problem stringlengths 0 45.3k | solution stringlengths 0 26.8k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 9
values | question_type stringclasses 5
values | problem_is_valid stringclasses 5
values | solution_is_valid stringclasses 5
values | source stringclasses 11
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|---|---|---|---|---|---|---|---|---|
A3 Let $a, b$ be positive real numbers. Prove that $\sqrt{\frac{a^{2}+a b+b^{2}}{3}}+\sqrt{a b} \leq a+b$.
|
Solution 1
Applying $x+y \leq \sqrt{2\left(x^{2}+y^{2}\right)}$ for $x=\sqrt{\frac{a^{2}+a b+b^{2}}{3}}$ and $y=\sqrt{a b}$, we will obtain $\sqrt{\frac{a^{2}+a b+b^{2}}{3}}+\sqrt{a b} \leq \sqrt{\frac{2 a^{2}+2 a b+2 b^{2}+6 a b}{3}} \leq \sqrt{\frac{3\left(a^{2}+b^{2}+2 a b\right)}{3}}=a+b$.
| proof | Inequalities | proof | Yes | Yes | olympiads | false |
A4 Let $x, y$ be positive real numbers such that $x^{3}+y^{3} \leq x^{2}+y^{2}$. Find the greatest possible value of the product $x y$.
| ## Solution 1
We have $(x+y)\left(x^{2}+y^{2}\right) \geq(x+y)\left(x^{3}+y^{3}\right) \geq\left(x^{2}+y^{2}\right)^{2}$, hence $x+y \geq x^{2}+y^{2}$. Now $2(x+y) \geq(1+1)\left(x^{2}+y^{2}\right) \geq(x+y)^{2}$, thus $2 \geq x+y$. Because $x+y \geq 2 \sqrt{x y}$, we will obtain $1 \geq x y$. Equality holds when $x=y... | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
A5 Determine the positive integers $a, b$ such that $a^{2} b^{2}+208=4\{l c m[a ; b]+g c d(a ; b)\}^{2}$.
| ## Solution
Let $d=\operatorname{gcd}(a, b)$ and $x, y \in \mathbb{Z}_{+}$such that $a=d x, b=d y$. Obviously, $(x, y)=1$. The equation is equivalent to $d^{4} x^{2} y^{2}+208=4 d^{2}(x y+1)^{2}$. Hence $d^{2} \mid 208$ or $d^{2} \mid 13 \cdot 4^{2}$, so $d \in\{1,2,4\}$. Take $t=x y$ with $t \in \mathbb{Z}_{+}$.
Cas... | (,b)\in{(2,12);(4,6);(6,4);(12;2)} | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A6 Let $x_{i}>1$, for all $i \in\{1,2,3, \ldots, 2011\}$. Prove the inequality $\sum_{i=1}^{2011} \frac{x_{i}^{2}}{x_{i+1}-1} \geq 8044$ where $x_{2012}=x_{1}$. When does equality hold?
| ## Solution 1
Realize that $\left(x_{i}-2\right)^{2} \geq 0 \Leftrightarrow x_{i}^{2} \geq 4\left(x_{i}-1\right)$. So we get:
$\frac{x_{1}^{2}}{x_{2}-1}+\frac{x_{2}^{2}}{x_{3}-1}+\ldots+\frac{x_{2011}^{2}}{x_{1}-1} \geq 4\left(\frac{x_{1}-1}{x_{2}-1}+\frac{x_{2}-1}{x_{3}-1}+\ldots+\frac{x_{2011}-1}{x_{1}-1}\right)$. ... | 8044 | Inequalities | proof | Yes | Yes | olympiads | false |
A7 Let $a, b, c$ be positive real numbers with $a b c=1$. Prove the inequality:
$$
\frac{2 a^{2}+\frac{1}{a}}{b+\frac{1}{a}+1}+\frac{2 b^{2}+\frac{1}{b}}{c+\frac{1}{b}+1}+\frac{2 c^{2}+\frac{1}{c}}{a+\frac{1}{c}+1} \geq 3
$$
| ## Solution 1
By $A M-G M$ we have $2 x^{2}+\frac{1}{x}=x^{2}+x^{2}+\frac{1}{x} \geq 3 \sqrt[3]{\frac{x^{4}}{x}}=3 x$ for all $x>0$, so we have:
$\sum_{\text {cyc }} \frac{2 a^{2}+\frac{1}{a}}{b+\frac{1}{a}+1} \geq \sum_{c y c} \frac{3 a}{1+b+b c}=3\left(\sum_{c y c} \frac{a^{2}}{1+a+a b}\right) \geq \frac{3(a+b+c)^{... | 3 | Inequalities | proof | Yes | Yes | olympiads | false |
A8 Decipher the equality $(\overline{L A R N}-\overline{A C A}):(\overline{C Y P}+\overline{R U S})=C^{Y^{P}} \cdot R^{U^{S}}$ where different symbols correspond to different digits and equal symbols correspond to equal digits. It is also supposed that all these digits are different from 0 .
| ## Solution
Denote $x=\overline{L A R N}-\overline{A C A}, y=\overline{C Y P}+\overline{R U S}$ and $z=C^{Y^{P}} \cdot R^{U^{S}}$. It is obvious that $1823-898 \leq x \leq 9187-121,135+246 \leq y \leq 975+864$, that is $925 \leq x \leq 9075$ and $381 \leq y \leq 1839$, whence it follows that $\frac{925}{1839} \leq \fr... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
A9 Let $x_{1}, x_{2}, \ldots, x_{n}$ be real numbers satisfying $\sum_{k=1}^{n-1} \min \left(x_{k} ; x_{k+1}\right)=\min \left(x_{1}, x_{n}\right)$.
Prove that $\sum_{k=2}^{n-1} x_{k} \geq 0$.
| ## Solution 1
Case I. If $\min \left(x_{1}, x_{n}\right)=x_{1}$, we know that $x_{k} \geq \min \left(x_{k} ; x_{k+1}\right)$ for all $k \in\{1,2,3, \ldots, n-1\}$. So $x_{1}+x_{2}+\ldots+x_{n-1} \geq \sum_{k=1}^{n-1} \min \left(x_{k} ; x_{k+1}\right)=\min \left(x_{1}, x_{n}\right)=x_{1}$, hence $\sum_{k=2}^{n-1} x_{k}... | proof | Inequalities | proof | Yes | Yes | olympiads | false |
A1 If for the real numbers $x, y, z, k$ the following conditions are valid, $x \neq y \neq z \neq x$ and $x^{3}+y^{3}+k\left(x^{2}+y^{2}\right)=y^{3}+z^{3}+k\left(y^{2}+z^{2}\right)=z^{3}+x^{3}+k\left(z^{2}+x^{2}\right)=2008$, find the product $x y z$.
|
Solution
$x^{3}+y^{3}+k\left(x^{2}+y^{2}\right)=y^{3}+z^{3}+k\left(y^{2}+z^{2}\right) \Rightarrow x^{2}+x z+z^{2}=-k(x+z):(1)$ and $y^{3}+z^{3}+k\left(y^{2}+z^{2}\right)=z^{3}+x^{3}+k\left(z^{2}+x^{2}\right) \Rightarrow y^{2}+y x+x^{2}=-k(y+x):(2)$
- From (1) $-(2) \Rightarrow x+y+z=-k:(*)$
- If $x+z=0$, then from $... | xy1004 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
A2 Find all real numbers $a, b, c, d$ such that $a+b+c+d=20$ and $a b+a c+a d+b c+b d+c d=$ 150 .
| ## Solution
$400=(a+b+c+d)^{2}=a^{2}+b^{2}+c^{2}+d^{2}+2 \cdot 150$, so $a^{2}+b^{2}+c^{2}+d^{2}=100$. Now $(a-b)^{2}+(a-c)^{2}+(a-d)^{2}+(b-c)^{2}+(b-d)^{2}+(c-d)^{2}=3\left(a^{2}+b^{2}+c^{2}+d^{2}\right)-2(a b+$ $a c+a d+b c+b d+c d)=300-300=0$. Thus $a=b=c=d=5$.
| =b===5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
A3 Let the real parameter $p$ be such that the system
$$
\left\{\begin{array}{l}
p\left(x^{2}-y^{2}\right)=\left(p^{2}-1\right) x y \\
|x-1|+|y|=1
\end{array}\right.
$$
has at least three different real solutions. Find $p$ and solve the system for that $p$.
| ## Solution
The second equation is invariant when $y$ is replaced by $-y$, so let us assume $y \geq 0$. It is also invariant when $x-1$ is replaced by $-(x-1)$, so let us assume $x \geq 1$. Under these conditions the equation becomes $x+y=2$, which defines a line on the coordinate plane. The set of points on it that s... | p=1orp=-1,0\leqx\leq1,withy=\x | Algebra | math-word-problem | Yes | Yes | olympiads | false |
A4 Find all triples $(x, y, z)$ of real numbers that satisfy the system
$$
\left\{\begin{array}{l}
x+y+z=2008 \\
x^{2}+y^{2}+z^{2}=6024^{2} \\
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2008}
\end{array}\right.
$$
| ## Solution
The last equation implies $x y z=2008(x y+y z+z x)$, therefore $x y z-2008(x y+y z+z x)+$ $2008^{2}(x+y+z)-2008^{3}=0$.
