problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
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values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
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Example 5 In the sequence $\left\{a_{n}\right\}$, $a_{1}=1, a_{n+1}=a_{n}+$ $\sqrt{a_{n}}+\frac{1}{4}$. Then $a_{99}=(\quad$.
(A) $2550 \frac{1}{4}$
(B) 2500
(C) $2450 \frac{1}{4}$
(D) 2401 | Solution: From the recursive formula, we have $a_{n+1}=\left(\sqrt{a_{n}}+\frac{1}{2}\right)^{2}$, which means $\sqrt{a_{n+1}}=\sqrt{a_{n}}+\frac{1}{2}$.
Therefore, $\left\{\sqrt{a_{n}}\right\}$ is an arithmetic sequence. Thus, we have
$$
\sqrt{a_{n}}=\sqrt{a_{1}}+(n-1) \times \frac{1}{2}=\frac{n+1}{2} \text {. }
$$
T... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,894 |
Example 6 Let the sequence of positive numbers $a_{0}, a_{1}, a_{2}, \cdots$ satisfy $\sqrt{a_{n} a_{n-2}}-\sqrt{a_{n-1} a_{n-2}}=2 a_{n-1}(n \geqslant 2)$, and $a_{0}=a_{1}=1$. Find the general term formula of this sequence. | Simplified Solution 1: Using the recursive formula, we get
$$
\begin{array}{l}
a_{0}=1, a_{1}=1=\left(2^{1}-1\right)^{2}, \\
a_{2}=9=\left(2^{2}-1\right)^{2}\left(2^{1}-1\right)^{2}, \\
a_{3}=21^{2}=\left(2^{3}-1\right)^{2}\left(2^{2}-1\right)^{2}\left(2^{1}-1\right)^{2} .
\end{array}
$$
Conjecture that when $n \in \m... | a_{n}=\prod_{k=1}^{n}\left(2^{k}-1\right)^{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,895 |
Example 7 In the sequence $\left\{a_{n}\right\}$, $a_{1}=\sqrt{2}, a_{n+1}=$ $\sqrt{2+a_{n}}(n \geqslant 1)$. Then $a_{n}=$ $\qquad$ . | Solution: In the sequence $\left\{a_{n}\right\}$, $a_{1}=\sqrt{2}=2 \cos \theta$, where $\theta=\frac{\pi}{4}$. From the known recurrence relation and trigonometric formulas, we get
$$
a_{2}=2 \cos \frac{\theta}{2}, a_{3}=2 \cos \frac{\theta}{2^{2}} \text {. }
$$
Conjecture $a_{n}=2 \cos \frac{\theta}{2^{n-1}}$, and p... | a_{n}=2 \cos \frac{\theta}{2^{n-1}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,896 |
Proposition 1 As shown in Figure 1, in quadrilateral $ABCD$, $F$ is any point on diagonal $AC$, $BF$ intersects $CD$ at point $E$, $DF$ intersects $BC$ at point $G$. $P$ is any point on $AC$, line $PG$ intersects $AB$ at point $R$, $PD$ intersects $AE$ at point $H$. Then $R$, $F$, and $H$ are collinear. | Prove: Since line $A H E$ intersects $\triangle D C P$, by Menelaus' theorem we have
$$
\frac{D H}{H P} \cdot \frac{P A}{A C} \cdot \frac{C E}{E D}=1 .
$$
By line $A B R$ intersecting $\triangle C P G$ we get
$$
\frac{P R}{R G} \cdot \frac{G B}{B C} \cdot \frac{C A}{A P}=1 \text {. }
$$
By line $B F E$ intersecting $... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,897 |
Proposition 2 As shown in Figure 2, in quadrilateral $ABCD$, $F$ is any point on diagonal $AC$, $BF$ intersects $CD$ at point $E$, and $DF$ intersects $BC$ at point $G$. $P$ is any point on $AC$, $GP$ intersects $AD$ at point $H$, and $CH$ intersects $AE$ at point $K$. Then points $B$, $P$, and $K$ are collinear. | Proof: Since line $A K E$ intersects $\triangle C D H$, by Menelaus' theorem we have
$$
\frac{H K}{K C} \cdot \frac{C E}{E D} \cdot \frac{A D}{A H}=1 .
$$
By line $B F E$ intersecting $\triangle D C G$ we get
$$
\frac{C B}{B G} \cdot \frac{G F}{F D} \cdot \frac{D E}{E C}=1 \text {. }
$$
By line $A P F$ intersecting $... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,898 |
Proposition 3 As shown in Figure 3, in quadrilateral $ABCD$, $F$ is any point on diagonal $AC$, $BF$ intersects $CD$ at point $E$, $DF$ intersects $BC$ at point $G$. $P$ is any point on $AC$, $PB$ intersects $AG$ at point $T$, $PD$ intersects $AE$ at point $H$. Then $BH$, $DT$, and $AC$ are concurrent (if the three lin... | Prove: Connect $B D$ intersecting $A C$ at $K$. Applying Ceva's Theorem to $\triangle C B D$ we get
$$
\frac{B K}{K D} \cdot \frac{D E}{E C} \cdot \frac{C G}{G B}=1 \text {. }
$$
By the line $A H E$ intersecting $\triangle D C P$ we get
$$
\frac{D H}{H P} \cdot \frac{P A}{A C} \cdot \frac{C E}{E D}=1 \text {. }
$$
By... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,899 |
Proposition 4 As shown in Figure 4, in quadrilateral $ABCD$, $F$ is a point on diagonal $AC$, $BF$ intersects $CD$ at point $E$, and $DF$ intersects $BC$ at point $G$. $P$ is any point on $AC$, $PG$ intersects $AB$ at point $R$, and $PE$ intersects $AD$ at point $Q$. Then $RE$, $QG$, and $AC$ are concurrent. | Proof: Let $R E$ and $Q G$ intersect at $O$. It suffices to prove that $O, F, A$ are collinear.
By line $G O Q$ intersecting $\triangle R P E$, we get
$$
\frac{E O}{O R} \cdot \frac{R G}{G P} \cdot \frac{P Q}{Q E}=1 \text{. }
$$
By line $A P C$ intersecting $\triangle R B G$, we get
$$
\frac{R A}{A B} \cdot \frac{B C}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,900 |
Proposition 5 As shown in Figure 5, in quadrilateral $ABCD$, $F$ is any point on diagonal $AC$, $BF$ intersects $CD$ at point $E$, $DF$ intersects $BC$ at point $G$, $AG$ intersects $BE$ at point $M$, $AE$ intersects $DG$ at point $N$, $CM$ intersects $AB$ at point $P$, $CN$ intersects $AD$ at point $Q$. Then $BN, DM, ... | Prove: Connect $B D$ intersecting $A C$ at point $K$. By Ceva's Theorem, we have
$$
\frac{B K}{K D} \cdot \frac{D E}{E C} \cdot \frac{C G}{G B}=1 .
$$
By the line $A N E$ intersecting $\triangle D F C$, we get
$$
\frac{D N}{N F} \cdot \frac{F A}{A C} \cdot \frac{C E}{E D}=1 \text {. }
$$
By the line $A M G$ intersect... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,901 |
Proposition 6 As
shown in Figure 6, in quadrilateral
$A B C D$, $F$ is any
point on diagonal $A C$, $B F$ intersects $C D$ at
point $E$, $D F$ intersects $B C$
at point $G$, $A G$ intersects $B E$
at point $M$, $A E$ intersects $D G$
at point $N$, $C M$ intersects $A B$ and $D G$ at points $P$ and $K$, respectively, an... | Proof: The proof will use the following lemma.
Lemma As shown in Figure 7, in convex quadrilateral $ABCD$, the extensions of opposite sides $AB$ and $DC$ intersect at point $P$, and the diagonals $AC$ and $BD$ intersect at point $O$. $PO$ intersects $BC$ and $AD$ at points $M$ and $N$ respectively. Then $PM \cdot ON =... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,902 |
Proposition 2 The number of incongruent scalene triangles inscribed in a regular $n$-sided polygon is $\left[\frac{n(n-6)}{12}\right]+1$, i.e., $\left\langle\frac{(n-3)^{2}}{12}\right\rangle$; the number of incongruent isosceles triangles inscribed in a regular $n$-sided polygon is $\left[\frac{n-1}{2}\right]$, of whic... | Proof: Similar to the proof of Proposition 1, the non-congruent non-equilateral triangles inscribed in a regular $n$-sided polygon $Q$ correspond one-to-one to the $n$-partitions with 3 parts, each part being distinct. By the principle of equality, the number of non-congruent non-equilateral triangles inscribed in $Q$ ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 713,903 |
Example 5 Given a convex quadrilateral with integer side lengths, and the length of any one side can divide the sum of the lengths of the other three sides. Prove: This quadrilateral must have two equal sides. | Explanation: Based on the characteristics of the problem, we can consider using proof by contradiction.
Assume that no two sides of the quadrilateral are equal. Without loss of generality, let $a_{1}>a_{2}>a_{3}>a_{4}$ represent the relationship of the four side lengths, and let $p=$ $a_{1}+a_{2}+a_{3}+a_{4}$ represen... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 713,904 |
Given $x \geqslant y \geqslant z>0$. Prove:
$$
\frac{x^{2} y}{z}+\frac{y^{2} z}{x}+\frac{z^{2} x}{y} \geqslant x^{2}+y^{2}+z^{2} .
$$ | Proof: By Cauchy's inequality, we have
$$
\begin{array}{l}
\left(\frac{x^{2} y}{z}+\frac{y^{2} z}{x}+\frac{z^{2} x}{y}\right)\left(\frac{x^{2} z}{y}+\frac{y^{2} x}{z}+\frac{z^{2} y}{x}\right) \\
\geqslant\left(x^{2}+y^{2}+z^{2}\right)^{2} .
