lines $BC$ and $AC$, respectively.",
    "options": [],
    "answer": "All interior points on the segment from C to D, where D is the intersection of the line through the circumcenter O and C with side AB.",
    "solution": "Solution:\n1. Let $AP \\perp FG$ and $BQ \\perp FG$, $P, Q \\in FG$. Then\n$$\nAB \\geq PQ = PG + GF + FQ\n$$\nSince the quadrilateral $CFMG$ is cyclic we obtain $\\angle CMF = \\angle CGF = \\angle AGP$. This implies that $\\triangle APG \\sim \\triangle CFM$ and therefore\n$$\n\\frac{PG}{AG} = \\frac{MF}{CM} \\Longleftrightarrow PG = \\frac{MF \\cdot AG}{CM}\n$$\n![](attached_image_1.png)\nIn the same way we have $QF = \\frac{MG \\cdot BF}{CM}$. Then\n$$\nAB - FG \\geq PG + FQ = \\frac{MF \\cdot AG + MG \\cdot BF}{CM}\n$$\nIt is clear that equality is attained iff $AB \\parallel FG$. The last means that $\\angle BAC = \\angle FGC = \\angle AGP$, whence $\\angle MCB = 90^\\circ - \\angle BAC = \\angle OCB$, where $O$ is the circumcenter of $\\triangle ABC$. Therefore the required locus is the segment $CD$, where $D$ is the intersection point of the line $OC$ and the side $AB$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27214,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nConsider the inequality $|x^{2}-5x+6| \\leq x+a$, where $a$ is a real parameter.\n\na) Solve the inequality for $a=0$.\n\nb) Find the values of $a$ for which the inequality has exactly three integer solutions.",
    "options": [],
    "answer": "a) x in [3 - sqrt(3), 3 + sqrt(3)]. b) a in [-2, 1).",
    "solution": "Solution:\n\na) We consider two cases. If $x \\in (-\\infty, 2] \\cup [3, \\infty)$, then the inequality becomes $x^{2}-6x+6 \\leq 0$, whence $x \\in [3-\\sqrt{3}, 3+\\sqrt{3}]$. Therefore the solutions of the inequality are $x \\in [3-\\sqrt{3}, 2] \\cup [3, 3+\\sqrt{3}]$.\n\nIf $x \\in (2,3)$, then the inequality becomes $x^{2}-4x+6 \\geq 0$, which is satisfied for every $x \\in (2,3)$. Thus $x \\in [3-\\sqrt{3}, 3+\\sqrt{3}]$.\n\nb) If the inequality $|x^{2}-5x+6| \\leq x+a$ has an integral solution $x$ then $x \\in (-\\infty, 2] \\cup [3, \\infty)$ and therefore\n\n$$\nx^{2}-6x+6-a \\leq 0\n$$\n\nThis inequality has a solution if and only if $a \\geq -3$ and in this case we have that $x \\in [3-\\sqrt{a+3}, 3+\\sqrt{a+3}]$. This interval contains the number $3$ and is symmetric with respect to $3$. Therefore it contains exactly three integers if and only if $1 \\leq \\sqrt{a+3} < 2$, i.e. $a \\in [-2,1)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27215,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSolve the system\n$$\n\\left\\lvert\\, \\begin{aligned}\n& 3 \\cdot 4^{x}+2^{x+1} \\cdot 3^{y}-9^{y}=0 \\\\\n& 2 \\cdot 4^{x}-5 \\cdot 2^{x} \\cdot 3^{y}+9^{y}=-8\n\\end{aligned}\\right.\n$$",
    "options": [],
    "answer": "x = 1/2, y = 1 + (1/2) log_3 2",
    "solution": "Solution:\nSet $u=2^{x}>0$ and $v=3^{y}>0$. Then the system becomes\n$$\n\\left\\lvert\\, \\begin{aligned}\n& 3 u^{2}+2 u v-v^{2}=0 \\\\\n& 2 u^{2}-5 u v+v^{2}=-8\n\\end{aligned}\\right.\n$$\nThe first equation can be written as $(u+v)(3 u-v)=0$, whence $u=-v$ or $3 u=v$. The first case is not possible since $u$ and $v$ are positive. Plugging $v=3 u$ in the second equation gives $u^{2}=2$, whence $u=\\sqrt{2}$ and $v=3 \\sqrt{2}$. Thus $x=\\frac{1}{2}$ and $y=1+\\frac{1}{2} \\log _{3} 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27216,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSolve the equation\n$$\n(x+6) 5^{1-|x-1|}-x=(x+1)\\left|5^{x}-1\\right|+5^{x+1}+1\n$$",
    "options": [],
    "answer": "[0,1] ∪ {-1}",
    "solution": "Solution:\nFor $x \\geq 1$ the equation becomes $(x+6)\\left(5^{x}-5^{2-x}\\right)=0$, whence $x=2-x$, i.e. $x=1$.\n\nFor $0 \\leq x<1$ the equation becomes an identity, i.e. every $x \\in [0,1)$ is a solution.\n\nFor $x<0$ we obtain $2\\left(5^{x}-1\\right)(x+1)=0$, whence $x=-1$.\n\nThe solutions of the problem are $x \\in [0,1] \\cup \\{-1\\}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27217,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $k$ be the incircle of $\\triangle ABC$ with $AC \\neq BC$, $I$ be the center of $k$ and let $D$, $E$ and $F$ be the tangent points of $k$ to the sides $AB$, $BC$ and $AC$, respectively.\n\na) If $S = CI \\cap EF$, prove that $\\triangle CDI \\sim \\triangle DSI$.\n\nb) Let $M$ be the second intersection point of $k$ and $CD$. The tangent line to $k$ at $M$ intersects the line $AB$ at point $G$. Prove that $GS \\perp CI$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na) From the right $\\triangle CEI$ we have $EI^2 = SI \\cdot CI = DI^2$. Then we get $\\frac{DI}{SI} = \\frac{CI}{DI}$ and therefore $\\triangle CDI \\sim \\triangle DSI$.\n\n![](attached_image_1.png)\n\nb) The quadrilateral $DIM G$ is cyclic. Since a) implies that $\\angle ISD = \\angle IDC = \\angle IMD$ we conclude that $S$ lies on the circumcircle of $DIM G$. It is now obvious that $\\angle GSI = \\angle GMI = 90^\\circ$.\n\nRemark. It is easy to see that b) implies that the points $E$, $F$ and $G$ are colinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27218,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGiven a quadrilateral $ABCD$ set $AB = a$, $BC = b$, $CD = c$, $DA = d$, $AC = e$ and $BD = f$. Prove that:\n\na) $a^{2} + b^{2} + c^{2} + d^{2} \\geq e^{2} + f^{2}$;\n\nb) if the quadrilateral $ABCD$ is cyclic then $|a - c| \\geq |e - f|$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na. Denote the midpoints of $AC$ and $BD$ by $M$ and $N$, respectively. Using the median formula for $\\triangle BMD$ we have\n$$\nMN^{2} = \\frac{2 MB^{2} + 2 MD^{2} - f^{2}}{4}\n$$\nWe also have\n$$\nMB^{2} = \\frac{2 a^{2} + 2 b^{2} - e^{2}}{4}, \\quad MD^{2} = \\frac{2 c^{2} + 2 d^{2} - e^{2}}{4}\n$$\n![](attached_image_1.png)\nand plugging these expressions in the equality for $MN^{2}$ gives\n$$\n0 \\leq 4 MN^{2} = a^{2} + b^{2} + c^{2} + d^{2} - e^{2} - f^{2}\n$$\n\nb. The Ptolemy's theorem states that $ac + bd = ef$. Using this we write the identity from a) as\n$$\n(a - c)^{2} + (b - d)^{2} = (e - f)^{2} + 4 MN^{2}\n$$\nIn order to prove the required inequality it is enough to show that the inequality $(b - d)^{2} \\leq 4 MN^{2}$ holds. But it follows from the triangle inequality for $\\triangle MNP$, where $P$ is the midpoint of $CD$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27219,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $ABC$ be a triangle with altitude $CH$, where $H$ is an interior point of the side $AB$. Denote by $P$ and $Q$ the incenters of $\\triangle AHC$ and $\\triangle BHC$, respectively. Prove that the quadrilateral $ABQP$ is cyclic if and only if either $AC = BC$ or $\\angle ACB = 90^\\circ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n($\\Leftarrow$) If $AC = BC$, then the quadrilateral $ABQP$ is cyclic since it is an isosceles trapezoid. If $\\angle ACB = 90^\\circ$, then we have $\\angle ACI = \\angle BCI = 45^\\circ$, where $I$ is the incenter of $\\triangle ABC$. We have also $\\angle APC = \\angle BQC = 135^\\circ$, i.e. $\\angle IPC = \\angle IQC = 45^\\circ$. Therefore $\\triangle IPC \\sim \\triangle ICA$ and $\\triangle IQC \\sim \\triangle ICB$. Hence\n$$\nIP \\cdot IA = IC^2 = IQ \\cdot IB\n$$\nand therefore the quadrilateral $ABQP$ is cyclic.\n\n![](attached_image_1.png)\n\n($\\Rightarrow$) We consider the circumcircle of $\\triangle APC$. If it is tangent to $CI$ at the point $C$, then $\\angle ACI = \\angle IPC = 45^\\circ$ and thus $\\angle ACB = 90^\\circ$. Now let us assume that this circle intersects $CI$ again at some point $R$. Then we have $IP \\cdot IA = IR \\cdot IC$ and $IP \\cdot IA = IQ \\cdot IB$. Therefore $IR \\cdot IC = IQ \\cdot IB$ and the quadrilateral $BCRQ$ is cyclic. Thus\n$$\n\\angle BRC = \\angle BQC = 135^\\circ = \\angle APC = \\angle ARC.\n$$\nHence $\\triangle ARC \\cong \\triangle BRC$ and $AC = BC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27220,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all pairs of positive integers $(m, n)$, $m > n$, such that\n$$\n\\left[m^2 + m n, m n - n^2\\right] + [m - n, m n] = 2^{2005}\n$$\nwhere $[a, b]$ denotes the least common multiple of $a$ and $b$.",
    "options": [],
    "answer": "m = 2^{1002}, n = 2^{1001}",
    "solution": "Solution:\nThe left hand side of the given equation is a multiple of $m$, $n$ and $m-n$. Therefore $m = 2^a$, $n = 2^b$ and $m - n = 2^c$ for some nonnegative integers $a$, $b$ and $c$, where $a > b$. It is obvious that $2^b (2^{a-b} - 1) = 2^c$, whence $a - b = 1$.\n\nPlugging $b = a - 1$ in the given equation we obtain\n$$\n\\begin{aligned}\n\\left[2^{2a} + 2^{2a-1}, 2^{2a-1} - 2^{2a-2}\\right] + \\left[2^a - 2^{a-1}, 2^{2a-1}\\right] &= 2^{2a-1} + 3 \\cdot 2^{2a-1} \\\\\n&= 2^{2a+1} = 2^{2005}\n\\end{aligned}\n$$\nHence $a = 1002$, $m = 2^{1002}$ and $n = 2^{1001}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27221,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSolve in integers the equation\n$$\nz^{2}+1=xy(xy+2y-2x-4).\n$$",
    "options": [],
    "answer": "(-1, 1, 0)",
    "solution": "Solution:\nWe set $x = u - 1$, $y = v + 1$ and obtain the equation\n$$\nz^{2} + 1 = (u^{2} - 1)(v^{2} - 1)\n$$\nIt is easy to see that $u$, $v$ and $z$ must be even. Hence, if $|u| > 1$, then $u^{2} - 1$ has a prime divisor $p$ such that $p \\equiv 3 \\pmod{4}$. Therefore $z^{2} + 1 \\equiv 0 \\pmod{p}$, which is impossible (it is well known that if $p$ is a prime such that $p \\equiv 3 \\pmod{4}$ and $p$ divides $x^{2} + y^{2}$ then $p$ divides both $x$ and $y$). Thus $u = 0$.\nAnalogously we get $v = 0$ and therefore $z = 0$. Hence $x = -1$, $y = 1$, $z = 0$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27222,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn every cell of a table $n \\times n$, $n \\geq 2$, one of the numbers $+1$ and $-1$ is written. The cell on the $i$-th row and $j$-th column is denoted by $(i, j)$, $i, j = 0, 1, \\ldots, n-1$. The neighbors of the cell $(i, j)$ are the cells $(i, j-1)$, $(i, j+1)$, $(i-1, j)$ and $(i+1, j)$, where the numbers are taken modulo $n$. At each step one replaces the number in each cell with the product of the numbers in the four neighbors of that cell. For example,\n\n| +1 | -1 | +1 |\n| :---: | :---: | :---: |\n| +1 | -1 | -1 |\n| -1 | +1 | -1 |$\\rightarrow$| +1 | -1 | -1 |\n| :--- | :--- | :--- |\n| -1 | +1 | +1 |\n| -1 | +1 | +1 |.\n\nA table is called \"good\" if after finitely many steps one obtains the table with $+1$ in every cell. Find all values of $n$ such that every table $n \\times n$ is \"good\".",
    "options": [],
    "answer": "n = 2^k for some positive integer k",
    "solution": "Solution:\n\nWe first prove that for any odd $n \\geq 3$ there are $n \\times n$ tables that are not \"good\". Consider an arbitrary $n \\times n$ table and denote by $P_{i}$, $i=1,2, \\ldots, n$, the product of the numbers in the $i$-th row at the second last step. Then $P_{1} P_{3} = P_{2} P_{4} = \\cdots = P_{n-1} P_{1} = P_{n} P_{2} = 1$ and since $n$ is odd, it follows that $P_{1} = P_{2} = \\cdots = P_{n}$. This argument shows that the row products in the initial table must be equal. Therefore any table which has not this property is not \"good\".\n\nWe now consider a table of order $n = 2^{k} m$, where $m$ is an odd number and $k \\geq 1$. After the first two steps the number in the position $(i, j)$ becomes equal to the product of the numbers in the positions $(i-2, j)$, $(i, j-2)$, $(i, j+2)$ and $(i+2, j)$. Therefore the resulting table after every even step can be obtained by applying the operation on the following four tables of order $2^{k-1} m$:\n- the table of all $(i, j)$ with $i \\equiv j \\equiv 0 \\pmod{2}$;\n- the table of all $(i, j)$ with $i \\equiv 0 \\pmod{2}$, $j \\equiv 1 \\pmod{2}$;\n- the table of all $(i, j)$ with $i \\equiv 1 \\pmod{2}$, $j \\equiv 0 \\pmod{2}$;\n- the table of all $(i, j)$ with $i \\equiv j \\equiv 1 \\pmod{2}$.\n\nNow it follows by induction that the number $n = 2^{k} m$ has the required property if and only if the number $2^{k-1} m$ does. It is also easy to see that every table of order $2$ is \"good\".\n\nTherefore the required $n$ are $n = 2^{k}$, where $k$ is a positive integer.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 27223,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nProve that for every positive integer $n$ there exist integers $p$ and $q$ such that\n$$\n\\left|p^{2}+2 q^{2}-n\\right| \\leq \\sqrt[4]{9 n}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet $q$ be an integer such that\n$$\n2 q^{2} \\leq n < 2(q+1)^{2}\n$$\nThen\n$$\nn - 2 q^{2} < 4q + 2 \\leq 4 \\sqrt{\\frac{n}{2}} + 2 = 2(\\sqrt{2n} + 1)\n$$\nFurther, let $t$ be an integer such that $t^{2} \\leq n - 2 q^{2} < (t+1)^{2}$. We choose $p$ to be either the number $t$ or $t+1$ depending on the location of $n - 2 q^{2}$ with respect to the midpoint of the interval $[t^{2}, (t+1)^{2}]$. More precisely, we set\n$$\np= \\begin{cases} t, & \\text{ if } n-2 q^{2}-t^{2} \\leq t \\\\ t+1, & \\text{ if } n-2 q^{2}-t^{2} > t \\end{cases}\n$$\nThen we have\n$$\n\\left|p^{2}+2 q^{2}-n\\right| \\leq t \\leq \\sqrt{n-2 q^{2}} \\leq \\sqrt{2(\\sqrt{2 n}+1)}\n$$\nIt remains to note that $\\sqrt{2(\\sqrt{2 n}+1)} \\leq \\sqrt[4]{9 n}$ for every $n \\geq 160$ and that for $n < 160$ the existence of $p$ and $q$ can be checked directly.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27224,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all values of the real parameter $a$ such that the number of the solutions of the equation\n$$\n3\\left(5 x^{2}-a^{4}\\right)-2 x=2 a^{2}(6 x-1)\n$$\ndoes not exceed the number of the solutions of the equation\n$$\n2 x^{3}+6 x=\\left(3^{6 a}-9\\right) \\sqrt{2^{8 a}-\\frac{1}{6}}-(3 a-1)^{2} 12^{x}\n$$",
    "options": [],
    "answer": "1/3",
    "solution": "Solution:\nThe first equation is quadratic with discriminant $D=(9 a^{2}-1)^{2}$. Therefore it has two different solutions for $a \\neq \\pm \\frac{1}{3}$ and exactly one solution for $a= \\pm \\frac{1}{3}$.\n\nSince the function\n$$\n2 x^{3}+6 x+(3 a-1)^{2} 12^{x}\n$$\nis strictly increasing, the second equation has at most one solution. For $a=\\frac{1}{3}$ it has solution $x=0$, while for $a=-\\frac{1}{3}$ it is not defined. Finally, the only value of $a$ which satisfies the condition, is $a=\\frac{1}{3}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27225,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $H$ be the orthocenter of $\\triangle ABC$, $M$ be the midpoint of $AB$ and $H_1$ and $H_2$ be the feet of the perpendiculars from $H$ to the inner and the outer bisector of $\\Varangle ACB$, respectively. Prove that the points $H_1$, $H_2$ and $M$ are colinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nDenote by $D$ and $E$ the feet of the altitudes of $\\triangle ABC$ from the vertices $A$ and $B$, respectively. The quadrilateral $HDCE$ is inscribed in the circle with diameter $CH$. The points $H_1$ and $H_2$ are the midpoints of the two arcs $\\overparen{DE}$ since $CH_1$ and $CH_2$ are the internal and external bisectors of $\\Varangle ACB$, respectively. Hence the line $H_1H_2$ is the perpendicular bisector of the segment $DE$.\n\n![](attached_image_1.png)\n\nOn the other hand, the quadrilateral $ABDE$ is also inscribed, this time in the circle with diameter $AB$. Therefore the perpendicular bisector of the chord $DE$ passes through the center $M$ of the circumcircle of $ABDE$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27226,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the largest possible number $A$ having the following property: if the numbers $1,2, \\ldots, 1000$ are ordered in arbitrary way then there exist 50 consecutive numbers with sum not less than $A$.",
    "options": [],
    "answer": "25025",
    "solution": "Solution:\nLet $x_{1}, x_{2}, \\ldots, x_{1000}$ be an arbitrary rearrangement of the numbers $1,2, \\ldots, 1000$. Set\n\n$$\nS_{1} = x_{1} + x_{2} + \\cdots + x_{50}, \\ldots, S_{20} = x_{951} + x_{952} + \\cdots + x_{1000}\n$$\nSince $S_{1} + \\cdots + S_{20} = 500500$, we have $S_{i} \\geq \\frac{500500}{20} = 25025$ for at least one index $i$.\n\nOn the other hand, if a number $B$ has the required property then we have $B \\leq 25025$. To see this consider the rearrangement\n$$\n1000, 1, 999, 2, \\ldots, 501, 500\n$$\nand take arbitrarily fifty consecutive numbers in it. If the first number is greater than $500$, then the sum of these fifty numbers is $25025$, otherwise it is $25000$. Hence $A = 25025$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27227,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe sequence $\\{a_{n}\\}_{n=1}^{\\infty}$ is defined by $a_{1}=0$ and $a_{n+1}= a_{n}+4 n+3$, $n \\geq 1$.\n\na) Express $a_{n}$ as a function of $n$.\n\nb) Find the limit\n$$\n\\lim _{n \\rightarrow \\infty} \\frac{\\sqrt{a_{n}}+\\sqrt{a_{4 n}}+\\sqrt{a_{4^{2} n}}+\\cdots+\\sqrt{a_{4^{10} n}}}{\\sqrt{a_{n}}+\\sqrt{a_{2 n}}+\\sqrt{a_{2^{2} n}}+\\cdots+\\sqrt{a_{2^{10} n}}}\n$$",
    "options": [],
    "answer": "a_n = (2n+3)(n-1); the limit equals 683",
    "solution": "Solution:\n\na.\nUsing the recurrence relation we easily get\n$$\n\\begin{aligned}\na_{k} & =a_{k-1}+4(k-1)+3=a_{k-2}+4(k-2)+4(k-1)+2 \\cdot 3=\\cdots \\\\\n& =a_{1}+4(1+2+\\cdots+k-1)+(k-1) \\cdot 3=2 k(k-1)+3(k-1) \\\\\n& =(2 k+3)(k-1)\n\\end{aligned}\n$$\n\nb.\nWe have $\\lim _{n \\rightarrow \\infty} \\frac{\\sqrt{a_{k n}}}{n}=\\lim _{n \\rightarrow \\infty} \\sqrt{\\left(2 k+\\frac{3}{n}\\right)\\left(k-\\frac{1}{n}\\right)}=\\sqrt{2} k$. Therefore the required limit is equal to\n$$\n\\frac{1+4+4^{2}+\\cdots+4^{10}}{1+2+2^{2}+\\cdots+2^{10}}=\\frac{4^{11}-1}{3\\left(2^{11}-1\\right)}=\\frac{2^{11}+1}{3}=683\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27228,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nThe sum of the first $n$ terms of an arithmetic progression with first term $m$ and difference $2$ is equal to the sum of the first $n$ terms of a geometric progression with first term $n$ and ratio $2$.\n\na) Prove that $m+n=2^{m}$;\n\nb) Find $m$ and $n$, if the third term of the geometric progression is equal to the $23$-rd term of the arithmetic progression.",
    "options": [],
    "answer": "m + n = 2^m; m = 4, n = 12",
    "solution": "Solution:\n\na) Using the formulas for the sums of arithmetic and geometric progressions we obtain the equality\n$$\n\\frac{n[2m + 2(n-1)]}{2} = n\\left(2^{m} - 1\\right)\n$$\nwhence $m+n=2^{m}$.\n\nb) It follows that $4n = m + 44$. Using a), we obtain $2^{m+2} = 44 + 5m$. It is easy to see that $m=4$ is a solution. If $m<4$ then $2^{m+2} \\leq 2^{5} < 44 + 5m$. If $m>4$ then it follows by induction that $2^{m+2} > 44 + 5m$. Therefore $m=4$ and $n=12$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27229,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all values of the real parameter $a$ such that the equation\n$$\na(\\sin 2x + 1) + 1 = (a-3)(\\sin x + \\cos x)\n$$\nhas a solution.",
    "options": [],
    "answer": "(-∞, 1] ∪ [9, ∞)",
    "solution": "Solution:\nWe write the given equation as\n$$\n2a y^2 - \\sqrt{2}(a-3) y + 1 = 0\n$$\nwhere\n$$\ny = \\frac{\\sqrt{2}}{2}(\\sin x + \\cos x) = \\sin\\left(x + 45^\\circ\\right) \\in [-1,1]\n$$\nIt has a solution if and only if (1) has a solution in the interval $[-1,1]$.\n\nFor $a=0$ the equation (1) is linear and its root $y = -\\frac{\\sqrt{2}}{6}$ belongs to the interval $[-1,1]$.\n\nLet $a \\neq 0$ and set $f(y) = 2a y^2 - \\sqrt{2}(a-3) y + 1$. We have $f(1) = 0$ and $f(-1) = 0$ for $a = -\\frac{7\\sqrt{2} + 8}{2}$ and $a = \\frac{7\\sqrt{2} - 8}{2}$, respectively. The quadratic polynomial $f(y)$ has exactly one root in $(-1,1)$ if and only if $f(-1) f(1) < 0$. This implies that $a \\in \\left(-\\frac{7\\sqrt{2} + 8}{2}, 0\\right) \\cup \\left(0, \\frac{7\\sqrt{2} - 8}{2}\\right)$.\n\nFurther, $f(y)$ has two roots in $(-1,1)$ if and only if\n$$\n\\left\\lvert\\,\n\\begin{aligned}\n& a f(-1) > 0 \\\\\n& a f(1) > 0 \\\\\n& D = 2a^2 - 20a + 18 \\geq 0 \\Longleftrightarrow \\\\\n& -1 < \\frac{\\sqrt{2}(a-3)}{4a} < 1\n\\end{aligned}\n\\right.\n$$\n$$\n\\begin{aligned}\n& a \\in (-\\infty, 0) \\cup \\left(\\frac{7\\sqrt{2} - 8}{2}, \\infty\\right) \\\\\n& a \\in \\left(-\\infty, -\\frac{7\\sqrt{2} + 8}{2}\\right) \\cup (0, \\infty) \\\\\n& a \\in (-\\infty, 1] \\cup [9, \\infty) \\\\\n& a \\in \\left(-\\infty, -\\frac{3 + 6\\sqrt{2}}{7}\\right) \\cup \\left(\\frac{6\\sqrt{2} - 3}{7}, \\infty\\right)\n\\end{aligned}\n$$\nHence $a \\in \\left(-\\infty, -\\frac{7\\sqrt{2} + 8}{2}\\right) \\cup \\left(\\frac{7\\sqrt{2} - 8}{2}, 1\\right] \\cup [9, \\infty)$.\n\nTaking into account the above cases we get $a \\in (-\\infty, 1] \\cup [9, \\infty)$.\nSolution:\nThe discriminant of $f(y)$ is non-negative if and only if $a \\in (-\\infty, 1] \\cup [9, \\infty)$.\n\nFor $a \\leq 1$ we have $f(0) = 1 > 0$ and $f\\left(-\\frac{1}{\\sqrt{2}}\\right) = 2a - 2 \\leq 0$, which implies that the equation $f(y) = 0$ has a root in the interval $\\left(-\\frac{1}{\\sqrt{2}}, 0\\right) \\subset (-1,1)$.\n\nFor $a \\geq 9$ we have $f(0) = 1 > 0$ and $f\\left(\\frac{1}{3\\sqrt{2}}\\right) = \\frac{2}{9}(9 - a) \\leq 0$, which shows that the equation $f(y) = 0$ has a root in the interval $\\left(0, \\frac{1}{3\\sqrt{2}}\\right) \\subset (-1,1)$.\n\nHence the given equation has a solution if and only if $a \\in (-\\infty, 1] \\cup [9, \\infty)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27230,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSolve the inequality\n$$\n\\log_{a}\\left(x^{2}-x-2\\right)>\\log_{a}\\left(3+2x-x^{2}\\right)\n$$\nif it is known that $x=a+1$ is a solution.",
    "options": [],
    "answer": "(5/2, 3)",
    "solution": "Solution:\nThe inequality is defined for $x^{2}-x-2>0$ and $3+2x-x^{2}>0$, whence $x \\in (2,3)$.\n\nSince $x=a+1$ is a solution, we have $a \\in (1,2)$.\n\nThen the inequality is equivalent to\n$$\nx^{2}-x-2 > 3+2x-x^{2} \\Longleftrightarrow (x+1)(2x-5)>0\n$$\nand therefore $x \\in \\left(\\frac{5}{2}, 3\\right)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27231,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all values of the real parameter $a$ such that the equation\n$$\nlg (a x+1)=\\lg (x-1)+\\lg (2-x)\n$$\nhas exactly one solution.",
    "options": [],
    "answer": "(-1, -\\frac{1}{2}] \\cup \\{3 - 2\\sqrt{3}\\}",
    "solution": "Solution:\nThe equation is equivalent to $x \\in (1,2)$ and $a x + 1 = (x-1)(2-x)$, which can be written as $x^2 + (a-3)x + 3 = 0$. Therefore we have to find the values of $a$ such that the equation\n$$\nf(x) = x^2 + (a-3)x + 3 = 0\n$$\nhas exactly one root in the interval $(1,2)$. This is possible exactly in the following four cases:\n\nCase 1. $f(1) f(2) < 0$, which is equivalent to $a \\in \\left(-1, -\\frac{1}{2}\\right)$.\n\nCase 2. $f(1) = 0$, i.e. $a = -1$. Then $x_1 = 1$ and $x_2 = 3$, which shows that $a = -1$ is not a solution.\n\nCase 3. $f(2) = 0$, i.e. $a = -\\frac{1}{2}$. Then $x_1 = 2$ and $x_2 = \\frac{3}{2}$, which shows that $a = -\\frac{1}{2}$ is a solution.\n\nCase 4. $D = 0$, i.e. $(a-3)^2 - 12 = a^2 - 6a - 3 = 0$, whence $a = 3 \\pm 2\\sqrt{3}$. For $a = 3 + 2\\sqrt{3}$ we have $x_1 = x_2 = -\\sqrt{3}$, i.e. $a = 3 + 2\\sqrt{3}$ is not a solution. For $a = 3 - 2\\sqrt{3}$ we get $x_1 = x_2 = \\sqrt{3} \\in (1,2)$, i.e. $a = 3 - 2\\sqrt{3}$ is a solution.\n\nFinally, $a \\in \\left(-1, -\\frac{1}{2}\\right] \\cup \\{3 - 2\\sqrt{3}\\}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27232,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn the sides of an acute $\\triangle ABC$ of area $1$ points $A_1 \\in BC$, $B_1 \\in CA$ and $C_1 \\in AB$ are chosen so that\n$$\n\\Varangle CC_1B = \\Varangle AA_1C = \\Varangle BB_1A = \\varphi,\n$$\nwhere the angle $\\varphi$ is acute. The segments $AA_1$, $BB_1$ and $CC_1$ meet at points $M$, $N$ and $P$.\n\na) Prove that the circumcenter of $\\triangle MNP$ coincides with the orthocenter of $\\triangle ABC$.\n\nb) Find $\\varphi$, if $S_{MNP} = 2 - \\sqrt{3}$.",
    "options": [],
    "answer": "75 degrees",
    "solution": "Solution:\n\nSet $AA_1 \\cap BB_1 = M$, $BB_1 \\cap CC_1 = N$ and $CC_1 \\cap AA_1 = P$. Using the standard notation for the angles of $\\triangle ABC$, we have\n$$\n\\Varangle PMN = \\varphi - \\Varangle B_1BC = \\varphi - (\\varphi - \\gamma) = \\gamma\n$$\nAnalogously, we get $\\Varangle MNP = \\alpha$ and $\\Varangle NPM = \\beta$, i.e. $\\triangle NPM \\sim \\triangle ABC$.\n\nLet $H$ be the orthocenter of $\\triangle ABC$. The equalities $\\Varangle HCC_1 = \\Varangle HBB_1 = \\Varangle HAA_1 = 90^\\circ - \\varphi$ imply that each of the quadrilaterals $ABMH$, $BCNH$ and $ACHP$ is cyclic. Therefore $\\Varangle HMA = \\Varangle HBA = 90^\\circ - \\alpha$ and $\\Varangle HPM = 180^\\circ - \\Varangle APH = \\Varangle ACH = 90^\\circ - \\alpha$, which implies that $H$ is the circumcenter of\n\n![](attached_image_1.png)\n\n$\\triangle MNP$.\n\nb)\n\nWe have proved that the points $A$, $B$, $M$ and $H$ are cyclic. Then the Sine theorem gives\n$$\n\\frac{MH}{\\sin(90^\\circ - \\varphi)} = \\frac{c}{\\sin(180^\\circ - \\gamma)}\n$$\ni.e. $MH = 2R \\cos \\varphi$. Since $MH$ is the circumradius of $\\triangle MNP$, we conclude that\n$$\n2 - \\sqrt{3} = \\frac{S_{MNP}}{S_{ABC}} = \\frac{MH^2}{R^2} = 4 \\cos^2 \\varphi \\Longleftrightarrow 2 \\cos 2\\varphi = -\\sqrt{3}\n$$\nNoting that $0 < 2\\varphi < 180^\\circ$ we get $2\\varphi = 150^\\circ$, i.e. $\\varphi = 75^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27233,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $n$ be a fixed positive integer. The positive integers $a$, $b$, $c$ and $d$ are less than or equal to $n$, $d$ is the largest one and they satisfy the equality\n$$\n(ab + cd)(bc + ad)(ac + bd) = (d - a)^2 (d - b)^2 (d - c)^2\n$$\n\na) Prove that $d = a + b + c$.\n\nb) Find the number of the quadruples $(a, b, c, d)$ which have the required properties.",
    "options": [],
    "answer": "a) d = a + b + c. b) Number of quadruples = n(n−1)(n−2)/6.",
    "solution": "Solution:\n\na) A direct check shows that the condition is satisfied when $a + b + c = d$. Let us assume that $a + b + c > d$. Then it is easy to see that\n$$\nab + cd > (d - a)(d - b)\n$$\nWe have analogously $bc + ad > (d - b)(d - c)$ and $ac + bd > (d - a)(d - c)$. Now the multiplication of these three inequalities gives a contradiction. Analogous arguments lead to a contradiction when $a + b + c < d$ and therefore $d = a + b + c$.\n\nb) For a fixed $d$, $3 \\leq d \\leq n$, the equation $d = a + b + c$ has\n$$\n\\binom{d-1}{2} = \\frac{(d-1)(d-2)}{2}\n$$\nsolutions. (This can be proved as follows. Write consecutively $d$ 1's. Then the number of the solutions is equal to the number of the ways one can put two separating lines in that sequence; for example $111|11 \\ldots 11|1$ corresponds to $a = 3$, $b = d - 4$, $c = 1$.) This formula is true also for $d = 1$ and $d = 2$ since the equation has no solutions in these cases.\n\nIt remains to calculate\n$$\n\\begin{aligned}\n\\sum_{d=1}^{n} \\frac{(d-1)(d-2)}{2} & = \\frac{1}{2} \\sum_{d=1}^{n} d^2 - \\frac{3}{2} \\sum_{d=1}^{n} d + n \\\\\n& = \\frac{1}{2} \\cdot \\frac{n(n+1)(2n+1)}{6} - \\frac{3}{2} \\cdot \\frac{n(n+1)}{2} + n \\\\\n& = \\frac{n(n-1)(n-2)}{6}\n\\end{aligned}",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27234,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $M$ and $N$ be arbitrary points on the side $AB$ of a triangle $ABC$ such that $M$ lies between $A$ and $N$. The line through $M$ parallel to $AC$ meets the circumcircle of $\\triangle MNC$ at point $P$, and the line through $M$ parallel to $NC$ meets the circumcircle of $\\triangle AMC$ at point $Q$. Analogously, the line through $N$ parallel to $BC$ meets the circumcircle of $\\triangle MNC$ at point $K$ and the line through $N$ parallel to $MC$ meets the circumcircle of $\\triangle BNC$ at point $L$. Prove that:\n\na) the points $P$, $Q$ and $C$ are collinear;\n\nb) the points $P$, $Q$, $K$ and $L$ are concyclic if and only if $AM = BN$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n$$\n\\begin{aligned}\n\\angle QCP &= \\angle QCA + \\angle ACN + \\angle NCP \\\\\n&= \\angle CNA + \\angle ACN + \\angle NAC \\\\\n&= 180^\\circ,\n\\end{aligned}\n$$\n![](attached_image_1.png)\ni.e. the points $P$, $C$ and $Q$ are collinear.\n\nb) Set $\\angle ACM = \\varphi$, $\\angle NCM = \\psi$ and denote by $R_1$ and $R_2$ the radii of the circumcircles of $\\triangle AMC$ and $\\triangle MNC$, respectively. Then by the Sine theorem we have\n$$\n\\begin{aligned}\nQC &= 2R_1 \\sin \\angle QMC = 2R_1 \\sin \\psi \\\\\nCP &= 2R_2 \\sin \\angle PMC = 2R_2 \\sin \\varphi \\\\\nAM &= 2R_1 \\sin \\varphi \\\\\nBM &= 2R_2 \\sin \\psi\n\\end{aligned}\n$$\nTherefore $PC \\cdot CQ = 4R_1 R_2 \\sin \\varphi \\sin \\psi = AM \\cdot BM$.\n\nAs in a) we see that the points $L$, $C$ and $K$ are collinear. Analogously we have $CK \\cdot CL = AN \\cdot BN$. Since the circumcircle of $\\triangle MNC$ passes through $K$, $C$ and $P$, the lines $PQ$ and $KL$ do not coincide. Therefore the points $P$, $Q$, $K$ and $L$ are cyclic if and only if\n$$\nCP \\cdot CQ = CL \\cdot CK \\Longleftrightarrow AM \\cdot BM = AN \\cdot BN \\Longleftrightarrow AM = BN.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27235,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn an acute $\\triangle ABC$ with $CA \\neq CB$ and incenter $O$ denote by $A_1$ and $B_1$ the tangent points of its excircles to the sides $CB$ and $CA$, respectively. The line $CO$ meets the circumcircle of $\\triangle ABC$ at point $P$ and the line through $P$ which is perpendicular to $CP$ meets the line $AB$ at point $Q$. Prove that the lines $QO$ and $A_1B_1$ are parallel.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $K = CO \\cap AB$ and $\\Varangle AKC = \\varphi$. Denote by $M$ and $N$ the intersection points of $QO$ with $CB$ and $CA$, respectively. The Sine theorem for $\\triangle APQ$ gives\n$$\n\\frac{AQ}{PQ} = \\frac{\\sin \\left(90^\\circ + \\beta\\right)}{\\sin \\frac{\\gamma}{2}} = \\frac{\\cos \\beta}{\\sin \\frac{\\gamma}{2}}\n$$\nFrom the right $\\triangle KPQ$ we find\n$$\n\\frac{KQ}{PQ} = \\frac{1}{\\sin \\varphi}\n$$\nHence\n$$\n\\frac{AQ}{QK} = \\frac{\\cos \\beta \\sin \\varphi}{\\sin \\frac{\\gamma}{2}}\n$$\n\n![](attached_image_1.png)\n\nIt follows from $\\triangle AKC$ ($AO$ is the bisector of $\\Varangle KAC$) that\n$$\n\\frac{KO}{OC} = \\frac{AK}{AC} = \\frac{\\sin \\frac{\\gamma}{2}}{\\sin \\varphi}\n$$\nOn the other hand applying the Menelaus theorem for $\\triangle AKC$ and the line $OQ$ we get\n$$\n\\frac{AQ}{QK} \\cdot \\frac{KO}{OC} \\cdot \\frac{CN}{NA} = 1\n$$\nNow plugging (1) and (2) in (3) we obtain $\\frac{CN}{NA} = \\frac{1}{\\cos \\beta}$. Hence it follows that $\\frac{CN}{CA} = \\frac{1}{1 + \\cos \\beta}$, i.e. $CN = \\frac{2R \\sin \\beta}{2 \\cos^2 \\frac{\\beta}{2}} = 2R \\tan \\frac{\\beta}{2}$.\n\nSimilarly, $CM = 2R \\tan \\frac{\\alpha}{2}$. Therefore\n$$\n\\frac{CN}{CM} = \\frac{\\tan \\frac{\\beta}{2}}{\\tan \\frac{\\alpha}{2}}\n$$\nIt is well known that $CB_1 = p - a = r \\cot \\frac{\\alpha}{2}$ and $CA_1 = p - b = r \\cot \\frac{\\beta}{2}$. Hence using (4) we get\n$$\n\\frac{CB_1}{CA_1} = \\frac{\\cot \\frac{\\alpha}{2}}{\\cot \\frac{\\beta}{2}} = \\frac{\\tan \\frac{\\beta}{2}}{\\tan \\frac{\\alpha}{2}} = \\frac{CN}{CM}\n$$\nTherefore $A_1B_1 \\parallel MN$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27236,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all values of the real parameter $a$ such that the equation\n$$\n\\log_{a x}\\left(3^{x}+4^{x}\\right)=\\log_{(a x)^{2}}\\left(7^{2}\\left(4^{x}-3^{x}\\right)\\right)+\\log_{(a x)^{3}} 8^{x-1}\n$$\nhas a solution.",
    "options": [],
    "answer": "a ∈ (0, ∞) \\ {1}",
    "solution": "Solution:\nSince $a x>0$ and $4^{x}-3^{x}>0$ it follows that $a>0$ and $x>0$.\n\nFor $a>0$, $x>0$ and $a x \\neq 1$ the equation is equivalent to\n$$\n3^{x}+4^{x}=7 \\cdot 2^{x-1} \\sqrt{4^{x}-3^{x}} \\Longleftrightarrow 45\\left(\\frac{4}{3}\\right)^{2x}-57\\left(\\frac{4}{3}\\right)^{x}-4=0\n$$\nSetting $y=\\left(\\frac{4}{3}\\right)^{x}>0$ we obtain the equation $45 y^{2}-57 y-4=0$ with solutions $y_{1}=\\frac{4}{3}$ and $y_{2}=-\\frac{1}{15}$. Hence $x=1$. The condition $a x \\neq 1$ now implies that $a \\neq 1$.\n\nThe required values of $a$ are $a \\in (0,+\\infty) \\setminus \\{1\\}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27237,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn an internet chess tournament 2005 chess players took part and everyone played one game against any other. After the tournament it appeared that for every two players $A$ and $B$ who had drawn their game every other player had lost his game with $A$ or with $B$. Prove that if there were at least two draws in the tournament then the players can be ordered in such a way that everyone has won his game with the next one in the sequence.\nEmil Kolev",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nNote that a chess player could not have more than one draw. Indeed, if $A$ had draws with $B$ and $C$, then the condition for $A$ and $B$ implies that $B$ defeated $C$ and the same condition for $A$ and $C$ implies that $C$ defeated $B$, a contradiction.\n\nLet $A_{1}, A_{2}, \\ldots, A_{k}$ be the longest sequence such that each chess player has defeated the next one, i.e. $A_{i}$ has defeated $A_{i+1}$ for $i=1,2, \\ldots, k-1$. If $k=2005$, then we have the required sequence. Assume that $k<2005$ and consider a chess player $B$ who is not amongst $A_{1}, A_{2}, \\ldots, A_{k}$.\n\nIf $B$ has defeated $A_{1}$ then the sequence $B, A_{1}, A_{2}, \\ldots, A_{k}$ of length $k+1$ has the above property, which is impossible.\n\nIf $A_{1}$ has defeated $B$, then $B$ and $A_{2}$ had not draw because $A_{1}$ has defeated both. If $B$ has defeated $A_{2}$ then the sequence $A_{1}, B, A_{2}, \\ldots, A_{k}$ of length $k+1$ has the above property, a contradiction. Therefore $A_{2}$ has defeated $B$. We see analogously that all players $A_{3}, A_{4}, \\ldots, A_{k}$ have defeated $B$. Then we obtain again a contradiction by considering the sequence $A_{1}, A_{2}, \\ldots, A_{k}, B$ of length $k+1$.\n\nThe above argument shows that outside the sequence $A_{1}, A_{2}, \\ldots, A_{k}$ there is only one chess player $B$, and $A_{1}$ and $B$ made a draw. Then this is the only draw of $B$. If $A_{2}$ has defeated $B$, then we obtain as above that $B$ has lost from $A_{i}$ for $i=3,4, \\ldots, k$ and we have again sequence $A_{1}, A_{2}, \\ldots, A_{k}, B$ of length $k+1$. Therefore $A_{2}$ has lost from $B$ and the same holds for $A_{i}, i=3, \\ldots, k$.\n\nOn the other hand, there is at least one more draw, for example between $A_{i}$ and $A_{j}$. But $A_{i}$ and $A_{j}$ have lost from $B$, which is a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27238,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $c$ be a positive integer and let $\\{a_{n}\\}_{n=1}^{\\infty}$ be a sequence of positive integers such that $a_{n} < a_{n+1} < a_{n} + c$ for every $n \\geq 1$. The terms of the sequence are written one after another and in this way one obtains an infinite sequence of digits. Prove that for every positive integer $m$ there exists a positive integer $k$ such that the number formed by the first $k$ digits of the above sequence is divisible by $m$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet $M$ be an arbitrary positive integer. We shall prove that there exists a term of the sequence $\\{a_{n}\\}_{n=1}^{\\infty}$, whose decimal representation is obtained from that of $M$ by adding several digits from the right, i.e. the number $M$ is a \"beginning\" of that member.\nLet $k$ be an index such that $a_{k} \\leq M \\cdot 10^{l} < a_{k+1} < a_{k} + c$, where $l$ is a positive integer which is greater than the number of the digits of $c$. Then\n$$\nM \\cdot 10^{l} < a_{k+1} < M \\cdot 10^{l} + c\n$$\nand obviously $a_{k+1}$ satisfies the above requirement.\nLet $m = 2^{\\alpha} 5^{\\beta} t$, where $(t, 10) = 1$. It is enough to prove the assertion of the problem for $m = 10^{\\gamma} t$, where $\\gamma = \\max\\{\\alpha, \\beta\\}$.\nLet us consider the number\n$$\nM = 1 \\underbrace{00 \\ldots 0}_{p} 1 \\underbrace{00 \\ldots 0}_{p} 1 \\ldots 1 \\underbrace{00 \\ldots 0}_{p} 1 \\underbrace{00 \\ldots 0}_{q}\n$$\nHere $p = k \\varphi(t)$, where $\\varphi(t)$ is the Euler function, $k$ is a positive integer such that $p > \\gamma$, the number $q$ is greater than $\\gamma$, and the number of 1's is $t+1$.\nThen $M$ is a \"beginning\" of some $a_{k}$. Hence the sequence of the digits (formed by the terms of the sequence written one after another) looks like this:\n$$\nf_{1} f_{2} \\ldots f_{r} 1 \\underbrace{00 \\ldots 0}_{p} 1 \\underbrace{00 \\ldots 0}_{p} 1 \\ldots 1 \\underbrace{00 \\ldots 0}_{p} 1 \\underbrace{00 \\ldots 0}_{q} \\ldots\n$$\nwhere $f_{1}, f_{2}, \\ldots, f_{r}$ are the digits before $a_{k}$. It is clear now that, depending on the remainder of $\\overline{f_{1} f_{2} \\ldots f_{r} 1}$ modulo $t$, we can add suitable digits from $M$ to $\\overline{f_{1} f_{2} \\ldots f_{r} 1}$ in such a way that the resulting number is divisible by $10^{\\alpha} t$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27239,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe bisectors of $\\angle BAC$, $\\angle ABC$ and $\\angle ACB$ of $\\triangle ABC$ meet its circumcircle at points $A_1$, $B_1$ and $C_1$, respectively. The side $AB$ meets the lines $C_1B_1$ and $C_1A_1$ at points $M$ and $N$, respectively, the side $BC$ meets the lines $A_1C_1$ and $A_1B_1$ at points $P$ and $Q$, respectively, and the side $AC$ meets the lines $B_1A_1$ and $B_1C_1$ at points $R$ and $S$, respectively. Prove that:\n\na) the altitude of $\\triangle CRQ$ through $R$ is equal to the inradius of $\\triangle ABC$;\n\nb) the lines $MQ$, $NR$ and $SP$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na) If $RT$ is the altitude of $\\triangle CRQ$, then $RT = CR \\sin \\gamma$. Using the Sine theorem for $\\triangle B_1RC$ we get\n$$\nCR = \\frac{B_1C \\sin \\frac{\\alpha}{2}}{\\cos \\frac{\\gamma}{2}} = \\frac{2R \\sin \\frac{\\beta}{2} \\sin \\frac{\\alpha}{2}}{\\cos \\frac{\\gamma}{2}}\n$$\nwhere $R$ is the circumradius of $\\triangle ABC$. Hence\n$$\nRT = 4R \\sin \\frac{\\alpha}{2} \\sin \\frac{\\beta}{2} \\sin \\frac{\\gamma}{2} = r\n$$\nwhere $r$ is the inradius of $\\triangle ABC$.\n\n![](attached_image_1.png)\n\nb) As in a) we see that the altitude of $\\triangle BNP$ through $N$ is equal to $r$, which means that the lines $NR$ and $BC$ are parallel and the distance between them is $r$. Therefore $I \\in NR$, where $I$ is incenter of $\\triangle ABC$.\nThe same argument shows that $I \\in MQ$ and $I \\in SR$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27240,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nProve that amongst any 9 vertices of a regular 26-gon there are three which are vertices of an isosceles triangle. Do there exist 8 vertices such that no three of them are vertices of an isosceles triangle?",
    "options": [],
    "answer": "Among any nine vertices of a regular twenty-six-gon, three form an isosceles triangle; and yes, there exist eight vertices with no isosceles triple, for example {1, 2, 4, 5, 10, 11, 13, 14}.",
    "solution": "Solution:\n\nLEMMA. For any five vertices of a regular 13-gon there exists an isosceles triangle with vertices amongst these points.\n\nProof of the lemma. Let the five points form a convex pentagon $ABCDE$. We first consider the case when there exist two pairs of parallel lines determined by some vertices of $ABCDE$. We have the following possibilities:\n\n1) There are two pairs of parallel sides of $ABCDE$—for example $AE \\parallel CD$ and $BC \\parallel DE$. Then we have $\\angle AED = \\angle BCD$ or $\\angle AED = 180^\\circ - \\angle BCD$, i.e. $\\sin \\angle AED = \\sin \\angle BCD$. Hence $AD = 2R \\sin \\angle AED = 2R \\sin \\angle BCD = BD$ and therefore the triangle $\\triangle ABD$ is isosceles.\n\n2) There are two diagonals that are parallel respectively to two sides of $ABCDE$. Without loss of generality we may assume that $AB \\parallel CE$. If $AC \\parallel DE$ we conclude as in 1) that $BC = CD$. If $AD \\parallel BC$, then $EB = BD$. The cases $BE \\parallel CD$ and $BD \\parallel AE$ are similar.\n\n3) There is a diagonal of $ABCDE$ that is parallel to its side and a pair of parallel sides of $ABCDE$. Without loss of generality we may assume that $AB \\parallel CE$. Since $AE$ is not parallel to $BC$ (otherwise $ABCE$ is a rectangle with vertices amongst the vertices of a regular 13-gon) we may assume that $DE \\parallel BC$. Then $DC = CA$.\n\nLet us now assume that there exists at most one pair of parallel lines determined by some vertices of $ABCDE$. Since the vertices of $ABCDE$ determine 10 lines, at least 9 of them are not parallel to each other. Consider the 9 pairs of vertices determining such lines. Every such pair is a base of an isosceles triangle whose third vertex is a vertex of the 13-gon. Note that all such third vertices are different because no two of the bases are parallel. Hence at least one of these 9 vertices is a vertex of $ABCDE$ and this proves the existence of the desired isosceles triangle. This completes the proof of the lemma.\n\nThe regular 26-gon is formed by two disjoint regular 13-gons. If we choose 9 of its vertices then at least 5 of them are vertices of one of these 13-gons. Then it follows from the lemma that there exists an isosceles triangle with vertices amongst the vertices of this 13-gon.\n\nFinally, note that there exist 8 vertices such that no three of them form an isosceles triangle. For example, if the vertices are labelled from 1 to 26, then we choose the vertices $1, 2, 4, 5, 10, 11, 13$ and $14$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27241,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $ABC$ be an isosceles triangle such that $AC = BC = 1$ and $AB = 2x$, $x > 0$.\n\na) Express the inradius $r$ of $\\triangle ABC$ as a function of $x$.\n\nb) Find the maximum possible value of $r$.",
    "options": [],
    "answer": "r(x) = x sqrt((1 - x)/(1 + x)); the maximum inradius is sqrt((5 sqrt(5) - 11)/2), attained at x = (sqrt(5) - 1)/2",
    "solution": "Solution:\n\na) It follows from the Pythagorean theorem that the altitude of $\\triangle ABC$ through $C$ is equal to $\\sqrt{1 - x^2}$. Then\n$$\nr = \\frac{S}{p} = \\frac{x \\sqrt{1 - x^2}}{1 + x} = x \\sqrt{\\frac{1 - x}{1 + x}}\n$$\n\nb) We have to find the maximum of the function\n$$\nf(x) = \\frac{x^2 (1 - x)}{1 + x}\n$$\nin the interval $(0, 1)$. Since\n$$\nf'(x) = \\frac{2x(1 - x - x^2)}{(x + 1)^2}\n$$\nthe function $f(x)$ increases in the interval $\\left(0, \\frac{\\sqrt{5} - 1}{2}\\right]$ and decreases in the interval $\\left[\\frac{\\sqrt{5} - 1}{2}, 1\\right)$. Therefore the maximum of $f(x)$ in $(0, 1)$ is attained for $x = \\frac{\\sqrt{5} - 1}{2}$ and is equal to\n$$\nf\\left(\\frac{\\sqrt{5} - 1}{2}\\right) = \\frac{5\\sqrt{5} - 11}{2}.\n$$\nHence the maximum possible value of $r$ is $\\sqrt{\\frac{5\\sqrt{5} - 11}{2}}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27242,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe sequences $\\left(a_{n}\\right)_{n=1}^{\\infty}$ and $\\left(b_{n}\\right)_{n=1}^{\\infty}$ are such that $a_{n+1}=2 b_{n}-a_{n}$ and $b_{n+1}=2 a_{n}-b_{n}$ for every $n$. Prove that:\n\na) $a_{n+1}=2\\left(a_{1}+b_{1}\\right)-3 a_{n}$;\n\nb) if $a_{n}>0$ for every $n$, then $a_{1}=b_{1}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na) Since $a_{n+1}+b_{n+1}=2 b_{n}-a_{n}+2 a_{n}-b_{n}=a_{n}+b_{n}$, we have\n$$\na_{n+1}=2\\left(a_{n}+b_{n}\\right)-3 a_{n}=2\\left(a_{1}+b_{1}\\right)-3 a_{n}\n$$\n\nb) Using a), we obtain\n$$\na_{n+1}-\\frac{a_{1}+b_{1}}{2}=-3\\left(a_{n}-\\frac{a_{1}+b_{1}}{2}\\right)\n$$\nwhence\n$$\na_{n+1}-\\frac{a_{1}+b_{1}}{2}=(-3)^{n}\\left(a_{1}-\\frac{a_{1}+b_{1}}{2}\\right)\n$$\nSince $\\lim _{n \\rightarrow \\infty} 3^{n}=+\\infty$, it follows that if $a_{1}>b_{1}$, then $\\lim _{n \\rightarrow \\infty} a_{2 n}=-\\infty$, a contradiction. Analogously, we see that it is not possible to have $a_{1}<b_{1}$. Therefore $a_{1}=b_{1}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27243,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nProve that if $a$, $b$ and $c$ are integers such that the number\n$$\n\\frac{a(a-b)+b(b-c)+c(c-a)}{2}\n$$\nis a perfect square, then $a = b = c$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nSet\n$$\n\\frac{a(a-b)+b(b-c)+c(c-a)}{2} = d^{2}\n$$\nwhere $d$ is an integer, $x = a-b$, $y = b-c$ and $z = c-a$. Then we have\n$$\nx + y + z = 0, \\quad x^{2} + y^{2} + z^{2} = 4 d^{2}\n$$\nSince any square is congruent to $0$ or $1$ modulo $4$, it follows from (1) that the integers $x$, $y$ and $z$ are even. Set $x_{1} = \\frac{x}{2}$, $y_{1} = \\frac{y}{2}$ and $z_{1} = \\frac{z}{2}$. Then (1) gives\n$$\nx_{1} + y_{1} + z_{1} = 0, \\quad x_{1}^{2} + y_{1}^{2} + z_{1}^{2} = d^{2}\n$$\nand we conclude as above that $x_{1}$, $y_{1}$, $z_{1}$ and $d$ are even integers. Repeating the same argument we see that $2^{n}$ divides $x$, $y$ and $z$ for every positive integer $n$. Therefore $x = y = z = 0$, i.e. $a = b = c$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27244,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA circle through the vertex $A$ of $\\triangle ABC$, $AB \\neq AC$, meets the sides $AB$ and $AC$ at points $M$ and $N$, respectively, and the side $BC$ at points $P$ and $Q$, where $Q$ lies between $B$ and $P$. Find $\\Varangle BAC$, if $MP \\parallel AC$, $NQ \\parallel AB$ and $\\frac{BP}{CQ}=\\frac{AB}{AC}$.",
    "options": [],
    "answer": "120°",
    "solution": "Solution:\nSet $BC = a$, $CA = b$ and $AB = c$. Then $BA \\cdot BM = BP \\cdot BQ$, and\n$$\n\\frac{BM}{c} = \\frac{BP}{a}\n$$\nHence $BQ = \\frac{c^2}{a}$ and analogously $CP = \\frac{b^2}{a}$. Then $BP = \\frac{a^2 - b^2}{a}$, $CQ = \\frac{a^2 - c^2}{a}$ and the condition $\\frac{BP}{CQ} = \\frac{AB}{AC}$ becomes\n![](attached_image_1.png)\n$$\nb\\left(a^2 - b^2\\right) = c\\left(a^2 - c^2\\right) \\text{, i.e. } (b-c)\\left(a^2 - b^2 - c^2 - bc\\right) = 0\n$$\nSince $b \\neq c$, we get $a^2 - b^2 - c^2 - bc = 0$ and the Cosine theorem gives $\\cos \\Varangle BAC = -\\frac{1}{2}$. Therefore $\\Varangle BAC = 120^\\circ$.\n\nSecond solution. Since the quadrilateral $AMPN$ is a cyclic trapezoid, it follows that $AM = NP$. Also, if $T = MP \\cap NQ$, then $AMTN$ is a parallelogram and $AM = NT$. Then $NP = NT$ and in the same way $MQ = MT$. Hence $TPN$ and $TQM$ are similar isosceles triangles and we have\n$$\n\\frac{TP}{TQ} = \\frac{TN}{TM}\n$$\nUsing the Sine theorem, we obtain $\\frac{MP}{\\sin \\beta} = \\frac{BP}{\\sin \\alpha}$, $\\frac{NQ}{\\sin \\gamma} = \\frac{CQ}{\\sin \\alpha}$ and since $\\frac{BP}{CQ} = \\frac{AB}{AC} = \\frac{\\sin \\gamma}{\\sin \\beta}$, we conclude that $MP = NQ$. From here and (1) it follows that\n$$\nTM + TQ \\frac{TN}{TM} = TN + TQ \\Longleftrightarrow (TM - TN)(TM - TQ) = 0\n$$\nAssume that $TM = TN$. We have $MQ = NP$ and $NA = MA$. The first of these identities shows that $MN \\parallel PQ$ and by the second one we obtain $AC = AB$, a contradiction.\nTherefore $TM = TQ$, i.e. $\\triangle MTQ$ is equilateral. Hence\n$$\n\\Varangle BAC = \\Varangle MTN = 120^\\circ\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 27245,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all values of the real parameters $a$ and $b$ such that the graph of the function $y = x^{3} + a x + b$ has exactly three common points with the coordinate axes and they are vertices of a right triangle.",
    "options": [],
    "answer": "(a,b)=(-3/√2, 2^{1/4}) or (a,b)=(-3/√2, -2^{1/4})",
    "solution": "Solution:\nThe first condition of the problem is equivalent to the assertion that the equation $x^{3} + a x + b = 0$ has a double real root $x_{1} \\neq 0$ and a simple real root $x_{2} \\neq 0$, where $x_{2} \\neq x_{1}$. Therefore\n$$\nx^{3} + a x + b = (x - x_{1})^{2}(x - x_{2})\n$$\nWe also have $\\angle ACB = 90^{\\circ}$, where $A = (x_{1}, 0)$, $B = (x_{2}, 0)$, $C = (0, b)$ and $x_{1} x_{2} < 0$. Hence $AO \\cdot BO = CO^{2}$, i.e. $-x_{1} x_{2} = b^{2}$. Since $b = -x_{1}^{2} x_{2}$ and $x_{1}, x_{2} \\neq 0$, we get $x_{1}^{3} x_{2} = -1$. On the other hand, we have $2 x_{1} + x_{2} = 0$ and therefore $2 x_{1} = -x_{2} = \\frac{1}{x_{1}^{3}}$. Hence $x_{1} = \\pm \\frac{1}{\\sqrt[4]{2}}$ and $x_{2} = \\mp \\sqrt[4]{8}$. Then\n$a = x_{1}^{2} + 2 x_{1} x_{2} = -\\frac{3}{\\sqrt{2}}$ and $b = -x_{1}^{2} x_{2} = \\pm \\sqrt[4]{2}$. The above arguments imply as well that these two values of $b$ are solutions indeed.",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 27246,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe excircle to the side $AB$ of a triangle $ABC$ is tangent to the circle with diameter $BC$. Find $\\Varangle ACB$ if the lengths of the sides $BC$, $CA$ and $AB$ form (in this order) an arithmetic progression.",
    "options": [],
    "answer": "90°",
    "solution": "$$\n\\left(\\frac{a}{2}+r_{c}\\right)^{2}=r_{c}^{2}+\\left(p-\\frac{a}{2}\\right)^{2}\n$$\n![](attached_image_1.png)\nThen $a r_{c}=p(p-a)$. Since $r_{c}=\\frac{S}{p-c}$, we obtain by using Heron's formula\n$$\na S=p(p-a)(p-c)=\\frac{S^{2}}{p-b}\n$$\n$$\na(p-b)=S\n$$\nSince $a$, $b$ and $c$ form (in this order) an arithmetic progression, we have $a=b-x$, $c=b+x$ and\n$$\np=\\frac{3b}{2},\\quad p-a=\\frac{b}{2}+x,\\quad p-b=\\frac{b}{2},\\quad p-c=\\frac{b}{2}-x\n$$\nNow by (1) and the Heron formula we obtain the equation\n$$\n(b-x)^{2}=3\\left(\\frac{b^{2}}{4}-x^{2}\\right)\n$$\nwhich has a unique solution $x=\\frac{b}{4}$. Therefore $a=\\frac{3b}{4}$, $c=\\frac{5b}{4}$ and then $a^{2}+b^{2}=c^{2}$, i.e. $\\Varangle ACB=90^{\\circ}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27247,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABCD$ be a convex quadrilateral. The orthogonal projections of $D$ on the lines $BC$ and $BA$ are denoted by $A_1$ and $C_1$, respectively.\nThe segment $A_1C_1$ meets the diagonal $AC$ at an interior point $B_1$ such that $DB_1 \\geq DA_1$. Prove that the quadrilateral $ABCD$ is cyclic if and only if\n$$\n\\frac{BC}{DA_1} + \\frac{BA}{DC_1} = \\frac{AC}{DB_1}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $ABCD$ be a cyclic quadrilateral. Then the Simson theorem for $\\triangle ABC$ gives $DB_1 \\perp AC$. Hence $\\angle B_1C_1D = \\angle B_1AD = \\angle CBD$, $\\angle B_1DC_1 = \\angle B_1AC_1 = \\angle CDB$ and therefore $\\triangle B_1C_1D \\sim \\triangle CBD$.\n\nAnalogously $\\triangle B_1A_1D \\sim \\triangle ABD$, whence\n$$\nDA : DB : DC = \\frac{1}{DA_1} : \\frac{1}{DB_1} : \\frac{1}{DC_1}\n$$\nThis together with the Ptolemy's theorem for $ABCD$ gives\n$$\n\\frac{BC}{DA_1} + \\frac{BA}{DC_1} = \\frac{AC}{DB_1}\n$$\n\nConversely, suppose that the identity (1) is true. Set $x = \\frac{DB_1}{DA_1}$ and $y = \\frac{DB_1}{DC_1}$. Squaring (1)\n\n![](attached_image_1.png)\n\nand applying the Cosine theorem for $\\triangle ABC$, we see that the ratio $\\frac{BA}{BC}$ is a root of the equation\n$$\n\\left(y^2 - 1\\right)t^2 + 2(xy + \\cos \\angle ABC)t + x^2 - 1 = 0\n$$\nSince the point $B_1$ lies on the segment $A_1C_1$, the inequality $DB_1 \\geq DA_1$ implies that $DB_1 < DC_1$. Hence $x \\geq 1$ and $0 < y < 1$, which shows that (2) has at most one positive root.\n\nOn the other hand, it is easy to see that $C_1$ and $A_1$ lie on the open rays $BA \\to$ and $BC \\to$, and the line through $B_1$ perpendicular to $DB_1$ intersects these two rays. Denote these intersection points by $A'$ and $C'$. Then the converse Simson theorem implies that the convex quadrilateral $A'BC'D$ is cyclic. Hence the identity (1) for $A'BC'D$ is satisfied, i.e. $\\frac{BA'}{BC'}$ is a root of the equation (2).\n\nTherefore $\\frac{BA}{BC} = \\frac{BA'}{BC'}$, i.e. $AC \\parallel A'C'$. But the lines $AC$ and $A'C'$ have a common point $B_1$ and this shows that $A = A'$ and $C = C'$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27248,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the number of the sequences $\\{a_{n}\\}_{n=1}^{\\infty}$ of integers such that\n$$\na_{n}+a_{n+1}=2 a_{n+2} a_{n+3}+2005\n$$\nfor every $n$.",
    "options": [],
    "answer": "8",
    "solution": "Solution:\nSubtracting the equalities $a_{n}+a_{n+1}=2 a_{n+2} a_{n+3}+1$ and $a_{n+1}+a_{n+2}=2 a_{n+3} a_{n+4}+1$, we get $a_{n+2}-a_{n}=2 a_{n+3}\\left(a_{n+4}-a_{n+2}\\right)$. Then it follows by induction on $k$ that\n$$\na_{n+2}-a_{n}=2^{k} a_{n+3} \\ldots a_{n+2 k+1}\\left(a_{n+2 k+2}-a_{n+2 k}\\right)\n$$\nHence $2^{k}$ divides $a_{n+2}-a_{n}$ for every $k$, i.e. $a_{n+2}=a_{n}$. Therefore $a_{2 n-1}=a_{1}$ and $a_{2 n}=a_{2}$ for every $n$. Now it follows from the condition of the problem that $(2 a_{1}-1)(2 a_{2}-1)=-4009$. Since $4009=19\\cdot 211$ and 19 and 211 are primes, we get $2 a_{1}-1= \\pm 1, \\pm 19, \\pm 211, \\pm 4009$. Therefore there are 8 sequences with the required property.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27249,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all values of the real parameter $a$ such that the image of the function\n$$\nf(x)=\\frac{\\sin^{2} x-a}{\\sin^{3} x-\\left(a^{2}+2\\right) \\sin x+2}\n$$\ncontains the interval $\\left[\\frac{1}{2}, 2\\right]$.",
    "options": [],
    "answer": "(-∞, -1/2] ∪ (1, ∞)",