$(x-2008)(y-2008)(z-2008)=0$.
Thus one of the variable equals 2008. Let this be $x$. Then the first equation implies $y=-z$. From the second one it now follows that $2 y^{2}=6024^{2}-20... | (2008,4016,-4016) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
A5 Find all triples $(x, y, z)$ of real positive numbers, which satisfy the system
$$
\left\{\begin{array}{l}
\frac{1}{x}+\frac{4}{y}+\frac{9}{z}=3 \\
x+y+z \leq 12
\end{array}\right.
$$
| ## Solution
If we multiply the given equation and inequality $(x>0, y>0, z>0)$, we have
$$
\left(\frac{4 x}{y}+\frac{y}{x}\right)+\left(\frac{z}{x}+\frac{9 x}{z}\right)+\left(\frac{4 z}{y}+\frac{9 y}{z}\right) \leq 22
$$
From AM-GM we have
$$
\frac{4 x}{y}+\frac{y}{x} \geq 4, \quad \frac{z}{x}+\frac{9 x}{z} \geq 6,... | (2,4,6) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
A6 If the real numbers $a, b, c, d$ are such that $0a+b+c+d
$$
| ## Solution
If $1 \geq a+b+c$ then we write the given inequality equivalently as
$$
\begin{gathered}
1-(a+b+c)+d[(a+b+c)-1]+a b+b c+c a>0 \\
\Leftrightarrow[1-(a+b+c)](1-d)+a b+b c+c a>0
\end{gathered}
$$
which is of course true.
If instead $a+b+c>1$, then $d(a+b+c)>d$ i.e.
$$
d a+d b+d c>d
$$
We are going to pro... | proof | Inequalities | math-word-problem | Incomplete | Yes | olympiads | false |
A7 Let $a, b$ and $c$ be a positive real numbers such that $a b c=1$. Prove the inequality
$$
\left(a b+b c+\frac{1}{c a}\right)\left(b c+c a+\frac{1}{a b}\right)\left(c a+a b+\frac{1}{b c}\right) \geq(1+2 a)(1+2 b)(1+2 c)
$$
| ## Solution 1
By Cauchy-Schwarz inequality and $a b c=1$ we get
$$
\begin{gathered}
\sqrt{\left(b c+c a+\frac{1}{a b}\right)\left(a b+b c+\frac{1}{c a}\right)}=\sqrt{\left(b c+c a+\frac{1}{a b}\right)\left(\frac{1}{c a}+a b+b c\right)} \geq \\
\left(\sqrt{a b} \cdot \sqrt{\frac{1}{a b}}+\sqrt{b c} \cdot \sqrt{b c}+\s... | proof | Inequalities | proof | Yes | Yes | olympiads | false |
A8 Show that
$$
(x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) \geq 4\left(\frac{x}{x y+1}+\frac{y}{y z+1}+\frac{z}{z x+1}\right)^{2}
$$
for any real positive numbers $x, y$ and $z$.
| ## Solution
The idea is to split the inequality in two, showing that
$$
\left(\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{z}}+\sqrt{\frac{z}{x}}\right)^{2}
$$
can be intercalated between the left-hand side and the right-hand side. Indeed, using the Cauchy-Schwarz inequality one has
$$
(x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\fr... | proof | Inequalities | proof | Yes | Yes | olympiads | false |
A9 Consider an integer $n \geq 4$ and a sequence of real numbers $x_{1}, x_{2}, x_{3}, \ldots, x_{n}$. An operation consists in eliminating all numbers not having the rank of the form $4 k+3$, thus leaving only the numbers $x_{3}, x_{7}, x_{11}, \ldots$ (for example, the sequence $4,5,9,3,6,6,1,8$ produces the sequenc... | ## Solution
After the first operation 256 number remain; after the second one, 64 are left, then 16, next 4 and ultimately only one number.
Notice that the 256 numbers left after the first operation are $3,7, \ldots, 1023$, hence they are in arithmetical progression of common difference 4. Successively, the 64 number... | 683 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
C1 On a $5 \times 5$ board, $n$ white markers are positioned, each marker in a distinct $1 \times 1$ square. A smart child got an assignment to recolor in black as many markers as possible, in the following manner: a white marker is taken from the board; it is colored in black, and then put back on the board on an emp... |
Solution
a) Position 20 white markers on the board such that the left-most column is empty. This
positioning is good because the coloring can be realized column by column, starting with the second (from left), then the third, and so on, so that the white marker on position $(i, j)$ after the coloring is put on positi... | proof | Combinatorics | proof | Yes | Yes | olympiads | false |
C2 Kostas and Helene have the following dialogue:
Kostas: I have in my mind three positive real numbers with product 1 and sum equal to the sum of all their pairwise products.
Helene: I think that I know the numbers you have in mind. They are all equal to 1.
Kostas: In fact, the numbers you mentioned satisfy my con... | ## Solution
Kostas is right according to the following analysis:
If $x, y, z$ are the three positive real numbers Kostas thought about, then they satisfy the following equations:
$$
\begin{gathered}
x y+y z+z x=x+y+z \\
x y z=1
\end{gathered}
$$
Subtracting (1) from (2) by parts we obtain
$$
\begin{gathered}
x y z... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false |
C3 Integers $1,2, \ldots, 2 n$ are arbitrarily assigned to boxes labeled with numbers $1,2, \ldots, 2 n$. Now, we add the number assigned to the box to the number on the box label. Show that two such sums give the same remainder modulo $2 n$.
| ## Solution
Let us assume that all sums give different remainder modulo $2 n$, and let $S$ denote the value of their sum.
For our assumption,
$$
S \equiv 0+1+\ldots+2 n-1=\frac{(2 n-1) 2 n}{2}=(2 n-1) n \equiv n \quad(\bmod 2 n)
$$
But, if we sum, breaking all sums into its components, we derive
$$
S \equiv 2(1+\l... | proof | Number Theory | proof | Yes | Yes | olympiads | false |
C4 Every cell of table $4 \times 4$ is colored into white. It is permitted to place the cross (pictured below) on the table such that its center lies on the table (the whole figure does not need to lie on the table) and change colors of every cell which is covered into opposite (white and black). Find all $n$ such tha... | ## Solution
The cross covers at most five cells so we need at least 4 steps to change the color of every cell. If we place the cross 4 times such that its center lies in the cells marked below, we see that we can turn the whole square black in $n=4$ moves.
, a point $N$ on the ray $A B$ (after $B$ ) and a point $P$ on the ray $B C$ (after $C$ ) in a way such that $A M-B C=B N-A C=C P-A B$. Prove that the angles of triangle $M N P$ do not depend on the choice of $M, N, P$.
| ## Solution
Consider the points $M^{\prime}$ on the ray $B A$ (after $A$ ), $N^{\prime}$ on the ray $C B$ (after $B$ ) and $P^{\prime}$ on the ray $A C$ (after $C$ ), so that $A M=A M^{\prime}, B N=B N^{\prime}, C P=C P^{\prime}$. Since $A M-B C=B N-A C=B N^{\prime}-A C$, we get $C M=A C+A M=B C+B N^{\prime}=C N^{\pri... | proof | Geometry | proof | Yes | Yes | olympiads | false |
G3 The vertices $A$ and $B$ of an equilateral $\triangle A B C$ lie on a circle $k$ of radius 1 , and the vertex $C$ is inside $k$. The point $D \neq B$ lies on $k, A D=A B$ and the line $D C$ intersects $k$ for the second time in point $E$. Find the length of the segment $C E$.
| ## Solution
As $A D=A C, \triangle C D A$ is isosceles. If $\varangle A D C=\varangle A C D=\alpha$ and $\varangle B C E=\beta$, then $\beta=120^{\circ}-\alpha$. The quadrilateral $A B E D$ is cyclic, so $\varangle A B E=180^{\circ}-\alpha$. Then $\varangle C B E=$ $120^{\circ}-\alpha$ so $\varangle C B E=\beta$. Thus... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
G4 Let $A B C$ be a triangle, $(B C<A B)$. The line $\ell$ passing trough the vertices $C$ and orthogonal to the angle bisector $B E$ of $\angle B$, meets $B E$ and the median $B D$ of the side $A C$ at points $F$ and $G$, respectively. Prove that segment $D F$ bisect the segment $E G$.
| ## Solution
We will prove that desired covering is impossible.
Let assume the opposite i.e. a square with side length $a$, can be tiled with $k$ congruent right angled triangles, whose sides are of lengths $b, b \sqrt{3}$ and $2 b$.
Then the area of such a triangle is $\frac{b^{2} \sqrt{3}}{2}$.