\end{array}
$$
Observing the above inequality, if we have
$$
\frac{x^{2} y}{z}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 713,905 |
Proposition In any $\triangle A B C$, let $B C=a, A C$ $=b, A B=c, m_{a}$ be the length of the median on side $B C$, and $w_{a}$ be the length of the angle bisector of $\angle A$, then
$$
\frac{m_{a}}{w_{a}} \geqslant \frac{(b+c)^{2}}{4 b c} .
$$ | Proof: Let $p$ be the semi-perimeter of $\triangle ABC$, then equation (2) is equivalent to
$$
(b+c)^{4} w_{a}^{2} \leqslant 16 b^{2} c^{2} m_{a}^{2} .
$$
By the angle bisector formula $w_{a}=\frac{2 \sqrt{b c p(p-a)}}{b+c}$ and the median length formula $m_{a}=\frac{1}{2} \sqrt{2\left(b^{2}+c^{2}\right)-a^{2}}$, we h... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 713,906 |
Proposition 1 Let the 6 edge lengths of tetrahedron $ABCD$ be $a, b, c, d, e, f$, and the volume be $V$. Then for any natural number $n$ we have
$$
a^{n}+b^{n}+c^{n}+d^{n}+e^{n}+f^{n} \geqslant 6\left(72 V^{2}\right)^{\frac{n}{6}},
$$
with equality holding if and only if the tetrahedron is a regular tetrahedron. | Proof: According to the arithmetic-geometric mean inequality and the lemma, we have
$$
\begin{array}{l}
a^{n}+b^{n}+c^{n}+d^{n}+e^{n}+f^{n} \\
\geqslant 6\left(a^{n} \cdot b^{n} \cdot c^{n} \cdot d^{n} \cdot e^{n} \cdot f^{n}\right)^{\frac{1}{6}} \\
=6(a b c d e f)^{\frac{n}{6}} \geqslant 6\left(72 V^{2}\right)^{\frac{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 713,907 |
Proposition 3 Let the areas of the four faces of a tetrahedron be $S_{1}, S_{2}, S_{3}, S_{4}$, and the volume be $V$. Then
$$
S_{1}^{2}+S_{2}^{2}+S_{3}^{2}+S_{4}^{2} \geqslant 9\left(3 V^{4}\right)^{\frac{1}{3}} .
$$ | Proof: By Lemma 1 of [1] and the inequality of the arithmetic mean of four numbers, we get
$$
\begin{array}{l}
S_{1}^{2}+S_{2}^{2}+S_{3}^{2}+S_{4}^{2} \geqslant 4 \sqrt{S_{1} S_{2} S_{3} S_{4}} \\
\geqslant 4 \sqrt{\left(\frac{3^{14}}{2^{12}}\right)^{\frac{1}{3}} V^{\frac{8}{3}}}=9\left(3 V^{4}\right)^{\frac{1}{3}},
\e... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 713,909 |
Proposition 4 Let the areas of the four faces of a tetrahedron be $S_{1}, S_{2}, S_{3}, S_{4}$, and the volume be $V, t \in \mathbf{R}$ and $t \geqslant 2$. Then
$$
S_{1}^{t}+S_{2}^{t}+S_{3}^{t}+S_{4}^{t} \geqslant 2^{2-t} \times 3^{\frac{7}{6} t} V^{\frac{2}{3} t} .
$$ | Proof: According to Lemma 2 and Inequality (3) in [1], we get
$$
\begin{array}{l}
\left(\frac{S_{1}^{t}+S_{2}^{t}+S_{3}^{t}+S_{4}^{t}}{4}\right)^{\frac{1}{t}} \\
\geqslant\left(\frac{S_{1}^{2}+S_{2}^{2}+S_{3}^{2}+S_{4}^{2}}{4}\right)^{\frac{1}{2}} \\
\geqslant\left(\frac{9\left(3 V^{4}\right)^{\frac{1}{3}}}{4}\right)^{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 713,910 |
3. For any natural numbers $n$, $p$, $r$, there exists a natural number $m$ such that
$$
(\sqrt{p+r}-\sqrt{p})^{-n}=\frac{\sqrt{m+r^{n}}+\sqrt{m}}{r^{n}} .
$$ | Proof: Since $(\sqrt{p+r}-\sqrt{p})^{n}(\sqrt{p+r}-$ $\sqrt{p})^{-n}=1$, and by reference [1] we know
$$
(\sqrt{p+r}-\sqrt{p})^{n}=\sqrt{m+r^{n}}-\sqrt{m} \text{. }
$$
Therefore, $(\sqrt{p+r}-\sqrt{p})^{-n}$
$$
=\frac{1}{\sqrt{m+r^{n}}-\sqrt{m}}=\frac{\sqrt{m+r^{n}}+\sqrt{m}}{r^{n}} \text{. }
$$
By the extended theor... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 713,911 |
Let the three sides of $\triangle A B C$ be $a, b, c$. For any positive integer $n, n>1$, we have
$$
\frac{1}{3^{n-1}} \leqslant \frac{a^{n}+b^{n}+c^{n}}{(a+b+c)^{n}}<\frac{1}{2^{n-1}},
$$
with equality holding if and only if $a=b=c$. | Proof: According to [2], we have
$$
\frac{a^{n}+b^{n}+c^{n}}{3} \geqslant\left(\frac{a+b+c}{3}\right)^{n},
$$
with equality holding if and only if \(a=b=c\). This easily shows that the first inequality holds, and the condition for equality also holds.
Next, we prove the second inequality, which is equivalent to
$$
a^{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 713,912 |
Paper $[1]$ provides an upper bound estimate for $\sum \frac{1}{a^{2}}$, that is, let $a, b, c$ be the side lengths of $\triangle ABC$, and $R, r$ represent the circumradius and inradius of $\triangle ABC$, respectively, then we have
$$
\sum \frac{1}{a^{2}} \leqslant \frac{\left(R^{2}+r^{2}\right)^{2}+R r(2 R-3 r)^{2}}... | $$
\begin{array}{l}
\text { Prove: } \sum \frac{1}{a^{2}}=\frac{b^{2} c^{2}+a^{2} c^{2}+a^{2} b^{2}}{a^{2} b^{2} c^{2}} \\
\geqslant \frac{(b c)(a c)+(a c)(a b)+(b c)(a b)}{a^{2} b^{2} c^{2}} \\
=\frac{c+a+b}{a b c} .
\end{array}
$$
From the identity in a triangle \(a+b+c=2p\) (where \(p\) is the semi-perimeter), and ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 713,913 |
1. Let $S=\{1,2, \cdots, 1000000\}, A$ be a subset of $S$ containing exactly 101 elements. Prove that there exist numbers $t_{1}$, $t_{2}, \cdots, t_{100}$ in $S$, such that the following sets
$$
A_{j}=\left\{x+t_{j} \mid x \in A\right\}, j=1,2, \cdots, 100
$$
are pairwise disjoint. | 1. Consider the set $D=\{x-y \mid x, y \in A\}, D$ contains at most $101 \times 100+1=10101$ elements. It is easy to see that the necessary and sufficient condition for two sets $A_{i}$ and $A_{j}$ to have a non-empty intersection is $t_{i}-t_{j} \in D$. Therefore, we need to select 100 elements from $S$ such that thei... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 713,914 |
Example 6 Does there exist a quadrilateral $A B C D$ with the following two properties?
(1) The lengths of the two diagonals $A C$ and $B D$ are coprime integers;
(2) If the reflections of $\triangle A C D$ and $\triangle B C D$ across the lines $A C$ and $B D$ are $\triangle A C D^{\prime}$ and $\triangle B D C^{\prim... | Explanation: We can start the discussion by assuming the existence of a quadrilateral, using a diagram and two known conditions.
As shown in Figure 3, assume that a quadrilateral $\square A B C D$ exists, satisfying conditions (1) and (2), and that $A C$ and $B D$ intersect at point $P$. Connect $P C^{\prime}$ and $P D... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,915 |
2. Find all pairs of positive integers $(a, b)$ such that $\frac{a^{2}}{2 a b^{2}-b^{3}+1}$ is a positive integer. | 2. Let $(a, b)$ be a solution that satisfies the conditions.
Since $k=\frac{a^{2}}{2 a b^{2}-b^{3}+1}>0$, we have $2 a b^{2}-b^{3}+1>0$, which means $a>\frac{b}{2}-\frac{1}{2 b^{2}}$.
Therefore, $a \geqslant \frac{b}{2}$.
Given $k \geqslant 1$, i.e., $a^{2} \geqslant b^{2}(2 a-b)+1$, then $a^{2}>b^{2}(2 a-b) \geqslan... | (a, b)=(2 l, 1) \text{ or } (l, 2 l) \text{ or } \left(8 l^{4}-l, 2 l\right) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 713,916 |
3. Given a convex hexagon, each pair of opposite sides has the following property: the distance between the midpoints of these two sides is equal to $\frac{\sqrt{3}}{2}$ times the sum of their lengths. Prove: all the interior angles of the hexagon are equal. (A convex hexagon $A B C D E F$ has three pairs of opposite s... | 3. First, prove a lemma:
In $\triangle P Q R$, $\angle Q P R \geqslant 60^{\circ}$, and $L$ is the midpoint of $Q R$, then $P L \leqslant \frac{\sqrt{3}}{2} Q R$, with equality if and only if $\triangle P Q R$ is an equilateral triangle.
Proof: Let $S$ be a point such that $\triangle Q R S$ is an equilateral triangle... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,917 |
4. Let $A B C D$ be a cyclic quadrilateral. From point $D$, draw perpendiculars to the lines $B C$, $C A$, and $A B$, with feet of the perpendiculars being $P$, $Q$, and $R$ respectively. Prove that $P Q=Q R$ if and only if the angle bisectors of $\angle A B C$ and $\angle A D C$, and the line $A C$ intersect at a sing... | 4. As shown in Figure 2, by Simson's Theorem, points $P$, $Q$, and $R$ are collinear. Furthermore, since $\angle DPC = \angle DQC = 90^{\circ}$, points $D$, $P$, $Q$, and $C$ are concyclic, leading to
$$
\begin{array}{l}
\angle DCA = \angle DPQ \\
= \angle DPR.
\end{array}
$$
Additionally, since points $D$, $Q$, $R$, ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,918 |
5. Let $n$ be a positive integer, and real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfy $x_{1} \leqslant$ $x_{2} \leqslant \cdots \leqslant x_{n}$.
