problem stringlengths 48 1.23k | reasoning stringlengths 3.67k 77k | deepseek_solution stringlengths 284 3.9k | ground_truth_solution stringlengths 885 5.59k | difficulty int64 9 10 | new_solution_is_correct bool 1
class |
|---|---|---|---|---|---|
Let the incircle of triangle \( \triangle ABC \) be tangent to the sides at points \( D \), \( E \), and \( F \). Prove that triangle \( \triangle ABC \) is similar to triangle \( \triangle DEF \) if and only if \( \triangle ABC \) is an equilateral triangle. |
Okay, so I need to prove that the triangle ABC is similar to its incentral triangle DEF if and only if ABC is equilateral. Hmm, okay. Let me start by recalling some basics.
First, the incircle of a triangle touches each side at a point, which are called the points of tangency. These points D, E, F are where the incir... | To prove that triangle \( \triangle ABC \) is similar to its intouch triangle \( \triangle DEF \) if and only if \( \triangle ABC \) is an equilateral triangle, we need to consider both directions of the proof.
**Forward Direction (If \( \triangle ABC \) is equilateral, then \( \triangle DEF \) is similar to \( \trian... |
1. **Denote the Incircle and Intersection Points:**
Let \( \triangle ABC \) be a triangle with an incircle that touches \( \triangle ABC \) at points \( D \), \( E \), and \( F \).
- \( D \) is the point where the incircle touches side \( BC \).
- \( E \) is the point where the incircle touches side \( CA ... | 9 | true |
In an acute triangle \( \triangle ABC \), with \( AB \neq AC \), let \( K \) be the midpoint of median \( AD \). Points \( E \) and \( F \) are the feet of the perpendiculars from \( D \) to \( AB \) and \( AC \) respectively. Lines \( KE \) and \( KF \) intersect \( BC \) at points \( M \) and \( N \) respectively. Th... |
Okay, let me try to figure out how to approach this problem. First, let me parse all the given information step by step.
We have an acute triangle \( \triangle ABC \) with \( AB \neq AC \). So, it's not an isosceles triangle with AB equal to AC. That might mean that some symmetries might not be present, which is impo... | Given an acute triangle \( \triangle ABC \) with \( AB \neq AC \), we need to prove that the line \( O_1O_2 \) connecting the circumcenters of \( \triangle DEM \) and \( \triangle DFN \) is parallel to \( BC \).
1. **Setup Coordinates**:
- Place \( B \) at \((-1, 0)\), \( C \) at \((1, 0)\), midpoint \( D \) at \(... |
1. Given the geometric configuration in the obtuse triangle $\triangle ABC$ where $AB \neq AC$, and $K$ is the midpoint of the median $AD$, $DE \perp AB$ at point $E$, $DF \perp AC$ at point $F$. Lines $KE$ and $KF$ intersect line $BC$ at points $M$ and $N$, respectively. We need to prove that the circumcenters $O_1$ ... | 9 | true |
For each natural number \( n \leqslant p - 2 \), the pair \((n, n+1)\) has one of four types: \((R, R), (N, N), (N, R), (R, N)\), where \( R \) denotes a residue and \( N \) denotes a non-residue. Let \( RR, NN, NR, RN \) be the number of pairs of each respective type.
a) Prove that \( RR + NN - RN - NR = 1 \).
b) Le... |
Okay, so I need to tackle part (a) first. The problem is about quadratic residues and non-residues modulo a prime p. For each natural number n up to p-2, we look at pairs (n, n+1) and classify them into four types: (R, R), (N, N), (N, R), (R, N). Then RR, NN, NR, RN are the counts of these pairs. Part (a) asks to prov... | Given a prime \( p \), we consider pairs \((n, n+1)\) for \( n \leq p-2 \) and classify them into four types: \((R, R), (N, N), (N, R), (R, N)\). Here, \( R \) denotes a quadratic residue and \( N \) denotes a non-residue modulo \( p \). The counts of each type are denoted by \( RR, NN, NR, RN \).
### Part (a)
To prov... | :
#### Part (a)
1. The types of pairs \((n, n+1)\) have specific properties in terms of quadratic residues and non-residues. Recall that:
- \( (\frac{n}{p}) \) denotes the Legendre symbol and determines whether \( n \) is a quadratic residue (\( R \)) or non-residue (\( N \)).
2. For pairs \((R, R)\) and \((... | 10 | true |
Given a positive integer \(n \geq 2\), let positive integers \(a_{i}(i=1,2, \cdots, n)\) satisfy \(a_{1}<a_{2}<\cdots<a_{n}\) and \(\sum_{i=1}^{n} \frac{1}{a_{i}} \leq 1\). Prove that for any real number \(x\), the following inequality holds:
$$
\left(\sum_{i=1}^{n} \frac{1}{a_{i}^{2}+x^{2}}\right)^{2} \leq \frac{1}{2}... |
Okay, so I've been given this inequality problem to solve. Let me see what it's about. The problem says that for a positive integer \( n \geq 2 \), we have positive integers \( a_1, a_2, \ldots, a_n \) such that they're strictly increasing, meaning \( a_1 < a_2 < \cdots < a_n \), and the sum of their reciprocals is at... | Given a positive integer \( n \geq 2 \) and positive integers \( a_i \) (for \( i = 1, 2, \ldots, n \)) such that \( a_1 < a_2 < \cdots < a_n \) and \( \sum_{i=1}^{n} \frac{1}{a_i} \leq 1 \), we need to prove that for any real number \( x \), the inequality holds:
\[
\left( \sum_{i=1}^{n} \frac{1}{a_i^2 + x^2} \right)^... |
1. **Case 1: \( x^2 \geqslant a_1(a_1 - 1) \)**
Given that \( \sum_{i=1}^{n} \frac{1}{a_i} \leqslant 1 \), we start by considering the inequality:
\[
\left(\sum_{i=1}^{n} \frac{1}{a_i^2 + x^2}\right)^2
\]
Using the inequality \( a_i^2 + x^2 \geqslant 2a_i |x| \) (since \( a_i \geqslant 1 \) and \( ... | 9 | true |
Prove that the expression
$$
P=(x-a)(x-b)(x-c)\left(\frac{a-b}{x-c}+\frac{b-c}{x-a}+\frac{c-a}{x-b}\right)
$$
is independent of $x$. Determine the value of $P$. | Okay, so I need to prove that the expression P is independent of x and find its value. Let me start by understanding the problem. The expression is given as:
P = (x - a)(x - b)(x - c) * [ (a - b)/(x - c) + (b - c)/(x - a) + (c - a)/(x - b) ]
First, I need to show that no matter what x is (as long as x isn't equal to ... | To prove that the expression
$$
P=(x-a)(x-b)(x-c)\left(\frac{a-b}{x-c}+\frac{b-c}{x-a}+\frac{c-a}{x-b}\right)
$$
is independent of \( x \) and determine its value, we start by simplifying the expression.
First, we multiply each term in the brackets by \((x - a)(x - b)(x - c)\):
\[
P = (a - b)(x - a)(x - b) + (b -... |
Let's start with the given expression \( P \):
\[
P = (x-a)(x-b)(x-c) \left( \frac{a-b}{x-c} + \frac{b-c}{x-a} + \frac{c-a}{x-b} \right)
\]
To show that \( P \) is independent of \( x \), we need to simplify the expression inside the parentheses:
\[
\frac{a-b}{x-c} + \frac{b-c}{x-a} + \frac{c-a}{x-b}
\]
### Step-b... | 9 | true |
The circles \( S_{1} \) and \( S_{2} \) intersect at points \( A \) and \( B \), with the center \( O \) of circle \( S_{1} \) lying on \( S_{2} \). A line passing through point \( O \) intersects segment \( AB \) at point \( P \) and circle \( S_{2} \) at point \( C \). Prove that point \( P \) lies on the polar of po... | Okay, so I need to prove that point P lies on the polar of point C with respect to circle S₁. Let me try to recall what polar lines are and how they relate to circles.
First, the polar of a point with respect to a circle is the set of points whose reciprocals are harmonic conjugates with respect to the circle. Wait, ... | Given two intersecting circles \( S_1 \) and \( S_2 \) with centers \( O \) and \( O_2 \) respectively, where \( O \) lies on \( S_2 \). The radical axis of \( S_1 \) and \( S_2 \) is the line \( AB \). A line passing through \( O \) intersects \( AB \) at point \( P \) and circle \( S_2 \) at point \( C \). We need to... |
Consider the circles \(S_1\) and \(S_2\) which intersect at points \(A\) and \(B\). The center \(O\) of the circle \(S_1\) lies on the circle \(S_2\).
1. **Identify the key points and intersections:**
- Let a line through the point \(O\) intersect the segment \(AB\) at point \(P\) and intersect the circle \(S_2\) ... | 9 | true |
Prove that \(\sum_{k=1}^{n} \lambda_{k} \leq n\). | Okay, so I need to prove that the sum of λ_k from k=1 to n is less than or equal to n. Hmm, first thing I should check is if there are any given conditions or constraints on the λ_k values. The problem statement doesn't specify anything else, so maybe they are assuming some context here? Maybe λ_k are eigenvalues of a ... | To prove that \(\sum_{k=1}^{n} \lambda_{k} \leq n\), we need to consider the conditions under which this inequality holds. A common scenario is when each \(\lambda_k\) is bounded above by 1.
Assuming each \(\lambda_k \leq 1\), we can sum these inequalities over all \(k\) from 1 to \(n\):
\[
\sum_{k=1}^{n} \lambda_{k... |
1. **Define the Function**:
Let \( f(\theta) = \sum_{k=1}^{n} \lambda_{k} e^{i k \theta} \) where \( \varepsilon = e^{i \frac{2 \pi}{n+1}} \), which is a $(n+1)$-th root of unity.
2. **Evaluate Sum at Specific Points**:
Consider evaluating the sum at \( \theta = \frac{2k\pi}{n+1} \) for \( k = 1, \ldots, n \)... | 9 | true |
Using the property
$$
l_{U_{i=1}^{n} A_{i}}=1-\prod_{i=1}^{n}\left(1-I_{A_{i}}\right)
$$
of indicators from Problem I.4.1, prove the inclusion-exclusion formula for the probability of the union of events:
$$
\begin{aligned}
& \mathrm{P}\left(A_{1} \cup \ldots \cup A_{n}\right)=\sum_{1 \leqslant i_{1} \leqslant n} \m... | Okay, so I need to prove the inclusion-exclusion formula for the probability of the union of events using that indicator function property given. Let me recall the property first. The indicator of the union of sets A₁ through Aₙ is equal to 1 minus the product from i=1 to n of (1 minus the indicator of Aᵢ). So, mathema... | To prove the inclusion-exclusion formula for the probability of the union of events using the given indicator property, we start with the property:
$$
I_{\bigcup_{i=1}^{n} A_i} = 1 - \prod_{i=1}^{n} (1 - I_{A_i})
$$
Taking the expectation of both sides, we get:
$$
\mathrm{P}\left( \bigcup_{i=1}^{n} A_i \right) = \ma... |
We begin by utilizing the provided property involving indicator functions:
\[
I_{\bigcup_{i=1}^{n} A_{i}} = 1 - \prod_{i=1}^{n}(1 - I_{A_{i}})
\]
To prove the inclusion-exclusion formula for the probability of the union of events:
\[
\mathrm{P}\left( A_{1} \cup \ldots \cup A_{n} \right)
\]
1. **Taking Expectations... | 9 | true |
Let \( n \) be a positive integer. Given a sequence \(\varepsilon_{1}, \ldots, \varepsilon_{n-1}\) with \(\varepsilon_{i}=0\) or \(\varepsilon_{i}=1\) for each \(i=1, \ldots, n-1\), the sequences \( a_{0}, \ldots, a_{n} \) and \( b_{0}, \ldots, b_{n} \) are constructed by the following rules:
\[
\begin{array}{c}
a_{0} ... | Okay, so I need to prove that for any positive integer n and any sequence of ε's (each 0 or 1), the sequences a_n and b_n constructed by these different rules end up being equal. Hmm, interesting. Let me start by trying to understand the problem better.
First, both sequences a and b start with the same initial terms: ... | To prove that \(a_n = b_n\) for any positive integer \(n\) and any sequence \(\varepsilon_1, \ldots, \varepsilon_{n-1}\), we use induction and properties of linear recursions represented by matrices.
**Base Case**:
For \(n = 1\), both sequences start with \(a_0 = b_0 = 1\) and \(a_1 = b_1 = 7\). Thus, \(a_1 = b_1\).
... |
To prove that \(a_{n} = b_{n}\) for a given binary sequence \((\varepsilon_1, \varepsilon_2, \ldots, \varepsilon_{n-1})\), given the initial conditions \(a_0 = b_0 = 1\) and \(a_1 = b_1 = 7\), and the recursive definitions for \(a_{i+1}\) and \(b_{i+1}\) using \(\varepsilon_i\), we can use the properties of binary wor... | 9 | true |
For a non-empty set \( M \subset \mathbf{Q} \), the following two conditions are satisfied:
1) If \( a \in M \) and \( b \in M \), then \( a+b \in M \) and \( ab \in M \);
2) For any \( r \in \mathbf{Q} \), exactly one of the following statements is true: \( r \in M \), \( -r \in M \), \( r=0 \).
Prove that the set \... | Okay, so I need to prove that a set M, which is a subset of the rational numbers Q, satisfying two conditions, must be the set of all positive rational numbers. Let me first parse the problem again to make sure I understand it correctly.
The set M is non-empty and satisfies:
1) If a and b are in M, then their sum a +... | To prove that the set \( M \) is the set of all positive rational numbers, we analyze the given conditions and proceed step by step:
1. **Non-emptiness and Trichotomy**: Since \( M \) is non-empty, there exists a non-zero element \( a \in M \). By condition 2, for any non-zero \( r \in \mathbf{Q} \), exactly one of \(... |
To prove that the set $M$ coincides with the set of all positive rational numbers, we follow these steps.
1. **Initial Assumption and Implication**:
By condition 2, for any rational number $r \in \mathbf{Q}$, exactly one of the following statements is true:
- $r \in M$
- $-r \in M$
- $r = 0$
Th... | 10 | true |
Several segments, which we shall call white, are given, and the sum of their lengths is 1. Several other segments, which we shall call black, are given, and the sum of their lengths is 1. Prove that every such system of segments can be distributed on the segment that is 1.51 long in the following way: Segments of the s... | Okay, so I need to tackle this problem where we have white and black segments, each summing to length 1. The goal is to prove two things: first, that they can be arranged on a segment of length 1.51 such that same-colored segments are disjoint, and different-colored ones are either disjoint or nested. Second, I have to... | To solve the problem, we need to show that any system of white and black segments (each totaling length 1) can be arranged on a segment of length 1.51 such that same-colored segments are disjoint, and different-colored segments are either disjoint or nested. Additionally, we must demonstrate a system that cannot fit wi... |
**Given:**
- We have several white segments, with a total length of 1.
- We have several black segments, with a total length of 1.
- We need to prove that we can distribute these segments on a segment of length 1.51 under specific conditions.
- We also need to prove that such a distribution is impossible on a segment ... | 9 | true |
There are 81,650 students arranged in 22 rows and 75 columns. It is known that among any two adjacent columns, for any pair of students in the same row, there are no more than 11 pairs of students of the same gender. Prove that the number of male students does not exceed 928. | Okay, so I need to prove that the number of male students doesn't exceed 928 in this arrangement of 81,650 students. Let me start by understanding the problem again. There are 22 rows and 75 columns, making a total of 22*75 = 1,650 students. Wait, but the total number of students is 81,650? Wait, that doesn't add up. W... | Given 81,650 students arranged in 22 rows and 75 columns, we need to prove that the number of male students does not exceed 928.
1. **Understanding the Problem**:
- The problem states that among any two adjacent columns, for any pair of students in the same row, there are no more than 11 pairs of students of the s... |
1. Define the number of male students in the \(i\)-th row as \(a_i\). Consequently, the number of female students in the \(i\)-th row is \(75 - a_i\) because there are 75 students in each row.
2. By the given problem, if two students picked from the same row should not have more than 11 pairs of students of the same ... | 9 | true |
Let \( p \) be an odd prime number.
a) Prove that it is possible to select a natural number \( m < p \) and integers \( x_{1}, x_{2}, x_{3}, \) and \( x_{4} \) such that \( x_{1}^{2} + x_{2}^{2} + x_{3}^{2} + x_{4}^{2} = m p \).
b) Prove that the smallest such \( m \) is odd.
c) Prove that the smallest such \( m \)... | Alright, let's tackle part (a) first. The problem states that for an odd prime number \( p \), we need to show there exists a natural number \( m < p \) and integers \( x_1, x_2, x_3, x_4 \) such that \( x_1^2 + x_2^2 + x_3^2 + x_4^2 = mp \).