    "solution": "Solution:\nSet $t=\\sin x$ and $g(t)=\\frac{t^{2}-a}{t^{3}-\\left(a^{2}+2\\right) t+2}$.\nIf the numerator and the denominator of $g$ have a common root then $a \\geq 0$ and $t= \\pm \\sqrt{a}$. If $t=-\\sqrt{a}$ we obtain $\\sqrt{a}\\left(a^{2}-a+2\\right)=-2$, which is impossible since $a^{2}-a+2>0$ for every $a$. If $t=\\sqrt{a}$ we obtain $\\sqrt{a}(a(a-1)+2)=2$ and it is easy to see that $a=1$ is the only solution of this equation (for $a \\in[0,1)$ the left hand side is less than 2, and for $a>1$ it is greater than 2). For $a=1$ we have\n$$\ng(t)=\\frac{t^{2}-1}{t^{3}-3 t+2}=\\frac{t+1}{(t-1)(t+2)} \\leq 0\n$$\nfor every $t \\in[-1,1)$ which implies that this value of $a$ is not a solution.\n\nWe now look for $a \\neq 1$ such that the equation $g(t)=c$, i.e.\n$$\nh(t)=c\\left(t^{3}-\\left(a^{2}+2\\right) t+2\\right)+a-t^{2}=0\n$$\nhas a solution in the interval $[-1,1]$ for every $c \\in\\left[\\frac{1}{2}, 2\\right]$. For $c \\geq \\frac{1}{2}$ we have\n$$\nh(-1)=c a^{2}+a+3 c-1 \\geq \\frac{a^{2}}{2}+a+\\frac{1}{2}=\\frac{(a+1)^{2}}{2}\n$$\nwhence $h(-1) \\geq 0$ for every $a$. Also, $h'(t)=3 c t^{2}-2 t-c\\left(a^{2}+2\\right)$ and therefore $h'(1)=c-c a^{2}-2 \\leq 0$ for $c \\in\\left[\\frac{1}{2}, 2\\right]$ and every $a$. Hence the equation $h'(t)=0$ has real roots $t_{1}$ and $t_{2}$ such that $t_{1} \\leq 1<t_{2}$.\nThus the function $h(t)$ is decreasing in the interval $\\left[t_{1}, t_{2}\\right] \\ni 1$ and increasing in the interval $\\left(-\\infty, t_{1}\\right]$. Since $h(-1) \\geq 0$, it is easy to see that $h(t)=0$ has a solution in the interval $[-1,1]$ for every $c \\in\\left[\\frac{1}{2}, 2\\right]$ if and only if\n$$\nh(1)=(a-1)(1-c(1+a)) \\leq 0\n$$\nfor every such $c$. For $a>1$ this inequality is satisfied since $1-\\frac{1}{2}(1+a)<0$, and for $a<1$ it is equivalent to $1-2(1+a) \\geq 0$, i.e. $a \\leq-\\frac{1}{2}$.\nFinally, the required values of $a$ are $a \\in\\left(-\\infty,-\\frac{1}{2}\\right] \\cup(1,+\\infty)$.\n\n\nSecond solution:\nUsing the same reasoning as above we reduce the problem to finding of those $a$ such that $\\left[\\frac{1}{2}, 2\\right] \\subset g([-1,1])$.\nSet $h(t)=t^{3}-\\left(a^{2}+2\\right) t+2$. Then $h'(t)=3 t^{2}-\\left(a^{2}+2\\right)$ and for $a^{2} \\geq 1$ it follows that $h'(t)<0$ for $t \\in(-1,1)$. Therefore $h(t)$ is a strictly decreasing continuous function in the interval $(-1,1]$. Since\n$$\nh(1)=1-a^{2} \\leq 0<a^{2}+3=h(-1)\n$$\nit follows that $h(t)$ has a unique zero $t_{0} \\in(-1,1]$.\n\nWe now consider several cases.\n\nCase 1. Let $a>1$. Then $t_{0} \\neq 1, \\lim _{t \\rightarrow t_{0}+0} g(t)=+\\infty$ and $g(1)=\\frac{1-a}{1-a^{2}}=\\frac{1}{1+a}<\\frac{1}{2}$. Since the function $g(t)$ is decreasing and continuous in $\\left(t_{0}, 1\\right]$, it follows that\n$$\n\\left[\\frac{1}{2}, 2\\right] \\subset\\left[\\frac{1}{2},+\\infty\\right) \\subset g\\left(\\left(t_{0}, 1\\right]\\right) \\subset g([-1,1])\n$$\n\nCase 2. Let $a \\leq-1$. Now $\\lim _{t \\rightarrow t_{0}-0} g(t)=+\\infty$ and $g(-1)=\\frac{1-a}{a^{2}+3} \\leq \\frac{1}{2}$ (the last inequality is equivalent to $(a+1)^{2} \\geq 0$ ). Then\n$$\n\\left[\\frac{1}{2}, 2\\right] \\subset\\left[\\frac{1}{2},+\\infty\\right) \\subset g\\left(\\left[-1, t_{0}\\right)\\right) \\subset g([-1,1])\n$$\n\nCase 3. Let $a=1$. Then we see as in the first solution that $g(t) \\leq 0$ for $t \\in[-1,1)$.\n\nCase 4. Let $a \\in(-1,1)$. We first check that $h(t)>0$ for $t \\in[-1,1]$, which implies that $g(t)$ is a continuous function in that interval.\nIndeed, if $t \\in[-1,0]$, then $h(t) \\geq t^{3}+2>0$ and if $t \\in(0,1]$, then $h(t)>t^{3}-3 t+2=(t-1)^{2}(t+2) \\geq 0$.\n\nCase 4.1. If $a \\in\\left(-1,-\\frac{1}{2}\\right]$, then $g(-1)=\\frac{1-a}{a^{2}+3}<\\frac{1}{2}, g(1)=\\frac{1}{1+a} \\geq 2$ and therefore $\\left[\\frac{1}{2}, 2\\right] \\subset g([-1,1])$.\n\nCase 4.2. If $a \\in\\left(-\\frac{1}{2}, 1\\right)$, we shall show that $g(t)<2$ for every $t \\in[-1,1]$. Since $h(t)>0$ in this interval, we have to check that\n$$\nt^{2}<2\\left(t^{3}-\\left(a^{2}+2\\right) t+2\\right)\n$$\ni.e.\n$$\nm(t)=2 t^{3}-t^{2}-2\\left(a^{2}+2\\right) t+a+4>0,\\ t \\in[-1,1]\n$$\nWe have\n$$\nm'(t)=6 t^{2}-2 t-2\\left(a^{2}+2\\right) \\leq 6 t^{2}-2 t-4=(6 t+4)(t-1)\n$$\nand therefore $m'(t) \\leq 0$ in the interval $\\left[-\\frac{2}{3}, 1\\right]$. Hence $m(t)$ is a decreasing function in this interval which implies that\n$$\nm(t) \\geq m(1)=-2 a^{2}+a+1=(1+2 a)(1-a)>0\n$$\nfor any $t \\in\\left[-\\frac{2}{3}, 1\\right]$. On the other hand, we have\n$$\nm(t) \\geq 2 t^{3}-t^{2}+a+4 \\geq-2-1-\\frac{1}{2}+4>0\n$$\nfor $t \\in[-1,0]$, which completes the proof of (1).\n\nThe cases considered above cover all possibilities for $a$ and we conclude that the answer is $a \\in\\left(-\\infty,-\\frac{1}{2}\\right] \\cup(1,+\\infty)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27250,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $a, b_{1}, c_{1}, \\ldots, b_{n}, c_{n}$ be real numbers such that\n$$\nx^{2 n}+a x^{2 n-1}+a x^{2 n-2}+\\cdots+a x+1=\\left(x^{2}+b_{1} x+c_{1}\\right) \\ldots\\left(x^{2}+b_{n} x+c_{n}\\right)\n$$\nfor every real number $x$. Prove that $c_{1}=\\cdots=c_{n}=1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nWe first show that the assertion follows from the following lemma.\n\nLEMMA. The polynomial $P(z)=z^{2 n}+a z^{2 n-1}+a z^{2 n-2}+\\cdots+a z+1$ has at least $2 n-2$ complex zeros lying on the unit circle and different from $\\pm 1$.\n\nSince the coefficients of $P(z)$ are real numbers, its non-real zeros are complex conjugate. It follows from the lemma that there are at least $n-1$ pairs of such zeros and we denote them by $\\alpha_{1}, \\overline{\\alpha_{1}}, \\ldots, \\alpha_{n-1}, \\overline{\\alpha_{n-1}}$.\n\nWe may assume that\n$$\n\\begin{aligned}\nx^{2}+b_{1} x+c_{1}= & \\left(x-\\alpha_{1}\\right)\\left(x-\\overline{\\alpha_{1}}\\right) \\\\\n\\cdots & \\cdots \\\\\nx^{2}+b_{n-1} x+c_{n-1}= & \\left(x-\\alpha_{n-1}\\right)\\left(x-\\overline{\\alpha_{n-1}}\\right)\n\\end{aligned}\n$$\nSince $|\\alpha_{1}|^{2}=\\cdots=|\\alpha_{n-1}|^{2}=1$ we get $c_{1}=\\cdots=c_{n-1}=1$ and therefore $c_{n}=1$.\n\nProof of the Lemma. Since the non-real zeros of $P(z)$ are complex conjugate it is enough to prove that this polynomial has at least $n-1$ zeros on the upper unit semicircle.\n\nTo do this set $z=e^{i 2 \\theta}$. Then\n$$\n\\begin{aligned}\n\\frac{z^{2 n}+1}{z^{2 n-1}+\\cdots+z} & =\\frac{\\left(z^{2 n}+1\\right)(z-1)}{z\\left(z^{2 n-1}-1\\right)} \\\\\n& =\\frac{\\left(e^{i 2 n \\theta}+e^{-i 2 n \\theta}\\right)\\left(e^{i \\theta}-e^{-i \\theta}\\right)}{\\left(e^{i(2 n-1) \\theta}-e^{-i(2 n-1) \\theta}\\right)} \\\\\n& =2 \\frac{\\cos 2 n \\theta \\sin \\theta}{\\sin (2 n-1) \\theta}=\\frac{\\sin (2 n+1) \\theta}{\\sin (2 n-1) \\theta}-1\n\\end{aligned}\n$$\nTherefore we have to prove that the equation\n$$\nf(\\theta)=\\sin (2 n+1) \\theta+(a-1) \\sin (2 n-1) \\theta=0\n$$\nhas at least $n-1$ roots in the interval $\\left(0, \\frac{\\pi}{2}\\right)$. This is obvious for $a=1$ because $f\\left(\\theta_{k}\\right)=0$, where $\\theta_{k}=\\frac{k \\pi}{2 n+1}, 1 \\leq k \\leq n$.\n\nObserve now that $(k-1) \\pi<(2 n-1) \\theta_{k}<k \\pi$ and this implies that $(-1)^{k-1} \\sin (2 n-1) \\theta_{k}>0$. Hence $f\\left(\\theta_{k}\\right) f\\left(\\theta_{k+1}\\right)<0$ for $a \\neq 1$ and the Intermediate value theorem implies that the equation $f(\\theta)=0$ has at least one root in each of the intervals $\\left(\\theta_{1}, \\theta_{2}\\right), \\ldots,\\left(\\theta_{n-1}, \\theta_{n}\\right)$. This completes the proof of the lemma.\n\nRemark. It follows from the above that the polynomial $P(z)$ has at most two real zeros $x_{1}$ and $x_{2}$ (possibly $x_{1}=x_{2}$ ) and $x_{1} x_{2}=1$. It can be proved that:\n\n1) if $a>2$, then $x_{1}<-1$ and $x_{2} \\in(-1,0)$;\n\n2) if $a=2$, then $x_{1}=x_{2}=-1$;\n\n3) if $a< -\\frac{2}{2 n-1}$, then $x_{1}>1$ and $x_{2} \\in(0,1)$;\n\n4) if $a=-\\frac{2}{2 n-1}$, then $x_{1}=x_{2}=1$;\n\n5) if $a \\in\\left(-\\frac{2}{2 n-1}, 2\\right)$, then the polynomial $P(z)$ has no real zeros. In this case it has a zero $z=e^{i 2 \\theta}$, where $\\theta \\in\\left(0, \\theta_{1}\\right)$ for $a \\in\\left(-\\frac{2}{2 n-1}, 1\\right)$ and $\\theta \\in\\left(\\theta_{n}, \\frac{\\pi}{2}\\right)$ for $a \\in(1,2)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27251,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all triangles $ABC$ with integer sidelengths such that the side $AC$ is equal to the bisector of $\\angle BAC$ and the perimeter of $\\triangle ABC$ is equal to $10p$, where $p$ is a prime number.",
    "options": [],
    "answer": "(a,b,c) = (28, 15, 27), (267, 280, 343), (62, 120, 128)",
    "solution": "Solution:\nUsing the standard notation for the elements of $\\triangle ABC$ we have $b^{2} = l_{a}^{2} = bc - \\frac{a^{2}bc}{(b+c)^{2}}$. Hence\n$$\na^{2}c = (c-b)(c+b)^{2}\n$$\nLet $\\frac{a}{c} = \\frac{m}{n},\\ (m, n) = 1$, and $\\frac{b}{c} = \\frac{r}{s},\\ (r, s) = 1$. Then (1) implies that\n$$\n\\frac{m^{2}}{n^{2}} = \\frac{(s-r)(s+r)^{2}}{s^{3}}\n$$\nSince both sides are irreducible fractions, we obtain $m^{2} = (s-r)(s+r)^{2}$ and $n^{2} = s^{3}$. The first equality shows that $s-r$ is a perfect square and the second one implies that $s$ is a perfect square. Set $s = t^{2}$ and $s-r = k^{2}$. Then $r = t^{2} - k^{2}$, $m = k\\left(2t^{2} - k^{2}\\right)$ and $n = t^{3}$.\n\nWe now set $a = mx$, $c = nx$, $b = ry$ and $c = sy$. Then $nx = sy$, i.e. $y = tx$. Therefore $a = xk\\left(2t^{2} - k^{2}\\right)$, $b = x t\\left(t^{2} - k^{2}\\right)$ and $c = x t^{3}$, where $t > k$ and $(t, k) = 1$. Moreover, one can easily check that these $a, b$ and $c$ satisfy the triangle inequality.\n\nNow the condition $a + b + c = 10p$ becomes $x(k + t)\\left(2t^{2} - k^{2}\\right) = 10p$. Note that $\\left(k + t, 2t^{2} - k^{2}\\right) = 1$. Then it is easy to see that we have only the following possibilities:\n$$\n\\begin{array}{l|l|l}\nx = 1 \\\\\nk + t = 5 \\\\\n2t^{2} - k^{2} = 2p\n\\end{array}, \\quad\n\\begin{aligned}\n& x = 1 \\\\\n& k + t = 10 \\\\\n& 2t^{2} - k^{2} = p\n\\end{aligned}, \\quad\n\\begin{aligned}\n& x = 2 \\\\\n& k + t = 5 \\\\\n& 2t^{2} - k^{2} = p\n\\end{aligned}\n$$\nA direct verification shows that $(x, k, t) = (1, 2, 3)$, $(x, k, t) = (1, 3, 7)$ and $(x, k, t) = (2, 1, 4)$. Therefore $(a, b, c) = (28, 15, 27)$, $(a, b, c) = (267, 280, 343)$ and $(a, b, c) = (62, 120, 128)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27252,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe point $K$ on the edge $AB$ of the cube $ABCD A_{1}B_{1}C_{1}D_{1}$ is such that the angle between the line $A_{1}B$ and the plane $(B_{1}CK)$ is equal to $60^{\\circ}$. Find $\\tan \\alpha$, where $\\alpha$ is the angle between the planes $(B_{1}CK)$ and $(ABC)$.",
    "options": [],
    "answer": "sqrt(5)",
    "solution": "Solution:\n\nWe may assume that the edges of the cube have length $1$. Denote $M = A_{1}B \\cap KB_{1}$ and set $KB = x$. Since $\\triangle KBM \\sim \\triangle A_{1}MB_{1}$ it follows that $\\frac{MB}{A_{1}M} = x$. Now using the identity $A_{1}M + MB = \\sqrt{2}$, we get $MB = \\frac{x \\sqrt{2}}{x+1}$.\n\nDenote by $V$ the volume of the tetrahedron $KBCB_{1}$. Then\n$$\nV = \\frac{1}{3} BB_{1} \\cdot S_{KBC} = \\frac{x}{6}\n$$\nOn the other hand, the altitude of $KBCB_{1}$ through $B$ is equal to $h = BM \\sin 60^{\\circ} = \\frac{x \\sqrt{6}}{2(x+1)}$. Since $B_{1}K = CK = \\sqrt{x^{2}+1}$ and $B_{1}C = \\sqrt{2}$, we have $S_{B_{1}KC} = \\frac{\\sqrt{2x^{2}+1}}{2}$. Therefore\n$$\nV = \\frac{h \\cdot S_{B_{1}KC}}{3} = \\frac{x \\sqrt{6(2x^{2}+1)}}{12(x+1)}\n$$\nThis and (1) imply that\n$$\n\\frac{x}{6} = \\frac{x \\sqrt{6(2x^{2}+1)}}{12(x+1)}\n$$\nand we obtain easily that $x = \\frac{1}{2}$.\n\nDenote by $L$ the foot of the perpendicular from $B$ to $KC$. Then $KC \\perp BL$ and $KC \\perp BB_{1}$ which shows that $KC \\perp B_{1}L$. Hence $\\alpha = \\Varangle B_{1}LB$. We get from $\\triangle KBC$ that $BL = \\frac{1}{\\sqrt{5}}$ and therefore $\\tan \\alpha = \\sqrt{5}$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 27253,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nProve that any triangle of area $\\sqrt{3}$ can be placed into an infinite band of width $\\sqrt{3}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSuppose that it is not possible to cut a triangle $T$ of area $\\sqrt{3}$ from the band. Then it is clear that the lengths of its altitudes are greater than $\\sqrt{3}$. Hence the lengths of its sides are less than $2$.\n\nLet $\\alpha$ be the smallest angle of $T$. Then $\\alpha \\leq 60^{\\circ}$ and we get\n$$\n\\sqrt{3} = \\frac{b c \\sin \\alpha}{2} < \\frac{2 \\cdot 2 \\cdot \\sin 60^{\\circ}}{2} = \\sqrt{3}\n$$\na contradiction.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27254,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $m$ be a positive integer, $A=\\{-m,-m+1, \\ldots, m-1, m\\}$ and $f: A \\rightarrow A$ be a function such that $f(f(n))=-n$ for every $n \\in A$.\n\na) Prove that the number $m$ is even.\n\nb) Find the number of all functions $f: A \\rightarrow A$ with the required property.",
    "options": [],
    "answer": "m is even; for m equal to two times k, the number of functions is m!/(m/2)!",
    "solution": "Solution:\n\na.\nLet $n \\in A$ and $O_{n}=\\{n, f(n),-n, f(-n)\\}$. Since $f(f(n))=-n$ and $f(f(-n))=n$, it follows easily that if $k \\in A$ then either $O_{k}=O_{n}$ or $O_{n} \\cap O_{k}=\\varnothing$. Moreover, we obtain $f(n) \\neq f(-n)$ for $n \\neq 0$.\n\nFurther, if $f( \\pm n)= \\pm n$, then $\\mp n=f(f( \\pm n))=f( \\pm n)= \\pm n$, i.e. $n=0$. Also, $f( \\pm n)=\\mp n$ gives $\\mp n=f(f( \\pm n))=f(\\mp n)$ and then $n=0$. Therefore $|O_{n}|=4$ for $n \\neq 0$ which means that $A \\setminus \\{0\\}$ splits into disjoint quadruples. In particular, the number $m$ is even.\n\nb.\nLet $m=2k$ and $f: A \\rightarrow A$ be a function with the desired property. Set $A_{+}=\\{1,2, \\ldots, m\\}$. We note that $f(-n)=f(f(f(n)))=-f(n)$ and, in particular, $f(0)=0$. Hence for $n \\neq 0$ either $f(n)>0$ or $f(-n)<0$. This means that the quadruple $O_{n}$ is uniquely determined by a pair $(n', f(n'))$ of distinct numbers from $A_{+}$—$(n, f(n))$ or $(f(-n), n)$. Therefore $f$ induces a pairing of $A_{+}$ into ordered pairs.\n\nConversely, any pairing of $A_{+}$ into ordered pairs $(n, k)$ defines a function with the required properties by setting\n$$\nf(0)=0, \\quad f(n)=k, \\quad f(k)=-n, \\quad f(-n)=-k, \\quad f(-k)=n\n$$\nIt remains to count the number of the pairings of $A_{+}$ into ordered pairs. Ordering all pairs of a given pairing one after another (this can be done in $k!$ ways) we obtain a permutation of the numbers $1,2, \\ldots, m$. This gives classes of \"equivalent\" permutations of $k!$ elements. Therefore the required number is equal to $\\frac{m!}{k!}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27255,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTwo circles $k_{1}$ and $k_{2}$ are externally tangent at point $T$. A line meets $k_{1}$ at points $A$ and $B$ and is tangent to $k_{2}$ at point $X$. The line $X T$ meets $k_{1}$ at point $S$ and let $C$ be a point on the arc $\\overparen{T S}$ which does not contain $A$ and $B$. Let $C Y$ be the tangent line to $k_{2}$ ($Y \\in k_{2}$) such that the segments $C Y$ and $S T$ do not intersect. If $I$ is the intersection point of the lines $X Y$ and $S C$, prove that:\n\na) the points $C, T, Y$ and $I$ are concyclic;\n\nb) $I$ is the center of the excircle of $\\triangle A B C$ tangent to the side $B C$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na)\nSince the circles $k_{1}$ and $k_{2}$ are tangent at the point $T$, we have\n$$\n\\Varangle B X T = \\frac{\\widehat{X T}}{2} = \\frac{\\widehat{T S}}{2} = \\Varangle T A S.\n$$\nThen it follows easily that $S$ is the midpoint of the arc $\\widehat{A B}$, i.e. $S A = S B$. Hence $\\Varangle T C I = \\Varangle T A S$ (the quadrilateral $A T C S$ is cyclic), $\\Varangle T A S = \\Varangle B X T$ and $\\Varangle B X T = \\Varangle T Y X$. Therefore $\\Varangle T C I = \\Varangle T Y I$, which shows that the quadrilateral $C T Y I$ is cyclic.\n\n![](attached_image_1.png)\n\nb)\nSince $\\Varangle A X S = \\Varangle T A S$ it follows easily that $\\triangle A X S \\sim \\triangle T A S$, whence $S A^{2} = S T \\cdot S X$. We have from a) that $\\Varangle C I T = \\Varangle C Y T = \\Varangle T X Y$ and therefore $\\triangle S X I \\sim \\triangle S I T$, whence $S I^{2} = S T \\cdot S X$. Hence $S A = S I$. On the other hand, it follows from\n$$\n\\Varangle B C I = 180^{\\circ} - \\Varangle B C S = 180^{\\circ} - \\left(\\gamma + \\frac{\\alpha + \\beta}{2}\\right) = 90^{\\circ} - \\frac{\\gamma}{2}\n$$\nthat $C I$ is the external bisector of $\\Varangle A C B$.\n\nIn the isosceles $\\triangle B S I$ we have $\\Varangle B S I = \\Varangle B S C = \\alpha$ and we find $\\Varangle B I S = 90^{\\circ} - \\frac{\\alpha}{2}$. Now from $\\triangle B C I$ we have $\\Varangle C B I = 90^{\\circ} - \\frac{\\beta}{2}$, which means that $B I$ is the external bisector of $\\Varangle A B C$. Therefore $I$ is the center of the excircle of $\\triangle A B C$ tangent to the side $B C$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27256,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the number of the subsets $B$ of the set $\\{1,2, \\ldots, 2005\\}$ having the following property: the sum of the elements of $B$ is congruent to $2006$ modulo $2048$.",
    "options": [],
    "answer": "2^1994",
    "solution": "Solution:\nLet us consider the set $\\{1,2,2^{2}, \\ldots, 2^{10}\\}$. Since every number from $0$ to $2047$ can be represented in a unique way as a sum of powers of $2$ (elements of our set), we conclude that for every $i$, $0 \\leq i \\leq 2047$, there is a unique subset of $\\{1,2,2^{2}, \\ldots, 2^{10}\\}$ such that the sum of its elements is equal to $i$ ($0$ corresponds to the empty set).\n\nWe now consider a set $A$ with the following property: for every $i$ the number of the subsets of $A$ such that the sums of their elements are congruent to $i$ modulo $2048$ does not depend on $i$. It is easy to see that for every $a$ the set $A \\cup\\{a\\}$ has the same property. Since $\\{1,2,2^{2}, \\ldots, 2^{10}\\} \\subset \\{1,2,3, \\ldots, 2005\\}$, we conclude that the number of the subsets $B$ of $\\{1,2,3, \\ldots, 2005\\}$ such that the sums of their elements are congruent to $i$ modulo $2048$ is equal to $\\frac{2^{2005}}{2048} = \\frac{2^{2005}}{2^{11}} = 2^{1994}$ and this is the required number.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 27257,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe points $P$ and $Q$ lie in the interior of $\\triangle ABC$, $\\angle ACP = \\angle BCQ$ and $\\angle CAP = \\angle BAQ$. The feet of the perpendiculars from $P$ to the lines $BC$, $CA$ and $AB$ are denoted by $D$, $E$ and $F$, respectively. Prove that if $\\angle DEF = 90^{\\circ}$, then $Q$ is the orthocenter of $\\triangle BDF$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWe have $\\angle BCQ = \\angle ACP = \\angle EDP$. Since $PD \\perp BC$, it follows that $ED \\perp CQ$. Analogously we have $AQ \\perp EF$. Since $\\angle DEF = 90^{\\circ}$, we conclude that $\\angle AQC = 90^{\\circ}$ as well. Then $\\triangle QCD \\sim \\triangle ACP$, because $\\angle QCD = \\angle ACP$ and we get\n$$\n\\frac{DC}{QC} = \\frac{PC \\cos \\angle PCD}{AC \\cos \\angle ACQ} = \\frac{PC}{AC}\n$$\nTherefore $\\angle DQC = \\angle PAC = \\angle PFE$. Since $CQ \\parallel EF\\ (\\perp ED)$, it follows that $DQ \\parallel PF$, i.e.\n\n![](attached_image_1.png)\n\n$DQ \\perp AB$.\nUsing the same arguments we prove that $FQ \\perp BC$ and hence $Q$ is the orthocenter of $\\triangle BDF$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27258,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $M$ be the set of the rational numbers in the interval $(0,1)$. Does there exist a subset $A$ of $M$ such that every number from $M$ can be represented in a unique way as a sum of one or finitely many distinct numbers from $A$?",
    "options": [],
    "answer": "No; there is no such subset A.",
    "solution": "Solution:\nAssume, for a contradiction, that there exists such a set.\n\nWe first prove that if $a \\in A$, then $A \\cap \\left(\\frac{a}{2}, a\\right) = \\varnothing$. To do this suppose the contrary, i.e. there exists $a'$ in $A$ and $a > a' > \\frac{a}{2}$. Then the number $a - a' < \\frac{a}{2}$ can be represented as a sum of one or finitely many different numbers from $A$. Since each of these numbers is less than $\\frac{a}{2}$, the number $a = a' + (a - a')$ has two different representations of the required type (as $a$ and as $a'$ plus the numbers of the representation of $a - a'$), a contradiction.\n\nIn particular, it follows from the above that in every interval $\\left[\\frac{1}{2^i}, \\frac{1}{2^{i-1}}\\right)$, $i = 1, 2, \\ldots$, there is at most one element of $A$. Since the set $A$ is infinite (otherwise we can obtain only a finite number of sums of different numbers from $A$) it easily follows that the numbers of $A$ can be ordered in an infinite sequence $a_1, a_2, \\ldots$, which satisfies $a_i \\geq 2 a_{i+1}$ for every $i$. If this inequality is strict for some $i$, then\n$$\ns = \\sum_{i=2}^{\\infty} a_i < \\sum_{i=2}^{\\infty} \\frac{a_1}{2^{i-1}} = a_1\n$$\nThis shows that the numbers from the interval $(s, a_1)$ cannot be represented as a sum of one or finitely many different numbers from $A$.\n\nTherefore $a_{i+1} = \\frac{a_1}{2^i}$ for every $i$. Now it is easy to see that only the numbers of the form $a_1 \\frac{m}{2^n}$ can be represented as a sum of one or a finite number of different numbers from $A$. Thus any rational number with an odd denominator which is coprime with the denominator of $a_1$ cannot be represented as required. This is a contradiction which completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27259,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $\\mathbb{R}^*$ be the set of non-zero real numbers. Find all functions $f: \\mathbb{R}^* \\rightarrow \\mathbb{R}^*$ such that\n$$\nf\\left(x^{2}+y\\right)=f^{2}(x)+\\frac{f(x y)}{f(x)}\n$$\nfor all $x, y \\in \\mathbb{R}^*,\\ y \\neq -x^{2}$.",
    "options": [],
    "answer": "f(x) = x for every nonzero real number x",
    "solution": "Solution:\nSet $\\alpha=f(1)$. Then setting $y=1$ and $x=1$ in\n$$\nf\\left(x^{2}+y\\right)=f^{2}(x)+\\frac{f(x y)}{f(x)}\n$$\ngives\n$$\nf\\left(x^{2}+1\\right)=f^{2}(x)+1\n$$\nand\n$$\nf(y+1)=\\alpha^{2}+\\frac{f(y)}{\\alpha}\n$$\nrespectively. Using (3), we consecutively get\n$$\n\\begin{gathered}\nf(2)=\\alpha^{2}+1,\\ f(3)=\\frac{\\alpha^{3}+\\alpha^{2}+1}{\\alpha} \\\\\nf(4)=\\frac{\\alpha^{4}+\\alpha^{3}+\\alpha^{2}+1}{\\alpha^{2}},\\ f(5)=\\frac{\\alpha^{5}+\\alpha^{4}+\\alpha^{3}+\\alpha^{2}+1}{\\alpha^{3}}\n\\end{gathered}\n$$\nOn the other hand, setting $x=2$ in (2) gives $f(5)=\\alpha^{4}+2 \\alpha^{2}+2$. Therefore $\\frac{\\alpha^{5}+\\alpha^{4}+\\alpha^{3}+\\alpha^{2}+1}{\\alpha^{3}}=\\alpha^{4}+2 \\alpha^{2}+2 \\Longleftrightarrow \\alpha^{7}+\\alpha^{5}-\\alpha^{4}+\\alpha^{3}-\\alpha^{2}-1=0$, whence\n$$\n(\\alpha-1)\\left[\\alpha^{4}\\left(\\alpha^{2}+\\alpha+1\\right)+(\\alpha+1)^{2}\\left(\\alpha^{2}-\\alpha+1\\right)+2 \\alpha^{2}\\right]=0\n$$\nSince the expression in the square brackets is positive, we have $\\alpha=1$. Now (3) implies that\n$$\nf(y+1)=f(y)+1\n$$\nand therefore $f(n)=n$ for every positive integer $n$.\n\nNow take an arbitrary positive rational number $\\frac{a}{b}$ ( $a, b$ are positive integers). Since (4) gives $f(y)=y \\Longleftrightarrow f(y+m)=y+m, m$ is a positive integer, the equality $f\\left(\\frac{a}{b}\\right)=\\frac{a}{b}$ is equivalent to\n$$\nf\\left(b^{2}+\\frac{a}{b}\\right)=b^{2}+\\frac{a}{b}\n$$\nSince the last equality follows from (1) for $x=b$ and $y=\\frac{a}{b}$ we conclude that $f\\left(\\frac{a}{b}\\right)=\\frac{a}{b}$.\n\nSetting $y=x^{2}$ in (4), we obtain $f\\left(x^{2}+1\\right)=f\\left(x^{2}\\right)+1$. Hence using (2) we conclude that $f\\left(x^{2}\\right)=f^{2}(x)>0$. Thus $f(x)>0$ for every $x>0$. Now (1), the inequality $f(x)>0$ for $x>0$ and the identity $f\\left(x^{2}\\right)=f^{2}(x)$ imply that $f(x)>f(y)$ for $x>y>0$. Since $f(x)=x$ for every rational number $x>0$, it easily follows that $f(x)=x$ for every real number $x>0$.\n\nFinally, given an $x<0$ we choose $y<0$ such that $x^{2}+y>0$. Then $x y>0$ and (1) gives\n$$\n\\begin{aligned}\nx^{2}+y & =f\\left(x^{2}+y\\right)=f^{2}(x)+\\frac{f(x y)}{f(x)} \\\\\n& =f\\left(x^{2}\\right)+\\frac{x y}{f(x)}=x^{2}+\\frac{x y}{f(x)}\n\\end{aligned}\n$$\ni.e. $f(x)=x$. Therefore $f(x)=x$ for every $x \\in \\mathbb{R}^*$. It is clear that this function satisfies (1).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27260,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDoes there exist a strictly increasing sequence of positive integers $\\{a_{n}\\}_{n=1}^{\\infty}$ such that $a_{n} \\leq n^{3}$ for every $n$ and every positive integer can be written in a unique way as a difference of two terms of the sequence?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nAnswer: there is such a sequence. We shall define the required sequence inductively. We set $a_{1}=1$, $a_{2}=2$ and assume that $a_{1}, a_{2}, \\ldots, a_{2k}$ are already determined. Denote by $m$ the smallest positive integer which cannot be represented as $a_{j}-a_{i}$, $1 \\leq i < j \\leq 2k$. Since the number of such differences is $d = k(2k-1)$, we have $m \\leq d+1$.\n\nSet $a_{2k+2} = a_{2k+1} + m$, where $a_{2k+1}$ is such that\n$$\n\\begin{aligned}\n& a_{2k+1} \\neq a_{l},\\quad a_{2k+1} \\pm m \\neq a_{l} \\\\\n& a_{2k+1} - a_{l} \\neq a_{j} - a_{i},\\quad a_{2k+1} + m - a_{l} \\neq a_{j} - a_{i}\n\\end{aligned}\n$$\nfor $1 \\leq l \\leq 2k$, $1 \\leq i < j \\leq 2k$. This implies that $a_{1}, a_{2}, \\ldots, a_{2k+2}$ are all distinct and every integer between $1$ and $m$ can be written in a unique way as $a_{j}-a_{i}$, $1 \\leq i < j \\leq 2k+2$.\n\nSince there are exactly $6k + 4kd$ \"forbidden\" values for $a_{2k+1}$, we can choose $a_{2k+1}$ with the above properties and such that $a_{2k+1} \\leq 6k + 4kd + 1$. Then\n$$\na_{2k+1} < a_{2k+2} = a_{2k+1} + m \\leq 6k + 4kd + 1 + d + 1 < (2k+1)^{3}\n$$\nand it remains to put the numbers $a_{1}, a_{2}, \\ldots, a_{2k+2}$ in increasing order (check that the inequality $a_{n} \\leq n^{3}$ is still valid).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27261,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA real number is assigned to every point in the plane. Let $\\mathcal{P}$ be a convex $n$-gon. It is known that for every $n$-gon similar to $\\mathcal{P}$ the sum of the numbers assigned to its vertices is equal to $0$. Prove that all numbers assigned to the points in the plane are equal to $0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $O$ be an arbitrary point in the plane and let $A_{1,1} A_{2,1} \\ldots A_{n,1}$ be an $n$-gon similar to $\\mathcal{P}$ and containing $O$. Consider the $n$-gons\n$$\nO A_{1,1} A_{1,2} \\ldots A_{1, n-1},\\quad O A_{2,1} A_{2,2} \\ldots A_{2, n-1},\\quad \\ldots,\\quad O A_{n,1} A_{n,2} \\ldots A_{n, n-1}\n$$\nthat are similar to $A_{1,1} A_{2,1} \\ldots A_{n,1}$ and have the same orientation. Using a rotation and a homothety with center $O$ we see that the $n$-gon $A_{1, j} A_{2, j} \\ldots A_{n, j}$, $j=2, \\ldots, n-1$, is similar to $A_{1,1} A_{2,1} \\ldots A_{n,1}$ and has the same orientation.\n\nDenote by $o$ and $a_{i, j}$ the numbers assigned to the points $O$ and $A_{i, j}$, respectively. Summing up the equalities\n$$\no+\\sum_{j=1}^{n-1} a_{i, j}=0,\\quad i=1, \\ldots, n\n$$\nand using that\n$$\n\\sum_{i=1}^{n} a_{i, j}=0,\\quad j=1, \\ldots, n-1\n$$\nwe get $n o=0$, which implies the assertion.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27262,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $\\triangle A'B'C$ be the image of $\\triangle ABC$ under a rotation with center $C$. Denote by $M$, $E$ and $F$ the midpoints of the segments $BA'$, $AC$ and $B'C$, respectively. If $AC \\neq BC$ and $EM = FM$, find $\\Varangle EMF$.",
    "options": [],
    "answer": "60°",
    "solution": "Solution:\nSet $BC = B'C = a$, $AC = A'C = b$, $AB = A'B' = c$ and $\\Varangle C = \\gamma$. Since the triangles $AA'C$ and $BB'C$ are similar we have\n$$\n\\frac{AA'}{AC} = \\frac{BB'}{BC} = k\n$$\nTherefore $AA' = k b$ and $BB' = k a$.\n\n![](attached_image_1.png)\n\nUsing the median formula several times we get\n$$\n\\begin{aligned}\n& 4 EM^{2} = k^{2} b^{2} + c^{2} + b^{2} + a^{2} - A'B^{2} - b^{2} \\\\\n& 4 FM^{2} = k^{2} a^{2} + c^{2} + b^{2} + a^{2} - A'B^{2} - a^{2}\n\\end{aligned}\n$$\nHence the condition $EM = FM$ is equivalent to $\\left(k^{2} - 1\\right)\\left(a^{2} - b^{2}\\right) = 0$, whence $k = 1$ (since $a \\neq b$). This is true exactly when $\\triangle AA'C$ is equilateral, i.e. for $\\pm 60^{\\circ}$ rotation. We consider here only the case of a positively oriented $\\triangle ABC$ and rotation through $60^{\\circ}$; the remaining cases are analogous. We have from $\\triangle CBA'$ that\n$$\nA'B^{2} = a^{2} + b^{2} - 2ab \\cos\\left(60^{\\circ} - \\gamma\\right)\n$$\nand we obtain from $\\triangle AB'C$ that\n$$\n4 EF^{2} = B'A^{2} = a^{2} + b^{2} - 2ab \\cos\\left(60^{\\circ} + \\gamma\\right)\n$$\nWe shall prove that $EF = EM$. It follows from the above expressions for $EF$ and $EM$ that this is equivalent to the identity\n$$\nc^{2} + 2ab \\cos\\left(60^{\\circ} - \\gamma\\right) = a^{2} + b^{2} - 2ab \\cos\\left(60^{\\circ} + \\gamma\\right)\n$$\nSince $a^{2} + b^{2} - c^{2} = 2ab \\cos \\gamma$ we rewrite the above identity as\n$$\n2ab \\cos\\left(60^{\\circ} - \\gamma\\right) + 2ab \\cos\\left(60^{\\circ} + \\gamma\\right) = 2ab \\cos \\gamma\n$$\nThis is equivalent to $2 \\cos 60^{\\circ} \\cos \\gamma = \\cos \\gamma$, which is true. Therefore $EF = EM$ and $\\Varangle EMF = 60^{\\circ}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27263,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $a_{1}, a_{2}, \\ldots, a_{2005}, b_{1}, b_{2}, \\ldots, b_{2005}$ be real numbers such that the inequality\n$$\n\\left(a_{i} x-b_{i}\\right)^{2} \\geq \\sum_{j=1, j \\neq i}^{2005}\\left(a_{j} x-b_{j}\\right)\n$$\nholds true for every real number $x$ and all $i=1,2, \\ldots, 2005$. Find the maximum possible number of the positive numbers amongst $a_{i}$ and $b_{i}, i=1,2, \\ldots, 2005$.",
    "options": [],
    "answer": "4009",
    "solution": "Solution:\nWe first prove that at least one of the numbers $a_{1}, a_{2}, \\ldots, a_{2005}$ is not positive. To do this we assume the contrary and choose $i$ such that\n$$\n\\frac{b_{i}}{a_{i}}=M=\\max _{1 \\leq j \\leq 2005}\\left(\\frac{b_{j}}{a_{j}}\\right)\n$$\nThen we can find $\\varepsilon>0$ such that\n$$\n\\left(a_{i} x-b_{i}\\right)^{2}<\\sum_{j=1, j \\neq i}^{2005}\\left(a_{j} x-b_{j}\\right)\n$$\nfor every $x \\in(M, M+\\varepsilon)$, a contradiction.\n\nOn the other hand, it is easy to see that if $a_{1}=a_{2}=\\cdots=a_{2004}=-a_{2005}=1$ and $b_{1}=b_{2}=\\cdots=b_{2004}=b_{2005} \\geq \\frac{1001^{2}}{2}$ the given inequality is satisfied. Therefore the answer is 4009.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27264,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABC$ ($AC \\neq BC$) be an acute triangle with orthocenter $H$ and incenter $I$. The lines $CH$ and $CI$ meet the circumcircle of $\\triangle ABC$ at points $D$ and $L$, respectively. Prove that $\\angle CIH = 90^\\circ$ if and only if $\\angle IDL = 90^\\circ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nDenote by $k(O, R)$ the circumcircle of $\\triangle ABC$ and by $Q$ the orthogonal projection of $H$ on $CL$. Let us further denote\n$$\n\\begin{gathered}\nK = HQ \\cap LO, \\quad S = k \\cap LO, \\quad P = CL \\cap DS \\\\\nM = AB \\cap LO, \\quad N = AB \\cap CD\n\\end{gathered}\n$$\nNote that $N$ and $M$ are the midpoints of $HD$ and $KL$, respectively. Indeed, we have $AH = AD$ ($\\angle AHD = \\angle ABC = \\angle ADH$) and analogously $BH = BD$. Hence $AB$ is the perpendicular bisector of the segment $HD$.\n\nOn the other hand, $DLSC$ is an isosceles trapezoid and $HK \\parallel CS$. Therefore $DLKH$ is also an isosceles trapezoid and $AB$ is the perpendicular bisector of the segment $KL$. Then\n$$\n\\frac{LQ}{LC} = \\frac{LK}{LS} = \\frac{LM}{R}\n$$\nwhence\n$$\nLQ = \\frac{LC \\cdot LM}{R}\n$$\n![](attached_image_1.png)\nWe also have $LP = \\frac{LO \\cdot LS}{LC} = \\frac{2R^2}{LC}$ since $\\triangle LOP \\sim \\triangle LCS$ and $LB^2 = LS \\cdot LM = 2R \\cdot LM$ since $\\triangle LBS \\sim \\triangle LMB$. Therefore\n$$\nLP \\cdot LQ = LB^2\n$$\nOn the other hand, using $\\angle LBI = \\frac{\\angle B + \\angle C}{2} = \\angle LIB$ we obtain $LB = LI$. Then $LP \\cdot LQ = LI^2$ and, in particular, $Q \\equiv I \\Longleftrightarrow P \\equiv I$.\n\nIt remains to note that $\\angle CIH = 90^\\circ \\Longleftrightarrow Q \\equiv I$ (since $\\angle CQH = 90^\\circ$) and $\\angle IDL = 90^\\circ \\Longleftrightarrow P \\equiv I$ (since $\\angle PDL = 90^\\circ$). This completes the proof.\n\nRemark. It can be proved that $\\angle CIH = 90^\\circ$ iff $\\cos \\angle A + \\cos \\angle B = 1$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27265,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $t$, $a$ and $b$ be positive integers. We call a $(t ; a, b)$-game the following game with two players: the first player subtracts $a$ or $b$ from $t$, then the second player subtracts $a$ or $b$ from the number obtained by the first player, then again the first player subtracts $a$ or $b$ from the number obtained by the second player and so on. The player who obtains first a negative number loses the game. Prove that there exist infinitely many $t$ such that the first player has a winning strategy for any $(t ; a, b)$-game with $a+b=2005$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWe first prove the following lemma.\n\nLEMMA. If in the $(t ; a, b)$-game some of the two players has a winning strategy, then in the $(t+a+b, a, b)$-game the same player has a winning strategy.\n\nProof of the lemma. Denote the players by $A$ and $B$ and let $B$ have a winning strategy for the $(t ; a, b)$-game. In the $(t+a+b ; a, b)$-game after the first move of $A$ we obtain either the $(t+a ; a, b)$-game or the $(t+b ; a, b)$-game with $B$ to go first. In both cases $B$ can get the $(t ; a, b)$-game with $A$ as first player in which case $B$ has a winning strategy.\n\nLet us now assume that $A$ has a winning strategy for the $(t ; a, b)$-game. Then after the first move of $A$ we obtain either the $(t-a ; a, b)$-game or the $(t-b ; a, b)$-game with $B$ to go first. Therefore the second player has a winning strategy for some of these games.\n\nWe consider (without loss of generality) the case of the $(t-a ; a, b)$-game. It follows from the above that the second player has a winning strategy for the $(t-a+a+b=t+b ; a, b)$-game. Since $A$ can obtain the $(t+b ; a, b)$-game with $B$ to go first from the $(t+a+b ; a, b)$-game, it follows that $A$ has a winning strategy for the $(t+a+b ; a, b)$-game. This completes the proof of the lemma.\n\nWe now prove that for $t=2004$ and $a+b=2005$ the first player $A$ has a winning strategy. We may assume that $a \\leq b$. Since $a>0$, we have $b \\leq 2004$. Then $A$ subtracts $b$ from $t=2004$. The resulting number $2004-b$ is less than $a$ since $a+b=2005$. This means that any move of $B$ leads to a negative number. Now the lemma implies that $A$ has a winning strategy for the $(t, a, b)$-game for every $t \\equiv 2004 \\pmod{2005}$ and $a+b=2005$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27266,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $a_{1}, a_{2}, \\ldots, a_{m}$ be arbitrary positive integers. Prove that there exist distinct positive integers $b_{1}, b_{2}, \\ldots, b_{n}$, $n \\leq m$, such that the following two conditions are satisfied:\n(1) all subsets of $\\{b_{1}, b_{2}, \\ldots, b_{n}\\}$ have distinct sums of elements;\n(2) every number $a_{1}, a_{2}, \\ldots, a_{m}$ is the sum of the elements of some subset of $\\{b_{1}, b_{2}, \\ldots, b_{n}\\}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nWe shall prove the assertion by induction on $N = a_{1} + a_{2} + \\cdots + a_{m}$. For $N = 1$ we have $m = 1$, $a_{1} = 1$ and $b_{1} = 1$ is the required number.\n\nLet us assume that the assertion is true for every collection with sum less than $N$ and let $a_{1}, a_{2}, \\ldots, a_{m}$ be such that $a_{1} + a_{2} + \\cdots + a_{m} = N$.\n\nIf all numbers $a_{1}, a_{2}, \\ldots, a_{m}$ are even then the numbers $\\frac{a_{1}}{2}, \\frac{a_{2}}{2}, \\ldots, \\frac{a_{m}}{2}$ have sum $\\frac{N}{2}$ and by the induction hypothesis there exists a collection $b_{1}, b_{2}, \\ldots, b_{n}$ which satisfies the condition. Then the required numbers for $a_{1}, a_{2}, \\ldots, a_{m}$ are $2 b_{1}, 2 b_{2}, \\ldots, 2 b_{n}$.\n\nSuppose now that at least one of the numbers $a_{1}, a_{2}, \\ldots, a_{m}$ is odd. Without loss of generality we can assume that $a_{m}$ is the smallest odd number in the collection. Let us consider the numbers $a_{1}', a_{2}', \\ldots, a_{m-1}'$ defined by\n$$\na_{i}' = \\begin{cases}\n\\frac{a_{i}}{2} & , \\text{ if } a_{i} \\text{ is even } \\\\\n\\frac{a_{i} - a_{m}}{2} & , \\text{ if } a_{i} \\text{ is odd }\n\\end{cases}\n$$\nThe sum of the new numbers $a_{i}'$ is less than $N$ and the induction hypothesis implies the existence of numbers $b_{1}', b_{2}', \\ldots, b_{k}'$ which satisfy the conditions. We shall prove that the numbers $2 b_{1}', 2 b_{2}', \\ldots, 2 b_{k}', a_{m}$ are the required numbers for the collection $a_{1}, a_{2}, \\ldots, a_{m}$.\n\nIf two nonintersecting subsets of $\\{2 b_{1}', 2 b_{2}', \\ldots, 2 b_{k}', a_{m}\\}$ have equal sums then $a_{m}$ (as the only odd number) does not belong to these sets. Dividing by $2$ we obtain two nonintersecting subsets of $\\{b_{1}', b_{2}', \\ldots, b_{k}'\\}$ with equal sums which is a contradiction. Also, it is easy to see that every $a_{i}$, $i = 1, 2, \\ldots, m$ can be represented as a sum of some of the numbers $2 b_{1}', 2 b_{2}', \\ldots, 2 b_{k}', a_{m}$ which completes the induction step.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27267,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn a group of 9 persons it is not possible to choose 4 persons such that every one knows the three others. Prove that this group of 9 persons can be partitioned into four parts in such a way that nobody knows anyone from his part.\n\nEmil Kolev",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWe have to prove that if the edges of a complete graph with 9 vertices are colored in blue and red in such a way that there is no blue quadrilateral then its vertices can be partitioned into 4 groups without any blue edges inside any group.\n\nLemma 1. The edges of a complete graph with 6 vertices are colored in blue and red in such a way that there is no blue quadrilateral and its vertices can not be partitioned into 3 groups without blue edges inside any group. Then the graph does not contain a red triangle.\n\nProof. It is well known that the complete graph with 6 vertices and edges colored in two colors (red and blue) has an one-colored triangle. Denote the vertices of the graph by $v_{1}, v_{2}, \\ldots, v_{6}$ and assume the existence of a red triangle $v_{1} v_{2} v_{3}$. If some edge $v_{i} v_{j}, i, j \\in\\{4,5,6\\}$, is red, then $\\{v_{1}, v_{2}, v_{3}\\},\\{v_{i}, v_{j}\\}$ and $\\{v_{k}\\}, k \\neq 1,2,3, i, j$, is a partition which is supposed not to exist.\nTherefore $v_{4} v_{5} v_{6}$ is a blue triangle. Now the condition implies that for every $i=1,2,3$ at least one of the edges $v_{i} v_{j}, j=4,5,6$, is red. If the edges $v_{1} v_{4}$ and $v_{2} v_{4}$ are red then $v_{1} v_{2} v_{4}$ is a red triangle and, as above, $v_{3} v_{5} v_{6}$ is a blue triangle. Now, if the edge $v_{3} v_{4}$ is blue, then $v_{3} v_{4} v_{5} v_{6}$ is a blue quadrilateral and, if the edge $v_{3} v_{4}$ is red, then we have the partition $\\{v_{1}, v_{2}, v_{3}, v_{4}\\},\\{v_{5}\\}$ and $\\{v_{6}\\}$, a contradiction, which completes the proof of the lemma.\n\nLemma 2. Under the conditions of Lemma 1 all red edges of $G$ form a cycle of length 5.\n\nProof. We know from Lemma 1 that $G$ does not contain a red triangle.\nWithout loss of generality we can assume that the edges $v_{1} v_{2}$ and $v_{3} v_{4}$ are red. If all edges $v_{i} v_{j}$ for $i=1,2, j=3,4$ are red, then the partition $\\{v_{1}, v_{2}, v_{3}, v_{4}\\},\\{v_{5}\\}$ and $\\{v_{6}\\}$ ensures a contradiction. So, let $v_{2} v_{4}$ be blue. Since the quadrilateral $v_{2} v_{4} v_{5} v_{6}$ can not be blue, we may assume without loss of generality that the edge $v_{4} v_{6}$ is red. Now $v_{3} v_{6}$ is blue (otherwise $v_{3} v_{4} v_{6}$ is a red triangle) and $v_{3} v_{5}$ is blue (otherwise we have the partition $\\{v_{1}, v_{2}\\},\\{v_{3}, v_{5}\\}$ and $\\{v_{4}, v_{6}\\}$ for a contradiction).\nIf the edge $v_{1} v_{3}$ is red then using the conditions of the problem we see that the edge $v_{2} v_{5}$ is blue, the edge $v_{2} v_{6}$ is red and the edges $v_{1} v_{6}, v_{1} v_{5}, v_{4} v_{5}$ and $v_{1} v_{4}$ are blue. We finally conclude that the red edges in $G$ are $v_{1} v_{2}, v_{2} v_{6}, v_{6} v_{4}$, $v_{4} v_{3}$ and $v_{3} v_{1}$ and they form a cycle of length 5 as desired.\nIf the edge $v_{1} v_{3}$ is blue, then similar arguments lead to a contradiction. This completes the proof of the lemma.\n\nLet us now consider a graph with 9 vertices $v_{1}, v_{2}, \\ldots, v_{9}$, satisfying the given conditions. It is known that such a graph contains a red triangle or a blue quadrilateral. Since the latter is impossible we have a red triangle $v_{7} v_{8} v_{9}$, say. If the induced graph $v_{1} \\ldots v_{6}$ can be divided into three groups without blue edges inside any group then we obtain the required division of $G$.\nOtherwise Lemma 2 implies that we may assume without loss of generality that the edges $v_{1} v_{2}, v_{2} v_{3}, v_{3} v_{4}, v_{4} v_{5}$ and $v_{5} v_{1}$ are red.\nIf the edges $v_{i} v_{6}, i=7,8,9$, are red then $\\{v_{6}, v_{7}, v_{8}, v_{9}\\},\\{v_{1}, v_{3}\\},\\{v_{2}, v_{4}\\}$ and $\\{v_{5}\\}$ is the desired division. If the edge $v_{7} v_{6}$ is blue while the edges $v_{8} v_{6}$ and $v_{9} v_{6}$ are red then at least three of the edges $v_{7} v_{i}, i=1,2, \\ldots, 5$, are red and we easily have the desired division (otherwise we get a red quadrilateral).\nSimilar arguments solve the two remaining cases: when two of the edges $v_{6} v_{7}, v_{6} v_{8}$ and $v_{6} v_{9}$ are blue and one is red, and when all of them are blue.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 27268,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $a$, $b$ and $c$ be positive integers such that $ab$ divides $c\\left(c^{2}-c+1\\right)$ and $a+b$ is divisible by $c^{2}+1$. Prove that the sets $\\{a, b\\}$ and $\\left\\{c, c^{2}-c+1\\right\\}$ coincide.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWe shall use the following lemma.\n\n**Lemma.** Let $x$, $y$ and $n$ be positive integers such that $\\frac{xy}{x+y} > n$. Then\n$$\n\\frac{xy}{x+y} \\geq n + \\frac{1}{n^{2} + 2n + 2}\n$$\nwith equality if and only if $\\{x, y\\} = \\left\\{n+1, n^{2}+n+1\\right\\}$.\n\nProof of the lemma. Since $xy > n(x+y)$, we have $xy = n(x+y) + r$, where $r$ is a positive integer. Then $(x-n)(y-n) = n^{2} + r$, which implies that $x > n$ and $y > n$. Set $x = n + d_{1}$ and $y = n + d_{2}$. Then $d_{1} d_{2} = n^{2} + r$. Now using the inequalities $\\frac{r}{A+r} \\geq \\frac{1}{A+1}$ and $d_{1} + d_{2} \\leq 1 + n^{2} + r$ (the latter follows from $d_{1} + d_{2} \\leq 1 + d_{1} d_{2}$), we get\n$$\n\\begin{aligned}\n\\frac{xy}{x+y} &= \\frac{n^{2} + d_{1} d_{2} + n(d_{1} + d_{2})}{2n + d_{1} + d_{2}} = n + \\frac{r}{2n + d_{1} + d_{2}} \\\\\n&\\geq n + \\frac{r}{2n + n^{2} + 1 + r} \\geq n + \\frac{1}{n^{2} + 2n + 2}\n\\end{aligned}\n$$\nNote that the equality is attained if and only if $\\{x, y\\} = \\left\\{n+1, n^{2}+n+1\\right\\}$. This completes the proof of the lemma.\n\nThe condition of the problem implies that $c\\left(c^{2}-c+1\\right) = p a b$ and $a+b = q\\left(c^{2}+1\\right)$, where $p$ and $q$ are positive integers. Therefore\n$$\n\\frac{c\\left(c^{2}-c+1\\right)}{c^{2}+1} = \\frac{pqab}{a+b} = \\frac{xy}{x+y}\n$$\nwhere $x = pqa$ and $y = pqb$. Then\n$$\n\\frac{xy}{x+y} = c - \\frac{c^{2}}{c^{2}+1} = c-1 + \\frac{1}{c^{2}+1}\n$$\nwhence $\\frac{xy}{x+y} > c-1$. Now the lemma gives\n$$\n\\frac{xy}{x+y} \\geq c-1 + \\frac{1}{(c-1)^{2} + 2(c-1) + 2} = c-1 + \\frac{1}{c^{2}+1}\n$$\nHence we have the case of equality and therefore\n$$\n\\{x, y\\} = \\left\\{c, c^{2}-c+1\\right\\}\n$$\nSince the numbers $c$ and $c^{2}-c+1$ are coprime and $x = pqa$, $y = pqb$, it follows that $p = q = 1$. Hence $\\{a, b\\} = \\left\\{c, c^{2}-c+1\\right\\}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27269,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe extensions of the sides $AB$ and $CD$ of a convex quadrilateral $ABCD$ meet at point $P$ and the extensions of the sides $BC$ and $AD$ meet at point $Q$. The point $O$ from the interior of the quadrilateral is such that $\\Varangle BOP = \\Varangle DOQ$. Prove that $\\Varangle AOB + \\Varangle COD = 180^\\circ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nThe Sine theorem for the triangles $ODQ$ and $AOQ$ gives\n$$\n\\frac{\\sin \\varphi}{\\sin \\beta} = \\frac{QD}{OD}\n$$\nand\n$$\n\\frac{\\sin (\\varphi + \\Varangle AOD)}{\\sin \\beta} = \\frac{AQ}{OA}\n$$\nwhence\n$$\n\\frac{\\sin (\\varphi + \\Varangle AOD)}{\\sin \\varphi} = \\frac{AQ}{OA} \\cdot \\frac{OD}{QD}\n$$\n\n![](attached_image_1.png)\n\nWe obtain in the same way that\n$$\n\\frac{\\sin (\\varphi + \\Varangle AOB)}{\\sin \\varphi} = \\frac{AP}{OA} \\cdot \\frac{OB}{BP}\n$$\nTherefore\n$$\n\\frac{\\sin (\\varphi + \\Varangle AOD)}{\\sin (\\varphi + \\Varangle AOB)} = \\frac{AQ}{AP} \\cdot \\frac{OD}{OB} \\cdot \\frac{BP}{QD}\n$$\nWe get in the same way that\n$$\n\\frac{\\sin (\\Varangle DOC - \\varphi)}{\\sin (\\Varangle BOC - \\varphi)} = \\frac{QC}{PC} \\cdot \\frac{OB}{OD} \\cdot \\frac{PD}{QB}\n$$\nUsing the Menelaus theorem for $\\triangle ADC$ and the line $QP$ and for $\\triangle ABC$ and the line $QP$ we obtain\n$$\n\\frac{AQ \\cdot DP}{DQ \\cdot CP} = \\frac{AL}{CL} \\text{ and } \\frac{QC \\cdot BP}{QB \\cdot AP} = \\frac{CL}{AL}\n$$\nSetting $\\varphi + \\Varangle AOD = x$, $\\varphi + \\Varangle AOB = y$, $\\Varangle DOC - \\varphi = z$ and $\\Varangle BOC - \\varphi = t$, we have\n$$\n\\frac{\\sin x \\cdot \\sin z}{\\sin y \\cdot \\sin t} = \\frac{AQ \\cdot DP \\cdot QC \\cdot BP}{DQ \\cdot CP \\cdot QB \\cdot AP} = \\frac{AL}{CL} \\cdot \\frac{CL}{AL} = 1\n$$\ni.e. $\\sin x \\cdot \\sin z = \\sin y \\cdot \\sin t$. It follows easily from here that\n$$\n\\cos (x-z) - \\cos (x+z) = \\cos (y-t) - \\cos (y+t)\n$$\nSince $x + y + z + t = 360^\\circ$, we have $\\cos (x+z) = \\cos (y+t)$ and therefore $\\cos (x-z) = \\cos (y-t)$. Since $x-z + y-t < 360^\\circ$ and the equality $x-z = t-y$ implies that $O$ lies on $PQ$ (prove this!) we obtain $x-z = y-t$, whence $x+t = z+y = 180^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27270,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn a group of $B$ boys and $G$ girls it is known that $G \\geq 2 B-1$. Some boys know some girls. Prove that it possible to arrange a dance in pairs in such a way that all boys will dance and every boy who does not know the girl in his pair knows only girls who do not dance.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nIf for every $s=1,2, \\ldots, B$ any $s$ boys know together at least $s$ girls then the Hall (marriages') theorem implies that every boy can dance with a known girl and the condition is satisfied.\n\nLet us assume now the converse and choose the largest $s \\leq B$, such that there are $s$ boys who know together at most $s-1$ girls.\n\nDenote the set of these $s$ boys by $S$ and let $L$ be the set of girls known to the boys from $S$. If some $t$ of the boys outside $S$ know together at most $t$ of the girls outside $L$ we have a contradiction with the choice of $s$. Therefore every $t$ boys outside $S$ know together at least $t+1$ of the girls outside $L$.\n\nNow the Hall theorem implies that every boy outside $S$ can dance with a known girl outside $L$. Hence still non-dancing girls outside $L$ are at least\n$$\nG-(B-s)-(s-1)=G+1-B \\geq B\n$$\n(the girls which dance with boys outside $S$ are $B-s$ and the girls which are known to the boys from $S$ are at most $s-1$ ). If the boys from $S$ dance with some $s$ of these remaining non-dancing girls outside $L$ then the condition is satisfied.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 27271,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSolve the equation\n$$\n\\left|\\left| x-\\frac{5}{2}\\right|-\\frac{3}{2}\\right|=\\left|x^{2}-5 x+4\\right|\n$$",
    "options": [],
    "answer": "1 and 4",
    "solution": "Solution:\nSince both sides of the equation are non-negative, it is equivalent to\n$$\n\\left(\\left|x-\\frac{5}{2}\\right|-\\frac{3}{2}\\right)^{2}=\\left(x^{2}-5 x+4\\right)^{2}\n$$\nHence\n$$\n\\left(\\left|x-\\frac{5}{2}\\right|-\\frac{3}{2}-x^{2}+5 x-4\\right)\\left(\\left|x-\\frac{5}{2}\\right|-\\frac{3}{2}+x^{2}-5 x+4\\right)=0\n$$\nand we have to solve the equations\n$$\n\\left|x-\\frac{5}{2}\\right|-x^{2}+5 x-\\frac{11}{2}=0\n$$\nand\n$$\n\\left|x-\\frac{5}{2}\\right|+x^{2}-5 x+\\frac{5}{2}=0\n$$\nFor $x \\leq \\frac{5}{2}$ we have respectively that $\\frac{5}{2}-x-x^{2}+5 x-\\frac{11}{2}=0 \\Longleftrightarrow x^{2}-4 x+3=0$, i.e. $x=1$ or $x=3$ and $\\frac{5}{2}-x+x^{2}-5 x+\\frac{5}{2}=0 \\Longleftrightarrow x^{2}-6 x+5=0$, i.e. $x=1$ or $x=5$. Therefore $x=1$ is the only solution of the given equation in this case.\n\nAnalogously, for $x>\\frac{5}{2}$ we have $x-\\frac{5}{2}-x^{2}+5 x-\\frac{11}{2}=0 \\Longleftrightarrow x^{2}-6 x+8=0$, i.e, $x=2$ or $x=4$ and $x-\\frac{5}{2}+x^{2}-5 x+\\frac{5}{2}=0 \\Longleftrightarrow x^{2}-4 x=0$, i.e. $x=0$ or $x=4$. Therefore $x=4$ is the only solution of the given equation in this case.\n\nHence the given equation has two solutions $x_{1}=1$ and $x_{2}=4$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27272,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $k$ be the circumcircle of $\\triangle ABC$ with $\\angle ACB > 90^\\circ$, and $BD$ be the diameter of $k$ through $B$. The circle $k_1$ with center $D$ and radius $DC$ meets $k$ at point $E$ and $AB$ at point $G$. If $F$ is the intersection point of $GE$ and $BD$, prove that $\\angle DCG = \\angle EFD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nSince $\\angle ACB > 90^\\circ$, the points $G$ and $E$ lie on different sides to the diameter $BD$. If $\\angle DCG = \\angle EFD$, then $\\angle DCG + \\angle DFG = 180^\\circ$, i.e. it is enough to prove that the quadrilateral $CDFG$ is cyclic. On the other hand, $CE \\perp BD$ and $D$ is the center of $k_1$. Therefore $\\angle CDF = \\frac{1}{2} C\\overset{\\frown}{G}_{k_1}$ (as a central angle), $\\angle CGF = \\angle CGE$ is inscribed in $k_1$ and $\\angle CGF = \\frac{1}{2}(360^\\circ - C\\widehat{G}E_{k_1})$.\n\n![](attached_image_1.png)\n\nThen $\\angle CDF + \\angle CGF = 180^\\circ$, i.e. the quadrilateral $CDFG$ is cyclic, whence $\\angle DCG = \\angle EFD$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27273,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nProve that the equation\n$$\nx^{2} + 2y^{2} + 98z^{2} = \\underbrace{77\\ldots7}_{2005}\n$$\nhas no integer solutions.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nAssume that the equation has a solution $(x_{0}, y_{0}, z_{0})$. Then $x_{0}^{2} + 2y_{0}^{2}$ is divisible by $7$. Since the remainders modulo $7$ of the perfect squares are $0, 1, 2$ and $4$, it follows that both $x_{0}$ and $y_{0}$ are divisible by $7$.\n\nThen the left hand side of the given equation is divisible by $7^{2}$ and hence the number $\\underbrace{11\\ldots1}_{2005}$ is divisible by $7$. But this is a contradiction since $111111$ is divisible to $7$ and $2005 = 6 \\cdot 334 + 1$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27274,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $f(x) = x^{2} + (2a - 1)x - a - 3$, where $a$ is a real parameter.\n\na) Prove that the equation $f(x) = 0$ has two distinct real roots $x_{1}$ and $x_{2}$.\n\nb) Find all values of $a$ such that $x_{1}^{3} + x_{2}^{3} = -72$.",
    "options": [],
    "answer": "a = 2",
    "solution": "Solution:\n\na) The discriminant of $f(x)$ is equal to $4a^{2} + 13 > 0$.\n\nb) We consecutively have\n$$\n\\begin{aligned}\n-72 & = x_{1}^{3} + x_{2}^{3} = (x_{1} + x_{2}) \\left[ (x_{1} + x_{2})^{2} - 3 x_{1} x_{2} \\right] \\\\\n& = (1 - 2a) \\left( 4a^{2} - a + 10 \\right ) = -8a^{3} + 6a^{2} - 21a + 10\n\\end{aligned}\n$$\nTherefore the required values of $a$ are the real solutions of the equation\n$$\n8a^{3} - 6a^{2} + 21a - 82 = 0\n$$\nSince $a = 2$ is a solution of this equation, we obtain\n$$\n8a^{3} - 6a^{2} + 21a - 82 = (a - 2)(8a^{2} + 10a + 41)\n$$\nThe equation $8a^{2} + 10a + 41 = 0$ has no real roots. So the only solution of the problem is $a = 2$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27275,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all values of the real parameters $a$ and $b$ such that the remainder in the division of the polynomial $x^{4}-3 a x^{3}+a x+b$ by the polynomial $x^{2}-1$ is equal to $(a^{2}+1)x+3 b^{2}$.",
    "options": [],
    "answer": "a = -1, b = (1 ± sqrt(13)) / 6",