(a) Prove:
$\left(\sum_{i=1}^{n} \sum_{j=1}^{n}\left|x_{i}-x_{j}\right|\right)^{2}$
$\leqslant \frac{2\left(n^{2}-1\right)}{3} \sum_{i=1}^{n} \sum_{j=1}^{n}\left(x_{i}-x_{... | 5. (a) Since the transformation of $x_{i}$ (subtracting a certain value from all) does not change the inequality on both sides, without loss of generality, assume $\sum_{i=1}^{n} x_{i}=0$, then
$$
\begin{array}{l}
\sum_{i, j=1}^{n}\left|x_{i}-x_{j}\right|=2 \sum_{1<i<j<n}\left(x_{j}-x_{i}\right) \\
=2 \sum_{i=1}^{n}(2 ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 713,919 |
6. Let $p$ be a prime. Prove: There exists a prime $q$, such that for any integer $n$, the number $n^{p}-p$ cannot be divisible by $q$.
| 6. Since $\frac{p^{p}-1}{p-1}=1+p+p^{2}+\cdots+p^{p-1} \equiv p+1$ $\left(\bmod p^{2}\right)$, then $\frac{p^{p}-1}{p-1}$ has at least one prime factor $q$, satisfying $q \not \equiv 1\left(\bmod p^{2}\right)$.
Below we prove that $q$ is the desired prime.
Assume there exists an integer $n$, such that $n^{p}=p(\bmod q)... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 713,920 |
1. Find the smallest positive integer $n$, such that
$$
x_{1}^{3}+x_{2}^{3}+\cdots+x_{n}^{3}=2002^{2002}
$$
has integer solutions.
(Uzbekistan provided) | Solution: Since $2002 \equiv 4(\bmod 9), 4^{3} \equiv 1(\bmod 9), 2002$ $=667 \times 3+1$, therefore,
$$
2002^{2002} \equiv 4^{2002} \equiv 4(\bmod 9) \text {. }
$$
Also, $x^{3} \equiv 0, \pm 1(\bmod 9)$, where $x$ is an integer, thus,
$$
x_{1}^{3}, x_{1}^{3}+x_{2}^{3}, x_{1}^{3}+x_{2}^{3}+x_{3}^{3} \equiv 4(\bmod 9) ... | 4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 713,921 |
4. Does there exist a positive integer $m$ such that the equation
$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a b c}=\frac{m}{a+b+c}
$$
has infinitely many positive integer solutions $(a, b, c)$?
(Provided by Germany) | Solution: Exists.
If $a=b=c=1$, then $m=12$. Let
$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a b c}-\frac{12}{a+b+c}=\frac{p(a, b, c)}{a b c(a+b+c)},
$$
where $p(a, b, c)=a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b)$ $+a+b+c-9 a b c$.
Assume $(x, a, b)$ is a solution satisfying $p(x, a, b)=0$, and $xb$ is another solution.
L... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 713,923 |
5. Given positive integers $m, n \geqslant 2, a_{1}, a_{2}, \cdots, a_{n}$ are integers, and none of them is a multiple of $m^{n-1}$. Prove: There exist integers $e_{1}, e_{2}, \cdots, e_{n}$, not all zero, such that $e_{1} a_{1}+e_{2} a_{2}+\cdots+e_{n} a_{n}$ is a multiple of $m^{n}$. For all $i=1,2, \cdots, n$, $|e_... | Proof: Let $B$ be the set of all $b=\left(b_{1}, b_{2}, \cdots, b_{n}\right)$ where all $b_{i}$ satisfy $0 \leqslant b_{1}<m$. For $b \in B$, let
$$
f(b)=b_{1} a_{1}+b_{2} a_{2}+\cdots+b_{n} a_{n} .
$$
If there exist different $b, b^{\prime} \in B$ such that $f(b) \equiv f\left(b^{\prime}\right)\left(\bmod m^{n}\right... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 713,924 |
1. As shown in Figure 1, let $B$ be a point on circle $S_{1}$. Draw the tangent to circle $S_{1}$ at $B$, and let $A$ be a point on this tangent different from $B$. Let $C$ be a point not on circle $S_{1}$, and the line segment $A C$ intersects circle $S_{1}$ at two distinct points. Circle $S_{2}$ is tangent to $A C$ a... | Proof 1: Let $E$ and $F$ be the midpoints of $BD$ and $CD$ respectively, $K$ be the circumcenter of $\triangle BCD$, and $TDT'$ be the common internal tangent of circles $S_1$ and $S_2$. Then $EK$ is the perpendicular bisector of $BD$ and bisects the angle formed by $BA$ and $DT$. Therefore, the distances from point $K... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,925 |
Example 7 Given that there are exactly 600 integer-sided triangles with unequal sides and the longest side exactly $n$. Find the value of $n$. | Explanation: The key to solving this problem is to establish an equation about $n$. Note that one side of the triangle, $n$, is fixed, and the three sides $x, y, n$ are all integers. Therefore, the number of such triangles is equal to the number of lattice points $(x, y)$ in the coordinate plane.
Let the lengths of th... | 51 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 713,926 |
2. As shown in Figure 2, let $\triangle ABC$ contain a point $F$ such that $\angle AFB = \angle BFC = \angle CFA$. Lines $BF$ and $CF$ intersect $AC$ and $AB$ at $D$ and $E$ respectively. Prove: $AB + AC \geqslant 4DE$.
(Korea provided) | Proof 1: First, we prove a lemma:
Given $\triangle D E F$, points $P, Q$ are on the lines $F D, F E$ respectively, such that $P F \geqslant \lambda D F, Q F \geqslant \lambda E F, \lambda>0$. If $\angle P F Q \geqslant 90^{\circ}$, then $P Q \geqslant \lambda D E$.
Let $\angle P F Q=\theta$. Since $\theta \geqslant 90^... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,927 |
4. As shown in Figure 3, given that circle $S_{1}$ intersects circle $S_{2}$ at points $P$ and $Q$, $A_{1}$ and $B_{1}$ are two points on circle $S_{1}$ different from $P$ and $Q$. Lines $A_{1} P$ and $B_{1} P$ intersect circle $S_{2}$ at $A_{2}$ and $B_{2}$, respectively. Lines $A_{1} B_{1}$ and $A_{2} B_{2}$ intersec... | Proof: Since $\angle A_{1} C A_{2}+\angle A_{1} Q A_{2}=\angle A_{1} C A_{2}+$ $\angle A_{1} Q P+\angle P Q A_{2}=\angle B_{1} C B_{2}+\angle C B_{1} B_{2}+$ $\angle C B_{2} B_{1}=180^{\circ}$, then $A_{1} 、 C 、 A_{2} 、 Q$ are concyclic. Let $O$ be the circumcenter of $\triangle A_{1} A_{2} C$, and $O_{1} 、 O_{2}$ be t... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,928 |
5. For any set $S$ of 5 points in the plane, satisfying that no three points in $S$ are collinear, let $M(S)$ and $m(S)$ be the maximum and minimum areas of triangles formed by any 3 points in $S$, respectively. Find the minimum value of $\frac{M(S)}{m(S)}$.
(Australia provided) | Solution: When these 5 points are the vertices of a regular pentagon, it is easy to see that $\frac{M(S)}{m(S)}$ equals the golden ratio $\tau=\frac{1+\sqrt{5}}{2}$.
Let the 5 points in $S$ be $A, B, C, D, E$, and the area of $\triangle ABC$ be $M(S)$. We will prove that there exists a triangle whose area is less than... | \frac{1+\sqrt{5}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,929 |
7. Given an acute triangle $\triangle A B C$ with its incircle $\odot I$ touching side $B C$ at point $K, A D$ is the altitude of $\triangle A B C$, and $M$ is the midpoint of $A D$. If $N$ is the intersection of $\odot I$ and $K M$, prove that $\odot I$ is tangent to the circumcircle of $\triangle B C N$ at point $N$.... | Proof: When $AB = AC$, obviously, the distance between the centers of these two circles equals the difference of their radii. Without loss of generality, assume $AB < AC$. As shown in Figure 6, let the perpendicular bisector of $BC$ intersect $NK$ at point $P$, and intersect $BC$ at point $A'$. Let the circumcenter of ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,930 |
8. As shown in Figure 7, let circles $S_{1}$ and $S_{2}$ intersect at points $A$ and $B$. A line passing through $A$ intersects circle $S_{1}$ at $C$ and circle $S_{2}$ at $D$. Points $M$, $N$, and $K$ lie on segments $CD$, $BC$, and $BD$ respectively, and $MN \parallel BD$, $MK \parallel BC$. Perpendiculars from $N$ a... | Proof: First, prove the lemma:
Given $\angle P_{1} Q_{1} R_{1}=\angle P_{2} Q_{2} R_{2}, T_{1} 、 T_{2}$ are the projections of $Q_{1} 、 Q_{2}$ on $P_{1} R_{1} 、 P_{2} R_{2}$, respectively. If $\frac{P_{1} T_{1}}{T_{1} R_{1}}=\frac{P_{2} T_{2}}{T_{2} R_{2}}$, then $\triangle P_{1} Q_{1} R_{1} \backsim \triangle P_{2} Q_... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,931 |
1. Find a function $f: \mathbf{R} \rightarrow \mathbf{R}$, such that for any real numbers $x, y$, we have $f(f(x)+y)=2 x+f(f(y)-x)$.
(Czech Republic provided) | Let $y=-f(x)$, then
$$
f(0)=2 x+f(f(-f(x))-x),
$$
i.e., $f(0)-2 x=f(f(-f(x))-x)$,
holds for all $x$.
Thus, for any real number $f(0)-2 x=y$, there exists a real number $z$ such that $y=f(z)$, i.e., the function $f$ is surjective.
Therefore, there exists a real number $a$ such that $f(a)=0$.
Let $x=a$, then the origina... | f(x)=x-a | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,932 |
2. Let $a_{1}, a_{2}, \cdots$ be an infinite sequence of real numbers, such that for all positive integers $i$, there exists a real number $c$, satisfying $0 \leqslant a_{i} \leqslant c$, and $\left|a_{i}-a_{j}\right| \geqslant \frac{1}{i+j}$ for all positive integers $i, j (i \neq j)$. Prove: $c \geqslant 1$. $\quad$.... | Proof: For $n \geqslant 2$, let $\sigma(1), \sigma(2), \cdots, \sigma(n)$ be a permutation of $1,2, \cdots, n$, and satisfy
$$
0 \leqslant a_{\sigma(1)}<a_{\sigma(2)}<\cdots<a_{\sigma(n)} \leqslant c,
$$
then
$$
\begin{aligned}
c \geqslant & a_{\sigma(n)}-a_{\sigma(1)} \\
= & \left(a_{\sigma(n)}-a_{\sigma(n-1)}\right)... | c \geqslant 1 | Algebra | proof | Yes | Yes | cn_contest | false | 713,933 |
3. Let $P(x)=a x^{3}+b x^{2}+c x+d$, where $a, b, c, d$ are integers, and $a \neq 0$. If there are infinitely many pairs of integers $x, y (x \neq y)$, satisfying $x P(x)=y P(y)$, prove: the equation $P(x)=0$ has an integer root.
