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Four, (50 points) Let $k$ be a given positive integer. Find all positive integers $a$ such that there exists $n \in \mathbf{N}_{+}, n^{2} \mid \left(a^{n}-1\right)$, and $n$ has exactly $k$ distinct prime factors. | First, we present a lemma.
Lemma: If $p \nmid (a-1)$, then $\left(a-1, \frac{a^{p}-1}{a-1}\right)=p$.
Proof is omitted.
Returning to the original problem.
When $a=1$, it clearly satisfies the condition.
First, we prove: $n \times\left(2^{n}-1\right) (n \geqslant 2)$.
Let $p$ be the smallest prime factor of $n$.
If $n ... | a \neq 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,253 |
Example 2 Proof: The sequence
$$
a_{n}=[\sqrt{2} n]+[\sqrt{3} n](n=0,1, \cdots)
$$
contains infinitely many even numbers and infinitely many odd numbers, where $[x]$ denotes the greatest integer not exceeding the real number $x$. ${ }^{[2]}$
(57th Romanian Mathematical Olympiad)
[Analysis] The conclusion to be proven,... | $$
\begin{array}{l}
\text { Assume that in the sequence } \left\{a_{n}\right\}, \text { one type of number (even or odd) } \\
\text { appears only finitely many times. Then for sufficiently large } n, \text { the sequence } \\
\left\{a_{n}\right\} \text { contains only one type of number, and thus there exists a positi... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 724,254 |
Example 1 A square piece of paper, cut along a straight line that does not pass through any vertex to divide it into two parts; take out one of the parts, and cut it along a straight line that does not pass through any vertex to divide it into two parts; then from the three parts obtained, take out one of them, and sti... | According to the problem, when the front knife cuts along a straight line that does not pass through a vertex, dividing it into two parts, each cut increases the sum of the interior angles of each part by $360^{\circ}$. Thus, after $k$ cuts, there are $k+1$ polygons, and the sum of the interior angles of these polygons... | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 724,255 |
Example 2 As shown in Figure 1,
$$
\begin{array}{l}
\angle A+\angle B+\angle C+ \\
\angle D+\angle E+\angle F+ \\
\angle G=(\quad)
\end{array}
$$
(A) $360^{\circ}$
(B) $450^{\circ}$
(C) $540^{\circ}$
(D) $720^{\circ}$ | $$
\begin{array}{l}
\angle B+\angle B M N+\angle E+\angle G=360^{\circ}, \\
\angle F N M+\angle F+\angle A+\angle C=360^{\circ} . \\
\text { Also } \angle B M N+\angle F N M=\angle D+180^{\circ} \text {, then } \\
\angle A+\angle B+\angle C+\angle D+\angle E+\angle F+\angle G \\
=540^{\circ} .
\end{array}
$$ | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 724,256 |
4. The sum of the interior angles of a polygon formed by cutting off a corner of a convex polygon is $2520^{\circ}$. Then the number of sides of the original polygon is ( ).
(A) 14
(B) 15 or 16
(C) 15
(D) 15 or 16 or 17 | Answer: D.
The text above has been translated into English, preserving the original text's line breaks and format. | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 724,257 |
As shown in Figure 3, in $\triangle ABC$, let $D_{1}, D_{2}, \cdots, D_{m}$ be the internal division points on side $BC$ in sequence, and $E_{1}, E_{2}, \cdots, E_{n}$ be the internal division points on side $AC$ in sequence. Connect $A D_{1}, A D_{2}, \cdots, A D_{m}, B E_{1}, B E_{2}, \cdots, B E_{n}$. Find the numbe... | (1) As shown in Figure 4, the number of triangles in the figure is $\mathrm{C}_{m+2}^{2}$.
(2) As shown in Figure 5, after drawing a line segment $B E_{i}$, the number of triangles added above $B E_{i}$ (including the $A$ part) is $\mathrm{C}_{m+2}^{2}$, and the number of triangles added below $B E_{i}$ (including the ... | \frac{1}{2}(m+1)(n+1)(m+n+2) | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 724,258 |
Let $n$ be a natural number. Prove:
$$
\sum_{k=0}^{n} \frac{4 k+1}{\mathrm{C}_{n}^{k}} \geqslant \frac{(n+1)^{2}(2 n+1)}{2^{n}} \text {. }
$$ | Prove that when $n=0,1$, obviously, the equality in the inequality holds.
When $n>1$, note that
$$
k \mathrm{C}_{n}^{k}=n \mathrm{C}_{n-1}^{k-1}(k=1,2, \cdots, n) \text {. }
$$
Then by the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
\sum_{k=0}^{n} \frac{4 k+1}{\mathrm{C}_{n}^{k}}=\sum_{k=0}^{n} \frac{(4 k+1... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 724,259 |
Example 1 Proof: From any $m$ integers $a_{1}, a_{2}, \cdots$, $a_{m}$, we can certainly select several numbers whose sum (including a single addend) is divisible by $m$. | Consider $m$ numbers
$$
a_{1}, a_{1}+a_{2}, \cdots, a_{1}+a_{2}+\cdots+a_{m} \text {, }
$$
if one of these numbers is divisible by $m$, then the conclusion holds. Otherwise, these $m$ numbers belong to $m-1$ residue classes modulo $m$. Hence, there must be two integers belonging to the same residue class modulo $m$, a... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 724,260 |
Example 2 An even number of people are sitting around a round table for a discussion. After a break, they sit around the table again in a different order. Prove: there are at least two people such that the number of people between them is the same before and after the break.
Translating the text into English while p... | Prove that if the seats are numbered in a clockwise order as $1, 2, \cdots, 2n$, and the seat numbers before and after each person's break are denoted as $(i, j)$. Then $i$ and $j$ traverse the complete residue system modulo $2n$. If for any two people $\left(i_{1}, j_{1}\right)$ and $\left(i_{2}, j_{2}\right)$, the nu... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 724,261 |
Example 3 Let positive integers $n, k$ satisfy $k \geqslant n$ and $(3, n) = 1$. Prove: there exists a multiple $m$ of $n$, such that the sum of the digits of $m$ in decimal representation equals $k$. | Prove that when $(n, 10)=1$, by Euler's theorem, $10^{\varphi(n)}=1(\bmod n)$.
Let $d=\varphi(n)$. Then, for any $i, j \in \mathbf{N}_{+}$, we have $10^{i d+1} \equiv 10(\bmod n), 10^{i d} \equiv 1(\bmod n)$.
Since $(3, n)=1$, we know $(9, n)=1$.
Thus, $k, k+9, \cdots, k+9(n-1)$ form a complete residue system modulo $n... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 724,262 |
Example 4 Find all odd prime numbers $p$ such that
$$
p \mid \sum_{k=1}^{2011} k^{p-1} .
$$ | If $p>2011$, then for $k(1 \leqslant k \leqslant 2011)$ by Fermat's Little Theorem we have
$k^{p-1} \equiv 1(\bmod p)$.
Thus $\sum_{k=1}^{2011} k^{p-1} \equiv 2011 \not\equiv 0(\bmod p)$, a contradiction.
Therefore, $p \leqslant 2011$.