And the area of the... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false |
G6 Let $A B C$ be a triangle with $A<90^{\circ}$. Outside of a triangle we consider isosceles triangles $A B E$ and $A C Z$ with bases $A B$ and $A C$, respectively. If the midpoint $D$ of the side $B C$ is such that $D E \perp D Z$ and $E Z=2 \cdot E D$, prove that $\widehat{A E B}=2 \cdot \widehat{A Z C}$.
| ## Solution
Since $D$ is the midpoint of the side $B C$, in the extension of the line segment $Z D$ we take a point $H$ such that $Z D=D H$. Then the quadrilateral $B H C Z$ is parallelogram and therefore we have
$$
B H=Z C=Z A
$$
=a b \sin A$.
But it is also true that $(A B C D)=4(A O D)=4 \cdot \frac{O A \cdot O D}{2} \sin \theta=2 O A \cdot O D \sin \theta=$ $=2 \cdot \frac{x}{2}... | proof | Geometry | proof | Yes | Yes | olympiads | false |
G9 Let $O$ be a point inside the parallelogram $A B C D$ such that
$$
\angle A O B+\angle C O D=\angle B O C+\angle C O D
$$
Prove that there exists a circle $k$ tangent to the circumscribed circles of the triangles $\triangle A O B, \triangle B O C, \triangle C O D$ and $\triangle D O A$.
 point on $\Gamma$. Denote by $\gamma$ the circle of diameter $A M$, by $X$ the (other than $M$ ) intersection... | ## Solution
Consider the line $\rho$ tangent to $\gamma$ at $A$, and take the points $\{K\}=A M \cap X Y,\{L\}=$ $\rho \cap X M$, and $\{F\}=O A \cap X Y$.
(Remark: Moving $M$ into its reflection with respect to the line $O A$ will move $X Y$ into its reflection with respect to $O A$. These old and the new $X Y$ meet... | proof | Geometry | proof | Yes | Yes | olympiads | false |
G11 Consider $A B C$ an acute-angled triangle with $A B \neq A C$. Denote by $M$ the midpoint of $B C$, by $D, E$ the feet of the altitudes from $B, C$ respectively and let $P$ be the intersection point of the lines $D E$ and $B C$. The perpendicular from $M$ to $A C$ meets the perpendicular from $C$ to $B C$ at point... |
Solution
Let $F$ be the foot of the altitude from $A$ and let $S$ be the intersection point of $A M$ and $R C$. As $P C$ is an altitude of the triangle $P R S$, the claim is equivalent to $R M \perp P S$, since the latter implies that $M$ is the orthocenter of $P R S$. Due to $R M \perp A C$, we need to prove that $A... | proof | Geometry | proof | Yes | Yes | olympiads | false |
NT1 Find all the positive integers $x$ and $y$ that satisfy the equation
$$
x(x-y)=8 y-7
$$
| ## Solution 1:
The given equation can be written as:
$$
\begin{aligned}
& x(x-y)=8 y-7 \\
& x^{2}+7=y(x+8)
\end{aligned}
$$
Let $x+8=m, m \in \mathbb{N}$. Then we have: $x^{2}+7 \equiv 0(\bmod m)$, and $x^{2}+8 x \equiv 0(\bmod m)$. So we obtain that $8 x-7 \equiv 0(\bmod m) \quad(1)$.
Also we obtain $8 x+8^{2}=8(x... | (x,y)=(63,56) | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
NT2 Let $n \geq 2$ be a fixed positive integer. An integer will be called " $n$-free" if it is not a multiple of an $n$-th power of a prime. Let $M$ be an infinite set of rational numbers, such that the product of every $n$ elements of $M$ is an $n$-free integer. Prove that $M$ contains only integers.
| ## Solution
We first prove that $M$ can contain only a finite number of non-integers. Suppose that there are infinitely many of them: $\frac{p_{1}}{q_{1}}, \frac{p_{2}}{q_{2}}, \ldots, \frac{p_{k}}{q_{k}}, \ldots$, with $\left(p_{k}, q_{k}\right)=1$ and $q_{k}>1$ for each $k$. Let $\frac{p}{q}=\frac{p_{1} p_{2} \ldots... | proof | Number Theory | proof | Yes | Yes | olympiads | false |
NT3 Let $s(a)$ denote the sum of digits of a given positive integer $a$. The sequence $a_{1}, a_{2}, \ldots a_{n}, \ldots$ of positive integers is such that $a_{n+1}=a_{n}+s\left(a_{n}\right)$ for each positive integer $n$. Find the greatest possible $n$ for which it is possible to have $a_{n}=2008$.
| ## Solution
Since $a_{n-1} \equiv s\left(a_{n-1}\right)$ (all congruences are modulo 9 ), we have $2 a_{n-1} \equiv a_{n} \equiv 2008 \equiv 10$, so $a_{n-1} \equiv 5$. But $a_{n-1}<2008$, so $s\left(a_{n-1}\right) \leq 28$ and thus $s\left(a_{n-1}\right)$ can equal 5,14 or 23 . We check $s(2008-5)=s(2003)=5, s(2008-1... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
NT4 Find all integers $n$ such that $n^{4}+8 n+11$ is a product of two or more consecutive integers.
|
Solution
We will prove that $n^{4}+8 n+11$ is never a multiple of 3 . This is clear if $n$ is a multiple of 3 . If
$n$ is not a multiple of 3 , then $n^{4}+8 n+11=\left(n^{4}-1\right)+12+8 n=(n-1)(n+1)\left(n^{2}+1\right)+12+8 n$, where $8 n$ is the only term not divisible by 3 . Thus $n^{4}+8 n+11$ is never the prod... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
NT5 Is it possible to arrange the numbers $1^{1}, 2^{2}, \ldots, 2008^{2008}$ one after the other, in such a way that the obtained number is a perfect square? (Explain your answer.)
| ## Solution
We will use the following lemmas.
Lemma 1. If $x \in \mathbb{N}$, then $x^{2} \equiv 0$ or $1(\bmod 3)$.
Proof: Let $x \in \mathbb{N}$, then $x=3 k, x=3 k+1$ or $x=3 k+2$, hence
$$
\begin{aligned}
& x^{2}=9 k^{2} \equiv 0(\bmod 3) \\
& x^{2}=9 k^{2}+6 k+1 \equiv 1(\bmod 3), \\
& x^{2}=9 k^{2}+12 k+4 \eq... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
NT6 Let $f: \mathbb{N} \rightarrow \mathbb{R}$ be a function, satisfying the following condition:
for every integer $n>1$, there exists a prime divisor $p$ of $n$ such that $f(n)=f\left(\frac{n}{p}\right)-f(p)$. If
$$
f\left(2^{2007}\right)+f\left(3^{2008}\right)+f\left(5^{2009}\right)=2006
$$
determine the value o... | ## Solution
If $n=p$ is prime number, we have
$$
f(p)=f\left(\frac{p}{p}\right)-f(p)=f(1)-f(p)
$$
i.e.
$$
f(p)=\frac{f(1)}{2}
$$
If $n=p q$, where $p$ and $q$ are prime numbers, then
$$
f(n)=f\left(\frac{n}{p}\right)-f(p)=f(q)-f(p)=\frac{f(1)}{2}-\frac{f(1)}{2}=0
$$
If $n$ is a product of three prime numbers, we... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
NT7 Determine the minimal prime number $p>3$ for which no natural number $n$ satisfies
$$
2^{n}+3^{n} \equiv 0(\bmod p)
$$
| ## Solution
We put $A(n)=2^{n}+3^{n}$. From Fermat's little theorem, we have $2^{p-1} \equiv 1(\bmod p)$ and $3^{p-1} \equiv 1(\bmod p)$ from which we conclude $A(n) \equiv 2(\bmod p)$. Therefore, after $p-1$ steps
at most, we will have repetition of the power. It means that in order to determine the minimal prime num... | 19 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
NT8 Let $a, b, c, d, e, f$ are nonzero digits such that the natural numbers $\overline{a b c}, \overline{d e f}$ and $\overline{a b c d e f}$ are squares.
a) Prove that $\overline{a b c d e f}$ can be represented in two different ways as a sum of three squares of natural numbers.
b) Give an example of such a number.... |
Solution
a) Let $\overline{a b c}=m^{2}, \overline{d e f}=n^{2}$ and $\overline{a b c d e f}=p^{2}$, where $11 \leq m \leq 31,11 \leq n \leq 31$ are natural numbers. So, $p^{2}=1000 \cdot m^{2}+n^{2}$. But $1000=30^{2}+10^{2}=18^{2}+26^{2}$. We obtain the following relations
$$
\begin{gathered}
p^{2}=\left(30^{2}+10... | 225625=475^{2}=450^{2}+150^{2}+25^{2}=270^{2}+390^{2}+25^{2} | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
NT9 Let $p$ be a prime number. Find all positive integers $a$ and $b$ such that:
$$
\frac{4 a+p}{b}+\frac{4 b+p}{a}
$$
and
$$
\frac{a^{2}}{b}+\frac{b^{2}}{a}
$$
are integers.
| ## Solution
Since $a$ and $b$ are symmetric we can assume that $a \leq b$. Let $d=(a, b), a=d u, b=d v$ and $(u, v)=1$. Then we have:
$$
\frac{a^{2}}{b}+\frac{b^{2}}{a}=\frac{d\left(u^{3}+v^{3}\right)}{u v}
$$
Since,
$$
\left(u^{3}+v^{3}, u\right)=\left(u^{3}+v^{3}, v\right)=1
$$
we deduce that $u \mid d$ and $v \... | (,b)={(1,1),(2,2),(p,p),(2p,2p),(5,25),(6,18),(18,6),(25,5),(30,150),(150,30)} | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
NT10 Prove that $2^{n}+3^{n}$ is not a perfect cube for any positive integer $n$.
| ## Solution
If $n=1$ then $2^{1}+3^{1}=5$ is not perfect cube.