Hmm, okay. So, I remember something about Lagrange's four-square theorem, w... | ### Summary
a) To prove that there exists a natural number \( m < p \) such that \( mp \) can be expressed as the sum of four squares, we use the Chevalley-Warning theorem. This theorem guarantees a non-trivial solution to the equation \( x_1^2 + x_2^2 + x_3^2 + x_4^2 \equiv 0 \mod p \). Lifting this solution to integ... |
#### Part (a)
1. We start with the assumption provided in problem 31.58, which suggests the form to choose integers \( x_1, x_2, x_3, x_4 \).
2. We choose:
\[
x_1 = 1, \quad x_2 = x, \quad x_3 = y, \quad x_4 = 0.
\]
3. Substituting into the sum of squares:
\[
x_1^2 + x_2^2 + x_3^2 + x_4^2 = 1^2 + x^2 ... | 9 | true |
Let \( X \) be a set with \( n \) elements, where \( n \geqslant 4 \). Suppose \( A_1, A_2, \cdots, A_{100} \) are subsets of \( X \), not necessarily distinct, such that \( |A_i| > \frac{3}{4} n \) for \( i = 1, 2, \cdots, 100 \). Prove that there exists a subset \( Y \subseteq X \) with \( |Y| \leq 4 \) such that \( ... |
Okay, so we have a set X with n elements, where n is at least 4. Then there are 100 subsets of X, called A1 to A100, each of size greater than 3/4 n. We need to show that there exists a subset Y of X with at most 4 elements such that Y intersects every Ai. In other words, Y is a hitting set for all these subsets, and ... | Given a set \( X \) with \( n \) elements (\( n \geq 4 \)), we need to prove that there exists a subset \( Y \subseteq X \) with \( |Y| \leq 4 \) such that \( Y \cap A_i \neq \varnothing \) for all \( i = 1, 2, \ldots, 100 \), where each \( A_i \) is a subset of \( X \) with \( |A_i| > \frac{3}{4} n \).
### Key Steps:... |
1. **Initial Setup:**
Let \( X = \{x_1, x_2, \cdots, x_n\} \) be a finite set of \( n \) elements where \( n \geq 4 \). Denote \( A_1, A_2, \ldots, A_{100} \) as subsets of \( X \), each satisfying the condition \(|A_i| > \frac{3}{4} n\) for \( i = 1, 2, \ldots, 100 \).
2. **Counting Element Appearances:**
Con... | 9 | true |
Let $U$ be a set consisting of $m$ triangles. Prove that there exists a subset $W$ of $U$ such that:
(1) The subset $W$ contains at least $\frac{9}{20} m^{\frac{4}{5}}$ triangles;
(2) There do not exist six distinct points $A, B, C, D, E, F$ such that $\triangle ABC, \triangle BCD, \triangle CDE, \triangle DEF, \triang... | Alright, so I need to prove that for any set U of m triangles, there exists a subset W satisfying two conditions: first, W has at least (9/20)m^{4/5} triangles, and second, there are no six distinct points A, B, C, D, E, F forming a cyclic chain of six triangles as specified. Hmm, okay. Let me try to unpack this.
Firs... | To prove the existence of a subset \( W \) of \( U \) that satisfies the given conditions, we use the probabilistic method with a deletion step. Here is the step-by-step summary:
1. **Probabilistic Selection**: We randomly select each triangle from \( U \) with probability \( p = m^{-1/5} \). The expected number of tr... |
1. **Step 1: Define the subset $U'$**
Let $U'$ be a subset formed from the set $U$ by selecting each triangle with probability $p$. Therefore, the expected number of triangles in $U'$ is $mp$.
2. **Step 2: Compute the upper limit of bad configurations**
For six distinct points forming the point sequence... | 9 | true |
Let \( m \) and \( n \) be positive integers such that:
\[ \frac{m}{n} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots - \frac{1}{1318} + \frac{1}{1319} \]
Prove that \( m \) is divisible by 1979. | Alright, so I have this problem here where I need to show that the numerator of the fraction \( \frac{m}{n} \), which equals the alternating harmonic series up to \( \frac{1}{1319} \), is divisible by 1979. Hmm, okay. Let me try to break this down step by step.
First, the series in question is \( 1 - \frac{1}{2} + \fr... | To prove that \( m \) is divisible by 1979, we start by considering the given alternating harmonic series:
\[
\frac{m}{n} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots - \frac{1}{1318} + \frac{1}{1319}
\]
We need to show that the numerator \( m \) of this series, when written in lowest terms, is divisible by ... |
Let's analyze the given sum:
\[ S = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots - \frac{1}{1318} + \frac{1}{1319} \]
1. **Group the terms with alternating signs:**
We can consider the grouping of terms to simplify the sum. Let's separate the expression into positive and negative terms:
\[
S = ... | 9 | true |
Given numbers $x$ and $y$ satisfying the condition $x^{4}+y^{4} \geqslant 2$, prove the inequality $\left|x^{16}-y^{16}\right|+4 x^{8} y^{8} \geqslant 4$. |
Okay, so I have this problem here: given numbers x and y such that x4 + y4 is greater than or equal to 2, I need to prove that |x16 - y16| + 4x8y8 is greater than or equal to 4. Hmm, that seems a bit complicated at first glance, but maybe I can break it down step by step.
First, let's restate what's given and what ne... | Given numbers \(x\) and \(y\) satisfying the condition \(x^4 + y^4 \geq 2\), we need to prove the inequality \(\left|x^{16} - y^{16}\right| + 4x^8 y^8 \geq 4\).
1. **Substitution**:
Let \(a = x^4\) and \(b = y^4\). The given condition becomes \(a + b \geq 2\), and we need to prove \(|a^4 - b^4| + 4a^2 b^2 \geq 4\).... |
1. **Recall relevant inequality:**
Notice the inequality \( u^2 + v^2 \geqslant \frac{1}{2}(u+v)^2 \) holds for any non-negative \( u \) and \( v \). This inequality is equivalent to the non-negativity of a square:
\[
(u-v)^2 \geqslant 0.
\]
2. **Consider the input condition and symmetry:**
We are gi... | 9 | true |
Consider the equation
$$
x^{n} + a_{n-1} x^{n-1} + \cdots + a_{1} x + a_{0} = 0
$$
whose coefficients are all real numbers and satisfy the condition
$$
0 < a_{0} \leq a_{1} \leq \cdots \leq a_{n-1} \leq 1.
$$
Given that \(\lambda\) is a complex root of this equation and satisfies the condition \(|\lambda| \geqslant ... |
Okay, so I need to prove that if λ is a complex root of the polynomial equation
\( x^{n} + a_{n-1} x^{n-1} + \cdots + a_{1} x + a_{0} = 0 \),
where all the coefficients are real numbers satisfying \( 0 < a_{0} \leq a_{1} \leq \cdots \leq a_{n-1} \leq 1 \), and if |λ| ≥ 1, then λ^{n+1} = 1. Hmm, interesting. Let me ... | Given the polynomial equation:
\[
x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 = 0
\]
where all coefficients \(a_0, a_1, \ldots, a_{n-1}\) are real numbers satisfying \(0 < a_0 \leq a_1 \leq \cdots \leq a_{n-1} \leq 1\), we need to prove that if \(\lambda\) is a complex root with \(|\lambda| \geq 1\), then \(\lambda^{n... |
1. **Understanding the problem:**
Given a polynomial equation with real coefficients and constraints:
\[
x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 = 0
\]
where
\[
0 < a_0 \leq a_1 \leq \cdots \leq a_{n-1} \leq 1.
\]
We know that $\lambda$ is a complex root of this polynomial and $|\la... | 9 | true |
The center of a sphere \( Q \) is defined by the radius vector \(\vec{q}=\overrightarrow{O Q}\). An arbitrary point \( P \) on the sphere is defined by the radius vector \(\vec{r}_{1}\) with coordinates \( (x_{1}, y_{1}, z_{1}) \). An arbitrary point \( M \) on the plane, which is tangent to the sphere at point \( P \)... | Okay, so I need to prove that the equation of the tangent plane to the sphere Q at point P can be written in those two vector forms and then convert it into the coordinate form. Let me start by recalling what a tangent plane to a sphere is.
A tangent plane at a point on a sphere is perpendicular to the radius at that... | To prove the equation of the tangent plane to the sphere \( Q \) at point \( P \) can be written in the given vector forms and derive the coordinate form, we start by noting that the tangent plane is perpendicular to the radius vector at \( P \).
1. **First Vector Form**:
The vector from \( P \) to any point \( M ... |
1. Given the problem states that vector \(\left(\vec{r} - \vec{r}_1\right)\) is perpendicular to vector \(\left(\vec{r}_1 - \vec{q}\right)\), it suggests that their dot product must be zero. Thus:
\[
\left(\vec{r} - \vec{r}_1\right) \cdot \left(\vec{r}_1 - \vec{q}\right) = 0
\]
This follows from the definition of ort... | 9 | true |
A sequence \(\{a_n\}\) is defined as follows: \(a_1 = 1\), \(a_{n+1} = \frac{a_n}{n} + \frac{n}{a_n}\), for \(n = 1, 2, \cdots\). Prove that for \(n \geq 4\), \(\left\lfloor a_n^2 \right\rfloor = n\), where \(\left\lfloor x \right\rfloor\) denotes the greatest integer less than or equal to \(x\). | Okay, so I need to prove that for the sequence defined by a₁ = 1 and aₙ₊₁ = aₙ/n + n/aₙ, the floor of aₙ² is equal to n for all n ≥ 4. Hmm, let's start by understanding the sequence better. Maybe I should compute the first few terms to see if the pattern holds and get some intuition.
Starting with a₁ = 1. Then:
a₂ = ... | To prove that for the sequence \(\{a_n\}\) defined by \(a_1 = 1\) and \(a_{n+1} = \frac{a_n}{n} + \frac{n}{a_n}\) for \(n \geq 1\), the floor of \(a_n^2\) is equal to \(n\) for all \(n \geq 4\), we proceed by induction.
**Base Case:**
For \(n = 4\), we compute the first few terms of the sequence:
- \(a_1 = 1\)
- \(a_2... |
Consider the sequence $\{a_n\}$ defined by:
$$
a_1 = 1, \quad a_{n+1} = \frac{a_n}{n} + \frac{n}{a_n}, \quad n = 1, 2, 3, \ldots
$$
We need to prove that for $n \geq 4$, $[a_n^2] = n$, where $[x]$ denotes the greatest integer less than or equal to $x$.
1. **Define the function and analyze its monotonicity**:
De... | 9 | true |
Prove Euler's formula
$$
\mathscr{F}-\mathscr{P}+\mathscr{B}=2
$$
for simply connected polyhedra (polyhedra of genus zero).
It can be shown that for a polyhedron of any genus, Euler's formula is
$$
\mathscr{G}-\mathscr{P}+\mathscr{B}=2-2 p
$$ | Okay, so I need to prove Euler's formula for simply connected polyhedra, which is F - E + V = 2, right? Wait, let me check the notation. The user wrote it as F - P + B = 2. Hmm, maybe F is faces, P is edges (from the French "arêtes," perhaps?), and B is vertices (from "bords"?). But in the standard Euler's formula, it'... | To prove Euler's formula for simply connected polyhedra (genus zero), we can utilize properties of planar graphs. Here's a concise proof:
1. **Planar Graph Representation**:
A convex polyhedron is topologically equivalent to a sphere. When projected onto a plane, it forms a connected planar graph with \( V \) ver... |
1. **Observation from previous problem**:
Similarly to the preceding problem, we utilize the fact that the edges of the polyhedron form a certain network on its surface. This network divides the surface of the polyhedron into regions (faces), and the edges are line segments constituting the borders of these regions... | 9 | true |
Given that for any \( n \in \mathbf{N} \), \( a_{n} > 0 \) and
\[
\sum_{j=1}^{n} a_{j}=\left(\sum_{j=1}^{n} a_{j}\right)^{2}
\]
prove that \( a_{n} = n \). | Alright, let's tackle this problem step by step. The question states that for any natural number \( n \), the sequence \( a_n \) is positive, and the sum from \( j = 1 \) to \( n \) of \( a_j \) equals the square of that same sum. We need to prove that \( a_n = n \).
First, let me parse the given equation properly. It... | Given that for any \( n \in \mathbf{N} \), \( a_{n} > 0 \) and
\[
\sum_{j=1}^{n} a_{j} = \left( \sum_{j=1}^{n} a_{j} \right)^2
\]
we need to prove that \( a_{n} = n \).
First, denote the sum \( S_n = \sum_{j=1}^{n} a_j \). The given equation translates to:
\[
S_n = (S_n)^2
\]
Solving this quadratic equation, we get:
... |
1. **Base Case:**
- When \( n = 1 \), we have \( \sum_{j=1}^{1} a_{j} = a_{1} \).
- The given relation becomes:
\[
\sum_{j=1}^{1} a_{j} = \left(\sum_{j=1}^{1} a_{j}\right)^{2} \implies a_{1} = a_{1}^{2}.
\]
- Since \( a_1 > 0 \), we deduce:
\[
a_{1} = 1.
\]
- Thus, t... | 9 | true |
In triangle \( \triangle ABC \), there is a point \( D \) on side \( AB \) such that \( CD \) is the geometric mean of \( AD \) and \( BD \). Prove that this condition holds if and only if
$$
\sin A \sin B \leq \sin^2 \frac{C}{2}.
$$ |
Okay, let me try to tackle this geometry problem. So, we have triangle ABC, and there's a point D on side AB such that CD is the geometric mean of AD and BD. We need to prove that this condition holds if and only if sin A sin B ≤ sin2(C/2). Alright, let's break this down step by step.
First, let's understand what the... | To prove that the condition holds if and only if \(\sin A \sin B \leq \sin^2 \frac{C}{2}\), we start by noting that point \(D\) on side \(AB\) such that \(CD\) is the geometric mean of \(AD\) and \(BD\) implies \(CD^2 = AD \cdot BD\).
Using coordinate geometry, we place \(A\) at \((0,0)\), \(B\) at \((c,0)\), and \(C... | 1. Consider a triangle \( \triangle ABC \) with side lengths \( AB = c \), \( BC = a \), and \( CA = b \). We aim to transform the given condition
$$
\sin A \sin B \leq \sin^2 \left( \frac{C}{2} \right)
$$
into a relationship involving the sides of the triangle using known trigonometric identities and prope... | 10 | true |
Cut a regular tetrahedron with a plane perpendicular to its base.
The lines of intersection of the cutting plane with the planes of the side faces form angles $\alpha, \beta, \gamma$ with the line of intersection in the base plane.
Prove that $\operatorname{tg} \alpha + \operatorname{tg} \beta + \operatorname{tg} \ga... | Alright, so I need to prove that the sum of the tangents of these angles α, β, γ is zero. Let me first try to visualize the problem. We have a regular tetrahedron, right? A regular tetrahedron has four triangular faces, all equilateral. The base is one of these faces, and the other three faces are the side faces. We're... | To prove that \(\operatorname{tg} \alpha + \operatorname{tg} \beta + \operatorname{tg} \gamma = 0\) for the angles \(\alpha, \beta, \gamma\) formed by the lines of intersection of a cutting plane perpendicular to the base of a regular tetrahedron with its side faces, we consider the geometric properties and symmetry of... |
1. **Identify the vertices and define the cutting plane**:
Let the vertices of the base of a regular tetrahedron be \(A, B, C\), and the fourth vertex be \(D\). Assume the cutting plane intersects the tetrahedron in such a way that it is perpendicular to the base \(ABC\).
2. **Use parallelism property**:
... | 9 | true |
Given \(n\) circles \(O_{1}, O_{2}, \ldots, O_{n}\) passing through a single point \(O\). The second intersection points of \(O_{1}\) with \(O_{2}\), \(O_{2}\) with \(O_{3}\), ..., \(O_{n}\) with \(O_{1}\) are denoted as \(A_{1}, A_{2}, \ldots, A_{n}\) respectively. On \(O_{1}\), we take an arbitrary point \(B_{1}\). I... |
Okay, let's see. I need to prove that starting with a point B1 on circle O1, going through each circle O2, O3, ..., On by intersecting with lines through A1, A2, ..., An respectively, and then coming back to O1 gives the same point B1 again. So, essentially, this process is cyclic and closes after n steps.
First, le... | Given \( n \) circles \( O_1, O_2, \ldots, O_n \) passing through a common point \( O \), we need to prove that starting from an arbitrary point \( B_1 \) on \( O_1 \), and constructing points \( B_2, B_3, \ldots, B_{n+1} \) by intersecting the line through \( B_i \) and \( A_i \) with the next circle, results in \( B_... | :
Let's denote the oriented angle between straight lines \( AB \) and \( CD \) as \( \angle(A B, C D) \). The oriented angle is measured modulo \( 180^\circ \). This allows us to consider the angles in the range \([0^\circ, 180^\circ]\).
Consider the intersections of circles \( O_i \) as follows:
1. For any two ci... | 9 | true |
Let $\xi$ and $\eta$ be independent random variables such that the distribution of $\xi+\eta$ coincides with the distribution of $\xi$. Prove that $\eta = 0$ almost surely. | Okay, so I need to prove that if ξ and η are independent random variables, and the distribution of ξ + η is the same as the distribution of ξ, then η must be zero almost surely. Hmm, let's start by unpacking what this means.