Let $2011=p q+r(0 \leqslant r < q)$, then if $r>q$, we have $q=r$, so
$$
2011=p q+r=... | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,263 |
Example 5 Find all prime numbers $p$, such that
$$
\left.p^{3}\right]\left[\left(\mathrm{C}_{p}^{1}\right)^{2}+\left(\mathrm{C}_{p}^{2}\right)^{2}+\cdots+\left(\mathrm{C}_{p}^{p-1}\right)^{2}\right]
$$ | Solution: Clearly, $p \neq 2,3$.
$$
\begin{array}{l}
\text { Since } p-1 \equiv-1(\bmod p), \\
p-2 \equiv-2(\bmod p), \\
\cdots \cdots \\
k=-(p-k)(\bmod p),
\end{array}
$$
Therefore, multiplying all the equations, we get
$$
\frac{(p-1)!}{(k-1)!} \equiv \pm(p-k)!(\bmod p),
$$
which means $\mathrm{C}_{p-1}^{k-1} \equiv... | p \geqslant 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,264 |
Example 6 Let $n$ be an odd number greater than 1, and $k_{1}, k_{2}, \cdots, k_{n}$ be $n$ given integers. For each permutation $a=\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ of $1,2, \cdots, n$, let $S(a)=\sum_{i=1}^{n} k_{i} a_{i}$. Prove: there exist two permutations $b 、 c(b \neq c)$ of $1,2, \cdots, n$ such that $n... | Prove that if for any two different permutations $b, c (b \neq c)$ of $1, 2, \cdots, n$, we have
$$
n!X(S(b)-S(c)),
$$
then when $a$ takes all permutations (a total of $n!$), $S(a)$ traverses a complete residue system modulo $n!$. Therefore,
$$
\begin{array}{l}
\sum_{a} S(a) \equiv 1+2+\cdots+n! \\
=\frac{n!(n!+1)}{2}... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 724,265 |
Example 7 Let $a_{1}, a_{2}, \cdots$ be a sequence of integers, with infinitely many positive integers and infinitely many negative integers. Suppose that for each positive integer $n$, the numbers $a_{1}, a_{2}, \cdots, a_{n}$ have distinct remainders when divided by $n$. Prove that each integer appears exactly once i... | Proof: Since subtracting the same integer from each term of the sequence does not change the conditions and conclusions of the problem, we may assume without loss of generality that \(a_{1}=0\). Then for each positive integer \(k\), we have \(\left|a_{k}\right|<k\) (if \(\left|a_{k}\right| \geqslant k\), take \(n=\left... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 724,266 |
Example 8 The sequence $\left\{a_{n}\right\}$ is defined as
$$
a_{0}=a\left(a \in \mathbf{N}_{+}\right), \dot{a}_{n+1}=a_{n}+40^{n!}(n \in \mathbf{N}) \text {. }
$$ | Prove: There are infinitely many terms in the sequence $\left\{a_{n}\right\}$ that are divisible by 2011.
Proof Since 2011 is a prime number, and $(40,2011)=1$, therefore,
$40^{2010} \equiv 1(\bmod 2011)$.
When $n>2010$, $2010 \mid n!$. Then
$a_{n+1}=a_{n}+40^{n!} \equiv a_{n}+1(\bmod 2011)$.
Thus, the sequence $\left... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,267 |
Example 9 For any positive integer $n$, let $S(n)$ denote the sum of all elements in the set $\{1,2, \cdots, n\}$ that are coprime with $n$. Prove: $2 S(n)$ is not a perfect square. | Prove first calculate $2 S(n)$.
When $n=1$, $S(n)=1$.
When $n>1$, if $(a, n)=1$, then $(n-a, n)=1$, i.e., the numbers in $2 S(n)$ appear in pairs, totaling $\varphi(n)$ pairs, with each pair summing to $n$, so,
$$
2 S(n)=\left\{\begin{array}{ll}
2, & n=1 ; \\
n \varphi(n), & n>1 .
\end{array}\right.
$$
Thus, when $n=1... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 724,268 |
Example 3 As shown in Figure 2, the vertices of a large hexagram (thick solid line) are the centers of six congruent shaded small hexagrams around it, and each pair of adjacent small hexagrams share one vertex.
If the distance from vertex $C$ to the center $A$ of a small hexagram is $a$, find:
(1) the distance from ve... | (1) As shown in Figure 2, connect $C O$. It is easy to see that
$$
\angle A C O=90^{\circ}, \angle A O C=30^{\circ} \text {. }
$$
Therefore, $A O=2 A C=2 a$.
(2) Observing Figure 2, we know that the area of the large hexagram is 12 times the area of the equilateral $\triangle A M N$.
From $A M^{2}=\left(\frac{A M}{2}\... | 4 \sqrt{3} a^{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,269 |
Example 10 Let integer $n \geqslant 3$. For the set
$$
M=\{1,2, \cdots, n-1\}
$$
of numbers, color them according to the following rules:
(1) For $1 \leqslant i \leqslant n-1$, the numbers $i$ and $n-i$ are the same color;
(2) There exists $k \in M, (k, n)=1$, such that for any $i \in M, i \neq k$, the numbers $i$ and... | Given $(k, n)=1$, we know
$$
0,1, \cdots, n-1 \text { and } 0, k, 2 k, \cdots,(n-1) k
$$
both form a complete residue system modulo $n$.
Let $a_{i} \equiv i k(\bmod n)\left(0 \leqslant a_{i} \leqslant n-1,0 \leqslant i \leqslant n-1\right)$.
Then $\left\{a_{0}, a_{1}, \cdots, a_{n-1}\right\}=\{0,1, \cdots, n-1\}$.
Thu... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 724,270 |
1. Let $m>1$ be an integer. Prove: There must exist a natural number $a$ composed only of 0s and 1s that is a multiple of $m$. | Consider the integers consisting entirely of the digit 1:
$1, 11, 111, 1111, \cdots$.
Since there must be two of them in the same residue class modulo $m$, the difference between these two is the desired $a$. | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 724,271 |
2. Let the integer $n \geqslant 2$. The integers $a_{1}, a_{2}, \cdots, a_{n}$ sum to 1, and the sequence $\left\{b_{m}\right\}$ is defined as:
$$
\begin{aligned}
b_{m}= & a_{m}+2 a_{m+1}+\cdots+(n-m+1) a_{n}+ \\
& (n-m+2) a_{1}+\cdots+n a_{m-1},
\end{aligned}
$$
where $m=1,2, \cdots, n$. Prove: $b_{1}, b_{2}, \cdots,... | 提示: Due to $b_{m+1}-b_{m}$
$$
\begin{array}{l}
=-a_{1}-a_{2}-\cdots-a_{m-1}+(n-1) a_{m}-a_{m+1}-\cdots-a_{n} \\
=-\left(a_{1}+a_{2}+\cdots+a_{n}\right) \equiv-1(\bmod n),
\end{array}
$$
Therefore, $b_{1}, b_{2}, \cdots, b_{n}$ form a complete residue system modulo $n$. | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 724,272 |
3. Let $A=\left\{a_{1}, a_{2}, \cdots, a_{\varphi(n)}\right\}$ be a reduced residue system modulo $n$. If the equation $x^{2} \equiv 1(\bmod n)$ has $N$ solutions in $A$, prove:
$$
a_{1} a_{2} \cdots a_{\varphi(n)} \equiv(-1)^{\frac{\kappa}{2}}(\bmod n) .
$$ | ```
Since the integer $x$ is a solution to the equation $x^{2} \equiv 1(\bmod n)$, $-x$ is also a solution to the equation. Therefore, $N$ is even. Let $a_{1}, a_{2}, \cdots, a_{\frac{N}{2}},-a_{1},-a_{2}, \cdots,-a_{\frac{N}{2}}$ be the $N$ solutions to the equation $x^{2} \equiv 1(\bmod n)$.