Perfect cube gives residues $-1,0$ and 1 modulo 9 . If $2^{n}+3^{n}$ is a perfect cube, then $n$ must be divisible with 3 (congruence $2^{n}+3^{n}=x^{3}$ modulo 9 ).
If $n=3 k$ then $2^{3 k}+3^{2 k}>\left(3^{k}\right)^{3}$. Also, $\left(3^{k}+1\right)^{3... | proof | Number Theory | proof | Yes | Yes | olympiads | false |
NT11 Determine the greatest number with $n$ digits in the decimal representation which is divisible by 429 and has the sum of all digits less than or equal to 11 .
| ## Solution
Let $A=\overline{a_{n} a_{n-1} \ldots a_{1}}$ and notice that $429=3 \cdot 11 \cdot 13$.
Since the sum of the digits $\sum a_{i} \leq 11$ and $\sum a_{i}$ is divisible by 3 , we get $\sum a_{i}=3,6$ or 9. As 11 divides $A$, we have
$$
11 \mid a_{n}-a_{n-1}+a_{n-2}-a_{n-3}+\ldots
$$
in other words $11 \m... | 30030000\ldots | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
NT12 Solve the equation $\frac{p}{q}-\frac{4}{r+1}=1$ in prime numbers.
| ## Solution
We can rewrite the equation in the form
$$
\begin{gathered}
\frac{p r+p-4 q}{q(r+1)}=1 \Rightarrow p r+p-4 q=q r+q \\
p r-q r=5 q-p \Rightarrow r(p-q)=5 q-p
\end{gathered}
$$
It follows that $p \neq q$ and
$$
\begin{gathered}
r=\frac{5 q-p}{p-q}=\frac{4 q+q-p}{p-q} \\
r=\frac{4 q}{p-q}-1
\end{gathered}
... | (p,q,r)=(3,2,7), | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
ALG 1. A number $A$ is written with $2 n$ digits, each of whish is 4 , and a number $B$ is written with $n$ digits, each of which is 8 . Prove that for each $n, A+2 B+4$ is a total square.
| ## Solution.
$$
\begin{aligned}
A & =\underbrace{44 \ldots 44}_{2 n}=\underbrace{44 \ldots 4}_{n} \underbrace{44 \ldots 4}_{n}=\underbrace{44 \ldots 4}_{n} \underbrace{400 \ldots 0}_{n}-\underbrace{44 \ldots 4}_{n}+\underbrace{88 \ldots 8}_{n}=\underbrace{44 \ldots 4}_{n} \cdot\left(10^{n}-1\right)+B \\
& =4 \cdot \un... | proof | Number Theory | proof | Yes | Yes | olympiads | false |
ALG 2. Let $a, b, c$ be lengths of triangle sides, $p=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$ and $q=\frac{a}{c}+\frac{c}{b}+\frac{b}{a}$.
Prove that $|p-q|<1$.
|
Solution: One has
$$
\begin{aligned}
a b c|p-q| & =a b c\left|\frac{c-b}{a}+\frac{a-c}{b}+\frac{b-a}{c}\right| \\
& =\left|b c^{2}-b^{2} c+a^{2} c-a c^{2}+a b^{2}-a^{2} b\right|= \\
& =\left|a b c-a c^{2}-a^{2} b+a^{2} c-b^{2} c+b c^{2}+a b^{2}-a b c\right|= \\
& =\left|(b-c)\left(a c-a^{2}-b c+a b\right)\right|= \\
... | proof | Inequalities | proof | Yes | Yes | olympiads | false |
ALG 3: Let $a, b, c$ be real numbers such that $a^{2}+b^{2}+c^{2}=1$. Prove that. $P=a b+b c+c a-2(a+b+c) \geq-\frac{5}{2}$. Are there values of $a, b, c$, such that $P=-\frac{5}{2}$.
|
Solution: We have $a b+b c+c a=\frac{(a+b+c)^{2}-c^{2}-b^{2}-a^{2}}{2}=\frac{(a+b+c)^{2}-1}{2}$.
If put $t=a+b+c$ we obtain
$$
P=\frac{t^{2}-1}{2}-2 t=\frac{t^{2}-4 t-1}{2}=\frac{(t-2)^{2}-5}{2} \geq-\frac{5}{2}
$$
Obviously $P=-\frac{5}{2}$ when $t=2$, i.e. $a+b+c=2$, or $c=2-a-b$. Substitute in $a^{2}+b^{2}+c^{2}... | P\geq-\frac{5}{2} | Inequalities | proof | Yes | Yes | olympiads | false |
## ALG 4.
Let $a, b, c$ be rational numbers such that
$$
\frac{1}{a+b c}+\frac{1}{b+a c}=\frac{1}{a+b}
$$
Prove that $\sqrt{\frac{c-3}{c+1}}$ is also a rational number
|
Solution. By cancelling the denominators
$$
(a+b)^{2}(1+c)=a b+c\left(a^{2}+b^{2}\right)+a b c^{2}
$$
and
$$
a b(c-1)^{2}=(a+b)^{2}
$$
If $c=-1$, we obtrin the contradiction
$$
\frac{1}{a-b}+\frac{1}{b-a}=\frac{1}{a+b}
$$
Furtherrdore,
$$
\begin{aligned}
(c-3)(c+1) & =(c-1)^{2}-4=\frac{(a+b)^{2}}{a b}-4 \\
& =\... | proof | Algebra | proof | Yes | Yes | olympiads | false |
ALG 5. Let $A B C$ be a scalene triangle with $B C=a, A C=b$ and $A B=c$, where $a_{r} b, c$ are positive integers. Prove that
$$
\left|a b^{2}+b c^{2}+c a^{2}-a^{2} b-b^{2} c-c^{2} a\right| \geq 2
$$
|
Solution. Denote $E=a b^{2}+b c^{2}+c a^{2}-a^{2} b-b^{2} c-c^{2} a$. We have
$$
\begin{aligned}
E= & \left(a b c-c^{2} a\right)+\left(c a^{2}-a^{2} b\right)+\left(b c^{2}-b^{2} c\right)+\left(a b^{2}-a b c\right)= \\
& (b-c)\left(a c-a^{2}-b c+a b\right)=(b-c)\left(a a^{2}-b\right)(c-a)
\end{aligned}
$$
So, $|E|=|a... | 2 | Algebra | proof | Yes | Yes | olympiads | false |
## ALG 6.
Let $a, b, c$ be positive numbers such that $a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}=3$. Prove that
$$
a+b+c \geq a b c+2
$$
|
Solution. We can consider the case $a \geq b \geq c$ which implies $c \leq 1$. The given inequality writes
$$
a+b-2 \geq(a b-1) c \geq(a b-1) c^{2}=(a b-1) \frac{3-a^{2} b^{2}}{a^{2}+b^{2}}
$$
Put $x=\sqrt{a b}$. From the inequality $3 a^{2} b^{2} \geq a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}=3$ we infer $x \geq 1$ and f... | proof | Inequalities | proof | Yes | Incomplete | olympiads | false |
ALG 6'. Let $a, b, c$ be positive numbers such that $a b+b c+c a=3$. Prove that
$$
a+b+c \geq a b c+2
$$
|
Solution. Eliminating $c$ gives
$$
a+b+c-a b c=a+b+(1-a b) c=a+b+\frac{(1-a b)(3-a b)}{a+b}
$$
Put $x=\sqrt{a b}$. Then $a+b \geq 2 x$, and since $1<x^{2}<3, \frac{(1-a b)(3-a b)}{a+b} \geq \frac{\left(1-x^{2}\right)\left(3-x^{2}\right)}{2 x}$.
It then suffices to prove that
$$
2 x+\frac{\left(1-x^{2}\right)\left(... | proof | Inequalities | proof | Yes | Yes | olympiads | false |
ALG 7 .
Let $x, y, z$ be real numbers greater than -1 . Prove that
$$
\frac{1+x^{2}}{1+y+z^{2}}+\frac{1+y^{2}}{1+z+x^{2}}+\frac{1+z^{2}}{1+x+y^{2}} \geq 2
$$
|
Solution. We have $y \leq \frac{1+y^{2}}{2}$, hence $\quad$
$$
\frac{1+x^{2}}{1+y+z^{2}} \geq \frac{1+x^{2}}{1+z^{2}+\frac{1+\dot{y}^{2}}{2}}
$$
and the similar inequalities.