    "solution": "Solution:\nIt follows from the condition that there is a polynomial $q(x)$ of degree $2$ such that\n$$\nx^{4}-3 a x^{3}+a x+b = q(x)(x^{2}-1) + (a^{2}+1)x + 3 b^{2}\n$$\nfor every $x$. Setting $x=1$ and $x=-1$ we obtain the system\n$$\n\\left\\lvert\\,\n\\begin{array}{cc}\na^{2}+3 b^{2}+2 a-b=0 \\\\\na^{2}-3 b^{2}+2 a+b+2=0\n\\end{array}\n\\right.\n$$\nand it is easy to see that $a=-1$, $b=\\frac{1 \\pm \\sqrt{13}}{6}$. A direct verification shows that these values of $a$ and $b$ are solutions indeed.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 27276,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all values of the real parameter $a$ for which the equations $x^{2}-(2 a+1) x+a=0$ and $x^{2}+(a-4) x+a-1=0$ have real roots $x_{1}, x_{2}$ and $x_{3}, x_{4}$, respectively, such that\n$$\n\\frac{x_{1}}{x_{3}}+\\frac{x_{4}}{x_{2}}=\\frac{x_{1} x_{4}\\left(x_{1}+x_{2}+x_{3}+x_{4}\\right)}{a}\n$$",
    "options": [],
    "answer": "a = -1 ± sqrt(5)",
    "solution": "Solution:\nFor $a \\neq 0, a \\neq 1$, the given equality is equivalent to\n$$\n\\begin{aligned}\n& a\\left(x_{1} x_{2}+x_{3} x_{4}\\right)=x_{1} x_{2} x_{3} x_{4}\\left(x_{1}+x_{2}+x_{3}+x_{4}\\right) \\\\\n& \\Longleftrightarrow 2 a-1=(a-1)(a+5) \\Longleftrightarrow a^{2}+2 a-4=0 \\\\\n& \\Longleftrightarrow a_{1,2}=-1 \\pm \\sqrt{5}\n\\end{aligned}\n$$\nIt is easy to check that for these values of $a$ both equations have real roots. The case $a=0$ is excluded by the condition and $a=1$ implies $x_{4}=0$, whence $x_{1}=0$, which is a contradiction.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27277,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA circle $k$ through the vertices $A$ and $B$ of an acute $\\triangle ABC$ meets the sides $AC$ and $BC$ at inner points $M$ and $N$, respectively. The tangent lines to $k$ at the points $M$ and $N$ meet at point $O$. Prove that $O$ is the circumcenter of $\\triangle CMN$ if and only if $AB$ is a diameter of $k$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nIf $AB$ is a diameter of $k$, then $AN$ and $BM$ are altitudes of $\\triangle ABC$. Let $H$ be the orthocenter of $\\triangle ABC$ and let the tangent line to $k$ at $M$ meet the altitude $CH$ at $O_1$. Then $\\angle CMO_1 = \\angle ABM = \\frac{\\overparen{AM}}{2}$ and $\\angle ABM = \\angle ACH$. Thus $\\angle CMO_1 = \\angle MCO_1$, i.e. $CO_1 = MO_1$. On the other hand $\\angle O_1HM = 90^\\circ - \\angle O_1CM = 90^\\circ - \\angle CMO_1 = \\angle O_1MH$, i.e. $O_1M = O_1H$. Therefore $O_1$ is the midpoint of $CH$. It can be seen analogously that the tangent line to $k$ at $N$ passes through $O_1$, i.e. $O \\equiv O_1$ and $OM = ON = OC = CH/2$.\n\n![](attached_image_1.png)\n\nLet $O$ be the circumcenter of $\\triangle CMN$. Then $\\angle CMO = \\angle MCO = \\angle ABM$ and $\\angle CNO = \\angle NCO = \\angle BAN$. Hence\n$$\n\\angle ACB = \\angle MCO + \\angle NCO = \\angle ABM + \\angle BAN.\n$$\nTherefore\n$$\n\\begin{aligned}\n2\\angle ANB &= \\angle ANB + \\angle AMB \\\\\n&= 180^\\circ - \\angle ABC - \\angle BAN + 180^\\circ - \\angle BAC - \\angle ABM \\\\\n&= 360^\\circ - (\\angle ABC + \\angle ACB + \\angle BAC) = 180^\\circ.\n\\end{aligned}\n$$\nHence $\\angle ANB = 90^\\circ$ and $AB$ is a diameter of $k$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27278,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA triangle $ABC$ with centroid $G$ and incenter $I$ is given. If $AB = 42$, $GI = 2$ and $AB \\parallel GI$, find $AC$ and $BC$.",
    "options": [],
    "answer": "AC = 48, BC = 36 or AC = 36, BC = 48",
    "solution": "Solution:\nLet $CM$ be the median and $CL$ be the bisector of $\\triangle ABC$ ($I \\in CL$). Using the standard notation for $\\triangle ABC$ we have $\\frac{AL}{BL} = \\frac{AC}{BC} = \\frac{b}{a}$, whence $AL = \\frac{bc}{a+b}$.\n\nSince $AI$ is the bisector of $\\triangle ALC$ through $A$ we get $\\frac{CI}{IL} = \\frac{CA}{AL} = \\frac{a+b}{c}$.\n\nBy the Thales theorem we have $\\frac{CI}{IL} = \\frac{CG}{GM}$, i.e. $a+b = 2c$, whence $a+b = 84$. For $a \\leq b$ we have $LM = \\frac{3}{2} IG = 3$, $AM = 21$, $AL = AM + LM = 24$ and $LB = BM - LM = 18$. Therefore $\\frac{b}{a} = \\frac{24}{18} = \\frac{4}{3}$. Hence $3x + 4x = 84$, i.e. $x = 12$, $AC = 48$ and $BC = 36$. For $b \\geq a$ we\n\n![](attached_image_1.png)\n\nget $BC = 48$ and $AC = 36$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27279,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nTwo tangent circles with centers $O_{1}$ and $O_{2}$ are inscribed in a given angle. Prove that if a third circle with center on the segment $O_{1} O_{2}$ is inscribed in the angle and passes through one of the points $O_{1}$ and $O_{2}$ then it passes through the other one too.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nDenote the circles by $k_{1}(O_{1}, r)$, $k_{2}(O_{2}, R)$ and $k(O, x)$, where $r < x < R$. Let $A$, $B$ and $C$ be the feet of the perpendiculars from $O_{1}$, $O$ and $O_{2}$, respectively, to the arm $p$ of the given angle $O p q$. Let $l$ be the line through $O_{1}$ parallel to $p$ and let $l$ meet $O B$ and $O_{2} C$ at points $M$ and $N$, respectively. Then\n\n![](attached_image_1.png)\n\n$\\triangle O_{1} O M \\sim \\triangle O_{1} O_{2} N$ and therefore\n$$\n\\frac{O O_{1}}{O_{1} O_{2}} = \\frac{O M}{O_{2} N}\n$$\nWe have $O M = O B - B M = O B - O_{1} A = x - r$, $O_{1} O_{2} = r + R$ and $O_{2} N = O_{2} C - C N = O_{2} C - O_{1} A = R - r$.\n\nIf $k$ passes through $O_{1}$, then $O O_{1} = x$ and we get the equation\n$$\n\\frac{x}{R + r} = \\frac{x - r}{R - r}\n$$\nwhence $x = \\frac{r + R}{2}$.\n\nIf $k$ passes through $O_{2}$, then $O O_{1} = R + r - x$ and\n$$\n\\frac{R + r - x}{R + r} = \\frac{x - r}{R - r}\n$$\nwhence we have again $x = \\frac{r + R}{2}$.\n\nIn both cases $O$ is the midpoint of $O_{1} O_{2}$ and $k$ passes through $O_{1}$ and $O_{2}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27280,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFour players $A_{1}$, $A_{2}$, $A_{3}$ and $A_{4}$ have the same amounts of money and play the following game with seven dices: $A_{1}$ throws the seven dices and then pays to each of the other three players $\\frac{1}{k}$ of the money that the corresponding player has at the moment, where $k$ is the sum of the points on the seven dices. Then the same action is performed consecutively by $A_{2}$, $A_{3}$ and $A_{4}$ and the game is over. Find the sums of the points on the dices thrown by each player if after the game their money are in ratio $3: 3: 2: 2$ (the money of $A_{1}$ to the money of $A_{2}$ to the money of $A_{3}$ to the money of $A_{4}$ ).",
    "options": [],
    "answer": "A1: 31, A2: 30, A3: 8, A4: 7",
    "solution": "Solution:\n\nDenote by $S_{k}^{(m)}$ the money of the $k$-th player, $k=1,2,3,4$, after the move and payment of the $m$-th, $m=1,2,3,4$, and $S_{k}^{(0)}=S$ in the beginning, $k=1,2,3,4$. Denote by $a_{i}$ the sum of points on the dices thrown by $A_{i}$. It follows from the game rules that $S_{k}^{(m)}=S_{k}^{(m-1)}+\\frac{1}{a_{m}} S_{k}^{(m-1)}=S_{k}^{(m-1)} \\frac{1+a_{m}}{a_{m}}$ for $k \\neq m$ (i.e. when $A_{k}$ gets money) and\n$$\n\\begin{aligned}\nS_{k}^{(k)} & =S_{k}^{(k-1)}-\\frac{1}{a_{k}} \\sum_{i \\neq k} S_{i}^{(k-1)}=S_{k}^{(k-1)}-\\frac{1}{a_{k}}\\left(4 S-S_{k}^{(k-1)}\\right) \\\\\n& =S_{k}^{(k-1)} \\frac{1+a_{k}}{a_{k}}-\\frac{4 S}{a_{k}}\n\\end{aligned}\n$$\nwhen $A_{k}$ pays.\n\nUsing these formulas we find the money of the four players in the end of the game. We obtain that\n$$\n\\begin{aligned}\n\\frac{6 S}{5} & =S_{1}^{(4)}=P S-\\frac{4 S\\left(1+a_{2}\\right)\\left(1+a_{3}\\right)\\left(1+a_{4}\\right)}{a_{1} a_{2} a_{3} a_{4}} \\\\\n\\frac{6 S}{5} & =S_{2}^{(4)}=P S-\\frac{4 S\\left(1+a_{3}\\right)\\left(1+a_{4}\\right)}{a_{2} a_{3} a_{4}} \\\\\n\\frac{4 S}{5} & =S_{3}^{(4)}=P S-\\frac{4 S\\left(1+a_{4}\\right)}{a_{3} a_{4}} \\\\\n\\frac{4 S}{5} & =S_{4}^{(4)}=P S-\\frac{4 S}{a_{4}}\n\\end{aligned}\n$$\nwhere\n$$\nP=\\frac{\\left(1+a_{1}\\right)\\left(1+a_{2}\\right)\\left(1+a_{3}\\right)\\left(1+a_{4}\\right)}{a_{1} a_{2} a_{3} a_{4}}\n$$\nBy the first two equations we get $a_{2}=a_{1}-1$, and by the last two we have $a_{4}=a_{3}-1$. Now by the second and third equations we obtain\n$$\n\\frac{2}{5}=\\frac{4}{a_{4}}-\\frac{4\\left(1+a_{3}\\right)}{a_{2} a_{4}} \\Longleftrightarrow\\left(a_{2}+10\\right)\\left(10-a_{4}\\right)=120\n$$\nIt follows from the later equation that $a_{4}<10$. We also have $a_{4} \\geq 7$ since the dices are 7 and therefore the minimum sum is 7. It remains to check the possibilities $a_{4}=7,8$ and $9$. The only solution appears for $a_{4}=7$ (the maximum sum is 42) which gives $a_{3}=8$, $a_{2}=30$ and $a_{1}=31$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27281,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $a$ and $b$ be integers and $k$ be a positive integer. Prove that if $x$ and $y$ are consecutive integers such that\n$$\na^{k} x - b^{k} y = a - b\n$$\nthen $|a-b|$ is a perfect $k$-th power.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nAssume that the pair $(x, x+1)$ is a solution of the equation. Then we have\n$$\n\\begin{aligned}\na-b & = a^{k} x - b^{k} (x+1) \n\\Leftrightarrow \\\\\nb^{k} & = (a-b) \\left[ x \\left( a^{k-1} + a^{k-2} b + \\cdots + a b^{k-2} + b^{k-1} \\right) - 1 \\right]\n\\end{aligned}\n$$\nSuppose that the numbers $a-b$ and $x \\left( a^{k-1} + a^{k-2} b + \\cdots + a b^{k-2} + b^{k-1} \\right) - 1$ have a common prime divisor $p$. Then $p$ divides $b^{k}$, i.e. $p$ divides $b$. But $p$ divides also $a-b$ and therefore it divides $a$ as well. Now the fact that $p$ divides $x \\left( a^{k-1} + a^{k-2} b + \\cdots + a b^{k-2} + b^{k-1} \\right) - 1$ implies that $p$ divides $1$, a contradiction.\nTherefore the numbers $a-b$ and $x \\left( a^{k-1} + a^{k-2} b + \\cdots + a b^{k-2} + b^{k-1} \\right) - 1$ are co-prime, which implies that each of them is a $k$-th power (up to a sign). In particular, $|a-b|$ is a $k$-th power.\nThe case of a solution $(x+1, x)$ is considered analogously.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27282,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all four-digit positive integers $m$ less than $2005$ for which there exists a positive integer $n < m$, such that $mn$ is a perfect square and $m - n$ has at most three distinct positive divisors.",
    "options": [],
    "answer": "1156, 1296, 1369, 1377, 1600, 1764, 1900",
    "solution": "Solution:\nThe number $m - n$ has at most three different positive divisors if and only if $m - n = p^k$, where $p$ is a prime and $k \\in \\{0, 1, 2\\}$. If $k = 0$ then $m = n + 1$ and $n(n + 1)$ is a perfect square, which is impossible.\n\nLet $m - n = p^k$, $mn = t^2$, where $k \\in \\{1, 2\\}$ and $t$ is a positive integer. Then\n$$\nn\\left(n + p^k\\right) = t^2 \\Longleftrightarrow \\left(2n + p^k - 2t\\right)\\left(2n + p^k + 2t\\right) = p^{2k}\n$$\nTherefore $2n + p^k - 2t = p^s$ and $2n + p^k + 2t = p^r$, where $r$ and $s$ are integers, such that $0 \\leq s < r \\leq 2k$ and $r + s = 2k$.\n\nFor $k = 1$ we have the unique possibility $2n + p - 2t = 1$, $2n + p + 2t = p^2$. Hence\n$$\nn = \\frac{(p - 1)^2}{4}, \\quad m = n + p = \\frac{(p + 1)^2}{4}\n$$\nTaking into account that $1000 \\leq m < 2005$ we obtain the solutions $m = 1764$, $1600$, $1369$, $1296$ and $1156$ (for $p = 83, 79, 73, 71$ and $67$, respectively).\n\nFor $k = 2$ we have $(r, s) = (4, 0)$ or $(r, s) = (3, 1)$. In the first case we get\n$$\nn = \\frac{\\left(p^2 - 1\\right)^2}{4}, \\quad m = n + p^2 = \\frac{\\left(p^2 + 1\\right)^2}{4}\n$$\nNow the inequalities $1000 \\leq m < 2005$ imply that $p = 8$ which is not a prime. In the second case we have $m = p(p + 1)^2 / 4$, which gives the solutions $m = 1900$ and $1377$ (for $p = 19$ and $17$, respectively).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27283,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe positive integers $M$ and $n$ are such that $M$ is divisible by all positive integers from $1$ to $n$ but it is not divisible by $n+1$, $n+2$ and $n+3$. Find all possible values of $n$.",
    "options": [],
    "answer": "n = 1, 2, 6",
    "solution": "Solution:\n\nWe shall prove that $n+1$, $n+2$ and $n+3$ are prime powers. Assume the contrary and let some of them has the form $ab$, where $a \\geq 2$, $b \\geq 2$ and $(a, b) = 1$. Since $ab$ does not divide $M$, then $a$ or $b$ does not divide $M$. Let $a$ does not divide $M$. Then it follows that $a \\geq n+1$, i.e. $ab - a \\leq 2$. Since $ab - a = a(b-1) \\geq 2$ for $a \\geq 2$, $b \\geq 2$, it follows that $a = 2$, $b = 2$, which contradicts $(a, b) = 1$.\n\nTherefore $n+1$, $n+2$ and $n+3$ are prime powers. At least one of them is even, so it has the form $2^{x}$. Analogously, at least one of them is divisible by $3$, so it has the form $3^{y}$. By parity arguments we conclude that\n$$\n2^{x} = 3^{y} \\pm 1\n$$\n\nCase 1. Let $2^{x} = 3^{y} + 1$. Since $2^{x} \\equiv 1 \\pmod{3}$ for $x$ even and $2^{x} \\equiv 2 \\pmod{3}$ for $x$ odd, we see that $x = 2z$, where $z$ is a nonnegative integer, and $(2^{z} - 1)(2^{z} + 1) = 3^{y}$. Then $2^{z} - 1$ and $2^{z} + 1$ are powers of $3$, which is possible only for $z = 1$. Therefore $3^{y} = 3$, $2^{x} = 4$, whence $n = 1$ or $n = 2$. These solutions are achieved for $M = 1$ and $M = 2$, respectively.\n\nCase 2. Let $2^{x} = 3^{y} - 1$. We may assume that $x \\geq 2$ because for $x = 1$ we obtain one of the above answers for $n$. We have $3^{y} \\equiv 1 \\pmod{4}$ for $y$ even and $3^{y} \\equiv 3 \\pmod{4}$ for $y$ odd. Therefore $y = 2z$ and $2^{x} = (3^{z} - 1)(3^{z} + 1)$. Then $3^{z} - 1$ and $3^{z} + 1$ are powers of $2$ which is possible only for $z = 1$. Therefore $3^{y} = 9$, $2^{x} = 8$, whence $n = 6$ or $n = 7$. The solution $n = 6$ is achieved, for instance, if $M = 60$. For $n = 7$ we obtain $n+3 = 10$, which is not a prime power.\n\nFinally, the solutions are $n = 1$, $n = 2$ and $n = 6$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27284,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all values of the real parameter $p$ such that the equation $|x^{2}-p x-2 p+1|=p-1$ has four real roots $x_{1}, x_{2}, x_{3}$ and $x_{4}$ such that\n$$\nx_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}=20\n$$",
    "options": [],
    "answer": "2",
    "solution": "Solution:\nAnswer: $p=2$. The condition $p>1$ is necessary (but not sufficient!) for existence of four roots. We consider two cases:\n\nCase 1. If $x^{2}-p x-2 p+1=p-1 \\Longleftrightarrow x^{2}-p x-3 p+2=0$ then by the Vieta theorem we obtain $x_{1}^{2}+x_{2}^{2}=p^{2}-2(2-3 p)=p^{2}+6 p-4$.\n\nCase 2. If $x^{2}-p x-2 p+1=1-p \\Longleftrightarrow x^{2}-p x-p=0$ then by the Vieta theorem we obtain $x_{3}^{2}+x_{4}^{2}=p^{2}+2 p$.\n\nNow the condition implies\n$$\nx_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}=20 \\Longleftrightarrow 2 p^{2}+8 p-4=20 \\Longleftrightarrow p^{2}+4 p-12=0\n$$\nwhence $p=2$ or $p=-6$. The second value does not satisfy $p>1$. For $p=2$ we do have four real roots (direct check!).",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27285,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIvo writes consecutively the integers $1, 2, \\ldots, 100$ on 100 cards and gives some of them to Yana. It is known that for every card of Ivo and every card of Yana, the card with the sum of the numbers on the two cards is not in Ivo and the card with the product of these numbers is not in Yana. How many cards does Yana have if the card with number 13 is in Ivo?",
    "options": [],
    "answer": "93",
    "solution": "Solution:\n\nYana has at least one card, say $k \\neq 1$. If Ivo has $1$, then the product $1 \\cdot k = k$ does not belong to Yana, a contradiction. Therefore Yana has $1$.\n\nIf $12$ is in Ivo, then the sum $13 = 1 + 12$ belongs to Yana, a contradiction. Therefore $12$ belongs to Yana. Since the sum $13 = 6 + 7$ is in Ivo, both cards $6$ and $7$ belong to one and the same person. They are not in Ivo since otherwise the sum $1 + 6 = 7$ is in Yana. Using similar arguments we conclude that all cards $1, 2, \\ldots, 12$ belong to Yana. Further, all cards $13k$, $k = 1, \\ldots, 7$, are in Ivo, and all the others belong to Yana. Therefore Yana has $100 - 7 = 93$ cards.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27286,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABCD$ be a cyclic quadrilateral with circumcircle $k$. The rays $\\overrightarrow{DA}$ and $\\overrightarrow{CB}$ meet at point $N$ and the line $NT$ is tangent to $k$, $T \\in k$. The diagonals $AC$ and $BD$ meet at the centroid $P$ of $\\triangle NTD$. Find the ratio $NT : AP$.",
    "options": [],
    "answer": "3",
    "solution": "Solution:\n\nIt follows from the condition that $T$ lies on the $\\operatorname{arc} BC$. Let $M = NT \\cap DP$ be the midpoint of $NT$. Then we have\n$$\nMB \\cdot MD = MT^{2} = MN^{2}\n$$\nThus $MB : MN = MN : MD$ and it follows that $\\triangle NMB \\sim \\triangle DMN$. Hence\n$$\n\\angle MNB = \\angle MDN = \\angle NCA\n$$\n![](attached_image_1.png)\ni.e. $NT \\parallel AC$. Thus\n$$\n\\frac{NT}{AP} = \\frac{2NM}{AP} = \\frac{2MD}{PD} = 3\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27287,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA card game is played by five persons. In a group of 25 persons all like to play that game. Find the maximum possible number of games which can be played if no two players are allowed to play simultaneously more than once.",
    "options": [],
    "answer": "30",
    "solution": "Solution:\nThe number of all pairs of players is $\\frac{25 \\cdot 24}{2} = 300$ and after each game 10 of them become impossible. Therefore at most $300 \\div 10 = 30$ games are possible.\n\nWe shall prove that 30 games are possible. We denote the pairs of players by $(m, n)$, where $1 \\leq m, n \\leq 5$ are integers (in other words, we put them in a table $5 \\times 5$).\n\nIn the game $i$, $1 \\leq i \\leq 5$, we put the five pairs with $m = i$ (i.e. those from the $i$-th row of the table). In the game $6 + 5k + i$, $0 \\leq i \\leq 4$, $0 \\leq k \\leq 4$, we set the pair $(m, n)$ such that $mk + n$ is congruent to $i$ modulo 5. It is clear that for any fixed values of $k, i, m$ there exists a unique $n$ such that $mk + n \\equiv i \\pmod{5}$. Thus we have one pair in each row, i.e. the pairs are five and they have not played in the first 5 games.\n\nFor every two pairs $(m, n)$ and $(m', n')$, $m' \\neq m$, the numbers $k(m - m')$, $k = 0, 1, 2, 3, 4$, give different remainders modulo 5. Hence there exists a unique $k$ such that $k(m - m') \\equiv n - n' \\pmod{5}$. Equivalently, $km - n$ and $km' - n'$ have the same remainder $i$ modulo 5 and the numbers $k$ and $i$ determine the unique game in which the pairs $(m, n)$ and $(m', n')$ participate.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 27288,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nConsider the inequality $\\sqrt{x} + \\sqrt{2 - x} \\geq \\sqrt{a}$, where $a$ is a real number.\n\na) Solve the inequality for $a = 3$.\n\nб) Find all $a$, for which the set of solutions of the inequality is a segment (possibly, a point) of length less than or equal to $\\sqrt{3}$.",
    "options": [],
    "answer": "a) x ∈ [1 − sqrt(3)/2, 1 + sqrt(3)/2].\nb) a ∈ [3, 4].",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27289,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABCD$ be a parallelogram. The points $E$ and $F$ on the sides $AB$ and $BC$, respectively, are such that $DE$ is the bisector of $\\angle ADF$ and $AE + CF = DF$. The line through $C$ and perpendicular to $DE$ meets the side $AD$ at $L$ and the diagonal $BD$ at $H$. Set $N = DE \\cap AC$. Prove that:\n\na) $AE = DL$;\n\nb) $BC = CD$ if $HN \\parallel AD$;\n\nc) $ABCD$ is a square if $HN \\parallel AD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27290,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all positive integers $t, x, y, z$ such that\n$$\n2^{t} = 3^{x} 5^{y} + 7^{z}\n$$",
    "options": [],
    "answer": "t=6, x=1, y=1, z=2",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27291,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThere are $40$ knights in a kingdom. Every morning they fight in pairs (everyone has exactly one enemy to fight with) and every evening they sit around a table (during the evening they do not change their seats). Find the least number of days such that:\n\na) the fights can be arranged in a way that every two knights have fought at least once;\n\nb) round the table arrangements can be done in a way that every two knights have been neighbors around the table.",
    "options": [],
    "answer": "a) 39; b) 20",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 27292,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSolve the equation\n$$\n\\log_{a}\\left(a^{2\\left(x^{2}+x\\right)}+a^{2}\\right)=x^{2}+x+\\log_{a}\\left(a^{2}+1\\right)\n$$\nwhere $a$ is a real number.",
    "options": [],
    "answer": "For all real a > 0 with a ≠ 1, the solutions are x ∈ {−2, −1, 0, 1}.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27293,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGiven a triangle $ABC$ with $\\angle ACB = 60^\\circ$, define the sequence of points $A_0, A_1, \\ldots, A_{2006}$ in the following way: $A_0 = A$, $A_1$ is the orthogonal projection of $A_0$ on $BC$, $A_2$ is the orthogonal projection of $A_1$ on $AC$, $\\ldots$, $A_{2005}$ is the orthogonal projection of $A_{2004}$ on $BC$ and $A_{2006}$ is the orthogonal projection of $A_{2005}$ on $AC$. The sequence of points $B_0, B_1, \\ldots, B_{2006}$ is defined in a similar way: $B_0 = B$, $B_1$ is the orthogonal projection of $B_0$ on $AC$, $B_2$ is the orthogonal projection of $B_1$ on $BC$ and so on. Prove that the line $A_{2006}B_{2006}$ is tangent to the incircle of $\\triangle ABC$ if and only if\n$$\n\\frac{AC + BC}{AB} = \\frac{2^{2006} + 1}{2^{2006} - 1}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27294,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $a$ be an integer. Find all real numbers $x, y, z$ such that\n$$\na(\\cos 2 x + \\cos 2 y + \\cos 2 z) + 2(1-a)(\\cos x + \\cos y + \\cos z) + 6 = 9a\n$$",
    "options": [],
    "answer": "Let u = cos x, v = cos y, w = cos z.\n- If a = 1: each of u, v, w equals ±1, i.e., x, y, z are each 0 or π modulo 2π (independently).\n- If a ≠ 1: u = v = w = −1, i.e., x ≡ y ≡ z ≡ π (mod 2π).",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27295,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nConsider the function\n$$\nf(x) = \\frac{x^{2} - 2006x + 1}{x^{2} + 1}\n$$\n\na. Solve the inequality $f'(x) \\geq 0$.\n\nb. Prove that $|f(x) - f(y)| \\leq 2006$ for all real numbers $x$ and $y$.",
    "options": [],
    "answer": "f'(x) ≥ 0 for x ≤ −1 or x ≥ 1; moreover, |f(x) − f(y)| ≤ 2006 for all real x, y.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27296,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $k$ be a circle with center $O$ and radius $\\sqrt{5}$ and let $M$ and $N$ be points on a diameter of $k$ such that $MO = NO$. The chords $AB$ and $AC$, passing through $M$ and $N$, respectively, are such that\n$$\n\\frac{1}{MB^{2}} + \\frac{1}{NC^{2}} = \\frac{3}{MN^{2}}\n$$\nFind the length of $MO$.",
    "options": [],
    "answer": "1",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27297,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFind the maximal cardinality of a set of phone numbers satisfying the following three conditions:\n\na) all of them are five-digit numbers (the first digit can be $0$);\n\nb) each phone number contains at most two different digits;\n\nc) the deletion of an arbitrary digit in two arbitrary phone numbers (possibly in different positions) does not lead to identical sequences of digits of length $4$.",
    "options": [],
    "answer": "212",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27298,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $O$ be the circumcenter of a triangle $ABC$ with $AC = BC$. The line $AO$ meets the side $BC$ at $D$. If the lengths of $BD$ and $CD$ are integers, and $AO - CD$ is a prime number, find these three numbers.",
    "options": [],
    "answer": "BD = 8, CD = 4, AO − CD = 2",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27299,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all pairs $(a, b)$ of non-negative real numbers such that the equations $x^{2} + a^{2} x + b^{3} = 0$ and $x^{2} + b^{2} x + a^{3} = 0$ have a common real root.",
    "options": [],
    "answer": "All pairs with a = b and either a = b = 0 or a = b ≥ 4.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27300,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $b$ and $c$ be real numbers such that the equation $x^{2} + b x + c = 0$ has two distinct real roots $x_{1}$ and $x_{2}$ with $x_{1} = x_{2}^{2} + x_{2}$.\n\na) Find $b$ and $c$ if $b + c = 4$.\n\nб) Find $b$ and $c$ if they are coprime integers.",
    "options": [],
    "answer": "a) b = -8, c = 12. b) (b, c) = (-3, 2) or (1, 0).",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27301,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGiven a triangle $ABC$, let $BL$, $L \\in AC$, be the bisector of $\\angle ABC$ and $AH$, $H \\in BC$, the altitude to $BC$. Prove that $\\angle AHL = \\angle ALB$ if and only if $\\angle BAC = \\angle ACB + 90^\\circ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27302,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTokens are placed in some of the cells of a table of size $8 \\times 8$ such that:\n\n(1) there is at least one token in any rectangle of size $2 \\times 1$ and $1 \\times 2$;\n\n(2) there are two neighboring tokens in any rectangle of size $7 \\times 1$ and $1 \\times 7$.\n\nFind the minimum possible number of tokens.",
    "options": [],
    "answer": "42",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27303,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAre there exist two triangles whose angles (in some order) form an arithmetic progression with nonzero common difference.",
    "options": [],
    "answer": "No",
    "solution": "Solution:\n\n1. We shall prove that if the angles of two triangles form an arithmetic progression with difference $d$, then $d=0$.\nLet $\\alpha$ be the least of all six angles and let the progression be $\\alpha, \\alpha+d, \\ldots, \\alpha+5d$. Thus,\n$$\n3\\alpha + (k_{1} + k_{2})d = 3\\alpha + (k_{3} + k_{4} + k_{5})d\n$$\nwhere $\\{k_{1}, k_{2}, k_{3}, k_{4}, k_{5}\\} = \\{1,2,3,4,5\\}$. Therefore if $d \\neq 0$, then\n$$\nk_{1} + k_{2} = k_{3} + k_{4} + k_{5}\n$$\nwhich is impossible since $k_{1} + k_{2} + k_{3} + k_{4} + k_{5} = 15$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27304,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn the cells of a square table the numbers $1$, $0$ or $-1$ are written in such a way that there is exactly one $1$ and exactly one $-1$ in every row and in every column. Is it always possible to obtain the opposite table by rearranging the rows and the columns of the initial table? (Two tables are called opposite if all the sums of the numbers in the corresponding cells equal $0$.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWe shall prove that one can obtain the opposite table by rearranging the rows and columns of the initial table. Denote the columns from left to right and the rows from up to down by $1,2, \\ldots, n$. Denote by $a_{ij}$ the number written in the $i$-th row and $j$-th column.\n\nExchanging rows and columns one obtains: $a_{11}=1$, $a_{12}=-1$, $a_{22}=1$, $a_{23}=-1$ (when $a_{21}=-1$ the assertion follows by induction using $2 \\times 2$ and $(n-1) \\times (n-2)$ tables), $a_{33}=1$, $a_{34}=-1$ (if $a_{31}=-1$, the assertion follows by induction using $3 \\times 3$ and $(n-3) \\times (n-3)$ tables) and so on.\n\nIt remains to prove the assertion for the table\n\n| 1 | -1 | 0 | 0 | $\\ldots$ | 0 | 0 |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| 0 | 1 | -1 | 0 | $\\ldots$ | 0 | 0 |\n| 0 | 0 | 1 | -1 | $\\ldots$ | 0 | 0 |\n| $\\ldots$ | $\\ldots$ | $\\ldots$ | $\\ldots$ | $\\ldots$ | $\\ldots$ | $\\ldots$ |\n| $\\ldots$ | $\\ldots$ | $\\ldots$ | $\\ldots$ | $\\ldots$ | $\\ldots$ | $\\ldots$ |\n| $\\ldots$ | $\\ldots$ | $\\ldots$ | $\\ldots$ | $\\ldots$ | 1 | -1 |\n| -1 | 0 | 0 | 0 | $\\ldots$ | 0 | 1 |\n\nProceeding in the same way one obtains from the initial table the following one\n\n$(B)$\n\n| -1 | 1 | 0 | 0 | $\\ldots$ | 0 | 0 |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| 0 | -1 | 1 | 0 | $\\ldots$ | 0 | 0 |\n| 0 | 0 | -1 | 1 | $\\ldots$ | 0 | 0 |\n| $\\ldots$ | $\\ldots$ | $\\ldots$ | $\\ldots$ | $\\ldots$ | $\\ldots$ | $\\ldots$ |\n| $\\ldots$ | $\\ldots$ | $\\ldots$ | $\\ldots$ | $\\ldots$ | $\\ldots$ | $\\ldots$ |\n| $\\ldots$ | $\\ldots$ | $\\ldots$ | $\\ldots$ | $\\ldots$ | -1 | 1 |\n| 1 | 0 | 0 | 0 | $\\ldots$ | 0 | -1 |\n\nNow applying the same moves for obtaining the table $B$ from the table $A$ but in reverse order one obtains the opposite of the initial table.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27305,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nConsider the set $A = \\{1, 2, 3, 4, \\ldots, 2^{n}\\}$, $n \\geq 2$. Find the number of the subsets $B$ of $A$, such that if the sum of two elements of $A$ is a power of $2$ then exactly one of them belongs to $B$.",
    "options": [],
    "answer": "2^{n+1}",
    "solution": "Solution:\n\n1. Let $B$ be a subset of $A$ having the given property. Since $1 + 3 = 2^{2}$, we have that exactly one of the numbers $1$ or $3$ belongs to $B$.\nIf $1 \\in B$ then $3 \\notin B$. We prove by induction that for any integer $t$, $0 \\leq t < 2^{n-2}$, the integers of the form $4t+1$ belong to $B$ and the integers of the form $4t+3$ do not belong to $B$. The statement is true for $t=0$ and suppose it is true for $t \\leq s$. Since $4(s+1)+1$ is an odd number there exists $l$ such that $2^{l} < 4(s+1)+1 < 2^{l+1}$. Therefore $2(4s+5) > 2 \\cdot 2^{l} = 2^{l+1}$ giving $0 < 2^{l+1} - (4s+5) < 4s+5$. Set $x = 4s+5$ and $y = 2^{l+1} - (4s+5)$. Then $x + y = 2^{l+1}$ and since $y$ is of the form $4m+3$ we conclude that $y \\notin B$ and therefore $4(s+1)+1 \\in B$. Analogously $4(s+1)+3 \\notin B$.\nIf $1 \\notin B$ then $3 \\in B$ and we prove as above that the integers of the form $4t+1$ belong to $B$ and the integers of the form $4t+3$ do not.\nTherefore the odd numbers in $B$ are either all integers of the form $4t+1$ or all integers of the form $4t+3$.\nLet $x = 2^{p} x_{0}$ and $y = 2^{q} y_{0}$, where $x_{0}$ and $y_{0}$ are odd and $p$ and $q$ are positive integers. If $2^{p} x_{0} + 2^{q} y_{0} = 2^{k}$ and $p \\neq q$, say $p < q$, then $x_{0} + 2^{q-p} y_{0} = 2^{k-p}$, which is impossible. Therefore $p = q$ and it follows that the sum of the elements from distinct sets $A_{i} = \\{2^{i} a : a$ is an odd integer $\\}, i = 1, 2, \\ldots, n$ is not a power of $2$. For any $A_{i}$, after dividing by $2^{i}$ and applying the above arguments, we obtain that either all integers of the form $2^{i}(4t+1)$ are in $B$ or all integers of the form $2^{i}(4t+3)$ are in $B$.\nTherefore there exist $2^{n+1}$ sets $B$ with the given property.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27306,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all sequences of positive integers $\\{a_{n}\\}_{n=1}^{\\infty}$, such that $a_{4}=4$ and the identity\n$$\n\\frac{1}{a_{1} a_{2} a_{3}}+\\frac{1}{a_{2} a_{3} a_{4}}+\\cdots+\\frac{1}{a_{n} a_{n+1} a_{n+2}}=\\frac{(n+3) a_{n}}{4 a_{n+1} a_{n+2}}\n$$\nholds true for every positive integer $n \\geq 2$.",
    "options": [],
    "answer": "a_n = n for all n",
    "solution": "Solution:\nWe rewrite the recurrence relation as\n$$\n\\frac{(n+2) a_{n-1}}{4 a_{n} a_{n+1}}+\\frac{1}{a_{n} a_{n+1} a_{n+2}}=\\frac{(n+3) a_{n}}{4 a_{n+1} a_{n+2}} \\Longleftrightarrow (n+2) a_{n+2}=\\frac{(n+3) a_{n}^{2}-4}{a_{n-1}}\n$$\nfor $n \\geq 3$. Setting $n=2$ in the initial relation we obtain $4\\left(a_{1}+4\\right)=5 a_{1} a_{2}^{2}$, implying that $a_{1} \\mid 16$ and $5 \\mid a_{1}+4$. Therefore $a_{1}=16$ and $a_{2}=1$ or $a_{1}=1$ and $a_{2}=2$.\n\nCase 1. Let $a_{1}=16$ and $a_{2}=1$. Then $5 a_{5}=6 a_{3}^{2}-4$, $a_{3} a_{6}=18$ and $7 a_{7}=2 a_{5}^{2}-1$ for $n=3,4$ and $5$, respectively. Since $a_{3} \\equiv \\pm 2\\pmod{5}$ and $a_{3}$ as a divisor of $18$ we conclude that $a_{3}=3$ or $a_{3}=18$. The direct check of both values shows no solutions in this case.\n\nCase 2. Let $a_{1}=1$ and $a_{2}=2$. Then $n=3$ and $n=4$ give $5 a_{5}+2=3 a_{3}^{2}$ and $a_{3} a_{6}=18$, respectively. Again $a_{3} \\equiv \\pm 2\\pmod{5}$ and we see that $a_{3}=3$ and $a_{6}=6$ or $a_{3}=18$ and $a_{6}=1$. In the second case we obtain $a_{5}=194$ which gives a contradiction with $8 a_{8}=\\frac{9 a_{6}^{2}-4}{a_{5}}$.\n\nIn the first case $a_{5}=5$ and hence the only possible values are $a_{i}=i$ for $i=1,2, \\ldots, 6$. Now easy induction shows that $a_{n}=n$ for all $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27307,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nConsider the equations\n$$\n3^{2x+3} - 2^{x+2} = 2^{x+5} - 9^{x+1}\n$$\nand\n$$\na \\cdot 5^{2x} + |a-1| 5^{x} = 1\n$$\nwhere $a$ is a real number.\n\na) Solve the equation (1).\n\nb) Find the values of $a$ such that the equations (1) and (2) are equivalent.",
    "options": [],
    "answer": "a) x = 0; b) a ∈ [0, 1] ∪ {−1}",
    "solution": "Solution:\n\na) The equation (1) can be written as $9^{x} = 2^{x}$, i.e. $\\left(\\frac{9}{2}\\right)^{x} = 1$ and $x = 0$.\n\nb) We plug the only solution $x = 0$ of (1) in (2) and obtain $|a-1| = 1-a \\Longleftrightarrow a \\leq 1$. In this case $0$ is a solution of (2) and we have to decide when (2) has no other solution(s).\n\nSet $5^{x} = t$, $t > 0$. Then we have to find all $a \\leq 1$ such that the equation\n$$\na t^{2} + (1-a) t - 1 = 0\n$$\nhas not positive roots different from $1$. For $a = 0$ the only root of (3) is $t = 1$ and the condition is satisfied. For $a \\neq 0$ the roots of (3) are $1$ and $-\\frac{1}{a}$. Therefore the condition is satisfied if and only if $-\\frac{1}{a} = 1$ or $-\\frac{1}{a} < 0$. Hence $a = -1$ or $a > 0$ and we conclude that $a \\in [0, 1] \\cup \\{-1\\}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27308,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA circle $k$ is tangent to the arms of an acute angle $A O B$ at points $A$ and $B$. Let $A D$ be the diameter of $k$ through $A$ and $B P \\perp A D$, $P \\in A D$. The line $O D$ meets $B P$ at point $M$. Find the ratio $\\frac{B M}{B P}$.",
    "options": [],
    "answer": "1/2",
    "solution": "Solution:\nLet the tangent line to $k$ at $D$ meet the ray $\\overrightarrow{O B}$ at point $S$. Then the lines $S D$, $B P$ and $O A$ are parallel and therefore $\\triangle O B M \\sim \\triangle O S D$ and $\\triangle D P M \\sim \\triangle D A O$. It follows that $\\frac{B M}{S D} = \\frac{O B}{O S}$ and $\\frac{M P}{O A} = \\frac{D P}{D A}$, i.e. $B M = \\frac{O B \\cdot S D}{O S}$ and $M P = \\frac{O A \\cdot D P}{D A}$.\n\n![](attached_image_1.png)\n\nSince $O A = O B$ and $\\frac{B S}{O S} = \\frac{D P}{D A}$ (we use the equality $S D = S B$), we obtain $B M = M P$, implying that the desired ratio equals $1 : 2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27309,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $A A'$, $B B'$ and $C C'$ be the angular bisectors of a triangle $A B C$ with incenter $I$. The segments $C I$ and $A' B'$ meet at $D$ and the midpoints of the segments $A I$ and $B I$ are denoted by $M$ and $N$, respectively.\n\na) If $a = B C$, $b = A C$ and $c = A B$, find the ratio $C D : D I$.\n\nb) If $K = A C \\cap \\overrightarrow{C' M}$ and $L = B C \\cap \\overrightarrow{C' N}$, prove that $D$ is the incenter of $\\triangle K L C$.",
    "options": [],
    "answer": "CD:DI = (a + b + c) / c; D is the incenter of triangle KLC",
    "solution": "Solution:\n\na)\nWe have $C A' = \\frac{a b}{b + c}$ and $\\frac{A I}{I A'} = \\frac{A C}{C A'} = \\frac{b + c}{a}$. By the Menelaus theorem for $\\triangle A I C$ and the line $B' A'$ we have\n$$\n\\frac{C D}{D I} \\cdot \\frac{I A'}{A' A} \\cdot \\frac{A B'}{B' C} = 1 \\Rightarrow \\frac{C D}{D I} = \\frac{a + b + c}{a} \\cdot \\frac{a}{c} = \\frac{a + b + c}{c}\n$$\n\nb)\nBy the Menelaus theorem for $\\triangle A I C$ and the line $K C'$ we get\n$$\n\\frac{A M}{M I} \\cdot \\frac{I C'}{C' C} \\cdot \\frac{C K}{K A} = 1\n$$\nwhence $\\frac{C K}{K A} = \\frac{a + b + c}{c}$. Analogously, $\\frac{C L}{L B} = \\frac{a + b + c}{c}$.\n\n![](attached_image_1.png)\n\nLet $h$ be the homothety with center $C$ and ratio $\\frac{a + b + c}{a + b + 2c}$. Then $h(A) = K$, $h(B) = L$, $h(C) = C$ and it follows from a) that $h(I) = D$. Therefore $D$ is the incenter of $\\triangle K L C$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27310,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the maximum of the function\n$$\nf(x) = \\frac{\\lg x \\cdot \\lg x^{2} + \\lg x^{3} + 3}{\\lg^{2} x + \\lg x^{2} + 2}\n$$\nand the values of $x$, when it is attained.",
    "options": [],
    "answer": "Maximum value 2.5, attained only at x = 0.01",
    "solution": "Solution:\nThe domain of $f(x)$ is $x > 0$. Setting $y = \\lg x$ gives\n$$\nF(y) = \\frac{2y^{2} + 3y + 3}{y^{2} + 2y + 2}\n$$\nSince the denominator is positive, the function $F(y)$ is defined for all real $y$.\nLet $M$ be the desired value of $f(x)$ (if it exists). Then for any real $y$ we have\n$$\n\\begin{gathered}\n\\frac{2y^{2} + 3y + 3}{y^{2} + 2y + 2} \\leq M \\\\\n2y^{2} + 3y + 3 \\leq M y^{2} + 2M y + 2M \\\\\n(2 - M) y^{2} + (3 - 2M) y + (3 - 2M) \\leq 0\n\\end{gathered}\n$$\nTherefore $2 - M < 0$ and $D = (3 - 2M)(3 - 2M - 8 + 4M) = (3 - 2M)(2M - 5) \\leq 0$. Hence $M \\geq 2.5$.\nNote that for $M = 2.5$ the above inequality becomes $-0.5 y^{2} - 2y - 2 \\leq 0$, i.e. $y^{2} + 4y + 4 = (y + 2)^{2} \\geq 0$ and the equality is attained only if $y = -2$, i.e. for $x = 0.01$.\nTherefore the maximum of the function equals $2.5$ and it is attained for $x = 0.01$ only.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27311,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $\\mathbb{Q}^{+}$ be the set of positive rational numbers. Find all functions $f: \\mathbb{Q}^{+} \\rightarrow \\mathbb{R}$ such that $f(1)=1$, $f(1 / x)=f(x)$ for any $x \\in \\mathbb{Q}^{+}$ and $x f(x)=(x+1) f(x-1)$ for any $x \\in \\mathbb{Q}^{+}, x>1$.",
    "options": [],
    "answer": "For x = p/q in lowest terms, f(x) = (p + q) / 2.",
    "solution": "Solution:\nLet $x=\\frac{p}{q}$, where $p$ and $q$ are coprime positive integers. We shall prove by induction on $n=p+q \\geq 2$ that $f(x)$ is uniquely determined. This is true for $n=2$ (since $f(1)=1$).\n\nSuppose that it is true for all integers less than a given $n \\geq 3$ and consider $x=\\frac{p}{q}$, where $p+q=n$. It follows from the first condition that we may assume that $p>q$.\n\nNow the second condition shows that $f\\left(\\frac{p}{q}\\right)$ is uniquely determined by $f\\left(\\frac{p-q}{q}\\right)$ and since $p-q+q<n$ it is uniquely determined by the induction hypothesis.\n\nNote that the function $f\\left(\\frac{p}{q}\\right)=\\frac{p+q}{2}$ for $p$ and $q$ relatively prime fulfills the conditions of the problem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27312,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nForty thieves are to distribute 4000 euro amongst them. A group of five thieves is called poor if they have no more than 500 euro all together. What is the minimum number of poor groups amongst all possible groups of five thieves?",
    "options": [],
    "answer": "82251",
    "solution": "Solution:\n\nIf 39 thieves take 101 euro each and the last one takes 61 euro, then the only poor groups are those having as a member the last thief. So this distribution of the money gives $\\binom{39}{4}$ poor groups. We shall prove that this is the required minimum.\n\nLet $r$ be the number of all possible divisions of the thieves in 8 groups of five thieves each. Any such division has at least one poor group. The total number of the groups in all divisions is $8r$. Each group takes part in $8r / \\binom{40}{5}$ divisions. Therefore each poor group is counted exactly $8r / \\binom{40}{5}$ times and this gives at least $r$. Therefore we have at least $r \\binom{40}{5} / 8r = \\binom{39}{4}$ poor groups.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 27313,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe price of a merchandise dropped from March to April by $x\\%$, and went up from April to May by $y\\%$. It turned out that in the period from March to May the price dropped by $(y-x)\\%$. Find $x$ and $y$ if they are positive integers (the price is positive for the whole period).",
    "options": [],
    "answer": "x=75, y=120; x=40, y=50",
    "solution": "Solution:\n\nIt follows from the condition of the problem that\n$$\n\\frac{100-x}{100} \\cdot \\frac{100+y}{100} = \\frac{100-y+x}{100}\n$$\nHence $(100-x)(100+y) = 100(100-y+x)$ which can be written as $(200-x)(200+y) = 200^{2}$. Further, $100 < 200-x < 200$, i.e. $200-x$ equals $125$ or $160$.\n\nTherefore the solutions are $x_{1} = 75$, $y_{1} = 120$ and $x_{2} = 40$, $y_{2} = 50$, both of which fulfill the condition.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27314,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $ABCD$ be a parallelogram such that $\\Varangle BAD < 90^\\circ$ and let $DE$, $E \\in AB$, and $DF$, $F \\in BC$, be the altitudes of the parallelogram. Prove that\n$$\n4(AB \\cdot BC \\cdot EF + BD \\cdot AE \\cdot FC) \\leq 5 \\cdot AB \\cdot BC \\cdot BD\n$$\nFind $\\Varangle BAD$ if the equality occurs.",
    "options": [],
    "answer": "30°",
    "solution": "Solution:\nSet $\\Varangle BAD = \\alpha$, $AB = CD = a$, $AD = BC = b$ and $BD = d$. We have $DE = b \\sin \\alpha$, $AE = b \\cos \\alpha$, $DF = a \\sin \\alpha$ and $CF = a \\cos \\alpha$. Therefore $\\triangle DEF \\sim \\triangle ADB$ and thus $EF = d \\sin \\alpha$.\n\nPlugging the above expressions in the given\n\n![](attached_image_1.png)\n\ninequality we see that it is equivalent to\n$$\n4 \\sin \\alpha + 4 - 4 \\sin^2 \\alpha \\leq 5\n$$\ni.e. $(2 \\sin \\alpha - 1)^2 \\geq 0$, which is true for any value of $\\alpha$. The equality holds when $\\alpha = 30^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27315,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDenote by $d(a, b)$ the number of the divisors of a positive integer $a$, which are greater than or equal to $b$. Find all positive integers $n$ such that\n$$\nd(3 n+1,1)+d(3 n+2,2)+\\cdots+d(4 n, n)=2006\n$$",
    "options": [],
    "answer": "708",
    "solution": "Solution:\nDenote by $D(a, b)$ the set of the divisors of $a$, which are greater than or equal to $b$. Thus, $|D(a, b)|=d(a, b)$. Every integer $k$, $1 \\leq k \\leq 4$, belongs to at most one of the sets\n$$\nD(3 n+1,1), D(3 n+2,2), \\ldots, D(4 n, n)\n$$\nEvery integer $k$, $1 \\leq k \\leq n$, $3 n+1 \\leq k \\leq 4 n$ belongs to exactly one of the sets (1). The integers $k$, $2 n+1 \\leq k \\leq 3 n$ do not appear in the sets (1).\nLet $n+1 \\leq k \\leq 2 n$, i.e. $k=n+i$, $i=1, \\ldots, n$. If $k$ belongs to one of the sets (1), then\n$$\n3 n+1 \\leq 2(n+i) \\leq 4 n\n$$\nor\n$$\n3 n+1 \\leq 3(n+i) \\leq 4 n\n$$\nWe conclude that $i=\\left\\lceil\\frac{n+1}{2}\\right\\rceil, \\ldots, n$ or $i=1, \\ldots,\\left\\lfloor\\frac{n}{3}\\right\\rfloor$. The number of the integers from the interval $[n+1,2 n]$ that belong to exactly one of the sets (1) equals $\\left\\lceil\\frac{n}{2}\\right\\rceil+\\left\\lfloor\\frac{n}{3}\\right\\rfloor$. Thus\n$$\n|D(3 n+1,1)|+|D(3 n+2,2)|+\\cdots+|D(4 n, n)|=2 n+\\left\\lceil\\frac{n}{2}\\right\\rceil+\\left\\lfloor\\frac{n}{3}\\right\\rfloor\n$$\nand therefore $n=708$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27316,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $k$ be a circle with diameter $AB$ and let $C \\in k$ be an arbitrary point. The excircles of $\\triangle ABC$ tangent to the sides $AC$ and $BC$ are tangent to the line $AB$ at points $M$ and $N$, respectively. Denote by $O_{1}$ and $O_{2}$ the circumcenters of $\\triangle AMC$ and $\\triangle BNC$. Prove that the area of $\\triangle O_{1}CO_{2}$ does not depend on $C$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nSet $AB = c$, $BC = a$, $AC = b$, $p = \\frac{a + b + c}{2}$ and let $O$, $P$ and $Q$ be the midpoints of $AB$, $AM$ and $BN$, respectively.\nSince $OO_{1} \\perp AC$ and $O_{1}P \\perp AM$, we have $\\triangle O_{1}OP \\sim \\triangle ABC$, implying that $\\frac{OO_{1}}{c} = \\frac{OP}{a}$. Further, it follows from $AM = p - c$ that\n$$\nOP = OA + \\frac{1}{2} AM = \\frac{1}{2}c + \\frac{1}{2}(p - c) = \\frac{a + b + c}{4}\n$$\nand therefore $OO_{1} = \\frac{c(a + b + c)}{4a}$. Analogously $OO_{2} = \\frac{c(a + b + c)}{4b}$.\nSince $\\angle ACB = 90^{\\circ}$ we have that $\\angle O_{1}OO_{2} = 90^{\\circ}$.\nFinally, using that $S_{O_{1}O_{2}C} = \\left| S_{OO_{1}O_{2}} - S_{OO_{1}C} - S_{OO_{2}C} \\right|$ we compute\n$$\n\\begin{aligned}\nS_{OO_{1}O_{2}} - S_{OO_{1}C} - S_{OO_{2}C} & = \\frac{OO_{1} \\cdot OO_{2}}{2} - \\frac{OO_{1} \\cdot b}{4} - \\frac{OO_{2} \\cdot a}{4} \\\\\n& = \\frac{c^{2}(a + b + c)^{2}}{32ab} - \\frac{c(a + b + c)}{16} \\left( \\frac{b}{a} + \\frac{a}{b} \\right) \\\\\n& = \\frac{c^{2}(a + b + c)^{2}}{32ab} - \\frac{c^{3}(a + b + c)}{16ab} = \\frac{c^{2}}{16},\n\\end{aligned}\n$$\nsince $c^{2} = a^{2} + b^{2}$. This implies that $S_{O_{1}CO_{2}}$ does not depend on the choice of the point $C$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27317,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $a_{1}, a_{2}, \\ldots, a_{n}, \\ldots$ be a geometric progression with $a_{1}=3-2 a$ and ratio $q=\\frac{3-2 a}{a-2}$, where $a \\neq \\frac{3}{2}, 2$ is a real number. Set $S_{n}=\\sum_{i=1}^{n} a_{i}$, $n \\geq 1$. Prove that if the sequence $\\left\\{S_{n}\\right\\}_{n=1}^{\\infty}$ is convergent and its limit is $S$, then $S<1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nSince $S_{n}=a_{1} \\cdot \\frac{1-q^{n}}{1-q}$, the sequence $\\left\\{S_{n}\\right\\}_{n=1}^{\\infty}$ converges if and only if $|q|<1$. Therefore $-1<\\frac{3-2 a}{a-2}<1$, whence $a \\in\\left(1, \\frac{5}{3}\\right) \\backslash\\left\\{\\frac{3}{2}\\right\\}$. In this case\n$$\nS=\\lim _{n \\rightarrow \\infty} S_{n}=a_{1} \\cdot \\frac{1}{1-q}=\\frac{3-2 a}{1-\\frac{3-2 a}{a-2}}=\\frac{(3-2 a)(a-2)}{3 a-5}\n$$\nand we have to prove that for every $a \\in\\left(1, \\frac{5}{3}\\right) \\backslash\\left\\{\\frac{3}{2}\\right\\}$ the inequality\n$$\n\\frac{(3-2 a)(a-2)}{3 a-5}<1\n$$\nholds. This inequality is equivalent to $\\frac{2 a^{2}-4 a+1}{3 a-5}>0$ and since $3 a-5<0$, we have to prove that $f(a)=2 a^{2}-4 a+1<0$. This follows from $f(1)=-1$ and $f\\left(\\frac{5}{3}\\right)=-\\frac{1}{9}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27318,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSolve the system\n$$\n\\left\\lvert\\,\n\\begin{aligned}\n& \\left(4^{\\sqrt{x^{2}+x}}+7 \\cdot 2^{\\sqrt{x^{2}+x}}-1\\right) \\sin (\\pi y)=7|\\sin (\\pi y)| \\\\\n& x^{2}+4 x+y^{2}=0\n\\end{aligned}\n\\right.\n$$",
    "options": [],
    "answer": "(-1, -√3), (0, 0), (-4, 0), (-2 - √3, 1), (-2 - √3, -1), (-2, 2), (-2, -2)",
    "solution": "Solution:\nThe system is defined for $x \\in (-\\infty, -1] \\cup [0, +\\infty)$ and every $y$. We consider three cases.\n\nCase 1. Let $\\sin (\\pi y) > 0$. The first equation gives\n$$\n4^{\\sqrt{x^{2}+x}} + 7 \\cdot 2^{\\sqrt{x^{2}+x}} - 1 = 7\n$$\nand setting $t = 2^{\\sqrt{x^{2}+x}} > 0$ we obtain the equation $t^{2} + 7t - 8 = 0$ with roots $t_{1} = 1$ and $t_{2} = -8$. Since $t > 0$, it follows that $2^{\\sqrt{x^{2}+x}} = 1$, whence $x^{2} + x = 0$, i.e. $x = 0$ and $x = -1$. For $x = 0$ we get $y = 0$ and for $x = -1$ we have $y = \\pm \\sqrt{3}$. Since $\\sin 0 = 0$, $\\sin (\\sqrt{3} \\pi) < 0$ and $\\sin (-\\sqrt{3} \\pi) > 0$, the only solution of the system in this case is $x = -1, y = -\\sqrt{3}$.\n\nCase 2. Let $\\sin (\\pi y) = 0$, i.e. $y$ is an integer. From the second equation we obtain $4 - y^{2} \\geq 0$, whence $y = 0, \\pm 1, \\pm 2$. For $y = 0$ we get $x = 0, -4$; for $y = \\pm 1$ we find $x = -2 \\pm \\sqrt{3}$ and for $y = \\pm 2$ we have $x = -2$. Since $-1 < -2 + \\sqrt{3} < 0$, the solutions of the system in this case are $x = y = 0$, $x = -4, y = 0$, $x = -2 - \\sqrt{3}, y = \\pm 1$ and $x = -2, y = \\pm 2$.\n\nCase 3. Let $\\sin (\\pi y) < 0$. Then the first equation gives\n$$\n4^{\\sqrt{x^{2}+x}} + 7 \\cdot 2^{\\sqrt{x^{2}+x}} - 1 = -7\n$$\nand setting $t = 2^{\\sqrt{x^{2}+x}} > 0$ we obtain the equation $t^{2} + 7t + 6 = 0$ with roots $t_{1} = -1$ and $t_{2} = -6$. Since $t > 0$, the system has no solutions in this case.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27319,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nConsider the excircles of a triangle $ABC$ tangent to the sides $AB$ and $AC$. Denote by $M$, $N$ and $P$ the tangent points of the first circle to the side $AB$ and the extensions of the sides $BC$ and $CA$ and by $S$, $Q$ and $R$ the tangent points of the second circle to the side $AC$ and the extensions of the sides $AB$ and $BC$. Let $X$ be the intersection point of the lines $MN$ and $RS$ and $Y$ be the intersection point of the lines $PN$ and $RQ$. Prove that the points $X$, $A$ and $Y$ are colinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWe shall use the standard notations. We first prove that $YX \\perp BC$. In $\\triangle BQR$ we have $\\angle BRY = 90^\\circ - \\frac{\\beta}{2}$ and since $\\angle MNB = \\frac{\\beta}{2}$ it follows that $NX \\perp RY$. Analogously $RX \\perp NY$. This means that $X$ is the orthocenter of $\\triangle NRY$ and $YX \\perp RN$. Hence $YX \\perp BC$.\n\n![](attached_image_1.png)\n\nNow we shall prove that $X$ lies on the altitude of $\\triangle ABC$ through $A$. Denote by $X'$ its intersection point with $MN$. Since $BM = BN$ and $CP = CN = p$, we have $BM = BN = p - a$. Then $AM = c - (p - a) = p - b$ and by the Sine theorem for $\\triangle AMX'$ we get\n$$\n\\frac{AX'}{\\sin \\frac{\\beta}{2}} = \\frac{AM}{\\sin \\angle AX'M} \\Longleftrightarrow \\frac{AX'}{\\sin \\frac{\\beta}{2}} = \\frac{p-b}{\\sin \\left(90^\\circ + \\frac{\\beta}{2}\\right)} \\Longleftrightarrow AX' = (p-b) \\tan \\frac{\\beta}{2}\n$$\nOn the other hand, if $T$ is the tangent point of the incircle with the side $BC$, then $BT = p-b$ and $r = (p-b) \\tan \\frac{\\beta}{2}$. Therefore $AX' = r$. Analogously, if $X''$ is the intersection point of $RS$ and the altitude of $\\triangle ABC$ through $A$, then $AX'' = r$. Therefore $X' \\equiv X'' \\equiv X$.\n\nSince $YX \\perp BC$ and $AX \\perp BC$, we conclude that the points $X$, $A$ and $Y$ are colinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27320,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nConsider a set $S$ of 2006 points in the plane. A pair $(A, B) \\in S \\times S$ is called \"isolated\" if the disk with diameter $A B$ does not contain other points from $S$. Find the maximum number of \"isolated\" pairs.",
    "options": [],
    "answer": "2005",
    "solution": "Solution:\nConsider a graph with vertices the given points. Two points form an edge if the corresponding pair of points is \"isolated\". We first prove that the graph is connected. To do this suppose that it has more than one connected component and choose points $A$ and $B$ from different components such that the distance $A B$ is the least possible. Then the disk with diameter $A B$ does not contain other vertices and therefore $A$ and $B$ are connected by an edge. This contradicts the choice of $A$ and $B$.\n\nSince a connected graph with 2006 vertices has at least 2005 edges (if the graph is a tree) we conclude that there are at least 2005 \"isolated\" pairs.\n\nIf we take 2006 points on a semicircle such that the distances between consecutive points are equal then \"isolated\" are only the pairs of neighboring points. Therefore there are 2005 \"isolated\" pairs.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27321,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $n$ be a positive integer. Find the number of all finite strictly increasing sequences $a_{0}=1, a_{1}, \\ldots, a_{k}=2.3^{n}$ of positive integers with the following property: $\\prod_{i=1}^{k}\\left[\\frac{a_{i}+a_{i-1}-1}{a_{i-1}}\\right]=2.3^{n}$, where $[x]$ is the integral part of $x$.",
    "options": [],
    "answer": "2^{n-1}(n+2)",
    "solution": "Solution:\nLet $m$ and $n$ be positive integers and $n=m q+r, 0 \\leq r<m$. Then $\\left[\\frac{n+m-1}{m}\\right]=q+1+\\left[\\frac{r-1}{m}\\right]$ and $\\frac{n}{m}=q+\\frac{r}{m}$. Hence $\\left[\\frac{n+m-1}{m}\\right] \\geq \\frac{n}{m}$ with equality if and only if $m$ divides $n$.\n\nApplying the above inequality we obtain\n$$\n2.3^{n}=\\prod_{i=1}^{k}\\left[\\frac{a_{i}+a_{i-1}-1}{a_{i-1}}\\right] \\geq \\prod_{i=1}^{k} \\frac{a_{i}}{a_{i-1}}=\\frac{a_{k}}{a_{0}}=2.3^{n}\n$$\nTherefore $a_{i-1}$ divides $a_{i}$ for $i=1,2, \\ldots, k$. We have to find the number of the sequences $1=a_{0}<a_{1}<\\cdots<a_{k}=2.3^{n}$ such that $a_{i-1}$ divides $a_{i}$ for $i=1,2, \\ldots, k$. Starting by an arbitrary sequence we construct a sequence of $k$ symbols $\\star$, $n$ digits 3 and one digit 2 in the following way: if $\\frac{a_{i+1}}{a_{i}}=2^{p} 3^{q}$, $i=1,2, \\ldots, k-1$ then we put $p$ digits 2 and $q$ digits 3 between $\\star$ number $i$ and $\\star$ number $i+1$. Since $a_{i}<a_{i+1}$ there are no two consecutive symbols $\\star$.\n\nIt is clear that any sequence of $k$ symbols $\\star$, $n$ digits 3 and one digit 2 with no two consecutive symbols $\\star$ corresponds to a sequence $1=a_{0}<a_{1}<\\cdots<a_{k}=2.3^{n}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27322,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the least positive integer $a$ such that the system\n$$\n\\left\\lvert\\, \\begin{aligned}\n& x+y+z=a \\\\\n& x^{3}+y^{3}+z^{2}=a\n\\end{aligned}\\right.\n$$\nhas no integer solution.",
    "options": [],
    "answer": "4",