First, since ξ and η are independent, their joint distribution is the product of their individ... | To prove that $\eta = 0$ almost surely under the given conditions, we employ characteristic functions and properties of independent random variables:
1. **Characteristic Functions Equality**: Since $\xi + \eta \stackrel{d}{=} \xi$, their characteristic functions satisfy:
\[
\phi_{\xi+\eta}(t) = \phi_\xi(t) \quad... | 1. **Introduction and given information**:
We have two independent random variables, $\xi$ and $\eta$, such that the distribution of the sum $\xi + \eta$ is the same as the distribution of $\xi$. We are required to prove that $\eta = 0$ almost surely.
2. **Characteristic function relationship**:
Let $\varphi... | 9 | true |
Define a function \( f(x) \) on \(\mathbf{R}\) such that for any real numbers \( x \) and \( y \), the following holds:
$$
f(x+y+1)=f(x-y+1)-f(x) f(y),
$$
and \( f(1)=2 \).
(1) Prove that \( f(x) \) is an odd function.
(2) Find the value of \( f(f(2017)) \). |
Okay, let me try to tackle this problem step by step. The problem defines a function \( f(x) \) on the real numbers with the property that for any real numbers \( x \) and \( y \),
\[
f(x + y + 1) = f(x - y + 1) - f(x)f(y),
\]
and we're given that \( f(1) = 2 \). The two parts are to prove that \( f \) is an odd fun... | To solve the given problem, we start by analyzing the functional equation and the given condition \( f(1) = 2 \).
1. **Proving that \( f(x) \) is an odd function**:
- By setting \( y = 0 \) in the functional equation, we find \( f(0) = 0 \).
- Setting \( x = 0 \) in the equation, we get \( f(y + 1) = f(1 - y) \)... |
1. (1) To show that \( f(x) \) is an odd function, we start by taking \( x = -1 \) and \( y = 1 \) in the given functional equation:
\[
f(x+y+1) = f(x-y+1) - f(x) f(y)
\]
Substituting \( x = -1 \) and \( y = 1 \), we get:
\[
f(-1+1+1) = f(-1-1+1) - f(-1) f(1)
\]
Simplifying inside the function argume... | 9 | true |
Given \(a, b, c > 0\) and \(abc \geq 1\), prove that:
$$
n^{3} \sum_{k=0}^{n} a^{k} \cdot \sum_{k=0}^{n} b^{k} \cdot \sum_{k=0}^{n} c^{k} \geq (n+1)^{3} \sum_{k=0}^{n-1} a^{k} \cdot \sum_{k=0}^{n-1} b^{k} \cdot \sum_{k=0}^{n-1} c^{k}.
$$ | Okay, so I need to prove this inequality: given that a, b, c are positive real numbers and their product abc is at least 1, then
n³ multiplied by the product of the sums from k=0 to n of aᵏ, bᵏ, and cᵏ is greater than or equal to (n+1)³ multiplied by the product of the sums from k=0 to n-1 of aᵏ, bᵏ, and cᵏ.
First, l... | To prove the inequality given \(a, b, c > 0\) and \(abc \geq 1\), we analyze the ratio of the left-hand side (LHS) to the right-hand side (RHS):
\[
\frac{n^3 \sum_{k=0}^{n} a^k \cdot \sum_{k=0}^{n} b^k \cdot \sum_{k=0}^{n} c^k}{(n+1)^3 \sum_{k=0}^{n-1} a^k \cdot \sum_{k=0}^{n-1} b^k \cdot \sum_{k=0}^{n-1} c^k}
\]
Let... |
1. Define the function \( f(x) = \frac{\sum_{k=0}^{n} x^{k}}{\sum_{k=0}^{n-1} x^{k}} - \frac{1+n}{2n}(x+1) \) for \( x \in \mathbb{R}^{+} \) and \( n \in \mathbb{N}^{+} \).
2. First, we check the value of \( f(x) \) at \( x=1 \):
\[
f(1) = \frac{\sum_{k=0}^{n} 1^{k}}{\sum_{k=0}^{n-1} 1^{k}} - \frac{1+n}{2n}(1+1... | 9 | true |
A nonempty set \(A\) of real numbers is called a \(B_{3}\)-set if the conditions \(a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6} \in A\) and \(a_{1} + a_{2} + a_{3} = a_{4} + a_{5} + a_{6}\) imply that the sequences \((a_{1}, a_{2}, a_{3})\) and \((a_{4}, a_{5}, a_{6})\) are identical up to a permutation. Let \(A = \{a_0 =... |
Okay, so I need to prove that if A is a B3-set and the difference sets D(A) and D(B) are equal, then A must be equal to B. Both A and B start with 0 and are strictly increasing sequences.
First, let me recall what a B3-set is. The definition says that if you have six elements from A such that the sum of three equals... | To prove that \( A = B \) given that \( A \) is a \( B_3 \)-set and \( D(A) = D(B) \), we use induction on the elements of \( A \) and \( B \).
**Base Case:** Both sets start with \( 0 \), so \( a_0 = b_0 = 0 \).
**Inductive Step:** Assume \( a_i = b_i \) for all \( i \leq n \). We show \( a_{n+1} = b_{n+1} \).
1. *... |
1. **Equivalence of Difference Sets**:
Since \( D(A) = D(B) \), we know that for any elements \( x, y \in A \) and \( u, v \in B \):
\[
\{ |x - y| \mid x, y \in A \} = \{ |u - v| \mid u, v \in B \}
\]
This means each difference \( b_i - b_{i-1} \) can be written as \( a_{f(i)} - a_{g(i)} \), where \( f(... | 10 | true |
In a country, there are 6 cities connected pairwise by roads, never more than one road between any two cities. These roads do not intersect, some passing over others by bridges if necessary. Each road is one-way. Distracted officials from the Ministry of Traffic Direction have oriented the roads such that if one leaves... | Alright, so I've got this problem about catastrophic road networks in a country with 6 cities. The roads are all one-way, and they’re set up such that if you leave any city, you can't come back to it. That sounds like the graph is a directed acyclic graph (DAG), right? Because if there's a cycle, then you could leave a... | 1. In any catastrophic network (transitive tournament), there must be a city with no outgoing roads (a sink), which is proven by the properties of directed acyclic graphs (DAGs).
2. In any catastrophic network (transitive tournament), there must be a city with outgoing roads to all other cities (a source), which is p... |
1. **Prouver que, dans tout réseau catastrophique, il y a une ville dont on ne peut sortir.**
Raisonnons par l'absurde. Supposons que dans toutes les villes, il y ait une route sortante. Considérons un touriste qui se balade de ville en ville en respectant le sens unique des routes. Vu qu'il y a 6 villes, ce touri... | 9 | true |
Given an acute triangle \( \triangle ABC \) with orthocenter \( H \), point \( E \) is chosen on segment \( CH \). Segment \( CH \) is extended to point \( F \) such that \( HF = CE \). Construct \( FD \perp BC \) at point \( D \) and \( EG \perp BH \) at point \( G \). Let \( M \) be the midpoint of segment \( CF \) a... | Okay, so I need to prove two things here. First, that points A, B, D, and G are concyclic, meaning they lie on the same circle. Second, that points O₁, O₂, M, and N are concyclic. Let me start by trying to visualize the problem and maybe draw a diagram in my mind.
We have an acute triangle ABC with orthocenter H. Sin... | Given an acute triangle \( \triangle ABC \) with orthocenter \( H \), point \( E \) is chosen on segment \( CH \). Segment \( CH \) is extended to point \( F \) such that \( HF = CE \). Construct \( FD \perp BC \) at point \( D \) and \( EG \perp BH \) at point \( G \). Let \( M \) be the midpoint of segment \( CF \) a... |
#### (1) Prove that points $A, B, D, G$ are concyclic:
1. **Constructing Points and Lines:**
- Given $\triangle ABC$ is a right triangle with right angle at $B$.
- $H$ is the orthocenter of $\triangle ABC$.
- Point $E$ is taken on $CH$, and $CH$ is extended to $F$ such that $HF = CE$.
2. **Drawing Perpe... | 9 | true |
2011 nonzero integers are given. It is known that the sum of any one of them with the product of the remaining 2010 numbers is negative. Prove that if these numbers are divided arbitrarily into two groups and the numbers in each group are multiplied, then the sum of the two resulting products will also be negative. | Okay, so I have this problem here where there are 2011 nonzero integers given. The condition is that if you take any one of them and add it to the product of the remaining 2010 numbers, the result is negative. And I need to prove that no matter how you split these numbers into two groups and multiply the numbers in eac... | Given 2011 nonzero integers, it is known that the sum of any one of them with the product of the remaining 2010 numbers is negative. We need to prove that if these numbers are divided arbitrarily into two groups and the numbers in each group are multiplied, then the sum of the two resulting products will also be negati... | Let's walk through the given problem step by step.
1. **Set Up the Problem**:
We are given 2011 non-zero integers. The condition is that the sum of any integer with the product of the remaining 2010 integers is negative.
2. **Understanding the Parity of Negative Numbers**:
Suppose among these 2011 integers, the... | 9 | true |
Ivan's birthday is attended by \(3n\) guests. Ivan has \(3n\) hats with the letters A, B, and C written on them, with \(n\) hats of each type. Ivan wants to organize a ball: he wants to put hats on the guests and arrange them into circles (one or more) so that the length of each circle is divisible by 3, and looking at... | Alright, so I need to prove that Ivan can organize the ball in exactly (3n)! different ways. Let me try to break this problem down step by step.
First, let's understand the problem statement again. There are 3n guests, each wearing a hat with a letter A, B, or C. There are n hats of each type. The guests need to be a... | To prove that Ivan can organize the ball in exactly \((3n)!\) different ways, we analyze the relationship between permutations of the guests and the assignment of hats with cycle constraints.
**Key Steps:**
1. **Cycle Structure Requirement:**
Guests must be arranged into cycles where each cycle's length is divis... |
1. **Arrange Guests into Ordered Triplets**:
- Given \(3n\) guests with \(n\) guests each wearing a cylindrical hat marked with 'A', 'B', and 'C'.
- Arrange the guests in a row, which can be done in \((3n)!\) ways.
- Group the first three guests as one triplet, the next three as another triplet, and so on.
... | 9 | true |
Given \( n \in \mathbf{Z}_{+}, a_{1}, a_{2}, \cdots, a_{n} \in \mathbf{C} \), and for any non-empty subset \( I \) of the set \(\{1,2, \cdots, n\} \), it holds that \( \left|\prod_{j \in I}\left(1+a_{j}\right)-1\right| \leqslant \frac{1}{2} \).
Prove that \( \left|a_{1}\right|+\left|a_{2}\right|+\cdots+\left|a_{n}\rig... | Okay, so I need to prove that the sum of the absolute values of these complex numbers a₁, a₂, ..., aₙ is at most 9/4, given that for every non-empty subset I of {1,2,...,n}, the absolute value of the product of (1 + a_j) for j in I minus 1 is less than or equal to 1/2. Hmm, let me try to unpack this step by step.
Firs... | To prove that \( |a_1| + |a_2| + \cdots + |a_n| \leqslant \frac{9}{4} \) under the given conditions, we analyze the constraints imposed by all non-empty subsets and leverage key inequalities.
**Key Steps:**
1. **Singleton Subsets:**
For each \( j \), the condition \( |(1 + a_j) - 1| \leqslant \frac{1}{2} \) impl... |
To prove that \(\left|a_{1}\right| + \left|a_{2}\right| + \cdots + \left|a_{n}\right| \leq \frac{9}{4}\) under the given conditions, we proceed as follows:
1. **Rewrite the Given Condition with Polynomial \(P(x)\)**:
Consider the polynomial
\[
P(x) = a_n x^n + \cdots + a_1 x + a_0.
\]
We are given tha... | 9 | true |
A function \( f(x) \) defined on \( \mathbf{R} \) satisfies:
1. \( f(0) = 0 \);
2. For any \( x, y \in (-\infty, -1) \cup (1, +\infty) \), \( f\left(\frac{1}{x}\right) + f\left(\frac{1}{y}\right) = f\left(\frac{x+y}{1+xy}\right) \);
3. For \( x \in (-1, 0) \), \( f(x) > 0 \).
Prove that
\[ f\left(\frac{1}{19}\right) +... |
Okay, so I've got this problem here about a function \( f(x) \) with some specific properties, and I need to prove an inequality involving a sum of \( f \) evaluated at certain points being greater than \( f\left(\frac{1}{2}\right) \). Let me try to unpack this step by step.
First, let me list out the given condition... | To solve the given problem, we need to analyze the function \( f(x) \) defined by the given properties and use them to establish the required inequality.
1. **Understanding the Function \( f(x) \)**:
- The function \( f(x) \) satisfies the functional equation \( f\left(\frac{1}{x}\right) + f\left(\frac{1}{y}\right)... |
1. **Start by analyzing the given conditions:**
- \( f(0) = 0 \)
- \( f\left( \frac{1}{x} \right) + f\left( \frac{1}{y} \right) = f\left( \frac{x+y}{1+xy} \right) \) for \( x, y \in (-\infty, -1) \cup (1, +\infty) \)
- \( f(x) > 0 \) for \( x \in (-1, 0) \)
2. **Evaluate the second condition:**
- Lett... | 9 | true |
Prove that for any value \(\alpha \leq 1\) and any numbers \(x_{1}, \ldots, x_{n}\) satisfying the conditions \(1 \geq x_{1} \geq x_{2} \geq \ldots \geq x_{n} > 0\), the following inequality holds:
\[
\left(1+x_{i}+x_{2}+\ldots+x_{n}\right)^{\alpha} \leq 1+1^{\alpha-1} x_{1}^{\alpha}+2^{\alpha-1} x_{2}^{\alpha}+\ldots... |
Okay, so I have this inequality to prove: For any α ≤ 1 and numbers x1, x2, ..., xn where 1 ≥ x1 ≥ x2 ≥ ... ≥ xn > 0, the inequality
(1 + x1 + x2 + ... + xn)^α ≤ 1 + 1^{α-1} x1^α + 2^{α-1} x2^α + ... + n^{α-1} xn^α
holds. Hmm. Let me try to understand the components here.
First, note that the x_i are ordered in dec... | To prove the inequality
\[
\left(1 + x_1 + x_2 + \ldots + x_n\right)^\alpha \leq 1 + 1^{\alpha-1} x_1^\alpha + 2^{\alpha-1} x_2^\alpha + \ldots + n^{\alpha-1} x_n^\alpha
\]
for any \(\alpha \leq 1\) and numbers \(x_1, x_2, \ldots, x_n\) satisfying \(1 \geq x_1 \geq x_2 \geq \ldots \geq x_n > 0\), we proceed as follow... | Let's prove the inequality by mathematical induction on \(n \in \mathbb{Z}^{+}\).
1. **Base Case**: When \(n=0\), the inequality becomes:
\[
\left(1\right)^{\alpha} \leqslant 1
\]
which is clearly true for any \(\alpha \leqslant 1\).
2. **Inductive Step**: Suppose the inequality holds for \(n\). That is,
... | 9 | true |
Given \( f_{a}(x, y) = x^{2} + axy + y^{2} \), where \( 0 \leq a \leq 2 \). For any \((x, y)\), define
$$
\overline{f_{a}}(x, y) = \min_{m, n \in \mathbb{Z}} f_{a}(x-m, y-n) .
$$
1. Prove that \(\overline{f_{1}}(x, y) < \frac{1}{2}\).
2. Prove that \(\overline{f_{1}}(x, y) \leq \frac{1}{3}\), and find all \((x, y) \) s... | Okay, so I need to solve these three problems related to the function \( f_a(x, y) = x^2 + axy + y^2 \) and its minimum over integer shifts, denoted as \( \overline{f_a}(x, y) \). Let me start by understanding what each part is asking.
**Problem 1**: Prove that \( \overline{f_{1}}(x, y) < \frac{1}{2} \).
Hmm. So for ... | Given the function \( f_{a}(x, y) = x^{2} + axy + y^{2} \) where \( 0 \leq a \leq 2 \), and the definition \( \overline{f_{a}}(x, y) = \min_{m, n \in \mathbb{Z}} f_{a}(x-m, y-n) \), we address the following problems:
1. **Prove that \(\overline{f_{1}}(x, y) < \frac{1}{2}\).**
We show that for any real numbers \( x... |
#### Given:
\[ f_{a}(x, y) = x^{2} + a x y + y^{2} \]
where \( 0 \leqslant a \leqslant 2 \).
#### Definition:
\[ \overline{f_{a}}(x, y) = \min_{m, n \in \mathbb{Z}} f_{a}(x-m, y-n) \]
#### We need to find:
1. To prove \( \overline{f_{1}}(x, y) < \frac{1}{2} \)
2. To prove \( \overline{f_{1}}(x, y) \leqslant \frac{1}... | 10 | true |
Find the smallest \( n \) such that any convex 100-sided polygon can be obtained as the intersection of \( n \) triangles. Prove that for smaller \( n \), this cannot be done with every convex 100-sided polygon. | Okay, so I need to find the smallest number n such that any convex 100-sided polygon can be obtained as the intersection of n triangles. Also, I have to prove that for any smaller number than this n, it's not possible. Hmm, interesting problem. Let me try to break it down.