Setting $a=1+x^{2}, b=1+y^{2}, c=1+z^{2}$, it sufices to prove that
$$
\frac{a}{2 c+b}+\frac{b}{2 a+c}+\frac{c}{2 b+a} \geq 1
$$
for all $a... | proof | Inequalities | proof | Yes | Yes | olympiads | false |
ALG 8. Prove that there exist two sets $A=\{x, y, z\}$ and $B=\{m, n, p\}$ of positive integers greater than 2003 such that the sets have no common elements and the equalities $x+y+z=m+n+p$ and $x^{2}+y^{2}+z^{2}=m^{2}+n^{2}+p^{2}$ hold.
|
Solution. Let $A B C$ be a triangle with $B C=a, A C=b, A B=c$ and $ak+3=c
$$
a triangle with such length sides there exist. After the simple calculations we have
$$
\begin{gathered}
A=\left\{3(k+1)^{2}-2,3(k+2)^{2}+4,3(k+3)^{2}-2\right\} \\
B=\left\{3(k+1)^{2}, 3(k+2)^{2}, 3(k+3)^{2}\right\}
\end{gathered}
$$
It e... | proof | Number Theory | proof | Yes | Yes | olympiads | false |
COM 1. In a group of 60 students: 40 speak English; 30 speak French; 8 speak all the three languages; the number of students that speak English and French but not German is equal to "the sum of the number of students that speak English and German but not French plus the number of students that speak French and German ... |
Solution: We use the following notation.
$E=\#$ students that speak English, $F=\#$ students that speak French,
$G=\#$ students that speak German; $m=$ \# students that speak all the three languages,
$x=\#$ students that speak English and French but not German,
$y=\#$ students that speak German and French but not ... | 26 | Combinatorics | math-word-problem | Yes | Incomplete | olympiads | false |
COM 2 Natural numbers 1,2,3, .., 2003 are written in an arbitrary sequence $a_{1}, a_{2}, a_{3}, \ldots a_{2003}$. Let $b_{1}=1 a_{1}, b_{2}=2 a_{2}, b_{3}=3 a_{3}, \ldots, b_{2003}=2003 a_{2003}$, and $B$ be the maximum of the numbers $b_{1}, b_{2}, b_{3}, \ldots, b_{2003}$.
a) If $a_{1}=2003, a_{2}=2002, a_{3}=2001... |
Solution: a) Using the inequality between the arithmetical and geometrical mean, we obtain that $b_{n}=n(2004-n) \leq\left(\frac{n+(2004-n)}{2}\right)^{2}=1002^{2}$ for $n=1,2,3, \ldots, 2003$. The equality holds if and only if $n=2004-n$, i.e. $n=1002$. Therefore, $B=b_{1002}=1002 \times(2004-1002)=1002^{2}$. b) Let ... | 1002^2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
COM 3. Prove that amongst any 29 natural numbers there are 15 such that sum of them is divisible by 15 .
|
Solution: Amongst any 5 natural numbers there are 3 such that sum of them is divisible by 3 . Amongst any 29 natural numbers we can choose 9 groups with 3 numbers such that sum of numbers in every group is divisible by 3. In that way we get 9 natural numbers such that all of them are divisiblc by 3. It is easy to see ... | proof | Number Theory | proof | Yes | Yes | olympiads | false |
COM 5. If $m$ is a number from the set $\{1,2,3,4\}$ and each point of the plane is painted in red or blue, prove that in the plane there exists at least an equilateral triangle with the vertices of the same colour and with length side $m$.
|
Solution. Suppose that in the plane there no exists an equilateral triangle with the vertices of the same colour and length side $m=1,2,3,4$.
First assertion: we shall prove that in the plane there no exists a segment with the length 2 such that the ends and the midpint of this segment have the same colour. Suppose t... | proof | Combinatorics | proof | Yes | Yes | olympiads | false |
GEO 1. Is there a convex quadrilateral, whose diagonals divide it into four triangles, such that their areas are four distinct prime integers.
|
Solution. No. Let the areas of those triangles be the prime numbers $p, q, r$ and $t$. But for the areas of the triangles we have $\mathrm{pq}=\mathrm{rt}$, where the triangles with areas $\mathrm{p}$ and $\mathrm{q}$ have only a common vertex. This is not possible for distinct primes.
| proof | Geometry | math-word-problem | Yes | Yes | olympiads | false |
GEO 2. Is there a triangle whose area is $12 \mathrm{~cm}^{2}$ and whose perimeter is $12 \mathrm{~cm}$.
|
Solution. No. Let $\mathrm{r}$ be the radius of the inscribed circle. Then $12=6 \mathrm{r}$, i.e. $\mathrm{r}=2 \mathrm{~cm}$. But the area of the inscribed circle is $4 \pi>12$, and it is known that the area of any triangle is bigger than the area of its inscribed circle.
| proof | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## GEO 3.
Let $G$ be the centroid of the the triangle $A B C$. Reflect point $A$ across $C$ at $A^{\prime}$. Prove that $G, B, C, A^{\prime}$ are on the same circle if and only if $G A$ is perpendicular to $G C$.
|
Solution. Observe first that $G A \perp G C$ if and only if $5 A C^{2}=A B^{2}+B C^{2}$. Indeed,
$$
G A \perp G C \Leftrightarrow \frac{4}{9} m_{a}^{2}+\frac{4}{9} m_{c}^{2}=b^{2} \Leftrightarrow 5 b^{2}=a^{2}+c^{2}
$$
Moreover,
$$
G B^{2}=\frac{4}{9} m_{b}^{2}=\frac{2 a^{2}+2 c^{2}-b^{2}}{9}=\frac{9 b^{2}}{9}=b^{2... | proof | Geometry | proof | Yes | Yes | olympiads | false |
GEO 5. Let three congruent circles intersect in one point $M$ and $A_{1}, A_{2}$ and $A_{3}$ be the other intersection points for those circles. Prove that $M$ is a.orthocenter for a triangle $A_{1} A_{2} A_{3}$.
|
Solution: The quadrilaterals $\mathrm{O}_{3} M O_{2} A_{1}, \mathrm{O}_{3} M O_{1} A_{2}$ and $O_{1} M O_{2} A_{3}$ are rombes. Therefore, $O_{2} A_{1} \| M O_{3}$ and $M O_{3} \| O_{1} A_{2}$, which imply $O_{2} A_{1} \| O_{1} A_{2}$. Because $O_{2} A_{1}=O_{3}{ }^{*} M=O_{1} A_{2}$ the quadrilateral $O_{2} A_{1} A_{... | proof | Geometry | proof | Yes | Yes | olympiads | false |
GEO 7. Through a interior point of a triangle, three lines parallel to the sides of the triangle are constructed. In that way the triangle is divided on six figures, areas equal $a, b, c, \alpha, \beta, \gamma$ (see the picture).
=n$ or $\sigma(n)=n$. Also let us denote by $b_{n}$ the diffe... | Combinatorics | proof | More than one problem | NaN | olympiads | false | |
87.1. Nine journalists from different countries attend a press conference. None of these speaks more than three languages, and each pair of the journalists share a common language. Show that there are at least five journalists sharing a common language.
|
Solution. Assume the journalists are $J_{1}, J_{2}, \ldots, J_{9}$. Assume that no five of them have a common language. Assume the languages $J_{1}$ speaks are $L_{1}, L_{2}$, and $L_{3}$. Group $J_{2}, J_{3}$, $\ldots, J_{9}$ according to the language they speak with $J_{1}$. No group can have more than three members... | proof | Combinatorics | proof | Yes | Yes | olympiads | false |
87.2. Let $A B C D$ be a parallelogram in the plane. We draw two circles of radius $R$, one through the points $A$ and $B$, the other through $B$ and $C$. Let $E$ be the other point of
intersection of the circles. We assume that $E$ is not a vertex of the parallelogram. Show that the circle passing through $A, D$, and... |
Solution. (See Figure 1.) Let $F$ and $G$ be the centers of the two circles of radius $R$ passing through $A$ and $B$; and $B$ and $C$, respectively. Let $O$ be the point for which the the rectangle $A B G O$ is a parallelogram. Then $\angle O A D=\angle G B C$, and the triangles $O A D$ and $G B C$ are congruent (sas... | proof | Geometry | proof | Yes | Yes | olympiads | false |
87.3. Let $f$ be a strictly increasing function defined in the set of natural numbers satisfying the conditions $f(2)=a>2$ and $f(m n)=f(m) f(n)$ for all natural numbers $m$ and $n$. Determine the smallest possible value of $a$.