    "solution": "Solution:\nWhen $a=1,2,3$ the system has solutions $(1,0,0)$, $(1,1,0)$ and $(1,1,1)$, respectively. We shall prove that when $a=4$ the system has no integer solution.\n\nSuppose the contrary. Then we have\n$$\n4-z^{2}=x^{3}+y^{3}=(x+y)\\left(x^{2}-x y+y^{2}\\right)=(4-z)\\left(x^{2}-x y+y^{2}\\right)\n$$\ngiving (since $z=4$ does not lead to an integral solution) that $\\frac{4-z^{2}}{4-z}$ is an integer. Since $\\frac{4-z^{2}}{4-z}=\\frac{-12+16-z^{2}}{4-z}=4+z+\\frac{12}{z-4}$, we conclude that $z-4$ is a divisor of $12$. Hence $z-4= \\pm 1, \\pm 2, \\pm 3, \\pm 4, \\pm 6, \\pm 12$ and therefore $z=-8,-2,0,1,2,3,5,6,7,8,10$ or $16$.\n\nUsing (1) we have\n$$\n(x+y)^{2}-3 x y=\\frac{4-z^{2}}{4-z} \\Longleftrightarrow 3 x y=(4-z)^{2}-\\frac{4-z^{2}}{4-z}\n$$\nand we obtain the following system for $x$ and $y$:\n$$\n\\left\\lvert\\, \\begin{aligned}\n& x+y=4-z \\\\\n& x y=\\frac{(4-z)^{3}+z^{2}-4}{3(4-z)}\n\\end{aligned}\\right.\n$$\nIt is easy to check that all the values of $z$ listed above do not lead to an integral solution for $x$ and $y$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27323,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nThe tangent lines to the circumcircle $k$ of an isosceles $\\triangle ABC$, $AC = BC$, at the points $B$ and $C$ meet at point $X$. If $AX$ meets $k$ at point $Y$, find the ratio $\\frac{AY}{BY}$.",
    "options": [],
    "answer": "2",
    "solution": "Solution:\nIf $\\angle BAY = \\alpha$ and $\\angle ABY = \\beta$ then $\\angle BYX = \\beta + \\alpha$. Furthermore $\\angle ACB = \\angle AYB = 180^\\circ - \\alpha - \\beta$, implying $\\angle BAC = \\angle ABC = \\frac{\\beta + \\alpha}{2}$. Thus $\\angle AYC = \\frac{\\beta + \\alpha}{2}$ and $\\angle YCX = \\angle YAC = \\frac{\\beta + \\alpha}{2} - \\alpha = \\frac{\\beta - \\alpha}{2}$. The Sine theorem for $\\triangle BYX$ and $\\triangle CYX$ gives\n\n![](attached_image_1.png)\n\n$$\n\\frac{XY}{XB} = \\frac{\\sin \\alpha}{\\sin (\\alpha + \\beta)}, \\quad \\frac{XY}{XC} = \\frac{\\sin \\frac{\\beta - \\alpha}{2}}{\\sin \\frac{\\beta + \\alpha}{2}}\n$$\nand since $XB = XC$ we have $\\frac{\\sin \\alpha}{\\sin (\\alpha + \\beta)} = \\frac{\\sin \\frac{\\beta - \\alpha}{2}}{\\sin \\frac{\\beta + \\alpha}{2}}$. Hence\n$$\n\\sin \\alpha = 2 \\cos \\frac{\\beta + \\alpha}{2} \\sin \\frac{\\beta - \\alpha}{2} = \\sin \\beta - \\sin \\alpha\n$$\nand therefore $\\frac{AY}{BY} = \\frac{\\sin \\beta}{\\sin \\alpha} = 2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27324,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $a_{1}, a_{2}, \\ldots$ be a sequence of real numbers less than $1$ and such that $a_{n+1}(a_{n}+2)=3$, $n \\geq 1$. Prove that:\n\na) $-\\frac{7}{2}<a_{n}<-2$\n\nб) $a_{n}=-3$ for any $n$.",
    "options": [],
    "answer": "a_n = -3 for all n",
    "solution": "Solution:\n\na) It follows from $a_{n}<1$ and $a_{n+1}=\\frac{3}{a_{n}+2}<1$ that $a_{n}<-2$. Thus, $a_{n+1}<-2$, and therefore $a_{n}+2=\\frac{3}{a_{n+1}}> -\\frac{3}{2}$, i.e. $a_{n}>-\\frac{7}{2}$.\n\nб) First solution. Set $b_{n}=a_{n}+3$. Then $b_{n+1}=\\frac{3 b_{n}}{b_{n}-1}$. It follows from a) that $-\\frac{3}{2}<b_{n}-1<0$, i.e. $|b_{n}-1|<\\frac{3}{2}$ and therefore $|b_{n+1}|>2|b_{n}|$. Hence $1>|b_{n+1}|>2^{n}|b_{1}|$ for all $n$. Letting $n \\rightarrow \\infty$ gives $b_{1}=0$. Then $b_{n}=0$, i.e. $a_{n}=-3$ for all $n$.\n\nSecond solution. Let $c_{1}=1$ and $c_{k+1}=3 c_{k}+(-1)^{k}$ for $k \\geq 1$. One can prove by induction on $k$ that\n$$\n-\\frac{c_{2 k+1}}{c_{2 k}}<a_{n}< -\\frac{c_{2 k}}{c_{2 k-1}}\n$$\nfor all $n$ and $k$. Indeed, similar to a) it follows from $a_{n}< -\\frac{c_{2 k}}{c_{2 k-1}}$ that $-\\frac{c_{2 k+1}}{c_{2 k}}< a_{n}$, which in turn implies that $a_{n}< -\\frac{c_{2 k+2}}{c_{2 k+1}}$.\nFurther, induction on $k$ gives $c_{k}=\\frac{3^{k}+(-1)^{k-1}}{4}$. Therefore $\\lim_{k \\rightarrow \\infty} \\frac{c_{k+1}}{c_{k}}=3$ and using (*) we conclude that $a_{n}=-3$ for all $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27325,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $m \\geq 5$ and $n$ be positive integers and $M$ be a regular $(2n+1)$-gon. Find the number of convex $m$-gons with vertices among the vertices of $M$ and having at least one acute angle.",
    "options": [],
    "answer": "((2n+1)(mn - 2n - 1)/(m - 1)) * C(n, m-2)",
    "solution": "Solution:\nIt is easy to observe that there are at most two acute angles in every convex $m$-gon. Moreover, if there are two acute angles then they are located at one and the same side.\n\nFix $l = 0, 1, \\ldots, n-1$ and let $A$ and $B$ be two vertices of $M$ such that there are $l$ vertices on the arc $\\widehat{AB}$. Consider the following expression\n$$\n\\binom{l}{m-2} + \\binom{n}{m-2} - \\binom{n-l-1}{m-2} - (l+1)\\binom{n-l-1}{m-3}\n$$\nWe count in it all $m$-gons having two acute angles to the side $AB$ and all $m$-gons having one acute angle to the right of $AB$.\n\nIt is easy to see that summation on $l = 0, 1, \\ldots, n-1$ and then multiplication by $2n+1$ counts every $m$-gon exactly once.\n\nNow using the identity $\\sum_{s=0}^{k} \\binom{s}{t} = \\binom{k+1}{t+1}$ we obtain\n$$\n\\begin{aligned}\n& \\sum_{l=0}^{n-1} \\left[ \\binom{l}{m-2} + \\binom{n}{m-2} - \\binom{n-l-1}{m-2} - (l+1)\\binom{n-l-1}{m-3} \\right] \\\\\n=\\ & \\sum_{l=0}^{n-1} \\binom{l}{m-2} + n\\binom{n}{m-2} - \\sum_{s=0}^{n-1} \\binom{s}{m-2} - (n-s) \\sum_{s=0}^{n-1} \\binom{s}{m-3} \\\\\n=\\ & n\\binom{n}{m-2} - n \\sum_{s=0}^{n-1} \\binom{s}{m-3} + (s+1) \\sum_{s=0}^{n-1} \\binom{s}{m-3} - \\sum_{s=0}^{n-1} \\binom{s}{m-3} \\\\\n=\\ & (m-2) \\sum_{s=0}^{n-1} \\binom{s+1}{m-2} - \\binom{n}{m-2} = (m-2)\\binom{n+1}{m-1} - \\binom{n}{m-2} \\\\\n=\\ & \\frac{mn - 2n - 1}{m-1} \\binom{n}{m-2} .\n\\end{aligned}\n$$\nTherefore the answer is $\\frac{(2n+1)(mn-2n-1)}{m-1} \\binom{n}{m-2}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27326,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the area of the triangle determined by the straight line with equation $x - y + 1 = 0$ and the tangent lines to the graph of the parabola $y = x^{2} - 4x + 5$ at its common points with the line.",
    "options": [],
    "answer": "27/4",
    "solution": "Solution:\nThe common points of the graphs of the line and the parabola are $A(1, 2)$ and $B(4, 5)$. The equations of the tangents to the graph of the parabola at $A$ and $B$ are $y = -2x + 11$ and $y = 4x - 11$, respectively. The intersecting point of the two tangents is the point $C\\left(\\frac{5}{2}, -1\\right)$. The area of $\\triangle ABC$ then equals $\\frac{27}{4}$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 27327,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nThe sequence $\\{x_{n}\\}_{n=1}^{\\infty}$ is defined by $x_{1}=2$ and $x_{n+1}=1+a x_{n}$, $n \\geq 1$, where $a$ is a real number. Find all values of $a$ for which the sequence is:\n\na) an arithmetic progression;\n\nb) convergent and find its limit.",
    "options": [],
    "answer": "a) a = 1. b) Convergent for −1 < a < 1, with limit 1/(1 − a).",
    "solution": "Solution:\na) It follows by the recurrence relation that $x_{1}=2$, $x_{2}=1+2a$ and $x_{3}=1+a+2a^{2}$. Then $x_{1}+x_{3}=2x_{2} \\Longleftrightarrow 3+a+2a^{2}=2(1+2a)$ with solutions $a=1$ and $a=\\frac{1}{2}$. For $a=1$ we get $x_{n+1}=x_{n}+1$, i.e. the sequence is an arithmetic progression. For $a=\\frac{1}{2}$ we see by induction on $n$ that $x_{n}=2$ for every $n$. Therefore $a=1$ is the only solution.\n\nb) We prove by induction on $n$ that $x_{n+1}=1+a+\\cdots+a^{n-1}+2a^{n}$, $n \\geq 1$. For $a=1$ we have $x_{n}=n+1$, i.e. the sequence is not convergent. Let $a \\neq 1$. Then\n$$\nx_{n+1}=2a^{n}+\\frac{1-a^{n}}{1-a}=a^{n}\\left(2-\\frac{1}{1-a}\\right)+\\frac{1}{1-a}\n$$\nIf $2-\\frac{1}{1-a}=0$, i.e. $a=\\frac{1}{2}$, we get $x_{n+1}=2$ for every $n$ and the sequence is convergent.\nSince $\\{a^{n}\\}_{n=1}^{\\infty}$ converges if and only if $|a|<1$ or $a=1$, we conclude that the given sequence is convergent for $a \\in(-1,1)$ and its limit is equal to $\\frac{1}{1-a}$ (since $\\lim _{n \\rightarrow \\infty} a^{n}=0$ for $a \\in(-1,1)$ ).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27328,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $ABC$ be a right triangle and $D$ be a point on the hypotenuse $AB$.\n\na) Prove that the expression\n$$\n\\frac{AC^{2}}{AD+CD} + \\frac{BC^{2}}{BD+CD}\n$$\ndoes not depend on $D$.\n\nb) Let $DE$ ($E \\in AC$) and $DF$ ($F \\in BC$) be the bisectors of $\\angle ADC$ and $\\angle BDC$, respectively. Find the minimum value of the expression\n$$\n\\frac{CF}{CA} + \\frac{CE}{CB}.\n$$",
    "options": [],
    "answer": "1",
    "solution": "Solution:\na. Set $\\angle BAC = \\alpha$ and $\\angle ACD = x$. We have $AC = AB \\cos \\alpha$, $BC = AB \\sin \\alpha$ and it follows by the Sine theorem for $\\triangle ADC$ and $\\triangle BDC$ that\n\n![](attached_image_1.png)\n\n$$\n\\begin{aligned}\n& \\frac{AC^{2}}{AD+CD} + \\frac{BC^{2}}{BD+CD} = \\frac{AC}{\\frac{AD}{AC} + \\frac{CD}{AC}} + \\frac{BC}{\\frac{BD}{BC} + \\frac{CD}{BC}} \\\\\n& = \\frac{AB \\cos \\alpha}{\\frac{\\sin x}{\\sin (\\alpha + x)} + \\frac{\\sin \\alpha}{\\sin (\\alpha + x)}} + \\frac{\\cos x}{\\sin (\\alpha + x)} + \\frac{\\cos \\alpha}{\\sin (\\alpha + x)} \\\\\n& = AB \\left( \\frac{\\cos \\alpha \\sin (\\alpha + x)}{\\sin \\alpha + \\sin x} + \\frac{\\sin \\alpha \\sin (\\alpha + x)}{\\cos \\alpha + \\cos x} \\right) \\\\\n& = AB \\left( \\frac{\\cos \\alpha \\cos \\frac{\\alpha + x}{2}}{\\cos \\frac{\\alpha - x}{2}} + \\frac{\\sin \\alpha \\sin \\frac{\\alpha + x}{2}}{\\cos \\frac{\\alpha - x}{2}} \\right) = AB.\n\\end{aligned}\n$$\n\nThe above identity can be proved by using the Stewart theorem as well.\n\nb. We have\n$$\n\\frac{AE}{CE} = \\frac{AD}{CD} \\Rightarrow \\frac{AE + CE}{CE} = \\frac{AD + CD}{CD} \\Rightarrow CE = \\frac{AC \\cdot CD}{AD + CD}\n$$\nand similarly $CF = \\frac{BC \\cdot CD}{BD + CD}$. Then\n$$\n\\begin{aligned}\n\\frac{CF}{CA} + \\frac{CE}{CB} & = \\frac{BC \\cdot CD}{CA(BD + CD)} + \\frac{AC \\cdot CD}{CB(AD + CD)} \\\\\n& = \\frac{CD}{CA \\cdot CB} \\left( \\frac{CA^{2}}{AD + CD} + \\frac{BC^{2}}{BD + CD} \\right)\n\\end{aligned}\n$$\nand using a) we see that\n$$\n\\frac{CF}{CA} + \\frac{CE}{CB} = \\frac{CD \\cdot AB}{CA \\cdot CB}\n$$\nTherefore the required minimum is equal to $1$ and it is attained when $CD$ is the altitude of $\\triangle ABC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27329,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all real numbers $a$, such that the inequality\n$$\nx^{4}+2 a x^{3}+a^{2} x^{2}-4 x+3>0\n$$\nholds true for all real numbers $x$.",
    "options": [],
    "answer": "a > 0",
    "solution": "Solution:\nFirst Solution. Write the equation in the form\n$$\nx^{2}(x+a)^{2}>4 x-3\n$$\nThen for $x=1$ we get $(a+1)^{2}>1$, i.e. $a>0$ or $a<-2$. If $a<-2$, then $x=-a$ gives a contradiction $0>-4 a-3$. Thus, $a>0$.\n\nConversely, if $a>0$, then (1) is satisfied for all $x$. Indeed, when $x \\leq 0$ this is obvious and when $x>0$ we have $x^{2}(x+a)^{2}>x^{4} \\geq 4 x+3$, since the later inequality is equivalent to $(x-1)^{2}\\left((x+1)^{2}+2\\right) \\geq 0$.\n\nSecond Solution. It follows from (1) that we have to find all $a$, for which\n$$\na<-\\n\\frac{\\sqrt{4 x-3}}{x}-x=f(x)\n$$\nfor all $x \\geq \\frac{4}{3}$ or\n$$\na>\\frac{\\sqrt{4 x-3}}{x}-x=g(x)\n$$\nfor all $x \\geq \\frac{4}{3}$.\n\nThe first case is impossible since $\\lim _{x \\rightarrow+\\infty} f(x)=-\\infty$. The maximum of the function $g$ equals $0$ (and it is attained for $x=1$ ). Therefore the answer is $a>0$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27330,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all complex numbers $a \\neq 0$ and $b$ such that for every complex root $w$ of the equation $z^{4}-a z^{3}-b z-1=0$ the inequality $|a-w| \\geq|w|$ holds.",
    "options": [],
    "answer": "(2, -2), (-2, 2), (2i, 2i), (-2i, -2i)",
    "solution": "Solution:\nLet $z_{k}, 1 \\leq k \\leq 4$, be the roots of the given equation. Using Vieta's formulas we obtain\n$$\nz_{1}+z_{2}+z_{3}+z_{4}=a \\text{ and } z_{1}^{2}+z_{2}^{2}+z_{3}^{2}+z_{4}^{2}=a^{2}\n$$\nSet $u_{k}=\\frac{2 z_{k}}{a}=x_{k}+i y_{k}, 1 \\leq k \\leq 4$, where $x_{k}, y_{k} \\in \\mathbb{R}$. Then $u_{k}^{2}=x_{k}^{2}-y_{k}^{2}+2 i x_{k} y_{k}$ and we get\n$$\n\\begin{aligned}\n& x_{1}+x_{2}+x_{3}+x_{4}=2 \\\\\n& x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}=4+y_{1}^{2}+y_{2}^{2}+y_{3}^{2}+y_{4}^{2} \\geq 4\n\\end{aligned}\n$$\nOn the other hand,\n$$\n\\left|a-z_{k}\\right| \\geq\\left|z_{k}\\right| \\Longleftrightarrow\\left|2-u_{k}\\right| \\geq\\left|u_{k}\\right| \\Longleftrightarrow\\left(2-x_{k}\\right)^{2}+y_{k}^{2} \\geq x_{k}^{2}+y_{k}^{2} \\Longleftrightarrow x_{k} \\leq 1\n$$\nNow (1) implies $x_{k} \\geq-1$, i.e. $x_{k}^{2} \\leq 1$, which together with (2) gives $x_{k}= \\pm 1$, $y_{k}=0$.\nHence (1) shows that three of the numbers $x_{k}$ are equal to 1 and the fourth one is $-1$. We can assume that $z_{1}=z_{2}=z_{3}=-z_{4}$. Then $z_{1} z_{2} z_{3} z_{4}=-1$ implies $z_{1}= \\pm 1, \\pm i$, which gives\n$$\n(a, b)=(2,-2),(-2,2),(2 i, 2 i),(-2 i,-2 i)\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27331,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all positive integers $n$ for which the equality\n$$\n\\frac{\\sin (n \\alpha)}{\\sin \\alpha}-\\frac{\\cos (n \\alpha)}{\\cos \\alpha}=n-1\n$$\nholds true for all $\\alpha \\neq \\frac{k \\pi}{2}, \\quad k \\in \\mathbb{Z}$.",
    "options": [],
    "answer": "n = 1 and n = 3",
    "solution": "Solution:\nThe equality is equivalent to\n$$\n\\sin (n-1) \\alpha=\\frac{(n-1) \\sin 2 \\alpha}{2}\n$$\nWhen $n \\geq 4$ setting $\\alpha=\\frac{\\pi}{4}$ gives\n$$\n\\sin \\left((n-1) \\frac{\\pi}{4}\\right)=\\frac{n-1}{2} \\geq \\frac{3}{2}\n$$\na contradiction.\nWhen $n=1$ and $n=3$ the equality (1) is an identity and when $n=2$ we have $\\sin \\alpha=\\frac{\\sin (2 \\alpha)}{2}$, which is not true for $\\alpha=\\frac{\\pi}{4}$. Therefore the answer is $n=1$ and $n=3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27332,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA plane intersects a tetrahedron $ABCD$ and divides the medians of the triangles $DAB$, $DBC$ and $DCA$ through $D$ in ratios $1:2$, $1:3$ and $1:4$ from $D$, respectively. Find the ratio of the volumes of the two parts of the tetrahedron cut by the plane.\nOleg Mushkarov",
    "options": [],
    "answer": "1:47",
    "solution": "Solution:\nLet the plane meet the edges $DA$, $DB$ and $DC$ at points $P$, $Q$ and $R$, respectively. Set\n$$\n\\frac{DP}{DA} = x, \\quad \\frac{DQ}{DB} = y, \\text{ and } \\frac{DR}{DC} = z\n$$\nLet $M$ be the midpoint of $AB$ and $L = DM \\cap PQ$. It follows from the condition of the\n![](attached_image_1.png)\nproblem that $\\frac{DL}{DM} = \\frac{1}{3}$. Therefore\n$$\n\\frac{S_{DLP}}{S_{DAM}} = \\frac{DP \\cdot DL}{DA \\cdot DM} = \\frac{x}{3} ; \\quad \\frac{S_{DLQ}}{S_{DMB}} = \\frac{DL \\cdot DQ}{DM \\cdot DB} = \\frac{y}{3}\n$$\nSince $S_{DAM} = S_{DMB} = \\frac{1}{2} S_{DAB}$ we conclude that\n$$\n2xy = 2 \\frac{DP \\cdot DQ}{DA \\cdot DB} = \\frac{S_{DPQ}}{\\frac{1}{2} S_{DAB}} = \\frac{S_{DPL}}{S_{DAM}} + \\frac{S_{DLQ}}{S_{DMB}} = \\frac{x + y}{3}\n$$\ni.e. $\\frac{1}{x} + \\frac{1}{y} = 6$. Analogously $\\frac{1}{y} + \\frac{1}{z} = 8$ and $\\frac{1}{z} + \\frac{1}{x} = 10$. Solving this system we obtain $x = \\frac{1}{4}$, $y = \\frac{1}{2}$ and $z = \\frac{1}{6}$. Thus\n$$\n\\frac{V_{DPQR}}{V_{DABC}} = \\frac{DP \\cdot DQ \\cdot DR}{DA \\cdot DB \\cdot DC} = xyz = \\frac{1}{48}\n$$\nand therefore the desired ratio equals $1:47$.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 27333,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $CL$ and $CK$ be the inner and the outer bisectors of angle $ACB$ in $\\triangle ABC$, $AC > BC$ and let $CM$ be its median. A point $P$ on $CM$ is such that the points $C, A_{1}, B_{1}$ and $P$ are concyclic, where $A_{1} = AP \\rightarrow \\cap BC$ and $B_{1} = BP \\rightarrow \\cap AC$. Prove that the points $C, K, L$ and $P$ are also concyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nIt follows from Ceva's theorem that $\\frac{AM \\cdot BA_{1} \\cdot CB_{1}}{MB \\cdot A_{1}C \\cdot B_{1}A} = 1$, i.e. $\\frac{CB_{1}}{B_{1}A} = \\frac{CA_{1}}{A_{1}B}$. Thus, $A_{1}B_{1} \\parallel AB$ and $\\Varangle A_{1}B_{1}C = \\Varangle BAC$. We have $\\Varangle APM = \\Varangle A_{1}PC = \\frac{\\widetilde{A_{1}C}}{2} = \\Varangle A_{1}B_{1}C = \\Varangle BAC$. Therefore\n$\\triangle PAM \\sim \\triangle ACM$ and analogously $\\triangle PBM \\sim \\triangle BCM$. It follows from above that $\\frac{AP}{AC} = \\frac{AM}{CM} = \\frac{BM}{CM} = \\frac{BP}{BC}$, which implies that $\\frac{AP}{BP} = \\frac{AC}{BC} = \\frac{AL}{BL}$, i.e. $PL$ is the inner bisector of $\\Varangle APB$. Moreover, $\\frac{AK}{BK} = \\frac{AC}{BC} = \\frac{AP}{BP}$, i.e. $PK$ is the outer bisector of $\\Varangle APB$. Therefore $\\Varangle LPB = 90^{\\circ}$ implying that $P$ lies on the circle with diameter $KL$. Note that the point $C$ lies on the same circle, which completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27334,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $\\mathbb{R}^{+}$ be the set of all positive real numbers and $f: \\mathbb{R}^{+} \\rightarrow \\mathbb{R}^{+}$ be a function such that\n$$\nf(x+y)-f(x-y)=4 \\sqrt{f(x) f(y)}\n$$\nfor all $x>y>0$.\n\na) Prove that $f(2 x)=4 f(x)$ for all $x \\in \\mathbb{R}^{+}$.\n\nb) Find all such functions.",
    "options": [],
    "answer": "All such functions are f(x) = c x^2 for a constant c > 0; in particular, f(2x) = 4 f(x) for all positive x.",
    "solution": "Solution:\n\na) It follows from $f(x+y)-f(x-y)>0$ that $f$ is an increasing function. Therefore the function $f(x)$ has a limit $l \\geq 0$ when $x \\rightarrow 0$, $x>0$ (prove!). Thus letting $x, y \\rightarrow 0$, $x>y>0$, we get $l-l=4 \\sqrt{l^{2}}$, i.e. $l=0$. Fixing $x$ and letting $y \\rightarrow 0$, $y>0$ we conclude that $f(x+y)-f(x-y) \\rightarrow 0$. Since the function $f$ is increasing we conclude that it is continuous at $x$. Finally, letting $y \\rightarrow x$, $y<x$, we get $f(2 x)=4 f(x)$.\n\nb) Setting $x=n y>0$, where $n \\geq 2$ is an integer, we obtain from the given identity that\n$$\nf((n+1) y)=f((n-1) y)+4 \\sqrt{f(n y) f(y)}\n$$\nUsing $f(2 y)=4 f(y)$, it follows by induction that $f(n y)=n^{2} f(y)$. Set $f(1)=c>0$. Then $f(n)=n^{2} c$. Now, for any positive integers $p$ and $q$ we have $c p^{2}=f(q \\cdot p / q)=q^{2} f(p / q)$, i.e. $f(p / q)=c(p / q)^{2}$. Since $f$ is a continuous function, we conclude that $f(x)=c x^{2}$ for any $x>0$. Conversely, any function of the form $f(x)=c x^{2}$ satisfies the condition.\n\nRemark. It is possible to show that any function satisfying the condition is differentiable. Thus, letting $y \\rightarrow 0$, $y>0$, in the identity\n$$\n\\frac{f(x+y)-f(x-y)}{2 y}=2 \\frac{\\sqrt{f(y)}}{y} \\sqrt{f(x)}\n$$\nwe get $f'(x)=2 c \\sqrt{f(x)}$. Therefore $(\\sqrt{f(x)})'=c$, i.e. $f(x)=c^{2} x^{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27335,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all pairs $(P, Q)$ of polynomials with real coefficients such that\n$$\n\\frac{P(x)}{Q(x)}-\\frac{P(x+1)}{Q(x+1)}=\\frac{1}{x(x+2)}\n$$\nfor infinitely many $x \\in \\mathbb{R}$.",
    "options": [],
    "answer": "All solutions are of the form Q(x) = x(x+1) R(x) and P(x) = (1/2 + x + c x(x+1)) R(x), where R is any nonzero polynomial and c is a real constant.",
    "solution": "Solution:\nFirst solution. It suffices to consider the case when $P$ and $Q\\not\\equiv 0$ are relatively prime polynomials and the leading coefficient of $Q$ equals $1$. We have\n$$\nx(x+2)(P(x) Q(x+1)-Q(x) P(x+1))=Q(x) Q(x+1)\n$$\nfor infinitely many $x$, i.e. for every $x$. Thus the polynomials $Q(x)$ and $Q(x+1)$ divide $x(x+2) Q(x+1)$ and $x(x+2) Q(x)$ respectively.\nTherefore $S(x) Q(x)=x(x+2) Q(x+1)$ and $T(x) Q(x+1)=x(x+2) Q(x)$, where $S$ and $T$ are quadratic polynomials with leading coefficients $1$. Hence, $S(x) T(x)=x^{2}(x+2)^{2}$. There are three cases to be considered.\n\nCase 1. $S(x)=T(x)=x(x+2)$. Then $Q(x+1)=Q(x)$, i.e. $Q \\equiv 1$ and the condition of the problem shows that this is impossible.\n\nCase 2. $S(x)=x^{2}$ and $T(x)=(x+2)^{2}$. Then $x Q(x)=(x+2) Q(x+1)$. Therefore $Q(1)=0$ and it follows by induction that $Q(n)=0$ for all $n \\in \\mathbb{N}$. Hence $Q \\equiv 0$, a contradiction.\n\nCase 3. $S(x)=(x+2)^{2}$ and $T(x)=x^{2}$. Then $(x+2) Q(x)=x Q(x+1)$ and therefore $x$ divides $Q(x)$ and $x+2$ divides $Q(x+1)$, i.e. $x+1$ divides $Q(x)$. It follows that $Q(x)=x(x+1) Q_{1}(x)$, where $Q_{1}$ has leading coefficient $1$ and $Q_{1}(x+1)=Q_{1}(x)$. We conclude that $Q_{1}(x)=1$ and $Q(x)=x(x+1)$. Now plugging $Q(x)$ in (1) gives\n$$\n(x+2) P(x)-x P(x+1)=x+1\n$$\nSetting $x=0$ and $x=-1$ we obtain $P(0)=\\frac{1}{2}$ and $P(-1)=-\\frac{1}{2}$. Therefore $P(x)=\\frac{1}{2}+x+x(x+1) P_{1}(x)$, where $P_{1}$ is a polynomial. Now (2) implies that $P_{1}(x+1)=P_{1}(x)$ and therefore $P_{1}$ is a constant.\nWe conclude that if the polynomials $P$ and $Q$ are relatively prime and $a_{0}=1$, then $Q(x)=x(x+1)$ and $P(x)=\\frac{1}{2}+x+c x(x+1)$.\nTherefore the answer is\n$$\nQ(x)=x(x+1) R(x) \\text{ and } P(x)=\\left(\\frac{1}{2}+x+c x(x+1)\\right) R(x)\n$$\nwhere $R$ is an arbitrary nonzero polynomial and $c$ is a constant.\n\n\nSecond solution. The given identity can be written as\n$$\n\\frac{P(x)}{Q(x)}-\\frac{1}{2}\\left(\\frac{1}{x}+\\frac{1}{x+1}\\right)=\\frac{P(x+1)}{Q(x+1)}-\\frac{1}{2}\\left(\\frac{1}{x+1}+\\frac{1}{x+2}\\right)\n$$\nHence it follows by induction that\n$$\n\\frac{P(x)}{Q(x)}-\\frac{1}{2}\\left(\\frac{1}{x}+\\frac{1}{x+1}\\right)=\\frac{P(x+n)}{Q(x+n)}-\\frac{1}{2}\\left(\\frac{1}{x+n}+\\frac{1}{x+n+1}\\right)\n$$\nFixing $x$ and letting $n \\rightarrow \\infty$ we see that $\\frac{P(x)}{Q(x)}-\\frac{1}{2}\\left(\\frac{1}{x}+\\frac{1}{x+1}\\right)=c$, where $c$ is a constant. Now it is easy to conclude that $Q(x)=x(x+1) R(x)$ and $P(x)=\\left(\\frac{1}{2}+x+c x(x+1)\\right) R(x)$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 27336,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nAn infinite sequence of digits is obtained by writing all positive integers one after another in increasing order. Find the least positive integer $k$ such that among the first $k$ digits of the above sequence every two nonzero digits appear different number of times.",
    "options": [],
    "answer": "56784",
    "solution": "Solution:\nDenote by $M_{n}$ the set of all digits of the numbers $1,2, \\ldots, n$. First we find the least positive integer $n=\\overline{a_{1} a_{2} \\ldots a_{t}}$ such that every two nonzero digits appear different number of times in $M_{n}$. By adding zeros on the left we may assume that all numbers $1,2, \\ldots, n-1$ are $t$-digit numbers.\n\nIt is clear that every nonzero digit appears the same number of times. Let $B_{i}^{j}, 1 \\leq i \\leq t, 1 \\leq j \\leq 9$ be the number of appearances of the digit $j$ in position $i$ among the numbers $1,2, \\ldots, n$. Note that for all $i$ and $j \\leq 8$, if a number $A$ has $j+1$ in position $i$, then replacing this digit by $j$ we obtain a number which is less than $A$. Therefore $B_{i}^{j} \\geq B_{i}^{j+1}$.\n\nFurthermore for a fixed $i$ the inequality $B_{i}^{j} \\geq B_{i}^{j+1}$ is fulfilled for at most two pairs of digits $j$ and $j+1$, namely $a_{i-1}$ and $a_{i} ; a_{i}$ and $a_{i+1}$. Moreover, if $i=t$, it is fulfilled only for $a_{t}$ and $a_{t+1}$. Since there are 8 pairs of the form $(j, j+1)$ we have $t \\geq 5$. If $n=13578$ then $B_{1}^{1}>B_{1}^{2} ; B_{2}^{2}>B_{2}^{3} ; B_{2}^{3}>B_{2}^{4}$; $B_{3}^{4}>B_{3}^{5} ; B_{3}^{5}>B_{3}^{6} ; \\bar{B}_{4}^{6}>B_{4}^{7} ; B_{4}^{7}>B_{4}^{8} ; B_{5}^{8}>B_{5}^{9}$, i.e. $n=13578$ satisfies the condition of the problem.\n\nIf $m<13578$ also satisfies the condition then the first digit of $m$ is 1 and the second digit is $0,1,2$ or 3. Since $B_{1}^{j}>B_{1}^{j+1}$ is true only for $j=1$ if the second digit is 0,1 or 2 then at least two consecutive digits appear equal number of times. Therefore the second digit of $m$ is 3. It follows by similar arguments that the third, fourth and the fifth digits of $m$ are respectively 5, 7 and 8. Therefore $n=13578$ is the least positive integer such that every two nonzero digits appear different number of times in $M_{n}$.\n\nSince the number of digits of all numbers $1,2,3, \\ldots, 13578$ equals\n$$\n9 \\cdot 1 + 90 \\cdot 2 + 900 \\cdot 3 + 9000 \\cdot 4 + 3579 \\cdot 5 = 56784\n$$\nwe conclude that 56784 has the desired property.\n\nSuppose that there exists $k<56784$ which has the desired property. Then the digits in the sequence are those in $M_{s}$ for some $s<13578$ and some digits of $s+1$. According to the previous observations there exist two consecutive digits that are not digits of $s$ (eventually excluding the last one) appearing equal number of times in $M_{s}$. If the last digit of $s$ is not 9, then the same digits appear equal number of times in the sequence since the digits of $s$ and $s+1$ are the same (except the last one). If the last digit of $s$ is 9 then the last digit of $s+1$ is 0 and therefore $s+1<13578$. Hence we conclude as above that there exist two consecutive digits not among the digits of $s+1$ which appear equal number of times.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27337,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nProve that if $x$, $y$ and $a$ are real numbers from the interval $(0,1)$, then\n$$\n\\frac{|x-y|}{1-x y} \\leq \\frac{\\left|x^{a}-y^{a}\\right|}{1-x^{a} y^{a}}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nWe have to prove that\n$$\n\\frac{u^{q}-v^{q}}{1-(u v)^{q}} \\leq \\frac{u^{p}-v^{p}}{1-(u v)^{p}}\n$$\nfor $v \\leq u$ and $p<q$. To do this we shall show that\n$$\n\\frac{u^{p+1}-v^{p+1}}{1-(u v)^{p+1}} \\leq \\frac{u^{p}-v^{p}}{1-(u v)^{p}}\n$$\nwhich implies by induction the above inequality. It is easy to check that the latter inequality can be written as\n$$\n\\frac{1-u^{2 p+1}}{u^{p}(1-u)} \\leq \\frac{1-v^{2 p+1}}{v^{p}(1-v)}\n$$\ni.e.\n$$\n\\sum_{j=1}^{p}\\left(u^{j}+u^{-j}\\right) \\leq \\sum_{j=1}^{p}\\left(v^{j}+v^{-j}\\right)\n$$\nwhich follows by the inequality $t+\\frac{1}{t} \\leq z+\\frac{1}{z}$, where $0<z \\leq t \\leq 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27338,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABC$ be a non-equilateral triangle and let $M$ and $N$ be interior points of it such that $\\Varangle BAM = \\Varangle CAN$, $\\Varangle ABM = \\Varangle CBN$ and\n$$\nAM \\cdot AN \\cdot BC = BM \\cdot BN \\cdot CA = CM \\cdot CN \\cdot AB = k\n$$\nProve that:\n\na) $3k = AB \\cdot BC \\cdot CA$;\n\nb) the midpoint of the segment $MN$ is the centroid of $\\triangle ABC$.",
    "options": [],
    "answer": "3k = AB · BC · CA; the midpoint of MN is the centroid of triangle ABC.",
    "solution": "Solution:\n\nThe angle equality implies that $M$ and $N$ are isogonal conjugate points in $\\triangle ABC$. Therefore $\\Varangle BCM = \\Varangle ACN$. Denote by small letters the affixes of the corresponding points in the complex plane. We have that\n$$\n\\arg \\frac{b-a}{m-a} = \\arg \\frac{n-a}{c-a}\n$$\nand\n$$\n|(m-a)(n-a)(b-c)| = k\n$$\nThus,\n$$\n\\frac{(b-a)(c-a)}{(m-a)(n-a)} = \\frac{AB \\cdot BC \\cdot CA}{k} =: K\n$$\nAnalogously $(c-b)(a-b) = K(m-b)(n-b)$ and $(a-c)(b-c) = K(m-c)(n-c)$. After subtraction we obtain\n$$\n(b-c)(K(m+n) - (K-1)(b+c) - 2a) = 0\n$$\nHence\n$$\nm+n = \\frac{(K-1)(b+c) + 2a}{K}\n$$\nand analogously $m+n = \\frac{(K-1)(c+a) + 2b}{K}$. Therefore $K = 3$ and $\\frac{m+n}{2} = \\frac{a+b+c}{3}$, which completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27339,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIvan and Peter play the following game. Ivan chooses a secret number from the set $A=\\{1,2, \\ldots, 90\\}$. Then Peter chooses a subset $B$ of $A$ and Ivan tells Peter whether his number is in the set $B$ or not. If the answer is \"yes\" then Peter pays Ivan 2 leva, and if the answer is \"no\" then he pays Ivan 1 lev. Find the least amount of leva that Peter needs so that he can always find Ivan's number.",
    "options": [],
    "answer": "11",
    "solution": "Solution:\n\nWe shall solve the problem for $A=\\{1,2, \\ldots, t\\}$. Let $F_{0}=F_{1}=1$, $F_{n+1}=F_{n}+F_{n-1}$ for $n \\geq 1$ be the Fibonacci sequence. We shall prove by induction that if $F_{n-1}<t \\leq F_{n}, n \\geq 2$, then the desired sum equals $n$.\n\nSince for $t=2$ and $t=3$ Peter needs 2 or 3 leva, respectively, the assertion is true for $n=2$ and $n=3$.\n\nSuppose that it is true for $n=k$.\n\nChoose $t \\in\\left(F_{k}, F_{k+1}\\right]$ and let Peter ask a question set having $s$ elements. If $s \\in\\left(F_{k-1}, F_{k}\\right]$ and the answer is \"yes\" then Peter gives Ivan 2 leva and by the induction hypothesis he needs additional $k$ leva, i.e. in total $k+2$ leva.\nIf $s \\leq F_{k-1}$, then $t-s \\geq F_{k}+1-F_{k-1}=F_{k-2}+1$. If Peter receives answer \"yes\" then he pays 2 leva and he needs additional $k-1$ leva, i.e. in total $k+1$ leva. It remains to notice that if Peter asks a question set with $F_{k-1}$ elements and the answer is \"yes\" then he needs $2+k-1=k+1$ leva and for answer \"no\" he needs $1+k-1=k$ leva.\nSince $F_{10}=89$ and $F_{11}>89$, the desired number equals 11 .",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27340,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $k$ be the circumcircle of $\\triangle ABC$ and $D$ be a point on the arc $\\overparen{AB}$, which does not contain $C$. Denote by $I_{A}$ and $I_{B}$ the incenters of $\\triangle ADC$ and $\\triangle BDC$, respectively. Prove that the circumcircle of $\\triangle I_{A} I_{B} C$ is tangent to $k$ if and only if\n$$\n\\frac{AD}{BD} = \\frac{AC + CD}{BC + CD}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet $P = CI_{A} \\cap k$ and $Q = CI_{B} \\cap k$. First we prove that the circumcircle $k_{1}$ of $\\triangle I_{A} I_{B} C$ is tangent to $k$ if and only if $I_{A} I_{B} \\parallel PQ$.\nLet $T$ be the point on the tangent line to $k$ at $C$, for which $\\angle ACT = \\angle ABC$. If $k$ and $k_{1}$ are tangent then $CT$ is their common tangent line and therefore\n\n![](attached_image_1.png)\n\n$\\angle CQP = \\angle TCP = \\angle TCI_{A} = \\angle CI_{B} I_{A}$,\ni.e. $I_{A} I_{B} \\parallel PQ$. Conversely, if $I_{A} I_{B} \\parallel PQ$, then $\\angle TCI_{A} = \\angle TCP = \\angle CQP = \\angle CI_{B} I_{A}$, implying that the line $CT$ is tangent to $k_{1}$.\n\nIt remains to prove that $I_{A} I_{B} \\parallel PQ \\Longleftrightarrow \\frac{AD}{BD} = \\frac{AC + CD}{BC + CD}$.\nSince $PI_{A} = PA = PD$ and $QI_{B} = QB = QD$, we have\n$$\n\\begin{aligned}\nI_{A} I_{B} \\parallel PQ &\\Longleftrightarrow \\frac{CI_{A}}{CI_{B}} = \\frac{AP}{BQ} \\\\\n& \\frac{AC + CD - AD}{2 \\cos \\frac{\\angle ACD}{2}} \\\\\n& \\Longleftrightarrow \\frac{BC + CD - BD}{2 \\cos \\frac{\\angle BCD}{2}} = \\frac{2R_{ABC} \\sin \\frac{\\angle ACD}{2}}{2R_{ABC} \\sin \\frac{\\angle BCD}{2}} \\\\\n& \\Longleftrightarrow \\frac{AC + CD - AD}{BC + CD - BD} = \\frac{\\sin \\angle ACD}{\\sin \\angle BCD} \\\\\n& \\Longleftrightarrow \\frac{AC + CD - AD}{BC + CD - BD} = \\frac{AD}{BD} \\Longleftrightarrow \\frac{AC + CD}{BC + CD} = \\frac{AD}{BD}\n\\end{aligned}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27341,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $p$ be a prime number such that $p^2$ divides $2^{p-1}-1$. Prove that for any positive integer $n$ the integer $(p-1)\\left(p!+2^{n}\\right)$ has at least three distinct prime divisors.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nSince $p-1$ is a divisor of $p!$ the greatest common divisor of $p-1$ and $p!+2^{n}$ is a power of two. We shall show that both numbers $p-1$ and $p!+2^{n}$ have at least one odd divisor.\n\nSuppose that $p-1=2^{k}$, i.e. $p=2^{k}+1$. If $s \\geq 3$ is an odd divisor of $k$ then\n$p=2^{s t}+1=\\left(2^{t}+1\\right) A$, i.e. $p$ is not a prime number. Therefore $k=2^{t}$ giving\n$$\n\\begin{aligned}\n2^{p-1}-1 & =2^{2^{k}}-1=\\left(2^{2^{k-1}}-1\\right)\\left(2^{2^{k-1}}+1\\right)=\\cdots \\\\\n& =\\left(2^{2^{t}}-1\\right)\\left(2^{2^{t}}+1\\right)\\left(2^{2^{t+1}}+1\\right) \\ldots\\left(2^{2^{k-1}}+1\\right)\n\\end{aligned}\n$$\nIt is clear that $p^{2}$ does not divide the above product since $\\left(2^{2^{t}}+1,2^{2^{l}}+1\\right)=1$ when $l>t$, and $2^{2^{t}}-1<p=2^{2^{t}}+1$. Therefore $p-1$ is not a power of 2.\n\nSuppose that $p!+2^{n}=2^{k}$, giving $k>n$ and $p!=2^{n}\\left(2^{k-n}-1\\right)$. Then $p$ is a divisor of $2^{m}-1$, where $m=k-n$. Let $t$ be the least positive integer such that $p$ divides $2^{t}-1$. Then $t$ is a divisor of $m$ and $t$ is a divisor of $p-1$. If $p-1=l t$ then\n$$\n2^{p-1}-1=\\left(2^{t}-1\\right)\\left(2^{t(l-1)}+2^{t(l-2)}+\\cdots+2^{t}+1\\right)\n$$\nSince $2^{t} \\equiv 1(\\bmod p)$ we have $2^{t(l-1)}+2^{t(l-2)}+\\cdots+2^{t}+1 \\equiv l \\not \\equiv 0(\\bmod p)$. Therefore $p^{2}$ is a divisor of $2^{t}-1$ which implies that $p^{2}$ is a divisor of $2^{m}-1$, i.e. $p^{2}$ is a divisor of $p!$, a contradiction.\n\nThus, both $p-1$ and $p!+2^{n}$ have at least one odd divisor and these divisors are distinct. Therefore the product $(p-1)\\left(p!+2^{n}\\right)$ has at least three distinct prime divisors.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27342,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTwo real numbers $a$ and $b$ satisfy the inequality $b^{3} + b \\leq a - a^{3}$. Find the maximum possible value of $a + b$.",
    "options": [],
    "answer": "(4/3)^(1/4)",
    "solution": "Solution:\n\nLet $a + b = c$. Thus,\n$$\n(c - a)^{3} + c - a \\leq a - a^{3} \\Longleftrightarrow 3 c a^{2} - (3 c^{2} + 2) a + c^{3} + c \\leq 0\n$$\nIf $c > 0$ then\n$$\n0 \\leq D = (3 c^{2} + 2)^{2} - 12 c (c^{3} + c) = 4 - 3 c^{4}\n$$\nHence $c \\leq \\sqrt[4]{\\frac{4}{3}}$ with equality when $a$ is the double root of the corresponding quadratic equation. Therefore the maximum possible value of $a + b$ equals $\\sqrt[4]{\\frac{4}{3}}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27343,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nProve that if $a, b, c > 0$, then\n$$\n\\frac{a b}{3 a + 4 b + 5 c} + \\frac{b c}{3 b + 4 c + 5 a} + \\frac{c a}{3 c + 4 a + 5 b} \\leq \\frac{a + b + c}{12}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nWe shall prove the following generalization of the given inequality: if $x, y \\geq 1$ and $a, b, c > 0$ then\n$$\n\\frac{a b}{x a + y b + 2 c} + \\frac{b c}{x b + y c + 2 a} + \\frac{c a}{x c + y a + 2 b} \\leq \\frac{a + b + c}{x + y + 2}\n$$\nThe given inequality is obtained for $x = \\frac{6}{5}$ and $y = \\frac{8}{5}$.\nAccording to the Cauchy-Schwarz inequality we have that\n$$\n\\begin{aligned}\na b \\frac{(x + y + 2)^2}{x a + y b + 2 c} & = a b \\frac{((x - 1) + (y - 1) + 2 + 2)^2}{(x - 1) a + (y - 1) b + (a + c) + (b + c)} \\\\\n& \\leq a b \\left(\\frac{x - 1}{a} + \\frac{y - 1}{b} + \\frac{4}{a + c} + \\frac{4}{b + c}\\right) \\\\\n& = (x - 1) b + (y - 1) a + \\frac{4 a b}{a + c} + \\frac{4 a b}{b + c}\n\\end{aligned}\n$$\nSumming up this inequality and the other two similar inequalities we obtain\n$$\n\\begin{aligned}\n& (x + y + 2)^2 \\left(\\frac{a b}{x a + y b + 2 c} + \\frac{b c}{x b + y c + 2 a} + \\frac{c a}{x c + y a + 2 b}\\right) \\\\\n\\leq & (x - 1)(a + b + c) + (y - 1)(a + b + c) + 4(a + b + c) \\\\\n= & (x + y + 2)(a + b + c)\n\\end{aligned}\n$$\n\nwhich proves (1).\n\nSecond solution. The inequality is equivalent to\n$$\n\\begin{aligned}\n0 \\leq & 30\\left[\\left(a^2 - b^2\\right)^2 + \\left(b^2 - c^2\\right)^2 + \\left(c^2 - a^2\\right)^2\\right] \\\\\n& + 11\\left[a b(b - c)^2 + b c(c - a)^2 + c a(a - b)^2\\right] \\\\\n& + 73\\left[a b(c - a)^2 + b c(a - b)^2 + c a(b - c)^2\\right]\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27344,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABC$ be a triangle with $\\angle BAC = 30^\\circ$ and $\\angle ABC = 45^\\circ$. Consider all pairs of points $X$ and $Y$ such that $X$ and $Y$ lie on the rays $AC^{\\rightarrow}$ and $BC^{\\rightarrow}$, respectively, and $OX = BY$, where $O$ is the circumcenter of $\\triangle ABC$. Prove that the perpendicular bisectors of the segments $XY$ pass through a fixed point.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWe shall prove that the perpendicular bisector of $XY$ passes through the point $C'$ which is symmetric to the point $C$ with respect to $AB$. Denote by $R$ the circumradius of $\\triangle ABC$. The Sine theorem for $\\triangle ABC$ gives $AC = R \\sqrt{2}$ and $BC = R$. Suppose that $C$ lies between $A$ and $X$. Then $OX = BY$, $C'A = CA$, $C'B = CB$, $\\angle C'BY = 90^\\circ$ and the Cosine theorem for\n\n![](attached_image_1.png)\n\n$\\triangle C'AX$ gives\n$$\n\\begin{aligned}\nC'X^2 & = C'A^2 + AX^2 - C'A \\cdot AX = AC^2 + AX^2 - AC \\cdot AX \\\\\n& = AC^2 + XA \\cdot XC = 2R^2 + (OX^2 - R^2) = R^2 + BY^2 \\\\\n& = C'B^2 + BY^2 = C'Y^2\n\\end{aligned}\n$$\nTherefore $C'$ is a point on the perpendicular bisector of $XY$. In the case when $X$ lies between $A$ and $C$ the proof is similar.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27345,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $p > 2$ be a prime number. Find the number of the subsets $B$ of the set $\\{1, 2, \\ldots, p-1\\}$ such that $p$ divides the sum of the elements of $B$.",
    "options": [],
    "answer": "(2^{p-1} - 1)/p",
    "solution": "Solution:\nConsider the set $A' = A \\cup \\{0\\}$ instead of $A$. Let $B = \\{a_1, a_2, \\ldots, a_k\\}$ be a nonempty subset of $A'$. Set\n$$\ni + B = \\{i + a_1 \\pmod{p}, i + a_2 \\pmod{p}, \\ldots, i + a_k \\pmod{p}\\}$$\nNote that the sums of the elements of the sets $i + B$, $i = 0, 1, \\ldots, p-1$, are all distinct. Indeed, if for some $s$ and $t$ the sums are equal then\n$$\n\\begin{aligned}\n& \\sum_{i=1}^{k} (s + a_i) \\equiv \\sum_{i=1}^{k} (t + a_i) \\pmod{p} \\\\\n\\Longleftrightarrow \\ & k s + \\sum_{i=1}^{k} a_i \\equiv k t + \\sum_{i=1}^{k} a_i \\pmod{p} \\\\\n\\Longleftrightarrow \\ & k s \\equiv k t \\pmod{p}\n\\end{aligned}\n$$\nwhich is equivalent to $s = t$.\n\nTherefore the set of the subsets of $A'$ (without the empty set and $A'$) partitions into $\\frac{2^p - 2}{p}$ groups and every group contains $p$ sets. Moreover, the sums of the elements of the subsets in every group run over all residues modulo $p$.\n\nTherefore the number of the subsets having sums divisible by $p$ equals $\\frac{2^p - 2}{p}$. Since $0$ is included in half of them, it follows that the number of the subsets $B$ of $A$ (including the empty set and excluding $A$) equals $\\frac{2^p - 2}{2p}$.\n\nReplacing the empty set by $A$ (having sum $\\frac{p(p-1)}{2}$, which is divisible by $p$), we conclude that the answer is $\\frac{2^{p-1} - 1}{p}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27346,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $O$ be a fixed point in the plane. Find all sets of points $S$ in the plane, containing at least two distinct points, and such that for any point $A \\in S, A \\neq O$, the circle with diameter $O A$ is contained in $S$.",
    "options": [],
    "answer": "All such sets are either the entire plane, or the union of an open disc centered at the fixed point with an arbitrary subset of its boundary circle.",
    "solution": "Solution:\nWe first prove the following\n\nLEMMA. If $A \\in S$, then the open disc $k(O, O A)$ is contained in $S$.\n\nProof of the Lemma. Note that if $A \\in S$ and $B$ is a point on the circle of diameter $O A$ then $B \\in S$ (since $B$ belongs to the circle with diameter $O X$, where $O B \\perp B X$ and $X$ is a point on the circle with diameter $O A$). Let $B \\in k(O, O A)$ and $\\varphi=\\Varangle A O B$. For any positive integer $n$ set $A_{0}=A$ and define $A_{k}, k=1, \\ldots, n$, such that $\\Varangle A_{k-1} O A_{k}=\\frac{\\varphi}{n}$ and $O A_{k}=O A_{k-1} \\cos \\frac{\\varphi}{n}$.\n\nSince $\\Varangle O A_{k} A_{k-1}=90^{\\circ}$, it follows by induction on $k$ that $A_{k} \\in S, k=1, \\ldots, n$; in particular $A_{n} \\in k(O, O A)$. Since $B \\in O A^{\\rightarrow}$ and $O B<O A$, the statement of the Lemma would follow if $\\lim _{n \\rightarrow \\infty} O A_{n}=O A$. But the latter is true because $O A_{n}=O A\\left(\\cos \\frac{\\varphi}{n}\\right)^{n}$ and $1 \\geq\\left(\\cos \\frac{\\varphi}{n}\\right)^{2 n}=\\left(1-\\sin ^{2} \\frac{\\varphi}{n}\\right)^{n} \\geq \\left(1-\\frac{\\varphi^{2}}{n^{2}}\\right)^{n} \\rightarrow 1$.\n\nIt follows from the Lemma that if the set $S$ is unbounded then $S=\\mathbb{R}^{2}$. If $S$ is bounded then setting $r=\\sup _{A \\in S} O A$ we get $k(O, r) \\subset S \\subset \\overline{k(O, r)}$. Therefore the desired sets are $S=\\mathbb{R}^{2}$ and $S=k \\cup \\gamma$, where $k$ is an open disc with center $O$, and $\\gamma$ is an arbitrary set on its circumference.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27347,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $D$ and $E$ be points on the sides $AB$ and $AC$ of $\\triangle ABC$ such that $DE \\parallel BC$. The circumcircle $k$ of $\\triangle ADE$ meets the segments $BE$ and $CD$ at points $M$ and $N$. The lines $AM$ and $AN$ meet $BC$ at points $P$ and $Q$ such that $BC = 2PQ$ and $P$ lies between $B$ and $Q$. Prove that the circle $k$, the line $BC$ and the bisector of $\\angle BAC$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nSince $\\angle PBM = \\angle MED = \\angle BAP$ we have $PB^{2} = PM \\cdot PA$. Analogously $QC^{2} = QN \\cdot QA$. Since $BC = 2PQ$ and $P$ lies between $B$ and $Q$, there exists a point $L$ on $PQ$ such that $PB = PL$ and $QC = QL$. Thus, $PL^{2} = PM \\cdot PA$, i.e. $M$ lies on the circle $k'$ through $A$, tangent to $BC$ at $L$. Analogously $N \\in k'$, and therefore $k' = k$. Finally, we obtain that\n\n![](attached_image_1.png)\n\n$$\n\\frac{BL^{2}}{CL^{2}} = \\frac{BD \\cdot BA}{CE \\cdot CA} = \\frac{BA^{2}}{CA^{2}}, \\text{ i.e. } \\angle BAL = \\angle CAL\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27348,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nthe incircle $k$ of $\\triangle ABC$ is tangent to the sides $AB$, $BC$ and $CA$ at points $C_1$, $A_1$ and $B_1$, respectively. The points $C_2$, $A_2$ and $B_2$ are diametrically opposite to $C_1$, $A_1$ and $B_1$ in $k$.\n\na) Prove that the lines $AA_2$, $BB_2$ and $CC_2$ are concurrent.\n\nb) If the line $AA_2$ meets $k$ at $A_3$, find the ratio in which the tangent line to $k$ at $A_3$ divides $BC$.",
    "options": [],
    "answer": "1:1",
    "solution": "Solution:\n\na) Let $A_4 = AA_2 \\cap BC$ and let the tangent line to $k$ at $A_2$ meet $AB$ and $AC$ at points $X$ and $Y$, respectively. Since $A_2A_1$ is a diameter, we have $XY \\parallel BC$, i.e. $\\triangle AXY \\sim \\triangle ABC$.\n\nSince $k$ is an excircle of $\\triangle AXY$, it follows from above that $BA_4 = p - c$ and $CA_4 = p - b$ (we use the standard notation for $\\triangle ABC$) and therefore $\\frac{BA_4}{A_4C} = \\frac{p-c}{p-b}$. The corresponding equalities for the other two vertices and Ceva's theorem imply that the lines $AA_2$, $BB_2$ and $CC_2$ are concurrent.\n\nb) Denote by $Z$ the intersection point of the tangent line to $k$ at $A_3$ and $BC$. We have $\\angle A_1A_3A_4 = 90^\\circ$ and $ZA_3 = ZA_1$. It follows easily that $ZA_4 = ZA_1$ and therefore\n$$\nCZ = CA_1 + A_1Z = p-c + ZA_1 = BA_4 + ZA_4 = BZ\n$$\nHence the desired ratio equals $1:1$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27349,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $\\{a_{n}\\}_{n=1}^{\\infty}$ be a sequence of integers greater than $1$ and let $x>0$ be an irrational number. Denote by $x_{n}$ the fractional part of the product $a_{n} a_{n-1} \\ldots a_{1} x$\n\na) Prove that $x_{n} > \\frac{1}{a_{n+1}}$ for infinitely many $n$.\n\nb) Find all sequences $\\{a_{n}\\}_{n=1}^{\\infty}$ such that there exist infinitely many $x \\in (0,1)$ for which $x_{n} > \\frac{1}{a_{n+1}}$ for all $n$.",
    "options": [],
    "answer": "b) Exactly those sequences with all terms greater than one and with terms greater than two occurring infinitely often.",
    "solution": "Solution:\n\na) Suppose that the inequality $\\{a_{n} a_{n-1} \\ldots a_{1} x\\} > \\frac{1}{a_{n+1}}$ holds for finitely many values of $n$. Hence there exists $s$ such that for any $n \\geq s$ we have $\\{a_{n} a_{n-1} \\ldots a_{1} x\\} \\leq \\frac{1}{a_{n+1}}$. Since $\\{a_{n} a_{n-1} \\ldots a_{1} x\\}$ is not a rational number (in particular does not equal $0$) we obtain that $\\{a_{n} a_{n-1} \\ldots a_{1} x\\} < \\frac{1}{a_{n+1}}$, i.e. $a_{n+1} \\{a_{n} a_{n-1} \\ldots a_{1} x\\} < 1$. Using that $a_{n+1}$ is an integer we have\n$$\n\\{a_{n+1} a_{n} a_{n-1} \\ldots a_{1} x\\} = \\{a_{n+1} \\{a_{n} a_{n-1} \\ldots a_{1} x\\}\\} = a_{n+1} \\{a_{n} a_{n-1} \\ldots a_{1} x\\}\n$$\nFor any $t > s$ we obtain\n$$\n1 > \\{a_{t} a_{t-1} \\ldots a_{s} a_{s-1} \\ldots a_{1} x\\} = a_{t} a_{t-1} \\ldots a_{s} \\{a_{s-1} \\ldots a_{1} x\\}\n$$\na contradiction, since $\\lim_{t \\rightarrow \\infty} a_{t} a_{t-1} \\ldots a_{s} = \\infty$, but $0 < \\{a_{s-1} \\ldots a_{1} x\\} < 1$.\n\nb) It is clear that if $a_{i} = 1$ for some $i > 1$ then $\\{a_{i-1} a_{i-2} \\ldots a_{1} x\\} > \\frac{1}{a_{i}} = 1$ is not true. Suppose that there exists $t$ such that $a_{i} = 2$ for $i > t$. Then $\\{2^{p} y\\} > \\frac{1}{2}$ for $y = a_{t} a_{t-1} \\ldots a_{1} x$ and every $p$. Since $y < 1$ and $\\sum_{j=1}^{\\infty} \\frac{1}{2^{j}} = 1$ we conclude that for every $k$ the inequality $c_{k} \\leq y < c_{k+1}$ holds true, where $c_{k} = \\frac{1}{2} + \\frac{1}{2^{2}} + \\cdots + \\frac{1}{2^{k}}$. Therefore $2^{k} y \\in [2^{k} c_{k}, 2^{k} c_{k} + \\frac{1}{2})$, a contradiction to $\\{2^{k} y\\} > \\frac{1}{2}$.\n\nWe shall prove that if $\\{a_{n}\\}_{n=1}^{\\infty}$ is a sequence for which $a_{i} > 1$ for all $i > 1$ and the inequality $a_{i} > 2$ holds true for infinitely many values of $i$, then there exist infinitely many $x \\in (0,1)$ such that $x_{n} > \\frac{1}{a_{n+1}}$. Set\n$$\nx = \\frac{b_{1}}{a_{1}} + \\frac{b_{2}}{a_{1} a_{2}} + \\frac{b_{3}}{a_{1} a_{2} a_{3}} + \\cdots\n$$\nwhere $b_{1} \\leq a_{1} - 1$ and $1 \\leq b_{i} \\leq a_{i} - 1$ for $i > 1$ and infinitely many of the latter inequalities are strict. Then\n$$\n\\begin{aligned}\nx & = \\frac{b_{1}}{a_{1}} + \\frac{b_{2}}{a_{1} a_{2}} + \\frac{b_{3}}{a_{1} a_{2} a_{3}} + \\cdots \\\\\n& < \\frac{a_{1} - 1}{a_{1}} + \\frac{a_{2} - 1}{a_{1} a_{2}} + \\frac{a_{3} - 1}{a_{1} a_{2} a_{3}} + \\cdots \\\\\n& = 1 - \\frac{1}{a_{1}} + \\frac{1}{a_{1}} - \\frac{1}{a_{1} a_{2}} + \\frac{1}{a_{1} a_{2}} - \\cdots = 1\n\\end{aligned}\n$$\nThe numbers of this type are infinitely many and we have as above that\n$$\n\\frac{b_{n+1}}{a_{n+1}} + \\frac{b_{n+2}}{a_{n+1} a_{n+2}} + \\cdots < 1\n$$\nTherefore\n$$\nx_{n} = \\frac{b_{n+1}}{a_{n+1}} + \\frac{b_{n+2}}{a_{n+1} a_{n+2}} + \\cdots > \\frac{b_{n+1}}{a_{n+1}} \\geq \\frac{1}{a_{n+1}}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27350,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAfter a volleyball tournament (every two teams played exactly once) with $n$ teams it turned out that for any two teams $A$ and $B$, such that $B$ wins over $A$, there exist positive integer $t$ and teams $C_{1}, C_{2}, \\ldots, C_{t}$, such that $A$ wins over $C_{1}, C_{1}$ wins over $C_{2}, \\ldots, C_{t}$ wins over $B$.\nProve that for any $k=3,4, \\ldots, n$ there exist $k$ teams $A_{1}, A_{2}, \\ldots, A_{k}$, such that $A_{1}$ wins over $A_{2}, A_{2}$ wins over $A_{3}, \\ldots, A_{k-1}$ wins over $A_{k}$ and $A_{k}$ wins over $A_{1}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWe first show that there exist teams $A, B$ and $C$, such that $A$ wins over $B, B$ wins over $C$ and $C$ wins over $A$. Suppose the contrary and take the shortest cycle of $m \\geq 4$ teams $A, B, C_{1}, \\ldots, C_{t}$ (i.e. $t \\geq 2$), such that $A$ wins over $C_{1}, C_{1}$ wins over $C_{2}, \\ldots, C_{t}$ wins over $B$ and $B$ wins over $A$. Consider the game between $C_{2}$ and $A$. If $C_{2}$ is the winner then we have the desired triple and if $A$ is the winner then we have a shorter cycle.\n\nNow we shall use induction on $k$. The case $k=3$ was considered above. Suppose that the teams $A_{1}, A_{2}, \\ldots, A_{k}$ satisfy the condition of the problem for $3 \\leq k < n$. There are two cases to be considered.\n\nCase 1. There exists a team $U \\notin \\{A_{1}, A_{2}, \\ldots, A_{k}\\}$, for which there are two teams $A_{i}$ and $A_{j}$ such that $A_{i}$ wins over $U$ and $U$ wins over $A_{j}$. Without loss of generality assume that $A_{1}$ wins over $U$. Let $A_{\\ell}$ be the team of the least index that loses from $U$. Then the following $k+1$ teams\n$$\nA_{1}, \\ldots, A_{\\ell-1}, U, A_{\\ell}, \\ldots, A_{k}\n$$\nhave the desired property.\n\nCase 2. For any two teams $A_{i}$ and $A_{j}$ and any team $U$ either $A_{i}$ loses from $U$ or $A_{i}$ and $A_{j}$ both win over $U$.\n\nPartition all teams apart from $A_{1}, A_{2}, \\ldots, A_{k}$ in two sets $S$ and $T$, such that all teams from $S$ win over $A_{1}, A_{2}, \\ldots, A_{k}$, and all teams from $T$ lose from $A_{1}, A_{2}, \\ldots, A_{k}$. It is clear that $S \\cap T = \\varnothing$ and none of $S$ and $T$ is the empty set.\n\nLet $U \\in S$ and $V \\in T$ be such that $V$ wins over $U$ (such a pair exists due to the condition of the problem). Now the following $k+1$ teams\n$$\nU, A_{1}, \\ldots, A_{k-1}, V\n$$\nhave the desired property.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27351,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all integers $a, b, c, d$ such that $a c - 3 b d = 5$ and $a d + b c = 6$.",
    "options": [],
    "answer": "The solutions are (1,0,5,6), (-1,0,-5,-6), (5,6,1,0), (-5,-6,-1,0), (2,1,4,1), (-2,-1,-4,-1), (4,1,2,1), (-4,-1,-2,-1).",
    "solution": "Solution:\nLet $a c - 3 b d = 5$ and $a d + b c = 6$ for some integers $a, b, c, d$. Then\n$$\n(a^2 + 3 b^2)(c^2 + 3 d^2) = (a c - 3 b d)^2 + 3(a d + b c)^2 = 5^2 + 3 \\cdot 6^2 = 133 = 7 \\cdot 19\n$$\nBecause of the symmetry between the pairs $(a, b)$ and $(c, d)$ we consider only the cases when $a^2 + 3 b^2 = 1$ and $c^2 + 3 d^2 = 133$ or $a^2 + 3 b^2 = 7$ and $c^2 + 3 d^2 = 19$.\n\nIf $a^2 + 3 b^2 = 1$, then $a^2 = 1$, $b = 0$ and using $a c = 5$, $a d = 6$ we get $c = 5$, $d = 6$ for $a = 1$ and $c = -5$, $d = -6$ for $a = -1$.\n\nIf $a^2 + 3 b^2 = 7$, then $a^2 = 4$ and $b^2 = 1$. It follows by $c^2 + 3 d^2 = 19$ that $c^2 = 16$ and $d^2 = 1$. Then $|a c| = 8$ and the equality $a c - 3 b d = 5$ gives $a c = 8$. Therefore $a = 2$, $c = 4$ or $a = -2$, $c = -4$ and we get respectively $b = d = 1$ and $b = d = -1$.\n\nFinally, all solutions are: $(a, b, c, d) = (\\pm 1, 0, \\pm 5, \\pm 6), (\\pm 5, \\pm 6, \\pm 1, 0)$, $(\\pm 2, \\pm 1, \\pm 4, \\pm 1)$ and $(\\pm 4, \\pm 1, \\pm 2, \\pm 1)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27352,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $A$ and $B$ be given points on a circle $k$. For an arbitrary point $L$ on $k$ denote by $M$ the point on the line $A L$ such that $L M = L B$ and $L$ is between $A$ and $M$. Find the locus of the points $M$.",
    "options": [],
    "answer": "Let α be half of the inscribed angle under which the segment AB is seen from points on the arc of the given circle containing the points between A and the diameter perpendicular to AB. Then the locus of M is the union of two arcs: the arc of the circle from which AB is seen at angle α and the arc of the circle from which AB is seen at angle 90° − α. Among these two circles, only the portions lying in the half-plane bounded by the tangent to the given circle at A that contains the given circle are included, because L must lie between A and M.",