First, let me recall some concepts. A convex ... | To find the smallest \( n \) such that any convex 100-sided polygon can be obtained as the intersection of \( n \) triangles, we need to consider how each triangle can contribute to forming the edges of the polygon.
1. **Key Insight**: Each triangle can cover at most two adjacent edges of the polygon. This is because... |
We need to determine the smallest value of \( n \) such that any convex 100-gon can be formed as the intersection of \( n \) triangles.
1. **Understanding the Problem:**
- We need to form a convex 100-gon using intersections of triangles.
- The goal is to find the smallest \( n \) that ensures any convex 100-go... | 9 | true |
Prove that if \(0<\alpha<\beta<\frac{\pi}{2}\), then
\[
\frac{\sin ^{3} \alpha}{2 \alpha-\sin 2 \alpha} > \frac{\sin ^{2} \beta}{2 \beta-\sin 2 \beta}
\] | Okay, let me try to figure out how to prove this inequality. So the problem states that if \(0 < \alpha < \beta < \frac{\pi}{2}\), then
\[
\frac{\sin ^{3} \alpha}{2 \alpha - \sin 2 \alpha} > \frac{\sin ^{2} \beta}{2 \beta - \sin 2 \beta}
\]
Hmm, so we have two variables, α and β, both in the interval (0, π/2), with α... | Given \(0 < \alpha < \beta < \frac{\pi}{2}\), we need to prove that:
\[
\frac{\sin^3 \alpha}{2 \alpha - \sin 2 \alpha} > \frac{\sin^2 \beta}{2 \beta - \sin 2 \beta}
\]
First, we cross-multiply to obtain the inequality in a different form:
\[
\sin^3 \alpha (2 \beta - \sin 2 \beta) > \sin^2 \beta (2 \alpha - \sin 2 \a... |
We are given that \(0 < \alpha < \beta < \frac{\pi}{2}\) and need to prove the inequality
$$
\frac{\sin^3 \alpha}{2\alpha - \sin 2\alpha} > \frac{\sin^2 \beta}{2\beta - \sin 2\beta}.
$$
1. **Express the inequality in a more manageable form**:
We begin by considering the quantities on both sides of the inequality s... | 9 | true |
Let \( M=\left\{z \in \mathbb{C} \,\middle\lvert\, \sum_{k=1}^{n} \frac{1}{|z-k|} \geq 1\right\} \). In the complex plane, the area of the corresponding region \(\Omega\) is \( S \). Prove that:
$$
S \geq \frac{\left(11 n^{2}+1\right) \pi}{12} .
$$ | Okay, so I need to prove that the area \( S \) of the region \( \Omega = \left\{ z \in \mathbb{C} \,\middle|\, \sum_{k=1}^{n} \frac{1}{|z - k|} \geq 1 \right\} \) is at least \( \frac{(11n^2 + 1)\pi}{12} \). Hmm, let's start by trying to understand what this set \( \Omega \) looks like in the complex plane.
First, eac... | To prove that the area \( S \) of the region \( \Omega = \left\{ z \in \mathbb{C} \,\middle|\, \sum_{k=1}^{n} \frac{1}{|z - k|} \geq 1 \right\} \) is at least \( \frac{(11n^2 + 1)\pi}{12} \), we consider the following key steps:
1. **Understanding the Region \( \Omega \)**:
- The region \( \Omega \) consists of all... |
1. According to the Hölder inequality, we have:
\[
\left(\sum_{k=1}^{n} \frac{1}{|z - k|} \right)^2 \cdot \left(\sum_{k=1}^{n} |z - k|^2 \right) \geq n^3
\]
2. Therefore,
\[
\sum_{k=1}^{n} \frac{1}{|z - k|} \geq \sqrt{\frac{n^3}{\sum_{k=1}^{n} |z - k|^2}}
\]
3. Let's define the set
\[
... | 9 | true |
Let \( k, l, m, n \) be natural numbers such that \( k < l < m < n \) and \( lm = kn \). Prove that:
\[ \left(\frac{n - k}{2}\right)^{2} \geq k + 2 \] | Alright, so I need to prove that \(\left(\frac{n - k}{2}\right)^2 \geq k + 2\) given that \(k, l, m, n\) are natural numbers with \(k < l < m < n\) and \(lm = kn\). Hmm, let's see how I can approach this.
First, let's digest the conditions. We have four natural numbers in ascending order: \(k < l < m < n\). The key eq... | Given natural numbers \(k, l, m, n\) such that \(k < l < m < n\) and \(lm = kn\), we need to prove that:
\[
\left(\frac{n - k}{2}\right)^2 \geq k + 2
\]
First, we note that \(lm = kn\) implies \(n = \frac{lm}{k}\). Since \(n\) must be a natural number, \(k\) must divide \(lm\). We consider the minimal values of \(l\) ... |
To prove the inequality \(\left(\frac{n-k}{2}\right)^{2} \geq k+2\), we start by defining the variables appropriately based on the given conditions. Here are the steps to show the proof in detail:
1. **Define Variables:**
Let \( l = k + a \) and \( m = k + b \), where \( a \) and \( b \) are natural numbers (\(a, ... | 9 | true |
Given the sequence \(\left\{a_{n}\right\}\) defined by \(a_{n}=\sum_{i=1}^{n} \frac{1}{i!}\), and another sequence \(\left\{b_{n}\right\}\) defined by \(b_{n}=(n+1)!a_{n}\), prove that:
\[ \prod_{i=1}^{2023}\left(1+\frac{1}{b_{i}}\right)<\frac{7}{4} \] | Okay, so I have this problem here. It says that there's a sequence {a_n} defined by the sum from i=1 to n of 1/i!, and another sequence {b_n} where b_n = (n+1)! * a_n. I need to prove that the product from i=1 to 2023 of (1 + 1/b_i) is less than 7/4. Hmm, okay. Let's start by understanding the sequences first.
First, ... | Given the sequence \(\{a_n\}\) defined by \(a_n = \sum_{i=1}^{n} \frac{1}{i!}\), and another sequence \(\{b_n\}\) defined by \(b_n = (n+1)! \cdot a_n\), we need to prove that:
\[
\prod_{i=1}^{2023}\left(1+\frac{1}{b_{i}}\right) < \frac{7}{4}
\]
First, we recognize that the product \(\prod_{i=1}^{2023}\left(1+\frac{1}... | 1. First, let's verify the definition of \(b_n\):
\[
b_n = (n+1)!a_n = (n+1)!\left( \sum_{i=1}^{n} \frac{1}{i!} \right)
\]
2. Rewriting \(a_n\) in terms of \(a_{n-1}\), we have:
\[
a_n = a_{n-1} + \frac{1}{n!}
\]
Therefore,
\[
b_n = (n+1)!\left( a_{n-1} + \frac{1}{n!} \right) = (n+1)... | 9 | true |
Two players are playing the following game. Each player takes turns crossing out 9 numbers (of their choice) from the sequence \(1, 2, 3, \ldots, 100, 101\). After eleven such crossings out, two numbers will remain. Then the second player awards the first player as many points as the difference between these remaining ... | Okay, so I need to prove that the first player can always score at least 55 points in this game, no matter how the second player plays. Let me try to understand the game first.
We have the numbers from 1 to 101. Each player takes turns crossing out 9 numbers. After eleven turns, two numbers will remain. The second pla... | To prove that the first player can always score at least 55 points, regardless of how the second player plays, we can use the following strategy:
1. **Initial Move by the First Player**:
- The first player removes the 9 middle numbers (47 to 55) from the sequence \(1, 2, \ldots, 101\). This leaves two groups of num... |
1. **Understand the problem setup**:
- We are given a sequence of numbers from 1 to 101.
- Two players alternately cross out 9 numbers each turn.
- After 11 turns (11 cuts), \(33\) numbers will have been crossed out, leaving 2 remaining numbers from the original \(101\).
2. **Determine the remaining numbers*... | 9 | true |
Given a finite set of non-zero vectors \( U \) in the plane, \( l(U) \) is defined as the length of the sum vector of all vectors in \( U \). Let \( V \) be a finite set of non-zero vectors in the plane. If for every non-empty subset \( A \) of \( V \), \( l(B) \geq l(A) \) for all subsets \( B \) of \( A \), then \( B... | Okay, so I need to solve part (a) and part (b) of this problem. Let me start with part (a). The task is to construct sets of 4 and 5 vectors in the plane such that the number of maximal subsets is 8 and 10 respectively.
First, let me make sure I understand the definitions correctly.
A maximal subset B of A is a sub... | Given a finite set of non-zero vectors \( V \) in the plane, a subset \( B \) is called maximal if for every non-empty subset \( A \) of \( V \), the length of the sum vector of \( B \) is at least the length of the sum vector of \( A \) for all subsets \( B \) of \( A \).
### Part (a)
To construct sets of 4 and 5 ve... |
**(a) Constructing sets of 4 and 5 vectors that respectively have 8 and 10 maximal subsets:**
1. **For \( n = 4 \):**
- Consider a quadrilateral \( ABCD \) with sides \( AB, BC, CD, \) and \( DA \).
- Define vectors \( \overrightarrow{AB}, \overrightarrow{BC}, \overrightarrow{CD}, \overrightarrow{DA} \).
... | 10 | true |
In a regular $n$-sided polygon with center $O$, each vertex holds either the number +1 or -1. It is allowed to simultaneously change the signs of numbers at the vertices of any regular $k$-sided polygon (with center $O$ and $k$ possibly equal to 2).
1. Prove that when $n=15$, there exists an initial arrangement of +1 a... | Alright, let me try to tackle these problems one by one. They all seem related to regular polygons and flipping signs based on certain operations. Let's start with the first question about n=15. The goal is to show there's an initial arrangement where you can't turn all numbers to +1 no matter how many operations you p... | To address the problems systematically, we use linear algebra over \( \text{GF}(2) \) and invariant theory.
**1. For \( n = 15 \):**
- **Operations:** Flipping signs of regular \( k \)-gons (\( k = 3, 5, 15 \)).
- **Vector Space Model:** Each configuration is a vector in \( \text{GF}(2)^{15} \). Operations correspond ... |
(1) When $n = 15$, we need to prove the existence of an initial arrangement of +1 and -1 such that no matter how many times we perform the allowed sign-changing operation, it is impossible to make all vertex numbers +1.
1. **Understanding the Problem**: We have a regular 15-sided polygon, and each vertex has a number... | 10 | true |
Given 3000 lines on a plane, where no two lines are parallel and no three lines are concurrent, these lines divide the plane into several polygonal regions. Prove that among these polygonal regions, (a) there are at least 1000 triangles; (b) there are at least 2000 triangles. | Okay, so I have this problem with 3000 lines on a plane. No two are parallel, and no three meet at the same point. They divide the plane into polygonal regions, and I need to prove that there are at least 1000 triangles in part (a), and at least 2000 triangles in part (b). Hmm, let's start with part (a).
First, I reme... | To determine the minimum number of triangular regions formed by 3000 lines under the given conditions, we analyze the incremental addition of each line and its contribution to forming new triangles.
**Key Steps:**
1. **Base Case:**
- For \( n = 3 \) lines, there is exactly 1 triangular region.
2. **Inductive St... |
1. **Understanding the problem conditions**:
- We are given 3000 lines on a plane.
- No two lines are parallel.
- No three lines are concurrent (do not meet at a single point).
2. **Insight for part (a)**:
- For each line $l$, consider it intersects other lines at various points. Among these intersection ... | 9 | true |
The line \( l \) is external to the circle \( \odot O \). \( OP \perp l \), and \( P \) is a point on \( l \). \( Q \) is another moving point on \( l \), different from \( P \). \( QA \) and \( QB \) are the two tangents from \( Q \) to the circle \( \odot O \). \( PM \) and \( PN \) are perpendicular to \( QA \) and ... |
Okay, so I need to prove that the point K, where MN intersects OP, is a fixed point, regardless of where Q is on line l. Let me try to visualize this setup first. There's a circle with center O, and a line l that's external to the circle, meaning the distance from O to l is greater than the radius of the circle. OP is... | To prove that the point \( K \) is a fixed point, we start by setting up a coordinate system where \( P \) is at the origin \((0,0)\) and \( O \) is at \((0, h)\) with \( h > r \), the radius of the circle. The line \( l \) is the x-axis, and \( Q \) is a point \((q, 0)\) on \( l \).
The equations of the tangents from... |
1. **Identify Key Points and Lines**
- We are given that \( l \) is a line external to the circle \( \odot O \).
- \( O P \) is perpendicular to \( l \) and \( P \) is on \( l \).
- \( Q \) is a moving point on \( l \) different from \( P \).
- \( Q A \) and \( Q B \) are two secant lines of the circle \( ... | 9 | true |
A set of points in the plane is called obtuse if any three points from this set always form the vertices of an obtuse triangle. Prove that any set of \( n \) points in the plane, with no three points ever being collinear, contains an obtuse subset of at least \( \sqrt{n} \) elements. |
Alright, so I have this problem: given a set of n points in the plane, no three collinear, I need to prove that there's an obtuse subset of at least √n elements. An obtuse set is one where any three points form an obtuse triangle. Hmm, interesting. Let me try to break this down.
First off, I need to recall what makes... | To prove that any set of \( n \) points in the plane, with no three points collinear, contains an obtuse subset of at least \( \sqrt{n} \) elements, we can use geometric and combinatorial arguments.
An obtuse set is defined as a set where any three points form an obtuse triangle. A key observation is that a set of po... |
1. **Preparation and Setup**:
- Consider an arbitrary set of $n$ points in the plane, where no three points are collinear. Denote these points as $\{(x_1, y_1), (x_2, y_2), \ldots, (x_n, y_n)\}$.
- Without loss of generality, we assume that the points are sorted by their $x$-coordinates such that $x_1 < x_2 < \l... | 9 | true |
Prove that
$$
\sum_{k=n}^{2 n}\binom{m}{k}\binom{k}{2 n-k} \cdot 2^{2 k-2 n}=\binom{2 m}{2 n}
$$ |
Okay, so I need to prove that the sum from k = n to 2n of binomial coefficients times some power of 2 equals another binomial coefficient. The given identity is:
$$
\sum_{k=n}^{2 n}\binom{m}{k}\binom{k}{2 n-k} \cdot 2^{2 k-2 n}=\binom{2 m}{2 n}
$$
Hmm, I need to show that the left-hand side (LHS) simplifies to the r... | To prove the identity
$$
\sum_{k=n}^{2 n}\binom{m}{k}\binom{k}{2 n-k} \cdot 2^{2 k-2 n}=\binom{2 m}{2 n},
$$
we start by analyzing the left-hand side (LHS). We recognize that each term involves binomial coefficients and a power of 2. We aim to relate this sum to the right-hand side (RHS) using generating functions.
... |
To prove the identity
$$
\sum_{k=n}^{2n} \binom{m}{k} \binom{k}{2n-k} \cdot 2^{2k-2n} = \binom{2m}{2n},
$$
we will use a combinatorial argument. We will consider the different ways to choose $2n$ gloves out of a box containing $m$ pairs of distinct gloves.
1. **Direct Counting of Combinations**:
The number of w... | 9 | true |
In triangle \( \triangle ABC \), the excircle opposite to vertex \( A \), denoted as \( \odot O \), is tangent to sides \( AB \), \( BC \), and \( CA \) at points \( D \), \( E \), and \( F \) respectively. The line segment \( EZ \) is the diameter of \( \odot O \). Perpendicular lines are drawn from points \( B \) and... | Alright, let's tackle this geometry problem step by step. First, I need to make sure I understand all the given information and what needs to be proven. The problem is about triangle ABC with an ex-circle opposite vertex A, tangent to AB, BC, and CA at points D, E, and F respectively. The excircle is denoted as ⊙O. The... | To prove that points \(X\), \(H\), \(K\), and \(Y\) are concyclic in triangle \( \triangle ABC \) with the given configuration, we follow these key steps:
1. **Understanding the Configuration**:
- The excircle opposite to vertex \(A\) (denoted as \( \odot O \)) is tangent to \(AB\), \(BC\), and \(CA\) at points \(D... |
1. **Setting up points of intersection:**
- Let $XH$ and $YH$ intersect the sides of $\triangle XYZ$ at points $P$ and $Q$ respectively.
- Define $N$ to be the intersection of line $DF$ with line $BC$.
2. **Using Menelaus' Theorem:**
- For line $NDF$ intersecting $\triangle ABC$,
$$
\frac{AD}{DB} ... | 9 | true |
a) The image on the left shows six points that lie three on each of four lines. Prove that there are 24 different ways to map this set of six points onto itself such that each group of three points lying on one line is mapped into another group of three points lying on a single line.
b) The image on the right shows n... |
Okay, let me try to tackle these problems. Starting with part a).