|
Solution. Since $f(n)=n^{2}$ is a function satisfying the conditions of the problem, the smallest posiible $a$ is at most 4. Assume $a=3$. It is easy to prove by induction that $f\left(n^{k}\right)=f(n)^{k}$ for all $k \geq 1$. So, taking into account that $f$ is strictly increasing, we get
$$
\begin{gathered}
f(3)^{... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
87.4. Let $a, b$, and $c$ be positive real numbers. Prove:
$$
\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \leq \frac{a^{2}}{b^{2}}+\frac{b^{2}}{c^{2}}+\frac{c^{2}}{a^{2}}
$$
|
Solution. The arithmetic-geometric inequality yields
$$
3=3 \sqrt[3]{\frac{a^{2}}{b^{2}} \cdot \frac{b^{2}}{c^{2}} \cdot \frac{c^{2}}{a^{2}}} \leq \frac{a^{2}}{b^{2}}+\frac{b^{2}}{c^{2}}+\frac{c^{2}}{a^{2}}
$$
or
$$
\sqrt{3} \leq \sqrt{\frac{a^{2}}{b^{2}}+\frac{b^{2}}{c^{2}}+\frac{c^{2}}{a^{2}}}
$$
On the other ha... | proof | Inequalities | proof | Yes | Yes | olympiads | false |
88.1. The positive integer $n$ has the following property: if the three last digits of $n$ are removed, the number $\sqrt[3]{n}$ remains. Find $n$.
|
Solution. If $x=\sqrt[3]{n}$, and $y, 0 \leq y1000$, and $x>31$. On the other hand, $x^{3}<1000 x+1000$, or $x\left(x^{2}-1000\right)<1000$. The left hand side of this inequality is an increasing function of $x$, and $x=33$ does not satisfy the inequality. So $x<33$. Since $x$ is an integer, $x=32$ and $n=32^{3}=32768... | 32768 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
88.2. Let $a, b$, and $c$ be non-zero real numbers and let $a \geq b \geq c$. Prove the inequality
$$
\frac{a^{3}-c^{3}}{3} \geq a b c\left(\frac{a-b}{c}+\frac{b-c}{a}\right)
$$
When does equality hold?
|
Solution. Since $c-b \leq 0 \leq a-b$, we have $(a-b)^{3} \geq(c-b)^{3}$, or
$$
a^{3}-3 a^{2} b+3 a b^{2}-b^{3} \geq c^{3}-3 b c^{2}+3 b^{2} c-b^{3}
$$
On simplifying this, we immediately have
$$
\frac{1}{3}\left(a^{3}-c^{3}\right) \geq a^{2} b-a b^{2}+b^{2} c-b c^{2}=a b c\left(\frac{a-b}{c}+\frac{b-c}{a}\right)
$... | proof | Inequalities | proof | Yes | Yes | olympiads | false |
88.3. Two concentric spheres have radii $r$ and $R, r<R$. We try to select points $A, B$ and $C$ on the surface of the larger sphere such that all sides of the triangle $A B C$ would be tangent to the surface of the smaller sphere. Show that the points can be selected if and only if $R \leq 2 r$.
|
Solution. Assume $A, B$, and $C$ lie on the surface $\Gamma$ of a sphere of radius $R$ and center $O$, and $A B, B C$, and $C A$ touch the surface $\gamma$ of a sphere of radius $r$ and center $O$. The circumscribed and inscribed circles of $A B C$ then are intersections of the plane $A B C$ with $\Gamma$ and $\gamma$... | R\leq2r | Geometry | proof | Yes | Yes | olympiads | false |
88.4. Let $m_{n}$ be the smallest value of the function
$$
f_{n}(x)=\sum_{k=0}^{2 n} x^{k}
$$
Show that $m_{n} \rightarrow \frac{1}{2}$, as $n \rightarrow \infty$.
|
Solution. For $n>1$,
$$
\begin{gathered}
f_{n}(x)=1+x+x^{2}+\cdots \\
=1+x\left(1+x^{2}+x^{4}+\cdots\right)+x^{2}\left(1+x^{2}+x^{4} \cdots\right) \\
=1+x(1+x) \sum_{k=0}^{n-1} x^{2 k}
\end{gathered}
$$
From this we see that $f_{n}(x) \geq 1$, for $x \leq-1$ and $x \geq 0$. Consequently, $f_{n}$ attains its minimum ... | \frac{1}{2} | Algebra | proof | Yes | Yes | olympiads | false |
89.2. Three sides of a tetrahedron are right-angled triangles having the right angle at their common vertex. The areas of these sides are $A, B$, and $C$. Find the total surface area of the tetrahedron.
|
Solution 1. Let $P Q R S$ be the tetrahedron of the problem and let $S$ be the vertex common to the three sides which are right-angled triangles. Let the areas of $P Q S, Q R S$, and $R P S$ be $A, B$, and $C$, respectively. Denote the area of $Q R S$ by $X$. If $S S^{\prime}$ is the altitude from $S$ (onto $P Q R$ ) ... | A+B+C+\sqrt{A^{2}+B^{2}+C^{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
89.3. Let $S$ be the set of all points $t$ in the closed interval $[-1,1]$ such that for the sequence $x_{0}, x_{1}, x_{2}, \ldots$ defined by the equations $x_{0}=t, x_{n+1}=2 x_{n}^{2}-1$, there exists a positive integer $N$ such that $x_{n}=1$ for all $n \geq N$. Show that the set $S$ has infinitely many elements.
|
Solution. All numbers in the sequence $\left\{x_{n}\right\}$ lie in the interval $[-1,1]$. For each $n$ we can pick an $\alpha_{n}$ such that $x_{n}=\cos \alpha_{n}$. If $x_{n}=\cos \alpha_{n}$, then $x_{n+1}=2 \cos ^{2} \alpha_{n}-1=\cos \left(2 \alpha_{n}\right)$. The nuber $\alpha_{n+1}$ can be chosen as $2 \alpha_... | notfound | Algebra | proof | Yes | Yes | olympiads | false |
90.1. Let $m, n$, and $p$ be odd positive integers. Prove that the number
$$
\sum_{k=1}^{(n-1)^{p}} k^{m}
$$
is divisible by $n$.
|
Solution. Since $n$ is odd, the sum has an even number of terms. So we can write it as
$$
\sum_{k=1}^{\frac{1}{2}(n-1)^{p}}\left(k^{m}+\left((n-1)^{p}-k+1\right)^{m}\right)
$$
Because $m$ is odd, each term in the sum has $k+(n-1)^{p}-k+1=(n-1)^{p}+1$ as a factor. As $p$ is odd, too, $(n-1)^{p}+1=(n-1)^{p}+1^{p}$ has... | proof | Number Theory | proof | Yes | Yes | olympiads | false |
90.2. Let $a_{1}, a_{2}, \ldots, a_{n}$ be real numbers. Prove
$$
\sqrt[3]{a_{1}^{3}+a_{2}^{3}+\ldots+a_{n}^{3}} \leq \sqrt{a_{1}^{2}+a_{2}^{2}+\ldots+a_{n}^{2}}
$$
When does equality hold in (1)?
|
Solution. If $0 \leq x \leq 1$, then $x^{3 / 2} \leq x$, and equality holds if and only if $x=0$ or $x=1$. - The inequality is true as an equality, if all the $a_{k}$ 's are zeroes. Assume that at least one of the numbers $a_{k}$ is non-zero. Set
$$
x_{k}=\frac{a_{k}^{2}}{\sum_{j=1}^{n} a_{j}^{2}}
$$
Then $0 \leq x_... | proof | Inequalities | proof | Yes | Yes | olympiads | false |
90.3. Let $A B C$ be a triangle and let $P$ be an interior point of $A B C$. We assume that a line $l$, which passes through $P$, but not through $A$, intersects $A B$ and $A C$ (or their extensions over $B$ or $C$ ) at $Q$ and $R$, respectively. Find $l$ such that the perimeter of the triangle $A Q R$ is as small as ... |
Solution. (See Figure 2.) Let
$$
s=\frac{1}{2}(A R+R Q+Q A)
$$
Let $\mathcal{C}$ be the excircle of $A Q R$ tangent to $Q R$, i.e. the circle tangent to $Q R$ and the extensions of $A R$ and $A Q$. Denote the center of $\mathcal{C}$ by $I$ and the measure of $\angle Q A R$ by $\alpha . I$ is on the bisector of $\ang... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false |
90.4. It is possible to perform three operations $f, g$, and $h$ for positive integers: $f(n)=$ $10 n, g(n)=10 n+4$, and $h(2 n)=n$; in other words, one may write 0 or 4 in the end of the number and one may divide an even number by 2. Prove: every positive integer can be constructed starting from 4 and performing a fi... |
Solution. All odd numbers $n$ are of the form $h(2 n)$. All we need is to show that every even number can be obtained fron 4 by using the operations $f, g$, and $h$. To this end, we show that a suitably chosen sequence of inverse operations $F=f^{-1}, G=g^{-1}$, and $H=h^{-1}$ produces a smaller even number or the num... | proof | Number Theory | proof | Yes | Yes | olympiads | false |
91.1. Determine the last two digits of the number
$$
2^{5}+2^{5^{2}}+2^{5^{3}}+\cdots+2^{5^{1991}}
$$
written in decimal notation.
|
Solution. We first show that all numbers $2^{5^{k}}$ are of the form $100 p+32$. This can be shown by induction. The case $k=1$ is clear $\left(2^{5}=32\right)$. Assume $2^{5^{k}}=100 p+32$. Then, by the binomial formula,
$$
2^{5^{k+1}}=\left(2^{5^{k}}\right)^{5}=(100 p+32)^{5}=100 q+32^{5}
$$
and
$$
\begin{gathere... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
91.2. In the trapezium $A B C D$ the sides $A B$ and $C D$ are parallel, and $E$ is a fixed point on the side $A B$. Determine the point $F$ on the side $C D$ so that the area of the intersection of the triangles $A B F$ and $C D E$ is as large as possible.