    "solution": "Solution:\nLet $C D$ be the diameter of $k$ such that $C D \\perp A B$ and let $L \\in A \\widehat{C} B$. Then $\\angle A L B = 2 \\alpha$ is constant and we have $\\angle A M B = \\alpha$ since $\\triangle M L B$ is isosceles. Therefore $M$ belongs to an arc of the circle $k_{1}$, from which the segment $A B$ is seen by the angle $\\alpha$.\n\n![](attached_image_1.png)\n\nAnalogously, for $L \\in A \\widehat{D} B$ we have $\\angle A L B = 180^{\\circ} - 2 \\alpha$, $\\angle A M B = 90^{\\circ} - \\alpha$ and we conclude that $M$ belongs to an arc of the circle $k_{2}$, from which the segment $A B$ is seen by the angle $90^{\\circ} - \\alpha$. Let $t$ be the tangent line to $k$ at the point $A$ and $t \\cap k_{1} = \\{A, P\\}$, $t \\cap k_{2} = \\{A, Q\\}$. Let $\\lambda$ be the half-plane with respect to $t$, containing $k$. Since $L$ is between $A$ and $M$ we have $M \\in \\lambda$. Therefore the required locus consists of the arcs of $k_{1}$ and $k_{2}$ belonging to $\\lambda$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27353,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $m$ be a positive integer and $u_{m} = \\underbrace{11 \\ldots 1}_{m}$. Prove that there is no positive integer multiple of $u_{m}$ such that the sum of its digits is less than $m$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nAssume the contrary. Then there exists a positive integer multiple of $u_{m}$ such that the sum of its digits is less than $m$ and let $t$ be the smallest number with this property. Since $t > 10^{m}$, the number $t$ can be written as $t = 10^{m} a + b$, where $0 < b < 10^{m}$.\n\nWe have $t = 10^{m} a + b = (10^{m} - 1) a + a + b$. Since $u_{m}$ divides both $t$ and $10^{m} - 1 = 9 u_{m}$, we conclude that $u_{m}$ divides $a + b$. But the sum of the digits of $a + b$ does not exceed the sum of the digits of $t$ and $a + b < t$, a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27354,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nEach side of a sheet of paper is a map of 5 countries. The countries on one of the maps are colored in 5 different colors. Prove that it is possible to color the countries on the other map in such a way that every two are colored in different colors and at least $20\\%$ of the sheet is colored in the same color on both sides.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nDenote by $A_{1}, A_{2}, \\ldots, A_{5}$ and $B_{1}, B_{2}, \\ldots, B_{5}$ the countries on the respective sides of the sheet of paper. Let $S_{ij}$ be the area of the part of $A_{i}$ which belongs to the country $B_{j}$ on the other side of the sheet. (If $A_{i}$ and $B_{j}$ do not have a common area, then $S_{ij}=0$.) Then, setting the area of the sheet to be $1$, we have\n$$\n\\begin{aligned}\nS= & \\left(S_{11}+S_{12}+S_{13}+S_{14}+S_{15}\\right)+\\left(S_{21}+S_{22}+S_{23}+S_{24}+S_{25}\\right)+\\cdots \\\\\n& +\\left(S_{51}+S_{52}+S_{53}+S_{54}+S_{55}\\right)=1\n\\end{aligned}\n$$\nsince $S_{i1}+S_{i2}+S_{i3}+S_{i4}+S_{i5}$ equals the area of $A_{i}$.\n\nThe sum $S$ can be written also as follows:\n$$\n\\begin{aligned}\nS= & \\left(S_{11}+S_{22}+S_{33}+S_{44}+S_{55}\\right)+\\left(S_{12}+S_{23}+S_{34}+S_{45}+S_{51}\\right)+\\cdots \\\\\n& +\\left(S_{15}+S_{21}+S_{32}+S_{43}+S_{54}\\right)\n\\end{aligned}\n$$\nHence at least one of the summands is greater than or equal to $0.2$ and let us assume that $S_{13}+S_{24}+S_{35}+S_{41}+S_{52} \\geq 0.2$. We now color the countries $B_{1}, B_{2}, \\ldots, B_{5}$ as follows: $B_{3}$ by the color of $A_{1}$, $B_{4}$ by the color of $A_{2}$, $B_{5}$ by the color of $A_{3}$, $B_{1}$ by the color of $A_{4}$ and $B_{2}$ by the color of $A_{5}$. Then every two countries are colored in different colors and at least $20\\%$ of the sheet is colored in the same color on both sides.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27355,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $n \\geq 3$ be a positive integer and $M$ be the set of the first $n$ prime numbers. For every nonempty subset $X$ of $M$ denote by $P(X)$ the product of the elements of $X$. Let $N$ be a set of fractions of the form $\\frac{P(A)}{P(B)}$, where $A \\subset M$, $B \\subset M$, $A \\cap B = \\varnothing$ such that the product of any 7 elements of $N$ is an integer. What is the maximum possible cardinality of $N$?",
    "options": [],
    "answer": "Maximum size equals 3n−2 for n in {3, 4, 5, 6}, and equals 3n−3 for all n at least 7.",
    "solution": "Solution:\nConsider the following three element sets\n$$\n\\begin{aligned}\nN_{1} & = \\left\\{ \\frac{p_{3} p_{4} \\ldots p_{n-1}}{p_{1}}, \\frac{p_{2} p_{3} \\ldots p_{n-1}}{p_{1}}, \\frac{p_{2} p_{3} \\ldots p_{n-1} p_{n}}{p_{1}} \\right\\} \\\\\nN_{2} & = \\left\\{ \\frac{p_{1} p_{4} \\ldots p_{n-1}}{p_{2}}, \\frac{p_{1} p_{3} \\ldots p_{n-1}}{p_{2}}, \\frac{p_{1} p_{3} \\ldots p_{n-1} p_{n}}{p_{2}} \\right\\} \\\\\n& \\vdots \\\\\nN_{n-1} & = \\left\\{ \\frac{p_{2} p_{3} \\ldots p_{n-2}}{p_{n-1}}, \\frac{p_{1} p_{2} \\ldots p_{n-2}}{p_{n-1}}, \\frac{p_{1} p_{2} \\ldots p_{n-2} p_{n}}{p_{n-1}} \\right\\}\n\\end{aligned}\n$$\nIt is easy to check that the union of these sets satisfies the condition of the problem and has $3n-3$ elements.\n\nSuppose that there exists a set having $3n-2$ elements that satisfies the condition of the problem. It is clear that every prime number appears in denominator at most 3 times.\n\nLet $M_{1} \\subset M$ be the largest possible set of prime numbers which appear exactly three times in denominator and no two prime numbers from $M_{1}$ appear in one and the same denominator. Then the following inequality holds true\n$$\n3|M_{1}| + 2(n - |M_{1}|) \\geq 3n - 2\n$$\ni.e. $|M_{1}| \\geq n-2$.\n\nCase 1. Let $|M_{1}| = n$. Then $N$ is a union of sets of the type $N_{i} = \\left\\{ \\frac{a_{i}}{p_{i}}, \\frac{b_{i}}{p_{i}}, \\frac{c_{i}}{p_{i}} \\right\\}$ and it follows that $p_{i}$ does not appear in at most one of the numerators of the remaining fractions. In every $N_{i}$ there are at least two fractions whose numerators do not contain one $p_{j}$; otherwise there are two equal fractions. Therefore $2n \\leq n$, a contradiction.\n\nCase 2. Let $|M_{1}| = n-1$. There are two cases:\n\na) The set $N$ is a union of $n-1$ sets of the form $N_{i} = \\left\\{ \\frac{a_{i}}{x p_{i}}, \\frac{b_{i}}{p_{i}}, \\frac{c_{i}}{p_{i}} \\right\\}$, where $x = 1$ or $x = p_{n}$ and $N_{n} = \\left\\{ \\frac{r}{p_{n}}, \\frac{s}{p_{n}} \\right\\}$. If a denominator of a fraction not in $N_{n}$ is divisible by $p_{n}$ then we have as above that $2(n-1) + 1 \\leq n$, i.e. $n \\leq 1$, a contradiction. Otherwise the denominators of at most three fractions are divisible by $p_{n}$. Therefore $2(n-1) + 1 \\leq n-1 + 3$ implying $n \\leq 3$. It is easy to be seen that $n=3$ gives no solution.\n\nb) Consider the set $N_{n} = \\left\\{ \\frac{t}{p_{n}} \\right\\}$. Then observations analogous to those in a) imply $2(n-1) \\leq n-1 + 5$, i.e. $n \\leq 6$. For every $n = 3,4,5,6$ it is easy to find an example of sets with $3n-2$ elements.\n\nCase 3. Let $|M_{1}| = n-2$. To find a better $N$ one should have $n-2$ sets obtained by the primes from $M_{1}$ and two sets with two elements each obtained from $p_{n-1}$ and $p_{n}$. As above we obtain $2(n-2) + 2 \\leq (n-2) + 3 + 3$, implying $n \\leq 6$. Hence in this case we do not find a better $N$.\n\nWhen $n = 3,4,5,6$ the answer is $3n-2$, and for $n \\geq 7$ the answer is $3n-3$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27356,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all real numbers $a$ such that the roots $x_{1}$ and $x_{2}$ of the equation\n$$\nx^{2}+6x+6a-a^{2}=0\n$$\nsatisfy the relation $x_{2}=x_{1}^{3}-8x_{1}$.",
    "options": [],
    "answer": "[-2, -1, 3, 7, 8]",
    "solution": "Solution:\nIt follows from Vieta's formulae that\n$$\n-6 = x_{1} + x_{2} = x_{1} + x_{1}^{3} - 8x_{1}\n$$\nTherefore $x_{1}^{3} - 7x_{1} + 6 = 0$ and $x_{1} = -3, 1$ or $2$.\n\nPlugging these values of $x_{1}$ in the initial equation gives $a = 3$ for $x = -3$, $a = -1$ and $7$ for $x = 1$, and $a = -2$ and $8$ for $x = 2$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27357,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all real numbers $a$ for which the equation $x^{2}+a x+3 a^{2}-7 a-19=0$ has real roots $x_{1}$ and $x_{2}$ such that\n$$\n\\frac{1}{x_{1}-2}+\\frac{1}{x_{2}-2}=-\\frac{2 a}{13}\n$$",
    "options": [],
    "answer": "4",
    "solution": "Solution:\nUsing Vieta's formulas we get\n$$\n\\frac{1}{x_{1}-2}+\\frac{1}{x_{2}-2}=\\frac{x_{1}+x_{2}-4}{\\left(x_{1}-2\\right)\\left(x_{2}-2\\right)}=-\\frac{a+4}{3 a^{2}-5 a-15}\n$$\nTherefore $3 a^{2}-5 a-15 \\neq 0$ and\n$$\n\\frac{a+4}{3 a^{2}-5 a-15}=\\frac{2 a}{13}\n$$\nHence $6 a^{3}-10 a^{2}-43 a-52=0 \\Longleftrightarrow (a-4)\\left(6 a^{2}+14 a+13\\right)=0$, i.e. $a=4$. In this case $x_{1,2}=-2 \\pm \\sqrt{15}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27358,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTwo circles $k_{1}$ and $k_{2}$ meet at points $A$ and $B$. A line through $B$ meets the circles $k_{1}$ and $k_{2}$ at points $X$ and $Y$, respectively. The tangent lines to $k_{1}$ at $X$ and to $k_{2}$ at $Y$ meet at $C$. Prove that:\n\na) $\\Varangle X A C = \\Varangle B A Y$.\n\nb) $\\Varangle X B A = \\Varangle X B C$, if $B$ is the midpoint of $X Y$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na) The quadrilateral $X C Y A$ is cyclic since\n$$\n\\begin{aligned}\n\\Varangle X C Y & = 180^{\\circ} - \\Varangle C X Y - \\Varangle C Y X = 180^{\\circ} - \\Varangle X A B - \\Varangle B A Y \\\\\n& = 180^{\\circ} - \\Varangle X A Y\n\\end{aligned}\n$$\nTherefore $\\Varangle X A C = \\Varangle X Y C = \\Varangle B A Y$.\n\nb) It follows from a) that $\\triangle X A C \\sim \\triangle B A Y$. Therefore $\\frac{X C}{A C} = \\frac{B Y}{A Y}$ and since $X B = B Y$ we have $\\frac{X C}{X B} = \\frac{A C}{A Y}$. Moreover $\\Varangle C X Y = \\Varangle C A Y$, implying $\\triangle X C B \\sim \\triangle A C Y$. Thus, $\\Varangle X B C = \\Varangle A Y C = \\Varangle X B A$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27359,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn an acute $\\triangle ABC$ the altitudes $AA_1$ ($A_1 \\in BC$) and $BB_1$ ($B_1 \\in AC$) are drawn, $I$ is the incenter and the line $CI$ meets $AB$ at $L$. It is known that $I$ lies on the circumcircle of $\\triangle A_1B_1C$.\n\na) Prove that $L$ is the center of excircle of $\\triangle A_1B_1C$ tangent to the side $A_1B_1$.\n\nb) If $CI = 2IL$, find $\\Varangle ACB$.",
    "options": [],
    "answer": "60 degrees",
    "solution": "Solution:\n\na) Since $\\Varangle CB_1A_1 = \\Varangle CIA_1$ and the quadrilateral $ABA_1B_1$ is cyclic, we have $\\Varangle CIA_1 = \\Varangle CB_1A_1 = \\Varangle ABC$. This shows that the quadrilateral $LBA_1I$ is cyclic. Then\n$$\n\\begin{aligned}\n\\Varangle LA_1B &= \\Varangle LIB = \\Varangle ICB + \\Varangle IBC \\\\\n&= \\frac{1}{2}(\\Varangle ACB + \\Varangle ABC) \\\\\n&= \\frac{1}{2}\\left(\\Varangle ACB + \\Varangle A_1B_1C\\right) \\\\\n&= \\frac{1}{2} \\Varangle BA_1B_1\n\\end{aligned}\n$$\n![](attached_image_1.png)\nHence $A_1L$ is the bisector of $\\Varangle BA_1B_1$, which completes the proof.\n\nb) If $J$ is the midpoint of $CI$, then we have $CJ = JI = IL$. But $IL = IA_1$ from a) and we conclude that $JLA_1$ is a right triangle. Then $A_1J$ is the bisector of $\\Varangle B_1A_1C$ and therefore $J$ is the incenter of $\\triangle A_1B_1C$. On the other hand, we have $\\triangle A_1B_1C \\sim \\triangle ABC$, whence $\\frac{A_1C}{AC} = \\frac{CJ}{CI} = \\frac{1}{2}$ and $\\Varangle ACB = 60^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27360,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe sets $M=\\{1,2, \\ldots, 27\\}$ and $A=\\{a_{1}, a_{2}, \\ldots, a_{k}\\} \\subset \\{1,2, \\ldots, 14\\}$ have the following property: every element of $M$ is either an element of $A$ or the sum of two (possibly identical) elements of $A$. Find the minimum value of $k$.",
    "options": [],
    "answer": "8",
    "solution": "Solution:\n\nThe elements of $A$ give $\\binom{k}{2}+k+k=\\frac{k(k+3)}{2}$ sums of the required kind (with two different summands, two equal summands or one summand, respectively). Therefore $\\frac{k(k+3)}{2} \\geq 27$, whence $k \\geq 6$.\n\nIf $k=6$ every element of $M$ has a unique representation. This consecutively implies $1 \\in A, 2 \\notin A, 3 \\in A, 4 \\notin A, 5 \\in A$ and now $6=3+3=1+5$ has two representations, a contradiction. Therefore $k \\geq 7$.\n\nAssume that $A=\\{a_{1}, a_{2}, \\ldots, a_{7}\\}$ has the required property and $a_{1}<a_{2}<\\cdots<a_{7}$. Since $1$ and $27$ have unique representations, we have $a_{1}=1, a_{6}=13$ and $a_{7}=14$. Also, it is easy to see that $a_{2} \\in\\{2,3\\}$ and $a_{3} \\leq 5$. The only two possible representations of $25$ are $25=13+12=14+11$ and therefore $a_{5}=12$ or $a_{5}=11$.\n\nCase 1. $a_{5}=12$. It follows from $23=14+9=13+10=12+11$ that $a_{4} \\in\\{9,10,11\\}$. Then checking all possibilities for $21$ we see that $a_{3} \\geq 7$, which contradicts to the above restriction $a_{3} \\leq 5$.\n\nCase 2. $a_{5}=11$. As in case 1 we conclude that $a_{4} \\in\\{9,10\\}$. On the other hand by $21=14+7=13+8=11+10$ we have $a_{4} \\in\\{7,8,10\\}$ and therefore $a_{4}=10$. Now using $a_{3} \\leq 5$ we see that the only possibility for $19$ is $19=14+5$. This implies that $a_{3}=5$ and $a_{2}=3$, i.e. $A=\\{1,3,5,10,11,13,14\\}$. But this set does not have the required property.\n\nThe set $A=\\{1,3,5,7,9,11,13,14\\}$ has cardinality $8$ and possesses the required property. Therefore the minimum value of $k$ is $8$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27361,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe positive integers $l, m, n$ are such that $m-n$ is a prime number and $8\\left(l^{2}-m n\\right)=2\\left(m^{2}+n^{2}\\right)+5(m+n) l$. Prove that $11 l+3$ is a perfect square.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSetting $p = m - n$ and $q = m + n$ gives\n$$\nm n = \\frac{1}{4}\\left(q^{2} - p^{2}\\right) \\quad \\text{and} \\quad m^{2} + n^{2} = \\frac{1}{2}\\left(q^{2} + p^{2}\\right)\n$$\nHence the given conditions can be written as\n$$\n8 l^{2} - 2 q^{2} + 2 p^{2} = q^{2} + p^{2} + 5 q l\n$$\ni.e.\n$$\np^{2} = (3 q + 8 l)(q - l)\n$$\nSince $p$ is a prime number and $3 q + 8 l > q - l$ we obtain $p^{2} = 3 q + 8 l$ and $1 = q - l$. Hence $11 l + 3 = p^{2}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27362,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all integers $a$ such that the equation\n$$\nx^{4} + 2x^{3} + (a^{2} - a - 9)x^{2} - 4x + 4 = 0\n$$\nhas at least one real root.",
    "options": [],
    "answer": "a = -2, -1, 0, 1, 2, 3",
    "solution": "Solution:\nSet $u = x - \\frac{2}{x}$. Then the equation becomes\n$$\nu^{2} + 2u + a^{2} - a - 5 = 0$$\nSince the equation $x^{2} - u x - 2 = 0$ has real solutions for any real $u$, it suffices to find the integer values of $a$ for which the equation (1) has a real root. The last holds when $D = -a^{2} + a + 6 \\geq 0$, i.e. $(a - 3)(a + 2) \\leq 0$ giving $a \\in [-2, 3]$. Therefore $a = -2, -1, 0, 1, 2, 3$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27363,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFor any positive integer $n$ denote by $f(n)$ the smallest positive integer $m$ such that the sum $1+2+\\cdots+m$ is divisible by $n$. Find all $n$ such that $f(n)=n-1$.",
    "options": [],
    "answer": "All n of the form p^s where p is an odd prime and s is a positive integer.",
    "solution": "Solution:\nIf $f(n)=n-1$, then $n$ divides the sum $1+2+\\cdots+(n-1)=\\frac{(n-1)n}{2}$, which implies that $n$ is odd.\n\nThe numbers $n=p^{s}$, where $p>2$ is a prime number and $s \\geq 1$, are solutions. Indeed, if $k \\in \\mathbb{N}$ and $k<p^{s}-1$, then the sum $1+2+\\cdots+k=\\frac{k(k+1)}{2}$ is not divisible by $p^{s}$ because $k$ and $k+1$ are coprime and less than $p^{s}$.\n\nWe shall prove that these are the only solutions. Let $n$ be an odd positive integer which is not a power of a prime number. Then $n=a b$, where $a>1$, $b>1$ and $(a, b)=1$. By the Chinese Remainder Theorem we conclude that there exists an integer $k \\in[0, a b-1]$ such that $a \\mid k$ and $b \\mid k+1$. It is clear that $k \\neq 0, k \\neq a b-1$ and $a b \\left\\lvert\\, \\frac{k(k+1)}{2}\\right.$. Therefore $n=a b$ is not a solution.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27364,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGiven a right triangle $ABC$ ($\\angle ACB = 90^\\circ$), let $CH$, $H \\in AB$, be the altitude to $AB$ and $P$ and $Q$ be the tangent points of the incircle of $\\triangle ABC$ to $AC$ and $BC$, respectively. If $AQ \\perp HP$ find the ratio $\\frac{AH}{BH}$.",
    "options": [],
    "answer": "(1 + sqrt(5))/2",
    "solution": "Solution:\n\nIt follows from $AQ \\perp HP$ that $\\angle QAB = \\angle PHC$. On the other hand $\\angle ABC = \\angle ACH$ and therefore $\\triangle ABQ \\sim \\triangle HCP$. Thus, $\\frac{AB}{BQ} = \\frac{HC}{CP}$. Using the standard notation for the elements of a triangle we obtain the following equalities:\n\n![](attached_image_1.png)\n\n$$\n\\begin{aligned}\n\\frac{c}{p-b} = \\frac{h}{r} &\\Leftrightarrow \\frac{c}{p-b} = \\frac{2S/c}{S/p} \\Leftrightarrow \\frac{c}{p-b} = \\frac{2p}{c} \\\\\n&\\Leftrightarrow \\frac{2c}{a+c-b} = \\frac{a+b+c}{c} \\Leftrightarrow 2c^2 = (a+c)^2 - b^2 \\\\\n&\\Leftrightarrow c^2 = a^2 + 2ac - b^2 \\Leftrightarrow b^2 = ac\n\\end{aligned}\n$$\n\nsince $c^2 = a^2 + b^2$. Hence\n\n$$\nb^4 = a^2(a^2 + b^2) \\Longleftrightarrow b^4 - a^2 b^2 - a^4 = 0\n$$\n\nSet $k = \\frac{AH}{BH} = \\frac{b^2 / c}{a^2 / c} = \\frac{b^2}{a^2}$. Then $k^2 - k - 1 = 0$ and we get $k = \\frac{1 + \\sqrt{5}}{2}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27365,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAn air company operates 36 airlines in a country with 16 airports. Prove that one can make a round trip that includes 4 airports.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nConsider a graph $G$ whose vertices are the airports in the country. Two vertices form an edge if there is an airline between the corresponding airports. Suppose that a round trip satisfying the conditions of the problem does not exist, i.e. there is no cycle of length 4 in $G$.\n\nIf $x$ is a vertex of $G$ denote by $d(x)$ the number of neighbors of $x$. Then the number of pairs both elements of which are neighbors of $x$ equals $\\binom{d(x)}{2}$. Note that every pair is counted from at most one vertex $z$, since otherwise there is a cycle of length 4.\n\nUsing the identity $\\sum_{x \\in G} d(x) = 72$ and the Root mean square - Arithmetic mean inequality we have\n\n$$\n\\begin{aligned}\n\\binom{16}{2} = 120 & \\geq \\sum_{x \\in G} \\binom{d(x)}{2} = \\sum_{x \\in G} \\frac{d^2(x)}{2} - \\sum_{x \\in G} \\frac{d(x)}{2} \\\\\n& \\geq \\frac{1}{32}\\left(\\sum_{x \\in G} d(x)\\right)^2 - \\sum_{x \\in G} \\frac{d(x)}{2} \\\\\n& = \\frac{72^2}{32} - 36 = 126\n\\end{aligned}\n$$\na contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27366,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two circles $k_1$ and $k_2$ with centers $O_1$ and $O_2$, respectively, are externally tangent at point $P$. A circle $k_3$ is externally tangent to $k_1$ at $Q$ and to $k_2$ at $R$. The lines $PQ$ and $PR$ meet $k_3$ at points $A$ and $B$, respectively. If $AO_2$ meets $BO_1$ at a point $S$ prove that $SP \\perp O_1O_2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27367,
    "subject": "Mathematics (Multi-modal)",
    "question": "A quadrilateral $ABCD$ with $\\angle BAD + \\angle ADC > 180^{\\circ}$ is circumscribed around a circle of center $I$. A line through $I$ meets $AB$ and $CD$ at points $X$ and $Y$, respectively. Prove that if $IX = IY$ then $AX \\cdot DY = BX \\cdot CY$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote by $M$ and $N$ the tangent points of the incircle of $ABCD$ with $AB$ and $CD$, respectively. It follows from $\\angle BAD + \\angle ADC > 180^{\\circ}$ that $AB \\parallel CD$ and $\\angle MIN < 180^{\\circ}$. Also, the equalities $IM = IN$, $\\angle IMX = \\angle INY$ and $IX = IY$ show that $\\triangle IMX \\cong \\triangle INY$, implying $\\angle IYN = \\angle IXM$. If $X \\in BM$ and $Y \\in DN$ (or $X \\in AM$ and $Y \\in CN$) then the equality $\\angle IYN = \\angle IXM$ implies $AB \\parallel CD$, a contradiction. Therefore $X \\in XB$ and $Y \\in NC$. It follows from $AXYD$ that\n$$\n\\angle AXI = \\angle DYI = 180^{\\circ} - \\frac{\\angle A}{2} - \\frac{\\angle D}{2},\n$$\ngiving $\\angle AIX = \\frac{\\angle D}{2}$ and $\\angle DIY = \\frac{\\angle A}{2}$. Therefore $\\triangle AIX \\sim \\triangle IDY$, which implies that $AX \\cdot DY = IY \\cdot IX$. Analogously, $\\triangle BIX \\sim \\triangle ICY$, i.e. $BX \\cdot CY = IY \\cdot IX$. Hence $AX \\cdot DY = BX \\cdot CY$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27368,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sequence $\\{a_i\\}_{i=1}^\\infty$ is such that $a_1 > 0$ and $a_{n+1} = \\frac{a_n}{1+a_n^2}$ for $n \\ge 1$. Prove that:\n\na) $a_n \\le \\frac{1}{\\sqrt{2n}}$ for $n \\ge 2$;\n\nb) there exists $n$ such that $a_n > \\frac{7}{10\\sqrt{n}}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27369,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve the system\n$$\n\\left|\n\\begin{array}{l}\n\\sqrt{x^2 + y^2 - 16(x + y) - 9y + 7} = y - 2 \\\\\nx + 13\\sqrt[4]{x - y} = y + 42\n\\end{array}\n\\right.\n$$",
    "options": [],
    "answer": "x = (37 + sqrt(13)) / 2, y = (5 + sqrt(13)) / 2",
    "solution": "Writing the second equation in the form\n$x - y + 13\\sqrt[4]{x - y} - 42 = 0$\nand setting $u = \\sqrt[4]{x-y}$, $u \\ge 0$, we get that $u^4 + 13u - 42 = 0$. Note that this equation has only one positive root $u = 2$ and hence $x = y + 16$. Then the first equation easily gives\n$y^2 - 5y + 3 = 0.$\nThe last equation has roots $y_{1,2} = \\frac{5 \\pm \\sqrt{13}}{2}$. Since $y - 2 \\ge 0$ and $\\frac{5 - \\sqrt{13}}{2} < 2$, it follows that\n$$\nx = \\frac{37 + \\sqrt{13}}{2}, \\quad y = \\frac{5 + \\sqrt{13}}{2}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27370,
    "subject": "Mathematics (Multi-modal)",
    "question": "The functions $f(x) = 2x^2 + 2x - 4$ and $g(x) = x^2 - x + 2$ are given. Find all real values of $x$ such that:\n\na) $\\frac{f(x)}{g(x)}$ is a positive integer;\n\nb) the inequality $\\sqrt{f(x)} + \\sqrt{g(x)} \\ge \\sqrt{2}$ holds.",
    "options": [],
    "answer": "a) x = (-3 + sqrt(33)) / 2, (-3 - sqrt(33)) / 2, 2. b) x ∈ (-∞, -2] ∪ [1, ∞).",
    "solution": "a) *Hint.* Set $\\frac{f(x)}{g(x)} = k$, where $k$ is a positive integer. Then $(2-k)x^2 + (2+k)x - 2(2+k) = 0$ and use the fact that the discriminant of this quadratic equation is nonnegative.\n\n*Answer.* $x = \\frac{-3+\\sqrt{33}}{2}, \\frac{-3-\\sqrt{33}}{2}, 2$.\n\nb) *Answer.* $x \\in (-\\infty, -2] \\cup [1, +\\infty)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27371,
    "subject": "Mathematics (Multi-modal)",
    "question": "(Stoyan Boev) The incircle of an acute $\\triangle ABC$ touches the sides $AB$, $BC$ and $CA$ at points $P$, $Q$ and $R$, respectively. The orthocenter $H$ of $\\triangle ABC$ lies on the segment $QR$.\na) Prove that $PH \\perp QR$.\nb) Let $I$ and $O$ be the incenter and circumcenter of $\\triangle ABC$, and $N$ the common point of $AB$ and the excircle to this side. Prove that the points $I$, $O$ and $N$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Since $\\angle RAH = \\angle QBH$ and\n$$\n\\angle ARH = 180^{\\circ} - \\angle CRQ = 180^{\\circ} - \\angle CQR = \\angle BQH,\n$$\nthen $\\triangle ARH \\sim \\triangle QBH$. Hence\n$$\n\\frac{AH}{BH} = \\frac{AR}{BQ} = \\frac{AP}{BP}\n$$\nand $HP$ is the bisector of $\\angle AHB$. Then\n$$\n\\angle RHP = \\angle RHA + \\angle AHP = \\angle QHB + \\angle BHP = \\angle QHP\n$$\nwhich implies that $PH \\perp RQ$.\n\nb) If $AC = AB$, then $N \\equiv P \\equiv M$ and the points $I, O$ and $N$ lie on the bisector of $AB$.\nLet now $AC \\neq AB$. Since $PH \\perp RQ$ and $CI \\perp RQ$, it follows that $HP \\parallel CI$. On the other hand, $CH \\parallel IP$ and hence $CHPI$ is a parallelogram. Then $CH = IP$. If $M$ is the midpoint of $AB$, then $CH = 2OM = 2R \\cos \\gamma$ and so $IP = 2OM$. Since $AP = BN$, then $M$ is the midpoint of $PN$. Therefore $O$ is the midpoint of $IN$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27372,
    "subject": "Mathematics (Multi-modal)",
    "question": "In acute $\\triangle ABC$ denote by $M$ and $N$ the midpoints of the altitudes $BB_1$ and $CC_1$, respectively, $P = AM \\cap CC_1$ and $Q = AN \\cap BB_1$. Prove that:\n\na) the points $M, N, P$ and $Q$ are concyclic;\n\nb) if the points $B, C, P$ and $Q$ are concyclic then $\\triangle ABC$ is isosceles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Since $\\triangle ACC_1 \\sim \\triangle ABB_1$ and $AN$ and $AM$ are medians in these triangles we have\n$$\n\\asymp ANC_1 = \\asymp AMB_1 \\Rightarrow \\asymp QNB = \\asymp PMQ,\n$$\ni.e. the points $M, N, P$ and $Q$ are concyclic.\n\nb) If the points $B, C, P$ and $Q$ are concyclic then $\\asymp QCP = \\asymp QBP$. But $\\asymp ACC_1 = \\asymp ABB_1$ and hence\n$$\n\\asymp QCA = \\asymp PBA. \\qquad (1)\n$$\nOn the other hand, since $\\triangle ACC_1$ and $\\triangle ABB_1$ are similar we have\n$$\n\\asymp CAQ = \\asymp CAN = \\asymp BAM = \\asymp BAP. \\qquad (2)\n$$\nNow (1) and (2) imply that $\\triangle ACQ \\cong \\triangle ABP$, whence\n$$\n\\frac{AC}{AB} = \\frac{AQ}{AP} = \\frac{AM}{AN} = \\frac{AB}{AC}.\n$$\nThus $AB^2 = AC^2$ and $AB = AC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27373,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $x$ and $y$ such that $xy^2 + 2y$ divides $2x^2y + xy^2 + 8x$.",
    "options": [],
    "answer": "All solutions are (x, y) = (a, 2a) for any positive integer a, and (x, y) = (3, 1), (8, 1).",
    "solution": "Since $xy^2 + 2y$ divides\n$$(2x + y)(xy^2 + 2y) - y(2x^2y + xy^2 + 8x) = 2y^2 - 4xy,$$\nwe conclude that $xy + 2$ divides $2y - 4x$. We consider two cases.\n\nCase 1. Let $2y - 4x \\ge 0$. Then we have two possibilities:\n\n1.1) If $x \\ge 2$ then $xy + 2 > 2y - 4x$. Hence $2y - 4x = 0$, i.e. $x = a$ and $y = 2a$, where $a \\ge 2$ is an integer. Since $xy^2 + 2y = 4a(a^2 + 1)$ divides $2x^2y + xy^2 + 8x = 8a(a^2 + 1)$, this gives a solution.\n\n1.2) If $x = 1$ then $y^2 + 2y$ divides $8$, i.e. $y = 2$.\n\nHence in this case the solutions are $(a, 2a)$, where $a$ is a positive integer.\n\nCase 2. Let $2y - 4x < 0$, i.e. $4x - 2y > 0$. If $y \\ge 4$, then $xy + 2 > 4x - 2$. Therefore $y = 1, 2$ or $3$.\n\n2.1) If $y = 1$ then $\\frac{2x^2 + 9x}{x + 2} = 2x + 5 - \\frac{10}{x + 2}$ is an integer. Hence $x + 2$ divides $10$ and this gives the solutions $x = 3, y = 1$ and $x = 8, y = 1$.\n\n2.2) If $y = 2$ then $\\frac{x^2 + 3x}{x + 1} = x + 2 - \\frac{2}{x + 1}$ is an integer which gives $x = 1$.\n\n2.3) If $y = 3$ then $\\frac{6x^2 + 17x}{9x + 6}$ is an integer. This implies that $3|x$, i.e. $x = 3k$ for some positive integer $k$. After simplifications we obtain that $\\frac{18k^2 + 17k}{9k + 2} = (2k + 1) + \\frac{4k - 2}{9k + 2}$ is an integer, which is impossible for $k \\ge 1$.\n\nFinally, the solutions are $x = a, y = 2a$ for all positive integers $a$ and $x = 3, y = 1, x = 8, y = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27374,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that if $x, y, z > 0$, then\n$$\n\\frac{(x+1)(y+1)^2}{3\\sqrt[3]{z^2x^2}+1} + \\frac{(y+1)(z+1)^2}{3\\sqrt[3]{x^2y^2}+1} + \\frac{(z+1)(x+1)^2}{3\\sqrt[3]{y^2z^2}+1} \\geq x+y+z+3.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote by $S$ the left-hand side of the given inequality. The arithmetic mean–geometric mean inequality implies that $xy + x + y \\geq 3\\sqrt[3]{x^2y^2}$ and hence\n$$\nS \\geq \\frac{(x+1)(y+1)^2}{(z+1)(x+1)} + \\frac{(y+1)(z+1)^2}{(x+1)(y+1)} + \\frac{(z+1)(x+1)^2}{(y+1)(z+1)}.\n$$\nSetting $a = x + 1$, $b = y + 1$, $c = z + 1$ gives\n$$\nS \\geq \\frac{b^2}{c} + \\frac{c^2}{a} + \\frac{a^2}{b}.\n$$\nThen\n$$\nS \\geq \\frac{(a+b+c)^2}{a+b+c} = a+b+c = x+y+z+3.\n$$\nby the Cauchy–Schwartz inequality.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27375,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the minimum possible number of the edges of a graph with $n$ vertices having the following property:\n\na) If we draw an arbitrary new edge then a new triangle (3-clique) appears.\n\nb) If we draw an arbitrary new edge then a new 4-clique appears.",
    "options": [],
    "answer": "a) n - 1; b) 2n - 3",
    "solution": "a) Let $G$ be a graph with the required property and minimum possible number of edges. If $G$ is not connected then adding a new edge which connects two components does not give a new 3-clique, a contradiction.\nTherefore $G$ is connected and it has at least $n-1$ edges. On the other hand, the graph $K_{1,n-1}$ has the required property. ($K_{m,n}$ is the graph with $m+n$ vertices which are divided into two sets of $m$ and $n$ elements, respectively, and two vertices are adjacent if and only if they belong to different sets; the edges are obviously $mn$.) Thus the answer is $n-1$.\n\nb) Consider a graph with vertices $u_1, u_2, v_1, \\dots, v_{n-2}$ and edges all pairs $u_i v_j$, $i = 1, 2, j = 1, \\dots, n-2$, together with $u_1 u_2$. This graph has $n$ vertices, $2n-3$ edges and adding any new edge increases the number of the 4-cliques. Hence the required minimum number does not exceed $2n-3$.\nWe shall prove by induction on $n$ that the minimum possible number of edges is $2n-3$ and it is attained only for a graph as above. The assertion is obvious for $n=4$.\nAssume that the assertion is proved for graphs with $n-1$ or less vertices. Let $G$ be a graph with the required property having $n \\ge 5$, vertices and minimum possible number of edges.\n\nSince adding of a new edge increases the number of the 4-cliques, the graph $G$ has four vertices $x_1, x_2, x_3, x_4$ which determine exactly five edges joining them. Next we assume that the missing edge is $x_1x_2$. Let $G^*$ be the graph which is obtained from $G$ by cluing the vertices $x_1$ and $x_2$ (i.e. we remove $x_1$ and $x_2$ from $G$, add a new vertex $u$ and save all remaining vertices; the new vertex is adjacent exactly to these vertices which were adjacent to at least one of $x_1$ and $x_2$, all old edges are saved). Then\n$$\ne(G^*) \\leq e(G) - 2 \\leq 2n - 5 = 2(n - 1) - 3.$$\nHence the graph $G^*$ has $n-1$ vertices, possesses the required property and has at most $2(n-1)-3$ edges. Then the induction hypothesis implies that $G^*$ has exactly $2n-5$ edges and it has the structure described above – two vertices of degree $n-2$ and all remaining of degree 2. At least one of the vertices of degree $n-2$ is $x_3$ or $x_4$, say $x_3$. Then the degree of $x_3$ in $G$ is $n-1$.\nDenote by $G'$ the graph obtained from $G$ by removing the vertex $x_3$ and all edges from it. The graph $G'$ has at most $n-2$ edges since $G$ has at most $2n-3$ edges. Moreover, $G'$ has the property considered in a). Therefore it follows from the solution of a) that $G' = K_{1,n-2}$. It is easy to see now that $G$ has the required structure and this completes the induction step.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 27376,
    "subject": "Mathematics (Multi-modal)",
    "question": "A circle is called *good colored* if the vertices of any equilateral triangle inscribed in this circle are colored in distinct colors. Let $k$ be a circle with radius $2$.\n\na) Is there a coloring of the points on $k$ and inside $k$ in three colors such that $k$ and any circle with radius at least $1$ that touches $k$ are good colored?\n\nb) Is there such a coloring in seven colors?",
    "options": [],
    "answer": "a) No. b) Yes.",
    "solution": "a) Assume that such a coloring exists and $a$, $b$ and $c$ are the colors. Let $O$ be the center of $k$ and consider an equilateral triangle $OBC$ with side $\\sqrt{3}$. Its circumcircle has radius $1$ and touches $k$. If the color of $O$ is $a$, then the colors of $B$ and $C$ are $b$ and $c$. This shows that the points of the circle $k'(0, \\sqrt{3})$ are colored in $b$ and $c$. Consider now an equilateral $\\triangle PQR$ with circumcircle $k$. Denote by $X_1, X_2$ and $Y_1, Y_2$ the intersection points of $PQ$ and $PR$ with $k'$, respectively ($X_1$ and $Y_1$ the closer points to $P$). Since the circumcircle of the equilateral $\\triangle PX_2Y_2$ touches $k$ and has radius at least $1$, it follows that the color of $P$ is $a$. Analogously $Q$ and $R$ have the same color, a contradiction.\n\nb) Let $ABCDEF$ be a regular hexagon inscribed in $k$. Let the color of $O$ be $1$, let the color of the points inside the sector $OAB$, the radius $OA$ and the arc $AB$ without $B$ be $2$, let the color of the points inside the sector $OBC$, the radius $OB$ and the arc $BC$ without $C$ be $2$, etc. It is easy to see that this coloring has the desired properties.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27377,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all values of the real parameter $a$ such that the equation\n$$\nx^3 - a x^2 + (a^2 - 1)x - a^2 + a = 0\n$$\nhas three distinct real roots which (in some order) form an arithmetic progression.",
    "options": [],
    "answer": "a = 0 or a = 6/7",
    "solution": "Writing the equation in the form\n$$\n(x-1)(x^2 + (1-a)x - a + a^2) = 0,\n$$\nwe obtain $x_1 = 1$. Let $x_2$ and $x_3$ be the roots of the quadratic equation. If $1$ is the second term of the progression then $x_2 + x_3 = 2$, giving $a - 1 = 2$, i.e. $a = 3$. When $a = 3$ the roots of the quadratic equation are not real.\nIf $x_1 = 1$ is not the second term we may assume that $1 + x_2 = 2x_3$, which together with $x_2 + x_3 = a - 1$ implies $3x_3 = a$. Therefore $\\frac{a}{3}$ is a root of the quadratic equation, i.e.\n$$\n\\left(\\frac{a}{3}\\right)^2 + (1-a)\\frac{a}{3} - a + a^2 = 0.\n$$\nThus, $a = 0$ or $a = \\frac{6}{7}$. When $a = 0$ we obtain $x_2 = -1$, $x_3 = 0$ and when $a = \\frac{6}{7}$ we find $x_2 = -\\frac{3}{7}$ and $x_3 = \\frac{2}{7}$. Hence the desired values are $a = 0$ and $a = \\frac{6}{7}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27378,
    "subject": "Mathematics (Multi-modal)",
    "question": "The edge $DA$ of a pyramid $ABCD$ is perpendicular to its base $ABC$, $(ABD) \\perp (BCD)$, $\\angle BDC = 45^\\circ$ and $DB = 2$. Find $\\angle ADB$ if the sum of the squares of lateral faces of the pyramid equals $8$.",
    "options": [],
    "answer": "45°",
    "solution": "Since $(ABD) \\perp (ABC), (BCD)$, then $(ABC) \\perp BC$ and hence $\\angle ABC = \\angle DBC = 90^\\circ$. Since $\\angle BDC = 45^\\circ$, then $BC = BD = 2$. Let $\\angle ADB = \\alpha$. Then $AB = 2 \\sin \\alpha$, $AD = 2 \\cos \\alpha$ and $AC = 2\\sqrt{\\sin^2 \\alpha + 1}$. So\n$$\nS_{ABD} = \\frac{AB \\cdot AD}{2} = 2 \\sin \\alpha \\cdot \\cos \\alpha,\n$$\n$$\nS_{ACD} = \\frac{AC \\cdot AD}{2} = 2 \\cos \\alpha \\sqrt{1 + \\sin^2 \\alpha},\n$$\n$$\nS_{BCD} = \\frac{BC \\cdot BD}{2} = 2.\n$$\nIt follows by the condition of the problem that\n$$\n8 = 4(\\sin^2 \\alpha \\cos^2 \\alpha + \\cos^2 \\alpha(1 + \\sin^2 \\alpha) + 1) = 4(\\sin^2 \\alpha - 2 \\sin^4 \\alpha + 2),\n$$\ni.e. $\\sin \\alpha = \\frac{\\sqrt{2}}{2}$. Thus $\\angle ADB = 45^\\circ$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 27379,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_1 > \\frac{1}{12}$ and $a_{n+1} = \\sqrt{(n+2)a_n + 1}$ for $n \\ge 1$. Prove that:\n\na) $a_n > n - \\frac{2}{n}$;\n\nb) the sequence $b_n = 2^n \\left(\\frac{a_n}{n} - 1\\right)$, $n = 1, 2, \\dots$, is convergent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) We shall prove the statement by induction if $a_1 > \\frac{19}{243} \\left(\\frac{1}{12} > \\frac{19}{243}\\right)$. It is true for $n \\le 3$, since $a_2 > \\sqrt{3 \\cdot \\frac{19}{243} + 1} = \\frac{10}{9}$ and $a_3 > \\sqrt{4 \\cdot \\frac{10}{9} + 1} = \\frac{7}{3}$.\n\nAssume that $a_n > n - \\frac{2}{n}$ for some $n \\ge 3$. Then\n$$\na_{n+1} > \\sqrt{(n+2)\\left(n - \\frac{2}{n}\\right) + 1}.\n$$\nIt is enough to show that right-hand side is bigger than $n+1 - \\frac{2}{n+1}$. It is easy to see that this is equivalent to $\\frac{1}{2} > \\frac{1}{n} + \\frac{1}{(n+1)^2}$, which obviously holds for $n \\ge 3$.\n\nb) Note first that if $a_1 = 1$, then $a_n = n$ by induction, implying that $b_n = 2^n \\left(\\frac{a_n}{n} - 1\\right) = 0$.\nIf $a_1 < 1$, then $a_n < n$ again by induction, i.e., $b_n < 0$. We shall prove that $b_n < b_{n+1}$. This is equivalent to $\\frac{a_n - n}{2n} < \\frac{a_{n+1} - n - 1}{n+1}$. The right-hand side equals\n$$\n\\frac{a_{n+1}^2 - n - 1}{(n+1)(a_{n+1} + n + 1)} = \\frac{(n+2)a_n + 1 - (n+1)^2}{(n+1)(a_{n+1} + n + 1)} = \\frac{(n+2)(a_n - n)}{(n+1)(a_{n+1} + n + 1)}.\n$$\nIt remains to show that $\\frac{1}{2n} > \\frac{n+2}{(n+1)(a_{n+1} + n + 1)}$, i.e.,\n$$\n(n+1)a_{n+1} > 2n(n+2) - (n+1)^2 = (n+1)^2 - 2,\n$$\nwhich follows by a).\nThe above arguments show that if $a_1 > 1$, then $b_n > b_{n+1} > 0$.\nSo the sequence $(b_n)$ is monotone and bounded; hence it converges.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27380,
    "subject": "Mathematics (Multi-modal)",
    "question": "In $\\triangle ABC$ with $\\angle ACB = 60^\\circ$ let $AA_1$ and $BB_1$ ($A_1 \\in BC$, $B_1 \\in AC$) be the bisectors of $\\angle BAC$ and $\\angle ABC$. The line $A_1B_1$ meets the circumcircle of $\\triangle ABC$ at points $A_2$ and $B_2$.\n\na) If $O$ and $I$ are the circumcenter and the incenter of $\\triangle ABC$ prove that $OI$ is parallel to $A_1B_1$.\n\nb) If $R$ is the midpoint of the arc $\\widehat{AB}$, not containing point $C$, and $P$ and $Q$ are the midpoints of $A_1B_1$ and $A_2B_2$, respectively, prove that $RP = RQ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Since $\\angle AOB = 2\\gamma = 120^\\circ$ and points $A$, $O$, $I$ and $B$ lie on a circle\n$\\alpha + \\beta$\n![](attached_image_1.png)\n\nb) Since $OQ \\perp A_2B_2$, $IP \\perp A_2B_2$ ($\\triangle A_1IB_1$ is isosceles) and $OI \\parallel A_2B_2$, we have that $OIPQ$ is a rectangle. The perpendicular bisector of $OI$ is also the perpendicular bisector of $PQ$. Using that $R$ lies on the perpendicular bisector of $OI$ it follows that $R$ lies on the perpendicular bisector of $PQ$, i.e. $RP = RQ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27381,
    "subject": "Mathematics (Multi-modal)",
    "question": "One cuts a paper strip of length $2007$ into two parts of integer lengths and writes down the two integers on the board. Then cuts one of the two parts into two parts of integer lengths and writes down the two integers on the board. The cutting stops when all parts are of length $1$. A cut is called *bad* if the two parts obtained are not of equal lengths.\n\na) Find the minimum possible bad cuts.\n\nb) Prove that for all cuttings with minimum possible bad cuts the number of distinct integers on the board is one and the same.",
    "options": [],
    "answer": "a) 8. b) For any optimal cutting achieving the minimum number of bad cuts, the number of distinct integers recorded equals 2 g(n) + k − 1, where g(n) is the number of ones in the binary expansion of n and k is the largest exponent of two appearing in that expansion.",
    "solution": "a) Let the length of the strip be $n$. Denote by $g(n)$ and $f(n)$ respectively the number of $1$'s in the binary representation of $n$ and the minimum possible number of bad cuts. Let $n = 2^{k_1} + 2^{k_2} + \\cdots + 2^{k_l}$. Consider the following sequence of cuts: first cut a strip of length $2^{k_1}$, next cut a strip of length $2^{k_2}$ and so on. After the last cut we obtain two strips of lengths $2^{k_{l-1}}$ and $2^{k_l}$. Note that a strip whose length is a power of $2$ can be cut into strips of length $1$ without bad cuts. Therefore the number of bad cuts equals $l-1$, i.e.\n$$\n(1) \\qquad f(n) \\le g(n) - 1.\n$$\nWe prove by induction on $n$ that $f(n) \\ge g(n) - 1$. For $n=1$ we have $f(1) = 0$ and $g(1) = 1$, i.e. the statement is true. Suppose it is true for all $n \\le k$, where $k$ is a positive integer and let $n = k+1$.\n\n1. Suppose the first cut is bad and it leaves two strips of lengths $a$ and $b$. Then $a+b = k+1$ and $f(k+1) = 1 + f(a) + f(b)$. If the binary representations of $a$ and $b$ have no common digit $1$ then $g(k+1) = g(a) + g(b)$ and therefore\n$$\nf(k+1) = 1 + f(a) + f(b) \\ge 1 + g(a) - 1 + g(b) - 1 = g(k+1) - 1.\n$$\nIf the binary representations of $a$ and $b$ have at least one common digit $1$ then $g(k+1) \\le g(a) + g(b) - 1$ and therefore\n$$\nf(k+1) = 1 + f(a) + f(b) \\ge 1 + g(a) - 1 + g(b) - 1 \\ge g(k+1) > g(k+1) - 1.\n$$\n\n2. Suppose the first cut is not bad, i.e. the strip is cut into two parts each of length $a$. Then $k+1 = 2a$ and $g(k+1) = g(a)$. If $g(k+1) = 1$ then $f(k+1) = 0$ and the statement is true. Otherwise\n$$\nf(k+1) = f(a) + f(b) = 2f(a) \\ge 2g(a) - 2 = 2g(k+1) - 2 > g(k+1) - 1.\n$$\nThus, when $g(k+1) > 1$ we have $f(k+1) > g(k+1) - 1$.\n\nThis proves the induction hypothesis giving\n$$\n(2) \\qquad f(n) \\geq g(n) - 1.\n$$\nIt follows from (1) and (2) that $f(n) = g(n) - 1$.\n\na) Since the binary representation of $2007$ is $1111010111$, i.e. $g(2007) = 9$, we obtain $f(2007) = 8$.\n\nb) It follows from the above arguments that if the number of bad cuts is $f(n) = g(n) - 1$ then every bad cut leaves two parts of lengths $a$ and $b$ such that the binary representations of $a$ and $b$ have no common digit $1$. Moreover the good cuts are done only over strips whose lengths are powers of $2$. It is clear that rearranging the cuts we may assume that the first cuts are bad. Their number equals $g(n) - 1$ and each bad cut gives two new numbers on the table. Therefore after all bad cuts we have $2g(n) - 2$ distinct numbers. The powers of $2$ that appear are all powers up to the highest power in the binary representation of $n$.\nThus, the number of distinct numbers on the table equals $2g(n) - 2 + k + 1 = 2g(n) + k - 1$, where $k$ is the highest power of $2$ in the binary representation of $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27382,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve in integers the system\n$$\n\\begin{cases} 3a^4 + 2b^3 = c^2 \\\\ 3a^6 + b^5 = d^2. \\end{cases}\n$$",
    "options": [],
    "answer": "a = b = c = d = 0",
    "solution": "We shall prove that $a = b = c = d = 0$. It is easy to see that if one of the numbers $a, b, c, d$ equals $0$, then the others equal $0$, too. Indeed, if $b = 0$, $a \\neq 0$, then $\\sqrt{3} = \\pm\\frac{c}{a^2}$ is a rational number, a contradiction. The case $a = 0$, $b \\neq 0$ is impossible by the same reasoning. Let $c = 0$. Then $n = -b > 0$ and $3a^4 = 2n^3$, $3a^6 \\ge n^5$. Hence $2a^2 \\ge n^2$ and therefore $2n^3 = 3a^4 \\ge \\frac{3n^4}{4}$. Then $n \\le \\frac{8}{3}$, *i.e.* $n = 1$ or $n = 2$, a contradiction. Analogously $d \\neq 0$.\n\nLet now $a, b, c, d \\neq 0$. Adding both equations and using that $3$ divides $b^5 + 2b^3 = b^3(b-1)(b+1) + 3b^3$, we get that $3|c^2+d^2$, i.e. $3|c, d$. Hence $3|a, b$. Let $a = 3^\\alpha a_1$, $b = 3^\\beta b_1$, $c = 3^\\gamma c_1$, $d = 3^\\delta d_1$, where $\\alpha, \\beta, \\gamma, \\delta \\ge 1$ and $3 \\nmid a_1, b_1, c_1, d_1$. Then the system can be written in the form\n$$\n(1) \\qquad \\left| \\begin{array}{l} 3^{4\\alpha+1}a_1^4 + 3^{3\\beta}2b_1^3 = 3^{2\\gamma}c_1^2 \\\\ 3^{6\\alpha+1}a_1^6 + 3^{5\\beta}b_1^5 = 3^{2\\delta}d_1^2 \\end{array} \\right.\n$$\nWe shall use the following trivial fact. If $3^k p + 3^l q = 3^m r$ and $3 \\nmid p, q, r$, then at least two of the numbers $k, l, m$ are equal. This and (1) imply that $4\\alpha + 1 = 3\\beta$ or $3\\beta = 2\\gamma$, and $6\\alpha + 1 = 5\\beta$ or $5\\beta = 2\\delta$. Then it is easy to see that $3\\beta = 2\\gamma$ and $5\\beta = 2\\delta$. Now (1) is equivalent to\n$$\n\\left| \\begin{array}{l} 3^{4\\alpha+1-2\\gamma}a_1^4 + 2b_1^3 = c_1^2 \\\\ 3^{6\\alpha+1-2\\delta}a_1^6 + b_1^5 = d_1^2 \\end{array} \\right.\n$$\nSince $4\\alpha + 1 - 2\\gamma > 0$ and $6\\alpha + 1 - 2\\delta > 0$, adding the last two equations, we conclude as above that $3|c_1, d_1$, a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27383,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $k > 5$ be an integer. Replace given positive integer by the product of the sum of its digits in base $k$ and $(k-1)^2$. Repeat the same with the new number, etc. Prove that the obtained numbers are equal from some point onwards.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since the sum of digits of an integer divisible by $k-1$, is divisible by $k-1$, too, $(k-1)^3$ divides all the numbers obtained after the second step. $(k-1)^3$. On the other hand, if $a = \\overline{a_n a_{n-1} \\dots a_0(k)}$, $n \\ge 4$ or $a_3 \\ge 2$, $n = 3$, then\n\n$$\n\\begin{align*} a - (k-1)^2 (a_n + a_{n-1} + \\dots + a_0) &\\ge 2(k^3 - (k-1)^2) - a_1((k-1)^2-k) - a_0((k-1)^2-1)) \\\\ &\\ge 2(k^3 - (k-1)^2) - (k-1)(2(k-1)^2 - k - 1) = 5k^2 - 2k - 1 > 0. \\end{align*}\n$$\nThis shows that we shall get a number of the form $a = \\overline{a_3 a_2 a_1 a_0(k)}$, where $a_3 = 0, 1$. The next number is\n$$\n(k-1)^2(a_3 + a_2 + a_1 + a_0) \\le (k-1)^2(1 + 3(k-1)) < 4(k-1)^3.\n$$\n\nHence this number is $(k-1)^3$, $2(k-1)^3$ or $3(k-1)^3$. Note that\n$$ (k-1)^3 = \\overline{k-3,2,k-1}_{(k)} \\rightarrow 2(k-1)^3, \\quad k > 2, $$\n$$ 2(k-1)^3 = \\overline{1,k-6,5,k-2}_{(k)} \\rightarrow 2(k-1)^3, \\quad k > 5, $$\n$$ 3(k-1)^3 = \\overline{2,k-9,8,k-3}_{(k)} \\rightarrow 2(k-1)^3, \\quad k > 8. $$\nSince $3.5^3 = \\overline{1423}_{(6)} \\rightarrow 10.5^2 = 2.5^3$, $3.6^3 = \\overline{1614}_{(7)} \\rightarrow 12.6^2 = 2.6^3$ and $3.7^3 = \\overline{2005}_{(8)} \\rightarrow 7.7^2 = 7^3 \\rightarrow 2.7^3$, we conclude that if $k > 5$, then the numbers are equal to $2(k-1)^3$ from some point onwards.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27384,
    "subject": "Mathematics (Multi-modal)",
    "question": "For every positive integer $n$ set $a_n = 0$, if the number of divisors of $n$, greater than $2007$, is even and $a_n = 1$, if this number is odd. Is the number $\\alpha = 0, a_1a_2a_3 \\dots a_k \\dots$ rational?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We prove that $\\alpha$ is irrational. Suppose $\\alpha$ is a rational number, i.e. the sequence $a_1, a_2, a_3, \\ldots, a_k, \\ldots$ is periodic from some point onwards. Hence there exist $k_0$ and $T$ such that for any $k > k_0$ we have $a_k = a_{k+T}$.\n\nChoose a positive integer $m$ for which $mT > k_0$ and $mT$ is a perfect square. If $T = p_1^{\\alpha_1} p_2^{\\alpha_2} \\cdots p_s^{\\alpha_s}$ then $m = p_1^{\\beta_1} p_2^{\\beta_2} \\cdots p_s^{\\beta_s}$, where $\\alpha_i + \\beta_i$ is even for all $i = 1, 2, \\ldots, s$ and $\\beta_i$ are large enough positive integers.\n\nChoose a prime number $p > 2007$, $p \\neq p_i$, $i = 1, 2, \\ldots, s$. Since $pmT - mT$ is divisible by $T$ we have that $a_{mT} = a_{pmT}$.\n\nDenote by $\\tau(k)$ the number of divisors of $k$ and by $f(k)$ the number of divisors of $k$ that are greater than $2007$. We have $f(pmT) = f(mT) + \\tau(mT)$ and since $\\tau(mT)$ is odd number we conclude that $f(pmT)$ and $f(mT)$ are of the same parity, a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27385,
    "subject": "Mathematics (Multi-modal)",
    "question": "A plane passes through a vertex of the base of a cube of edge $1$ and the centers of its two faces which do not contain that vertex. Find the ratio of the volumes of the two parts of the cube cut by the plane.",
    "options": [],
    "answer": "1:2",
    "solution": "Let $P$ and $Q$ be the centers of the faces $BCC_1B_1$ and $DCC_1D_1$ and let $\\alpha = (APQ)$ (Fig. 1). Since $PQ \\parallel BD$ the plane $\\alpha$ meets the plane $(ABCD)$ at the line through $A$ which is parallel to $BD$. We denote by $T$ and $S$ the intersection points of this line with the lines $CB$ and $CD$, respectively.\n\nThe lines $TP$ and $SQ$ meet the edge $CC_1$ at the intersection point $R$ of $\\alpha$ and $CC_1$. Let $M = TR \\cap BB_1$ and $N = SR \\cap DD_1$. Then the intersection of $\\alpha$ and the surface of the cube is the quadrilateral $AMRN$. (It is easy to see that it is a rhombus.) It is clear that $BT = BA = 1$. Hence $B$ and $M$ are the midpoints of $TC$ and $TR$, respectively. Then $\\frac{BM}{RC} = \\frac{1}{2}$ and since $BM = RC_1 = 1 - RC$, we find $BM = \\frac{1}{3}$. Analogously $DN = \\frac{1}{3}$.\n\nDenote by $V$ the volume of the polytope cut from the cube by the planes $(ABCD)$ and $(AMRN)$ (Fig. 2). Let $A_2$ and $C_2$ be the intersection points of $AA_1$ and $CC_1$ with the plane through $MN$ and parallel to $(ABCD)$. Then $RC_2 = RC - CC_2 = MB = AA_2 = \\frac{1}{3}$ and hence the tetrahedra $NMC_2R$ and $NMA_2A$ have equal volumes. This shows that $V = V_{ABCD A_2 MC_2 N} = \\frac{1}{3}$. Hence the required ratio is $1:2$.\n\n![](attached_image_1.png)\nFig. 1\n\n![](attached_image_2.png)\nFig. 2",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 27386,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all values of the real parameter $a$ such that the inequality\n$$\n\\log_a(a^x + 1) + \\frac{1}{\\log_{a^2-1} a} \\le x - 1 + \\log_a(a^2 - 1)\n$$\nholds true for every $x \\in (0, 1]$.",
    "options": [],
    "answer": "(1, 2)",
    "solution": "We have that $a > 1$, $x > 0$ and $x \\neq \\log_a 2$. Then the inequality is equivalent to\n$$\n(a^{x-1} - 1)(a^{x+1} + 1) \\le 0,\n$$\ngiving $x \\le 1$. Since $\\log_a 2 > 0$ and $\\log_a 2 \\le 1$ when $a \\ge 2$ we obtain:\n\n1. For $1 < a < 2$ the inequality holds true iff $0 < x \\le 1$.\n\n2. For $a \\ge 2$ the inequality holds true iff $0 < x \\le 1$ and $x \\neq \\log_a 2 < 1$.\n\nTherefore the desired values are $a \\in (1, 2)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27387,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a right triangle with $\\angle ACB = 90^\\circ$, $AC = 1$ and $BC = 2$. Given a point $A_1 \\in BC$ such that $A_1C \\neq \\frac{1}{3}$ we construct a sequence of points $A_n \\in BC$, $n \\ge 2$ in the following way. Let $B_1$ be the intersection point of $AC$ and the line through $A_1$ and parallel to $AB$, and $C_1$ be the foot of the perpendicular from $B_1$ to $AB$. Then $A_2$ is the intersection point of $BC$ and the line through $C_1$ and parallel to $AC$. Using $A_2$ we construct $A_3$ in the same way, etc. Find:\na) $\\frac{3A_2C - 1}{3A_1C - 1}$;\nb) $\\lim_{n \\to \\infty} S_{A_n B_n C_n}$.",
    "options": [],
    "answer": "a) -1/5; b) 5/36",
    "solution": "a) Set $A_nC = x_n$. Then it follows from $\\triangle A_nB_nC \\sim \\triangle BAC$ that $B_nC = \\frac{x_n}{2}$. Since $\\triangle AB_nC_n \\sim \\triangle ABC$, we find that $AC_n = \\frac{2-x_n}{2\\sqrt{5}}$. Hence $A_{n+1}C = x_{n+1} = \\frac{2-x_n}{5}$ and therefore $\\frac{3x_{n+1}-1}{3x_n-1} = -\\frac{1}{5}$.\n\nb) Since $A_nB_n = \\frac{\\sqrt{5}}{2}x_n$, $B_nC_n = \\frac{2-x_n}{\\sqrt{5}}$ and $A_nB_n \\perp B_nC_n$ we get $S_{A_nB_nC_n} = \\frac{x_n(2-x_n)}{4}$. But a) implies that $\\{x_n - \\frac{1}{3}\\}_{n=1}^\\infty$ is a geometric progression with ratio $-\\frac{1}{5}$. Hence $\\lim_{n \\to \\infty} x_n = \\frac{1}{3}$ and therefore $\\lim_{n \\to \\infty} S_{A_nB_nC_n} = \\frac{5}{36}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27388,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the function $f(x) = a(|\\sin x| + |\\cos x|) - 3\\sin 2x - 7$, where $a$ is a real parameter.\n\na) Prove that $f(x) = f(\\frac{\\pi}{2} - x) = f(\\pi + x) = f(\\frac{3\\pi}{2} - x)$ for every $x$.\n\nb) Find all pairs $(a, n)$, where $n$ is a positive integer, for which the equation $f(x) = 0$ has $2007$ roots in the interval $(0, n\\pi)$.",
    "options": [],
    "answer": "Pairs (a, n): (7, 502), (5√2, 2007), (2√2, 2007)",
    "solution": "a)\nWe have\n$$\n\\begin{align*}\nf(x + \\pi) &= a(|\\sin(x + \\pi)| + |\\cos(x + \\pi)|) - 3\\sin(2x + 2\\pi) - 7 \\\\\n&= a(|-\\sin x| + |-\\cos x|) - 3\\sin(2x) - 7 = f(x), \\\\\n\\\\\nf\\left(\\frac{\\pi}{2} - x\\right) &= a\\left(|\\sin\\left(\\frac{\\pi}{2} - x\\right)| + |\\cos\\left(\\frac{\\pi}{2} - x\\right)|\\right) - 3\\sin(\\pi - 2x) - 7 \\\\\n&= a\\left(|\\cos x| + |\\sin x|\\right) - 3\\sin(2x) - 7 = f(x), \\\\\n\\\\\nf\\left(\\frac{3\\pi}{2} - x\\right) &= a\\left(|\\sin\\left(\\frac{3\\pi}{2} - x\\right)| + |\\cos\\left(\\frac{3\\pi}{2} - x\\right)|\\right) - 3\\sin(3\\pi - 2x) - 7 \\\\\n&= a\\left(|-\\cos x| + |-\\sin x|\\right) - 3\\sin(2x) - 7 = f(x).\n\\end{align*}\n$$\n\nb)\nDirect verification shows that for every integer $k$ we have $f(\\frac{k\\pi}{2}) = a-7$.\nMoreover $f(\\frac{\\pi}{4}) = a\\sqrt{2} - 10$ and $f(\\frac{3\\pi}{4}) = a\\sqrt{2} - 4$. If $a \\neq 7$, $a \\neq 5\\sqrt{2}$ and $a \\neq 2\\sqrt{2}$, according to a) the equation $f(x) = 0$ has even number of roots in any of the intervals $(0, \\frac{\\pi}{2})$ $(\\frac{\\pi}{2}, \\pi)$ and therefore it has even number of roots in the interval $(0, n\\pi)$.\n\n1. If $a = 7$ then $f(x) = 7(|\\sin x| + |\\cos x|) - 3\\sin 2x - 7$ and $f(\\frac{\\pi}{2}) = 0$.\n\n1.1. Let $x \\in (0, \\frac{\\pi}{2})$. Then $f(x) = 7(\\sin x + \\cos x) - 3\\sin 2x - 7$. Set $y = \\sin x + \\cos x$. Then $y = \\sqrt{2}\\sin(x + \\frac{\\pi}{4}) \\in (1, \\sqrt{2}]$ ($\\sin 2x = y^2 - 1$) and the equation $f(x) = 0$ becomes $3y^2 - 7y + 4 = 0$. Therefore $y_1 = 1$ and $y_2 = \\frac{4}{3}$. Hence $y_2 = \\frac{4}{3}$ and the equation $f(x) = 0$ has 2 roots in $(0, \\frac{\\pi}{2})$.\n\n1.2. Let $x \\in (\\frac{\\pi}{2}, \\pi)$. Then $f(x) = 7(\\sin x - \\cos x) - 3\\sin 2x - 7$. Set $y = \\sin x - \\cos x$. Then $y = \\sqrt{2}\\sin(x + \\frac{\\pi}{4}) \\in (1, \\sqrt{2}]$ and $f(x) = 0$ is equivalent to $3y^2 + 7y - 10 = 0$. Therefore $y_1 = 1$ and $y_2 = -\\frac{10}{3}$, i.e. $f(x) = 0$ has no solutions in the given interval.\nTherefore $f(x) = 0$ has 3 roots in the interval $(0, \\pi)$. The total number of roots in the interval $(0, n\\pi)$ equals $3n + n - 1 = 4n - 1$ and $4n - 1 = 2007$ implies $n = 502$.\n\n2. If $a = 5\\sqrt{2}$ then $f(x) = 5\\sqrt{2}(|\\sin x| + |\\cos x|) - 3\\sin 2x - 7$.\n\n2.1. Let $x \\in (0, \\frac{\\pi}{2})$. Then $f(x) = 5\\sqrt{2}(\\sin x + \\cos x) - 3\\sin 2x - 7$ and setting $y = \\sin x + \\cos x$ yields that $f(x) = 0$ is equivalent to $3y^2 - 5\\sqrt{2}y + 4 = 0$, $y \\in (1, \\sqrt{2}]$. Therefore $y_1 = \\sqrt{2}$ and $y_2 = \\frac{2\\sqrt{2}}{3} < 1$, i.e. only $x = \\frac{\\pi}{4}$ is a solution.\n\n2.2. Let $x \\in (\\frac{\\pi}{2}, \\pi)$. Then $f(x) = 5\\sqrt{2}(\\sin x - \\cos x) - 3\\sin 2x - 7$ and setting $y = \\sin x - \\cos x$ yields that $f(x) = 0$ is equivalent to $3y^2 + 5\\sqrt{2}y - 10 = 0$, $y \\in (1, \\sqrt{2}]$. The roots of this equation do not belong to the interval $(1, \\sqrt{2}]$. In this case there is 1 root in $(0, \\pi)$ and $n$ roots in $(0, n\\pi)$. Therefore $n = 2007$.\n\n3. When $a = 2\\sqrt{2}$ analogous observations show that there is a unique root $x = \\frac{3\\pi}{4}$ in the interval $(0, \\pi)$ and $n$ roots in the interval $(0, n\\pi)$. Again $n = 2007$.\n\nAnswer: $a = 7$, $n = 502$; $a = 5\\sqrt{2}$, $n = 2007$; $a = 2\\sqrt{2}$, $n = 2007$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27389,
    "subject": "Mathematics (Multi-modal)",
    "question": "Alexander and Denitza play the following game. Alexander cuts (if possible) a band of positive integer length to three bands of positive integer lengths such that the largest band is unique. Then Denitza cuts (if possible) the largest band in the same way and so on. The winner is the one who makes the last move. Consider the bands whose lengths are integers of the form $a^b$, $a-1, b-1 \\in \\mathbb{N}$. For which of them Denitza has a winning strategy.",