Since for any $a_{i} \in... | (-1)^{\frac{N}{2}}(\bmod n) | Number Theory | proof | Yes | Yes | cn_contest | false | 724,273 |
4. Connecting the $n$ equal division points of a circle, we get a closed broken line with $n$ sides (each division point is exactly the endpoint of two sides of this closed broken line). Prove: If $n$ is even, then there must be two parallel sides in this closed broken line. | Label the $n$ equally divided points in counterclockwise order: $0,1, \cdots, n-1$. Then the sides of the closed broken line can be uniquely represented by a permutation $a_{n}=a_{0}, a_{1}, \cdots, a_{n-1}$ of these $n$ numbers as $a_{0} a_{1}, a_{1} a_{2}, \cdots, a_{n-1} a_{0}$. Hence
$$
\begin{array}{l}
\Leftrighta... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,274 |
5. Let $n \geqslant 2$ be a positive integer. Find all sets $A$ containing $n$ integers such that the sum of all elements in any non-empty subset of $A$ cannot be divisible by $n+1$. | Let $A=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$,
$$
S_{k}=\sum_{i=1}^{k} a_{i}(k=1,2, \cdots, n) .
$$
According to the problem, $\left\{S_{k}\right\}$ contains all non-zero residues modulo $(n+1)$, and $a_{2}, S_{2}, S_{3}, \cdots, S_{n}$ also contain all non-zero residues modulo $(n+1)$. Therefore,
$$
\begin{array... | A=\left\{a+k_{i}(n+1) \mid k_{i} \in \mathbf{Z}, 1 \leqslant i \leqslant n\right\}, \text{ where } a \in \mathbf{Z}, (a, n+1)=1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,275 |
As shown in Figure 1, given $\triangle A B C$ with its circumcircle $\odot O$, $D$ is a point on side $A B$. $\odot I$ is tangent to segments $B D$, $C D$, and $\odot O$. $\odot J$ is tangent to segments $A D$, $C D$, and $\odot O$. Prove: If $A$, $B$, $I$, and $J$ are concyclic, then $D$ is the tangency point of the e... | Prove that: $IJ // AB$.
Assume $IJ \cap AB = T$. Then
$$
\frac{IT}{TJ} \cdot \frac{JL}{LO} \cdot \frac{OQ}{QI} = \frac{r_i}{r_j} \cdot \frac{r_j}{r} \cdot \frac{r}{r_i} = 1.
$$
By the converse of Menelaus' theorem, points $L, Q, T$ are collinear. Thus,
$$
TQ \cdot TL = TB \cdot TA = TT \cdot TJ.
$$
Therefore, points... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,276 |
1. Let the feet of the altitudes from $A$, $B$, and $C$ to the sides $BC$, $CA$, and $AB$ of acute triangle $\triangle ABC$ be $D$, $E$, and $F$, respectively. Let the line $EF$ intersect the circumcircle of $\triangle ABC$ at point $P$, and let the line $BP$ intersect $DF$ at point $Q$. Prove: $AP = AQ$. | 1. As shown in Figure 2, let the line $E F$ intersect the circumcircle of $\triangle A B C$ at points $P$ and $P^{\prime}$, where $P$ lies on arc $\overparen{A B}$ and $P^{\prime}$ lies on arc $\overparen{C A}$. The lines $B P$ and $B P^{\prime}$ intersect $D F$ at points $Q$ and $Q^{\prime}$, respectively. We need to ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,277 |
3. Let $P$ be a point inside the convex polygon $A_{1} A_{2} \cdots A_{n}$, such that the projections of $P$ onto the lines $A_{1} A_{2}, A_{2} A_{3}, \cdots, A_{n} A_{1}$, denoted as $P_{1}, P_{2}, \cdots, P_{n}$, all lie on the edges of the convex polygon. Prove: For any points $X_{1}, X_{2}, \cdots, X_{n}$ on the ed... | 3. Let $P_{n+1}=P_{1}, X_{n+1}=X_{1}, A_{n+1}=A_{1}$.
First, we prove a lemma.
Lemma: As shown in Figure 3, let $Q$ be a point inside the convex polygon $A_{1} A_{2} \cdots A_{n}$. Then, at least one of the circumcircles of $\triangle X_{i} A_{i+1} X_{i+1} (i=1,2, \cdots, n)$ contains point $Q$.
Proof: If point $Q$ i... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,278 |
5. Given a convex pentagon $A B C D E$ satisfying $B C \parallel A E$, $A B = B C + A E$, $\angle A B C = \angle C D E$, $M$ is the midpoint of side $C E$, $O$ is the circumcenter of $\triangle B C D$, and $\angle D M O = 90^{\circ}$. Prove: $2 \angle B D A = \angle C D E$. | 5. As shown in Figure 4, let $T$ be a point on ray $A E$ such that $A T = A B$. Connect $B E$ and $C T$.
Since $A E \parallel B C$, we know
$\angle C B T = \angle A T B = \angle A B T$.
Thus, $B T$ is the angle bisector of $\angle A B C$.
Since $E T = A T - A E = A B - A E = B C$, quadrilateral $B C T E$ is a parallelo... | 2 \angle B D A = \angle C D E | Geometry | proof | Yes | Yes | cn_contest | false | 724,279 |
Example 4 As shown in Figure 3, in the acute $\triangle A B C$, $D 、 E 、 F$ are the trisection points on sides $A B 、 B C 、 C A$, respectively, and $P 、 Q 、 R$ are the intersections of the three medians of $\triangle A D F 、 \triangle B D E 、 \triangle C E F$, respectively. Find:
(1) $S_{\triangle D E F}: S_{\triangle ... | (1) As shown in Figure 3, draw $D G \perp B C$ at point $G$, and draw $A H \perp B C$ at point $H$.
Then $D G / / A H$.
Therefore, $\triangle B D G \backsim \triangle B A H$.
Since $\frac{B D}{B A}=\frac{1}{3}, B E=\frac{2}{3} B C$, we have
$D G=\frac{1}{3} A H, S_{\triangle B D E}=\frac{2}{9} S_{\triangle A B C}$.
Sim... | \frac{5}{9} S_{\triangle A B C} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,280 |
6. Given that the vertices $X, Y, Z$ of the equilateral $\triangle XYZ$ lie on the sides $BC$, $CA$, $AB$ of the acute $\triangle ABC$ respectively. Prove: the incenter of $\triangle ABC$ lies inside $\triangle XYZ$.
保留源文本的换行和格式，直接输出翻译结果如下：
6. Given that the vertices $X, Y, Z$ of the equilateral $\triangle XYZ$ lie o... | 6. Prove a stronger conclusion:
The incenter $I$ of $\triangle ABC$ is inside the incircle of $\triangle XYZ$ (thus, $I$ is inside $\triangle XYZ$).
Let $d(U, VW)$ denote the distance from point $U$ to line $VW$, $O$ be the incenter of $\triangle XYZ$, and the inradii of $\triangle ABC$ and $\triangle XYZ$ be $r$ and ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,281 |
$6^{\prime}$. Given that the vertices $X, Y, Z$ of the equilateral $\triangle XYZ$ lie on the sides $BC, CA, AB$ of $\triangle ABC$, respectively. Prove: if the incenter of $\triangle ABC$ is outside $\triangle XYZ$, then $\triangle ABC$ has an angle greater than $120^{\circ}$. | $\mathbf{6}^{\prime}$. Suppose the incenter $I$ of $\triangle ABC$ is inside $\triangle AYZ$, and the incircle $\odot I$ of $\triangle ABC$ touches side $BC$ at point $A_1$, which lies on segment $CX$. Then,
$\angle YZA \leqslant 180^{\circ} - \angle YZX = 120^{\circ}$.
Furthermore, points $I$ and $Y$ are on the same s... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,282 |
1. The number of integer pairs $(x, y)$ that satisfy $x^{2}-4 y^{2}=2011$ is ( ).
(A) 0
(B) 1
(C) 2
(D) 3 | -1.A.
If the integer pair $(x, y)$ satisfies
$$
x^{2}-4 y^{2}=2011 \text{, }
$$
but the left side $x^{2}-4 y^{2}$ when divided by 4 leaves a remainder of 1 or 0, while the right side 2011 when divided by 4 leaves a remainder of 3, which is a contradiction. | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 724,284 |
2. Figure 1 is the "Yang Hui Triangle" from ancient China. According to the numerical pattern, please fill in all the " $O$ " in the eighth row with the appropriate numbers. Then, the sum of the 36 numbers in the first eight rows is ( ).
(A) 257
(B) 256
(C) 255
(D) 254 | 2. C.