 We assume $C D|A P E|+|E B S|+|B S V| \\
=|A P E|+|E B S|+|A U R|>|A P E|+|E B S|+|A P R| \\
=|A R E|+|E B S|
\end{gathered}
$$
A similar inequality can be established, when $F$ is between $G$ and $D$. So the choice $F=G$ minimizes $|A E P|+|E B Q|$ and maximizes $|E Q F P|$. - Proofs in t... | proof | Geometry | math-word-problem | Yes | Incomplete | olympiads | false |
91.3. Show that
$$
\frac{1}{2^{2}}+\frac{1}{3^{2}}+\ldots+\frac{1}{n^{2}}<\frac{2}{3}
$$
for all $n \geq 2$.
|
Solution. Since
$$
\frac{1}{j^{2}}<\frac{1}{j(j-1)}=\frac{1}{j-1}-\frac{1}{j}
$$
we have
$$
\begin{gathered}
\sum_{j=k}^{n} \frac{1}{j^{2}}<\left(\frac{1}{k-1}-\frac{1}{k}\right)+\left(\frac{1}{k}-\frac{1}{k+1}\right)+\cdots+\left(\frac{1}{n-1}-\frac{1}{n}\right) \\
=\frac{1}{k-1}-\frac{1}{n}<\frac{1}{k-1}
\end{gat... | \frac{2389}{3600}<\frac{2}{3} | Inequalities | proof | Yes | Yes | olympiads | false |
91.4. Let $f(x)$ be a polynomial with integer coefficients. We assume that there exists a positive integer $k$ and $k$ consecutive integers $n, n+1, \ldots, n+k-1$ so that none of the numbers $f(n), f(n+1), \ldots, f(n+k-1)$ is divisible by $k$. Show that the zeroes of $f(x)$ are not integers.
|
Solution. Let $f(x)=a_{0} x^{d}+a_{1} x^{d-1}+\cdots+a_{d}$. Assume that $f$ has a zero $m$ which is an integer. Then $f(x)=(x-m) g(x)$, where $g$ is a polynomial. If $g(x)=b_{0} x^{d-1}+b_{1} x^{d-2}+$ $\cdots+b_{d-1}$, then $a_{0}=b_{0}$, and $a_{k}=b_{k}-m b_{k-1}, 1 \leq k \leq d-1$. So $b_{0}$ is an integer, and ... | proof | Number Theory | proof | Yes | Yes | olympiads | false |
92.1. Determine all real numbers $x>1, y>1$, and $z>1$, satisfying the equation
$$
\begin{aligned}
x+y+z+\frac{3}{x-1} & +\frac{3}{y-1}+\frac{3}{z-1} \\
& =2(\sqrt{x+2}+\sqrt{y+2}+\sqrt{z+2})
\end{aligned}
$$
|
Solution. Consider the function $f$,
$$
f(t)=t+\frac{3}{t-1}-2 \sqrt{t+2}
$$
defined for $t>1$. The equation of the problem can be written as
$$
f(x)+f(y)+f(z)=0
$$
We reformulate the formula for $f$ :
$$
\begin{aligned}
f(t) & =\frac{1}{t-1}\left(t^{2}-t+3-2(t-1) \sqrt{t+2}\right) \\
& =\frac{1}{t-1}\left(t^{2}-... | \frac{3+\sqrt{13}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
92.2. Let $n>1$ be an integer and let $a_{1}, a_{2}, \ldots, a_{n}$ be $n$ different integers. Show that the polynomial
$$
f(x)=\left(x-a_{1}\right)\left(x-a_{2}\right) \cdots\left(x-a_{n}\right)-1
$$
is not divisible by any polynomial with integer coefficients and of degree greater than zero but less than $n$ and s... |
Solution. Suppose $g(x)$ is a polynomial of degree $m$, where $1 \leq m<n$, with integer coefficients and leading coefficient 1 , such that
$$
f(x)=g(x) h(x)
$$
whre $h(x)$ is a polynomial. Let
$$
\begin{aligned}
& g(x)=x^{m}+b_{m-1} x^{m-1}+\cdots+b_{1} x+b_{0} \\
& h(x)=x^{n-m}+c_{n-m-1} x^{n-m-1}+\cdots+c_{1} x+... | proof | Algebra | proof | Yes | Yes | olympiads | false |
92.4. Peter has many squares of equal side. Some of the squares are black, some are white. Peter wants to assemble a big square, with side equal to $n$ sides of the small squares, so that the big square has no rectangle formed by the small squares such that all the squares in the vertices of the rectangle are of equal... |
Solution. We show that Peter only can make a $4 \times 4$ square. The construction is possible, if $n=4$ :
Now consider the case $n=5$. We may assume that at least 13 of the 25 squares are ... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
93.1. Let $F$ be an increasing real function defined for all $x, 0 \leq x \leq 1$, satisfying the conditions
$$
\begin{aligned}
& F\left(\frac{x}{3}\right)=\frac{F(x)}{2} \\
& F(1-x)=1-F(x)
\end{aligned}
$$
Determine $F\left(\frac{173}{1993}\right)$ and $F\left(\frac{1}{13}\right)$.
|
Solution. Condition (i) implies $F(0)=\frac{1}{2} F(0)$, so $F(0)=0$. Because of condition (ii), $F(1)=1-F(0)=1$. Also $F\left(\frac{1}{3}\right)=\frac{1}{2}$ and $F\left(\frac{2}{3}\right)=1-F\left(\frac{1}{3}\right)=\frac{1}{2}$. Since $F$ is an increasing function, this is possible only if $F(x)=\frac{1}{2}$ for al... | \frac{3}{16},\frac{1}{7} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
93.2. A hexagon is inscribed in a circle of radius $r$. Two of the sides of the hexagon have length 1, two have length 2 and two have length 3. Show that $r$ satisfies the equation
$$
2 r^{3}-7 r-3=0
$$
 We join the vertices of the hexagon to the center $O$ of its circumcircle. We denote by $\alpha$ the central angles corresponding the chords of length 1 , by $\beta$ those corresponding the chords of length 2 , and by $\gamma$ those corresponding the chords of length 3. Clearly $\alpha+\beta+... | 2r^{3}-7r-3=0 | Geometry | proof | Yes | Yes | olympiads | false |
93.3. Find all solutions of the system of equations
$$
\left\{\begin{aligned}
s(x)+s(y) & =x \\
x+y+s(z) & =z \\
s(x)+s(y)+s(z) & =y-4
\end{aligned}\right.
$$
where $x, y$, and $z$ are positive integers, and $s(x), s(y)$, and $s(z)$ are the numbers of digits in the decimal representations of $x, y$, and $z$, respect... |
Solution. The first equation implies $x \geq 2$ and the first and third equation together imply
$$
s(z)=y-x-4
$$
So $y \geq x+5 \geq 7$. From (1) and the second equation we obtain $z=2 y-4$. Translated to the values of $s$, these equation imply $s(x) \leq s(2 y) \leq s(y)+1$ and $s(x) \leq s(y)$. We insert these ine... | (2,8,12) | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
93.4. Denote by $T(n)$ the sum of the digits of the decimal representation of a positive integer $n$.
a) Find an integer $N$, for which $T(k \cdot N)$ is even for all $k, 1 \leq k \leq 1992$, but $T(1993 \cdot N)$ is odd.
b) Show that no positive integer $N$ exists such that $T(k \cdot N)$ is even for all positive in... |
Solution. a) If $s$ has $n$ decimal digits and $m=10^{n+r} s+s$, then $T(k m)$ is even at least as long as $k s<10^{n+r}$, because all non-zero digits appear in pairs in $k m$. Choose $N=5018300050183$ or $s=50183, n=5, r=3$. Now $1992 \cdot s=99964536<10^{8}$, so $T(k N)$ is even for all $k \leq 1992$. But $1993 \cdo... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
94.1. Let $O$ be an interior point in the equilateral triangle $A B C$, of side length $a$. The lines $A O, B O$, and $C O$ intersect the sides of the triangle in the points $A_{1}, B_{1}$, and $C_{1}$. Show that
$$
\left|O A_{1}\right|+\left|O B_{1}\right|+\left|O C_{1}\right|<a
$$
|
Solution. Let $H_{A}, H_{B}$, and $H_{C}$ be the orthogonal projections of $O$ on $B C, C A$, and $A B$, respectively. Because $60^{\circ}\left|O A_{1}\right| \frac{\sqrt{3}}{2}
$$
In the same way,
$$
\left|O H_{B}\right|>\left|O B_{1}\right| \frac{\sqrt{3}}{2} \quad \text { and } \quad\left|O H_{C}\right|>\left|O C... | proof | Geometry | proof | Yes | Yes | olympiads | false |
94.2. We call a finite plane set $S$ consisting of points with integer coefficients a twoneighbour set, if for each point $(p, q)$ of $S$ exactly two of the points $(p+1, q),(p, q+1)$, $(p-1, q),(p, q-1)$ belong to $S$. For which integers $n$ there exists a two-neighbour set which contains exactly $n$ points?
|
Solution. The points $(0,0),(1,0),(1,1),(0,1)$ clearly form a two-neighbour set (which we abbreviate as $2 \mathrm{NS})$. For every even number $n=2 k \geq 8$, the set $S=\{(0,0), \ldots$, $(k-2,0),(k-2,1),(k-2,2), \ldots,(0,2),(0,1)\}$ is a $2 \mathrm{NS}$. We show that there is no $2 \mathrm{NS}$ with $n$ elements f... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
94.3. A piece of paper is the square $A B C D$. We fold it by placing the vertex $D$ on the point $H$ of the side $B C$. We assume that $A D$ moves onto the segment $G H$ and that $H G$ intersects $A B$ at $E$. Prove that the perimeter of the triangle $E B H$ is one half of the perimeter of the square.