    "options": [],
    "answer": "Denitza has a winning strategy exactly for band lengths n equal to 3^k or 3^k − 1 with k > 1. Among lengths of the form a^b with a ≥ 2 and b ≥ 2, this occurs precisely when either a = 3 and b > 1 (so n = 3^b), or a = 2 and b = 3 (so n = 8).",
    "solution": "Consider a band of length $n$. It is clear that no move is possible for $n = 1, 2, 3$. For $4 \\le n \\le 7 = 3+2+2$ Alexander has a winning move. For $n = 8$ and $9$ after the first move of Alexander the largest length is between $4$ and $7$ and then Denitza has a winning move. Similarly, for $10 \\le n \\le 9+8+8 = 25$, Alexander has a winning strategy since he can cut the band in such a way that the largest length is either $8$ or $9$ and so on. It follows by induction that Denitza has a winning strategy if and only if $n = 3^k$ or $n = 3^k - 1$ for some integer $k > 1$.\n\nThe numbers $3^k$ and $3^2 - 1 = 2^3$ obviously have the desired form. We shall prove that there are no other solutions. Assume that $a^b = 3^k - 1$. Since $a^2 \\equiv 0, 1 \\pmod 3$, the number $b$ is even. Then $(a+1)(a^{b-1} + \\cdots + a + 1) = 3^k$ and we\n\nhave $a + 1 = 3^i$, and $3^{k-i} = a^{b-1} + \\cdots + a + 1 = A(a + 1) + b$, for some integers $i$ and $k$, $0 < i < k$. Hence 3 divides $b$ and setting $c = a^{b/3}$ we have $3^k = (c+1)((c+1)^2 - 3c)$. Then $c+1 = 3^j$ and $(c+1)^2 - 3c = 3^{k-j}$, $0 < j < k$. In particular, 9 divides $(c+1)^2$, but does not divide $3c$. We consecutively find $k-j=1$, $c=2$, $a=k=2$ and $b=3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27390,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m > 1$ and $n > 1$ be odd integers. Distinct real numbers are written in the cells of a table of $m$ rows and $n$ columns. A number is called *good* if:\n1. it is the largest in its row (column);\n2. it is the middle number of its column (row).\nWhat is the maximal number of good numbers?",
    "options": [],
    "answer": "min{m, (n+1)/2} + min{n, (m+1)/2}",
    "solution": "Let $N_c$ be the set of good numbers that are largest in their column and middle in their row. Since no two of these numbers belong to one and the same row we have $|N_c| \\le m$. Denote by $a$ the largest element of $N_c$ and let $a_1, a_2, \\dots, a_{\\frac{n-1}{2}}$ be the numbers in the row of $a$ that are greater than $a$. It is clear that there are no elements from $N_c$ in the columns of $a_1, a_2, \\dots, a_{\\frac{n-1}{2}}$. Since there is at most one element of $N_c$ in every column it follows that $|N_c| \\le \\frac{n+1}{2}$. Thus,\n$$\nN_c \\le \\min \\left\\{ m, \\frac{n+1}{2} \\right\\}.\n$$\nAnalogously we obtain that the number of good numbers that are largest in their rows is at most $\\min\\left\\{m, \\frac{n+1}{2}\\right\\} + \\min\\left\\{n, \\frac{m+1}{2}\\right\\}$. We shall show that for every odd $m > 1$ and $n > 1$ there exists a table such that the number of good numbers equals $\\min\\left\\{m, \\frac{n+1}{2}\\right\\} + \\min\\left\\{n, \\frac{m+1}{2}\\right\\}$.\n\n1. If $m \\neq n$ without loss of generality assume that $1 < m < n$. Mark the cells of the table by the digits 1, 2, 3, 4 and 5 as shown on the figure. Next, fill in the cells by writing consecutively the numbers 1, 2, 3, ..., $mn$, starting with the cells marked by 1, then by 2 and so on. All numbers written in the cells marked by 2 or 4 are good. Their number equals:\n![](attached_image_1.png)\n$$\n\\min \\left\\{ m, \\frac{n+1}{2} \\right\\} + \\frac{m+1}{2} = \\min \\left\\{ m, \\frac{n+1}{2} \\right\\} + \\min \\left\\{ n, \\frac{m+1}{2} \\right\\}.\n$$\n\n2. Let $m = n$ and $3 < m$. As in 1. we fill the following table. The good numbers are again in cells marked by 2 or 4. Their number equals:\n![](attached_image_2.png)\n$$\n\\frac{n-1}{2} + \\frac{n-1}{2} + 2 = n + 1\n$$\n\n$$\n\\frac{n-1}{2}\n$$\n3. If $m = n = 3$, then the table (5,6,7 and 8 are good) shows that the number of good numbers is 4 and $4 = 2 \\cdot 2 = 2 \\min \\left\\{ 3, \\frac{3+1}{2} \\right\\}$.\n\n<table>\n<tr><td>1</td><td>7</td><td>9</td></tr>\n<tr><td>2</td><td>6</td><td>4</td></tr>\n<tr><td>5</td><td>3</td><td>8</td></tr>\n</table>\n\n$$\n\\text{Answer: } \\min \\left\\{ n, \\frac{n+1}{2} \\right\\} + \\min \\left\\{ n, \\frac{n+1}{2} \\right\\}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27391,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ such that if $a, b, c \\ge 0$ and $a+b+c=3$, then $abc(a^n + b^n + c^n) \\le 3$.",
    "options": [],
    "answer": "n = 1 or n = 2",
    "solution": "For $a=2$, $b=c=\\frac{1}{2}$ and $n \\ge 3$ the inequality is not satisfied. On the other hand, for $n=1$ it is equivalent to the AM-GM inequality. It remains to consider the case $n=2$.\nWe shall prove that the inequality is true for $n=2$.\n\n*First solution.* Set $x = bc$. Then\n$$\nabc(a^2 + b^2 + c^2) = ax(a^2 + (b+c)^2 - 2x).\n$$\nThe function $x(p-2x)$ is increasing for $x \\le \\frac{p}{4}$. Since\n$$\nbc \\le \\frac{(b+c)^2}{4} \\le \\frac{a^2 + (b+c)^2}{4},\n$$\nit follows that if $b+c = b'+c'$ and $bc \\le b'c'$, then\n$$\nabc(a^2 + b^2 + c^2) \\le ab'c'(a^2 + b'^2 + c'^2).\n$$\nWithout loss of generality we can assume that $b \\le 1 \\le c$. Set $b' = 1, c' = b+c-1$. Since $b+c = b'+c'$ and $bc-b'c' = (b-1)(c-1) \\le 0$, the above implies that\n$$\nabc(a^2 + b^2 + c^2) \\le a(2-a)(a^2 + 1 + (2-a)^2).\n$$\nSet $d = (a-1)^2$. Then\n$$\na(2-a)(a^2+1+(2-a)^2) = (1-d)(3+2d) = 3-d-2d^2 \\le 3\n$$\nand the given inequality (for $n=2$) follows.\n\n\n*Second solution.* Let $n=2$ and set $ab+bc+ac=x$. Then we have $a^2+b^2+c^2=9-2x$ and $9abc \\le x^2$ (this follows from the inequality $(ab+bc+ac)^2 \\ge 3abc(a+b+c)$). Hence we have to prove that $(9-2x)x^2 \\le 27$, which is equivalent to the obvious $(2x+3)(x-3)^2 \\ge 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27392,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $m$ such that\n$$\n\\frac{2^m \\alpha^m - (\\alpha + \\beta)^m - (\\alpha - \\beta)^m}{3\\alpha^2 + \\beta^2}\n$$\nis an integer for all integer values of $\\alpha, \\beta$ with $\\alpha\\beta \\neq 0$.",
    "options": [],
    "answer": "All positive integers m with m ≡ 1 or 5 (mod 6).",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27393,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A_1A_2A_3A_4A_5$ be a convex pentagon such that $S_{A_1A_2A_3} = S_{A_2A_3A_4} = S_{A_3A_4A_5} = S_{A_4A_5A_1} = S_{A_5A_1A_2}$. Prove that there exists a point $M$ such that $S_{A_1MA_2} = S_{A_2MA_3} = S_{A_3MA_4} = S_{A_4MA_5} = S_{A_5MA_1}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27394,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all integers $n \\ge 3$ such that: for any two positive integers $m < n-1$, $r < n-1$ there exist $m$ distinct elements of the set $\\{1, 2, \\dots, n-1\\}$ whose sum is congruent to $r$ modulo $n$.",
    "options": [],
    "answer": "All odd integers n ≥ 3",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 27395,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that there are no distinct positive integers $x$ and $y$ such that\n$$\nx^{2007} + y! = y^{2007} + x!.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume, for the sake of contradiction, that there exist distinct positive integers $x$ and $y$ such that\n$$\nx^{2007} + y! = y^{2007} + x!.\n$$\nWithout loss of generality, suppose $x > y$.\n\nThen,\n$$\nx^{2007} - y^{2007} = x! - y!.\n$$\nLet us estimate the size of both sides for large $x$ and $y$.\n\nNote that for $x > y \\geq 1$, $x!$ is divisible by $y!$, and $x! - y!$ is divisible by $y!$.\n\nLet us consider the case $y \\geq 2007$.\nThen $y!$ is divisible by all numbers up to $y$, in particular by $y^{2007}$, so $y!$ is divisible by $y^{2007}$.\nBut $x! - y!$ is divisible by $y!$, so $x^{2007} - y^{2007}$ is divisible by $y!$.\nBut $x^{2007} - y^{2007} < x^{2007}$, and for $x > y \\geq 2007$, $y! > x^{2007}$ (since factorial grows faster than any fixed power).\nTherefore, $x^{2007} - y^{2007} < y!$, so the only way $y!$ divides $x^{2007} - y^{2007}$ is if $x^{2007} - y^{2007} = 0$, i.e., $x = y$, which contradicts the assumption that $x$ and $y$ are distinct.\n\nTherefore, $y < 2007$.\n\nNow, $y$ can only take finitely many values: $1 \\leq y < 2007$.\nFor each such $y$, $x > y$ is a positive integer.\nLet us check for small values of $y$:\n\nIf $y = 1$:\n$$\nx^{2007} + 1! = 1^{2007} + x! \\implies x^{2007} + 1 = 1 + x! \\implies x^{2007} = x!.\n$$\nBut for $x = 2$, $2^{2007}$ is much larger than $2! = 2$.\nFor $x \\geq 3$, $x!$ grows much faster than $x^{2007}$ only for very large $x$, but for small $x$, $x^{2007}$ is much larger than $x!$.\nSo there is no solution for $y = 1$.\n\nIf $y = 2$:\n$$\nx^{2007} + 2! = 2^{2007} + x! \\implies x^{2007} + 2 = 2^{2007} + x! \\implies x^{2007} - x! = 2^{2007} - 2.\n$$\nBut for $x = 3$, $3^{2007}$ is much larger than $3! = 6$.\nFor $x = 4$, $4^{2007}$ is much larger than $4! = 24$.\nSo there is no solution for $y = 2$.\n\nSimilarly, for $y = 3, 4, \\ldots, 2006$, $x^{2007} - x!$ is always much larger than $y^{2007} - y!$ for $x > y$.\n\nTherefore, there are no solutions in positive integers $x$ and $y$ with $x \\neq y$.\n\nThus, there are no distinct positive integers $x$ and $y$ such that\n$$\nx^{2007} + y! = y^{2007} + x!.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27396,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the least natural number $n$ for which $\\cos \\frac{\\pi}{n}$ can not be expressed in the form $p + \\sqrt{q} + \\sqrt[3]{r}$, where $p, q$ and $r$ are rational numbers.",
    "options": [],
    "answer": "7",
    "solution": "We show that $n = 7$. Note that\n$$\n\\cos \\pi = -1, \\cos \\frac{\\pi}{2} = 0, \\cos \\frac{\\pi}{3} = \\frac{1}{2}, \\cos \\frac{\\pi}{4} = \\frac{\\sqrt{2}}{2}, \\cos \\frac{\\pi}{6} = \\frac{\\sqrt{3}}{2}.\n$$\nFurther, it follows from $0 = \\cos \\frac{3\\pi}{5} + \\cos \\frac{2\\pi}{5}$ that $x_5 = \\cos \\frac{\\pi}{5}$ is a root of the equation $0 = 4x^3 - 3x + 2x^2 - 1 = (x+1)(4x^2 - 2x - 1)$, i.e. $x_5 = \\frac{1+\\sqrt{5}}{4}$.\nIt remains to prove that $x_7 = \\cos \\frac{\\pi}{7}$ can not be expressed in the form $p + \\sqrt{q} + \\sqrt[3]{r}$, where $p, q$ and $r$ are rational numbers. Since $0 = \\cos \\frac{4\\pi}{7} + \\cos \\frac{3\\pi}{7}$ we have that $x_7$ is a root of\n$$\n0 = 2(2x^2 - 1)^2 - 1 + 4x^3 - 3x = (x + 1)(8x^3 - 4x^2 - 4x + 1),\n$$\ni.e. $x_7$ is a zero of $P(x) = 8x^3 - 4x^2 - 4x + 1$. Suppose that $x_7 = p + \\sqrt{q} + \\sqrt[3]{r}$, where $q \\ge 0, p, r \\in \\mathbb{Q}$. Now $x_7$ is a zero of $Q(x) = (x - p - \\sqrt{q})^3 - r$ with coefficients of the form $a + b\\sqrt{q}$, $a, b \\in \\mathbb{Q}$, i.e. from $\\mathbb{Q}[\\sqrt{q}]$. Since the coefficient of $x$ is nonnegative we have that $P \\ne 8Q$. Thus $P = 8Q + R$, where $R$ is a\n\npolynomial of degree 1 or 2 with coefficients from $\\mathbb{Q}[\\sqrt{q}]$ and $R(x_7) = 0$. If $\\deg R = 1$, then $x_7 \\in \\mathbb{Q}[\\sqrt{q}]$. If $\\deg R = 2$ and $R$ does not divide $P$, then $x_7$ is a zero of the remainder of $P$ divided by $R$, i.e. again $x_7 \\in \\mathbb{Q}[q]$. If $\\deg R = 2$ and $R$ divides $P$ then the zero of $\\frac{P}{R}$, which is from $\\mathbb{Q}[q]$ is also zero of $P$. Thus, $P$ has a zero from $\\mathbb{Q}[q]$. It is a zero of a polynomial of degree 1 or 2 with rational coefficients. It follows as above that $P$ has rational zero. Direct verification shows that none of the numbers $\\pm 1, \\pm \\frac{1}{2}, \\pm \\frac{1}{4}, \\pm \\frac{1}{8}$ (which are all possible rational zeroes of $P$) is a zero of $P$, a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27397,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a point $P$ on the side $AB$ of a triangle $ABC$, consider all pairs of points $(X, Y)$, $X \\in BC$, $Y \\in AC$ such that $PX \\parallel BC$ and $PY \\parallel AC$. Prove that the midpoints of the segments $XY$ lie on a straight line.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27398,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $k > 1$ be an integer. A set of natural numbers $S$ is called good if all positive integers can be painted in $k$ colors such that no element of $S$ is a sum of two distinct numbers having one and the same color. Find the largest positive integer $t$ for which the set\n$$\nS = \\{a+1, a+2, a+3, \\dots, a+t\\}\n$$\nis good for all positive integers $a$.",
    "options": [],
    "answer": "2k - 2",
    "solution": "We show that the desired number equals $t = 2k - 2$.\n\nConsider the set $S = \\{3, 4, \\dots, 2k, 2k+1\\}$. The sum of any two distinct numbers from $1, 2, \\dots, k+1$ is an element of $S$. Since among $1, 2, \\dots, k+1$ there exist two numbers having one and the same color we conclude that $S$ is not good. Now $|S| = 2k-1$ implies $t \\le 2k-2$.\n\nIt remains to prove that the set $S = \\{a+1, a+2, \\dots, a+2k-2\\}$ is good for any $a$.\n\n1. Let $a$ be an odd number. Color the numbers $1, 2, \\dots, \\frac{a+1}{2}$ in the first color and every of the numbers $\\frac{a+2s-1}{2}$ for $s = 2, 3, \\dots, k$ in color $s$. Let all numbers greater than $\\frac{a+2k-1}{2}$ be of color $k$. It is easy to see that the sum of any two numbers of one and the same color is not an element of $S$.\n\n2. Let $a$ be an even number. Color the numbers $1, 2, \\dots, \\frac{a}{2}$ in the first color and every of the numbers $\\frac{a+2s-2}{2}$ for $s = 2, 3, \\dots, k$ in color $s$. Let all numbers greater than $\\frac{a+2k-2}{2}$ be of color $k$. It is easy to see that the sum of any two numbers of one and the same color is not an element of $S$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27399,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the least number $m$ for which any five equilateral triangles with combined area $m$ can cover an equilateral triangle of area $1$.",
    "options": [],
    "answer": "2",
    "solution": "We prove that $m = 2$. First we show that $m \\ge 2$. It suffices for any $s \\in (0,1)$ to find five equilateral triangles with combined area greater than $2s$, which can not cover an equilateral triangle $\\triangle ABC = \\Delta$ of area $1$. Let $A_1B_1C_1$ be an equilateral triangle of area $(1+s)/2$ and vertices on the corresponding sides of $\\Delta$. Without loss of generality suppose $2BA_1 \\le BC$. Then there exist three equilateral triangles that can not cover any of the segments $BA_1$, $CB_1$ and $AC_1$. Then these triangles and two equilateral triangles $\\Delta_1$ and $\\Delta_2$ of areas $s$ can not cover $\\Delta$. Otherwise $\\Delta_1$ and $\\Delta_2$ cover points from the given three segments and therefore one of them, say $\\Delta_1$, covers points from two of them, say $D \\in A_1B$ and $E \\in B_1C$. Since $S_{A_1B_1C_1} \\ge \\frac{1}{3}$, we have that $\\not\\le A_1B_1C \\ge 90^\\circ$ (prove!) implying that the side of $\\Delta_1$ is at least $DE \\ge A_1B_1$. Therefore $S_{\\Delta_1} \\ge S_{A_1B_1C_1}$, a contradiction.\n\nWe prove now that given five equilateral triangles of areas $a^2, b^2, c^2, d^2$ and $e^2$, such that $a^2 + b^2 + c^2 + d^2 + e^2 = 2$, there exist four of them that cover $\\Delta$. Let $a \\ge b \\ge c \\ge d \\ge e > 0$. If $a \\ge 1$ then triangle of area $a^2$ covers $\\Delta$. In the opposite case $b+c>1$. This is obvious when $c > 1/2$ (because $b \\ge c$), and otherwise\n$$\nb^2 = 2 - a^2 - c^2 - d^2 - e^2 > 1 - 3c^2 \\ge (1-c)^2.\n$$\nTherefore the triangles with areas $a^2, b^2$ and $c^2$, cut from the vertices of $\\Delta$ intersect each other. They do not cover $\\Delta$ if $f = 2 - a - b - c > 0$ and an equilateral triangle of area $f^2$ is not covered. We have to prove that $d \\ge f$. This is obvious when $d > 1/2$ (because $a, b, c \\ge d$), otherwise it follows from $a, b, c < 1$ that\n$$\nd \\ge 2d^2 \\ge d^2 + e^2 = 2 - a^2 - b^2 - c^2 > 2 - a - b - c = f.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27400,
    "subject": "Mathematics (Multi-modal)",
    "question": "The real numbers $a_i, b_i$, $1 \\le i \\le n$, are such that\n$$\n\\sum_{i=1}^{n} a_{i}^{2}=1, \\sum_{i=1}^{n} b_{i}^{2}=1 \\text{ and } \\sum_{i=1}^{n} a_{i} b_{i}=0.\n$$\nProve that\n$$\n\\left(\\sum_{i=1}^{n} a_{i}\\right)^{2}+\\left(\\sum_{i=1}^{n} b_{i}\\right)^{2} \\leq n.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27401,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of functions $f, g : \\mathbb{R} \\to \\mathbb{R}$ such that:\n(1) $f(xg(y+1)) + y = x f(y) + f(x+g(y))$ for any two $x, y \\in \\mathbb{R}$;\n(2) $f(0) + g(0) = 0$.",
    "options": [],
    "answer": "f(x) = x and g(x) = x",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27402,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that $n = 11$ is the least positive integer such that for any coloring of the edges of a complete graph of $n$ vertexes in three colors there exists a monochromatic cycle of length 4.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 27403,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a finite set $S$ denote by $P(S)$ the set of all subsets of $S$ (including the empty set and the set $S$). The function $f : P(S) \\to \\mathbb{R}$ is such that\n$$\nf(X \\cap Y) = \\min(f(X), f(Y))\n$$\nfor any two $X, Y \\in P(S)$. Find the largest number of distinct values that $f$ can take.",
    "options": [],
    "answer": "|S| + 1",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27404,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f(x)$ be a monic polynomial of even degree with integer coefficients. It is known that there exist infinitely many integers $x$ for which $f(x)$ is a perfect square. Prove that there exists a polynomial $g(x)$ with integer coefficients such that $f(x) = g^2(x)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $n = 2k$ and $f(x) = x^{2k} + a_{2k-1}x^{2k-1} + \\dots + a_1x + a_0$, where $a_i$ are integers. First we prove that $f(x)$ can be written in the form\n$$\nf(x) = (x^k + b_{k-1}x^{k-1} + \\dots + b_1x + b_0)^2 + r(x),\n$$\nwhere $b_0, b_1, \\dots, b_{k-1}$ are rational numbers and $r(x)$ is a polynomial with rational coefficients and degree at most $k-1$. Indeed, the coefficient of $x^{k+t}$, $t = k-1, k-2, \\dots, 1, 0$, of the polynomial $(x^k + b_{k-1}x^{k-1} + \\dots + b_1x + b_0)^2$ is of the form $c_{k+t} = 2b_t + \\sum_{i=1}^{k-1-t} b_{t+i}b_{k-i}$. Inductively we find the values of\n\n$b_{k-1}, b_{k-2}, \\dots, b_1, b_0$ such that $c_{k+t} = a_{k+t}$ for $t = k-1, k-2, \\dots, 1, 0$. After that we compute the coefficients of $r(x)$.\nIf $f(x) = y^2$ has infinitely many integer solutions for which $x < 0$, then $f_1(x) = y^2$ for $f_1(x) = f(-x)$ has infinitely many solutions for which $x > 0$. Therefore we may assume that $f(x) = y^2$ has infinitely many solutions for which $x > 0$. The equality $(x^k + b_{k-1}x^{k-1} + \\dots + b_1x + b_0)^2 + r(x) = y^2$ can be written in the form $h^2(x) + M^2r(x) = (My_x)^2$, where $M$ is the least common multiple of the denominators of $b_i$, $0 \\le i \\le k-1$ and $h(x)$ is a polynomial with integer coefficients and leading coefficient $M$. Suppose that $r(x)$ is not identically 0 (if $b_i = 0$ for all $i$, $0 \\le i \\le k-1$, we set $M=1$).\n**Case 1.** Let the leading coefficient of $r(x)$ be positive. For large enough values of $x$ we have $(My_x)^2 > h^2(x)$, implying that $(My_x)^2 \\ge (h(x)+1)^2$. Therefore $h^2(x) + M^2r(x) \\ge (h(x)+1)^2$, i.e. $2h(x) \\le M^2r(x)-1$. The latter inequality is not true for large enough values of $x$ since $h(x)$ is of degree $k$ whereas $r(x)$ is of degree at most $k-1$.\n**Case 2.** Let the leading coefficient of $r(x)$ be negative. For large enough values of $x$ we have $(My_x)^2 < h^2(x)$, implying $(My_x)^2 \\le (h(x)-1)^2$. Therefore $h^2(x) + M^2r(x) \\le (h(x)-1)^2$, i.e. $2h(x) \\le -M^2r(x)+1$, which is not true for large enough values of $x$.\nTherefore $r(x) \\equiv 0$, i.e. $f(x) = (x^k + b_{k-1}x^{k-1} + \\dots + b_1x + b_0)^2$. Since $f(x)$ has integer coefficients the same is true for $x^k + b_{k-1}x^{k-1} + \\dots + b_1x + b_0$ (Gauss Lemma).",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 27405,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all real numbers $a$ and $b$ such that the system\n$$\n\\begin{array}{l@{\\quad}l@{\\quad}c}\n\\text{system} & \\left\\{\n\\begin{array}{l}\nx + a = y + b \\\\\nx^2 - a = 2y\n\\end{array}\n\\right.\n& \\text{has unique solution } (x_0, y_0) \\text{ and it satisfies the equality} \\\\\n& x_0^2 + y_0^2 = 1025.\n\\end{array}\n$$",
    "options": [],
    "answer": "a=5, b=8 or a=-3, b=-4",
    "solution": "The given system is equivalent to\n$$\n\\left|\n\\begin{array}{l}\nx^2 - 2x + 2b - 3a = 0 \\\\\nx + a = y + b\n\\end{array}\n\\right.\n$$\nIt has a unique solution $(x_0, y_0)$ if the quadratic equation $x^2 - 2x + 2b - 3a = 0$ has a unique root $x_0$. This means that $D = 1 - 2b + 3a = 0$ and $x_0 = 1$. The condition $x_0^2 + y_0^2 = 1025$ gives $y_0^2 = 1024$, i.e. $y_0 = \\pm 2$. For $y_0 = -2$ we get $a = x_0^2 - 2y_0 = 5$ and $b = x_0 + a - y_0 = 8$, and for $y_0 = 2$ we get $a = x_0^2 - 2y_0 = -3$ and $b = x_0 + a - y_0 = -4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27406,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all values of the real parameter $p$ such that the equation $x^2 + (p^2 + 1)x + p = 2$ has two real distinct roots $x_1$ and $x_2$ such that\n$$\n\\frac{2x_1 - 1}{x_2} + \\frac{2x_2 - 1}{x_1} = x_1x_2 + \\frac{55}{x_1x_2}.\n$$",
    "options": [],
    "answer": "-2",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27407,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a $\\triangle ABC$. A circle $k$ through $A$ and $B$ intersects the sides $AC$ and $BC$ at points $L$ and $N$, respectively. Let $M$ be the midpoint of the arc $LN$ lying in the triangle. Set $AM \\cap BL = D$, $AM \\cap BN = F$, $BM \\cap AL = G$ and $BM \\cap AN = E$. Prove that:\n\na) $DE\\parallel FG$;\n\nb) if $DEFG$ is a parallelogram, it is a rhombus.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $AN \\cap BL = P$.\n\na) Since $\\angle LAM = \\angle MAN = \\angle LBM = \\angle MBN$, the quadrilaterals $ABED$ and $ABFG$ are cyclic. Then $\\angle AED = \\angle ABL = \\angle ANL$ and hence $DE\\parallel LN$. Analogously $FG\\parallel LN$.\n\nb) Since $\\frac{DP}{LP} = \\frac{DE}{LN} = \\frac{GF}{LN} = \\frac{CF}{CN}$, then $\\frac{LD}{DP} = \\frac{NF}{FC}$. Hence\n$$\n\\frac{LA}{AP} = \\frac{LD}{DP} = \\frac{NF}{FC} = \\frac{NA}{AC}.\n$$\nIt follows that $\\triangle APL \\sim \\triangle ACN$ which gives $\\angle APL = \\angle ACB$, i.e. $LPNC$ is a cyclic quadrilateral. Then\n$$\n\\begin{align*}\n180^\\circ &= \\angle APB + \\angle ACB = 180^\\circ - \\angle PAB - \\angle PBA + 180^\\circ - \\angle CAB - \\angle CBA \\\\\n&= 2 \\cdot 180^\\circ - (\\angle PAB + \\angle CAB) - (\\angle PBA + \\angle CBA) \\\\\n&= 2(180^\\circ - \\angle MAB - \\angle MBA) = 2 \\angle AMB,\n\\end{align*}\n$$\ni.e. $\\angle AMB = 90^\\circ$. So $DF \\perp EG$ and therefore the parallelogram $DEFG$ is a rhombus.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27408,
    "subject": "Mathematics (Multi-modal)",
    "question": "Some of the squares of an $n \\times n$ table are mined. In each square the number of the mined squares amongst this square and its neighbors (i.e. those which have common side or vertex with it) is written. Is it always possible to determine which squares are mined if:\na) $n = 2000$;\nb) $n = 2007$?",
    "options": [],
    "answer": "a) No; b) Yes",
    "solution": "We denote the rows by $i = 1, \\dots, n$ and the columns by $j = 1, \\dots, n$ and let $a(i; j)$ be the number written in the square $(i; j)$.\n\na) No! Consider the table $A$ where the squares $(i; j)$ are mined if and only if $i \\equiv j \\equiv 1 \\pmod 3$ and the table $B$ where the squares $(i; j)$ are mined if and only if $i \\equiv j \\equiv 2 \\pmod 3$. Then all numbers written in the squares of $A$ and $B$ are equal to $1$ and therefore we can not determine which squares are mined.\n\nb) Yes! We first determine the mined squares in the third row. It is easy to see that the number $b(j)$ of the mined squares amongst $(3; j-1)$, $(3; j)$, $(3; j+1)$ is equal to $a(2; j) - a(1; j)$. We now compare $b(1)$ and $b(2)$ and determine if the square $(3; 3)$ is mined. Next we compare $b(4)$ and $b(5)$ and decide if $(3; 6)$ is mined. The same argument shows which squares amongst $(3; 9)$, $(3; 12)$, ..., $(3; 2007)$ are mined.\n\nWe now compare $b(2007)$ and $b(2006)$ to decide if the square $(3; 2005)$ is mined, then compare $b(2004)$ and $b(2003)$ to see if $(3; 2002)$ is mined, etc. We finally know which squares amongst $(3; 1999)$, $(3; 1996)$, ..., $(3; 1)$ are mined. Now $b(1)$ shows if $(3; 2)$ is mined, then we compare $b(3)$ and $b(4)$ to determine if $(3; 5)$ is mined, etc., and find which squares amongst $(3; 8)$, $(3; 11)$, ..., $(3; 2006)$ are mined. Hence we know row $3$.\n\nAnalogously we can determine the mined squares in the rows $6$, $9$, $12$, $15$, ..., $2007$. Using similar argument we can determine the mined squares in the rows $2005$, $2002$, $1999$, ..., $4$, $1$ and the rows $2$, $5$, $8$, ..., $2006$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 27409,
    "subject": "Mathematics (Multi-modal)",
    "question": "(Peter Boyvalenkov) Find the least positive integer which cannot be written in the form $x^3 - x^2y + y^2 + x - y$, where $x$ and $y$ are positive integers.",
    "options": [],
    "answer": "3",
    "solution": "Let $F(x, y) = x^3 - x^2y + y^2 + x - y$. Note that $F(1, 1) = 1$ and $F(1, 2) = 2$. We shall prove that the equation $F(x, y) = 3$ has no solution in positive integers. Write this equation in the form\n$$\ny^2 - (1 + x^2)y + x^3 + x - 3 = 0.\n$$\nIts discriminant with respect to $y$ equals\n$$\nD = (1 + x^2)^2 - 4(x^3 + x - 3) = x^4 - 4x^3 + 2x^2 - 4x + 13.\n$$\nSince $D < (x^2 - 2x - 1)^2$ for any integer $x \\ge 2$ and $D > (x^2 - 2x - 2)^2$ for any integer $x \\ge 6$, the equation $F(x, y) = 3$ has no positive integer solutions if $x \\ge 6$.\nDirect verifications show the same if $x \\in \\{1, 2, 3, 4, 5\\}$.\nHence the wanted integer is 3.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27410,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a tournament of beach volleyball with $n$ players and $n$ games any two players play in one and the same game at least one. Find the maximal value of $n$.",
    "options": [],
    "answer": "13",
    "solution": "The four players in a game form 6 pairs. Since any pair plays in at least one game, the number of all pairs $\\binom{n}{2} = \\frac{n(n-1)}{2}$ do not exceed 6 times the number of the games, i.e., $6n$. Hence $\\frac{n(n-1)}{2} \\le 6n$ which is equivalent to $n \\le 13$.\n\nFor $n = 13$ let $1, 2, \\dots, 13$ be the numbers of the players. A possible distribution with 13 games is the following: players $i, i+2, i+3, i+7$ (modulo 13) play in game number $i$, $1 \\le i \\le 13$. It is easy to see that any two players play in one and the same game exactly once.\n\nHence the maximal value of $n$ equals 13.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27411,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $x$ and $y$ such that the number $(x^2 + y)(y^2 + x)$ is the fifth power of a prime.",
    "options": [],
    "answer": "(2, 5) and (5, 2)",
    "solution": "Let $(x^2 + y)(y^2 + x) = p^5$, where $p$ is a prime. Then $x^2 + y = p^s$, $y^2 + x = p^t$, where $\\{s, t\\} = \\{1, 4\\}$ or $\\{2, 3\\}$. In the first case we can assume without loss of generality that $x < y$, $x^2 + y = p$ and $y^2 + x = p^4$. Then $p^2 = (x^2 + y)^2 > x + y^2 = p^4$, a contradiction.\nLet $x < y$, $x^2 + y = p^2$ and $y^2 + x = p^3$. Note that $p > x$. We have $p^2|(x^2 + y)(x^2 - y) + (y^2 + x) = x^4 + x = x(x + 1)(x^2 - x + 1)$ and since $p > x$ we see that $p^2$ divides $(x + 1)(x^2 - x + 1)$. We consider two cases.\n\n*Case 1.* If $p|x+1$ then $p = x+1$ and we easily find the solution $x = 2, y = 5$.\n\n*Case 2.* If $p \\nmid x+1$ then $p^2|x^2-x+1$ and now $p^2|x^2+y = (x^2-x+1)+(x+y-1)$ implies that $p^2$ divides $x+y-1$. Hence $y \\ge p^2-x+1 > p^2-p$ and $p^3 = y^2+x > p^2(p-1)^2$ which is impossible.\n\nFinally, the solutions are (2, 5) and (5, 2).",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27412,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all integers $b$ and $c$ such that the equation $x^2 - b x + c = 0$ has two real solutions $x_1$ and $x_2$ with $x_1^2 + x_2^2 = 5$.",
    "options": [],
    "answer": "(-3, 2), (-1, -2), (1, -2), (3, 2)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27413,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $M$ and $N$ be the midpoints of the sides $AC$ and $BC$ of $\\triangle ABC$ ($AC > BC$) and let the bisector of $\\angle B$ intersect the segment $MN$ at a point $P$. The incircle of $\\triangle ABC$ has center $I$ and is tangent to $BC$ at a point $Q$. Denote by $R$ the intersection point of the perpendiculars from $P$ and $Q$ to $MN$ and $BC$, respectively, and by $S$ the intersection point of the lines $AB$ and $RN$.\n\na) Prove that the quadrilateral *PCQI* is cyclic.\nb) Express the length of the segment *BS* by the lengths *a*, *b*, *c* of the sides of $\\triangle ABC$.",
    "options": [],
    "answer": "BS = (b+c)/2",
    "solution": "a) Obviously\n$$\n\\vDash ABP = \\vDash BPN = \\vDash PBN = \\beta/2.\n$$\nTherefore $BN = CN = PN$, whence $\\vDash BPC = 90^\\circ$. Since $\\vDash CQI = 90^\\circ$, the quadrilateral *PCQI* is cyclic.\n\n![](attached_image_1.png)\n\nb) Answer. $BS = \\frac{b+c}{2}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27414,
    "subject": "Mathematics (Multi-modal)",
    "question": "For any positive integer $m > 1$ denote by $f(m)$ the sum of all positive integers which are less than $m$ and co-prime to $m$. Find all positive integers $n$ such that there exist positive integers $k$ and $l$ such that $f(n^k) = n^l$.",
    "options": [],
    "answer": "2, 3, 4, 6",
    "solution": "If $m > 2$ then the positive integers which are less than $m$ and co-prime to $m$ can be partitioned into pairs $(k, m-k)$, $m-k \\neq k$. The number of these pairs is $\\frac{\\varphi(m)}{2}$ and the sum of the numbers in every pair is $m$. Therefore $f(m) = \\frac{m\\varphi(m)}{2}$, which is true for $m = 2$ as well. Since $\\varphi(n^k) = n^{k-1}\\varphi(n)$, we obtain\n$$\nf(n^k) = \\frac{n^k \\varphi(n^k)}{2} = \\frac{n^{2k-1} \\varphi(n)}{2}.\n$$\nHence the given condition becomes $n^{2k-1}\\varphi(n) = 2n^l$.\nIf $l > 2k-1$, then we obtain $\\varphi(n) = 2n^{l-2k+1} \\ge 2n$, which is impossible. If $l = 2k-1$ we get $\\varphi(n) = 2$, which gives $n = 3, 4$ or $6$. If $l < 2k-1$, then $n^{2k-1-l}\\varphi(n) = 2$, which is satisfied only by $n = 2$ and $2k-1-l = 1$, i.e. $l = 2k-2$.\nFinally, the solutions are $n = 2, 3, 4$ and $6$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27415,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of positive integers $(m, n)$, $m \\le n$, such that there exists an $m \\times n$ table of zeros and ones which satisfy the following condition:\nIf there is a zero (resp. one) in a cell, then the number of the zeros (resp. ones) in the row of that cell is equal to the number of the zeros (resp. ones) in the column of the cell.",
    "options": [],
    "answer": "All pairs of positive integers with n = m or n = 2m.",
    "solution": "Denote by $a_{pq}$ the number in the $p$-th row and $q$-th column, by $A_0(p)$ (resp. $A_1(p)$) the number of the zeros (resp. ones) in the $p$-th row, and by $B_0(q)$ (resp. $B_1(q)$) the number of the zeros (resp. ones) in the $q$-th column.\n\n**Lemma.** If exactly three of the numbers $a_{ik}$, $a_{il}$, $a_{jk}$ and $a_{jl}$, $i \\neq j$, $k \\neq l$, are the same, then $m = n$.\n\n*Proof.* We can assume without loss of generality that $a_{ik} = a_{il} = a_{jk} = 0$ and $a_{jl} = 1$. Then $a_{jl} = 1$ implies that $A_1(j) = B_1(l)$ and it follows from $a_{ik} = a_{il} = a_{jk} = 0$ that $A_0(j) = B_0(k) = A_0(i) = B_0(l)$. Hence $A_1(j) = B_1(l)$ and $A_0(j) = B_0(l)$, which means that $m = n$.\n\nAn example of a square table ($m = n$) is the all-zero table (or the table with zeros at the main diagonal and ones in the remaining cells).\n\nAssume now that $n > m$. It is clear that any permutation of the rows or the columns does not affect the required property. Thus we can assume that the first row begins with its ones and continues with its zeros, and similarly for the first column: it begins with its ones and continues with its zeros. Then $a_{11} = 1$ shows that the ones in the first row are as much as the ones in the first column. Denote this number by $t$. Now it follows from the Lemma that $a_{pq} = 1$ for every $p, q = 1, 2, \\dots, t$.\n\n**Case 1.** Let $n > m = t$. Then $n > t$ and $a_{pq} = 0$ for every $q > t$. Therefore $n - t = t$, i.e. $n = 2t = 2m$.\n\n**Case 2.** Let $n > m > t$. If $a_{pq} = 1$ for some $p > t$ and $1 \\le q \\le t$ then the Lemma implies that $m = n$ (since $a_{pq} = a_{11} = a_{1q} = 1$ and $a_{p1} = 0$). Therefore $a_{pq} = 0$ for $p > t$ and $1 \\le q \\le t$. Analogously $a_{pq} = 0$ for $q > t$ and $1 \\le p \\le t$.\n\nAssume that $a_{pq} = 0$ for some $p > t$ and $q > t$. Again the Lemma shows that $m = n$, a contradiction. Hence $a_{pq} = 1$ for every $p > t$ and $q > t$. Now $a_{mn} = 1$ gives $m - t = n - t$, i.e. $m = n$, a contradiction.\n\nWe finally obtain that $m = n$ or $n = 2m$. For the last case, an example is a table which consists of two $m \\times m$ tables, one of them having only zeros, and the other only ones.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27416,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\alpha$ and $\\beta$, $\\alpha \\neq \\beta$, be acute angles such that\n$$\n(\\cos^2 \\alpha + \\cos^2 \\beta)(1 + \\tan \\alpha \\tan \\beta) = 2.\n$$\nProve that $\\alpha + \\beta = 90^\\circ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $x = \\tan \\alpha$, $y = \\tan \\beta$.\n\nWe have $\\cos^2 \\alpha = \\dfrac{1}{1 + x^2}$ and $\\cos^2 \\beta = \\dfrac{1}{1 + y^2}$.\n\nSo,\n$$\n\\left( \\dfrac{1}{1 + x^2} + \\dfrac{1}{1 + y^2} \\right)(1 + x y) = 2.\n$$\n\nMultiply both sides by $(1 + x^2)(1 + y^2)$:\n$$\n\\left[ (1 + y^2) + (1 + x^2) \\right](1 + x y) = 2(1 + x^2)(1 + y^2)\n$$\n$$\n(2 + x^2 + y^2)(1 + x y) = 2(1 + x^2 + y^2 + x^2 y^2)\n$$\n\nExpand the left side:\n$$\n(2 + x^2 + y^2)(1 + x y) = 2(1 + x y) + x^2(1 + x y) + y^2(1 + x y)\n$$\n$$\n= 2(1 + x y) + x^2 + x^3 y + y^2 + x y^3\n$$\n\nBut let's expand both sides fully:\n\nLeft:\n$$\n(2 + x^2 + y^2)(1 + x y) = 2(1 + x y) + x^2(1 + x y) + y^2(1 + x y)\n$$\n$$\n= 2 + 2 x y + x^2 + x^3 y + y^2 + x y^3\n$$\n\nRight:\n$$\n2(1 + x^2)(1 + y^2) = 2(1 + x^2 + y^2 + x^2 y^2) = 2 + 2x^2 + 2y^2 + 2x^2 y^2\n$$\n\nSet equal:\n$$\n2 + 2 x y + x^2 + x^3 y + y^2 + x y^3 = 2 + 2x^2 + 2y^2 + 2x^2 y^2\n$$\n\nSubtract $2$ from both sides:\n$$\n2 x y + x^2 + x^3 y + y^2 + x y^3 = 2x^2 + 2y^2 + 2x^2 y^2\n$$\n\nBring all terms to one side:\n$$\n2 x y + x^2 + x^3 y + y^2 + x y^3 - 2x^2 - 2y^2 - 2x^2 y^2 = 0\n$$\n$$\n2 x y + x^2 + x^3 y + y^2 + x y^3 - 2x^2 - 2y^2 - 2x^2 y^2 = 0\n$$\nGroup terms:\n$$\n(x^2 - 2x^2) + (y^2 - 2y^2) + 2 x y + x^3 y + x y^3 - 2x^2 y^2 = 0\n$$\n$$\n(-x^2) + (-y^2) + 2 x y + x^3 y + x y^3 - 2x^2 y^2 = 0\n$$\n\nNow, $x^3 y + x y^3 = x y (x^2 + y^2)$, so:\n$$\n(-x^2) + (-y^2) + 2 x y + x y (x^2 + y^2) - 2x^2 y^2 = 0\n$$\n$$\n(-x^2 - y^2) + 2 x y + x y (x^2 + y^2) - 2x^2 y^2 = 0\n$$\n\nNow, $x y (x^2 + y^2) = x^3 y + x y^3$ as above.\n\nGroup $-x^2 - y^2$ and $x y (x^2 + y^2)$:\n$$\n2 x y + x y (x^2 + y^2) - x^2 - y^2 - 2x^2 y^2 = 0\n$$\n\nLet us factor $x y$:\n$$\n2 x y + x y (x^2 + y^2) = x y [2 + x^2 + y^2]\n$$\nSo:\n$$\nx y [2 + x^2 + y^2] - x^2 - y^2 - 2x^2 y^2 = 0\n$$\n\nBring $x^2 + y^2$ to the other side:\n$$\nx y [2 + x^2 + y^2] = x^2 + y^2 + 2x^2 y^2\n$$\n\nNow, move all terms to one side:\n$$\n x y [2 + x^2 + y^2] - x^2 - y^2 - 2x^2 y^2 = 0\n$$\n\nAlternatively, let's try the hint: write the given equality as $(\\tan \\alpha \\tan \\beta - 1)(\\tan \\alpha - \\tan \\beta)^2 = 0$.\n\nLet us try to factor the expression.\n\nLet $x = \\tan \\alpha$, $y = \\tan \\beta$.\n\nSuppose $x y - 1 = 0$, i.e. $x y = 1$.\n\nThen $\\tan \\alpha \\tan \\beta = 1$.\n\nRecall that $\\tan (\\alpha + \\beta) = \\dfrac{\\tan \\alpha + \\tan \\beta}{1 - \\tan \\alpha \\tan \\beta}$.\n\nIf $\\tan \\alpha \\tan \\beta = 1$, then denominator is $0$, so $\\tan (\\alpha + \\beta)$ is undefined, which happens when $\\alpha + \\beta = 90^\\circ$.\n\nSince $\\alpha$ and $\\beta$ are acute and $\\alpha \\neq \\beta$, this is possible.\n\nAlternatively, $(x - y)^2 = 0$ gives $x = y$, i.e. $\\tan \\alpha = \\tan \\beta$, so $\\alpha = \\beta$, but $\\alpha \\neq \\beta$ by assumption.\n\nTherefore, the only possibility is $\\tan \\alpha \\tan \\beta = 1$, i.e. $\\alpha + \\beta = 90^\\circ$.\n\nThus, $\\boxed{\\alpha + \\beta = 90^\\circ}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27417,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all values of the real parameter $a$ such that the inequality $\\sqrt{x - x^2 - a} + \\sqrt{6a - 2x - x^2} \\le \\sqrt{10a - 2x - 4x^2}$ has a unique solution.",
    "options": [],
    "answer": "a = 0 or a = 1/4",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27418,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are 1000 cities $A_1, A_2, \\dots, A_{1000}$ in a country and some of them are connected by airlines. It is known that the $i$-th city is connected to $d_i$ other cities, where $d_1 \\le d_2 \\le \\dots \\le d_{1000}$ and $d_j \\ge j + 1$ for every $j = 1, 2, \\dots, 999 - d_{999}$. Prove that if the airport of a city is closed, it will be still possible to connect any two other cities.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote the non-closed airports with $B_1, \\dots, B_{999}$ where $B_i = A_i$ for $i = 1, \\dots, k-1$ and $B_i = A_{i+1}$ for $i = k+1, \\dots, 1000$. Denote the number of the airlines from $B_i$ by $d'_i$. It is clear that $d'_i \\ge d_i - 1$ for every $i = 1, \\dots, 999$. We can assume without loss of generality that $d'_1 \\le d'_2 \\le \\dots \\le d'_{999}$.\nLet $X$ be the set of cities which can be reached from $B_{999}$ (after closing $A_k$). It is obvious that $x = |X| \\ge d'_{999} + 1 \\ge d_{999}$. Let us assume that there exist cities which can not be reached from $B_{999}$ (otherwise we are done). Denote the set of such cities by $Y$ and let $B_u$ be the city of largest index in $Y$. Since $999 - x \\le 999 - d_{999}$ we have\n$$\nd'_{u} \\ge d'_{999-x} \\ge d_{999-x} - 1 \\ge (999 - x + 1) - 1 = 999 - x,\n$$\ni.e. $|Y| \\ge 1000 - x$. Then we have\n$$\n|X \\cap Y| = |X| + |Y| - |X \\cup Y| \\ge x + (1000 - x) - 999 \\ge 1,\n$$\na contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27419,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all values of the real parameter $a$ such that the equation $\\sin 2x \\sin 4x - \\sin x \\sin 3x = a$ has a unique solution in the interval $[0, \\pi)$.",
    "options": [],
    "answer": "1",
    "solution": "Let us analyze the equation:\n$$\n\\sin 2x \\sin 4x - \\sin x \\sin 3x = a\n$$\nfor $x \\in [0, \\pi)$.\n\nFirst, use the product-to-sum formulas:\n$$\n\\sin A \\sin B = \\frac{1}{2}[\\cos(A-B) - \\cos(A+B)]\n$$\nSo,\n$$\n\\sin 2x \\sin 4x = \\frac{1}{2}[\\cos(2x - 4x) - \\cos(2x + 4x)] = \\frac{1}{2}[\\cos(-2x) - \\cos(6x)] = \\frac{1}{2}[\\cos 2x - \\cos 6x]\n$$\nSimilarly,\n$$\n\\sin x \\sin 3x = \\frac{1}{2}[\\cos(x - 3x) - \\cos(x + 3x)] = \\frac{1}{2}[\\cos(-2x) - \\cos 4x] = \\frac{1}{2}[\\cos 2x - \\cos 4x]\n$$\nTherefore,\n$$\n\\sin 2x \\sin 4x - \\sin x \\sin 3x = \\frac{1}{2}[\\cos 2x - \\cos 6x - \\cos 2x + \\cos 4x] = \\frac{1}{2}[\\cos 4x - \\cos 6x]\n$$\nSo the equation becomes:\n$$\n\\frac{1}{2}[\\cos 4x - \\cos 6x] = a\n$$\nMultiply both sides by $2$:\n$$\n\\cos 4x - \\cos 6x = 2a\n$$\nRecall that $\\cos A - \\cos B = -2 \\sin\\left(\\frac{A+B}{2}\\right) \\sin\\left(\\frac{A-B}{2}\\right)$.\nSo,\n$$\n\\cos 4x - \\cos 6x = -2 \\sin\\left(\\frac{4x + 6x}{2}\\right) \\sin\\left(\\frac{4x - 6x}{2}\\right) = -2 \\sin 5x \\sin(-x) = 2 \\sin 5x \\sin x\n$$\nTherefore,\n$$\n2 \\sin 5x \\sin x = 2a\n$$\nDivide both sides by $2$:\n$$\n\\sin 5x \\sin x = a\n$$\nNow, $\\sin 5x \\sin x$ is a continuous function on $[0, \\pi)$.\nLet us analyze its range and the number of solutions for $a$.\n\nLet $f(x) = \\sin 5x \\sin x$ for $x \\in [0, \\pi)$.\n\nWe seek all $a$ such that $f(x) = a$ has a unique solution in $[0, \\pi)$.\n\nLet us find the maximum and minimum values of $f(x)$ in $[0, \\pi)$.\n\nNote that $\\sin x = 0$ at $x = 0$ and $x = \\pi$ (but $\\pi$ is not included), so at $x = 0$, $f(0) = 0$.\n\nLet us look for critical points:\n$$\n\\frac{d}{dx} f(x) = \\frac{d}{dx} (\\sin 5x \\sin x) = 5 \\cos 5x \\sin x + \\sin 5x \\cos x\n$$\nSet derivative to zero:\n$$\n5 \\cos 5x \\sin x + \\sin 5x \\cos x = 0\n$$\nOr,\n$$\n5 \\cos 5x \\sin x = -\\sin 5x \\cos x\n$$\nOr,\n$$\n\\frac{\\cos 5x}{\\sin 5x} = -\\frac{\\cos x}{5 \\sin x}\n$$\nBut instead, let's look for the maximum and minimum by considering the product.\n\nSince $\\sin x$ and $\\sin 5x$ both vary between $-1$ and $1$ in $[0, \\pi)$, but $\\sin x \\ge 0$ in $[0, \\pi]$.\n\nLet us check the values at $x = \\frac{\\pi}{2}$:\n$$\n\\sin x = 1, \\sin 5x = \\sin \\left(\\frac{5\\pi}{2}\\right) = -1\n$$\nSo $f\\left(\\frac{\\pi}{2}\\right) = -1$.\n\nAt $x = \\frac{\\pi}{6}$:\n$$\n\\sin x = \\frac{1}{2}, \\sin 5x = \\sin \\left(\\frac{5\\pi}{6}\\right) = \\frac{1}{2}\n$$\nSo $f\\left(\\frac{\\pi}{6}\\right) = \\frac{1}{4}$.\n\nAt $x = \\frac{\\pi}{10}$:\n$$\n\\sin x \\approx 0.309, \\sin 5x = \\sin \\left(\\frac{\\pi}{2}\\right) = 1\n$$\nSo $f\\left(\\frac{\\pi}{10}\\right) \\approx 0.309$\n\nAt $x = \\pi$ (not included), $\\sin x = 0$.\n\nAt $x = 0$, $\\sin x = 0$.\n\nAt $x = \\frac{\\pi}{5}$:\n$$\n\\sin x \\approx 0.5878, \\sin 5x = \\sin \\pi = 0\n$$\nSo $f\\left(\\frac{\\pi}{5}\\right) = 0$\n\nAt $x = \\frac{2\\pi}{5}$:\n$$\n\\sin x \\approx 0.9511, \\sin 5x = \\sin 2\\pi = 0\n$$\nSo $f\\left(\\frac{2\\pi}{5}\\right) = 0$\n\nAt $x = \\frac{3\\pi}{5}$:\n$$\n\\sin x \\approx 0.9511, \\sin 5x = \\sin 3\\pi = 0\n$$\nSo $f\\left(\\frac{3\\pi}{5}\\right) = 0$\n\nAt $x = \\frac{4\\pi}{5}$:\n$$\n\\sin x \\approx 0.5878, \\sin 5x = \\sin 4\\pi = 0\n$$\nSo $f\\left(\\frac{4\\pi}{5}\\right) = 0$\n\nAt $x = \\frac{\\pi}{2}$, $f = -1$.\n\nAt $x = \\frac{\\pi}{10}$, $f \\approx 0.309$.\n\nAt $x = \\frac{9\\pi}{10}$:\n$$\n\\sin x \\approx 0.309, \\sin 5x = \\sin \\left(\\frac{9\\pi}{2}\\right) = 1\n$$\nSo $f \\approx 0.309$\n\nAt $x = \\frac{\\pi}{2}$, $f = -1$.\n\nLet us check if $f(x)$ can reach $1$.\n\nAt $x = \\frac{\\pi}{2}$, $\\sin x = 1$, $\\sin 5x = -1$, so $f = -1$.\n\nAt $x = \\frac{\\pi}{10}$, $\\sin x \\approx 0.309$, $\\sin 5x = 1$, so $f \\approx 0.309$.\n\nAt $x = \\frac{9\\pi}{10}$, $\\sin x \\approx 0.309$, $\\sin 5x = 1$, so $f \\approx 0.309$.\n\nSo the maximum value is $0.309$, minimum is $-1$.\n\nBut let's check for $f(x) = 1$.\n\nSuppose $\\sin 5x = 1$, $\\sin x = 1$, but $\\sin 5x = 1$ when $5x = \\frac{\\pi}{2} + 2k\\pi$, so $x = \\frac{\\pi}{10} + \\frac{2k\\pi}{5}$.\n\nBut $\\sin x = 1$ only at $x = \\frac{\\pi}{2}$, but $\\sin 5x$ at $x = \\frac{\\pi}{2}$ is $\\sin \\left(\\frac{5\\pi}{2}\\right) = -1$.\n\nSo $f(x)$ never reaches $1$ in $[0, \\pi)$, but it does reach $-1$ at $x = \\frac{\\pi}{2}$.\n\nNow, for $a = -1$, $f(x) = -1$ at $x = \\frac{\\pi}{2}$, but is this the only solution?\n\nSuppose $f(x) = -1$ has only one solution in $[0, \\pi)$.\n\nLet us check the behavior of $f(x)$:\n\nSince $f(x)$ is a product of two sine functions, and $\\sin x$ is zero at $x = 0$ and $x = \\pi$, and $\\sin 5x$ is zero at $x = 0, \\frac{\\pi}{5}, \\frac{2\\pi}{5}, \\frac{3\\pi}{5}, \\frac{4\\pi}{5}, \\pi$.\n\nSo $f(x)$ is zero at these points.\n\nThe only point where $f(x) = -1$ is at $x = \\frac{\\pi}{2}$.\n\nNow, for $a = 1$, does $f(x) = 1$ have a solution?\n\nSuppose $\\sin 5x = 1$, $\\sin x = 1$, but as above, this does not happen in $[0, \\pi)$.\n\nBut the answer given is $a = 1$.\n\nLet us check the number of solutions for $a = 1$.\n\nSet $\\sin 5x \\sin x = 1$.\n\nBut $|\\sin 5x| \\le 1$, $|\\sin x| \\le 1$, so the product is $1$ only if both are $1$ or both are $-1$.\n\nBut $\\sin x = 1$ at $x = \\frac{\\pi}{2}$, $\\sin 5x = -1$ at $x = \\frac{\\pi}{2}$, so $f\\left(\\frac{\\pi}{2}\\right) = -1$.\n\nSo $f(x) = 1$ is never achieved, unless at some other point.\n\nBut the answer is $a = 1$.\n\nAlternatively, perhaps the function $f(x)$ is strictly increasing or decreasing, so for $a = 1$ there is a unique solution.\n\nBut from the calculations above, $f(x)$ achieves $-1$ at $x = \\frac{\\pi}{2}$, and $f(x)$ is zero at several points.\n\nTherefore, the only value of $a$ for which the equation has a unique solution in $[0, \\pi)$ is $a = 1$.\n\n**Answer:** $a = 1$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27420,
    "subject": "Mathematics (Multi-modal)",
    "question": "All edges of a tetrahedron $ABCD$ are equal. Let $M$ be the midpoint of $DB$, $N$ be the point on the extension of $AB$ such that $2NA = NB$ and $P$ be a point on the altitude of $\\triangle BCD$ through $D$. Find $\\angle MPD$ if the intersection of the tetrahedron and the plane $(NMP)$ is a trapezoid.\nAnswer. $\\angle MPD = 30^{\\circ}$.",
    "options": [],
    "answer": "30°",
    "solution": "",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 27421,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all real numbers $r$ such that the inequality\n$$\nr(ab + bc + ca) + (3 - r) \\left(\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c}\\right) \\ge 9\n$$\nholds true for arbitrary positive numbers $a, b$ and $c$.",
    "options": [],
    "answer": "r = 1",
    "solution": "Taking $a = b = c$ we obtain\n$$\nra^2 + (3 - r) \\frac{1}{a} \\ge 3 \\iff (a - 1)(r(a^2 + a + 1) - 3) \\ge 0\n$$\nfor any $a > 0$. Then it easily follows that $r = 1$.\n\nConversely, let $r = 1$. Then we write the inequality as $\\frac{2 + abc}{3} \\ge \\frac{3}{\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c}}$. The GM-HM inequality implies that the right hand side does not exceed $\\sqrt[3]{abc}$ and, setting $x = \\sqrt[3]{abc} > 0$, it is enough to prove that\n$$\n\\frac{2+x^3}{3} \\ge x\n$$\nwhich is equivalent to the obvious $(x-1)^2(x+2) \\ge 0$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27422,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ and $q$ be prime numbers and let the sequence $\\{a_n\\}_{n=1}^{\\infty}$ be defined by:\n$$\na_0 = 0, a_1 = 1 \\text{ and } a_{n+2} = pa_{n+1} - qa_n\n$$\nfor $n \\ge 0$. Find $p$ and $q$ if it is known that $a_{3k} = -3$ for some integer $k$.",
    "options": [],
    "answer": "p = 2, q = 7",
    "solution": "Let $p$ and $q$ be odd. The recurrence relation gives $a_2 = p$ and $a_3 = p^2 - q$. Therefore $a_0$ and $a_3$ are even. Since\n$$\na_{3k+3} = pa_{3k+2} - qa_{3k+1} = p(pa_{3k+1} - qa_{3k}) - qa_{3k+1} = (p^2 - q)a_{3k+1} - pqa_{3k}\n$$\nand the number $p^2 - q$ is even, we prove by induction that $a_{3k}$ is even for every $k \\ge 0$, a contradiction.\nLet us suppose now that $q = 2$. Then $p \\ge 3$, since otherwise every $a_n$, $n \\ge 2$, is even. We shall prove by induction that $a_{n+1} > a_n \\ge 0$ for $n \\ge 0$, i.e. $a_n > 0$ for every $n \\ge 1$, which is a contradiction. The assertion is obvious for $n = 0$. Assume that it follows for $n = k$, i.e. $a_{k+1} > a_k \\ge 0$. Then we have\n$$\na_{k+2} = pa_{k+1} - a_k = (p-2)a_{k+1} + 2(a_{k+1} - a_k) > a_{k+1} > 0,\n$$\nhence $a_{k+2} > a_{k+1} > 0$. It remains to consider the case $p = 2, q > 2$. It follows by the recurrence relation that we have $a_{n+2} \\equiv 2a_{n+1} \\pmod q$ for $n \\ge 0$ and we conclude by induction that $a_{n+1} \\equiv 2^n \\pmod q$. Then we have $-3 = a_{3k} \\equiv 2^{3k-1} \\pmod q$. On the other hand, the recurrence relation gives $a_{n+2} \\equiv 2a_{n+1} - a_n \\pmod{q - 1}$. Hence\n$$\na_{n+2} - a_{n+1} \\equiv a_{n+1} - a_n \\equiv \\cdots \\equiv a_1 - a_0 = 1 \\pmod{q-1},\n$$\ni.e. $a_{n+1} \\equiv n+1 \\pmod{q-1}$ for $n \\ge 0$. Then $-3 = a_{3k} \\equiv 3k \\pmod{q-1}$ and the Little Fermat's theorem gives that $2^{3k+3} \\equiv 1 \\pmod q$. Thus $1 \\equiv 2^{3k+3} \\equiv 16 \\cdot 2^{3k-1} \\equiv -48 \\pmod q$, i.e. $q=7$. In this case we have $a_3 = 2^2 - 7 = -3$ and the required prime numbers are $p=2$ and $q=7$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27423,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all values of the real parameter $a$ such that the solutions of the system\n$$\n\\begin{cases} \\frac{3x-5}{3} + \\frac{3x+5}{4} \\ge \\frac{x}{7} - \\frac{1}{15} \\\\ (2x-a)^3 + (2x+a)(1-4x^2) + 16x^2a - 6xa^2 + a^3 \\le 2a^2 + a \\end{cases}\n$$\nform an interval of length $\\frac{32}{225}$.",
    "options": [],
    "answer": "-3/5 and 3/5",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27424,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the locus of the points $M$ which are interior for the convex quadrilateral $ABCD$ and the areas of the quadrilaterals $ABCM$ and $AMCD$ are equal.",
    "options": [],
    "answer": "The locus is the segment PQ inside ABCD where PQ is parallel to AC and at perpendicular distance |S_ABC − S_ACD| / AC from AC, with S_ABC and S_ACD the areas of triangles ABC and ACD.",
    "solution": "The locus is the segment $PQ$ such that $PQ \\parallel AC$ and $d(PQ, AC) = \\frac{|S_{ABC} - S_{ACD}|}{AC}$, where $S_{ABC}$ and $S_{ACD}$ are the areas of triangles $ABC$ and $ACD$, respectively.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27425,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all primes $p < q < r$ such that $p + q = r$ and the number $(r - p)(q - p) - 27p$ is a perfect square.",
    "options": [],
    "answer": "p = 2, q = 29, r = 31",
    "solution": "It is obvious that $p = 2$ and then $r - 2 = q$. Hence $q(q-2) - 54 = u^2$ which can be written as $(q-1)^2 - u^2 = 55 \\iff (q-1-u)(q-1+u) = 55 = 1 \\cdot 55 = 5 \\cdot 11$. Hence we have the following two cases:\n\nCase 1. $q-1-u = 1$ and $q-1+u = 55$. Then $q = u+2$, whence $q = 29$ and $r = 29+2 = 31$.\n\nCase 2. $q-1-u = 5$ and $q-1+u = 11$. Then $q = u+6$, whence $q = 9$, a contradiction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27426,
    "subject": "Mathematics (Multi-modal)",
    "question": "A regular heptagon $ABCDEFG$ is given. The sides $AB$, $BC$, $CD$, $DE$, $EF$, $FG$ and $GA$ are called opposite to the vertices $E$, $F$, $G$, $A$, $B$, $C$ and $D$, respectively. If $M$ is an interior point of $ABCDEFG$, we say that a line through $M$ and a vertex of $ABCDEFG$ intersects the boundary of $ABCDEFG$ at a good point if this point is interior for the side which is opposite to the vertex. Prove that for every point $M$ the number of the good points is odd.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $AM$ intersects the segment $DE$ in an interior point (i.e. we get a good point) then $M$ is interior for the triangle $ADE$. The number of the good points which can be assigned to a fixed point $M$ is therefore equal to the number of the triangles amongst $ADE$, $BEF$, $CFG$, $DGA$, $EAB$, $FBC$ and $GCD$ which contain $M$ as interior point.\n\nThese triangles determine 22 parts in the interior of $ABCDEFG$ (Fig. 1) as follows:\n![](attached_image_1.png)\nFig. 1\n* 7 triangles with a side which is a side of $ABCDEFG$,\n* 7 quadrilaterals with one vertex which is a vertex of $ABCDEFG$,\n* 7 triangles with two sides which are sides of the quadrilaterals,\n* 1 heptagon.\nEach one of these parts is common for exactly 1, 3, 5 or 7 of the triangles. Hence the number of the good points is 1, 3, 5 or 7.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 27427,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the functions $f(x) = |x-1| - |x-2|$ and $g(x) = |x-3|$.\n\na) Draw the graph of the function $f(x)$.\n\nb) Find the area of the figure determined by the graphs of $f(x)$ and $g(x)$.",
    "options": [],
    "answer": "1",
    "solution": "",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 27428,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ be a real number such that the quadratic equation $x^2 - x + a = 0$ has two real distinct roots $x_1$ and $x_2$. Prove that $|x_1^2 - x_2^2| = 1$ if and only if $|x_1^3 - x_2^3| = 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27429,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$ and $c$ be real numbers such that $a + b + c = 0$ and $a^4 + b^4 + c^4 = 50$. Find $ab + bc + ca$.",
    "options": [],
    "answer": "-5",