(Polish contribution) | Proof: Let $x, y$ be distinct integers satisfying $x P(x) = y P(y)$. Then,
$$
x\left(a x^{3} + b x^{2} + c x + d\right) = y\left(a y^{3} + b y^{2} + c y + d\right),
$$
which simplifies to
$$
a\left(x^{4} - y^{4}\right) + b\left(x^{3} - y^{3}\right) + c\left(x^{2} - y^{2}\right) + d(x - y) = 0.
$$
Since $x - y \neq 0$... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 713,934 |
5. Given a positive integer $n$ that is not a perfect cube, define $a=\sqrt[3]{n}$, $b=\frac{1}{a-[a]}$, $c=\frac{1}{b-[b]}$, where $[x]$ denotes the greatest integer not exceeding $x$. Prove: There are infinitely many such integers $n$ for which there exist integers $r, s, t$, not all zero, such that $r a + s b + t c ... | To prove: It is sufficient to prove that there exist rational numbers $r, s, t$ not all zero such that $r a + s b + t c = 0$.
Let $m = [a]$, $k = n - m^3$, then
$1 \leqslant k \leqslant [(m+1)^3 - 1] - m^3 = 3m(m+1)$.
From $a^3 - m^3 = (a - m)(a^2 + am + m^2)$, we get
$b = \frac{1}{a - m} = \frac{a^2 + am + m^2}{k}$.
S... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 713,935 |
1. Let $a, b, c$ be positive integers, satisfying $a \leqslant b \leqslant c$, and $a +$ $b+c=15$. Then, the number of triangles with side lengths $a, b, c$ is ( ) .
(A) 5
(B) 7
(C) 10
(D) 15 | (Tip: Discuss in 5 categories according to $1 \leqslant a \leqslant 5$. Answer: (B).) | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 713,937 |
1. Find all real functions $f, g, h: \mathbf{R} \rightarrow \mathbf{R}$, such that for any real numbers $x, y$, we have
$$
\begin{array}{l}
(x-y) f(x)+h(x)-x y+y^{2} \leqslant h(y) \\
\leqslant(x-y) g(x)+h(x)-x y+y^{2} .
\end{array}
$$
(53rd Romanian Mathematical Olympiad (First Round)) | Solution: From equation (1), we have
$$
(x-y) f(x) \leqslant (x-y) g(x).
$$
It is easy to see that $f(x) = g(x)$ for all real numbers $x$. Therefore, we have
$$
(x-y) f(x) + h(x) - x y + y^{2} = h(y).
$$
Let $x = 0$, we get $h(y) = y^{2} - f(0) y + h(0)$, which means $h$ is a quadratic function. Define $f(0) = a, h(0... | f(x) = g(x) = -x + a, h(x) = x^{2} - a x + b | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,938 |
2. (1) Let $a$ be a real number greater than 1, and let $f, g, h: \mathbf{R} \rightarrow \mathbf{R}$ be real functions such that $f(x) + g(x) + h(x) \geqslant 0$ for all $x \in \mathbf{R}$. Prove that $a^{f(x)} + a^{g(x)} + a^{h(x)} = 3$ has a solution if and only if the functions $f, g, h$ have a common root.
(2) Solv... | (1) Proof: By the AM-GM inequality,
$$
\begin{aligned}
1 & =\frac{1}{3}\left(a^{f(x)}+a^{g(x)}+a^{h(x)}\right) \\
& \geqslant \sqrt[3]{a^{f(x)+g(x)+h(x)}} \geqslant \sqrt[3]{a^{0}}=1 .
\end{aligned}
$$
Therefore, $x$ is a solution to the original equation if and only if
$$
a^{f(x)}=a^{g(x)}=a^{h(x)}, f(x)+g(x)+h(x)=0,... | x=1 \text{ or } -1 | Algebra | proof | Yes | Yes | cn_contest | false | 713,939 |
3. Let the arithmetic sequence $a_{n}(n \geqslant 1)$ contain 1 and $\sqrt{2}$. Prove: any three terms of $\left\{a_{n}\right\}$ do not form a geometric sequence.
(53rd Romanian Mathematical Olympiad (First Round)) | Proof: Let the arithmetic sequence $a_{1}, a_{2}, \cdots, a_{n}, \cdots$ have a common difference of $r$, then there exist positive integers $k, l$ such that $a_{k}=1, a_{l}=\sqrt{2}$. Thus,
$$
\sqrt{2}-1=a_{l}-a_{k}=(l-k) r,
$$
which implies $r=\frac{\sqrt{2}-1}{l-k}$.
Assume $a_{m}, a_{n}, a_{p}$ form a geometric se... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 713,940 |
4. Given the sequence $\left\{x_{n}\right\}$ satisfies $x_{1}=1$, for $n \geqslant 1$, we have
$$
\begin{array}{l}
4\left(x_{1} x_{n}+2 x_{2} x_{n-1}+3 x_{3} x_{n-2}+\cdots+n x_{n} x_{1}\right) \\
=(n+1)\left(x_{1} x_{2}+x_{2} x_{3}+\cdots+x_{n} x_{n+1}\right) .
\end{array}
$$
Find the general term formula for $\left\... | For $n=1$, substituting into equation (1) yields $4 x_{1}^{2}=2 x_{1} x_{2}$, thus $x_{2}=2$.
For $n=2$, substituting into equation (1) yields
$4\left(x_{1} x_{2}+2 x_{2} x_{1}\right)=3\left(x_{1} x_{2}+x_{2} x_{3}\right)$, thus $x_{3}=3$.
Assume $x_{k}=k$, where $1 \leqslant k \leqslant n$. From
$$
4 \sum_{k=1}^{n} k ... | x_{n}=n | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,941 |
5. Find the complex solutions of the system of equations $\left\{\begin{array}{l}x(x-y)(x-z)=3, \\ y(y-x)(y-z)=3 \\ z(z-x)(z-y)=3\end{array}\right.$.
(53rd Romanian Mathematical Olympiad (Second Round)) | Solution: It is easy to know that $x \neq 0, y \neq 0, z \neq 0, x \neq y, y \neq z, z \neq x$. From the first two equations, we can get $x^{2}+y^{2}=y z+z x$.
Similarly, we can get $y^{2}+z^{2}=x y+z x, z^{2}+x^{2}=x y+y z$.
Adding the above three equations, we get
$$
x^{2}+y^{2}+z^{2}=x y+y z+z x \text {. }
$$
Subtr... | (x, y, z) \text{ is a permutation of } \{1, \varepsilon, \varepsilon^2\} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,942 |
7. For a positive integer $n(n \geqslant 2)$, define $f(n)$ as the minimum number of elements in a set $S$ that satisfies the following conditions: the smallest element $1 \in S$, the largest element $n \in S$, and every element other than 1 is the sum of two elements in $S$ (which can be the same). Prove:
(1) $f(n) \g... | Solution: (1) For $n \geqslant 2$, let $f(n)=k$, then $S$ has $k$ elements, and $1=a_{1}<a_{2}<\cdots<a_{k}=n$, so $a_{2}=a_{1}+a_{1}=2$, $a_{3}=a_{1}+a_{2}$ or $a_{2}+a_{2}$. Therefore, $a_{3}=3$ or 4, i.e., $a_{3} \leqslant 2^{2}$.
Assume $a_{i} \leqslant 2^{i-1}$, then $a_{i+1}=a_{r}+a_{s} \leqslant 2 a_{i} \leqslan... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 713,944 |
8. For all non-negative integers $x, y$, find all functions $f: \mathbf{N} \rightarrow \mathbf{N}$, satisfying $f(3 x+2 y)=f(x) f(y)$, where $\mathbf{N}$ is the set of non-negative integers.
(53rd Romanian Mathematical Olympiad (Final)) | Solution: Let $x=y=0$, we get $f(0)=f(0)^{2}$. Therefore, $f(0)=0$ or $f(0)=1$.
If $f(0)=0$, for $x=0$ or $y=0$, we get $f(2 y)=$ $f(3 x)=0$ for all $x, y \in \mathrm{N}$. Let $f(1)=a$, then
$$
\begin{array}{l}
f(5)=f(3 \times 1+2 \times 1)=f(1) f(1)=a^{2}, \\
f(25)=f(3 \times 5+2 \times 5)=f(5) f(5)=a^{4} .
\end{array... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,945 |
9. Find all real polynomials $f$ and $g$ such that for all $x \in \mathbf{R}$, we have $\left(x^{2}+x+1\right) f\left(x^{2}-x+1\right)=\left(x^{2}-x+1\right) g\left(x^{2}+x+1\right)$.
(53rd Romanian Mathematical Olympiad (Final)) | Let $w$ be a non-real cube root of 1, satisfying $w^{2}+w+1=0$, then $g\left(w^{2}+w+1\right)=g(0)=0$.
Let $\alpha$ be a non-real cube root of -1, then $f\left(\alpha^{2}-\alpha+1\right)=$ $f(0)=0$. Therefore, we can set $f(x)=x a(x), g(x)=x b(x)$. Thus, the original condition becomes
$$
a\left(x^{2}-x+1\right)=b\left... | f(x) = kx, g(x) = kx | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,946 |
10. Find all real numbers $a, b, c, d, e \in [-2, 2]$, such that
$$
\begin{array}{l}
a+b+c+d+e=0 \\
a^{3}+b^{3}+c^{3}+d^{3}+e^{3}=0 \\
a^{5}+b^{5}+c^{5}+d^{5}+e^{5}=10
\end{array}
$$
(53rd Romanian Mathematical Olympiad (Final)) | Let $a=2 \cos x, b=2 \cos y, c=2 \cos z, d=2008 t$, $e=2 \cos u$. Since $2 \cos 5 x=(2 \cos x)^{5}-5(2 \cos x)^{3}+$ $5(2 \cos x)=a^{5}-5 a^{3}+5 a$, we have,
$$
\sum 2 \cos 5 x=\sum a^{5}-5 \sum a^{3}+5 \sum a=10,
$$
which means $\sum \cos 5 x=5$.