Notice that each number is the sum of the two adjacent numbers in the row above it, so the eighth row should be: $1, 7, 21, 35, 35, 21, 7, 1$.
Also, the sum of the numbers in each row is exactly twice the sum of the numbers in the row above it. Therefore, the sum of the first eight rows, which consist of 36 num... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 724,285 |
3. In quadrilateral $ABCD$, $AD \parallel BC, \angle ABC$ $=80^{\circ}, AB=AD=\frac{1}{2} BC, CH \perp AB$ at point $H$, and connect $DH$. Then $\angle CHD=(\quad)$.
(A) $30^{\circ}$
(B) $35^{\circ}$
(C) $40^{\circ}$
(D) $45^{\circ}$ | 3. C.
As shown in Figure 5, take the midpoint $E$ of $B C$, and connect $H E$ and $D E$. It is easy to see that $A D \Perp B E$.
Then $H E=B E$
$$
=A B=D E \text {, }
$$
and $D E \perp C H$.
Therefore, $\angle C H D=90^{\circ}-\angle H D E$
$$
=\frac{1}{2} \angle H E D=\frac{1}{2} \angle H B E=40^{\circ} \text {. }
$... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 724,286 |
4. Simplify:
\[
\begin{array}{l}
\frac{1}{\sqrt{8}+\sqrt{11}}+\frac{1}{\sqrt{11}+\sqrt{14}}+\frac{1}{\sqrt{14}+\sqrt{17}}+ \\
\frac{1}{\sqrt{17}+\sqrt{20}}+\frac{1}{\sqrt{20}+\sqrt{23}}+\frac{1}{\sqrt{23}+\sqrt{26}}+ \\
\frac{1}{\sqrt{26}+\sqrt{29}}+\frac{1}{\sqrt{29}+\sqrt{32}}
\end{array}
\]
The result is ( ).
(A) 1... | 4. B.
Notice that $\frac{1}{\sqrt{n}+\sqrt{n+3}}=\frac{\sqrt{n}-\sqrt{n+3}}{-3}$. Then the original expression is
$$
\begin{aligned}
= & \frac{\sqrt{8}-\sqrt{11}}{-3}+\frac{\sqrt{11}-\sqrt{14}}{-3}+\frac{\sqrt{14}-\sqrt{17}}{-3}+ \\
& \frac{\sqrt{17}-\sqrt{20}}{-3}+\frac{\sqrt{20}-\sqrt{23}}{-3}+\frac{\sqrt{23}-\sqrt{... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,287 |
5. As shown in Figure 2, a square $E F G H$ with side length 1 is moving on the plane of a square $A B C D$ with side length 3, always keeping $E F / / A B$. The midpoints of segments $C F$ and $D H$ are $M$ and $N$, respectively. Then the length of segment $M N$ is ( ).
(A) $\frac{\sqrt{10}}{2}$
(B) $\frac{\sqrt{17}}{... | 5. B.
As shown in Figure 6, connect $C G$, take the midpoint $T$ of $C G$, and connect $M T$ and $T N$.
Since $M$ is the midpoint of $C F$, $M T$ is the midline of $\triangle G C F$, so,
$$
\begin{array}{l}
M T / / F G, \\
M T=\frac{F G}{2}=\frac{1}{2} .
\end{array}
$$
Similarly,
$$
\begin{array}{l}
N T / / G H, \\
... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 724,288 |
1. Given that $a$, $b$, and $c$ are non-zero rational numbers, and satisfy
$$
\begin{array}{l}
a b^{2}=\frac{c}{a}-b . \\
\text { Then }\left(\frac{a^{2} b^{2}}{c^{2}}-\frac{2}{c}+\frac{1}{a^{2} b^{2}}+\frac{2 a b}{c^{2}}-\frac{2}{a b c}\right) \div \\
\left(\frac{2}{a b}-\frac{2 a b}{c}\right) \div \frac{101}{c}=
\end... | 2. 1. $-\frac{1}{202}$.
From $a b^{2}=\frac{c}{a}-b$, we get
$$
\frac{c}{a b}-a b=1 \text {. }
$$
Notice that
$$
\begin{array}{l}
\frac{a^{2} b^{2}}{c^{2}}-\frac{2}{c}+\frac{1}{a^{2} b^{2}}=\left(\frac{a b}{c}-\frac{1}{a b}\right)^{2} . \\
\text { Then the original expression }=\left[\frac{1}{2}\left(\frac{1}{a b}-\fr... | -\frac{1}{202} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,289 |
2. Given that $A D$ is the median of $\triangle A B C$, $\angle A B C=$ $30^{\circ}, \angle A D C=45^{\circ}$. Then $\angle A C B=$ $\qquad$ | 2. $105^{\circ}$.
As shown in Figure 7, draw
the altitude $C H$. Then
$$
\begin{array}{c}
H D=D C \\
=\frac{1}{2} B C \\
=H C .
\end{array}
$$
Since $\angle H A D$
$$
\begin{array}{l}
\quad=\angle B A D \\
= \angle A D C-\angle A B D=45^{\circ}-30^{\circ}=15^{\circ}, \\
\angle H D A=\angle H D C-\angle A D C \\
= 60... | 105^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,290 |
Example 5 Given a regular hexagon $A B C D E F$ with side length $1, Q R$ is any line segment inside the hexagon parallel to $A B$. Find the maximum area of $\triangle P Q R$ inscribed in the regular hexagon $A B C D E F$ with $Q R$ as the base. | To obtain the maximum area of $\triangle PQR$, point $P$ should be on $DE$, and points $Q, R$ should be on $AF, BC$ respectively. Draw $PH \perp QR$ at point $H$, intersecting $AB$ at point $G$. Draw $AM \perp QR$ and $BN \perp QR$ at points $M, N$ respectively. Let $PH = x$. Then
$$
\begin{array}{l}
H G = \sqrt{3} - x... | \frac{9 \sqrt{3}}{16} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,291 |
3. The equation concerning $x, y$
$$
\frac{1}{x}+\frac{1}{y}+\frac{1}{x y}=\frac{1}{2011}
$$
has $\qquad$ groups of positive integer solutions $(x, y)$. | 3. 12 .
From $\frac{1}{x}+\frac{1}{y}+\frac{1}{x y}=\frac{1}{2011}$, we get
$$
\begin{array}{l}
x y-2011 x-2011 y-2011=0 \\
\Rightarrow(x-2011)(y-2011) \\
\quad=2011 \times 2012 \\
=2^{2} \times 503 \times 2011 .
\end{array}
$$
Thus, the positive integer solutions of the original equation are
$$
(2+1)(1+1)(1+1)=12 \t... | 12 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,292 |
4. Two sheets of the same size, each divided into seven rectangles of the same size, are placed as shown in Fig. 3, with the overlapping vertex marked as $A$, and point $C$ on the dividing line of the other sheet. If $B C=\sqrt{28}$, then the length of $A B$ is $\qquad$ . | 4. $7 \sqrt{2}$.
Let $B D=x$. Then
$$
A B=A C=7 x, A D=6 x \text {. }
$$
Thus, $C D=\sqrt{A C^{2}-A D^{2}}=\sqrt{13} x$,
$$
B C=\sqrt{B D^{2}+C D^{2}}=\sqrt{14} x=\sqrt{28} \text {. }
$$
Therefore, $x=\sqrt{2}$.
So, $A B=7 \sqrt{2}$. | 7 \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,293 |
5. For $n$ consecutive natural numbers, if each number is written in its standard prime factorization form, and each prime factor appears an odd number of times, such $n$ consecutive natural numbers are called a “consecutive $n$ strange group” (for example, when $n=3$, $22=2^{1} \times 11^{1}$, $23=23^{1}$, $24=2^{3} \... | 5.7.
For example: $29, 30, 31, 32, 33, 34, 35$ is a consecutive 7 peculiar group.