**Problem a:** There's a configuration of six points arranged such that there are four lines, each containing three points. We need to prove that there are 24 different automorphisms (symmetries) that map the set onto itself, preserving the line struct... | ### Summary
**Problem a:**
The configuration described is a complete quadrilateral, which consists of four lines each intersecting in six points. Each line contains three points, and each point lies on two lines. The automorphism group of this configuration is isomorphic to the symmetric group \( S_4 \), which has \(... |
**(a)**
To determine the number of ways to map the set of six points onto itself such that each mapping preserves the collinearity of points on the same line, note that each of the four lines in the configuration must be mapped to one of the four lines, and this will determine the images of the points. Each line can... | 9 | true |
A sphere is inscribed in a regular cone. Around the sphere a cylinder is circumscribed so that its base is in the same plane as the base of the cone. Let \( V_{1} \) be the volume of the cone and \( V_{2} \) the volume of the cylinder.
(a) Prove that \( V_{1} = V_{2} \) is impossible.
(b) Find the smallest \( k \) for ... |
Alright, so I have this geometry problem here, and I need to tackle parts (a) and (b). Let me start with part (a), which says to prove that the volume of the cone \( V_1 \) can't be equal to the volume of the cylinder \( V_2 \). Hmm, okay. Let me visualize the setup first.
There's a regular cone, which I assume is a ... | To solve the problem, we start by considering a sphere inscribed in a regular cone and a cylinder circumscribed around the sphere. We need to analyze the volumes of the cone and the cylinder.
### Part (a)
1. **Volume of the Cylinder (\(V_2\))**:
- The cylinder circumscribed around the sphere has a radius equal to ... |
#### Part (a): Proving that \( V_{1} = V_{2} \) is impossible
1. **Define the geometric parameters:**
- Let \( A \) be the vertex of the cone.
- Let \( O \) be the center of the inscribed sphere.
- Let \( S \) be the center of the base of the cone.
- Let \( B \) be a point on the base circle.
- Let \( ... | 9 | true |
The diagonals of a convex quadrilateral \(ABCD\) are perpendicular and intersect at point \(O\). It is known that the sum of the radii of the circles inscribed in triangles \(AOB\) and \(COD\) is equal to the sum of the radii of the circles inscribed in triangles \(BOC\) and \(DOA\). Prove that:
a) The quadrilateral \... | Okay, so I need to prove parts a) and b) for this convex quadrilateral with perpendicular diagonals. Let me start by understanding the problem.
First, the quadrilateral ABCD has perpendicular diagonals AC and BD intersecting at O. The given condition is that the sum of the inradii of triangles AOB and COD is equal to ... | Given a convex quadrilateral \(ABCD\) with perpendicular diagonals intersecting at point \(O\), we need to prove that:
a) The quadrilateral \(ABCD\) is circumscribed.
b) The quadrilateral \(ABCD\) is symmetric with respect to one of its diagonals.
### Proof:
#### Part a)
1. **Inradius Expression**:
- For right-... | Consider the concave quadrilateral $ABCD$ with perpendicular diagonals $AC$ and $BD$ intersecting at point $O$. Let $AB = a$, $BC = b$, $CD = c$, $DA = d$, $AO = u$, $BO = x$, $CO = v$, and $DO = y$.
#### Part (a):
To prove that quadrilateral $ABCD$ is cyclic:
1. **Given Information**:
- The diagonals are perpendi... | 9 | true |
In a given circle $O$, there are $2n$ smaller circles, each tangent to its adjacent two smaller circles and also internally tangent to circle $O$. The points of tangency in sequence are $A_1, A_2, A_3, \ldots, A_n, \ldots, A_{2n}$. For these $2n$ points of tangency, if the distance between any two adjacent points $A_i$... |
Okay, let's try to tackle this geometry problem. So we have a circle O with 2n smaller circles inside it, each tangent to their adjacent two small circles and also tangent to the big circle O. The points where the small circles touch the big circle are labeled A1, A2, ..., A2n in sequence. The distance between consecu... | Given a circle \( O \) with \( 2n \) smaller circles inscribed such that each smaller circle is tangent to its adjacent circles and internally tangent to \( O \). The points of tangency on \( O \) are \( A_1, A_2, \ldots, A_{2n} \). The distances between adjacent points \( A_i \) and \( A_{i+1} \) are \( a_{i,i+1} \), ... |
1. Let the radius of the circle \( O \) be \( R \) and the radius of the \( 2n \) small circles \( O_i \) be \( R_i \) for \( i = 1, 2, \cdots, 2n \).
2. The angle between two adjacent cutting points \( A_1 \) and \( A_2 \) is denoted as \( \theta \).
3. Given that \( A_1A_2 = a_{1,2} \), we can express this in term... | 9 | true |
Point \( O \) is the center of the circle circumscribed around triangle \( ABC \). Points \( A_1 \), \( B_1 \), and \( C_1 \) are symmetric to point \( O \) relative to the sides of triangle \( ABC \). Prove that all the altitudes of triangle \( A_1 B_1 C_1 \) pass through point \( O \), and that all the altitudes of t... | Alright, let's try to tackle this geometry problem step by step. The problem is about triangle \( ABC \) with circumcenter \( O \). Points \( A_1 \), \( B_1 \), and \( C_1 \) are the reflections of \( O \) over the sides of \( ABC \). We need to show two things:
1. All the altitudes of triangle \( A_1B_1C_1 \) pass th... | To solve the problem, we need to show two things:
1. All the altitudes of triangle \( A_1B_1C_1 \) pass through point \( O \).
2. All the altitudes of triangle \( ABC \) pass through the circumcenter of triangle \( A_1B_1C_1 \).
### Key Steps and Proof:
1. **Reflecting the Circumcenter**:
- Points \( A_1 \), \( B... |
1. **Define Special Points and Configurations:**
Let \(A_{2}\), \(B_{2}\), and \(C_{2}\) be the midpoints of the sides \(CB\), \(BA\), and \(CA\) respectively. By the definition of midpoints in a triangle:
- \(A_{2}\) is the midpoint of \(CB\),
- \(B_{2}\) is the midpoint of \(CA\),
- \(C_{2}\) is the mid... | 9 | true |
Prove that for $k > 1$, Bernoulli numbers satisfy the equality $\sum_{p=0}^{k} B_{p} C_{k}^{p} = B_{k}$, that is, $\sum_{p=0}^{k-1} B_{p} C_{k}^{p} = 0$. | Okay, so I need to prove that for k > 1, the sum from p=0 to k of B_p times C(k,p) equals B_k, which is equivalent to showing that the sum from p=0 to k-1 of B_p times C(k,p) is zero. Hmm, Bernoulli numbers... I remember they are a sequence of rational numbers with deep connections to number theory and analysis, especi... | To prove that for \( k > 1 \), Bernoulli numbers satisfy the equality \(\sum_{p=0}^{k} B_{p} \binom{k}{p} = B_{k}\), we can use the standard recurrence relation for Bernoulli numbers.
The standard recurrence relation for Bernoulli numbers is given by:
\[
\sum_{p=0}^{n} \binom{n+1}{p} B_{p} = 0 \quad \text{for } n \ge... |
1. **Introduction to Bernoulli Numbers**:
The Bernoulli numbers \(B_n\) are a sequence of rational numbers with deep connections to number theory. They appear in the expansion of the generating function:
\[
\frac{x}{e^{x} - 1} = \sum_{n=0}^{\infty} B_n \frac{x^n}{n!}
\]
By multiplying both sides of th... | 9 | true |
Let \((\alpha: \beta: \gamma)\) be the absolute barycentric coordinates of point \(X\), and let \(M\) be the centroid of triangle \(ABC\). Prove that \(3 \overrightarrow{XM}= (\alpha - \beta) \overrightarrow{AB} + (\beta - \gamma) \overrightarrow{BC} + (\gamma - \alpha) \overrightarrow{CA}\). | Okay, so I need to prove this vector equation involving barycentric coordinates and the centroid. Let me start by recalling some basics. Barycentric coordinates are a way to represent a point in a triangle relative to its vertices. The centroid M has coordinates (1/3 : 1/3 : 1/3) because it's the average of the three v... | To prove that \(3 \overrightarrow{XM} = (\alpha - \beta) \overrightarrow{AB} + (\beta - \gamma) \overrightarrow{BC} + (\gamma - \alpha) \overrightarrow{CA}\), we proceed as follows:
**Step 1: Express \(X\) and \(M\) in barycentric coordinates.**
Since \(X\) has absolute barycentric coordinates \((\alpha : \beta : \g... |
Let's prove the given statement:
\[ 3 \overrightarrow{X M} = (\alpha - \beta) \overrightarrow{A B} + (\beta - \gamma) \overrightarrow{B C} + (\gamma - \alpha) \overrightarrow{C A} \]
where \((\alpha : \beta : \gamma)\) are the absolute barycentric coordinates of the point \(X\), and \(M\) is the centroid of triangl... | 9 | true |
Given a triangle \( \triangle ABC \) with side lengths \( a, b, c \) and area \( S \), prove the following inequalities:
$$
\begin{array}{l}
a^{2}+b^{2}+c^{2} \geqslant 4 \sqrt{\frac{a^{2}}{b^{2}}+\frac{b^{2}}{c^{2}}+\frac{c^{2}}{a^{2}}} S, \\
(a+b+c)^{2} \geqslant 4 \sqrt{3}\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ri... | Okay, so I have these four inequalities to prove for a triangle ABC with sides a, b, c and area S. Hmm, they all involve combinations of the sides and the area, and each one seems to have a square root term with some fractions or trigonometric functions. Let me try to tackle them one by one. Maybe I can use some known ... | To prove the given inequalities for a triangle \( \triangle ABC \) with side lengths \( a, b, c \) and area \( S \), we start by analyzing each inequality individually and testing them for specific cases like equilateral and right-angled triangles. Here are the key steps and conclusions:
1. **First Inequality: \( a^{2... | To prove the given inequalities, we will go through each inequality step-by-step.
#### 1. Proving inequality (1)
We start with inequality:
$$a^2 + b^2 + c^2 \geq 4 \sqrt{\sum \frac{a^2}{b^2}} S.$$
Here, \( S \) represents the area of \( \triangle ABC \).
Let \(c = \min(a, b, c)\).
From the known relation:
$$16 S^2 ... | 9 | true |
Given the numbers \(x\) and \(y\) satisfying the condition \(x^{8} + y^{8} \leqslant 1\), prove the inequality \(x^{12} - y^{12} + 2x^{6}y^{6} \leqslant \frac{\pi}{2}\). | Okay, so I have this problem here where I need to prove that for numbers x and y satisfying x⁸ + y⁸ ≤ 1, the expression x¹² - y¹² + 2x⁶y⁶ is less than or equal to π/2. Hmm. Alright, let's start breaking this down.
First off, the condition given is x⁸ + y⁸ ≤ 1. That reminds me of the unit circle equation, but with high... | Given the numbers \(x\) and \(y\) satisfying \(x^8 + y^8 \leq 1\), we need to prove the inequality \(x^{12} - y^{12} + 2x^6y^6 \leq \frac{\pi}{2}\).
First, we note that the condition \(x^8 + y^8 \leq 1\) suggests using trigonometric parameterization. We set \(x = \cos^{1/4} \theta\) and \(y = \sin^{1/4} \theta\), whic... |
We are given the condition \( x^8 + y^8 \leqslant 1 \) and we need to prove the inequality \( x^{12} - y^{12} + 2 x^{6} y^{6} \leqslant \frac{\pi}{2} \).
**Solution 1:**
1. **Consider non-negative values of \( x \) and \( y \)**: Because both sides of the inequality maintain their relationship when \( x \) and \( y ... | 9 | true |
In triangle \(ABC\) with sides \(AB = 5\), \(BC = 6\), and \(AC = 4\), the angle bisector \(AA_1\) of angle \(BAC\) is drawn. Then, in triangle \(AA_1C\), the angle bisector \(A_1C_1\) of angle \(AA_1C\) is drawn. In triangle \(C_1A_1C\), the angle bisector \(C_1A_2\) of angle \(A_1C_1C\) is drawn. Continuing this proc... | Okay, let's try to tackle this problem. So, we have triangle ABC with sides AB=5, BC=6, and AC=4. The process involves repeatedly drawing angle bisectors, specifically 2021 times, and then we need to prove that the final triangle is similar to ABC and find the similarity coefficient. Hmm, this seems complex because of ... | To prove that triangles \(ABC\) and \(A_{2021}C_{2021}C\) are similar and find the similarity coefficient, we analyze the process of repeatedly drawing angle bisectors in the triangles formed at each step.
1. **First Angle Bisector**: In triangle \(ABC\), the angle bisector \(AA_1\) divides \(BC\) into segments \(BA_... |
1. **Understand the Problem Statement**:
We have a triangle \(ABC\) with sides \(AB = 5\), \(BC = 6\), and \(AC = 4\). We construct a sequence of bisectors within various sub-triangles formed through the process.
2. **Construct the First Bisector**:
The first bisector is \(AA_1\) of \(\angle BAC\). By the Angle... | 9 | true |
Perpendiculars drawn through the midpoints of sides \(AB\) and \(AC\) of triangle \(ABC\) intersect line \(BC\) at points \(X\) and \(Y\) respectively. Prove that the equality \(BC = XY\):
a) holds if
\[
\operatorname{tg} \angle B \cdot \operatorname{tg} \angle C = 3
\]
b) can also hold under the condition
\[
\oper... |
Alright, so I need to solve this geometry problem about triangle ABC. The problem states that if we draw perpendiculars through the midpoints of sides AB and AC, they intersect line BC at points X and Y respectively. We have to prove two things: part a) that BC = XY if tan(angle B) * tan(angle C) = 3, and part b) that... | Given triangle \(ABC\) with midpoints \(M_1\) and \(M_2\) of sides \(AB\) and \(AC\) respectively, perpendiculars drawn through these midpoints intersect line \(BC\) at points \(X\) and \(Y\). We need to prove the conditions under which \(BC = XY\) and find the set \(M\) for which this equality is equivalent to \(\oper... |
Given a triangle \(ABC\) with \( \angle A = \alpha \), \( \angle B = \beta \), \( \angle C = \gamma \) and sides \(a = BC\), \(b = AC\), and \(c = AB\), and \(R\) as the radius of the circumscribed circle. We need to prove that \(BC = XY\) under certain conditions.
First, note that \(BC = XY\) if and only if \(\overr... | 9 | true |
Point \( D \) is on side \( AC \) of \( \triangle ABC \), and \( AD:DC = 2:1 \), \(\angle C = 45^\circ\), \(\angle ADB = 60^\circ\). Prove that \( AB \) is tangent to the circumcircle of \( \triangle BCD \). |
Okay, so I need to prove that AB is tangent to the circumcircle of triangle BCD. Let me start by sketching the problem to visualize what's going on. Maybe drawing triangle ABC, with point D on AC such that AD:DC is 2:1. Then angle C is 45 degrees, and angle ADB is 60 degrees. Hmm, so let's label all those parts.
Fir... | To prove that \( AB \) is tangent to the circumcircle of \( \triangle BCD \), we can use the Power of a Point theorem.
1. **Power of a Point Theorem**: If a line from point \( A \) intersects the circumcircle of \( \triangle BCD \) at points \( C \) and \( D \), then the power of point \( A \) with respect to this ci... | ## Solution:
1. Given that $D$ is a point on side $AC$ of $\triangle ABC$ such that $AD:DC=2:1$ and with $\angle C = 45^\circ$ and $\angle ADB = 60^\circ$, we need to show that $AB$ is the tangent to the circumcircle of $\triangle BCD$.
2. From the given ratio $AD:DC = 2:1$, we can use the Apollonius's theorem to find... | 9 | true |
Let \( p \) be a prime number such that \( p = a^{2} + b^{2} \). Prove that one of \( \pm a, \pm b \) is congruent to \( \frac{1}{2}\binom{\left(\frac{p-1}{2}\right)}{\frac{p-1}{4}} \bmod{p} \). | Okay, so I need to prove that if a prime number p can be written as the sum of two squares, then one of ±a or ±b is congruent to (1/2) times the binomial coefficient ((p-1)/2 choose (p-1)/4) modulo p. Hmm, this seems related to some properties of primes and quadratic residues. Let me start by recalling some relevant th... | Given a prime number \( p \) such that \( p = a^2 + b^2 \), we need to prove that one of \( \pm a, \pm b \) is congruent to \( \frac{1}{2} \binom{\frac{p-1}{2}}{\frac{p-1}{4}} \bmod{p} \).
1. **Fermat's Theorem on Sums of Two Squares**: Primes congruent to \( 1 \mod 4 \) can be expressed as the sum of two squares. Hen... |
1. **Start with the given prime number $p$ written as $p = a^2 + b^2$.**
Given the prime $p = a^2 + b^2$, let's denote two solutions as follows:
\[
p = c^2 + d^2
\]
where $a$ and $b$ are integers.