Thus, $\cos 5 x=\cos 5 y=\cos 5 z=\cos 5 t=\cos 5 u=1$.
From $\cos 5 x... | a, b, c, d, e \in \left\{2, \frac{\sqrt{5}-1}{2}, -\frac{\sqrt{5}+1}{2}\right\} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,947 |
2. If the perimeter of a convex hexagon is 20, all side lengths are integers, and no three sides can form a triangle, then, such a hexagon ( ).
(A) does not exist
(B) exists uniquely
(C) has a finite number, but more than one
(D) has an infinite number
Translate the above text into English, please retain the original ... | (Construct a hexagon with side lengths of $1,1,2,3,5,8$. By the instability of an $n(n \geqslant 4)$-sided polygon, there are infinitely many. Answer:
(D).) | null | Number Theory | proof | Yes | Yes | cn_contest | false | 713,948 |
11. Let $I \subseteq \mathbf{R}$ be an interval, and $f: I \rightarrow \mathbf{R}$ be a function such that the inequality $|f(x)-f(y)| \leqslant|x-y|$ holds for all $x, y \in I$. Prove: $f$ is a monotonic function on $I$ if and only if: for all $x, y \in I$, either $f(x) \leqslant f\left(\frac{x+y}{2}\right) \leqslant ... | Proof: If $f$ is not a monotonic function on the interval $I$, then there exists $x < y < z$ such that $f(x) < f(y) > f(z)$ or $f(x) > f(y) < f(z)$.
Let $\lambda \in \mathbf{R}$, and suppose $f(x) < \lambda < f(z)$. Consider the midpoint of $I_{0} = [x, z]$, then $y$ belongs to either $\left[x, \frac{x+z}{2}\right]$ o... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 713,949 |
12. Find all pairs of sets $A, B$ such that $A, B$ satisfy:
(1) $A \cup B=\mathbf{Z}$;
(2) If $x \in A$, then $x-1 \in B$;
(3) If $x \in B, y \in B$, then $x+y \in A$.
(2002, Romania for IMO and Balkan Mathematical Olympiad Selection Test (First Round)) | Solution: If $0 \in B$, by (3) we know that for all $x \in B, 0+x \in A$, so, $B \subset A$. Also, by (1) we know $A=\mathbf{Z}$. By (2) we know that for $\forall x \in A$, $x-1 \in B$, so, $B=\mathbf{Z}$.
If $0 \notin B$, by (1) we know, $0 \in A$. By (2) we know, $-1 \in B$. By (3) we get $-2=(-1)+(-1) \in A$, and t... | A=2\mathbf{Z}, B=2\mathbf{Z}+1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 713,950 |
13. Let the sequence $\left\{a_{n}\right\}(n \geqslant 0)$ be defined as follows:
$$
a_{0}=a_{1}=1, a_{n+1}=14 a_{n}-a_{n-1}(n \geqslant 1) .
$$
Prove: For all positive integers $n, 2 a_{n}-1$ is a perfect square. | Proof: Consider the sequence $\left\{b_{n}\right\}(n \geqslant 0)$, where $b_{0}=-1, b_{1}$ $=1, b_{n+1}=4 b_{n}-b_{n-1}(n \geqslant 1)$. Using mathematical induction, it is easy to prove:
(1) $2 a_{n}-1=b_{n}^{2}$;
(2) $2 b_{n} b_{n-1}=a_{n}+a_{n-1}-4$.
Therefore, the conclusion holds. | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 713,951 |
14. Let the integer $n \geqslant 4, a_{1}, a_{2}, \cdots, a_{n}$ be positive real numbers such that $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}=1$. Prove:
$$
\begin{array}{l}
\frac{a_{1}}{a_{2}^{2}+1}+\frac{a_{2}}{a_{3}^{2}+1}+\cdots+\frac{a_{n-1}}{a_{n}^{2}+1}+\frac{a_{n}}{a_{1}^{2}+1} \\
\geqslant \frac{4}{5}\left(a_{1} \s... | Proof: By Cauchy's inequality, we have
$$
\frac{a_{1}^{2}}{x_{1}}+\frac{a_{2}^{2}}{x_{2}}+\cdots+\frac{a_{n}^{2}}{x_{n}} \geqslant \frac{\left(a_{1}+a_{2}+\cdots+a_{n}\right)^{2}}{x_{1}+x_{2}+\cdots+x_{n}},
$$
where $x_{1}, x_{2}, \cdots, x_{n}$ are any positive real numbers. Therefore, we have
$$
\begin{array}{l}
\fr... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 713,952 |
15. Let $m, n$ be positive integers of different parity, and satisfy $m < n < 5m$. Prove: There exists a partition that splits the set $\{1,2,3, \cdots, 4mn\}$ into several subsets, each containing exactly two elements, such that the sum of the two elements in each subset is a perfect square.
(2002, Romania for IMO and... | Proof: If there exist non-negative integers $a, b, a<b$, such that
(1) $(2 a+1)^{2}+4 m n=(2 b+1)^{2}$,
(2) $(2 a+1)^{2}<4 m n$,
then the set $\{1,2,3, \cdots, 4 m n\}$ can be decomposed as follows:
$$
\begin{array}{r}
\left\{1,(2 a+1)^{2}-1\right\},\left\{2,(2 a+1)^{2}-2\right\}, \cdots,\left\{2 a^{2}+\right. \\
\lef... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 713,953 |
16. Find all real values of $m$ such that the inequality
$$
m x^{2}+8(m-1) x+7 m-16 \leqslant 0
$$
has at most 6 integer solutions, and one of them is 2.
(2002, Bulgarian Spring Mathematical Competition) | Solution: Let $f(x)=m x^{2}+8(m-1) x+7 m-16$. Since the inequality $f(x) \leqslant 0$ has a finite number of integer solutions, it must be that $m>0$.
Since 2 is a solution, then $f(2) \leqslant 0$, which means $m \leqslant \frac{32}{27}$. Also, because $f(-1)=-80, f(3)>\frac{48}{55}$. Therefore, $m \in\left(\frac{48}... | m \in\left(\frac{48}{55}, 1\right) \cup\left(1, \frac{32}{27}\right] | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 713,954 |
17. Find all pairs of positive integers $(b, c)$ such that the sequence $a_{1}=b, a_{2}=c, a_{n+2}=\left|3 a_{n+1}-2 a_{n}\right|(n \geqslant 1)$, has only finitely many composite numbers.
(2002, Bulgaria National Mathematical Olympiad (Final)) | Let $(b, c)$ be the pair of numbers that satisfy the conditions.
If $a_{k}=a_{k+1}$, and $k$ is the smallest index that satisfies this condition. If $k \geqslant 4$, since $a_{k+1}=\left|3 a_{k}-2 a_{k-1}\right|$, and $a_{k} \neq a_{k-1}$, then $a_{k+1}=2 a_{k-1}-3 a_{k}$, i.e., $a_{k-1}=2 a_{k}$.
Also, because $a_{k}... | (b, c)=(5,4) \text{ or } (7,4) \text{ or } (2 p, p) \text{ or } (p, p), \text{ where } p \in P | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 713,955 |
18. Find all real polynomials $p(x)$, for all real numbers $x$, that satisfy the equation $p(2 p(x))=2 p(p(x))+2(p(x))^{2}$.
(19th Iranian Mathematical Olympiad (First Round)) | Prove: Let $p(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}$, then the leading terms of $p(2 p(x)), 2 p(p(x)), 2(p(x))^{2}$ are $2^{n} a_{n}^{n+1} x^{n^{2}}, 2 a_{n}^{n+1} x^{n^{2}}, 2 a_{n}^{2} x^{2 n}$, respectively.
If $n \geqslant 3$, then $n^{2}>2 n$, so $2^{n} a_{n}^{n+1}=2 a_{n}^{n+1}$. Since $a_{n} \neq ... | p(x) \equiv 0, p(x) \equiv-\frac{1}{2}, p(x)=x^{2}+b x, b \in \mathbf{R} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,956 |
19. Find all functions $f: \mathbf{R} \backslash\{0\} \rightarrow \mathbf{R}$, such that for all $x, y \in \mathbf{R} \backslash\{0\}$, we have
$$
x f\left(x+\frac{1}{y}\right)+y f(y)+\frac{y}{x}=y f\left(y+\frac{1}{x}\right)+x f(x)+\frac{x}{y} \text {. }
$$
(19th Iranian Mathematical Olympiad (Second Round)) | Let $g(x)=f(x)-x$, then the original functional equation becomes
$$
x g\left(x+\frac{1}{y}\right)+y g(y)=y g\left(y+\frac{1}{x}\right)+x g(x) \text {. }
$$
Let $y=1$, we get
$$
x g(x+1)+g(1)=g\left(1+\frac{1}{x}\right)+x g(x) \text {. }
$$
Substitute $x$ with $\frac{1}{x}$, we have
$$
\frac{1}{x} g\left(1+\frac{1}{x}... | f(x)=A+\frac{B}{x}+x | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,957 |
20. Consider a permutation $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ of $\{1,2, \cdots, n\}$. If at least one of $a_{1}, a_{1}+a_{2}, \cdots, a_{1}+a_{2}+\cdots+a_{n}$ is a perfect square, it is called a "square permutation". Find all positive integers $n$ such that every permutation of $\{1,2, \cdots, n\}$ is a "squa... | Solution: Let $a_{i}=i, 1 \leqslant i \leqslant n, b_{k}=\sum_{i=1}^{k} a_{i}$. If $b_{k}=m^{2}$, i.e., $\frac{k(k+1)}{2}=m^{2}$, then $\frac{k}{2}<m$, and $\frac{(k+1)(k+2)}{2}=$ $m^{2}+k+1<(m+1)^{2}$. Therefore, $b_{k+1}$ is not a perfect square.
If $a_{k}$ and $a_{k+1}$ are swapped, then the sum of the first $k$ te... | n=\frac{1}{4}\left[(3+2 \sqrt{2})^{k}+(3-2 \sqrt{2})^{k}-2\right], k=1,2, \cdots | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 713,958 |
3. Among all triangles with side lengths as consecutive positive integers and a perimeter not exceeding 100, the number of acute triangles is | (Tip: Establish an inequality relationship, determine the range of side lengths. Answer: 29.) | 29 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,959 |
21. Let $x_{1}, x_{2}, \cdots, x_{n}$ be positive real numbers, and satisfy $\sum_{i=1}^{n} x_{i}^{2}=n$, $\sum_{i=1}^{n} x_{i} \geqslant S>0$. For $0 \leqslant \lambda \leqslant 1$, prove: there are at least $\left[\frac{S^{2}(1-\lambda)^{2}}{n}\right]$ numbers greater than $\frac{\lambda S}{n}$.