Below is the proof: When $n \geqslant 8$, there does not exist a consecutive $n$ peculiar group.
Otherwise, one of them (denoted as $m$) is divisible by 8, so among these $n$ numbers, there must be $m-4$ or $m+4$, which is divisible by... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,294 |
Three, (10 points) In $\triangle A B C$, $\angle A B C=60^{\circ}$, $\angle A C B=40^{\circ}$, $P$ is the intersection of the angle bisector of $\angle A B C$ and the angle bisector of $\angle A C B$. Prove: $A B=P C$.
| Three, as shown in Figure 8, construct a regular $\triangle P D C$ with $P C$ as a side. Then,
$$
\angle A C D=80^{\circ} \text {. }
$$
Connect $A P$. Then $A P$ bisects $\angle B A C$, so $\angle P A C=40^{\circ}$.
Furthermore,
$$
\angle A P C=120^{\circ} \text {. }
$$
Since $\angle C P D=60^{\circ}$, we have,
$$
\a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,295 |
Four, (15 points) Regarding the equation of $m$ and $n$
$$
5 m^{2}-6 m n+7 n^{2}=2011
$$
Does there exist an integer solution? If it exists, write down one solution; if not, explain the reason. | Four, does not exist.
(1) If $m$ and $n$ have the same parity, then the left side of the given equation is even, which cannot equal 2011.
(2) If $m$ and $n$ have different parities, then $m+n$ and $m-n$ are both odd.
Rewrite the equation as
$$
4(m-n)^{2}+(m+n)^{2}+2 n^{2}=2011 \text{. }
$$
We will discuss two cases.
(... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,296 |
Five. (15 points) As shown in Figure 4, rectangle $ABCD$ is a cargo yard with a length of $1000 \, \text{m}$ and a width of $600 \, \text{m}$. $A$ and $D$ are entrances. It is planned to build a toll booth $P$ within the cargo yard and a loading platform $H$ on the railway segment $BC$. Let the total length of the road... | (1) As shown in Figure 9, the rectangle $ABCD$ is rotated $60^{\circ}$ clockwise around point $A$ to the position of rectangle $AB_{1}C_{1}D_{1}$. Thus, point $P$ inside the rectangle and point $H$ on side $BC$ are rotated to points $P_{1}$ and $H_{1}$, respectively, such that
$$
A P = A P_{1} = P P_{1}, \quad P_{1} H_... | 600 + 500 \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,297 |
1. Find all positive integers $n$ such that the equation in $x, y$
$$
\frac{1}{x}+\frac{1}{y}=\frac{1}{n}
$$
has exactly 2011 solutions in positive integers $(x, y)$ with $x \leqslant y$. | 1. From the given, we have
$$
x y-n x-n y=0 \Rightarrow(x-n)(y-n)=n^{2} \text {. }
$$
Therefore, apart from $x=y=2 n$, if $x-n$ takes a positive divisor of $n^{2}$ that is less than $n$, we can obtain a set of positive integer solutions $(x, y)$ that satisfy the condition.
Thus, the number of positive divisors of $n^... | n=p^{2010} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,298 |
2. As shown in Figure 1, the diagonals $AC$ and $BD$ of quadrilateral $ABCD$ intersect at point $E$. The perpendicular bisectors of sides $AB$ and $CD$ intersect at point $F$. Points $M$ and $N$ are the midpoints of sides $AB$ and $CD$, respectively. Line $EF$ intersects sides $BC$ and $AD$ at points $P$ and $Q$. If
$$... | 2. As shown in Figure 3, connect $A F$, $B F$, $C F$, and $D F$.
From the problem, we know that $\triangle A F B$ and $\triangle C F D$ are both isosceles triangles, with $FM$ and $FN$ being the altitudes to the bases of these isosceles triangles.
From $M F \cdot C D = N F \cdot A B$, we know
$\triangle A F B \backsim... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,299 |
4. There are $n(n \geqslant 3)$ table tennis players participating in a round-robin tournament, where each pair of players competes exactly once (there are no ties). After the tournament, it is found that the players can be arranged in a circle such that for any three players $A$, $B$, and $C$, if $A$ and $B$ are adjac... | 4. All possible values of $n$ are all odd numbers greater than or equal to 3.
Reason as follows.
When $n$ is an odd number greater than or equal to 3, let $n=2k+1$, and the $n$ players are numbered as $A_{1}, A_{2}, \cdots, A_{2k+1}$. Construct the competition results that satisfy the conditions as follows:
Player $A_... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 724,301 |
5. Given a non-negative real number $\alpha$. Find the smallest real number $\lambda=$
$\lambda(\alpha)$, such that for any complex numbers $z_{1}, z_{2}$ and real number $x \in$ $[0,1]$, if $\left|z_{1}\right| \leqslant \alpha\left|z_{1}-z_{2}\right|$, then
$\left|z_{1}-x z_{2}\right| \leqslant \lambda\left|z_{1}-z_{2... | 5. As shown in Figure 4, in the complex plane, points $A, B, C$ correspond to the complex numbers $z_{1}, z_{2}, x z_{2}$, respectively.
Obviously, point $C$ lies on line segment $OB$. The vectors $\overrightarrow{B A}$ and $\overrightarrow{C A}$ correspond to the complex numbers
$z_{1}-z_{2}$ and $z_{1}-x z_{2}$, re... | \max \{\alpha, 1\} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,303 |
6. Do there exist positive integers $m, n$ such that $m^{20} + 11^n$ is a perfect square? Please prove your conclusion.
Translate the above text into English, preserving the original text's line breaks and format, and output the translation result directly. | 6. Suppose there exist positive integers $m, n$, such that $m^{20}+11^{n}=k^{2}\left(k \in \mathbf{N}_{+}\right)$.
Then $11^{n}=k^{2}-m^{20}=\left(k-m^{10}\right)\left(k+m^{10}\right)$.
Therefore, there exist integers $\alpha, \beta \geqslant 0(\beta>\alpha)$, such that
$$
\left\{\begin{array}{l}
k-m^{10}=11^{\alpha}, ... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,304 |
7. There are $n$ boxes numbered from left to right as $B_{1}, B_{2}, \cdots, B_{n}$, containing a total of $n$ balls. Each time, you can choose a box $B_{k}$ and perform the following operation: if $k=1$, and $B_{1}$ contains at least one ball, then you can move 1 ball from $B_{1}$ to $B_{2}$; if $k=n$, and $B_{n}$ con... | For any two vectors
$$
x=\left(x_{1}, x_{2}, \cdots, x_{n}\right) \text { and } y=\left(y_{1}, y_{2}, \cdots, y_{n}\right) \text {, }
$$
if there exists $k(1 \leqslant k \leqslant n)$ such that
$$
x_{1}=y_{1}, x_{2}=y_{2}, \cdots, x_{k-1}=y_{k-1}, x_{k}>y_{k},
$$
then we denote $x>y$.
A non-negative integer vector $x... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 724,305 |
Given the sequence $\left\{a_{n}\right\}$ with the general term
$$
a_{n}=(\sqrt{3}+\sqrt{2})^{2 n}\left(n \in \mathbf{N}_{+}\right) \text {, }
$$
Let $b_{n}=a_{n}+\frac{1}{a_{n}}$.
(1) Find the recurrence relation between $b_{n+2} 、 b_{n+1} 、 b_{n}$;
(2) Find the unit digit of the integer part of $a_{2011}$. | -、(1) From the given, we have
$$
\begin{array}{l}
b_{n}=(\sqrt{3}+\sqrt{2})^{2 n}+(\sqrt{3}-\sqrt{2})^{2 n} \\
\Rightarrow b_{n}=(5+2 \sqrt{6})^{n}+(5-2 \sqrt{6})^{n} .