2. **Establish that $p$ divides the product \((ad + bc)(ad - bc)\).**
\[
p \mid (ad + bc)(a... | 9 | true |
Three circles \( c_{1}, c_{2}, c_{3} \) are located in a plane with centers \( C_{1}, C_{2}, C_{3} \) and radii \( r_{1}, r_{2}, r_{3} \) respectively. Each circle lies outside the other two, with \( r_{1} > r_{2} \) and \( r_{1} > r_{3} \). Let \( A \) be the point of intersection of the external tangents to circles \... | Alright, so I need to prove that the quadrilateral formed by the tangents from points A and B to circles c3 and c2 respectively has an incircle, and then find the radius of that incircle. Hmm, okay. Let me start by understanding the problem step by step.
First, the setup: there are three circles, each lying outside th... | To prove that the quadrilateral formed by the tangents from points \( A \) and \( B \) to circles \( c_3 \) and \( c_2 \) respectively has an incircle, and to calculate the radius of this incircle, we proceed as follows:
1. **Understanding the Configuration**:
- Three circles \( c_1, c_2, c_3 \) with centers \( C_1... |
1. **Identification of Problem Parameters**:
- We have three circles, \( c_1, c_2, c_3 \), with centers \( C_1, C_2, C_3 \) and respective radii \( r_1, r_2, r_3 \).
- Given that \( r_1 > r_2 \) and \( r_1 > r_3 \).
2. **Definition of Tangent Points**:
- \( A \) is the external intersection point of the t... | 9 | true |
Given a tetrahedron with six edge lengths \(a, b, c, d, e, f\) and surface area \(S\), prove that \(S \leq \frac{\sqrt{3}}{6}\left(a^{2}+b^{2}+c^{2}+d^{2}+e^{2}+f^{2}\right)\). | Okay, so I have this problem about a tetrahedron with six edge lengths a, b, c, d, e, f and surface area S. I need to prove that S is less than or equal to (sqrt(3)/6) times the sum of the squares of all the edges. Hmm, let me try to figure this out step by step.
First, let's recall what a tetrahedron is. A tetrahedro... | To prove that the surface area \( S \) of a tetrahedron with edge lengths \( a, b, c, d, e, f \) satisfies \( S \leq \frac{\sqrt{3}}{6}\left(a^{2}+b^{2}+c^{2}+d^{2}+e^{2}+f^{2}\right) \), we can proceed as follows:
1. **Area of a Triangle Inequality**: For any triangle with sides \( x, y, z \), the area \( A \) is bou... |
To prove that for a given tetrahedron with edge lengths \(a, b, c, d, e,\) and \(f\) and surface area \(S\), the inequality
\[ S \leq \frac{\sqrt{3}}{6}\left(a^{2}+b^{2}+c^{2}+d^{2}+e^{2}+f^{2}\right) \]
holds, we will use the inequality known as the Weitzenböck inequality.
1. **Step 1: Understand Weitzenböck's In... | 9 | true |
The diagonals of a cyclic quadrilateral are perpendicular. Prove that the midpoints of its sides and the bases of the perpendiculars dropped to the sides from the point of intersection of the diagonals lie on one circle. Find the radius of this circle if the radius of the given circle is $R$ and the distance from its c... | Okay, so I have this problem about a cyclic quadrilateral with perpendicular diagonals. I need to prove that the midpoints of its sides and the bases of the perpendiculars from the intersection point of the diagonals lie on a single circle. Then, I have to find the radius of that circle given the original circle's radi... | To prove that the midpoints of the sides and the bases of the perpendiculars dropped from the intersection point of the diagonals of a cyclic quadrilateral with perpendicular diagonals lie on one circle, and to find the radius of this circle, we proceed as follows:
1. **Setup and Coordinates**:
- Consider a cyclic ... |
1. **Definitions and Initial Setup:**
Let \( ABCD \) be a cyclic quadrilateral with diagonals intersecting at point \( P \). Let \( K \) be the midpoint of side \( BC \), and \( L \) the midpoint of side \( AD \).
2. **Diagonals Perpendicular:**
Given that the diagonals \( AC \) and \( BD \) are perpendicular, ... | 9 | true |
Let $ABCD$ be a cyclic quadrilateral. Let $E$ and $F$ be variable points on the sides $AB$ and $CD$, respectively, such that $AE:EB = CF:FD$. Let $P$ be the point on the segment $EF$ such that $PE:PF = AB:CD$. Prove that the ratio between the areas of triangles $APD$ and $BPC$ does not depend on the choice of $E$ and $... | Alright, let me try to figure out this geometry problem. It's about cyclic quadrilaterals and ratios of areas. Hmm. Let's start by understanding the problem statement again.
We have a cyclic quadrilateral ABCD. Points E and F are on sides AB and CD respectively, such that the ratio AE:EB is equal to CF:FD. Then there'... | Given a cyclic quadrilateral \(ABCD\) with points \(E\) and \(F\) on sides \(AB\) and \(CD\) respectively such that \(AE:EB = CF:FD\), and a point \(P\) on segment \(EF\) such that \(PE:PF = AB:CD\), we need to prove that the ratio between the areas of triangles \(APD\) and \(BPC\) is constant.
1. **Setup and Definiti... | 1. **Case 1: $AD$ and $BC$ are parallel**
- When $AD \parallel BC$, quadrilateral $ABCD$ is an isosceles trapezoid with $AB = CD$. In this scenario, the point $P$ is the midpoint of $EF$.
- Let's denote the midpoints of $AB$ and $CD$ by $M$ and $N$ respectively. Since $M$ and $N$ divide $AB$ and $CD$ into equal... | 9 | true |
The midpoint of each side of the base of a quadrilateral pyramid is connected by a segment to the intersection point of the medians of the opposite lateral face. Prove that: a) these segments intersect and are divided by their intersection point in a ratio of \(3:2\) from the side of the base; b) the midpoints of these... | Okay, so I've got this geometry problem about a quadrilateral pyramid. The base is a quadrilateral, and they're talking about connecting midpoints of each side of the base to the centroid (I think that's what they mean by the intersection point of the medians) of the opposite lateral face. Then I need to prove three th... | a) All four segments intersect at the centroid of the pyramid, dividing each segment in a \(3:2\) ratio from the base.
b) Midpoints of these segments form a parallelogram due to equal and parallel opposite sides.
c) The ratio of the parallelogram's area to the base area is \(\frac{1}{72}\).
**Final Answer**
a) \b... |
Given the problem, we need to establish and prove several points about a quadrilateral pyramid whose midpoints of the sides are connected with the intersection points of the medians of the opposite lateral faces.
#### Part (a) Prove that these segments intersect and are divided in the ratio \(3:2\), counting from the... | 9 | true |
Xiaoming plays a computer game where initially only the number 1 is present. The computer performs operations sequentially, and at each step, every existing number \( x \) is replaced by four numbers \( 10x + 1 \), \( 10x + 2 \), \( 10x + 3 \), and \( 10x + 4 \). For any \( n \geq 1 \), between the \( n \)-th and \( (n... | Okay, so I need to prove that after each operation and deletion process, the number of remaining numbers, a_n, is at least 2^n. Let me start by understanding the problem step by step.
First, the game starts with the number 1. Then, in each step, every existing number x is replaced by four numbers: 10x + 1, 10x + 2, 10... | To prove that \( a_n \geq 2^n \), we construct a specific set \( S \) of numbers that cannot all be deleted under the given rules. Here's the step-by-step reasoning:
1. **Construction of Set \( S \):**
- After \( n \) operations, all numbers have \( n+1 \) digits starting with 1.
- Restrict digits 2 to \( n+1 \)... |
1. **Initial Setup**: Define \(a_0 = 1\).
2. **Interpretation of Condition**: The condition \(\left\lfloor \frac{s}{10^k} \right\rfloor \neq \left\lfloor \frac{t}{10^k} \right\rfloor\) means that the first \(n + 1 - k\) digits of \(s\) and \(t\) are not the same.
3. **Maximum Deletions per Round**: Thus, in the \(k... | 9 | true |
Given that \( x \cdot y \in \mathbf{R}^{-} \). Prove that \( \frac{x^{4}}{y^{4}} + \frac{y^{4}}{x^{4}} - \frac{x^{2}}{y^{2}} - \frac{y^{2}}{x^{2}} + \frac{x}{y} + \frac{y}{x} \geqslant 2 \). | Okay, so I need to prove that this expression is greater than or equal to 2, given that the product of x and y is a negative real number. Let me write down the expression first to make sure I have it right:
\[
\frac{x^{4}}{y^{4}} + \frac{y^{4}}{x^{4}} - \frac{x^{2}}{y^{2}} - \frac{y^{2}}{x^{2}} + \frac{x}{y} + \frac{y... | Given that \( x \cdot y \in \mathbf{R}^{-} \), we need to prove that the expression
\[
\frac{x^{4}}{y^{4}} + \frac{y^{4}}{x^{4}} - \frac{x^{2}}{y^{2}} - \frac{y^{2}}{x^{2}} + \frac{x}{y} + \frac{y}{x} \geqslant 2
\]
is valid.
First, we substitute \( z = \frac{x}{y} \), noting that \( z \) is a negative real number... |
We are given the problem:
\[
\frac{x^4}{y^4} + \frac{y^4}{x^4} - \frac{x^2}{y^2} - \frac{y^2}{x^2} + \frac{x}{y} + \frac{y}{x} \geq 2
\]
where \( x \cdot y \in \mathbf{R}^{-} \).
**Step-by-step Proof:**
1. Begin by examining the sum and re-expressing the given terms in a helpful form:
\[
\frac{x^4}{y^4} + \frac{y^4... | 9 | true |
Prove that for any tetrahedron, there exists a triangle with side lengths equal to the products of the lengths of opposite edges of the tetrahedron.
Also, prove that the area of this triangle is equal to \( 6VR \), where \( V \) is the volume of the tetrahedron and \( R \) is the radius of its circumscribed sphere. (T... | Okay, so I need to prove two things about a tetrahedron. First, that there exists a triangle with side lengths equal to the products of the lengths of opposite edges of the tetrahedron. Second, that the area of this triangle is equal to 6VR, where V is the volume and R is the radius of the circumscribed sphere. Hmm, th... | To prove that for any tetrahedron, there exists a triangle with side lengths equal to the products of the lengths of opposite edges, we use Ptolemy's inequality. For any tetrahedron with vertices \(A, B, C, D\), the products of the lengths of opposite edges \(AB \cdot CD\), \(AC \cdot BD\), and \(AD \cdot BC\) satisfy ... |
1. Let's consider a given tetrahedron $ABCD$. We will use an inversion with the center at $D$ and with radius $r^2$.
2. In an inversion, the points $A, B, C$ transform to points $A^*, B^*, C^*$ such that:
\[
A^*B^* = \frac{AB \cdot r^2}{DA \cdot DB}, \quad B^*C^* = \frac{BC \cdot r^2}{DB \cdot DC}, \quad \te... | 9 | true |
In a regular quadrilateral pyramid \(P-ABCD\), with \(AB=3\) and the dihedral angle between the lateral faces \(PAD\) and \(CPD\) being \(\frac{2}{3} \pi\):
1. Prove that plane \(PAD \perp\) plane \(PBC\);
2. Determine whether there exists a circumscribed sphere for the quadrilateral pyramid. If it exists, find the vol... | Okay, so I have this problem about a regular quadrilateral pyramid \( P-ABCD \). The base is a square with side length \( AB = 3 \). The dihedral angle between the lateral faces \( PAD \) and \( PCD \) is given as \( \frac{2}{3}\pi \). There are two parts: first, I need to prove that the planes \( PAD \) and \( PBC \) ... | Given a regular quadrilateral pyramid \(P-ABCD\) with \(AB = 3\) and a dihedral angle of \(\frac{2}{3}\pi\) between the lateral faces \(PAD\) and \(PCD\):
1. **Prove that plane \(PAD \perp\) plane \(PBC\):**
- The dihedral angle between \(PAD\) and \(PCD\) is given as \(120^\circ\). By calculating the normal vecto... |
#### (1) Prove that plane \(PAD \perp\) plane \(PBC\):
Given, \( P-ABCD \) is a regular tetrahedron with \( AB = 3 \) and the dihedral angle between faces \( PAD \) and \( CPD \) is \( \frac{2}{3} \pi \).
1. Construct perpendiculars:
- Draw perpendiculars from points \( C \) and \( A \) to line \( PD \).
- L... | 9 | true |
Let \( \mathrm{S} \) be the set of all prime numbers \( \mathrm{p} \) that satisfy the following condition: \( \mathrm{S} = \left\{\mathrm{p} \mid \mathrm{p} \text{ is a prime, and the length of the smallest repeating cycle in the decimal expansion of } \frac{1}{\mathrm{p}} \text{ is a multiple of 3} \right\} \). This ... | Okay, let me try to tackle these two problems step by step. First, I need to understand what's being asked here.
The set S consists of all prime numbers p such that the length of the smallest repeating cycle in the decimal expansion of 1/p is a multiple of 3. So, for example, primes where the decimal repeats every 3, ... | To address the two problems, we proceed as follows:
1. **Proving that the set \( S \) is infinite:**
The set \( S \) consists of primes \( p \) for which the multiplicative order of 10 modulo \( p \) is a multiple of 3. Using Zsigmondy's theorem, we know that for integers \( a > b > 0 \) with \( \gcd(a, b) = 1 \),... | ### Problem:
Let \( \mathrm{S} \) be the set of all prime numbers \( \mathrm{p} \) such that the length of the smallest repeating cycle in the decimal representation of \( \frac{1}{\mathrm{p}} \) is a multiple of 3. Specifically, for each \( p \in S \), there exists the smallest positive integer \( r=r(p) \) such that... | 9 | true |
Let \( A \) be a point outside a circle \( \omega \). The tangents from \( A \) to the circle \( \omega \) touch \( \omega \) at points \( S \) and \( T \). Let \( X \) and \( Y \) be the midpoints of \( AT \) and \( AS \), respectively. Let \( R \) be the intersection of the second tangent from \( X \) to the circle \... | Okay, so I need to prove that quadrilateral KRLQ is cyclic. Hmm, cyclic quadrilaterals have the property that the sum of opposite angles is 180 degrees, or that the power of a point with respect to the circle is equal. Alternatively, if four points lie on a circle, then certain cross ratios are real, or there might be ... | To prove that \( KRLQ \) is a cyclic quadrilateral, we start by analyzing the given configuration and using coordinate geometry to verify the concyclicity of the points.
1. **Setup and Tangents**:
- Let \( \omega \) be the unit circle centered at the origin.
- Point \( A \) is at \( (0, 2) \) outside the circle.... | 1. **Consider the intersection $K$ of three radical axes:**
- Let's denote the radical axis involving $K$ for easier reference. In circle geometry, defining radical axes and their intersections is crucial for proving concyclic points.
2. **Identify $L'$:**
- Let $L'$ be the intersection of $SX$ with circle $\... | 9 | true |
Given two non-concentric circles $O_{1}$ and $O_{2}$ on a plane intersecting at points $A$ and $B$. The external tangents to both circles touch $O_{1}$ at points $P_{1}$ and $Q_{1}$, and $O_{2}$ at points $P_{2}$ and $Q_{2}$. Let $M_{1}$ and $M_{2}$ be the midpoints of the segments $P_{1}Q_{1}$ and $P_{2}Q_{2}$, respec... | Okay, so I need to prove that lines AB, EF, and CD are concurrent. Let me start by visualizing the problem. There are two non-concentric circles, O₁ and O₂, intersecting at points A and B. The external tangents touch O₁ at P₁ and Q₁, and O₂ at P₂ and Q₂. The midpoints of P₁Q₁ and P₂Q₂ are M₁ and M₂. Then, extending AM₁... | Given two non-concentric circles \( O_1 \) and \( O_2 \) intersecting at points \( A \) and \( B \). The external tangents to both circles touch \( O_1 \) at points \( P_1 \) and \( Q_1 \), and \( O_2 \) at points \( P_2 \) and \( Q_2 \). The midpoints of the segments \( P_1Q_1 \) and \( P_2Q_2 \) are \( M_1 \) and \( ... | :
1. Consider two distinct circles \( O_1 \) and \( O_2 \) intersecting at points \( A \) and \( B \).
2. Let two external common tangents intersect circle \( O_1 \) at \( P_1 \) and \( Q_1 \), and circle \( O_2 \) at \( P_2 \) and \( Q_2 \).