(19th Iranian Mathema... | Proof: Define $A=\left\{j \mid 1 \leqslant j \leqslant n, x_{j}>\frac{\lambda S}{n}\right\}$. By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
S \leqslant \sum_{i=1}^{n} x_{i}=\sum_{i \in A} x_{i}+\sum_{i \notin A} x_{i} \\
\leqslant \sqrt{|A| \sum_{i \in A} x_{i}^{2}}+\sum_{i \notin A} x_{i} \\
\leqslant ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 713,960 |
1. Find all positive integers $n$, such that $n^{4}-4 n^{3}+22 n^{2}-$ $36 n+18$ is a perfect square. | 1. Let $m \in \mathbf{N}_{+}$, such that $n^{4}-4 n^{3}+22 n^{2}-36 n+18=$ $m^{2}$, completing the square we get
$$
\left(n^{2}-2 n+9\right)^{2}-63=m^{2} \text {. }
$$
Thus, $63=\left(n^{2}-2 n+9\right)^{2}-m^{2}$
$$
=\left(n^{2}-2 n+9-m\right)\left(n^{2}-2 n+9+m\right) \text {. }
$$
Notice that, $n^{2}-2 n+9+m=(n-1)... | n=1 \text{ or } n=3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,961 |
2. Let $O$ be the circumcenter of acute $\triangle A B C$, and $P$ a point inside $\triangle A O B$. The projections of $P$ onto the sides $B C$, $C A$, and $A B$ of $\triangle A B C$ are $D$, $E$, and $F$, respectively. Prove that the parallelogram with $F E$ and $F D$ as adjacent sides lies inside $\triangle A B C$. | 2. As shown in Figure 1, construct square $EFDG$ with $FE$ and $FD$ as adjacent sides. To prove the proposition, it is only necessary to prove that
$$
\angle FDE < \angle CED, \text{ (1) }
$$
and $\angle FED < \angle EDC$. (2)
Noting that (1) and (2) are symmetric, it is only necessary to prove one of them, and the ot... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,962 |
3. Consider a square on the complex plane, whose 4 vertices correspond to the 4 roots of a certain monic quartic equation with integer coefficients $x^{4}+p x^{3}+q x^{2}+r x+s=0$. Find the minimum value of the area of such a square.
| 3. According to the problem, the 4 roots of the equation can only be in two scenarios: 2 real roots and 1 pair of conjugate complex roots; 2 pairs of conjugate complex roots.
(1) If the 4 roots of the equation are 2 real roots and 1 pair of conjugate complex roots, then we can set these 4 roots as $a \pm b, a \pm b \ma... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,963 |
4. Let $n$ be a positive integer, and let $A_{1}, A_{2}, \cdots, A_{n+1}$ be $n+1$ non-empty subsets of the set $\{1,2, \cdots, n\}$. Prove: There exist two non-empty disjoint subsets $\left\{i_{1}, i_{2}, \cdots, i_{k}\right\}$ and $\left\{j_{1}, j_{2}, \cdots, j_{m}\right\}$ of $\{1,2, \cdots, n+1\}$, such that
$$
A_... | 4. Prove by induction on $n$, using mathematical induction.
When $n=1$, we must have $A_{1}=A_{2}=\{1\}$, and the proposition is proved.
Assume the proposition holds for $n=l$, consider the case for $n=l+1$.
If there exists $A_{i}=A_{j}$, then the proposition holds. Therefore, we can assume that $A_{1}, A_{2}, \cdots,... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 713,964 |
5. In the given trapezoid $A B C D$, $A D / / B C, E$ is a moving point on side $A B$, $O_{1} 、 O_{2}$ are the circumcenters of $\triangle A E D 、 \triangle B E C$ respectively. Prove: $\mathrm{O}_{1} \mathrm{O}_{2}$ is a constant length. | 5. As shown in Figure 2, connect $O_{1} E$, $O_{1} D$, $O_{2} E$, and $O_{2} C$ (assuming $\angle \mathrm{A} \geqslant \angle B$). Note that
$$
\begin{array}{l}
\angle E O_{1} D=360^{\circ}-2 \angle A \\
= 2\left(180^{\circ}-\angle A\right) \\
= 2 \angle B . \text { Therefore, } \\
\angle O_{1} E D=\frac{1}{2}\left(18... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,965 |
6. Let $n(n \geqslant 2)$ be a given positive integer, find all integer tuples $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ that satisfy the conditions:
(1) $a_{1}+a_{2}+\cdots+a_{n} \geqslant n^{2}$;
(2) $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2} \leqslant n^{3}+1$. | 6. From the conditions, we have
$$
\begin{array}{l}
\sum_{k=1}^{n}\left(a_{k}-n\right)^{2}=\sum_{k=1}^{n} a_{k}^{2}-2 n \sum_{k=1}^{n} a_{k}+n^{3} \\
\leqslant\left(n^{3}+1\right)-2 n^{3}+n^{3}=1 . \\
\text { Therefore, } \sum_{k=1}^{n}\left(a_{k}-n\right)^{2}=0 \text { or } 1 . \\
\text { If } \sum_{k=1}^{n}\left(a_{k... | (n, n, \cdots, n) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 713,966 |
7. Let $\alpha, \beta$ be the roots of the equation $x^{2}-x-1=0$, and let $a_{n}=\frac{\alpha^{n}-\beta^{n}}{\alpha-\beta}, n=1,2, \cdots$.
(1) Prove: For any positive integer $n$, $a_{n+2}=a_{n+1}+a_{n}$;
(2) Find all positive integers $a, b, a<b$, such that for any positive integer $n$, $b$ divides $a_{n}-2 n a^{n}$... | 7. (1) From the conditions, we know that $\alpha^{2}=\alpha+1, \beta^{2}=\beta+1$. Therefore,
$$
\begin{array}{l}
a_{n+2}=\frac{\alpha^{n+2}-\beta^{n+2}}{\alpha-\beta}=\frac{\left(\alpha^{n+1}+\alpha^{n}\right)-\left(\beta^{n+1}+\beta^{n}\right)}{\alpha-\beta} \\
=\frac{\alpha^{n+1}-\beta^{n+1}}{\alpha-\beta}+\frac{\al... | a=3, b=5 | Algebra | proof | Yes | Yes | cn_contest | false | 713,967 |
8. Let $S=\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ be the longest sequence composed of 0s and 1s that satisfies the following condition: any two consecutive 5 terms in the sequence $S$ are different, i.e., for any $1 \leqslant i<j \leqslant n-4, a_{i}, a_{i+1}$, $a_{i+2}, a_{i+3}, a_{i+1}$ and $a_{j}, a_{j+1}, a_{j+2}... | 8. Proof by contradiction.
If the first 4 terms and the last 4 terms of $S$ are not the same, let the last 4 terms of $S$ be abod. Since $S$ is the longest sequence with the property mentioned in the problem, adding 0 or 1 to the end of $S$ would result in the 5-term segments $a b c d 0$ and $a b o d 1$ appearing in $... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 713,968 |
1. Positive real numbers $x, y$ satisfy $x y=1$. Then, $\frac{1}{x^{4}}+$ $\frac{1}{4 y^{4}}$ has the minimum value of ( ).
(A) $\frac{1}{2}$
(B) $\frac{5}{8}$
(C) 1
(D) $\sqrt{2}$ | $-1 .(\mathrm{C})$.
Since $x=\frac{1}{y}$, we have,
$$
\frac{1}{x^{4}}+\frac{1}{4 y^{4}}=\frac{1}{x^{4}}+\frac{x^{4}}{4}=\left(\frac{1}{x^{2}}-\frac{x^{2}}{2}\right)^{2}+1 \text {. }
$$
When $\frac{1}{x^{2}}=\frac{x^{2}}{2}$, i.e., $x=\sqrt[4]{2}$, $\frac{1}{x^{4}}+\frac{1}{4 y^{4}}$ is minimized, and the minimum
valu... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 713,969 |
Example 1 Find a point $P$ on the plane of equilateral $\triangle A B C$ such that $\triangle P A B$, $\triangle P B C$, and $\triangle P C A$ are all isosceles triangles. How many points $P$ with this property are there? | Solution: Let any side of the known triangle be denoted as $a$. When $a$ is the base of the sought isosceles triangle, point $P$ lies on the perpendicular bisector of $a$; when $a$ is one of the legs of the sought isosceles triangle, point $P$ lies on the circumference of the circle with the vertex of $\triangle ABC$ a... | 10 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,970 |
Proposition 2
As shown in Figure 2, let $I$ be the incenter of $\triangle ABC$, and points $B_{1}, C_{1}$ be the midpoints of sides $AC$ and $AB$, respectively. Ray $B_{1}I$ intersects side $AB$ at point $B_{2}$, and ray $C_{1}I$ intersects the extension of $AC$ at point $C_{2}$. Then $S_{\triangle ABC} = S_{\triangle ... | Necessity. From $S_{\triangle A B C}=r p$ we get
$S_{\triangle A B_{1} B_{2}}=\frac{A B_{1} \cdot A B_{2}}{A B \cdot \overline{A C}} \cdot S_{\triangle A B C}=\frac{A B_{2}}{2 c} \cdot r p$,
$S_{\triangle A B B_{2}}=\frac{r}{2} \cdot A B_{2}, S_{\triangle A B B_{1}}=\frac{b r}{4}$.
And $S_{\triangle A B_{1} B_{2}}-S_{\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,972 |
Text [1] provides a problem from the 2000 Asia Pacific Mathematical Olympiad:
As shown in Figure 1, let $A M$ and $A N$ be a median and an angle bisector of $\triangle A B C$, respectively. Draw a perpendicular line to $A N$ through point $N$, intersecting $A M$ and $A B$ at points $Q$ and $P$, respectively. Draw a pe... | This article proves the problem using pure plane geometry methods and provides several variations of the problem. First, we prove a lemma.