\end{array}
$$
Then $b_{n+2}=[(5+2 \sqrt{6})+(5-2 \sqrt{6})]$.
$$
\begin{array}{l}
{\left[(5+2 \sqrt{6})^{n+1}+(5-2 \sqrt{6})^{n+1}\right]-} \\
(5+2 \... | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,307 |
II. As shown in Figure $1, \triangle A B C$'s incircle touches $B C$, $C A$, and $A B$ at points
$D$, $E$, and $F$, respectively. $P$ is a point
inside the incircle, and line segments $P A$,
$P B$, and $P C$ intersect
the incircle at points $X$, $Y$, and $Z$,
respectively.
Prove: $X D$, $Y E$, and
$Z F$ are concurrent. | $$
\frac{X F}{X E}=\frac{\sin \angle X D F}{\sin \angle X D E} \text {. }
$$
Notice that
$$
\begin{array}{l}
\frac{X F}{\sin \angle X A F}=\frac{X A}{\sin \angle X D F}, \\
\frac{X E}{\sin \angle X A E}=\frac{X A}{\sin \angle X D E} . \\
\text { Then } \frac{X F}{X E}=\frac{\sin \angle X A F \cdot \sin \angle X D E}{\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,308 |
Three, find all positive integer solutions $(x, y, z)$ of the indeterminate equation $1+2^{x} \times 7^{y}=z^{2}$.
(Weng Shiyou supplied the problem) | Three, the original equation can be transformed into
$$
2^{x} \times 7^{y}=z^{2}-1=(z+1)(z-1) \text {. }
$$
It is easy to see that $z$ is an odd number, and $z+1$ and $z-1$ are two consecutive even numbers, with only one of them being a multiple of 4.
Since $(z+1)-(z-1)=2$, they cannot both be multiples of 7.
From th... | (5,1,15) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,309 |
Four, let $n$ sets $A_{1}, A_{2}, \cdots, A_{n}$ be a partition of the set $A=\{1,2, \cdots, 29\}$, and the sum of any elements in $A_{i}(i=1,2$, $\cdots, n)$ does not equal 30. Find the minimum possible value of $n$.
[Note] If the non-empty subsets $A_{1}, A_{2}, \cdots, A_{n}$ $\left(n \in \mathbf{N}_{+}, n \geqslan... | The minimum value of $n$ is 3.
First, we decompose
$$
\begin{array}{l}
A_{1}=\{1,2, \cdots, 7\}, \\
A_{2}=\{10,11, \cdots, 15,21,22\}, \\
A_{3}=\{8,9,16,17, \cdots, 20,23,24, \cdots, 29\}
\end{array}
$$
which satisfies the conditions.
Next, we prove that $n=2$ does not satisfy the conditions.
Assume, for contradiction... | 3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 724,310 |
Five, if positive integers $a, b, c$ satisfy $a^{2}+b^{2}=c^{2}$, then $(a, b, c)$ is called a Pythagorean triple. Find all Pythagorean triples that contain 30.
(Yang Chunhong, contributor) | Five, Solution 1 Since $30=2 \times 3 \times 5$, therefore, 30 has four even factors: $2,6,10,30$ and three odd factors: $3,5,15$.
(1) Let $30=2 \mathrm{mn}$.
Then $m n=15=1 \times 15=3 \times 5$.
By $m=15, n=1$, we do not get a basic group;
By $m=5, n=3$, we also do not get a basic group.
(2) Let $10=2 \mathrm{mn}$.
T... | (30,224,226),(30,72,78),(30,40,50),(30,16,34),(30,24,18) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,311 |
Six. As shown in Figure 2, draw the tangent line $PA$ and the secant line $PBC$ from point $P$ to circle $\odot O$, and let $AD \perp PO$, with the foot of the perpendicular being $D$. Prove:
$AC$ is the tangent to the circumcircle of $\triangle ABD$. | Six, as shown in Figure 3, connect $O A, O B, O C, C D$. Then $O A \perp P A$.
By the projection
theorem and the secant
line theorem, we have
$$
\begin{array}{l}
P D \cdot P O \\
=P A^{2} \\
=P B \cdot P C .
\end{array}
$$
Therefore, $B, C, O, D$ are concyclic.
$$
\begin{array}{l}
\text { Then } \angle P D B=\angle O... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,312 |
Example 7 As shown in Figure 6, in the regular nonagon ABCDEFGHI, it is known that $A E=1$. Then the length of $A B + A C$ is
The translation preserves the original text's line breaks and format. | Solving: Since the sum of the interior angles of a regular nonagon is
$$
(9-2) \times 180^{\circ}=1260^{\circ} \text {, }
$$
and each interior angle is $140^{\circ}$, therefore,
$$
\angle C A B=\left(180^{\circ}-140^{\circ}\right) \div 2=20^{\circ} \text {. }
$$
Connecting $A H$, and drawing $H M$ and $G N$ perpendic... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,313 |
Seven, in $\triangle A B C$, prove:
$$
\begin{array}{l}
\frac{1}{1+\cos ^{2} A+\cos ^{2} B}+\frac{1}{1+\cos ^{2} B+\cos ^{2} C}+ \\
\frac{1}{1+\cos ^{2} C+\cos ^{2} A} \leqslant 2 .
\end{array}
$$ | Proof 1 By Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
\sin ^{2} C=\sin ^{2}(A+B) \\
=(\sin A \cdot \cos B+\cos A \cdot \sin B)^{2} \\
\leqslant\left(\sin ^{2} A+\sin ^{2} B\right)\left(\cos ^{2} A+\cos ^{2} B\right) .
\end{array}
$$
Then $\cos ^{2} A+\cos ^{2} B \geqslant \frac{\sin ^{2} C}{\sin ^{2} A+\si... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 724,314 |
Eight, given that $n$ is a positive integer, and the real number $x$ satisfies
$$
|1-| 2-\cdots|(n-1)-| n-x|| \cdots \|=x \text {. }
$$
Find the value of $x$.
(Zhang Limin, problem contributor) | Conclusion 1 If $a \geqslant 0, b \geqslant 0$, then
$$
|a-b| \leqslant \max \{a, b\} \text {. }
$$
Conclusion 2 If $0 \leqslant x < k-1$, then
$$
\begin{array}{l}
|(k-1)-x|=x-(k-1) \\
<k-(k-1)=1 ;
\end{array}
$$
If $x \leqslant k-1$, then by Conclusion 1 we have
$$
|(k-1)-x| \leqslant k-1 \text {. }
$$
In summary, ... | x=\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,315 |
9. 1. Let $P(x)$ be a real quadratic polynomial with leading coefficient 1. It is known that $P(x)$ and $P(P(P(x)))$ have a common real root. Prove:
$$
P(0) P(1)=0 \text {. }
$$ | 9. 1. Let $P(x)=x^{2}+a x+b, t$ be a common root of $P(x)$ and $P(P(P(x)))$. Then
$$
\begin{array}{l}
0=P(P(P(t)))=P(P(0))=P(b) \\
=b^{2}+a b+b=P(0) P(1) .
\end{array}
$$ | P(0) P(1)=0 | Algebra | proof | Yes | Yes | cn_contest | false | 724,316 |
9.2. Given an acute triangle $\triangle A B C$, a circle passing through its vertex $B$ and circumcenter $O$ intersects sides $B C$ and $B A$ at points $P$ and $Q$ (distinct from point $B$). Prove that the orthocenter of $\triangle P O Q$ lies on the line $A C$. | 9.2. Let $\angle O B A=\angle O A B=\alpha$, $\angle O B C=\angle O C B=\gamma$.
Then $\angle A C B=\frac{1}{2} \cdot \angle A O B=90^{\circ}-\alpha$.
$H B 、 P 、 O 、 Q \mid Y$ points are concyclic, so
$$
\angle O P Q=\alpha, \angle O Q P=\gamma \text {. }
$$
Let $O O_{1}$ be the altitude of $\triangle O P Q$, and $H$ ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,317 |
9.3. There is a convex 2011-gon on the blackboard. Betya draws its diagonals one by one. It is known that each diagonal drawn intersects at most one of the previously drawn diagonals at an interior point. Question: What is the maximum number of diagonals Betya can draw? | 9.3.4016.