3. Let \( M_1 \) and \( M_2 \) be the midpoints of \( P_1 Q_1 \) and \( P_2... | 9 | true |
Given any integer \( n \geq 2 \), assume that the integers \( a_1, a_2, \ldots, a_n \) are not divisible by \( n \) and that \( n \) does not divide \( a_1 + a_2 + \cdots + a_n \). Prove that there exist at least \( n \) different sequences \( (e_1, e_2, \cdots, e_n) \) consisting of zeros or ones such that \( e_1 a_1 ... | Okay, so I have this problem here where I need to prove that if I have an integer n ≥ 2, and integers a₁, a₂, ..., aₙ none of which are divisible by n, and the sum a₁ + a₂ + ... + aₙ is also not divisible by n, then there are at least n different binary sequences (e₁, e₂, ..., eₙ) where each eᵢ is 0 or 1 such that the ... | Given any integer \( n \geq 2 \), we need to prove that if the integers \( a_1, a_2, \ldots, a_n \) are not divisible by \( n \) and their sum \( a_1 + a_2 + \cdots + a_n \) is not divisible by \( n \), then there exist at least \( n \) different binary sequences \( (e_1, e_2, \ldots, e_n) \) such that \( e_1 a_1 + e_2... |
Given any integer $n \geq 2$, let the integers $a_{1}, a_{2}, \ldots, a_{n}$ be such that none of them are divisible by $n$ and $n$ does not divide $a_{1} + a_{2} + \cdots + a_{n}$. We need to prove that there exist at least $n$ different sequences $\left(e_{1}, e_{2}, \ldots, e_{n}\right)$ consisting of zeros or ones... | 9 | true |
In the given problem, $\triangle ABC$ has an incircle $\odot I$ which touches $BC$ at $D$. $AE$ is perpendicular to $BC$ at $E$, and $F$ is the midpoint of $AE$. $DF$ intersects the incircle $\odot I$ at $G$. Construct the circumcircle of $\triangle BCG$, denoted as $\odot \circ$. Prove that $\odot \circ$ and $\odot I$... | Alright, so I have this geometry problem here. Let me try to make sense of it step by step. First, let me parse the problem again:
We have triangle ABC with an incircle ⊙I that touches BC at D. Then there's AE, which is perpendicular to BC at E, and F is the midpoint of AE. DF intersects the incircle ⊙I at G. We need ... | To prove that the circumcircle of triangle \( \triangle BCG \) (denoted as \( \odot \circ \)) and the incircle \( \odot I \) of \( \triangle ABC \) are tangent at point \( G \), we can follow these key steps:
1. **Setup and Coordinates**:
- Place \( \triangle ABC \) with \( BC \) on the x-axis, \( B \) at \( (0, 0)... |
#### Method 1 - Using Harmonic Conjugates:
1. Construct line \( AD \) intersecting circle \(\odot \mathrm{I}\) at \( J \).
2. Extend line \( DI \) to intersect \(\odot \mathrm{I}\) at \( K \).
3. Note that \(DK \parallel AE\).
4. Recognize \( DG \) bisects \( AE \), meaning \( D \) divides \( AE \) harmonically.
5. Co... | 9 | true |
p is a prime greater than 2. Prove that
\[ \sum_{0 \leq n \leq p} \binom{p}{n} \binom{p+n}{n} \equiv 2p + 1 \pmod{p^2}. \]
Here, \(\binom{a}{b}\) is the binomial coefficient given by \(\frac{a!}{b!(a-b)!}\). | Okay, so I need to prove that the sum of binomial coefficients from n=0 to p, specifically the sum of \(\binom{p}{n} \binom{p+n}{n}\), is congruent to 2p + 1 modulo p squared, where p is a prime greater than 2. Hmm, let's start by understanding the components of this problem.
First, I recall that binomial coefficients... | To prove that for a prime \( p > 2 \),
\[
\sum_{0 \leq n \leq p} \binom{p}{n} \binom{p+n}{n} \equiv 2p + 1 \pmod{p^2},
\]
we start by analyzing the sum modulo \( p^2 \).
1. **Terms with \( n = 0 \) and \( n = p \)**:
- For \( n = 0 \), the term is \( \binom{p}{0} \binom{p}{0} = 1 \).
- For \( n = p \), the te... | Given \( p \) is a prime greater than 2, we need to prove that
\[
\sum_{0 \leq n \leq p} \binom{p}{n} \binom{p+n}{p} \equiv 2p + 1 \pmod{p^{2}}.
\]
We define \( D_n = \binom{p}{n} \binom{p+n}{p} \). The required sum is \( \sum_{n=0}^p D_n \).
First, we analyze \( D_n \) for different values of \( n \).
#### Step 1: ... | 9 | true |
Let \( P \) be any point in the plane of triangle \( \triangle ABC \). Denote \( BC=a \), \( CA=b \), \( AB=c \), \( PA=u \), \( PB=v \), and \( PC=\omega \). Prove that:
\[ \frac{u}{a}+\frac{v}{b}+\frac{\omega}{c} \geqslant \sqrt{3}. \] |
Okay, so I need to prove that for any point P in the plane of triangle ABC, the expression u/a + v/b + ω/c is at least √3. Hmm, let me start by understanding the notation. They define BC = a, CA = b, AB = c. So, the sides of the triangle are a, b, c. Then PA = u, PB = v, PC = ω. So, u, v, ω are the distances from poin... | To prove the inequality \(\frac{u}{a} + \frac{v}{b} + \frac{\omega}{c} \geqslant \sqrt{3}\) for any point \(P\) in the plane of triangle \(\triangle ABC\) with sides \(BC = a\), \(CA = b\), \(AB = c\), and distances \(PA = u\), \(PB = v\), \(PC = \omega\):
1. **Equilateral Triangle Case:**
- For an equilateral tria... | 1. **Initialization of Variables with Complex Numbers:**
Let points \( P, A, B, C \) in the plane correspond to the complex numbers \( z, z_1, z_2, z_3 \), respectively. Given distances are:
\[
BC = a, \quad CA = b, \quad AB = c
\]
and
\[
PA = u, \quad PB = v, \quad PC = \omega
\]
2. **Fund... | 9 | true |
For any positive integer \( n \), prove that:
$$
\sum_{k=0}^{n} C_{n}^{k} \cdot 2^{k} \cdot C_{n-k}^{[(n-k) / 2]} = C_{2n+1}^{n}
$$
where \( C_{0}^{0} = 1 \), and \( [(n-k) / 2] \) denotes the integer part of \( (n-k) / 2 \). | Okay, so I need to prove that the sum from k=0 to n of C(n, k) * 2^k * C(n - k, floor((n - k)/2)) equals C(2n + 1, n). Hmm, let me start by understanding what each part of this equation represents.
First, let's recall that C(n, k) is the binomial coefficient, which counts the number of ways to choose k elements from a... | To prove the identity:
$$
\sum_{k=0}^{n} C_{n}^{k} \cdot 2^{k} \cdot C_{n-k}^{\left\lfloor (n-k)/2 \right\rfloor} = C_{2n+1}^{n}
$$
we use a combination of combinatorial interpretation, generating functions, and mathematical induction.
1. **Base Cases**:
- For \( n = 0 \):
\[
\sum_{k=0}^{0} C_{0}^{0} \cdo... |
Given the problem, we need to prove that
\[
\sum_{k=0}^{n} C_{n}^{k} \cdot 2^{k} \cdot C_{n-k}^{\left[\frac{n-k}{2}\right]} = C_{2 n+1}^{n}
\]
where \( C_0^0 = 1 \) and \(\left[\frac{n-k}{2}\right]\) represents the integer part of \(\frac{n-k}{2}\).
To do this, we will use generating functions.
1. **Rewrite the ... | 9 | true |
Prove that the distances \( d_{1}, d_{2}, d_{3} \) from the incenter of a triangle to its vertices satisfy the equality:
$$
\frac{d_{1}^{2}}{b c}+\frac{d_{2}^{2}}{c a}+\frac{d_{3}^{2}}{a b}=1
$$
Prove the converse theorem. |
Okay, so I need to prove that in a triangle, the distances from the incenter to the vertices satisfy the equation (d12)/(bc) + (d22)/(ca) + (d32)/(ab) = 1. And then also prove the converse theorem. Hmm, let me start by recalling some properties of the incenter.
First off, the incenter is the point where the angle bis... | To prove that the distances \( d_{1}, d_{2}, d_{3} \) from the incenter of a triangle to its vertices satisfy the equality:
$$
\frac{d_{1}^{2}}{b c}+\frac{d_{2}^{2}}{c a}+\frac{d_{3}^{2}}{a b}=1
$$
we start by noting that the incenter \( I \) of a triangle \( ABC \) is equidistant from all sides and the inradius \( r... |
To prove the reverse theorem, we begin by recalling the given formula in the original theorem and using it alongside other known geometrical concepts.
1. **Formula Recall:**
Given the distances \(d_1, d_2, d_3\) from the incenter \( I \) of a triangle to its vertices \( A, B, \) and \( C \) respectively, we have:
... | 9 | true |
Prove that there is only one triplet of natural numbers greater than one such that the product of any two of these numbers, increased by 1, is divisible by the third. | Okay, so I need to prove that there's only one triplet of natural numbers (all greater than one) where, if you take any two numbers, multiply them, and add 1, the result is divisible by the third number. Hmm, let's start by understanding the problem better.
First, let's denote the three numbers as \( a \), \( b \), an... | Given three natural numbers \(a\), \(b\), and \(c\) greater than one, we need to prove that there is only one triplet such that the product of any two numbers increased by 1 is divisible by the third. The conditions are:
1. \(ab + 1 \equiv 0 \pmod{c}\)
2. \(ac + 1 \equiv 0 \pmod{b}\)
3. \(bc + 1 \equiv 0 \pmod{a}\)
W... | :
1. Let the triple of numbers \( a, b, c \in \mathbf{N} \) satisfy the given conditions:
\[ \frac{ab + 1}{c}, \quad \frac{ac + 1}{b}, \quad \text{and} \quad \frac{bc + 1}{a} \]
2. Notice that the numbers \( a, b, c \) are pairwise coprime. To understand why consider the following:
- If \( \gcd(a, b) > 1 \), th... | 9 | true |
In triangle ABC, let A, B, and C represent its three internal angles. Prove that:
\[
\left(\frac{\cos A}{\cos B}\right)^{2} + \left(\frac{\cos B}{\cos C}\right)^{2} + \left(\frac{\cos C}{\cos A}\right)^{2} \geq 4\left(\cos^{2} A + \cos^{2} B + \cos^{2} C\right).
\] | Alright, so I came across this inequality in a triangle that needs proving:
\[
\left(\frac{\cos A}{\cos B}\right)^{2} + \left(\frac{\cos B}{\cos C}\right)^{2} + \left(\frac{\cos C}{\cos A}\right)^{2} \geq 4\left(\cos^{2} A + \cos^{2} B + \cos^{2} C\right).
\]
Hmm, triangle angles, so A, B, C are all between 0 and π r... | To prove the inequality in triangle \(ABC\):
\[
\left(\frac{\cos A}{\cos B}\right)^{2} + \left(\frac{\cos B}{\cos C}\right)^{2} + \left(\frac{\cos C}{\cos A}\right)^{2} \geq 4\left(\cos^{2} A + \cos^{2} B + \cos^{2} C\right),
\]
we start by considering the case of an equilateral triangle where all angles are \(\frac{... |
1. Given $\triangle ABC$ such that $A, B, C$ are its angles, we need to prove:
\[
\left(\frac{\cos A}{\cos B}\right)^{2} + \left(\frac{\cos B}{\cos C}\right)^{2} + \left(\frac{\cos C}{\cos A}\right)^{2} \geqslant 4(\cos^{2} A + \cos^{2} B + \cos^{2} C).
\]
2. We will utilize the Nesbitt’s inequality, which ... | 9 | true |
A picture gallery is a non-convex $n$-gon. Prove that $\left[\begin{array}{ll} \\ \text { " }\end{array}\right]$ guards are sufficient to monitor the entire gallery. | Okay, so I need to prove that a certain number of guards are sufficient to monitor a non-convex n-gon gallery. Hmm, I remember there's a theorem related to art galleries and the number of guards needed. Let me think... Oh right, it's the Art Gallery Theorem, sometimes called Chvásal's Theorem. The classic result says t... | To prove that \(\left\lfloor \frac{n}{3} \right\rfloor\) guards are sufficient to monitor a non-convex \(n\)-gon art gallery, we follow these key steps:
1. **Triangulation**:
Any simple polygon (convex or non-convex) can be partitioned into \(n - 2\) triangles via diagonals. This process, called triangulation, en... | To demonstrate that it is sufficient to have $\left\lfloor \frac{n}{3} \right\rfloor$ guards to cover a non-convex $n$-gon, we will follow these steps:
1. **Triangulate the Polygon**:
- Given a non-convex $n$-gon, we begin by dividing it into triangles through non-intersecting diagonals.
- This procedure is poss... | 10 | true |
In a $5 \times 5 \times 10$ cuboid, there are 2001 given points. Prove that we can choose two of these points such that the distance between them is less than 0.7. |
Okay, so I need to prove that in a 5x5x10 cuboid with 2001 points, there are two points within a distance less than 0.7. Hmm, this seems like a problem that can be approached using the pigeonhole principle. Let me recall that the pigeonhole principle states that if you have more pigeons than pigeonholes, at least one ... | To prove that in a \(5 \times 5 \times 10\) cuboid with 2001 points, there exist two points with a distance less than 0.7, we use the pigeonhole principle and a volume argument.
1. **Volume Argument**:
- If we assume all points are at least 0.7 units apart, we can place spheres of radius 0.35 (diameter 0.7) around... | 1. **Define the spheres:**
- Consider 2001 spheres with a radius of \(0.35\) such that their centers are the given points in the cuboid.
- The volume of each sphere is given by the formula for the volume of a sphere:
\[
V_{sphere} = \frac{4}{3} \pi r^3
\]
where \( r = 0.35 \).
2. **Calc... | 9 | true |
A lion is running around a circus arena, which is a circle with a radius of 10 m. Moving along a broken line, the lion ran 30 km. Prove that the sum of all the angles of its turns is not less than 2998 radians. |
Okay, so I have this problem where a lion is running around a circus arena, which is a circular area with a radius of 10 meters. The lion is moving along a broken line, meaning it's not a smooth curve but a series of straight segments connected by turns. The total distance the lion ran is 30 kilometers, which is 30,00... | To prove that the sum of all the angles of the lion's turns is not less than 2998 radians, we consider the following key points:
1. **Circular Arena Constraint**: The lion is confined within a circular arena with a radius of 10 meters. The maximum straight-line distance the lion can travel without exiting the arena is... |
1. **Restate Given Information:**
- A lion runs around a circus arena, which is a circle with a radius of \(10\) meters.
- The total distance run by the lion along a broken line is \(30\) kilometers (or \(30000\) meters).
- We need to show that the sum of all the turning angles of the lion is at least \(2998\... | 9 | true |
Prove that every non-negative integer can be uniquely expressed in the form
$$
\frac{(x+y)^{2}+3 x+y}{2}
$$
where \( x \) and \( y \) are non-negative integers. |
Alright, so I came across this problem that says every non-negative integer can be uniquely expressed in the form [(x + y)2 + 3x + y]/2 where x and y are non-negative integers. I need to prove this. Hmm, okay, let's try to unpack this.
First, let me restate the problem in my own words: For any non-negative integer n,... | To prove that every non-negative integer can be uniquely expressed in the form
$$
\frac{(x+y)^{2}+3 x+y}{2}
$$
where \( x \) and \( y \) are non-negative integers, we can relate this to the Cantor pairing function and use the properties of triangular numbers.
### Proof:
1. **Existence**:
- For a given non-nega... |
To prove that every non-negative integer can be uniquely expressed in the form:
\[
\frac{(x+y)^2 + 3x + y}{2}
\]
where \( x \) and \( y \) are non-negative integers, we'll proceed by construction and demonstrate its uniqueness using a step-by-step approach.
1. **Expression Transformation:**
Given the original e... | 9 | true |
A numerical set $M$ containing 2003 different positive numbers is such that for any three different elements $a, b, c$ from $M$, the number $a^2 + bc$ is rational. Prove that it is possible to choose a natural number $n$ such that for any $a$ from $M$, the number $a \sqrt{n}$ is rational. | Alright, so I have this problem here about a numerical set M with 2003 different positive numbers. The condition is that for any three different elements a, b, c from M, the number a² + bc is rational. I need to prove that there exists a natural number n such that for any a in M, a√n is rational. Hmm, okay. Let me try ... | Given a numerical set \( M \) containing 2003 different positive numbers such that for any three different elements \( a, b, c \) from \( M \), the number \( a^2 + bc \) is rational, we need to prove that there exists a natural number \( n \) such that for any \( a \) from \( M \), the number \( a \sqrt{n} \) is ration... |
1. **Choose four distinct numbers from M**:
Let $a, b, c, d \in M$. We start by selecting four distinct elements from the set \(M\).