 The fold gives rise to an isosceles trapezium $A D H G$. Because of symmetry, the distance of the vertex $D$ from the side $G H$ equals the distance of the vertex
$H$ from side $A D$; the latter distance is the side length $a$ of the square. The line $G H$ thus is tangent to the circle with c... | proof | Geometry | proof | Yes | Yes | olympiads | false |
94.4. Determine all positive integers $n<200$, such that $n^{2}+(n+1)^{2}$ is the square of an integer.
|
Solution. We determine the integral solutions of
$$
n^{2}+(n+1)^{2}=(n+p)^{2}, \quad p \geq 2
$$
The root formula for quadratic equations yields
$$
n=p-1+\sqrt{2 p(p-1)} \geq 2(p-1)
$$
Because $n<200$, we have $p \leq 100$. Moreover, the number $2 p(p-1)$ has to be the square of an integer. If $p$ is odd, $p$ and ... | 20,3,119 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
95.1. Let $A B$ be a diameter of a circle with centre $O$. We choose a point $C$ on the circumference of the circle such that $O C$ and $A B$ are perpendicular to each other. Let $P$ be an arbitrary point on the (smaller) arc $B C$ and let the lines $C P$ and $A B$ meet at $Q$. We choose $R$ on $A P$ so that $R Q$ and... |
Solution 1. (See Figure 7.) Draw $P B$. By the Theorem of Thales, $\angle R P B=\angle A P B=$ $90^{\circ}$. So $P$ and $Q$ both lie on the circle with diameter $R B$. Because $\angle A O C=90^{\circ}$, $\angle R P Q=\angle C P A=45^{\circ}$. Then $\angle R B Q=45^{\circ}$, too, and $R B Q$ is an isosceles right trian... | proof | Geometry | proof | Yes | Yes | olympiads | false |
95.2. Messages are coded using sequences consisting of zeroes and ones only. Only sequences with at most two consecutive ones or zeroes are allowed. (For instance the sequence 011001 is allowed, but 011101 is not.) Determine the number of sequences consisting of exactly 12 numbers.
|
Solution 1. Let $S_{n}$ be the set of acceptable sequences consisting of $2 n$ digits. We partition $S_{n}$ in subsets $A_{n}, B_{n}, C_{n}$, and $D_{n}$, on the basis of the two last digits of the sequence. Sequences ending in 00 are in $A_{n}$, those ending in 01 are in $B_{n}$, those ending in 10 are in $C_{n}$, an... | 466 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
95.3. Let $n \geq 2$ and let $x_{1}, x_{2}, \ldots x_{n}$ be real numbers satisfying $x_{1}+x_{2}+\ldots+x_{n} \geq 0$ and $x_{1}^{2}+x_{2}^{2}+\ldots+x_{n}^{2}=1$. Let $M=\max \left\{x_{1}, x_{2}, \ldots, x_{n}\right\}$. Show that
$$
M \geq \frac{1}{\sqrt{n(n-1)}}
$$
When does equality hold in (1)?
|
Solution. Denote by $I$ the set of indices $i$ for which $x_{i} \geq 0$, and by $J$ the set of indices $j$ for which $x_{j}<0$. Let us assume $M<\frac{1}{\sqrt{n(n-1)}}$. Then $I \neq\{1,2, \ldots, n\}$, since otherwise we would have $\left|x_{i}\right|=x_{i} \leq \frac{1}{\sqrt{n(n-1)}}$ for every $i$, and $\sum_{i=1... | proof | Inequalities | proof | Yes | Yes | olympiads | false |
95.4. Show that there exist infinitely many mutually non-congruent triangles $T$, satisfying
(i) The side lengths of $T$ are consecutive integers.
(ii) The area of $T$ is an integer.
|
Solution. Let $n \geq 3$, and let $n-1, n, n+1$ be the side lengths of the triangle. The semiperimeter of the triangle then equals on $\frac{3 n}{2}$. By Heron's formula, the area of the triangle is
$$
\begin{gathered}
T=\sqrt{\frac{3 n}{2} \cdot\left(\frac{3 n}{2}-n+1\right)\left(\frac{3 n}{2}-n\right)\left(\frac{3 ... | proof | Geometry | proof | Yes | Yes | olympiads | false |
96.1. Show that there exists an integer divisible by 1996 such that the sum of the its decimal digits is 1996 .
|
Solution. The sum of the digits of 1996 is 25 and the sum of the digits of $2 \cdot 1996=3992$ is 23 . Because $1996=78 \cdot 25+46$, the number obtained by writing 781996 's and two 3992 in succession satisfies the condition of the problem. - As $3 \cdot 1996=5998$, the sum of the digits of 5988 is 30 , and $1996=65 ... | proof | Number Theory | proof | Yes | Yes | olympiads | false |
96.2. Determine all real numbers $x$, such that
$$
x^{n}+x^{-n}
$$
is an integer for all integers $n$.
|
Solution. Set $f_{n}(x)=x^{n}+x^{-n}$. $f_{n}(0)$ is not defined for any $n$, so we must have $x \neq 0$. Since $f_{0}(x)=2$ for all $x \neq 0$, we have to find out those $x \neq 0$ for which $f_{n}(x)$ is an integer foe every $n>0$. We note that
$$
x^{n}+x^{-n}=\left(x+x^{-1}\right)\left(x^{n-1}+x^{1-n}\right)-\left... | \frac{}{2}\ | Algebra | math-word-problem | Yes | Yes | olympiads | false |
96.3. The circle whose diameter is the altitude dropped from the vertex $A$ of the triangle $A B C$ intersects the sides $A B$ and $A C$ at $D$ and $E$, respectively $(A \neq D, A \neq E)$. Show that the circumcentre of $A B C$ lies on the altitude dropped from the vertex $A$ of the triangle $A D E$, or on its extensi... |
Solution. (See Figure 8.) Let $A F$ be the altitude of $A B C$. We may assume that $\angle A C B$ is sharp. From the right triangles $A C F$ and $A F E$ we obtain $\angle A F E=\angle A C F . \angle A D E$ and $\angle A F E$ subtend the same arc, so they are equal. Thus $\angle A C B=\angle A D E$, and the triangles $... | proof | Geometry | proof | Yes | Yes | olympiads | false |
96.4. The real-valued function $f$ is defined for positive integers, and the positive integer a satisfies
$$
\begin{gathered}
f(a)=f(1995), \quad f(a+1)=f(1996), \quad f(a+2)=f(1997) \\
f(n+a)=\frac{f(n)-1}{f(n)+1} \quad \text { for all positive integers } n
\end{gathered}
$$
(i) Show that $f(n+4 a)=f(n)$ for all po... |
Solution. To prove (i), we the formula $f(n+a)=\frac{f(n)-1}{f(n)+1}$ repeatedly:
$$
\begin{gathered}
f(n+2 a)=f((n+a)+a)=\frac{\frac{f(n)-1}{f(n)+1}-1}{\frac{f(n)-1}{f(n)+1}+1}=-\frac{1}{f(n)} \\
f(n+4 a)=f((n+2 a)+2 a)=-\frac{1}{-\frac{1}{f(n)}}=f(n)
\end{gathered}
$$
(ii) If $a=1$, then $f(1)=f(a)=f(1995)=f(3+498... | 3 | Algebra | proof | Yes | Yes | olympiads | false |
97.1. Let A be a set of seven positive numbers. Determine the maximal number of triples $(x, y, z)$ of elements of A satisfying $x<y$ and $x+y=z$.
|
Solution. Let $0<a_{1}<a_{2}<\ldots<a_{7}$ be the elements of the set $A$. If $\left(a_{i}, a_{j}, a_{k}\right)$ is a triple of the kind required in the problem, then $a_{i}<a_{j}<a_{i}+a_{j}=a_{k}$. There are at most $k-1$ pairs $\left(a_{i}, a_{j}\right)$ such that $a_{i}+a_{j}=a_{k}$. The number of pairs satisfying... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
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