Lemma: $AM$ is a ray inside $\angle BAX$. Take any point $L$ on $AX$, draw $PL \perp AX$ intersecting $AB$ at point $P$ and $AM$ at point $Q$, then draw $PO \perp AB$ intersecting ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,973 |
Variation 1 In $\triangle A B C$, $B L$ is perpendicular to the angle bisector $A N$ of $\angle A$ at point $L$. Extend the median $A M$ to intersect $B L$ at point $Q, B O \perp A B$ intersects the extension of $A N$ at point $O$. Prove: $O Q \perp B C$. | Proof 1: In Figure 3, $O Q$ and $B N$ are corresponding sides of $\triangle B O Q$ and $\triangle A B N$, respectively. We will prove that
$\triangle B O Q \sim \triangle A B N$.
Extend $B L$ and $A C$ to intersect at point $D$, and extend $L M$ to intersect $A B$ at $T$. Since $M L$ is the midline of $\triangle B C D$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,974 |
Variation 2 As shown in Figure 4, through the midpoint $M$ of side $BC$ of $\triangle ABC$, draw a perpendicular to the angle bisector $AN$ at point $L$, and let it intersect $AB$ and $AC$ at points $P$ and $D$, respectively. $PO \perp AB$ intersects $AN$ at $O$. Prove: $OM \perp BC$. | Proof 2: Using line $PMD$ to intercept $\triangle ABC$, by Menelaus' theorem we have $\frac{AP}{PB} \cdot \frac{BM}{MC} \cdot \frac{CD}{DA}=1$. Therefore, $PB=DC$.
Since $O$ is a point on the angle bisector of $\angle A$, by symmetry it is easy to know that $OD \perp AC$, and $OP=OD$. Thus,
Rt $\triangle OPB \cong$ Rt... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,975 |
Proposition 1 Let $A$ be any point on the line segment $B C$, and construct two similar isosceles triangles $\triangle E B A$ and $\triangle F A C$ with $B A$ and $A C$ as their bases, respectively. Then, construct another isosceles triangle $\triangle D B C$ with $B C$ as its base, such that its vertex angle is the su... | Proof: As shown in Figure 6, extend $A E$ and $D B$ to intersect at point $M$, and extend $A F$ and $D C$ to intersect at point $N$. It is easy to see that $\triangle M B A \sim \triangle N C A$, and $B E$ and $C F$ are corresponding sides in these similar triangles, thus
$$
\frac{A E}{A M}=\frac{A F}{A N}.
$$
Therefo... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 713,976 |
In $\triangle A B C$, $A P$ bisects $\angle B A C$, intersecting $B C$ at $P$, and $B Q$ bisects $\angle A B C$, intersecting $C A$ at $Q$. It is known that $\angle B A C=$ $60^{\circ}$, and $A B+B P=A Q+Q B$. What are the possible values of the angles of $\triangle A B C$? | Solution: As shown in Figure 1, take point $D$ on the extension of $AB$ such that $BD = BP$; take point $E$ on the extension of $AQ$ such that $QE = QB$. Connect $PD$ and $PE$, then
$$
AD = AB + BP = AQ + QB = AE,
$$
and $\triangle ADP \cong \triangle AEP$.
Therefore, $\angle AEP = \angle ADP = \frac{1}{2} \angle ABC ... | \angle ABC = 80^{\circ}, \angle ACB = 40^{\circ}, \angle BAC = 60^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,977 |
Weisenböck Inequality:
Let the three sides and the area of $\triangle A B C$ be $a, b, c$ and $S$, respectively. Then we have
$$
a^{2}+b^{2}+c^{2} \geqslant 4 \sqrt{3} S .
$$ | Proof 1: $a^{2}+b^{2}+c^{2} \geqslant a^{2}+\frac{1}{2}(b+c)^{2}$
$$
\begin{array}{l}
=\frac{3}{2} a^{2}+\frac{1}{2}\left[(b+c)^{2}-a^{2}\right] \\
\geqslant \sqrt{3} \cdot \sqrt{\left[(b+c)^{2}-a^{2}\right] a^{2}} \\
\geqslant \sqrt{3} \cdot \sqrt{\left[(b+c)^{2}-a^{2}\right]\left[a^{2}-(b-c)^{2}\right]} .
\end{array}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 713,978 |
Proposition 1 Let the altitudes and exradii, circumradius, and inradius of $\triangle A B C$ be $h_{a}, h_{b}, h_{c}, r_{a}, r_{b}, r_{c}, R, r$. For $n \geqslant 1$, we have
$$
\frac{r_{a}^{n}}{h_{a}^{n}}+\frac{r_{b}^{n}}{h_{b}^{n}}+\frac{r_{c}^{n}}{h_{c}^{n}} \geqslant 3\left(\frac{2 R-r}{3 r}\right)^{n} \text {. }
$... | Proposition Proof: From the identity in a triangle $a h_{a} = 2 p r$ and similar expressions (4), and the inequality $\frac{a^{n}+b^{n}+c^{n}}{3} \geqslant \left(\frac{a+b+c}{3}\right)^{n}$, we have
$$
\begin{array}{l}
\frac{r_{a}^{n}}{h_{a}^{n}}+\frac{r_{b}^{n}}{h_{b}^{n}}+\frac{r_{c}^{n}}{h_{c}^{n}}=\sum \frac{r_{a}^... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 713,980 |
2. Given an $n \times n$ ($n$ is an odd number) chessboard where each unit square is colored in a checkerboard pattern, and the 4 corner unit squares are colored black. A figure formed by 3 connected unit squares in an L-shape is called a "domino". For what value of $n$ can all the black squares be covered by non-overl... | Solution: Let $n=2m+1$, consider the odd rows, then each row has $m+1$ black cells, with a total of $(m+1)^2$ black cells. Any two black cells cannot be covered by a single "domino", therefore, at least $(m+1)^2$ "dominoes" are needed to cover all the black cells on the chessboard. Since when $n=1,3,5$, we have $3(m+1)... | 7 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 713,981 |
Example 3 As shown in Figure 4, in a $4 \times 4$ grid square,
construct a grid point $\triangle A B C$
$(A B=\sqrt{5}, B C=$
$\sqrt{13}, C A=\sqrt{10}$ ). How many
grid point triangles congruent to
$\triangle A B C$ (including
$\triangle A B C)$ can be
constructed in Figure 4? | (1) As shown in Figure 5, $\triangle ABC$ is an inscribed lattice triangle in a $3 \times 3$ square, and there are 4 such $3 \times 3$ squares in Figure 4.
(2) As shown in Figure 5, one vertex of the triangle is at a vertex of the $3 \times 3$ square (point $C$, which has 4 possible positions), and the other two vertic... | 32 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 713,982 |
3. Let $n$ be a positive integer. If a sequence of $n$ positive integers (which can be the same) is called "full", then this sequence should satisfy the condition: for each positive integer $k (k \geqslant 2)$, if $k$ is in this sequence, then $k-1$ is also in this sequence, and the first occurrence of $k-1$ is before ... | Solution: There are $n$! "full" sequences.
To prove this conclusion, we construct a "full" sequence that is in bijection with the permutations of the set $\{1,2, \cdots, n\}$.
Let $a_{1}, a_{2}, \cdots, a_{n}$ be a "full" sequence, $r=$ $\max \left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$. Then, all integers from 1 to $r... | n! | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 713,983 |
4. Let $T$ be a set of ordered triples $(x, y, z)$, where $x, y, z$ are integers, and $0 \leqslant x, y, z \leqslant 9$. Two players, A and B, play the following game: A selects a triple $(x, y, z)$ from $T$, and B has to guess A's chosen triple using several "moves". One "move" consists of: B giving A a triple $(a, b,... | Solution: Two "movements" are not enough. Because each answer is an even number between 0 and 54, i.e., there are 28 possible values for each answer. The maximum number of possible results from two "movements" is $28^{2}$, which is less than the 1000 possible values for $(x, y, z)$.
Below, we prove that 3 "movements" ... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 713,984 |
5. Let $r(r \geqslant 2)$ be a fixed positive integer, and let $F$ be an infinite family of sets, each containing $r$ elements. If any two sets in $F$ have a non-empty intersection, prove: there exists a set with $r-1$ elements that has a non-empty intersection with every set in $F$.
(Brazil provided) | Proof: We prove the following proposition:
If set $A$ is a set with fewer than $r$ elements, and is included in infinitely many sets of $F$, then either $A$ has a non-empty intersection with all sets in $F$, or there exists an $x \notin A$, such that $A \cup\{x\}$ is included in infinitely many sets of $F$.
Of course,... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 713,985 |
6. Given that $n$ is a positive even number. Prove: there exists a permutation $x_{1}, x_{2}, \cdots, x_{n}$ of $1, 2, \cdots, n$ such that for every $1 \leqslant i \leqslant n, x_{i+1}$ is one of the four numbers $2 x_{4}, 2 x_{i}-1, 2 x_{4}-n, 2 x_{i}-n-1$, where $x_{n+1}=x_{1}$.
(Poland provided) | Proof: Let $n=2 m$, we define a directed graph $G: G$ has $m$ vertices, labeled as $1,2, \cdots, m$; and $2 m$ edges, labeled as $1,2, \cdots, 2 m$. The edges emanating from vertex $i$ are labeled $2 i-1$ and $2 i$; the edges entering vertex $i$ are labeled $i$ and $i+m$.
We inductively define the $j$-th edge, where $... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 713,986 |
7. There are 120 people. Any two people are either friends or not friends. A set of four people that contains exactly one pair of friends is called a "weak quartet." Find the maximum number of "weak quartets." (Provided by New Zealand)
| Solution: Let the 120 people be 120 points in graph $G$. If two people know each other, then connect the points corresponding to these two people with a line segment. Let $Q(G)$ be the number of "weak quartets" in graph $G$. If $x, y$ are two points in $G$, and there is a line segment between $x, y$, let $G^{\prime}$ s... | 4769280 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 713,987 |
22. Let real numbers $\alpha, \beta, \gamma$ satisfy $\beta \gamma \neq 0$, and $\frac{1-\gamma^{2}}{\beta \gamma} \geqslant 0$. Prove: $10\left(\alpha^{2}+\beta^{2}+\gamma^{2}-\beta \gamma^{3}\right) \geqslant 2 \alpha \beta+5 \alpha \gamma$.