2. **Examine rationality conditions**:
Given the problem's condition, for any three distinct elements from \(M\):
\[
a^2 + bc \quad \text{is rational}
\]
Specificall... | 9 | true |
Let $\xi, \eta, \zeta$ be random variables such that $|\xi| \leq 1$, $|\eta| \leq 1$, and $|\zeta| \leq 1$. Prove the validity of Bell's inequality: $|\mathrm{E}[\zeta \zeta] - \mathrm{E}[\eta \zeta]| \leq 1 - \mathrm{E}[\xi \eta]$. |
Okay, so I need to prove Bell's inequality: |E[ξζ] - E[ηζ]| ≤ 1 - E[ξη], given that |ξ| ≤ 1, |η| ≤ 1, and |ζ| ≤ 1. Hmm, alright. Let me start by recalling that E[XY] is the expected value of the product of random variables X and Y. Since all variables are bounded by 1 in absolute value, their products will also be bou... | To prove Bell's inequality \( |\mathrm{E}[\xi \zeta] - \mathrm{E}[\eta \zeta]| \leq 1 - \mathrm{E}[\xi \eta] \) given that \( |\xi| \leq 1 \), \( |\eta| \leq 1 \), and \( |\zeta| \leq 1 \), we proceed as follows:
Key Steps and Reasoning:
1. **Pointwise Inequality**: For all real numbers \( a, b, c \in [-1, 1] \), the ... |
Given that \( \xi, \eta, \zeta \) are random variables such that \( |\xi| \leq 1 \), \( |\eta| \leq 1 \), and \( |\zeta| \leq 1 \), we aim to prove the Bell inequality:
\[|\mathrm{E}[\zeta \xi]-\mathrm{E}[\eta \zeta]| \leq 1 - \mathrm{E}[\xi \eta].\]
Let's break down the proof step-by-step:
1. **Establishing the In... | 10 | true |
Given the function \( y = f(x) \), which is an odd function on \(\mathbf{R}\) and satisfies \( f(x) = \frac{3^x}{9^x+1} - \frac{1}{2} \) when \( x \leqslant 0 \):
1. Determine and prove the monotonicity of \( y = f(x) \) on \( (-\infty, 0) \);
2. Find the range of \( y = f(x) \);
3. Solve the inequality \( f(x) > \frac... |
Okay, so I need to solve this problem about the odd function \( y = f(x) \). Let me read the problem again and take notes.
First, it says that \( f(x) \) is an odd function defined on all real numbers. An odd function satisfies \( f(-x) = -f(x) \) for all \( x \). The function is given by \( f(x) = \frac{3^x}{9^x + 1... | Given the function \( y = f(x) \), which is an odd function on \(\mathbf{R}\) and satisfies \( f(x) = \frac{3^x}{9^x+1} - \frac{1}{2} \) when \( x \leqslant 0 \):
1. **Determine and prove the monotonicity of \( y = f(x) \) on \( (-\infty, 0) \)**:
- The function \( f(x) \) is given by \( \frac{3^x}{9^x + 1} - \frac... |
**Step 1: Proving Monotonicity of \( f(x) \) on \( (-\infty, 0) \)**
Given:
- \( y = f(x) \) is an odd function, and for \( x \leq 0 \), \( f(x) = \frac{3^x}{9^x + 1} - \frac{1}{2} \).
To determine the monotonicity:
1. We'll first find the derivative \( f'(x) \) for \( x \leq 0 \) to check if the function is monoton... | 9 | true |
Consider the ellipse \( C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \) with \( a > b > 0 \). The left and right foci are \( F_{1} \) and \( F_{2} \) respectively, and the right vertex is \( A \). Let \( P \) be any point on ellipse \( C \). It is given that the maximum value of \( \overrightarrow{P F_{1}} \cdot \overr... | Okay, so I have this problem about an ellipse. Let me start by understanding what's given and what needs to be found.
First, part (1) asks to find the equation of the ellipse C given that the maximum value of the dot product of vectors PF₁ and PF₂ is 3 and the minimum value is 2. The ellipse is given by x²/a² + y²/b² ... | Given the ellipse \( C: \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \) with \( a > b > 0 \), the left and right foci are \( F_1 \) and \( F_2 \) respectively, and the right vertex is \( A \). For any point \( P \) on the ellipse, the maximum value of \( \overrightarrow{PF_1} \cdot \overrightarrow{PF_2} \) is 3 and th... |
1. **Step 1: Understanding the parameters of the hyperbola \( C \)**
Given the equation of the hyperbola
\[
\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \quad \text{ where } a > b > 0,
\]
we know the left and right foci are \( F_1(-c, 0) \) and \( F_2(c, 0) \) respectively, and the right vertex is \(... | 9 | true |
Suppose \( x_{0}, x_{1}, \ldots, x_{n} \) are integers and \( x_{0}>x_{1}>\cdots>x_{n} \). Prove that at least one of the numbers \( \left|F\left(x_{0}\right)\right|,\left|F\left(x_{1}\right)\right|,\left|F\left(x_{2}\right)\right| \ldots, \left|F\left(x_{n}\right)\right| \), where
\[ F(x)=x^{n}+a_{1} x^{n-1}+\cdots+a_... | Okay, so I need to prove that among the absolute values of F(x₀), F(x₁), ..., F(xₙ), where these x's are strictly decreasing integers, at least one of them is greater than n! divided by 2ⁿ. Hmm, let's start by understanding the problem step by step.
First, F(x) is a polynomial of degree n with real coefficients. The x... | To prove that at least one of the numbers \( \left|F\left(x_{0}\right)\right|, \left|F\left(x_{1}\right)\right|, \ldots, \left|F\left(x_{n}\right)\right| \) is greater than \( \frac{n!}{2^n} \), we use the concept of divided differences and analyze the polynomial \( F(x) \).
1. **Divided Difference Approach**:
- Th... |
1. Recall the polynomial \( F(x) = x^n + a_1 x^{n-1} + \cdots + a_n \) with real coefficients \( a_i \) and the condition \( x_0 > x_1 > \cdots > x_n \) where each \( x_i \) is an integer.
2. Apply Lagrange's interpolation formula for the polynomial \( F(x) \). According to Lagrange's interpolation formula, for a ... | 9 | true |
Suppose \( x_{i}, y_{i} \in \mathbf{R}^{+} (i = 1, 2, \cdots, n) \). Prove that \( \sum_{i=1}^{n} \frac{x_{i}^{3}}{y_{i}} \geq \frac{\left( \sum_{i=1}^{n} x_{i} \right)^{3}}{n \sum_{i=1}^{n} y_{i}} \). | Alright, so I need to prove this inequality: the sum from i=1 to n of (x_i^3 / y_i) is greater than or equal to (sum of x_i)^3 divided by (n times the sum of y_i). All the x_i and y_i are positive real numbers. Hmm, okay. Let me think about what inequalities I know that might be applicable here.
First off, the left-ha... | To prove the inequality \( \sum_{i=1}^{n} \frac{x_{i}^{3}}{y_{i}} \geq \frac{\left( \sum_{i=1}^{n} x_{i} \right)^{3}}{n \sum_{i=1}^{n} y_{i}} \) for positive real numbers \( x_i \) and \( y_i \), we can use Hölder's inequality.
Hölder's inequality states that for sequences of non-negative real numbers \( a_i \), \( b... |
To prove the inequality
\[
\sum_{i=1}^{n} \frac{x_{i}^{3}}{y_{i}} \geq \frac{\left( \sum_{i=1}^{n} x_{i} \right)^{3}}{n \sum_{i=1}^{n} y_{i}}
\]
we will use two main tools: the Cauchy-Schwarz inequality and the convexity of a function. Here are the detailed steps:
1. **Rewriting the left-hand side**:
Notice t... | 9 | true |
Given that $A$ and $B$ are the left and right vertices of the ellipse $C: \frac{x^{2}}{25}+\frac{y^{2}}{9}=1$, the line $l$ intersects the ellipse $C$ at points $M$ and $N$. Let the slope of line segment $AM$ be $k_{1}$ and the slope of line segment $BN$ be $k_{2}$, with $k_{1}: k_{2} = 1: 9$.
(1) Prove that the line ... |
Alright, let's tackle this problem step by step. It's an ellipse problem with parts (1) and (2). Starting with part (1):
First, the ellipse equation is given as x2/25 + y2/9 = 1. That means the major axis is along the x-axis since 25 > 9. The semi-major axis length a is 5, and semi-minor axis length b is 3. So the ve... | Given the ellipse \( C: \frac{x^2}{25} + \frac{y^2}{9} = 1 \) with vertices \( A(-5, 0) \) and \( B(5, 0) \), the line \( l \) intersects the ellipse at points \( M \) and \( N \). The slopes of segments \( AM \) and \( BN \) are \( k_1 \) and \( k_2 \) respectively, with the ratio \( k_1 : k_2 = 1 : 9 \).
(1) To prov... |
**Part 1:** To prove that the line $l$ passes through the fixed point.
1. Assume that the equation of line $l$ is given by:
\[
l: y = kx + m
\]
Let the intersection points with the ellipse be denoted as \( M \left( x_{1}, y_{1} \right) \) and \( N \left( x_{2}, y_{2} \right) \).
2. Solve the system of e... | 9 | true |
Let \( A \) be the vertex of a spherical segment, and \( B \) be a point on the circumference of its base. Prove that the surface area of this segment is equal to the area of a circle with radius \( AB \). | Okay, so I need to prove that the surface area of a spherical segment (which I think is like a portion of a sphere, maybe like a spherical cap or something) is equal to the area of a circle with radius AB, where A is the vertex of the segment and B is a point on the circumference of its base. Hmm. Let me start by recal... | To prove that the surface area of a spherical segment (spherical cap) is equal to the area of a circle with radius \( AB \), follow these steps:
**1. Define Key Components:**
- Let the sphere have radius \( R \), center \( O \).
- The spherical cap has vertex \( A \) (top point) and a circular base with radius \( r \)... |
1. **Introduction and Setup**:
- Let $M$ be the center of the base of the spherical segment, $h$ be the height of the segment, $O$ be the center of the sphere, and $R$ be the radius of the sphere.
- Denote the points as follows:
- $A$ is the apex (or vertex) of the spherical segment.
- $B$ is a... | 9 | true |
Given numbers \( x \) and \( y \) that satisfy the condition \( x^{8} + y^{8} \leq 2 \), prove the inequality \( x^{2} y^{2} + \left| x^{2} - y^{2} \right| \leq \frac{\pi}{2} \). |
Okay, so I need to prove that for numbers x and y satisfying x8 + y8 ≤ 2, the inequality x2y2 + |x2 - y2| ≤ π/2 holds. Hmm, interesting. Let me try to break this down step by step.
First, I know that the given condition is x8 + y8 ≤ 2. That seems like a higher power constraint. Maybe I can start by thinking about pos... | Given numbers \( x \) and \( y \) that satisfy the condition \( x^8 + y^8 \leq 2 \), we need to prove the inequality \( x^2 y^2 + \left| x^2 - y^2 \right| \leq \frac{\pi}{2} \).
First, we substitute \( a = x^2 \) and \( b = y^2 \). The given condition translates to \( a^4 + b^4 \leq 2 \). The inequality to prove becom... | :
We are given the numbers \(x\) and \(y\) that satisfy the condition \(x^{8} + y^{8} \leq 2\). We need to prove that:
\[
x^{2} y^{2} + |x^{2} - y^{2}| \leq \frac{\pi}{2}
\]
Let's set \(u = x^2\) and \(v = y^2\). According to the given condition:
\[
x^8 + y^8 \leq 2 \implies u^4 + v^4 \leq 2
\]
#### Step 1: Analy... | 9 | true |
Let $I, O, G, H$ be the incenter, circumcenter, centroid, and orthocenter of $\triangle ABC$, respectively. Let $BC = a, CA = b, AB = c$, and $p = \frac{1}{2}(a + b + c)$. Let $R$ and $r$ be the radii of the circumcircle and incircle, respectively. Prove the following equations:
(I) $a \cdot IA^{2} + b \cdot IB^{2} + ... |
Alright, so I have these four geometry problems to solve, all related to triangle centers: incenter (I), circumcenter (O), centroid (G), orthocenter (H). Each part involves proving a specific formula involving distances between these centers or squared lengths weighted by the sides. Let me take them one by one and see... | To prove the given equations involving the incenter \(I\), circumcenter \(O\), centroid \(G\), and orthocenter \(H\) of \(\triangle ABC\), we use various geometric properties, coordinate geometry, and trigonometric identities.
**(I) Proving \(a \cdot IA^{2} + b \cdot IB^{2} + c \cdot IC^{2} = abc\):**
1. **Distance f... |
Given the problem:
Let \(I, O, G, H\) be the incenter, circumcenter, centroid, and orthocenter of \(\triangle ABC\) respectively, and let \(BC = a, CA = b, AB = c\), \(p = \frac{1}{2}(a+b+c)\), and \(R\) and \(r\) be the circumradius and inradius respectively. We need to prove the following formulas:
#### (I) \(a \cd... | 10 | true |
$P$ is inside $\triangle ABC$, with $AB = AC$. The projections of $P$ onto $BC$, $CA$, and $AB$ are $D$, $E$, and $F$ respectively. The altitude from $A$ to $BC$ is $h$. The circumcenter of $\triangle BPC$ is $O$, and the midpoint of $BC$ is $M$. Given $PD^2 = PE \cdot PF$ and $PD + PF = PE$, prove the following:
1. $... | Okay, so I have this geometry problem here with triangle ABC where AB equals AC, so it's an isosceles triangle with AB = AC. Point P is inside the triangle, and its projections onto BC, CA, and AB are D, E, and F respectively. The altitude from A to BC is h. Then there's the circumcenter O of triangle BPC, and M is the... | **1. $\angle BPC = \angle A + \angle B$**
In $\triangle ABC$, since $AB = AC$, $\angle B = \angle C = \frac{1}{2}(180^\circ - \angle A)$. Using the given conditions and properties of the circumcircle, we show that $\angle BPC = 180^\circ - (\angle PBC + \angle PCB)$. By analyzing the projections and symmetry, $\angle ... |
#### Part (1)
To prove: $\angle BPC = \angle A + \angle B$.
Given:
1. $P$ is within $\triangle ABC$ and lies on segment $BC$.
2. $AB = AC$.
3. $PD = PE \cdot PF$.
4. $PD + PF = PE$.
By the given conditions:
\[ \alpha + \beta = \alpha' + \beta' = C \]
Therefore:
\[ \alpha - \beta' = \alpha' - \beta \]
By the Sine R... | 10 | true |
For any positive integers \(a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}\), prove the following inequality:
$$
\begin{array}{l}
\left(a_{1} b_{2}+a_{1} b_{3}+a_{2} b_{1}+a_{2} b_{3}+a_{3} b_{1}+a_{3} b_{2}\right)^{2} \\
\geqslant 4\left(a_{1} a_{2}+a_{2} a_{3}+a_{3} a_{1}\right)\left(b_{1} b_{2}+b_{2} b_{3}+b_{3} b_{1}\rig... |
Okay, so I need to prove this inequality:
\[
\left(a_{1} b_{2}+a_{1} b_{3}+a_{2} b_{1}+a_{2} b_{3}+a_{3} b_{1}+a_{3} b_{2}\right)^{2} \geqslant 4\left(a_{1} a_{2}+a_{2} a_{3}+a_{3} a_{1}\right)\left(b_{1} b_{2}+b_{2} b_{3}+b_{3} b_{1}\right),
\]
for any positive integers \(a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}\),... | To prove the inequality:
\[
\left(a_{1} b_{2}+a_{1} b_{3}+a_{2} b_{1}+a_{2} b_{3}+a_{3} b_{1}+a_{3} b_{2}\right)^{2} \geqslant 4\left(a_{1} a_{2}+a_{2} a_{3}+a_{3} a_{1}\right)\left(b_{1} b_{2}+b_{2} b_{3}+b_{3} b_{1}\right),
\]
for any positive integers \(a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}\), we can use the iden... | To prove the inequality for any positive integers \(a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}\):
\[ \left( a_1 b_2 + a_1 b_3 + a_2 b_1 + a_2 b_3 + a_3 b_1 + a_3 b_2 \right)^2 \geq 4 \left( a_1 a_2 + a_2 a_3 + a_3 a_1 \right) \left( b_1 b_2 + b_2 b_3 + b_3 b_1 \right), \]
and show that the equality holds if and only if \... | 9 | true |
In a convex quadrilateral \(ABCD=N\), the midpoints of the sides \(BC, CD, DA, AB\) are \(A^{\prime}, B^{\prime}, C^{\prime}, D^{\prime}\) respectively. The intersections of the lines \(AA^{\prime}, BB^{\prime}, CC^{\prime}, DD^{\prime}\) taken pairwise (adjacent pairs) are \(A_1, B_1, C_1, D_1\). The area of the quadr... | Okay, so I have this problem about a convex quadrilateral \(ABCD\), and I need to prove that the area \(T\) of the quadrilateral is between \(5T_1\) and \(6T_1\), where \(T_1\) is the area of another quadrilateral formed by the intersections of certain lines. The midpoints of the sides are given as \(A', B', C', D'\), ... | To prove that \(5T_1 \leq T \leq 6T_1\) for a convex quadrilateral \(ABCD\) with midpoints \(A', B', C', D'\) of sides \(BC, CD, DA, AB\) respectively, and intersections \(A_1, B_1, C_1, D_1\) of lines \(AA', BB', CC', DD'\), we analyze the areas \(T\) and \(T_1\).
1. **Coordinate Geometry Approach**:
- Assign coor... | :
To prove the inequality \(5T_1 \leq T \leq 6T_1\), consider the given convex quadrilateral \(ABCD = N\).
#### Step-by-Step Solution
1. **Reflection and Construction**:
- Reflect quadrilateral \(ABCD\) through point \(B'\), the midpoint of side \(BC\), to form a new quadrilateral \(EFEHG\).
- Proceed to re... | 9 | true |
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