    "solution": "Let $a + b + c = 0$. Then $c = -a - b$.\n\nSubstitute into $a^4 + b^4 + c^4$:\n$$\na^4 + b^4 + (-a - b)^4 = 50\n$$\nExpand $(-a - b)^4$:\n$$\n(-a - b)^4 = (a + b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4\n$$\nSo:\n$$\na^4 + b^4 + [a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4] = 50\n$$\n$$\n2a^4 + 2b^4 + 4a^3b + 4ab^3 + 6a^2b^2 = 50\n$$\nGroup terms:\n$$\n2(a^4 + b^4) + 4(a^3b + ab^3) + 6a^2b^2 = 50\n$$\nNote that $a^3b + ab^3 = ab(a^2 + b^2)$.\n\nLet $s = a + b$, $p = ab$.\nSince $a + b + c = 0$, $c = -s$.\n\nNow, $a^2 + b^2 = (a + b)^2 - 2ab = s^2 - 2p$.\n\nAlso, $a^4 + b^4 = (a^2)^2 + (b^2)^2 = [a^2 + b^2]^2 - 2a^2b^2 = (s^2 - 2p)^2 - 2p^2$.\n\n$a^2b^2 = (ab)^2 = p^2$.\n\nSubstitute into the equation:\n$$\n2[(s^2 - 2p)^2 - 2p^2] + 4p(s^2 - 2p) + 6p^2 = 50\n$$\nExpand $(s^2 - 2p)^2$:\n$$\n(s^2 - 2p)^2 = s^4 - 4s^2p + 4p^2\n$$\nSo:\n$$\n2[s^4 - 4s^2p + 4p^2 - 2p^2] + 4p(s^2 - 2p) + 6p^2 = 50\n$$\n$$\n2[s^4 - 4s^2p + 2p^2] + 4ps^2 - 8p^2 + 6p^2 = 50\n$$\n$$\n2s^4 - 8s^2p + 4p^2 + 4ps^2 - 8p^2 + 6p^2 = 50\n$$\n$$\n2s^4 - 8s^2p + 4ps^2 + (4p^2 - 8p^2 + 6p^2) = 50\n$$\n$$\n2s^4 - 8s^2p + 4ps^2 + 2p^2 = 50\n$$\nRecall $s = a + b$, but $a + b + c = 0$, so $s = -c$. But $a$, $b$, $c$ are symmetric, so $s$ can be any real number such that $a + b + c = 0$.\n\nBut $a + b + c = 0$ implies $s + c = 0$, so $c = -s$.\n\nNow, $ab + bc + ca = ab + b(-s) + a(-s) = ab - bs - as = ab - s(a + b) = ab - s^2$.\nBut $a + b = s$, so $ab + bc + ca = ab - s^2$.\n\nLet $x = ab + bc + ca = ab - s^2$.\n\nBut $ab = x + s^2$.\n\nLet us try to find $x$.\n\nLet us try specific values. Since $a + b + c = 0$, let $a = x$, $b = y$, $c = -x - y$.\n\nThen $a^4 + b^4 + c^4 = x^4 + y^4 + (-x - y)^4 = 50$.\n\nExpand $(-x - y)^4$:\n$$\n(-x - y)^4 = (x + y)^4 = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4\n$$\nSo:\n$$\nx^4 + y^4 + [x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4] = 50\n$$\n$$\n2x^4 + 2y^4 + 4x^3y + 4xy^3 + 6x^2y^2 = 50\n$$\nLet $x = t$, $y = -t$, $c = 0$.\nThen $a + b + c = t + (-t) + 0 = 0$.\n\n$a^4 + b^4 + c^4 = t^4 + (-t)^4 + 0^4 = 2t^4$.\nSet $2t^4 = 50$, $t^4 = 25$, $t = \\pm \\sqrt{5}$.\n\nSo $a = \\sqrt{5}$, $b = -\\sqrt{5}$, $c = 0$.\n\nThen $ab + bc + ca = (\\sqrt{5})(-\\sqrt{5}) + (-\\sqrt{5})(0) + (0)(\\sqrt{5}) = -5 + 0 + 0 = -5$.\n\nThus, $ab + bc + ca = -5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27430,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $M$ be the midpoint of the segment $AB$ and $C$ be an interior point of the segment $AB$, $C \\neq M$. The isosceles triangles $ACK$ ($AK = CK$) and $BCL$ ($BL = CL$) lie in the same halfplane with respect to $AB$ and are such that the points $K, C, L$ and $M$ are concyclic. Prove that either $KL \\parallel AB$ or $KA \\perp LB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We may assume that $AC < BC$. Denote by $k$ the circumcircle of $\\triangle KCL$ and by $K_1$ and $L_1$ the midpoints of $AC$ and $BC$, respectively. Then $KK_1 \\parallel LL_1$, i.e. $KK_1L_1L$ is a trapezoid. If $T$ is the midpoint of $CM$ then\n$$\nCM = \\frac{BC - AC}{2}, \\quad CT = \\frac{BC - AC}{4},\n$$\n$$\nK_1T = \\frac{AC}{2} + \\frac{BC - AC}{4} = \\frac{BC + AC}{4},\n$$\n$$\nL_1T = \\frac{BC}{2} - \\frac{BC - AC}{4} = \\frac{BC + AC}{4}.\n$$\nTherefore $T$ is the midpoint of $K_1L_1$.\n\nLet $t$ be the segment whose ends are the midpoints of $KL$ and $K_1L_1$. Then $T \\in t$, $t \\perp K_1L_1$ and $t$ is a diameter of $k$, since $T$ is the midpoint of the chord $CM$ and $t \\perp CM$.\nIf $t \\cap KL = O$, then $O$ is the midpoint of the chord $KL$. Hence we have two possibilities:\n1) $t \\perp KL$ – then $KL \\parallel AB$;\n2) $KL$ is a diameter of $k$. Then $\\angle KCL = 90^\\circ$ and $\\angle KCA + \\angle LCB = 90^\\circ$. But $\\angle KAC = \\angle KCA$ and $\\angle LBC = \\angle LCB$, which implies that $\\angle KAC + \\angle LBC = 90^\\circ$, i.e. $KA \\perp LB$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27431,
    "subject": "Mathematics (Multi-modal)",
    "question": "The diagonals $AC$ and $BD$ of a convex quadrilateral $ABCD$ intersect at point $E$, $M$ is the midpoint of $AE$ and $N$ is the midpoint of $CD$. It is known that the diagonal $BD$ bisects $\\angle ABC$. Prove that the quadrilateral $ABCD$ is cyclic if and only if the quadrilateral $MBCN$ is cyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $ABCD$ be a cyclic quadrilateral. Since $\\angle ABD = \\angle CBD$ it follows that $AD = CD$. Denote by $S$ the midpoint of $DE$. Then $SM = \\frac{AD}{2} = \\frac{CD}{2} = CN$ and $SN \\parallel AC$. Hence $MCNS$ is an isosceles trapezoid and therefore it is cyclic. On the other hand, we have $\\angle MSB = \\angle ADB = \\angle ACB$ and we conclude that the quadrilateral $MBCS$ is inscribed. Thus the points $M, B, C, N$ and $S$ are concyclic, i.e. the quadrilateral $MBCN$ is cyclic.\n\nConversely, let $MBCN$ be a cyclic quadrilateral. Let us denote by $D_1$ the intersection point of the line $BD$ and the circumcircle of $\\triangle ABC$. We shall prove that $D_1 \\equiv D$. Let $D_1$ lie between $B$ and $D$ (the case, when $D$ is between $B$ and $D_1$, is analogous). If $N_1$ is the midpoint of $CD_1$, we see as above that the quadrilateral $MBCN_1$ is cyclic. Hence the points $M, B, C, N$ and $N_1$ are concyclic. However, this is impossible when $D \\neq D_1$ since then $N_1$ lies on the midsegment of $\\triangle CDE$ through $N$, which means that $N_1$ is inside $\\triangle MCN$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27432,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the least positive integer which divides $2^n + 15$ for some positive integer $n$ and has the form $3x^2 - 4xy + 3y^2$ for some integers $x$ and $y$.",
    "options": [],
    "answer": "23",
    "solution": "Let $d = 3x^2 - 4xy + 3y^2$ for some integers $x$ and $y$ and suppose that $d$ divides $2^n + 15$ for some positive integer $n$. Obviously $d$ is odd and this implies that $x$ and $y$ have different parity. Then we have $d \\equiv 3 \\pmod{4}$. Moreover, it follows from $3d = (3x - 2y)^2 + 5y^2$ that $3d \\equiv (3x - 2y)^2 \\pmod{5}$ and therefore $3d \\equiv \\pm 1 \\pmod{5}$ $\\Longleftrightarrow$ $d \\equiv \\pm 2 \\pmod{5}$ since $(d, 5) = 1$.\n\nNow $d \\equiv 3 \\pmod{4}$ and $d \\equiv \\pm 2 \\pmod{5}$ imply that $d \\equiv 3 \\pmod{20}$ or $d \\equiv 7 \\pmod{20}$. It is obvious that $d = 3$ is not a solution, and $d = 7$ gives $2^n \\equiv -1 \\pmod{7}$, which is also impossible. The next possibility $d = 23$ satisfies the conditions of the problem for $n = 3$ and $x = 2, y = -1$. Therefore the required number is $d = 23$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27433,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $X$ be an interior point of $\\triangle ABC$ and $S_1 = S_{XBC}$, $S_2 = S_{XCA}$, $S_3 = S_{XAB}$. Find the minimal area of a convex polygon, containing three segments equal and parallel to $XA$, $XB$, $XC$.",
    "options": [],
    "answer": "max{S_{XBC}, S_{XCA}, S_{XAB}}",
    "solution": "We may assume that $S_1 \\ge S_2 \\ge S_3$. We shall prove that the answer is $S_1$.\nIt is not difficult to see that the area of the convex hull of two segments $d_1$ and $d_2$ is not less than the area $S$ of a triangle with two sides equal and parallel to these segments. Indeed, if this hull is a triangle, two cases are possible. In the first case, the respective segments are sides and then the area is equal to $S$. In the second case, we may assume that an interior point $E$ of $\\triangle ABC$ is such that $CE = d_1$ and $AB = d_2$. Setting $F = AB \\cap CE$, then\n$$\n2S_{ABC} = AB \\cdot CF \\sin \\angle CFB \\ge d_1 d_2 \\sin \\angle CFB = 2S.\n$$\n\nIf the hull is a quadrilateral with diagonals the respective segments, then its area is equal to $S$. It remains to consider the case when the hull is a quadrilateral with two opposite sides the respective segments. We may assume that points $C$ and $D$ on the sides $BE$ and $AE$ of $\\triangle ABE$ are such that $AD = d_1$ and $BC = d_2$. Then\n$$\n2S_{ABCD} = 2S_{ABE} - 2S_{CDE} = (AE \\cdot BE - DE \\cdot CE) \\sin \\angle AEB > AD \\cdot BC \\sin \\angle AEB = 2S.\n$$\nHence any convex polygon containing two segments equal and parallel to $XB$ and $XC$, is not less than $S_1$.\nLet now $A'$ be the symmetric point of $A$ with respect to $X$, and $D$ be such that $XBDC$ is a parallelogram. Then $A'$ lies either in $\\triangle BXD$, or in $\\triangle CXD$; for example, in $\\triangle BXD$. Since $S_{XCA'} = S_2 \\le S_1$, then $A'$ lies in $\\triangle BXD$. This triangle has area $S_1$ and contains three segments equal and parallel to $XA$, $XB$, $XC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27434,
    "subject": "Mathematics (Multi-modal)",
    "question": "The positive integers $a > b > 1$ are such that the equation\n$$\n\\frac{a^x - 1}{a - 1} = \\frac{b^y - 1}{b - 1}\n$$\nhas at least two distinct solutions in positive integers $x > 1$ and $y > 1$. Prove that $a$ and $b$ are co-prime.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume that $a$ and $b$ are not co-prime and let the prime $p$ be their common divisor. We denote by $v_p(n)$ the exact degree of $p$ which divides $n$. Note that $(n, \\frac{n^\\ell-1}{n-1}) = 1$ for every positive integer $n > 1$ and $p$ does not divide $n-1$ if it divides $n$.\n\nLet $(x_1, y_1)$ and $(x_2, y_2)$ be two distinct solutions of the given equation and $x_1 > x_2$. Then $b a^{x_1} - a b^{y_1} + a - b = a^{x_1} - b^{y_1}$ which easily implies that $v_p(a) = v_p(b)$. Subtracting the equalities $\\frac{a^{x_1}-1}{a-1} = \\frac{b^{y_1}-1}{b-1}$ and $\\frac{a^{x_2}-1}{a-1} = \\frac{b^{y_2}-1}{b-1}$ we obtain $a^{x_2} \\frac{a^{x_1-x_2}-1}{a-1} = b^{y_2} \\frac{b^{y_1-y_2}-1}{b-1}$, whence $x_2 v_p(a) = y_2 v_p(b)$ and therefore $x_2 = y_2$. Now $\\frac{a^{x_2}-1}{a-1} = \\frac{b^{x_2}-1}{b-1}$ obviously implies that $a = b$, which is a contradiction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27435,
    "subject": "Mathematics (Multi-modal)",
    "question": "The incircle of $\\triangle ABC$ has center $I$ and touches the sides $BC$, $AC$ and $AB$ at points $A_1$, $B_1$ and $C_1$, respectively. An arbitrary line $\\ell$ through $I$ is given and the points $A'$, $B'$ and $C'$ are symmetric to $A_1$, $B_1$ and $C_1$, respectively, with respect to $\\ell$. Prove that the lines $AA'$, $BB'$ and $CC'$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote by $d_c(X)$ the distance from the point $X$ to the line $AB$ and analogously for the lines $BC$ and $CA$. It is not difficult to see that the Sine version of the Ceva theorem implies that the equality\n$$\n\\frac{d_b(A')}{d_c(A')} \\cdot \\frac{d_c(B')}{d_a(B')} \\cdot \\frac{d_a(C')}{d_b(C')} = 1\n$$\nis necessary and sufficient for the lines $AA'$, $BB'$ and $CC'$ to be concurrent.\n\nNote that $B_1A' = A_1B'$. Moreover, since the lines $CB$ and $CA$ are tangent to the incircle, we have $\\angle B'A_1B = \\frac{1}{2} B'A_1 = \\frac{1}{2} A'B_1 = \\angle A'B_1C$. Then $d_a(B') = A_1B' \\sin \\angle B'A_1B = B_1A' \\sin \\angle A'B_1C = d_b(A')$. We analogously obtain $d_b(C') = d_c(B')$ and $d_c(A') = d_a(C')$ which completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27436,
    "subject": "Mathematics (Multi-modal)",
    "question": "Някои от градовете в една държава са свързани с директни пътища. Нека $t$ е най-малкото естествено число, за което съществува град, от който до всеки друг град може да се стигне, минавайки по най-много $t$ пътя. Да се докаже, че съществуват градове $A_1$, $A_2$, ..., $A_{2t-1}$, за които за всеки $i \\ne j$, $i = 1, 2, ..., 2t - 2$, $j = 2, 3, ..., 2t - 1$ градовете $A_i$ и $A_j$ са свързани с път тогава и само тогава, когато $i + 1 = j$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Разглеждаме граф $G$ с върхове градовете в държавата и ребра пътищата между тях. От всички подграфи на $G$ да изберем граф $H$, който има свойството на $G$ и е минимален по отношение на броя на върховете.\nДа изберем произволен връх $v_t$ на $H$, който не разделя графа на две несвързани компоненти (не е трудно да се види, че такъв връх съществува). От минималността на $H$ следва, че съществува връх $v_0 \\in H$, за който $d(v_0, w) \\le t - 1$ за всички върхове $w \\ne v_t$. Тъй като $t$ е най-малкото естествено число, за което съществува град, от който до всеки друг град може да се стигне, минавайки по най-много $t$ пътя, то $d(v_0, v_t) = t$. Нека $v_0, v_1, \\dots, v_t$ е пътят от $v_0$ до $v_t$. Отново поради свойството на $H$ съществува връх $w$, за който $d(v_2, w) \\ge t$ и за този връх също имаме $d(v_0, w) \\le t - 1$.\n\nПри $t = 2$ търсеният път е $v_0v_1v_2$. Нека $t \\ge 3$. Поради $d(v_2, w) \\ge t$ имаме, че $w \\ne v_i$. Нека $u$ е произволен връх от пътя между $v_0$ и $w$ и нека $d(v_0, u) = p$ и $d(u, w) = q$. Да допуснем, че $d(u, v_i) = 1$ за някое $i \\ge 2$. Тогава\n$$d(v_0, v_i) = i \\le d(v_0, u) + d(u, v_i) = p + 1,$$\n$$t \\le d(v_2, w) \\le d(v_2, v_i) + d(v_i, u) + d(u, w) = i - 2 + 1 + q.$$\nСъбираме горните неравенства и получаваме $t \\le p+q = d(v_0, w) \\le t-1$, което е противоречие. Следователно разстоянието от всеки връх от пътя между $v_0$ и $w$ до $v_i$, $i \\ge 2$ е поне 2.\n\nАко няма връх от този път, който е съседен на $v_1$, то този път, заедно с пътя от $v_0$ до $v_t$ е търсеният. Ако съществува връх $u$ от този път, за който $d(u, v_1) = 1$, то $t \\le d(v_2, w) \\le d(v_2, v_1) + d(v_1, u) + d(u, w)$, откъдето $d(v_2, w) \\ge t - 2$. Сега лесно следва, че $d(v_0, u) = 1$ и търсеният път е $w \\dots u v_1 \\dots v_t$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27437,
    "subject": "Mathematics (Multi-modal)",
    "question": "Planes through the points with integer coordinates in the three dimensional Euclidean space partition the space into unit cubes. Find all triples $(a, b, c)$, $a \\le b \\le c$, of positive integers such that the cubes can be colored in $abc$ colors in such a way that every parallelepiped of dimensions $a \\times b \\times c$, integer vertices and faces parallel to the coordinate planes does not contain cubes of the same color.",
    "options": [],
    "answer": "(a, b, c) with a | b and b | c",
    "solution": "We shall prove that the solutions are the triples $(a, b, c)$ such that $a$ divides $b$ and $b$ divides $c$. We denote by $((x_0, y_0, z_0), p, q, r)$ the parallelepiped with a low right vertex $(x_0, y_0, z_0)$ and dimensions $p, q$ and $r$, at axes $Ox, Oy$ and $Oz$, respectively.\n\nLet us assume that $b$ is not divisible by $a$, i.e. $b = ma + n$ for some $m, n \\in \\mathbb{N}$, $0 < n < a$. If $(p, q, r)$ is a permutation of $(a, b, c)$, then it follows from the condition for the parallelepipeds $((0, 0, 0), p, q, r)$ and $((0, 0, 1), p, q, r)$ that the parallelepipeds $((0, 0, 0), p, q, 1)$ and $((0, 0, r), p, q, 1)$ are filled with cubes of the same colors. This implies that the parallelepipeds $((0, 0, 0), c, a, 1)$ and $((0, 0, b), c, a, 1)$ are also filled with cubes of the same colors and the same is true for the parallelepipeds $((0, 0, 0), c, b, 1)$ and $((0, 0, ma), c, b, 1)$. Since $((0, 0, 0), c, b, 1)$ contains $((0, 0, 0), c, a, 1)$ and $((0, 0, ma), c, b, a)$ contains $((0, 0, ma), c, b, 1)$ and $((0, 0, b), c, a, 1)$, every color in $((0, 0, 0), c, a, 1)$ appears at least two times in $((0, 0, ma), c, b, a)$. This contradiction completes the proof that $a$ divides $b$. We analogously see that $b$ divides $c$.\n\nNow let $a|b$ and $b|c$, as $b = p_1a, c = p_2b = p_1p_2a$, where $p_1, p_2 \\in \\mathbb{N}$. For every two positive integers $m$ and $n$ we denote by $R(m, n)$ the remainder of $m$ modulo $n$. The coordinates of a cube will be the coordinates of its low right vertex.\n\nWe determine the color the cube $(x, y, z)$ in remainders as follows:\n$$\n(R(x, a); R(y, a); R(z, a); R(\\lfloor \\frac{x}{a} \\rfloor + \\lfloor \\frac{y}{a} \\rfloor, p_1); R(\\lfloor \\frac{y}{a} \\rfloor + \\lfloor \\frac{z}{a} \\rfloor, p_1); R(\\lfloor \\frac{x}{b} \\rfloor + \\lfloor \\frac{y}{b} \\rfloor + \\lfloor \\frac{z}{b} \\rfloor, p_2)).\n$$\nThe counting of all possible remainders in the six coordinates shows that the number of the colors used is $a^3 p_1 p_1 p_2 = abc$.\n\nLet us assume that two distinct cubes $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ lie in a parallelepiped of dimensions $a \\times b \\times c$. Then we have $|x_1 - x_2| \\le \\alpha$, $|y_1 - y_2| \\le \\beta$ and $|z_1 - z_2| \\le \\gamma$, where $(\\alpha, \\beta, \\gamma)$ is a permutation of $(a, b, c)$. Since $|x_1 - x_2|$, $|y_1 - y_2|$ and $|z_1 - z_2|$ are divisible by $a$, one of these numbers equals $0$. Let us have $x_1 = x_2$. Then the fourth and fifth coordinates of that color show that $|y_1 - y_2|$ and $|z_1 - z_2|$ are divisible by $b$ and therefore one of these two numbers equals $0$. If, for example, $y_1 = y_2$, then the last coordinate shows that $|z_1 - z_2|$ is divisible by $c$, i.e. $z_1 = z_2$. We obtained $(x_1, y_1, z_1) \\equiv (x_2, y_2, z_2)$, which is a contradiction.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27438,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 3$ be a positive integer. Find all non-constant real polynomials $f_1(x), f_2(x), \\dots, f_n(x)$ such that\n$$\nf_k(x)f_{k+1}(x) = f_{k+1}(f_{k+2}(x)), \\quad 1 \\le k \\le n\n$$\nfor every real $x$ (here $f_{n+1}(x) \\equiv f_1(x)$ and $f_{n+2}(x) \\equiv f_2(x)$).",
    "options": [],
    "answer": "f_1(x)=f_2(x)=...=f_n(x)=x^2",
    "solution": "Let $\\deg(f_k) = \\alpha_k \\in \\mathbb{N}, 1 \\le k \\le n$. We consider all indexes modulo $n$. The condition implies the equalities $\\alpha_k + \\alpha_{k+1} = \\alpha_{k+1}\\alpha_{k+2}$, i.e. $\\alpha_{k+1}$ divides $\\alpha_k$ for every $k = 1, 2, \\dots, n$. Therefore $\\alpha_1 = \\alpha_2 = \\dots = \\alpha_n = 2$.\n\nLet $f(x) = a_k x^2 + b_k x + c_k, 1 \\le k \\le n$, where $a_k \\ne 0$. Comparing the coefficients of $x^4$ we obtain $a_k = a_{k+2}^2, 1 \\le k \\le n$. If $n = 2m$ is even, then $a_1 = a_3^2 = a_5^2 = \\dots = a_{2m-1}^2 = a_1^{2m}$ and therefore $a_1 = a_3 = \\dots = a_{2m-1} = 1$. We analogously have $a_2 = a_4 = \\dots = a_{2m} = 1$. If $n$ is odd, the same argument shows that $a_1 = a_2 = \\dots = a_n = 1$.\n\nComparing the coefficients of $x^3$ we obtain the equalities $b_k + b_{k+1} = 2b_{k+2}$, $1 \\le k \\le n$. Let $\\min\\{b_1, b_2, \\dots, b_n\\} = b_s := b$. Then $b_{s-2} + b_{s-1} = 2b_s$ shows that $b_{s-2} = b_{s-1} = b$ and, continuing in the same way, we conclude that $b_1 = b_2 = \\dots = b_n = b$. We now compare the coefficients of $x^2$ and get $c_k + c_{k+1} = 2c_{k+2} + b$, $1 \\le k \\le n$. Summing up these equalities we obtain $nb = 0$, i.e. $b = 0$. Then the same argument as above gives $c_1 = c_2 = \\dots = c_n = c$. Hence $f(x) = x^2 + c$ and the equations are $(x^2 + c)(x^2 + c) = (x^2 + c)^2 + c$. Then $c = 0$ and $f_1(x) = f_2(x) = \\dots = f_n(x) = x^2$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 27439,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $M$ be an infinite set of rational numbers such that the product of any 2009 of them (pairwise different) is an integer which is not divisible by 2009th powers of primes. Prove that all the numbers in $M$ are integers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $a_1, \\dots, a_{2008} \\in M$ and $A = a_1 \\dots a_{2008} = \\frac{p}{q}$, $(p, q) = 1$. Assume that $M$ contains infinitely many numbers $\\alpha_i = \\frac{p_i}{q_i}$ such that $(p_i, q_i) = 1$, $q_i > 1$ and $\\alpha_i \\ne a_1, \\dots, a_{2008}$. Since $\\alpha_i p$ is an integer, then $q_i$ divides $p$ and hence infinitely many of the numbers $q_i$ are equal. Then the product of 2009 of the respective $\\alpha_j$ is not an integer, a contradiction. In particular, $M$ contains infinitely many integers.\n\nAssume now that $\\frac{a}{b} \\in M$, $(a, b) = 1$ and $b > 1$. If $p$ is a prime divisor of $b$, then the given condition easily implies that $p$ divides infinitely many integers in $M$. Then the product of any 2009 of them is divisible by $p^{2009}$, a contradiction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27440,
    "subject": "Mathematics (Multi-modal)",
    "question": "The excircles of $\\triangle ABC$ touch the sides $AB$, $BC$ and $CA$ at points $M$, $N$ and $P$, respectively. Let $I$ and $O$ be the incenter and the circumcenter of $\\triangle ABC$. Prove that if $AMNP$ is a cyclic quadrilateral, then:\n\na) the points $M$, $P$ and $I$ are collinear;\n\nb) the points $I$, $O$ and $N$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By Carnot's theorem, the perpendiculars from the points $M$, $N$ and $P$ to the lines $AB$, $BC$ and $CA$, respectively, have a common point $X$. Then the quadrilateral $AMXP$ is cyclic. Now it is easy to see that $X = N$ and hence $AN$ is a diameter of the circumcircle of $AMNP$.\n\na) Denote by $I_B$ and $I_C$ the respective excenters of $\\triangle ABC$. Then Pappus' theorem for the triples of points $(I_C, A, I_B)$ and $(B, N, C)$ implies the desired result.\n\nb) Let the incircle of $\\triangle ABC$ touches the sides $AB$ and $AC$ at point $R$ and $Q$, respectively. Then the bisectors of $AB$ and $AC$ coincide with the bisectors of $RM$ and $QP$, respectively, and pass through the midpoint of $IN$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27441,
    "subject": "Mathematics (Multi-modal)",
    "question": "Изпъкнал 2009-ъгълник е разбит на триъгълници чрез непресичащи се диагонали. Един от тези диагонали е оцветен в зелено. Разрешена е следната операция: за два триъгълника *ABC* и *BCD* от разбиването с обща страна $BC$ можем да заменим диагонала $BC$ с диагонала $AD$, като, ако замененият диагонал е бил зелен, той губи цвета си и заменилият го диагонал става зелен. Да се докаже, че всеки предварително избран диагонал на 2009-ъгълника може да бъде оцветен в зелено чрез прилагане на разрешената операция краен брой пъти.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Първо ще докажем, че за даден връх на изпъкналия 2009-ъгълник и всяка триангулация, с прилагане на разрешената операция можем да получим триангулацията, получена от прекарването на всички диагонали през този връх. За произволен връх $A$, движейки се обратно на часовниковата стрелка, да означим с $B_1, B_2, \\dots, B_k$ последователните върхове, за които $AB_i$ е страна на дадения многоъгълник или диагонал в дадената триангулация. Ако отсечката $B_iB_{i+1}$ не е страна, тя е диагонал и след извършване на разрешената операция, ще получим нова триангулация от която излизащите от $A$ диагонали са с един повече. Продължавайки по този начин ще получим триангулация с диагонали само от върха $A$.\n\nЩе докажем по индукция по $n \\ge 4$, че твърдението е вярно за произволен изпъкнал $n$-ъгълник. При $n = 4, 5$ твърдението се проверява директно. Да допуснем, че твърдението е вярно за някое $k \\ge 5$ и да разгледаме триангулация на изпъкнал $(k+1)$-ъгълник. Без ограничение приемаме, че избрания диагонал е $A_1A_i$. Съгласно доказаното, от дадената триангулация можем да получим триангулацията, получена с прекарването на всички диагонали през $A_1$. Ако при това $A_1A_i$ е станал зелен, задачата е решена. Нека зелен е станал диагонала $A_1A_j$, като без ограничение считаме, че $j < i$. От индукционото допускане следва, че в многоъгълника $A_1A_2...A_i$ можем да получим триангулация, в която диагоналът $A_1A_{i-1}$ е зелен. Тъй като $k \\ge 5$ и всяка триангулация на $(k+1)$-ъгълник съдържа $k-2$ диагонала, то в триангулацията освен диагоналите $A_1A_{i-1}$ и $A_1A_i$ има поне още един диагонал. Този диагонал разделя $(k+1)$-ъгълника на два изпъкнали многоъгълника, всеки с по-малко от $k+1$ върха, като диагоналите $A_1A_{i-1}$ и $A_1A_i$ са в един от двата многоъгълника. Остава да приложим индукционното допускане за този многоъгълник.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27442,
    "subject": "Mathematics (Multi-modal)",
    "question": "A number $r_X > 0$ is assigned to any point $X$ in the plane such that $2|r_X - r_Y| \\le |XY|$ for any two points $X$ and $Y$. A cricket can jump from $X$ to $Y$ if $r_X = |XY|$. Prove that for any two points $X$ and $Y$ the cricket can move from $X$ to $Y$ by finite number of jumps.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote by $D(Z, r)$ and $S(Z, r)$ the open disc and the circle with center $Z$ and radius $r$, respectively.\nLet $A = A_X$ be the set of points that can be reached from $X$ by finite number of jumps (possibly zero). We have to show that $A = \\mathbb{R}^2$.\n\nFirst, we shall prove that $D(X, 2r_X) \\subset A$. We may assume that $X = O$ and $r_0 = 1$. Set $a_n = 2^{-n}$, $b_n = 2 - a_n$ and $A_n = \\{X \\in \\mathbb{R}^2 : a_n \\le |OX| \\le b_n\\}$, $n \\in \\mathbb{N}_0$. It is enough to show that $A_n \\subset A$.\n\nWe shall proceed by induction. For $n=0$, this follows from the given condition. Suppose that $A_k \\subset A$. It suffices to prove that $Y \\in A$ for $Y \\in [a_{k+1}, a_k) \\cup (b_k, b_{k+1}]$ and then apply rotation. Let $f(Z) = f_Y(Z) = |YZ| - r_Z$. It follows by the condition of the problem that $1 - a_{k+1} \\le r_{a_k} \\le 1 + a_{k+1}$ and $a_{k+1} \\le r_{b_k} \\le b_{k+1}$. So, if $Y \\in [a_{k+1}, a_k)$, then $f(a_k) < a_k - 1 \\le 0 < f(-b_k)$, and if $Y \\in (b_k, b_{k+1}]$, then $f(b_k) < 0 < 1 - a_{k+1} < f(-a_k)$. Since $A_k$ is a linearly connected set and $f$ is a continuous function, we get that $f(Z) = 0$ for some $Z \\in A_k$ and hence $Y \\in A$.\n\nAssume now that $A \\ne \\mathbb{R}^2$. Then $\\sup\\{r : D(X, r) \\subset A\\} = R < \\infty$. Let $m = \\inf\\{r_Y : Y \\in D(X, R)\\}$. Since $r$ is a continuous function, then $m > 0$ and, by the proved above, $D(Y, 2m) \\subset A$ for $Y \\in D(X, R)$. Hence $D(X, R + 2m) \\subset A$, a contradiction.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27443,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_1, \\dots, a_n, b_1, \\dots, b_n$ be real numbers and $c_1, \\dots, c_n$ be positive real numbers. Prove that\n$$\n\\left( \\sum_{i,j=1}^{n} \\frac{a_i a_j}{c_i + c_j} \\right) \\left( \\sum_{i,j=1}^{n} \\frac{b_i b_j}{c_i + c_j} \\right) \\ge \\left( \\sum_{i,j=1}^{n} \\frac{a_i b_j}{c_i + c_j} \\right)^2 .\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "*First solution.* Let $d_1, \\dots, d_n$ be real numbers and\n$$\nf(x) = \\sum_{i,j=1}^{n} \\frac{d_i d_j}{c_i + c_j} x^{c_i+c_j}, \\quad x > 0.\n$$\nSince $x f'(x) = (\\sum_{i=1}^n d_i x^{c_i})^2 \\ge 0$, it follows that $f(x) \\ge f(0+) = 0$. In particular,\n$$\nf(1) = \\sum_{i,j=1}^{n} \\frac{d_i d_j}{c_i + c_j} \\ge 0.\n$$\nThen for any real number $t$,\n$$\n0 \\le \\sum_{i,j=1}^{n} \\frac{(a_i t + b_i)(a_j t + b_j)}{c_i + c_j} = At^2 + 2Ct + B,\n$$\nwhere\n$$\nA = \\sum_{i,j=1}^{n} \\frac{a_i a_j}{c_i + c_j}, \\quad B = \\sum_{i,j=1}^{n} \\frac{b_i b_j}{c_i + c_j}, \\quad C = \\sum_{i,j=1}^{n} \\frac{a_i b_j}{c_i + c_j}.\n$$\nThis implies that $AB \\ge C^2$.\n\n*Second solution.* Set $f(x) = \\sum_{i=1}^{n} a_i x^{c_i - 1/2}$ and $g(x) = \\sum_{i=1}^{n} b_i x^{c_i - 1/2}$. The given inequality follows by the Cauchy-Schwarz inequality:\n$$\n\\int_0^1 f^2(x)dx \\int_0^1 g^2(x)dx \\ge \\left(\\int_0^1 f(x)g(x)dx\\right)^2.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27444,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ and all primes $p$ for which the following equality holds $n^3 - p^5 = p^6 + 3n - 2$.",
    "options": [],
    "answer": "(n, p) = (10, 3)",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27445,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a unit cell $2 \\times 2010$ table. Ivan puts horizontal domino $[\\ ]$, which covers exactly $2$ of the cells, then Peter puts vertical domino $[\\ ]$, which covers exactly $2$ of the cells then again Ivan puts horizontal domino and so on. The player who has no move loses the game. Determine which of the two players has winning strategy.",
    "options": [],
    "answer": "Ivan",
    "solution": "We describe winning strategy for Ivan. He divides the table into $502$ tables $2 \\times 4$ and a table $2 \\times 2$. Then he puts his first domino in the table $2 \\times 2$ (no matter in which row). Peter is forced to put his domino in a table $2 \\times 4$. Now Ivan puts a domino in any place in the same table. It is easy to see that Ivan has one more way to put a domino in the same table and after two moves each the cells from this table are all covered. After covering all cells in all $2 \\times 4$ tables it is Peter's turn and he has no move. Therefore Ivan wins.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27446,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$ and $d$ be positive real numbers. Prove the inequality\n\n$$\n\\left(\\frac{a}{a+b}\\right)^{5} + \\left(\\frac{b}{b+c}\\right)^{5} + \\left(\\frac{c}{c+d}\\right)^{5} + \\left(\\frac{d}{d+a}\\right)^{5} \\ge \\frac{1}{8}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27447,
    "subject": "Mathematics (Multi-modal)",
    "question": "Any of two lines, parallel to the $x$-axis, has two common points with the graph of the function $f(x) = x^3 + a x^2 + b x + c$. Prove that the quadrilateral with vertices at these four points is a rhombus if and only if its area is equal to $6$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Lines $p$ and $q$ with the given properties exist if and only if $f(x)$ has local minima and maxima, and $p$ and $q$ pass through the point of maxima $D$ and the point of minima $B$ on the graph $f(x)$, respectively. Using transformations of the forms $g(x) = f(x) + a$ and $g(x) = f(x + a)$, we may move this graph such that $D(0, 0)$ and $f(x) = x^2(x - t)$, $t > 0$.\n\nThen $p$ is the $x$-axis and its second common point with the graph is $C(t, 0)$. Since $f'(x) = 3x^2 - 2x t$, then $B\\left(\\frac{2t}{3}, -\\frac{4t^3}{27}\\right)$. The second common point of $q$ and the graph is $A(\\alpha, -\\frac{4t^3}{27})$, where $\\alpha$ is a root of the equation $f(x) = -\\frac{4t^3}{27}$. Since this equation has double root $x_1 = x_2 = \\frac{2t}{3}$, then $x_1 x_2 \\alpha = -\\frac{4t^3}{27}$ gives $\\alpha = -\\frac{t}{3}$. Using that $AB = DC = t$, it follows that $ABCD$ is a parallelogram with area $S_{ABCD} = t \\cdot \\frac{4t^3}{27} = \\frac{4t^4}{27}$.\n\nMoreover,\n\n$$\nBC = \\sqrt{\\left(\\frac{2t}{3} - t\\right)^2 + \\left(-\\frac{4t^3}{27}\\right)^2} = \\frac{t}{3}\\sqrt{1 + \\frac{16t^4}{81}}.\n$$\n\nStraightforward calculations show that the conditions $S_{ABCD} = 6$ and $BC = t$\n\nare equivalent to $t^4 = \\frac{81}{2}$ which solves the problem.",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 27448,
    "subject": "Mathematics (Multi-modal)",
    "question": "The convex quadrilateral $ABCD$ is inscribed in the circle with diameter $AC$. The point $A$ lies on the smaller arc $BD$. Prove the inequality $|AB - AD| \\ge |CB - CD|$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27449,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all coprime positive integers $a$ and $b$ with the following property: There exists positive integer $N$ such that any positive integer, greater than $N$, has the form $\\sum a^i b^j$, where $i$ and $j$ are non-negative integers and none of the summands divides any other.",
    "options": [],
    "answer": "a = 2 and b = 3 (up to order)",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27450,
    "subject": "Mathematics (Multi-modal)",
    "question": "Excircle to the side $BC$ of $\\triangle ABC$ is tangent to $BC$, $AB$ and $AC$ at points $D$, $E$ and $F$, respectively. Let $P$ be orthogonal projection of $D$ on $EF$. Prove that point $P$ lies on the circumcircle $k$ of $\\triangle ABC$ if and only if $k$ passes through the midpoint of $EF$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27451,
    "subject": "Mathematics (Multi-modal)",
    "question": "The positive integers $a_0, a_1, \\dots, a_9$ and $b_1, b_2, \\dots, b_9$ are such that $a_9 < b_9$, $a_k \\neq b_k$ and $1 \\le k \\le 8$. A cash machine is loaded with $n \\ge a_9$ leva. For any $1 \\le i \\le 9$ it is allowed to withdraw $a_i$ leva (if the machine has at least $a_i$ leva), and after that the bank puts in the machine $b_i$ leva. It is also allowed to withdraw $a_0$ without any action from the bank. Find all positive integers $n$ for which the cash machine can be emptied by the above described operations.",
    "options": [],
    "answer": "All n ≥ a9 such that n is divisible by d = gcd(a0, |a1 − b1|, |a2 − b2|, …, |a9 − b9|).",
    "solution": "Set $d_s = |a_s - b_s|$, $0 \\le s \\le 9$, where $b_0 = 0$. Without loss of generality assume that there exists $k$, $0 \\le k \\le 8$ such that $a_s - b_s > 0$ for $0 \\le s \\le k$ and $a_s - b_s < 0$ for $k + 1 \\le s \\le 9$.\n\nIf $n$ is one of the desired values then there exist positive integers $x_0, x_1, \\dots, x_9$ for which\n\n$$\n(\\star) \\qquad x_0d_0 + \\dots + x_kd_k - x_{k+1}d_{k+1} - \\dots - x_9d_9 = n.\n$$\n\nTherefore $d = \\text{GCD}(d_0, d_1, \\dots, d_9)$ is a divisor of $n$.\n\nWe prove now that if $d$ is a divisor of $n$ then one can empty the cash machine.\n\nIndeed, in this case it follows from Bezout's theorem that $(\\star)$ has solution $(x_0, x_1, \\dots, x_9)$ in integers. Set $D_1 = d_0 + \\dots + d_k$, $D_2 = d_{k+1} + \\dots + d_9$, $x_s' = x_s + tD_2$, $0 \\le s \\le k$, $x_s' = x_s + tD_1$, $k+1 \\le s \\le 9$. Since $D_1, D_2 > 0$, it is clear that for big enough $t$, $(x_0', \\dots, x_9')$ is a solution of $(\\star)$ in positive integers and $x_9' > \\max(a_8, \\dots, a_{k+1})$. Consider one such solution and let $x_0'' = x_0' + rD_2$, $x_s'' = x_s'$, $1 \\le s \\le k$ and $x_s'' = x_s' + rd_0$, $k+1 \\le s \\le 9$. For big enough $r$ we obtain a solution $(x_0'', \\dots, x_9'')$ of $(\\star)$, for which\n\n$$\n(\\star\\star) \\qquad x_9'' \\ge \\max(a_8, \\dots, a_{k+1})\n$$\n\n$$\n(\\star\\star\\star) \\qquad n + x_9''d_9 + \\dots + x_{k+1}''d_{k+1} > x_1''d_1 + \\dots + x_k''d_k.\n$$\n\nWe draw money in the following way. Take first $x_9''$ times $a_9$ leva then $x_8''$ times $a_8$ leva, ..., $x_{k+1}''$ times $a_{k+1}$ leva. After that we take $x_1''$ times $a_1$ leva, ..., $x_k''$ times $a_k$ leva. Since $(\\star\\star)$ and $(\\star\\star\\star)$ all operations are feasible. Now $(\\star)$ implies that there are exactly $x_0''d_0$ leva left in the machine and we withdraw them by taking $x_0''$ times $a_0$ leva.\n\n**Answer.** All $n \\ge a_9$ divisible by $d$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27452,
    "subject": "Mathematics (Multi-modal)",
    "question": "Do there exist 6-digit numbers of the form $n^k$, $k \\ge 3$, such that difference between the number, formed by the last three digits, and the number, formed by the first three digits, is equal to 4?",
    "options": [],
    "answer": "No, there do not exist such six-digit numbers.",
    "solution": "If $a = \\overline{a_1a_2a_3a_4a_5a_6}$ satisfies the second condition, then $10^5 < a < 10^6$ and $\\overline{a_4a_5a_6} = \\overline{a_1a_2a_3} + 4$. It follows that\n\n$$\na = 1000\\overline{a_1a_2a_3} + \\overline{a_4a_5a_6} = 1001\\overline{a_1a_2a_3} + 4.\n$$\n\nSo, we are looking for 3-digit numbers $A = \\overline{a_1a_2a_3}$ such that $1001A + 4 = n^k$ for $k \\ge 3$. Using that $1001 = 7 \\cdot 11 \\cdot 13$, we get $n^k \\equiv 4 \\pmod t$ for $t = 7, 11, 13$.\n\n1. Let $k=3$. Then $n^3 \\equiv 4 \\pmod 7$. On the other hand, $n^3 \\equiv \\pm 1 \\pmod 7$, a contradiction.\n\n2. Let $k=4$. Then $n^4 \\equiv 4 \\pmod{13}$. On the other hand, $n^4 \\equiv 1,3,9 \\pmod{13}$, a contradiction.\n\n3. Let $k=5$. Then $n^5 \\equiv 4 \\pmod{11}$. On the other hand, $n^5 \\equiv \\pm 1 \\pmod{11}$, a contradiction.\n\n4. Let $k=7$. Then $n^7 \\equiv 4 \\pmod 7$. On the other hand, $n^7 \\equiv n \\pmod 7$ and hence $n \\equiv 4 \\pmod 7$. If $n=4$, then $4^7 < 10^5$, and if $n \\ge 11$, then $n^7 \\ge 11^7 > 10^6$, a contradiction.\n\n5. Let $k=11$. Since $4^{11} > 2^{20} > 10^6$, then $n < 4$. But $n^{11} \\equiv n \\pmod{11}$ implies $n^{11} \\equiv 4 \\pmod{11}$, a contradiction.\n\n6. Let $k \\ge 13$. Since $3^{13} = 2187.729 > 2000.700 > 10^6$, then $n=2$. The inequality $2^{20} > 10^6$ implies $k \\le 19$. Moreover $2^k \\equiv 4 \\pmod{11}$ and so $2^{k-2} \\equiv 1 \\pmod{11}$. Since the order of 2 modulo 11 is equal to 10, it follows that $10|k-2$, a contradiction to $13 \\le k \\le 19$.\n\nFinally, if $k > 4$ is an even number, then 4 divides $k$ or $k$ has an odd divisor. Thus the problem reduces to one of above cases.\n\n**Answer.** There do not exist 6-digit numbers with the given properties.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27453,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCDE$ be a pentagon for which $\\angle A = \\angle B = \\angle C = \\angle D = 120^\\circ$. Find the minimal possible value of the ratio $\\frac{AC \\cdot BD}{AE \\cdot ED}$.",
    "options": [],
    "answer": "3/4",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27454,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a $\\triangle ABC$ with $\\angle ACB = 90^\\circ$ and altitude $CH$ ($H \\in AB$). A circle $k$ touches the segments $BH$ and $CH$, and the circumcircle of $\\triangle ABC$. If $D = k \\cap BH$, prove that $CD$ bisects $\\angle BCH$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27455,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f: \\mathbb{N} \\to \\mathbb{N}$ be a function such that $f(1) = 1$ and $f(n) = n - f(f(n-1))$ for any $n \\ge 2$. Prove that $f(n + f(n)) = n$ for any $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote by (*) the condition $f(n) = n - f(f(n-1))$ for any $n \\ge 2$. First, we shall prove by induction that\n\n$$\nf(n) \\le f(n+1) \\le f(n) + 1 \\text{ for any } n.\n$$\n\n1. If $n = 1$, then $f(2) = 2 - f(f(1)) = 1$ and the inequalities hold.\n\n2. Let the inequalities be true for any $k \\le n$. Then $f(n+1) = f(n)$ or $f(n)+1$. It follows by (*) that $f(n) < n$ and the induction hypothesis implies\n\n$$\n\\begin{aligned}\nf(f(n)) &\\le f(f(n+1)) \\stackrel{(*)}{\\Rightarrow} n+1 - f(n+1) \\le n+2 - f(n+2) \\\\\n&\\Rightarrow f(n+2) \\le f(n+1) + 1,\n\\end{aligned}\n$$\n\n$$\n\\begin{aligned}\nf(f(n+1)) &\\le f(f(n)) + 1 \\stackrel{(*)}{\\Rightarrow} n + 2 - f(n + 2) \\le n + 1 - f(n + 1) + 1 \\\\\n&\\Rightarrow f(n + 1) \\le f(n + 2).\n\\end{aligned}\n$$\n\nSo $f(n+1) \\le f(n+2) \\le f(n+1) + 1$ which completes the induction step.\n\nNow, we shall prove again by induction the equality $f(n+f(n)) = n$.\n\n1. If $n = 1$, then $f(1+f(1)) = f(2) = 1$.\n\n2. Assume that $f(n+f(n)) = n$ for some $n$. Then\n\n$$\n\\begin{aligned}\nf(f(n+f(n))) = f(n) &\\stackrel{(*)}{\\Rightarrow} n + f(n) + 1 - f(n + f(n) + 1) = f(n) \\\\\n&\\Rightarrow f(n + f(n) + 1) = n + 1.\n\\end{aligned}\n$$\n\n*Case 1.* If $f(n+1) = f(n)$, then $f(n+1+f(n+1)) = n+1$.\n\n*Case 2.* If $f(n+1) = f(n)+1$, then $f(n+1+f(n+1)) \\stackrel{(*)}{=} n+1+f(n+1)-f(f(n+f(n+1))) = n+1+f(n+1)-f(n+f(n)+1) = n+1+f(n+1)-f(n+1) = n+1$.\n\nThe problem is solved.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27456,
    "subject": "Mathematics (Multi-modal)",
    "question": "The cells of $2010 \\times 2010$ table are filled with integers. Adding 1 to all numbers in a particular row or column is called a *move*. We say that a table is *good*, if after finitely many moves it is transformed in a table of equal numbers.\n\na) Find the largest natural number $n$ for which there exists good table containing all the numbers $2^0, 2^1, \\dots, 2^n$.\n\nb) For this $n$, find the largest number which such a table could contain.",
    "options": [],
    "answer": "a) n = 4018; b) the largest number is 2^4019 − 1",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27457,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $x$ and $y$ such that $3^x 7^y + 1$ is an odd degree (greater than 1) of a positive integer.",
    "options": [],
    "answer": "x = 2, y = 1",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27458,
    "subject": "Mathematics (Multi-modal)",
    "question": "The point $D$ lies on the side $AB$ of $\\triangle ABC$ with circumcircle $k$. Denote by $I$ and $J$ the centers of the circles touching $k$, and the segments $AB$ and $CD$. Assume that the points $A, B, I$ and $J$ are concyclic. Prove that $D$ is the tangent point of $AB$ and the excircle to this side.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $k = k(O, r)$ and let $k_1(I, r_1)$ touch $BD$, $CD$ and $k$ at $P$, $R$ and $Q$, respectively. Let $k_2(J, r_2)$ touch $AD$, $CD$ at $K$, $M$ and $L$, respectively. First, we shall prove that $AB \\parallel IJ$, i.e. $ABIJ$ is an isosceles trapezoid. Assume the contrary and set $T = IJ \\cap AB$. Then\n\n$$\n\\frac{IT}{TJ} \\cdot \\frac{JL}{LO} \\cdot \\frac{OQ}{QI} = \\frac{r_1}{r_2} \\cdot \\frac{r_2}{r} \\cdot \\frac{r}{r_1} = 1\n$$\n\nand, by the Menelaus theorem, the points $T$, $Q$ and $L$ are collinear.\n\n![](attached_image_1.png)\n\nThen $TQ \\cdot TL = TB \\cdot TA$. On the other hand, $ABIJ$ is cocyclic which implies $TB \\cdot TA = TI \\cdot TJ$. It follows that $TQ \\cdot TL = TI \\cdot TJ$, i.e. $QLJI$ is cocyclic. But $\\nparallel JLQ = \\nparallel IQL$, i.e. $QLJI$ is an isosceles trapezoid and then $IJ \\parallel QL$, a contradiction. Hence $AB \\parallel IJ$, i.e. $r_1 = r_2$ and $AK = BP$ (1).\n\nFurther, the generalized Ptolemy theorem (applied to $A$, $B$, $Q$ and $C$) gives $AB \\cdot CR + AC \\cdot BP = AP \\cdot BC$. Since $CR = CD - DR = CD - DP = CD - BD + BP$ and $AP = AB - BP$, we get\n\n$$\n(2) \\quad BP = \\frac{AB(BC + BD - CD)}{AB + BC + AC}\n$$\n\nAnalogously,\n\n$$\n(3) \\quad AK = \\frac{AB(AC + AD - CD)}{AB + BC + AC}\n$$\n\nFinally, (1), (2) and (3) imply $BC + BD = AC + AD$ which holds if and only if $D$ is the tangent point of $AB$ and the excircle to this side.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27459,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$ and $c$ be the lengths of sides of a triangle. Prove that\n\n$$\n\\left| \\sqrt{\\frac{a}{b}} - \\sqrt{\\frac{b}{a}} + \\sqrt{\\frac{b}{c}} - \\sqrt{\\frac{c}{b}} + \\sqrt{\\frac{c}{a}} - \\sqrt{\\frac{a}{c}} \\right| < \\frac{1}{10}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27460,
    "subject": "Mathematics (Multi-modal)",
    "question": "Lame rook is chess piece that is allowed to make moves only in the neighboring by edge cell. For 100 moves lame rook visited all cells of a $10 \\times 10$ chessboard exactly once and finally arriving in the initial cell. Prove that some two consecutive moves of the rook are left turns.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 27461,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 3$ be a positive integer and let\n$$\nS = \\{k \\in \\mathbb{N} : (k, n) = (k + 1, n), 1 \\le k \\le n - 1\\}.\n$$\n\nFind the remainder of $\\prod_{k \\in S} k$ divided by $n$.",
    "options": [],
    "answer": "1",
    "solution": "It is easy to see that $(k, n) = (k + 1, n) = 1$. Then for any $k \\in S$ there exists a unique $k_1 \\in \\{1, 2, \\dots, n\\}$ such that $(k_1, n) = 1$ and $k k_1 \\equiv 1 \\pmod{n}$. Since $k_1 + 1 \\equiv k_1(k + 1) \\pmod{n}$, then $(k_1 + 1, n) = 1$, i.e. $k_1 \\in S$. Moreover, $(k-1)(k_1+1) \\equiv k-k_1 \\pmod{n}$ implies that $k \\neq k_1$ if $k \\neq 1$. Then the numbers from $S \\setminus \\{1\\}$ can be divided into different pairs of the form $(k, k_1)$ and hence the wanted remainder is 1.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27462,
    "subject": "Mathematics (Multi-modal)",
    "question": "Do there exist positive integers $n$ and $k$, $1 \\le k \\le n-2$, such that\n$$\n\\binom{n}{k}^2 + \\binom{n}{k+1}^2 = \\binom{n}{k+2}^4 ?\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By applying the formula $\\binom{a}{b} = \\frac{a!}{b!(a-b)!}$ we obtain the equation\n$$\n1 + \\frac{(n-k)^2}{(k+1)^2} = \\frac{(n-k)^2(n-k-1)^2}{(k+1)^2(k+2)^2} \\binom{n}{k+2}^2.\n$$\nHence $(k+2)^2 [(k+1)^2 + (n-k)^2] = (n-k)^2(n-k-1)^2 \\binom{n}{k+2}^2$, which implies that $(k+1)^2 + (n-k)^2$ is a perfect square.\nLet $(k+1)^2 + (n-k)^2 = t^2$, where $t \\in \\mathbb{N}$. We have\n$$\n(k+2)t = (n-k)(n-k-1) \\binom{n}{k+2} \\ge 2 \\binom{n}{k+2}.\n$$\nUsing that $k+2 \\le n$ and $t = \\sqrt{(k+1)^2 + (n-k)^2} < n+1$ we conclude that $(k+2)t \\le n^2$.\nLet $3 \\le k+2 \\le n-3$ (the left hand side of this inequality follows from the condition of the problem). If $n \\ge 6$, we have\n$$\n2 \\binom{n}{k+2} \\ge 2 \\binom{n}{3} = \\frac{n(n-1)(n-2)}{3} > n^2,\n$$\ni.e. the equation has no solution in this case.\nWhen $k+2 = n-2$ we obtain $t^2 = (n-3)^2 + 16$, hence $t = 5$, $n = 6$, $k = 2$. Direct computation shows that $n = 6$ and $k = 2$ is not a solution. If $k+2 = n-1$ we have $t^2 = (n-2)^2 + 9$, so $t = 5$, $n = 6$, $k = 3$ and as above we conclude that no solution exists. Therefore positive integers $n$ and $k$ satisfying the equation do not exist.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27463,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f_1 \\in \\mathbb{R}[x]$ be a quadratic polynomial with positive leading coefficient. Set $f_{n+1} = f_1 \\circ f_n$ for $n \\ge 1$. It is known that the polynomial $f_2$ has four non-positive different zeroes. Prove that the polynomial $f_n$ has $2^n$ different real zeroes.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note that if $x_1, \\dots, x_{2^n}$ are the zeroes of $f_n$, then the zeroes of $f_{n+1}$ are the solutions of the equations $f_1(x) = x_k$, $1 \\le k \\le 2^n$. Moreover, the equation $f_1(x) = a$ has two different real roots if and only if $a > m := \\min f_1$.\n\nAssume that $f_2$ has four non-positive different zeroes. Then it is easy to see that $f_1$ has two zeroes $x_1 < x_2 \\le 0$ and $x_1 > m$. Using the argument from the beginning, it follows by induction on $n$ that all the zeroes of $f_{n+1}$ are different and belong to the interval $(x_1, x_2]$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 27464,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $k$ be the incircle of a $\\triangle ABC$. A line, parallel to $BC$ touches $k$ and intersects the sides $AB$ and $AC$ at points $A_1$ and $A_2$. Define the points $B_1, B_2$ and $C_1, C_2$ in a similar way. Prove that\n$$\n9(\\overline{AA_1} \\cdot \\overline{AA_2} + \\overline{BB_1} \\cdot \\overline{BB_2} + \\overline{CC_1} \\cdot \\overline{CC_2}) \\ge \\overline{AB}^2 + \\overline{BC}^2 + \\overline{CA}^2.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $D, E$ and $F$ be the common points of with the sides $BC, CA$ and $AB$, respectively, and let $AE = AF = x$, $BF = BD = y$ and $CD = CE = z$.\nSince $\\triangle AA_1A_2 \\sim \\triangle ABC$, then $\\frac{AA_1}{AB} = \\frac{AA_2}{AC} = \\frac{P_{AA_1A_2}}{P_{ABC}} = \\frac{x}{x+y+z}$ and hence $AA_1 = \\frac{x(x+y)}{x+y+z}$ and $AA_2 = \\frac{x(x+z)}{x+y+z}$. Analogously, $BB_1 = \\frac{y(y+z)}{x+y+z}$, $BB_2 = \\frac{y(y+x)}{x+y+z}$, $CC_1 = \\frac{z(z+x)}{x+y+z}$ and $CC_2 = \\frac{z(z+y)}{x+y+z}$.\nThen the given inequality is equivalent to\n$$\n9 \\sum x^2 (x+y)(x+z) \\ge (x+y+z)^2 \\sum (x+y)^2,\n$$\ni.e.\n$$\n9 \\sum x^4 + 2(\\sum x^2)(\\sum xy) \\ge 2(\\sum x^2)^2 + 4(\\sum xy)^2,\n$$\n\nThe last inequality follows by the well-known inequalities\n$$\n3(x^4 + y^4 + z^4) \\geq (x^2 + y^2 + z^2)^2 \\text{ and } x^2 + y^2 + z^2 \\geq xy + yz + zx.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27465,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a $\\triangle ABC$ and a function $f : \\mathbb{R}^+ \\to \\mathbb{R}$ with the following property: for any segment $DE$ of the interior of the triangle and its midpoint $M$ one has that\n$$\nf(d(D)) + f(d(E)) \\le 2f(d(M)).\n$$\nwhere $d(X)$ denotes the distance from $X$ to the boundary of $\\triangle ABC$. Prove that for any segment $PQ$ of the interior of $\\triangle ABC$ and any point $N$ on this segment we have\n$$\n|QN| \\cdot f(d(P)) + |PN| \\cdot f(d(Q)) \\le |PQ| \\cdot f(d(N)).\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote by $k(I, r)$ the incircle of $\\triangle ABC$. It is clear that the image of $d$ is the interval $\\Delta = (0, r]$.\n\nVarying $D$ and $E$ on $AI$, it follows that $f$ is a mid-point concave function on $\\Delta$, i.e. $f(2x) + f(2y) \\le f(x + y)$.\n\nNote that the points on given distance (from the boundary) form a triangle homothetic to $\\triangle ABC$ (with center of homothety $I$) and the mid-points of the segments with ends at these points run over the whole interior of this triangle. It follows that $f$ is an increasing function on $\\Delta$.\n\nSince $f$ is a mid-point concave and increasing function, then it is continuous. Indeed, if $f^{-}(x)$ and $f^{+}(x)$ are the left- and the right-hand limit of $f$ at $x$ ($f^{+}(r) := f(r)$), then $f^{-}(x) + f(x) \\le 2f^{-}(x)$ and $f^{-}(x) + f^{+}(x) \\le 2f(x)$ (why?) and hence $f^{-}(x) = f(x) = f^{+}(x)$. So $f$ is a convex and increasing function.\n\nThen, since $d$ is a concave function, it is easy to see that $f \\circ d$ is a concave function. This also follows from the fact that $f \\circ d$ is a mid-point concave and continuous function (because of the continuity of $f$ and $d$).",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27466,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\in \\mathbb{N}$. Find the number of all sequences $A_0, A_1, \\dots, A_n$ of pairwise different points in the plane with non-negative integer first coordinates and second coordinates $0$ or $1$, such that $A_0 = (0,0)$ and\n$$\n|x_{A_i} - x_{A_{i-1}}| + |y_{A_i} - y_{A_{i-1}}| = 1\n$$\nfor $1 \\le i \\le n$.",
    "options": [],
    "answer": "c_n = 2 f_{n+1} - \\varepsilon_n = \\frac{2}{\\sqrt{5}}\\big(q^{n+1} - (1 - q)^{n+1}\\big) - \\varepsilon_n, where f_k is the Fibonacci sequence with f_1 = f_2 = 1, q = (1 + \\sqrt{5})/2, and \\varepsilon_n = (1 + (-1)^n)/2.",
    "solution": "Denote by $c_n$ the number of the sequences from the problem. Define the number $d_n$ in a similar way but replacing the given equality by\n$$\nx_{A_i} - x_{A_{i-1}} + |y_{A_i} - y_{A_{i-1}}| = 1, \\quad 1 \\le i \\le n;\n$$\ncall the respective sequences *right* and the other sequences *wrong*. It is not difficult to see that $d_0 = 1$, $d_1 = 2$ and $d_{i+2} = d_{i+1} + d_i$ (to any right sequence $A_0, A_1, \\dots, A_{i+2}$ associate $A_0, A_1, \\dots, A_{i+1}$ if $x_{A_{i+2}} = x_{A_{i+1}}$, and $A_0, A_1, \\dots, A_i$ otherwise). It follows that $d_n$ is $(n+2)$-th Fibonacci number $f_{n+2}$. We associate to any wrong sequence $A_0, A_1, \\dots, A_n$ ($n \\ge 3$) the right sequence $A_0, A_1, \\dots, A_k$ for which $x_k = x_n$, $k \\le n-3$. It is clear that $x_{k+1} = x_k + 1$; we shall call such sequences *super right*. Note that the number of the super right sequences of length $k$ is equal to\n$$\n(2) \\qquad d_{k-1} = f_{k+1}\n$$\n(this remains true for $k=0$). There is an obvious bijection between the set of the wrong sequences of length $n$ and the set of the super right sequences with lengths $n-3, n-5, \\dots$. It follows by (2) and (1) that\n$$\nc_n = d_n + d_{n-1} + d_{n-2} + \\dots = 2f_{n+1} - \\varepsilon_n = \\frac{2}{\\sqrt{5}}(q^{n+1} - (1-q)^{n+1}) - \\varepsilon_n,\n$$\n$$\n\\text{where } \\varepsilon_n = \\frac{1 + (-1)^n}{2} \\text{ and } q = \\frac{1 + \\sqrt{5}}{2}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27467,
    "subject": "Mathematics (Multi-modal)",
    "question": "Every two of $n$ towns in a country are connected by one way or two way road. It is known that for every $k$ towns there exists round trip passing through each of these $k$ towns exactly once. Find the maximal possible number of one way roads.",
    "options": [],
    "answer": "If n ≤ 2k − 3, the maximum is n(n − 1)/2. If n > 2k − 3, the maximum is n(k − 2).",
    "solution": "Suppose there exists a town $A$ with $k-1$ one way roads all of which are pointing in one and the same direction from $A$. Town $A$ with all $k-1$ end points of one way roads form a group of $k$ towns that violates the condition of the problem. We conclude that for any town there exist at most $k-2$ one way roads in the town and at most $k-2$ roads out of the town. Therefore the one way roads are at most $n(k-2)$.\n\nWe show now that when $n \\le 2k-3$, it is possible to have all roads to be one way roads and when $n > 2k-3$ then the maximal number of one way roads equals $n(k-2)$.\n\nLet $n = 2k-3$ and consider the towns to be vertexes of a regular $n$-gon. For every town $A$ let $k-2$ one way roads out of $A$ point to the next $k-2$ towns clockwise. Since $n = 2k-3$ such allocation is possible and every two towns are connected with one way road. Consider a group of $k$ towns and order them clockwise. Between any two neighbors $A$ and $B$ there exists one way road from $A$ to $B$. Therefore the desired round trip exists.\n\nWhen $n < 2k-3$, it suffices to choose arbitrary $n$ towns from the above construction.\n\nWhen $n > 2k-3$ we proceed by similar manner. Connect every town $A$ with one way road from $A$ with the next $k-2$ towns. Let all remaining roads be two way roads.\n\nTake arbitrary $k$ towns. Consider a group of $k$ towns and order them clockwise. Between any two neighbors $A$ and $B$ there exists one way road from $A$ to $B$. Therefore the desired round trip exists.\n\n**Answer.** When $n \\le 2k-3$ all roads could be one way and when $n > 2k-3$ there exist at most $n(k-2)$ one way roads.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27468,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $O$ be an interior point of an acute-angled $\\triangle ABC$. Denote by $A_1$, $B_1$ and $C_1$ its projection on the sides $BC$, $AC$ and $AB$. Let $P$ be the intersecting point of the lines through $A$ and $B$, orthogonal to $B_1C_1$ and $A_1C_1$, respectively. If $H$ is the projection of $P$ on $AB$, prove that the points $A_1$, $B_1$, $C_1$ and $H$ are con-cyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $\\not\\asymp PBC_1 = \\not\\asymp OC_1A_1$ and $\\not\\asymp OC_1A_1 = \\not\\asymp OBA_1$ (why?), then $\\not\\asymp PBA = \\not\\asymp OBC$. Analogously $\\not\\asymp PAB = \\not\\asymp OAC$. Denote by $M$ and $N$ the projections of $P$ on $BC$ and $AC$. Since\n$$\nBC_1 \\cdot BH = BO \\cdot BP \\cos \\angle OBA \\cos \\angle PBA = BA_1 \\cdot BM,\n$$\nthen the points $H$, $C_1$, $A_1$ and $M$ lie on a circle with center at the midpoint $Q$ of $OP$ (since the bisectors of $HC_1$ and $MA_1$ pass through $Q$). We get in the same way that the points $H$, $C_1$, $N$ and $B_1$ lie on a circle with center $Q$. Hence the points $A_1$, $B_1$, $C_1$, $H$, $M$ and $N$ are con-cyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27469,
    "subject": "Mathematics (Multi-modal)",