(19th Greek Mathematical Olympiad) | Proof: Since $\frac{1-\gamma^{2}}{\beta \gamma} \geqslant 0 \Leftrightarrow \beta \gamma\left(1-\gamma^{2}\right) \geqslant 0$, therefore,
$$
10\left(\alpha^{2}+\beta^{2}+\gamma^{2}-\beta \gamma^{3}\right) \geqslant 10\left(\alpha^{2}+\beta^{2}+\gamma^{2}-\beta \gamma\right) \text {. }
$$
Thus, it is sufficient to pro... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 713,988 |
23. Let $x$, $y$, and $a$ be real numbers. And satisfy $x+y=x^{3}+y^{3}=$ $x^{5}+y^{5}=a$. Find all possible values of $a$.
(Greece for the 43rd IMO selection exam) | If $x=-y$, then $a=0$.
Let $x, y$ be the two roots of the quadratic equation $z^{2}-a z+p=0$, then
$$
\begin{array}{l}
x+y=a, \\
x^{2}+y^{2}=(x+y)^{2}-2 x y=a^{2}-2 p, \\
x^{3}+y^{3}=a^{3}-3 a p, \\
x^{4}+y^{4}=\left(x^{2}+y^{2}\right)^{2}-2 x^{2} y^{2}=a^{4}-4 a^{2} p+2 p^{2}, \\
x^{5}+y^{5}=a^{5}-5 a^{3} p+5 a p^{2} ... | -2, -1, 0, 1, 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,989 |
24. Given the sequence $a_{1}=20, a_{2}=30, a_{n+2}=3 a_{n+1}-a_{n}$ $(n \geqslant 1)$. Find all positive integers $n$ such that $1+5 a_{n} a_{n+1}$ is a perfect square.
(19th Balkan Mathematical Olympiad) | Let $b_{n}=a_{n}+a_{n+1}, c_{n}=1+5 a_{n} a_{n+1}$, then
$5 a_{n+1}=b_{n+1}+b_{n}, a_{n+2}-a_{n}=b_{n+1}-b_{n}$.
Therefore, $c_{n+1}-c_{n}=5 a_{n+1}\left(a_{n+2}-a_{n}\right)=b_{n+1}^{2}-b_{n}^{2}$.
Thus, $c_{n+1}-b_{n+1}^{2}=c_{n}-b_{n}^{2}=\cdots=c_{1}-b_{1}^{2}$
$$
=501=3 \times 167 \text {. }
$$
Suppose there exis... | n=3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,990 |
25. For a natural number $n(n \geqslant 2)$, for what value of the parameter $p$ does the system of equations
$$
\left\{\begin{array}{l}
x_{1}^{4}+\frac{2}{x_{1}^{2}}=p x_{2}, \\
x_{2}^{4}+\frac{2}{x_{2}^{2}}=p x_{3}, \\
\cdots \cdots \\
x_{n-1}^{4}+\frac{2}{x_{n-1}^{2}}=p x_{n} \\
x_{n}^{4}+\frac{2}{x_{n}^{2}}=p x_{1}... | Solution: Clearly, $p \neq 0$, and if $\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ is a solution to the original system of equations, then changing $p$ to $-p$ makes $\left(-x_{1},-x_{2}, \cdots,-x_{n}\right)$ also a solution to the original system of equations.
Assume without loss of generality that $\rho>0$, and $\lef... | (-\infty,-2 \sqrt{2}) \cup(2 \sqrt{2},+\infty) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,991 |
26. Find all polynomials $P(x)$ with real coefficients, such that for all real numbers $x$, the equation $(x+1) P(x-1)+(x-1) P(x+1)=2 x P(x)$ holds.
(51st Czech and Slovak Mathematical Olympiad (Open Book)) | When $x=1$ and $x=-1$, we get
$$
P(1)=P(0)=P(-1) \text {. }
$$
Let $P(0)=d, P(x)=x(x-1)(x+1) Q(x)+d$, substituting into the original equation and eliminating $x(x-1)(x+1)$, we get
$$
(x-2) Q(x-1)+(x+2) Q(x+1)=2 x Q(x) \text {. }
$$
By the problem's condition, when $x \neq 0, \pm 1$, the above equation holds. Therefor... | P(x)=a x^{3}-a x+d | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,992 |
Example 1 Given $x_{i} \in \mathbf{R}, i=1,2, \cdots, n, n \geqslant 2$, satisfying
$$
\sum_{i=1}^{n}\left|x_{i}\right|=1, \sum_{i=1}^{n} x_{i}=0 .
$$
Prove: $\left|\sum_{i=1}^{n} \frac{x_{i}}{i}\right| \leqslant \frac{1}{2}-\frac{1}{2 n}$.
(1989, National High School Mathematics Competition) | Let the sum of all non-negative numbers among $x_{i}$ be $A$, and the sum of all negative numbers be $B$. Then, by the given conditions, we have $A-B=1$ and $A+B=0$. Therefore, it must be that $A=\frac{1}{2}$ and $B=-\frac{1}{2}$.
Let $S_{k}=\sum_{i=1}^{k} x_{i}, k=1,2, \cdots, n$, and let $S_{0}=0$. Then
$$
\left|S_{... | \frac{1}{2}-\frac{1}{2 n} | Inequalities | proof | Yes | Yes | cn_contest | false | 713,993 |
27. Find the solutions of the system of equations $\left\{\begin{array}{l}x^{2}-1=p(y+z) . \\ y^{2}-1=p(z+x), \\ z^{2}-1=p(x+y)\end{array}\right.$ and discuss the number of solutions. Here $x, y, z$ are real numbers, and $p$ is a parameter.
(51st Czech and Slovak Mathematical Olympiad (Second Round)) | Solution: From the first two equations, we get $x^{2}-y^{2}=p(y-x)$, which is
$(x-y)(x+y+p)=0$.
Similarly, we get $(x-z)(x+z+p)=0$.
Therefore, the solutions of the system of equations can only be in the form $(u, u, u)$ or $(u, u, -p-u)$.
(1) If $(u, u, u)$ is a solution to the original system of equations, then $u^{2}... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,994 |
28. Find $a$ and $b$ in the set of real numbers such that the equation
$$
\frac{a x^{2}-24 x+b}{x^{2}-1}=x
$$
has two roots, and the sum of these roots is 12, where a repeated root counts as one root.
(51st Czech and Slovak Mathematical Olympiad (Final)) | Solution: Transform and simplify the original equation to get
$$
x^{3}-a x^{2}+23 x-b=0 .
$$
Since a cubic equation has either one root or three roots in the real number range, these three roots are either one of 1 or -1, and the other two roots are not equal; or there is no root of $\pm 1$, but there is a double root... | (11, -35), (35, -5819) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,995 |
29. Let $\mathbf{R}_{+}$ denote the set of positive real numbers. Find the function $f: \mathbf{R}_{+} \rightarrow \mathbf{R}_{+}$, such that for all $x, y \in \mathbf{R}_{+}$, we have $f(x f(y))=f(x y)+x$.
(51st Czech and Slovak Mathematical Olympiad (Final)) | Solution: Replace $x$ in the original equation with $f(x)$, we get
$$
f(f(x) f(y))=f(f(x) y)+f(x) \text {. }
$$
By swapping $x$ and $y$ in the original equation, we get
$$
f(y f(x))=f(y x)+y \text {, }
$$
Thus, $f(f(x) f(y))=f(y x)+y+f(x)$.
Swapping $x$ and $y$, the value on the left side remains unchanged, hence,
$$... | f(x)=x+1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 713,996 |
30. Given positive real numbers $a, b, c, d$. Prove:
$$
\begin{array}{l}
\sqrt{(a+c)^{2}+(b+d)^{2}} \leqslant \sqrt{a^{2}+b^{2}}+\sqrt{c^{2}+d^{2}} \\
\leqslant \sqrt{(a+c)^{2}+(b+d)^{2}}+\frac{2|a d-b c|}{\sqrt{(a+c)^{2}+(b+d)^{2}}} .
\end{array}
$$
(52nd Belarusian Mathematical Olympiad (Final B Category)) | Proof: Let $u=(a, b), v=(c, d)$, then the inequality
$$
\sqrt{(a+c)^{2}+(b+d)^{2}} \leqslant \sqrt{a^{2}+b^{2}}+\sqrt{c^{2}+d^{2}}
$$
is a triangle inequality constructed by vectors $\boldsymbol{u}$ and $\boldsymbol{v}$.
From (1), to prove the inequality
$$
\begin{array}{l}
\sqrt{a^{2}+b^{2}}+\sqrt{c^{2}+d^{2}} \\
\leq... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 713,997 |
31. Given that $A$ and $B$ are finite sets composed of different positive real numbers, $n$ ($n>1$) is a given positive integer, and both $A$ and $B$ have at least $n$ elements. If the sum of any $n$ different real numbers in $A$ belongs to $B$, and the product of any $n$ different real numbers in $B$ belongs to $A$, f... | Assume set $A$ contains $m$ elements, denoted as $a_{1}, a_{2}$, $\cdots, a_{m}$, and satisfies $0n$. Let
$$
\begin{array}{l}
S=a_{1}+a_{2}+\cdots+a_{n+1}, \\
P=\left(S-a_{1}\right)\left(S-a_{2}\right) \cdots\left(S-a_{n+1}\right),
\end{array}
$$
Then $S-a_{k}(k=1,2, \cdots, n+1)$ belongs to $B$. Furthermore, let
$$
\... | 2n | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 713,998 |
32. Given rational numbers $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{n}$, for any positive integer $k$, it satisfies $\sum_{i=1}^{n}\left\{k \alpha_{i}\right\}<\frac{n}{2}$.
(1) Prove: $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{n}$ contain at least one integer;
(2) Can $\frac{n}{2}$ be replaced by a larger number, and the... | (1) Proof: Let $\alpha_{i}=\frac{p_{i}}{q_{i}}, p_{i} \in \mathbf{Z}, q_{i} \in \mathbf{N}, i=1,2, \cdots$, $n$. Let $d=q_{1} q_{2} \cdots q_{n}$, then $\alpha_{i}=\frac{P_{i}}{d}$, where $P_{i}=\frac{p_{i} d}{q_{i}}$ is an integer. If all $\alpha_{1}$ are not integers, then
$$
\left\{(d-1) \alpha_{i}\right\}=\left\{d ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 713,999 |
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