Use induction to prove: For a convex $n$-sided polygon, at most $2n-6$ diagonals can be drawn.
Let $A_{1} A_{2} \cdots A_{n}$ be a convex polygon. We can sequentially draw $2n-6$ diagonals as follows: $A_{2} A_{4}, A_{3} A_{5}$, $A_{4} A_{6}, \cdots, A_{n-2} A_{n}, A_{1} A_{3}, A_{1} A_{4}, \cdots, A_{1} A_... | 4016 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 724,318 |
9.4. Does there exist three pairwise coprime positive integers such that the square of each number is divisible by the sum of the other two? | 9.4. Not your business.
Suppose there exist integers $a, b, c$ such that:
By $(a+b) \mid c^{2}, (b+c) \mid a^{2}$, we get
$p \mid a, p \mid c$.
Thus, $p \mid (a+b)-a=b$. This contradicts the fact that $a, b, c$ are coprime.
$$
\begin{array}{l}
\text { By }(b+c) \mid a^{2} \\
=(a+b+c)^{2}-(b+c)(2 a+b+c),
\end{array}
$$... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 724,319 |
9.5. For 2011 positive integers, write their $\mathrm{C}_{2011}^{2}$ pairwise sums on the blackboard. Question: Is it possible that exactly half of these $\mathrm{C}_{2011}^{2}$ numbers are multiples of 3, and at the same time, exactly $=$ one-third of them are congruent to 1 modulo 3? | 9.5. Let the number of 2011 numbers that are congruent to $0, 1, 2$ modulo 3 be $a, b, c$ respectively. Then
$$
\begin{array}{l}
\frac{a(a-1)}{2}+b c=\frac{b(b-1)}{2}+a c \\
=\frac{c(c-1)}{2}+a b, \\
a+b+c=2011 .
\end{array}
$$
It is not difficult to verify that $a=b=670, c=671$ satisfies the above equations. | a=b=670, c=671 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,320 |
9.6. First, Betya writes the numbers $1, 2$ in her notebook, and Nik writes $3, 4$ in his notebook. Then, at the beginning of each minute, Nik and Betya each write down a quadratic polynomial with real coefficients, whose roots are the two numbers in their notebooks, denoted as $f(x)$ and $g(x)$, respectively. If the e... | 9.6. It is not hard to see that the two numbers in each person's exercise book are always the roots of the equation
$$
\alpha(x-1)(x-2)+\beta(x-3)(x-4)=0
$$
where $\alpha, \beta$ are parameters that depend on time.
At a certain moment, 5 is a root of this equation.
Substituting 5, we get $6 \alpha+\beta=0$.
Thus, $\al... | \frac{14}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,321 |
9.7. Given that $T$ is a point on side $AC$ of equilateral $\triangle ABC$, and $M, N$ are points on the circumcircle arcs $\overparen{AB}$ and $\overparen{BC}$ of $\triangle ABC$, respectively. Given that $MT \parallel BC$ and $NT \parallel AB$, line segments $AN$ and $MT$, as well as $CM$ and $NT$, intersect at point... | 9.7. Let the line $l$ be the tangent to the circumcircle at point $B$, and let $P$ and $Q$ be the intersections of the rays $T M$ and $T N$ with $l$, respectively. Let $K$ and $L$ be the intersections of $T M$ with $T N$ and $A B$ with $C B$, respectively.
From the quadrilaterals $A B Q T$ and $B C T P$ being parallel... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,322 |
9.8. In each cell of a $100 \times 100$ grid, fill in either 0 or 1. If the sum of the numbers in a cell and its adjacent cells (two cells are adjacent if they share a common edge) is even, then the cell is called "beautiful." Is it possible for there to be only one beautiful cell in the grid? | 9.8. Suppose there is a unique beautiful cell $X$ in the grid.
First, color the grid in a checkerboard pattern with alternating black and white cells. Without loss of generality, assume $X$ is a black cell. Draw a $45^{\circ}$ diagonal line from the lower left to the upper left through the center of $X$. Let the cente... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 724,323 |
1. If for all real numbers $x, x^{2}-2 \sqrt{a} x+b$ is always positive, and
$$
\begin{array}{c}
M=2 \sqrt{b^{2}+2 b+1}-\sqrt{(a+b+2)^{2}}- \\
\sqrt{(a-b-3)^{2}},
\end{array}
$$
then the value of $M$ is ( ).
(A) 3
(B) -3
(C) $-2 a+2 b-3$
(D) $4 b+7$ | - 1. B.
From the problem, we know
$$
\Delta=(-2 \sqrt{a})^{2}-4 \times 1 \times ba \text {. }
$$
Since $\sqrt{a}$ is defined, therefore, $a \geqslant 0$.
Thus, $M=2|b+1|-|a+b+2|-|a-b-3|$
$$
=2(b+1)-(a+b+2)+(a-b-3)=-3 .
$$ | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,325 |
2. As shown in Figure 1, quadrilateral $ABCD$ is an inscribed quadrilateral of $\odot O$, and $AC \perp BD$ at point $E$. If $AB=8, CD=6$, then the radius of $\odot O$ is ().
(A) 6
(B) $2 \sqrt{7}$
(C) 5
(D) $\sqrt{7}$ | 2. C.
As shown in Figure 4, draw the diameter $B F$ of $\odot O$ through point $B$, and connect $A F, D F$.
Since $B F$ is the diameter of $\odot O$, we know
$$
\begin{array}{l}
B D \perp D F . \\
\text { Since } A C \perp B D, \text { therefore, } \\
A C \parallel F D \\
\Rightarrow A F=C D=6 .
\end{array}
$$
Then ... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 724,326 |
3. Given the equation in terms of $x$
$$
\frac{1}{|x|}+x^{2}=2 x+\frac{3|x|}{x} \text {. }
$$
The number of real solutions to this equation is ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | 3. B.
From the original equation, it is easy to know that $x>0$. Then the original equation can be transformed into
$$
\begin{array}{l}
\frac{1}{x}=-x^{2}+2 x+3 . \\
\text { Let } f(x)=\frac{1}{x}, g(x)=-x^{2}+2 x+3 .
\end{array}
$$
Obviously, $f(1)=1<4=g(1)$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,327 |
4. Let four distinct real numbers $a, b, c, d$ satisfy
$$
\begin{array}{l}
\left(a^{2011}-c^{2011}\right)\left(a^{2011}-d^{2011}\right)=2011, \\
\left(b^{2011}-c^{2011}\right)\left(b^{2011}-d^{2011}\right)=2011 . \\
\text { Then }(a b)^{2011}-(c d)^{2011}=(\quad) .
\end{array}
$$
Then $(a b)^{2011}-(c d)^{2011}=(\quad... | 4. B.
From $a \neq b$, we know $a^{2011} \neq b^{2011}$.
Thus, $a^{2011}$ and $b^{2011}$ are the two distinct real roots of the quadratic equation in $x$:
$$
\left(x-c^{2011}\right)\left(x-d^{2011}\right)=2011,
$$
which is
$$
x^{2}-\left(c^{2011} \div d^{2011}\right) x+(c d)^{2011}-2011=0.
$$
By Vieta's formulas, we g... | -2011 | Algebra | MCQ | Yes | Yes | cn_contest | false | 724,328 |
5. Given that $I$ and $O$ are the incenter and circumcenter of $\triangle ABC$ respectively, and points $B, C, I, O$ are concyclic. Then the degree measure of $\angle BAC$ is ( ).
(A) $30^{\circ}$
(B) $45^{\circ}$
(C) $60^{\circ}$
(D) $75^{\circ}$ | 5. C.