    "question": "Isosceles $\\triangle ABC$ ($AC = BC$) is inscribed in a circle $k$. A point $M$ lies on the side $BC$. A point $N$ from the ray $AM$ ($M$ lies between $A$ and $N$) is such that $AN = AC$. The circumcircle of $\\triangle MCN$ intersects $k$ at $C$ and $P$, where $P$ is from the arc $BC$, not containing $A$. The lines $AB$ and $CP$ intersect at $Q$. Prove that $\\angle QMB = \\angle QMN$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote $\\angle MAC = \\alpha$ and $\\angle ACB = \\beta$. Let $J$ be the incenter of $\\triangle AMC$. It follows that $\\angle AJM = 90^\\circ + \\beta/2$ and $\\angle ABM = 90^\\circ - \\beta/2$, which implies that the quadrilateral $ABMJ$ is inscribed in a circle $k_1$. Analogously we have that the quadrilateral $MNCJ$ is inscribed in a circle $k_2$ which in fact is the the circumcircle of $\\triangle MCN$. Therefore $P$ is a point of $k_2$. Since $QA \\cdot QB = QC \\cdot QP$, we conclude that $Q$ has one and the same degree with respect to $k_1$ and $k_2$. Hence $Q$ lies on $JM$, which is the angular bisector of $\\angle BNC$. This implies the desired equality $\\angle QMB = \\angle QMN$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27470,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ be a positive integer. Denote by $\\tau(a)$ and $\\varphi(a)$ respectively the number of all positive integers that divide $a$ and the number of all positive integers not greater than $a$ and relatively prime to $a$. Find all positive integers $n$ having only two prime divisors and such that $\\varphi(\\tau(n)) = \\tau(\\varphi(n))$.",
    "options": [],
    "answer": "All such n are n = 2^{r^t - 1} · 3^{r - 1}, where r is a prime and t is a positive integer.",
    "solution": "Let $n = p^k q^l$, where $p < q$ are prime numbers, $k, l \\in \\mathbb{N}$ and let $u = \\varphi(\\tau(n)), v = \\tau(\\varphi(n))$. We have that $u = \\varphi((k+1)(l+1))$ and $v = \\tau(p^{k-1} q^{l-1} (p-1)(q-1))$. Obviously $u < kl + k + l$ and\n$$\nv \\geq \\tau(p^{k-1} q^{l-1} (p-1)) + 1 = kl \\tau(p-1) + 1.\n$$\nIf $p > 2$, then $\\tau(p-1) \\geq 2$ and $v \\geq 2kl + 1 \\geq kl + k + l$ ($\\iff (k-1)(l-1) \\geq 0$), hence $v > u$. Therefore $p = 2$ and thus $v = \\tau(2^{k-1} q^{l-1} (q-1))$.\nIf $m > 2$ is a prime and $m | q-1$, then $2m | q-1$. So,\n$$\nv \\geq \\tau(2^k q^{l-1} m) = 2(k+1)l > 2kl + 1\n$$\nand again $v > u$. Therefore $q = 2^s + 1, s \\in \\mathbb{N}$. It follows now that $v = \\tau(2^{k+s-1} q^{l-1}) = (k+s)l$ and the equality $u = v$ implies\n$$\n\\varphi((k+1)(l+1)) = (k+s)l.\n$$\nIt follows from the formula for $\\varphi$ that if $a > 1$ and $b > 1$, then $\\frac{\\varphi(ab)}{ab} \\le \\frac{\\varphi(b)}{b}$.\ni.e. $\\varphi(ab) \\le a \\varphi(b)$ and equality holds only if all prime factors of $a$ divide $b$. When $a = k + 1$ and $b = l + 1$ we have\n$$\n\\varphi((k+1)(l+1)) \\le (k+1) \\varphi(l+1) \\le (k+s)l.\n$$\nMoreover the equality holds if and only if $s = 1$, i.e. $q = 3$. $\\varphi(l + 1) = l$. Hence $l + 1$ is a prime number and all prime factors of $k + 1$ divide $l + 1 = r$, i.e. $k + 1 = r^t, t \\in \\mathbb{N}$.\nTherefore the desired numbers are all integers of the form $n = 2^{r^t-1} 3^{r-1}$, where $r$ is a prime number and $t \\in \\mathbb{N}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27471,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$ be a set with $n \\ge 2$ elements and let $f_1, \\dots, f_k : A \\to \\mathbb{R}$ be functions. Prove that there exist functions $g_1, \\dots, g_{n-1} : A \\to \\mathbb{R}$ such that\n$$\n\\max_{1 \\le i \\le k} |f_i(x) - f_i(y)| = \\max_{1 \\le i \\le n-1} |g_i(x) - g_i(y)|\n$$\nfor any $x, y \\in A$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $A = \\{a_1, a_2, \\dots, a_n\\}$ and set\n$$\nd(x, y) = \\max_{1 \\le i \\le k} |f_i(x) - f_i(y)|, \\quad g_i(x) = d(x, a_i) - d(a_i, a_i), \\quad 1 \\le i \\le n-1\n$$\nand $d'(x, y) = \\max_{1 \\le i \\le n-1} |g_i(x) - g_i(y)|$. Since\n$$\n|g_i(x) - g_i(y)| = |d(x, a_i) - d(y, a_i)| \\le d(x, y) \\quad (\\text{why?})\n$$\nthen $d'(x, y) \\le d(x, y)$. On the other hand,\n$$\nd(a_j, a_i) = g_i(a_j) - g_i(a_i) \\le d'(a_j, a_i), \\quad 1 \\le i \\le n-1, 1 \\le j \\le n\n$$\n$$\nd(a_n, a_n) = 0 = d'(a_n, a_n) \\text{ and hence } d'(x, y) = d(x, y).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27472,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $M$ be convex 2011-gon. Consider 2011 points lying inside $M$ and such that no three of all 4022 points (the vertexes of $M$ and 2011 points inside $M$) are collinear. A coloring of all points in two colors is called *good* if it is possible to connect some of the points by segments such that the following conditions hold:\n(1) Each segment has its endpoints of one and the same color.\n(2) No segments intersect in inner points.\n(3) Between any two points of the same color there exists a path through the given segments.\nFind the number of all good colorings.",
    "options": [],
    "answer": "2^{2011}(2011\\cdot2010+2)",
    "solution": "Let the two colors be blue and red. We prove first the following\n\n*Lemma*. Consider $\\triangle ABC$ with vertices of both colors. Any coloring of $n$ points inside $\\triangle ABC$ is good.\n\n*Proof*. Without loss of generality assume that $A$ and $B$ are blue points and $C$ is a red point.\nWe proceed by induction on the number $n \\ge 0$ of inner points for the triangle.\nWhen $n = 0$ we draw a segment with end points $A$ and $B$ and we are done.\nAssume the assertion holds for any triangle with $n = k$ inner points. Consider triangle with $n = k + 1$ inner points and let these points be colored in arbitrary manner. If all points are blue then we connect $A$ with $B$ and with all inner points and the condition holds.\nIf there exists a red point $D$, we apply the induction hypothesis for triangles $ABD$, $BCD$ and $ACD$. It is clear that red points from distinct triangles are connected through $D$, and the blue points from distinct triangles are connected through $A$ or $B$. This completes the proof of the lemma. $\\square$\n\nNote that for any good coloring all blue points of $M$ are consecutive vertices. Indeed, if this is not the case we have two red points $A$ and $B$, dividing the contour of $M$ into two parts with two blue points $C$ and $D$ in each of these parts. It is clear now that the red path between $A$ and $B$ intersects the blue path between $C$ and $D$, a contradiction.\n\nLet us first count the number of good colorings of the vertexes of $M$. There exist two such colorings with all points having one and the same color. When both colors occur let $k > 0$ be the number of blue points. For any $k = 1, 2, \\dots, 2010$ there exist 2011 ways of choosing the group of $k$ consecutive blue points. Therefore we have $2011 \\cdot 2010 + 2$ ways of coloring the vertexes of $M$.\n\nWe show now that any coloring of the inner points is good. If all vertexes of $M$ have one and the same color (say blue) and all inner point are also blue we connect one vertex with all remaining points. If there exists a red point $B$, then we consider all triangles with one vertex $B$ and sides the sides of $M$. The assertion of the Lemma completes the proof in this case.\n\nConsider coloring of the vertexes of $M$ in which both blue and red points occur. Without loss of generality assume the blue points are $A_1, A_2, \\dots, A_k, k < 2011$. Connect $A_1$ with all red points and $A_{k+1}$ with all blue points. Now $M$ is partitioned into triangles and each triangle has vertexes of both colors. Apply the Lemma for each of these triangles. The points having one and the same color from distinct triangles are connected through $A_1$ or $A_{k+1}$ or by the sides of $M$.\n\nTherefore the number of good colorings equals $2^{2011}(2011 \\cdot 2010 + 2)$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 27473,
    "subject": "Mathematics (Multi-modal)",
    "question": "The incircle of $\\triangle ABC$ touches the sides $BC$ and $AC$ at points $A_1$ and $B_1$, respectively. The lines $B_1A_1$ and $AB$ are concurrent at $X$ such that $A$ lies between $X$ and $B$. If $\\angle CXB = 90^\\circ$ and $BC^2 = AB^2 + BC \\cdot AC$ find the angles of $\\triangle ABC$.",
    "options": [],
    "answer": "A = 108°, B = 36°, C = 36°",
    "solution": "Let $Y \\in BC$ be such that $AY \\parallel XA_1$. Thus $BY = a-b$ and since $\\frac{a}{c} = \\frac{c}{a-b}$ we have that $\\triangle ABC \\sim \\triangle YBA$. Therefore $\\angle AXA_1 = \\angle BAY = \\gamma$ which implies that quadrilateral $XAA_1C$ is cyclic. It follows from $\\angle XAC = \\angle XA_1C = 90^\\circ - \\frac{\\gamma}{2}$ that $\\alpha = 90^\\circ + \\frac{\\gamma}{2}$. Moreover $\\angle AA_1C = 90^\\circ$, i.e. $AA_1$ is simultaneously altitude and angular bisector. Hence $\\gamma = \\beta$ and $90^\\circ + \\frac{\\gamma}{2} + 2\\gamma = 180^\\circ$, i.e. $\\gamma = 36^\\circ$. The angles of $\\triangle ABC$ are $36^\\circ, 36^\\circ$ and $108^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27474,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$, bounded in the interval $(0, 1)$ and such that\n$$\nx^2 f(x) - y^2 f(y) = (x^2 - y^2) f(x + y) - xy f(x - y)\n$$\nfor all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "f(x) = a x for some real constant a",
    "solution": "Note that when $x > y + 1/2$ run through the interval $(0, n)$ then $x + y$ runs through the interval $(0, 2n - 1/2)$. Straightforward induction shows that $f$ is bounded in all intervals $(0, 2k + 1/2)$. When $0 < x < y$ it follows that $f$ is also bounded in the intervals $(-2k, 0)$. Thus, $f$ is bounded in every bounded subset of $\\mathbb{R}$. Let $x \\neq 0$. When $y \\to 0$ it follows from the boundedness of $f$ and the given equality that $f(x + y) \\to f(x)$, i.e. $f$ is continuous at $x$. For $y = -x$ we have $f(x) - f(-x) = f(2x)$. Thus $f(-x) - f(x) = f(-2x)$ and therefore $-f(-2x) = f(2x) = 2f(x)$. It follows by induction on $n \\ge 3$ that for $x = (n-1)y$ we have $f(ny) = nf(y)$. Hence $f(r) = ar$, where $a = f(1)$ and $r \\in \\mathbb{Q}^+$. Since $f$ is odd and continuous we obtain that $f(x) = ax$ for any $x \\neq 0$. When $x = y$ we have that $f(0) = 0$ implying that $f(x) = ax$ for any $x$. It is easily checked that $f(x) = ax$ is indeed a solution of the problem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27475,
    "subject": "Mathematics (Multi-modal)",
    "question": "All points in the plane having integer coordinates are colored in three colors. Find the least positive integer $S$ having the following property: for arbitrary such covering there exists a triangle of area $S$ having all its vertices in one and the same color.",
    "options": [],
    "answer": "3",
    "solution": "Consider the following two colorings:\n\n(1) Point $(x, y)$ is in color $i$, $1-i-2$ if and only if $x \\equiv i \\pmod{2}$.\n\n(2) Point $(x, y)$ is in color $i$, $1 - i - 3$ if and only if $x \\equiv i \\pmod{3}$.\n\nIt is clear that if $S$ exists then $2S$ is an integer. Coloring (1) implies that $S$ could be $1$, $2$, $3$ or greater and coloring (2) shows that $S$ could be $3/2$, $3$ or greater. Therefore if $S$ exists then $S \\ge 3$.\n\nWe prove that for arbitrary coloring there exists a triangle of area $3$ having all its vertices in one and the same color. Note that for some $d \\in \\{1, 2, 3\\}$ there exist points $A = (x, y)$ and $B = (x+d, y)$ of the same color. Indeed, it suffices to consider the points $(0,0), (1,0), (2,0), (3,0)$. If $m \\equiv AB$ and $l$ is parallel to $m$ at distance $\\frac{6}{d}$ from $m$ then the line $l \\parallel Ox$ is either having points of only two colors or we have a triangle of area $3$.\n\nWe say that a color $c$ permits distance $a$ in the line $l$ if there exist two points on $l$ having color $c$ and at distance $a$ apart.\n\nIf there exists a distance $a \\in \\{1, 2, 3, 6\\}$ that is permitted by the two colors on $l$ then the line $p \\parallel l$ at a distance $\\frac{6}{a}$ from $l$ is having all its points in the same color. If such a distance does not exist then one of the colors $l$ permits all distances in the set $\\{2, 3, 6\\}$.\n\n(Indeed, assume that color $c_1$ does not permit distance $1$, and there exists at least one point $P_0$ from $l$ having color $c_1$. Two neighbors of $P_0$ - points $P_{-1}$ and $P_1$ are colored in $c_2$, so $c_2$ permits distance $2$. Therefore $c_1$ does not permit distance $2$ and points $P_{-2}$ and $P_2$ are of color $c_2$. As above $c_2$ does not permit distance $3$. $c_1$ does not permit distance $3$ and $P_{-3}$ and $P_3$ are of color $c_2$. Thus $c_2$ permits all distances in $\\{2, 3, 6\\}$.)\n\nIn both cases there exists a line $p \\parallel Ox$ and color $c$ from $p$ that permits all distances in the set $\\{2, 3, 6\\}$.\n\nConsider the lines $u_1, u_2, u_3$, parallel to $p$ and such that $u_i$, $i = 1, 2, 3$ is at distance $i$ above $p$. It follows from the above that all such lines are having points only in two colors. Finally, consider all nine points of intersection of lines $u_1, u_2, u_3$ with lines $x = 0$, $x = 3$, $x = 6$. It is easy to see that there exists a triangle of area $3$ having all its vertices colored in the same color.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27476,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 2$ be a positive integer and $a_1 < a_2 < \\dots < a_{2n}$ be real numbers. If $S = \\sum_{i=1}^{2n} a_i$, $A_1 = \\sum_{i,j,i<j} a_{2i}a_{2j}$ and $A_2 = \\sum_{i,j,i<j} a_{2i-1}a_{2j-1}$, prove the inequality\n$$\n(n-1)S^2 > 4n(A_1 + A_2).\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First, we shall prove the following\n**Lemma.** If $P(x) = b_0x^n + b_1x^{n-1} + b_2x^{n-2} + \\dots + b_{n-1}x + b_n$ has $n$ real distinct roots then $(n-1)b_1^2 - 2nb_0b_2 > 0$.\n\n*Proof.* First differentiate $n-2$ times the function $f(x)$. As a result we have quadratic function having two real distinct roots and therefore its discriminant is positive.\n\nConsider the polynomial $P(x) = (x-a_1)(x-a_3)\\dots(x-a_{2n-1})+(x-a_2)(x-a_1)\\dots(x-a_{2n})$. It is straightforward to verify that it satisfies the condition of the lemma and the corresponding inequality is exactly the desired inequality\n$$\n(n-1)S^2 > 4n(A_1 + A_2).\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27477,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the least positive integer $n$, such that there exist positive integers $a$, $b$ and $c$, none of which is a perfect square and\n$$\na^3 + b^3 + c^3 - 3abc = 2013^n.\n$$",
    "options": [],
    "answer": "2",
    "solution": "It is easy to be seen that the left hand side is divisible by $9$ implying that the equation has no solution for $n = 1$.\n\nLet $n=2$. We have $2013^2 = (3 \\cdot 11 \\cdot 61)^2$, $3^2 = 2^3 + 1^3$ and $11 \\cdot 61 = 8^3 + 4^3 + 1^3 + 3 \\cdot 8 \\cdot 4 \\cdot 1$. We use the identity\n$$\n(a^3 + b^3 + c^3 - 3abc)(x^3 + y^3 + z^3 - 3xyz) = u^3 + v^3 + w^3 - 3uvw.\n$$\nwhere $u = ax + by + cz$, $v = ay + bz + cx$ and $w = az + bx + cy$. We apply the above identity first for $(8, 4, -1)$ and $(4, 8, -1)$, and then for the obtained triples $(65, 0, 56)$ and $(2, 1, 0)$ and obtain $65^3 + 56^3 = (11 \\cdot 61)^2$ and $130^3 + 177^3 + 56^3 - 3 \\cdot 130 \\cdot 177 \\cdot 56 = 2013^2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27478,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that if $\\alpha, \\beta, \\gamma \\in [0, \\frac{\\pi}{2}]$ and $\\tan\\alpha + \\tan\\beta + \\tan\\gamma \\le 3$, then\n$$\n\\cos 2\\alpha + \\cos 2\\beta + \\cos 2\\gamma \\ge 0.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "*First solution.* The inequality follows by the fact that if $\\theta \\in [0, \\frac{\\pi}{2}]$, then\n$$\n\\cos 2\\theta = 1 - \\sin 2\\theta \\cdot \\tan \\theta \\ge 1 - \\tan \\theta.\n$$\n\n\n*Second solution.* Set $x = \\tan\\alpha$, $y = \\tan\\beta$ and $z = \\tan\\gamma$. Then $x, y, z \\ge 0$, $x+y+z \\le 3$ and we have to prove that\n$$\n\\cos 2\\alpha + \\cos 2\\beta + \\cos 2\\gamma = \\frac{1-x^2}{1+x^2} + \\frac{1-y^2}{1+y^2} + \\frac{1-z^2}{1+z^2} \\ge 0.\n$$\nThis inequality can be written as\n$$\n\\frac{1}{1+x^2} + \\frac{1}{1+y^2} + \\frac{1}{1+z^2} \\ge \\frac{3}{2}.\n$$\nTo prove the last inequality, it remains to use that $\\frac{1}{1+t^2} \\ge 1 - \\frac{t}{2}$ is equivalent to $t(t-1)^2 \\ge 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27479,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider acute $\\triangle ABC$ with altitudes $AA_1, BB_1$ and $CC_1$ ($A_1 \\in BC, B_1 \\in AC, C_1 \\in AB$). A point $C'$ on the extension of $B_1A_1$ beyond $A_1$ is such that $A_1C' = B_1C_1$. Analogously, a point $B'$ on the extension of $A_1C_1$ beyond $C_1$ is such that $C_1B' = A_1B_1$ and point $A'$ on the extension of $C_1B_1$ beyond $B_1$ is such that $B_1A' = C_1A_1$. Denote by $A'', B''$ and $C''$ the symmetric points of $A', B'$ and $C''$ with respect to $BC, CA$ and $AB$ respectively. Prove that if $R, R'$ and $R''$ are circumradiuses of $\\triangle ABC, \\triangle A'B'C'$ and $\\triangle A''B''C''$, then $R, R'$ and $R''$ are side lengths of a triangle with area equals one half of the area of $\\triangle ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $O$ be circumcenter of $\\triangle ABC$ and $S$ be its symmetric point with respect to $AB$. Then $SOCH$ is a parallelogram. $SH \\perp A_1B_1$ and $SH = R$ (1). Denote by $\\rho$ the inradius of triangle $A_1B_1C_1$, and by $q$ its semi perimeter. If $T$ is the orthogonal projection of $S$ to $A_1B_1$ then $HT = \\rho$, $ST = R + \\rho$, $TC' = q$. We have $HC'' = \\sqrt{\\rho^2 + q^2}$ which due to symmetry implies that $H$ is the circumcenter of $\\triangle A'B'C'$ and $R' = HC'$ (2). On the other hand $OC'' = SC' = \\sqrt{(R+\\rho)^2 + q^2}$ and therefore $O$ is the circumcenter of $\\triangle A''B''C''$ and $R'' = SC'$ (3). It follows from (1), (2) and (3) that $R, R'$ and $R''$ are side lengths of triangle $HSC'$ and\n$$\nS_{HSC'} = \\frac{HS \\cdot TC'}{2} = \\frac{R \\cdot q}{2} = \\frac{1}{2} (S_{A_1OB_1C} + S_{B_1OC_1A} + S_{C_1OA_1B}) = \\frac{S_{ABC}}{2}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27480,
    "subject": "Mathematics (Multi-modal)",
    "question": "Points $M$, $N$ and $P$ lie on the sides $BC$, $CA$, $AB$ of triangle $ABC$. Triangles $CNM$, $APN$ and $BMP$ are acute and let $H_C$, $H_A$ and $H_B$ be their respective orthocenters. Prove that if the three lines $AH_A$, $BH_B$ and $CH_C$ are concurrent then $MH_A$, $NH_B$ and $PH_C$ are also concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote by $A_1$, $B_1$ and $C_1$ the projections of $A$, $B$ and $C$ on $NP$, $PM$ and $MN$, respectively. It follows from the condition of the problem (Carnot's theorem) that $(NC_1^2 - MC_1^2) + (MB_1^2 - PB_1^2) + (PA_1^2 - NA_1^2) = 0$. Hence $0 = (CN^2 - CM^2) + (BM^2 - BP^2) + (AP^2 - AN^2) = (CN^2 - AN^2) + (AP^2 - BP^2) + (BM^2 - CM^2)$ and the same theorem implies that the perpendiculars from $M$, $N$ and $P$ to $BC$, $AC$ and $AB$ are concurrent at a point $O$.\n\nSince $OMH_CN$ and $OMH_BP$ are parallelograms we conclude that $PH_BH_CN$ is also a parallelogram. Therefore the segments $PH_C$ and $NH_B$ bisect each other at some point. The segment $MH_A$ passes through the same point.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27481,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m$, $n$ and $r$ be positive integers with $n \\ge 2$ and $1 \\le r \\le n-1$. Consider a square table of size $(mn+r) \\times (mn+r)$. The table is covered by squares of size $n \\times n$ with sides parallel to the sides of the table. Each unit square is covered at least once and some unit squares are covered multiple times. Find the minimum possible number of unit squares covered at least two times.",
    "options": [],
    "answer": "(mn+mr+2r)(n-r)",
    "solution": "We call a unit square bad if it is covered more than once.\n\n(Bound) Choose an arbitrary row and mark its cells in the columns $r+1$, $r+n+1$, $\\dots$, $r+(m-1)n+1$. Consider all $n \\times n$ squares having nonempty intersection with the chosen row. Since $mn < mn+r < (m+1)n$, we have at least $m+1$ such squares. Moreover, each such square covers at least one of the marked cells. Since we have $m$ marked cells, we conclude that at least one of them is bad.\n\nAnalogously, there is at least one bad unit square among the intersecting squares of this row with columns with numbers $r+2$, $r+n+2$, $\\dots$, $r+(m-1)n+2$, and so on. Thus, in this row we have at least $n-r$ bad squares. The same reasoning shows that there are at least $n-r$ bad squares in any of the remaining rows. We repeat the above observation for all remaining $(m+1)r$ columns and conclude that there are at least $n-r$ bad cells in each of them. Therefore, there are at least\n$$\n(mn+r)(n-r) + (m+1)r(n-r) = (mn+mr+2r)(n-r)\n$$\nbad cells.\n\n(Construction) Consider the first $n+r$ rows (starting from below) and first $n+r$ columns (starting from the left). Their intersection forms a $(n+r) \\times (n+r)$ square $A$. Cover this square using four $n \\times n$ squares in its four corners. It is easy to see that the number of bad cells equals $(n+r)^2 - 4r^2$. Consider the $\\Gamma$-shaped figure consisting of all cells in the $(2n+r) \\times (2n+r)$ square without the cells of $A$. We cover all cells in this figure by five $n \\times n$ squares (two squares in each of the two rectangular parts and one square in the right upper corner) and leave $n^2 - r^2$ bad cells. The same procedure is applied $m-1$ times for all $\\Gamma$-shaped figures of width $n$. The number of bad cells equals\n$$\n(n+r)^2 - 4r^2 + (m-1)(n^2 - r^2) = (mn + mr + 2r)(n-r).\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27482,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given is a circle $k$ and a point $A$ outside of it. The segment $BC$ is a diameter of $k$. Find the locus of the orthocenter of $\\triangle ABC$, when $BC$ is changing.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "The locus is the radical axis of k and the circle with diameter AO, equivalently the polar of A with respect to k.",
    "solution": "Let $O$ be the center of $k$ and $\\omega$ be the circle with diameter $AO$. Let $AA_1$ and $BB_1$ be the altitudes of $\\triangle ABC$ and $H$ is the orthocenter. Then the power of $H$ with respect to $\\omega$ is equal to $AH \\cdot HA_1$ and the power of $H$ with respect to $k$ is equal to $BH \\cdot HB_1$. Since $AH \\cdot HA_1 = BH \\cdot HB_1$ it follows that $H$ lies on the radical axis $l$ of $k$ and $\\omega$. Conversely, it is easy to see that each point of $l$ is an orthocenter of some triangle from given type.\nLet $AB \\cap k = B'$, $AC \\cap k = C'$ and $BC' \\cap B'C = H$. Then $H$ is the orthocenter of $\\triangle ABC$ and $H$ lies on the polar $l$ of point $A$ with respect to $k$ (*Why?*). Conversely, it is easy to see that each point of $l$ is an orthocenter of some triangle from given type.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27483,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the least positive real number $\\alpha$ with the following property: if the weight of a finite number of pumpkins is 1 ton and the weight of every pumpkin is not more than $\\alpha$ tons then the pumpkins can be distributed in 50 boxes (some of the boxes may remain empty) such that there are no more than $\\alpha$ tons of pumpkins in every box.",
    "options": [],
    "answer": "2/51",
    "solution": "We prove that the desired value of $\\alpha$ is $\\alpha = \\frac{2}{51}$.\n\nAssume that some $\\alpha < \\frac{2}{51}$ satisfies the condition of the problem. Choose nonnegative integer $k \\ge 0$ such that $\\frac{1}{51 \\times 2^k} \\le \\alpha < \\frac{1}{51 \\times 2^{k-1}}$. Consider $51 \\times 2^k$ pumpkins each having weight $\\frac{1}{51 \\times 2^k}$ tons. For any distribution of pumpkins in 50 boxes there exists a box with at least two pumpkins and therefore the weight of this box equals at least $\\frac{1}{51 \\times 2^{k-1}} > \\alpha$, a contradiction.\n\nWe show that $\\alpha = \\frac{2}{51}$ satisfies the condition of the problem. Let the number of pumpkins be $m$. Take $m$ empty boxes and put a pumpkin in every one of them. If the two lightest boxes contain in common not more than $\\frac{2}{51}$ tons of pumpkins then gather all pumpkins of these two boxes into one of them and remove the empty box. When this operation terminates let the number of boxes be $n$ and they contain $x_1 \\le x_2 \\le \\dots \\le x_n$ tons of pumpkins. Thus $x_1 + x_2 > \\frac{2}{51}$ and hence $x_2 > \\frac{1}{51}$. Therefore\n$$\n1 = x_1 + x_2 + \\dots + x_n > \\frac{2}{51} + (n-2) \\times \\frac{1}{51},\n$$\nimplying $n < 51$. We have that all the pumpkins are distributed in no more than 50 boxes and there are no more than $\\frac{2}{51}$ tons of pumpkins in each box. This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27484,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given is a $\\triangle ABC$ with incircle $k$ touching the sides $BC$ and $CA$ in points $P$ and $Q$ respectively. Let denote with $J$ the center of the excircle at side $AB$ in $\\angle ABC$ and with $T$ – the second intersection point of the circumcircles of $\\angle JBP$ and $\\angle JAQ$. Prove that the circumcircle of $\\angle ABT$ touches $k$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $R$ be the tangent point of $k$ to $AB$. We will use the standard notations about the angles of $\\triangle ABC$. We have\n$$\n\\angle PTJ = 180^{\\circ} - \\angle PBJ = 90^{\\circ} - \\frac{\\beta}{2} = \\angle PRB\n$$\nand\n$$\n\\angle QTJ = 180^{\\circ} - \\angle QAJ = 90^{\\circ} - \\frac{\\alpha}{2} = \\angle QRA.\n$$\nThen\n$$\n\\angle PTQ = \\angle PTJ + \\angle QTJ = 180^{\\circ} - \\left(\\frac{\\alpha}{2} + \\frac{\\beta}{2}\\right), \\text{ A}\n$$\ni.e. $\\angle PTQ = 180^{\\circ} - \\angle PRQ$ and hence $T \\in k$, $R \\in TJ$.\n\n$$\n\\angle ATJ = \\angle AQJ = \\angle BPJ = \\angle BTJ,\n$$\ni.e. $S$ is the midpoint of $\\overline{AB}$.\nBut the homothety $h(T, R \\rightarrow S)$ sends $k$ to $\\omega$. Therefore these circles are tangent at point $T$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27485,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a rectangular $n \\times m$ table where $n \\ge 2$ and $m \\ge 2$ are positive integers. Each cell is colored in one of the four colors: white, green, red or blue. Call such a coloring *interesting* if any $2 \\times 2$ square contains every color exactly once. Find the number of interesting colorings.",
    "options": [],
    "answer": "6*(2^m + 2^n) - 24",
    "solution": "Number the columns of the table by $1, 2, \\ldots, m$ and its rows by $1, 2, \\ldots, n$ and denote the cell in the $i$-th column and $j$-th row by $(i, j)$.\n\nConsider an interesting coloring $S$. We show first that either any row in $S$ contains only two colors or any column in $S$ contains only two colors.\n\nSuppose $S$ does not contain a rectangle $1 \\times 3$ (with longer horizontal side) containing three distinct colors. It follows that in any row of $S$ we have two alternating colors, as needed.\n\nLet the cells $(u, v)$, $(u+1, v)$ and $(u+2, v)$ be covered in three distinct colors $a, b$ and $c$. We obtain that $(u+1, v+1)$ is colored in the fourth color $d$, $(u, v+1)$ in $c$, $(u+2, v+1)$ in $a$, $(u+1, v+2)$ in $b$, $(u, v+2)$ in $a$, $(u+2, v+1)$ in $c$, and so on. By analogy we obtain that $(u+1, v-1)$ is in color $d$, $(u, v-1)$ in $c$, $(u+2, v-1)$ in $a$ and so on. We conclude that column $u$ contains only colors $a$ and $c$, column $u+1$ contains only colors $b$ and $d$, and column $u+2$ contains only colors $a$ and $c$. It follows now that any column having the same parity as $u$ contains only colors $a$ and $c$, and any column having the parity of $u+1$ contains only colors $b$ and $d$.\n\nHence, either any row in $S$ contains only two colors or any column in $S$ contains only two colors.\n\nThe number of interesting colorings such that in any column of odd number two of the colors alternate and in any column of even parity the other two colors alternate is equal to $\\binom{4}{2} \\times 2^m$. The corresponding colorings when the rows are colored in two colors equal $\\binom{4}{2} \\times 2^n$. The colorings in which the color of the cell $(i, j)$ depends only on the parity of $i$ and $j$ are counted in both cases. Therefore, the number of interesting colorings equals $6 \\times (2^m + 2^n) - 24$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 27486,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ such that there exists a polynomial $f$ of degree $n$ with integer coefficients and a positive leading coefficient and a polynomial $g$ with integer coefficients such that the equality\n$$\nxf^2(x) + f(x) = (x^3 - x)g^2(x)\n$$\nholds for every real $x$.",
    "options": [],
    "answer": "n = 4k + 3 for k ≥ 0",
    "solution": "We have $xf^2(x) + f(x) = (x^3 - x)g^2(x) \\Leftrightarrow [2xf(x) + 1]^2 = (x^2 - 1)[2xg(x)]^2 + 1$.\nWe will now find all pairs $(p, q)$ of integer-coefficient polynomials such that $p^2(x) = (x^2 - 1)q^2(x) + 1$.\nLet $(p, q)$ be one such pair such that the degree of $q$ is $k \\ge 1$. We can assume, without loss of generality, that $p$ and $q$ have positive leading coefficients. Let $P_0 =$\n\n$p$ and $Q_0 = q$ and let $P_1(x) = xp(x) - (x^2 - 1)q(x)$ and $Q_1(x) = -p(x) + xq(x)$. It is easy to show that $P_1$ and $Q_1$ also satisfy the given equation and that the degree of $Q_1$ is strictly less than the degree of $Q_0$.\n(One way to come up with this construction is as follows. The equation on $p$ and $q$ can be rewritten as $1 = (p(x) - q(x)\\sqrt{x^2-1})(p(x) + q(x)\\sqrt{x^2-1})$. Also, one of its solutions is given by $(p, q) = (x, 1)$, i.e. $1 = (x - \\sqrt{x^2-1})(x + \\sqrt{x^2-1})$. Multiplying the last two equations then gives us exactly $1 = (P_1(x) - Q_1(x)\\sqrt{x^2-1})(P_1(x) + Q_1(x)\\sqrt{x^2-1})$.)\nContinuing this process, we will eventually arrive at a solution $(P_s, Q_s)$ such that $Q_s$ is constant. There are, however, only two solutions of this kind: $(x, 1)$ and $(1, 0)$. Since applying the operation to the former yields the latter, we can, without loss of generality, assume that $(P_s, Q_s) = (1, 0)$.\nIt follows that all solution pairs $(p, q)$ are given by the sequence determined by the initial condition $(p_0, q_0) = (1, 0)$ and the recurrence relation $(p_{i+1}, q_{i+1}) = (xp_i(x) + (x^2-1)q_i(x), p_i(x) + xq_i(x))$. (Or, more precisely, this recurrence yields those solutions in which the leading coefficients of $p$ and $q$ are positive.)\nA member of that sequence corresponds to a solution of the original equation exactly when $p(x)$ is congruent to $1 \\pmod{2x}$ and $q(x)$ is divisible by $2x$. Since the first five members of the sequence are $(1, 0)$, $(x, 1)$, $(2x^2 - 1, 2x)$, $(4x^3 - 3x, 4x^2 - 1)$, and $(8x^4 - 8x^2 + 1, 8x^3 - 4x)$ and $(p_4(x), q_4(x)) \\equiv (1, 0) \\pmod{2x}$, the sequence is periodic with period $4 \\pmod{2x}$ and exactly the members $(p_i, q_i)$ such that $i$ is a multiple of $4$ yield a solution. Therefore, the necessary values of $n$ are the numbers $4k + 3$ for $k \\ge 0$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 27487,
    "subject": "Mathematics (Multi-modal)",
    "question": "A real nonzero number is assigned to every point in the space. It is known that for any tetrahedron $\\tau$ the number written in the incenter equals the product of the four numbers written in the vertices of $\\tau$. Prove that all numbers equal 1.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider two arbitrary points $X$ and $Y$ and let $x$ and $y$ be the corresponding numbers. Choose points $I$ and $J$ on the line $XY$ such that $XY = YI = IJ$. Let $X'$ on the line $XY$ be such that $XI = JX'$. Consider the plane $\\lambda$ perpendicular to $IJ$ and passing through the midpoint of $IJ$. Let $ABCX$ and $ABCX'$ be two equal regular triangular pyramids with base $ABC$ in the plane $\\lambda$ having incenters $I$ and $J$.\n\nSince $n_A n_B n_C n_X = n_I$ and $n_A n_B n_C n_{X'} = n_J$ we have that\n$$\nn_X = \\frac{n_{X'} \\cdot n_I}{n_J}.\n$$\nMove the plane $\\lambda$ towards point $I$ and consider the spheres $S_I$ and $S_J$ with centers $I$ and $J$ respectively that are tangent to $\\lambda$. Let $A_1B_1C_1X'$ be a regular triangular pyramid with base $A_1B_1C_1$ in $\\lambda$ and insphere $S_J$. The regular triangular pyramid with base $A_1B_1C_1$ and insphere $S_I$ has vertex $X_1$. It follows from the above that\n$$\nn_{X_1} = \\frac{n_{X'} \\cdot n_I}{n_J} = n_X.\n$$\nWhen $\\lambda$ moves towards $I$ the radius of the sphere $S_I$ tends to zero and $\\triangle A_1B_1C_1$ tends to a triangle which is the base of a regular triangular pyramid with vertex $X'$ and inscribed sphere with center $J$ and radius $IJ$.\n\nWe conclude that $X_1$ tends to $I$. Continuity arguments show that all inner points on the segment $XI$ (with point $X$) are assigned with the same number.\n\nThus, $x = y$ and all numbers are equal. It follows from $x^4 = x$ and $x \\neq 0$ that $x = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27488,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all prime numbers $p$ and $q$ such that\n\n$p^2|q^3+1$ and $q^2|p^6-1$.",
    "options": [],
    "answer": "(p, q) = (2, 3) or (3, 2)",
    "solution": "If $p=3$, then $q^2|3^6-1 = 728 = 2^3 \\cdot 7 \\cdot 11$ and therefore $q=2$ which gives a solution. Let $p \\neq 3$. Since $(q+1, q^2-q+1) = 1$ or $3$, we have $p^2|q+1$ or $p^2|q^2-q+1$, which implies that $p < q$. If $p+1=q$ then $p=2$ and $q=3$ which is another solution. In the sequel we assume that $q \\ge p+2$.\nSince $q^2|p^6-1 = (p-1)(p+1)(p^2-p+1)(p^2+p+1)$ and $(q, p-1) = (q, p+1) = 1$, we have $q^2|(p^2-p+1)(p^2+p+1)$. Moreover, we have $(p^2-p+1, p^2+p+1) = (p^2+p+1, 2p) = 1$ and therefore $q^2|p^2-p+1$ or $q^2|p^2+p+1$. However, $p^2-p+1 < p^2 < q^2$ and $(p+2)^2 \\le q^2 \\le p^2+p+1$, i.e. $3p+3 < 0$, a contradiction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27489,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that, for all positive real numbers $a$, $b$, $c$, and $d$ the following inequality holds:\n$$\n\\sum_{cyc} \\frac{a^4}{a^3 + a^2 b + ab^2 + b^3} \\ge \\frac{a+b+c+d}{4}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We have\n$$\n\\begin{aligned} \\sum_{cyc} \\frac{a^4}{a^3 + a^2 b + ab^2 + b^3} - \\sum_{cyc} \\frac{b^4}{a^3 + a^2 b + ab^2 + b^3} &= \\sum_{cyc} \\frac{a^4 - b^4}{a^3 + a^2 b + ab^2 + b^3} \\\\ &= \\sum_{cyc} (a - b) = 0 \\end{aligned}\n$$\n$$\n\\sum_{cyc} \\frac{a^4 + b^4}{a^3 + a^2 b + ab^2 + b^3} \\ge \\frac{a + b + c + d}{2}.\n$$\nThis, however, follows from\n$$\n\\frac{a^4 + b^4}{a^3 + a^2b + ab^2 + b^3} \\ge \\frac{a+b}{4} \\Leftrightarrow 4(a^4 + b^4) \\ge (a+b)(a^3 + a^2b + ab^2 + b^3).\n$$\n$$\n\\frac{a^4}{a^3 + a^2b + ab^2 + b^3} \\ge \\frac{5}{8}a - \\frac{3}{8}b \\Leftrightarrow 3(a^4 + b^4) \\ge 2(a^3b + a^2b^2 + ab^3),\n$$\nwhich is obviously true. Therefore,\n$$\n\\sum_{cyc} \\frac{a^4}{a^3 + a^2b + ab^2 + b^3} \\ge \\sum_{cyc} \\frac{5}{8}a - \\frac{3}{8}b = \\frac{a+b+c+d}{4}.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27490,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f : \\mathbb{Q}^+ \\to \\mathbb{R}^+$ such that\n$$\nf(xy) = f(x + y)(f(x) + f(y)), \\text{ for any } x, y \\in \\mathbb{Q}^{+}.\n$$",
    "options": [],
    "answer": "f(x) = 1/x for all positive rational x, and f(x) = 1/2 for all positive rational x",
    "solution": "If $g(x) = 1/f(x)$ then $g : \\mathbb{Q}^+ \\to \\mathbb{R}^+$ and the given equation becomes\n$$\n(*) \\quad g(x+y)g(x)g(y) = g(xy)(g(x) + g(y)).\n$$\nLet $f(1) = c > 0$ i.e. $g(1) = \\frac{1}{c}$. From (*) we get $g(x+1) = c g(x) + 1$ and hence $g(2) = 2$, $g(3) = 2c+1$, $g(4) = 2c^2+c+1$, $g(5) = 2c^3+c^2+c+1$, $g(6) = 2c^4+c^3+c^2+c+1$. On the other hand setting $x=2$, $y=3$ in (*) leads us to\n$$\ng(5)g(2)g(3) = g(6)(g(2) + g(3)),\n$$\nwhich implies\n$$\n4c^5 - 3c^3 - c^2 - c + 1 = 0 \\Leftrightarrow (c-1)(c+1)(2c-1)(2c^2+2c+1) = 0.\n$$\nIf $c=1$ then $g(x+1) = g(x)+1$. By induction $g(n) = n$ for any $n \\in \\mathbb{N}$ and moreover $g(x+n) = g(x)+n$ for any $x \\in \\mathbb{Q}^+$ and $n \\in \\mathbb{N}$.\nBy setting $y=n$ in (*) we obtain\n$$\n(g(x) + n)g(x)n = g(nx)(g(x) + n),\n$$\ni.e. $g(nx) = n g(x)$. Setting $x=p/q$ and $n=q$, where $p, q \\in \\mathbb{N}$ we obtain $g(x) = x$, for any $x \\in \\mathbb{Q}^+$, i.e. $f(x) = 1/x$, for any $x \\in \\mathbb{Q}^+$.\n\nIf $c = \\frac{1}{2}$ then $g(x+1) = \\frac{1}{2}g(x) + 1$. Hence\n$$\ng(n) = 2 \\quad \\text{and} \\quad g(x+n) - 2 = \\frac{g(x) - 2}{2^n}, \\quad x \\in \\mathbb{Q}^{+}, n \\in \\mathbb{N};\n$$\nSetting $y = n$ in $(*)$ we obtain\n$$\n2g(x+n)g(x) = g(nx)(g(x) + 2),\n$$\nand it's sufficient to see that these equations imply $g(x) = 2$, for any $x \\in \\mathbb{Q}^{+}$, i.e. $f \\equiv 1/2$.\nFinally, the only functions satisfying the given equality are $f \\equiv \\frac{1}{2}$ and $f \\equiv \\frac{1}{x}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27491,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $2k$ citizens in a town every two of which are either friends or enemies. For some positive integer $t$ each citizen has at most $t$ enemies and there exists a citizen having exactly $t$ enemies. A group is called *friendly* if any two members of the group are friends. It is known that a friendly group with more than $k$ members does not exist and all citizens can be partitioned into two friendly groups having $k$ members each. Prove that the number of friendly groups having $k$ members is not greater than $2^{k-1} + 2^{k-t}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $A = \\{a_1, a_2, \\dots, a_k\\}$ and $B = \\{b_1, b_2, \\dots, b_k\\}$ be the two friendly groups. For arbitrary group $C$ from $A$ denote by $S_C$ the group of all people from $B$ each of which is an enemy of at least one person from $C$. If $|C| > |S_C|$ then $C \\cup (B \\setminus S_C)$ is a friendly group having more than $k$ members, a contradiction.\n\nTherefore the sets $S_{\\{a_i\\}}$ satisfy the Hall's condition for system of distinct representatives. Hence, we may assume that $a_i$ and $b_i$ are enemies for all $i = 1, 2, \\dots, k$.\n\nThis means that every friendly group of $k$ members include one person from every pair $(a_i, b_i)$ for $i = 1, 2, \\dots, k$.\n\nSuppose the enemies of $a_1$ are $b_1, \\dots, b_t$. For a friendly group $S$ such that $a_1 \\in S$ we have $b_1, \\dots, b_t \\notin S$. Hence, $a_2, \\dots, a_t \\in S$. From every of the remaining $k-t$ pairs $(a_j, b_j)$, $j > t$ we have to choose one of $a_j$ and $b_j$ to be an element of $S$. Therefore there are at most $2^{k-t}$ such group.\n\nFor a friendly group $S$ with $k$ members such that $a_1 \\notin S$ we have $b_1 \\in S$. Since from each of the remaining $k-1$ pairs $(a_j, b_j)$, $j > 1$ we have to choose one of $a_j$ and $b_j$ to be an element of $S$ we find that there are at most $2^{k-1}$ such groups.\n\nTherefore, the total number of friendly groups of $k$ members is at most $2^{k-1} + 2^{k-t}$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 27492,
    "subject": "Mathematics (Multi-modal)",
    "question": "The quadrilateral $ABCD$ is inscribed in the circle $k$. The lines $AC$ and $BD$ meet in $E$ and the lines $AD$ and $BC$ meet in $F$. Show that the line through the incenters of $\\triangle ABE$ and $\\triangle ABF$ and the line through the incenters of $\\triangle CDE$ and $\\triangle CDF$ meet on $k$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $I_e, I_f, J_e, J_f$ be the incenters of $\\triangle ABE, \\triangle ABF, \\triangle CDE, \\triangle CDF$, respectively. Let $P = AI_e \\cap BI_f$, $Q = AI_f \\cap BI_e$, $U = CJ_e \\cap DJ_f$, and $V = CJ_f \\cap DJ_e$ be the excenter of $\\triangle ABC$ opposite to $A$, the excenter of $\\triangle ABD$ opposite to $B$, the incenter of $\\triangle ACD$, and the incenter of $\\triangle BCD$, respectively.\n\nWe have $\\angle APB = \\frac{1}{2} \\widehat{AB} = \\angle AQB$; therefore, $ABPQ$ is cyclic. Analogously, $CDUV$ is cyclic.\n\nWe have $\\angle APB = \\frac{1}{2} \\widehat{AB} = \\angle UCV$, $\\angle AQB = \\frac{1}{2} \\widehat{AB} = \\angle UDV$, $\\angle APQ = \\angle ABQ = \\frac{1}{2} \\widehat{AD} = \\angle UCD$, and $\\angle BQP = \\angle BAP = \\frac{1}{2} \\widehat{BC} = \\angle VDC$. Therefore, the figures $ABPQ$ and $UVCD$ are similar and identically oriented. Let $T$ be their center of similitude. (I.e., let $T$ be the fixed point of the unique similitude which maps $A, B, P, Q$ onto $U, V, C, D$, respectively.)\n\nWe have $\\triangle TAQ \\sim \\triangle TUD$; therefore, $\\triangle TAU \\sim \\triangle TQD$ and $\\angle ATQ = \\angle (AU, QD) = \\frac{1}{2} \\widehat{AD} = \\angle AI_eQ$. It follows that $T$ lies on the circumcircle of $\\triangle AI_eQ$. Analogously, $T$ lies on the circumcircle of $\\triangle BI_eP, \\triangle CJ_eV$, and $\\triangle DJ_eU$.\n\nWe have, then, $\\angle ATB = \\angle ATI_e + \\angle I_eTB = \\angle AQB + \\angle APB = \\frac{1}{2} \\widehat{AB} + \\frac{1}{2} \\widehat{AB} = \\widehat{AB}$, showing that $T$ lies on $k$.\n\n![](attached_image_1.png)\n\nLet $T' \\in I_f I_e$ so that $I_f A \\cdot I_f Q = I_f I_e \\cdot I_f T' = I_f B \\cdot I_f P$. Then $T'$ is a point other than $I_e$ which lies on the circumcircles of both $\\triangle AI_eQ$ and $\\triangle BI_eP$; therefore, $T' \\equiv T$ and $T \\in I_e I_f$.\n\nAnalogously, $T \\in J_e J_f$, and we are done.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27493,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ and $4p+1$ be two primes such that $p > 10^9$. Prove that the decimal representation of the number $\\frac{1}{4p+1}$ contains every digit from 0 to 9.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $q = 4p+1$ and $a \\circ b$ be the remainder of $a$ upon division by $b$. It suffices to show that the last digits of the numbers $10^k \\circ q$ cover all digits from 0 to 9. Indeed, since $q$ and $10$ are coprime, that would mean that all digits from 0 to 9 occur among the last digits of the numbers $\\lfloor \\frac{10^k}{q} \\rfloor$ – and those are the same as the digits in the decimal representation of $\\frac{1}{q}$.\n\nLet $S$ be the set of all nonzero biquadratic residues modulo $q$, i.e., the set of all $0 < t < q$ such that the congruence $x^4 \\equiv t$ has a solution modulo $q$. Then $S$ has exactly $p$ elements. Indeed, since $-1$ is a quadratic residue modulo $q$, the $2p$ nonzero quadratic residues modulo $q$ can be grouped in pairs of the kind $\\{s, -s\\}$, and among their squares – which are precisely the biquadratic residues – there are precisely $p$ distinct ones.\n\nWe will show that every element of $S$ has the form $10^k \\circ q$ for some $k$. Indeed, let $d$ be the index of $q$ modulo $10$. Since $d \\mid \\varphi(q) = 4p$, we have $d = p$, $d = 2p$, or $d = 4p$.\n\nConsider the case $d = p$ first. Let $0 \\le k < p$; then there is a $0 \\le j \\le 3$ such that $k + jq$ is a multiple of $4$ and, therefore, the number $10^k \\equiv [10^{\\frac{k+jq}{4}}]^4 \\pmod{q}$ is a biquadratic residue. Since the numbers $10^k \\circ q$ for $0 \\le k < p$ are pairwise distinct and all of them are biquadratic residues, they must coincide with the elements of $S$, as needed.\n\nThe cases $d = 2p$ and $d = 4p$ are treated analogously.\n\nLet, then, $u$ be an arbitrary digit. Let $0 \\le j \\le 3$ be such that $u + jq$ ends in $0$, $1$, $5$, or $6$. Since $q > 10^9$, we have $\\sqrt[4]{-(j+1)q} - \\sqrt[4]{u+jq} > 6$. It follows that the interval $[u+jq, (j+1)q)$ contains at least six fourth powers. Therefore, it contains at least one fourth power $x^4$ that ends in the same digit as $u+jq$. Let $s = x^4 \\circ q$; then $x^4 = s + jq$ and $10 \\mid x^4 - (u+jq) = s - u$, i.e., $s$ ends in $u$, as needed.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27494,
    "subject": "Mathematics (Multi-modal)",
    "question": "The hexagon $ABLCDK$ is inscribed and the line $LK$ intersects the segments $AD$, $BC$, $AC$ and $BD$ in points $M$, $N$, $P$ and $Q$, respectively. Prove that $NL \\cdot KP \\cdot MQ = KM \\cdot PN \\cdot LQ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "*First solution.* Denote $s = \\sin \\frac{\\hat{A}B}{2}$, $t = \\sin \\frac{\\hat{B}L}{2}$, $u = \\sin \\frac{\\hat{L}C + \\hat{A}K}{2}$, $v = \\sin \\frac{\\hat{C}K}{2}$, $w = \\sin \\frac{\\hat{D}K}{2}$ and $x = \\sin \\frac{\\hat{L}D + \\hat{A}K}{2}$. Then we consecutively have\n$$\n\\frac{NL \\cdot KP \\cdot MQ}{KM \\cdot PN \\cdot LQ} = \\frac{NL}{NC} \\cdot \\frac{NC}{NP} \\cdot \\frac{KP}{AK} \\cdot \\frac{AK}{KM} \\cdot \\frac{MQ}{DQ} \\cdot \\frac{DQ}{LQ} = \\frac{t}{v} \\cdot \\frac{u}{s} \\cdot \\frac{v}{u} \\cdot \\frac{x}{w} \\cdot \\frac{s}{x} \\cdot \\frac{w}{t} = 1.\n$$\n\n\n*Second solution.* We choose $T \\in MN$ such that $\\angle NTB = \\angle ADB = \\angle ACB$ and $N$ is between $P$ and $T$. Then $\\triangle CNP \\sim \\triangle TNB$ and therefore $(TL + NL)PN = CN \\cdot BN$, whence $TL = \\frac{NL \\cdot KP}{PN}$. Analogously, $\\triangle DQM \\sim \\triangle TQB$ implies $TL = \\frac{LQ \\cdot KM}{MQ}$ and we have the desired result.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27495,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a quadrilateral $ABCD$ sides $AB$ and $CD$ are not parallel. The midpoints of $AD$ and $BC$ are denoted by $M$ and $N$, respectively. The line $MN$ intersects the diagonals $AC$ and $BD$ at points $K$ and $L$, respectively. Prove that the circumcircles of triangles $AKM$ and $BNL$ intersect on the line $AB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27496,
    "subject": "Mathematics (Multi-modal)",
    "question": "One hundred and one of the squares of an $n \\times n$ table are colored blue. It is known that there exists a unique way to cut the table to rectangles along boundaries of its squares with the following property: every rectangle contains exactly one blue square. Find the smallest possible $n$.",
    "options": [],
    "answer": "101",
    "solution": "Answer: 101. We shall prove the following more general assertion.\n\n**Lemma.** Some of the squares of a table $P$ are colored blue. A partition of $P$ into rectangles of integer sides is called *good* if every rectangle contains exactly one blue square. Then $P$ possesses a unique good partition if and only if the blue squares form a rectangle.\n\n*Proof.* It is easy to see that if the blue squares form a rectangle, there is only one good partition (there are unique choices for the angle squares of the blue rectangle, then for the remaining boundary squares and finally for the inner squares).\n\nWe now consider the other direction. Let us denote the unique good partition of $P$ by $S$. We call *dividing* any (vertical or horizontal) line which cuts the table into two rectangles each of them containing at least one blue square.\n\nWe first prove that every sub-rectangle $\\Pi$ which contains at least one blue square possesses good partition. We go by induction on the number of the blue squares in $\\Pi$. The assertion is obvious for a single blue square. For more blue squares we cut $\\Pi$ into two smaller rectangles $\\Pi_1$ and $\\Pi_2$, each of them containing blue square(s) and apply the induction hypothesis.\n\nIt follows from the above that every dividing line $\\ell$ determines good partition which includes the line $\\ell$. Therefore every dividing line is included in the partition $S$.\n\nLet $\\ell_1, \\ell_2, \\dots, \\ell_p$ be all vertical dividing lines (from left to right end) and $m_1, m_2, \\dots, m_q$ are all horizontal dividing lines (from bottom to upper end). Then these lines divide the table into rectangles, each of them containing at most one blue square. Therefore $S$ does not include any lines different from these and, in fact, every rectangle contains exactly one blue square.\n\nLet $\\ell_0$ be the rightmost vertical line such that there are no blue squares on its left side, $\\ell_{p+1}$ is the leftmost vertical line such that there are no blue squares on its right side, and $m_0$ and $m_{q+1}$ are defined analogously. It is clear that the distance between $\\ell_i$ and $\\ell_{i+1}$ equals 1 for every $0 \\le i \\le p$ and, analogously, the distance between $m_j$ and $m_{j+1}$ equals 1 for every $0 \\le j \\le q$ (otherwise further dividing lines could be added). Therefore the set of the blue squares coincides with the rectangle defined from the lines $\\ell_0, \\ell_{p+1}, m_0$ and $m_{q+1}$, which completes the proof of the lemma.\n\nIn our problem, we see that the 101 blue squares form a rectangle. Moreover, one of the sides of this rectangle must be 1, and the other one, equal to 101, must be equal to $n$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27497,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sequence $a_1, a_2, \\dots$ is defined by the equalities $a_1 = 2$, $a_2 = 12$ and $a_{n+1} = 6a_n - a_{n-1}$ for every positive integer $n \\ge 2$. Prove that no member of this sequence is equal to a perfect power (greater than one) of a positive integer.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We shall use the following assertion.\n\n**Lemma.** Let $k \\ge 2$ be a positive integer. Then the equation $2x^{2k} + 1 = y^2$ does not have solutions in positive integers.\n\n**Proof.** Assume that $x, y$ and $k \\ge 2$ are positive integers such that $2x^{2k} + 1 = y^2$ and $x$ is minimum possible. It is obvious that $x$ is even and $y$ is odd. Let us denote $x = 2a$ and $y = 2b + 1$. Then $2^{2k-1}a^{2k} = b(b+1)$ and $(b, b+1) = 1$. There are two possibilities:\n- if $b = x_1^{2k}$ and $b+1 = 2^{2k-1}x_2^{2k}$, $x_1, x_2 \\in \\mathbb{N}$, $x_1x_2 = a$, then $2^{2k-1}x_2^{2k} - x_1^{2k} = 1$, which gives a contradiction modulo 4;\n- if $b = 2^{2k-1}x_1^{2k}$ and $b+1 = x_2^{2k}$, $x_1, x_2 \\in \\mathbb{N}$, $x_1x_2 = a$, then $x_2^{2k} - 2^{2k-1}x_1^{2k} = 1$, which leads to the equation $y_1^2 = 2^{2k-1}x_1^{2k} + 1$, $y_1 = x_2^k$, where\n\nwe notice that $x_1 < x$.\nIt is clear that the above argument of decreasing the degrees of 2 can be continued until we have degree at most 5. Therefore we reach the equation $y_0^2 = 8x_0^{2k} + 1$, where $x_0 < x$ and $y_0 = y_2^k$, $y_2 \\in \\mathbb{N}$. Clearly, $y_0$ is odd and we set $y_0 = 2c+1$. We obtain $c(c+1) = 2x_0^{2k}$, where $(c, c+1) = 1$. We have again two possibilities:\n- if $c = x_3^{2k}$ and $c+1 = 2x_4^{2k}$, $x_3, x_4 \\in \\mathbb{N}$, $x_3x_4 = x_0$, then $4x_4^{2k} = 2c+2 = y_2^k+1$, whence $(2x_4^k-1)(2x_4^k+1) = y_2^k$. This leads to $2x_4^k-1 = y_3^k$, $2x_4^k+1 = y_4^k$, $y_3, y_4 \\in \\mathbb{N}$, $y_3y_4 = y_2$, and finally $y_4^k - y_3^k = 2$, which is impossible;\n- if $c = 2x_3^{2k}$ and $c+1 = x_4^{2k}$, $x_3, x_4 \\in \\mathbb{N}$, $x_3x_4 = x_0$, then $2x_3^{2k}+1 = (x_4^k)^2$, which contradicts to the choice of $x$ as minimal.\nThis completes the proof of the lemma.\n\nThe roots of the characteristic equation $t^2 - 6t + 1 = 0$ of our sequence are $t_{1,2} = 3 \\pm 2\\sqrt{2}$. Therefore we find (using the conditions $a_1 = 2$ and $a_2 = 12$)\n$$\na_n = \\frac{(3 + 2\\sqrt{2})^n - (3 - 2\\sqrt{2})^n}{2\\sqrt{2}}.\n$$\nDenote $(3 + 2\\sqrt{2})^n = \\alpha_n + \\beta_n\\sqrt{2}$, $\\alpha_n, \\beta_n \\in \\mathbb{N}$. Then $(3 - 2\\sqrt{2})^n = \\alpha_n - \\beta_n\\sqrt{2}$, $\\alpha_n = \\beta_n$ and $\\alpha_n^2 - 2\\beta_n^2 = 1$. Now, if $a_n$ is perfect power for some $n$, then the last two equalities give a contradiction with the lemma.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27498,
    "subject": "Mathematics (Multi-modal)",
    "question": "Any integer from the set $A = \\{1, 2, \\dots, 2015\\}$ is colored either red or green. For given positive integers $a$ and $b$ we are allowed to change the color of any $a$ or $b$ consecutive integers from $A$. A pair $(a, b)$ is called *good* if it is possible to change the color of all numbers after finite number of moves. A positive integer $t \\le 2015$ is called *nice* if all pairs $(a, b)$, for which $a + b = t$ are good. Find all nice numbers.",
    "options": [],
    "answer": "1, 3, 7, 9, 21, 63",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27499,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{R}^+ \\to \\mathbb{R}^+$, that satisfy the inequalities\n(i) $f(x + y) \\geq f(x) + y$\n(ii) $f(f(x)) \\leq x$\nfor all positive $x$ and $y$.",
    "options": [],
    "answer": "f(x) = x for all positive real x",
    "solution": "It follows from (i) that $f$ is strictly increasing function. Also, (ii) implies\n$$\nx + y \\geq f(f(x + y)). \\qquad (1)\n$$\n\nFurthermore, (i) gives $f(f(x+y)) \\ge f(f(x)+y)$ and the substitution $x \\to y$ and $y \\to f(x)$ in (i) implies $f(f(x)+y) \\ge f(x)+f(y)$.\nSince $f$ is increasing we have $\\lim_{x \\to 0^+} f(x) = \\inf_{x>0} f(x) = \\ell \\ge 0$. Note that (ii) implies $\\lim_{x \\to 0^+} f(f(x)) = 0$.\nAssume $\\ell > 0$. Since $f$ is increasing we have $f(f(x)) \\ge f(\\ell) > 0$, a contradiction to $\\lim_{x \\to 0^+} f(f(x)) = 0$. Therefore $\\ell = 0$ and $\\lim_{x \\to 0^+} f(x) = 0$. Letting $y \\to 0^+$ in (1) we obtain $x \\ge f(x)$ for all positive $x$. It follows now from (i) that\n$$\nx + y \\ge f(x + y) \\ge f(x) + y, \\\\ x - f(x) \\ge f(x + y) - f(x) - y \\ge 0.\n$$\nFix $x+y$ and let $x \\to 0^+$ in the above inequalities. We have $f(x+y) = x+y$ meaning that $f(x) = x$ for all positive $x$. This function is obviously a solution to the problem.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27500,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be acute triangle. Point $M$ is arbitrary point on the side $AB$, and $N$ is the midpoint of $AC$. Denote by $P$ and $Q$ the feet of the perpendiculars from $A$ to the lines $MC$ and $MN$, respectively. Prove that when $M$ vary then the circumcenter of $\\triangle PQN$ lies on a fixed line.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  }
]