As shown in Figure 5, connect $I B$.
$$
I C, O B, O C \text {. }
$$
From 1.0, the incenter and circumcenter of $\triangle A B C$ respectively give
$$
\begin{array}{c}
\angle B I C \\
=90^{\circ}+\frac{1}{2} \angle B A C, \\
\angle B O C=2 \angle B A C .
\end{array}
$$
Since $B, C, I, O$ are concyclic,
$$
\angl... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 724,329 |
6. Figure 2 is a part of the HZ district map. A river runs through the district, with the two banks being the broken lines $A-B-C$ and $D-O-E$, and there are two locations $M$ and $N$. Two bridges perpendicular to the riverbanks and roads are to be built to connect $M$ and $N$ to both banks of the river, making the tot... | 6. B.
Translate point $M(-16,12)$ 3 units to the left to get point $M^{\prime}(-13,12)$, and translate point $N(11,-1)$ 3 units upwards to get point $N^{\prime}(11,2)$. Connect $M^{\prime} N^{\prime}$, which intersects $A B$ and $B C$ at points $P$ and $Q$ respectively.
Then the minimum length of the road is
$$
M^{\pr... | 32 | Geometry | MCQ | Yes | Yes | cn_contest | false | 724,330 |
1. In the right trapezoid $A B C D$, $A D / / B C, A B \perp$ $B C, P$ is a point on $A B$. If $A D=1.5, A B=B C=$ $3, A P=2$, then $\angle P C D=$ $\qquad$ | $=1.45^{\circ}$.
As shown in Figure 6, draw $C E \perp$
$A D$ at point $E$, extend $D E$ to point $F$ such that $E F=P B$, and connect $C F$ and $D P$.
It is easy to prove that quadrilateral $A B C E$ is a square and
$\triangle B C P \cong \triangle E C F$.
Thus, $P C=F C, \angle B C P=\angle E C F$.
Therefore, $P D=\... | 45^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,331 |
2. Given real numbers $x, y$ satisfy
$$
\begin{array}{l}
\sqrt{125 x-16}+\sqrt{25-8 y}=6, \\
\sqrt{25-80 x}+\sqrt{12.5 y-16}=6 .
\end{array}
$$
Then $\frac{x}{y}=$ | 2.0.1.
Subtracting the two equations yields
$$
\begin{array}{l}
\sqrt{125 x-16}-\sqrt{12.5 y-16}+ \\
(\sqrt{25-8 y}-\sqrt{25-80 x})=0 \\
\Rightarrow(x-0.1 y)\left(\frac{125}{\sqrt{125 x-16}+\sqrt{12.5 y-16}}+\right. \\
\left.\frac{80}{\sqrt{25-80 x}+\sqrt{25-8 y}}\right)=0 \\
\Rightarrow x-0.1 y=0 \Rightarrow \frac{x}... | 0.1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,332 |
3. Given that $A D$ is the angle bisector of $\triangle A B C$, $\angle C$ $=90^{\circ}, \frac{A B-A C}{B D-C D}=\sqrt{5}$. Then
$$
\frac{A B+A C}{A D+B C}=
$$
$-$ $\qquad$ | 3. $25 \sqrt{5}-10 \sqrt{30}$.
As shown in Figure 7, draw $D E \perp A B$ at point $E$. It is easy to see that
$$
\begin{array}{l}
A C=A E, \\
C D=D E .
\end{array}
$$
Let $B E=m, B D=n$, and $D E=k$.
In the right triangle $\triangle B D E$, by the Pythagorean theorem, we have
$$
m^{2}+k^{2}=n^{2} \text {. }
$$
Also... | 25 \sqrt{5}-10 \sqrt{30} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,333 |
4. Given a prime number $p$ such that $p^{3}-6 p^{2}+9 p$ has exactly 30 positive divisors. Then the smallest value of $p$ is $\qquad$ . | 4. 23 .
Obviously, when $p=2$ or 3, it does not meet the requirements of the problem.
Therefore, $p>3$.
Also, $p^{3}-6 p^{2}+9 p=p(p-3)^{2}$, at this point,
$(p, p-3)=(p, 3)=1$.
Since $p$ has two factors, $(p-3)^{2}$ has 15 factors.
And $15=5 \times 3$, to make $p$ the smallest, $p-3$ is also even, so it can only be
... | 23 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,334 |
One. (20 points) There are three stocks $A$, $B$, and $C$, and the purchasing situation of 28 investors is as follows: Each investor has bought at least one of the stocks; Among those who did not buy stock $A$, the number of people who bought stock $B$ is twice the number of people who bought stock $C$; Among those who... | Let the number of people who only buy stock $A$, stock $B$, and stock $C$ be $x$, $y$, and $z$ respectively. Among those who do not buy stock $A$, the number of people who buy both stocks $B$ and $C$ is $m$. Among those who buy stock $A$, the number of people who buy at least one of stocks $B$ or $C$ is $n$ (as shown i... | 25\% | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 724,336 |
II. (25 points) As shown in Figure 3, in $\triangle ABC$, $AB < AC$, the incircle $\odot I$ of $\triangle ABC$ touches $AB$ and $AC$ at points $D$ and $E$ respectively, the extensions of $BI$ and $CI$ intersect line $DE$ at points $F$ and $G$, and $BG$ intersects $CF$ at point $M$. Prove: $I$ is the orthocenter of $\tr... | II. As shown in Figure 9,
Connect $A I$, $D I$, and $E I$.
Obviously, $A$, $D$, $I$, and $E$ are concyclic.
Thus, $\angle I E D$
$=\angle I A D$
$=\frac{1}{2} \angle B A C$.
Therefore, $\angle C E G=\angle I E C+\angle I E D$
$$
=90^{\circ}+\frac{1}{2} \angle B A C=\angle B I C \text {. }
$$
Hence, $\triangle C E G \... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,337 |
$$
\text { Three. (25 points) Among the } 1,2, \cdots, 2008 \text { numbers, }
$$
add a negative sign in front of $n$ numbers so that the algebraic sum of these 2008 numbers equals zero. Find the maximum and minimum values of $n$.
| Three, find the minimum value of $n$.
To make $n$ as small as possible, the numbers to which negative signs are added should be as large as possible. Let these numbers be:
$$
2008, 2007, \cdots, 2010-n, k
$$
where, $00, \\
\frac{1}{2} n^{2}-\frac{4019}{2} n+10105270 \text{. }
$$
Since $n$ is a positive integer less t... | 589, 1419 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,338 |
1. Given real numbers $x, y$ satisfy
$$
17\left(x^{2}+y^{2}\right)-30 x y-16=0 \text {. }
$$
then the maximum value of $f(x, y)=\sqrt{16 x^{2}+4 y^{2}-16 x y-12 x+6 y+9}$ is . $\qquad$ | $-1.7$.
From $17\left(x^{2}+y^{2}\right)-30 x y-16=0$, we get
$$
\begin{array}{l}
(x+y)^{2}+16(x-y)^{2}=16 . \\
\text { Let }\left\{\begin{array}{l}
x+y=4 \cos \theta, \\
x-y=\sin \theta
\end{array}(\theta \in \mathbf{R})\right. \\
\Rightarrow\left\{\begin{array}{l}
x=2 \cos \theta+\frac{1}{2} \sin \theta, \\
y=2 \cos ... | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,339 |
2. Let $A$ be the set of values of $m$ for which the roots of the equation in $x$
$$
2(m+1) x^{2}-\left(m^{2}+m+16\right) x+8 m=0
$$
are both integers. Then $|A|=$
$\qquad$ . | 2. 2 .
If $m=-1$, then $x=-\frac{1}{2}$, which does not meet the requirement. If $m \neq-1$, then we can get $x_{1}=\frac{m}{2}, x_{2}=\frac{8}{m+1}$. According to the problem, $m$ is an even number, and $(m+1) \mid 8$, so $m=0$ or -2.
Therefore, $|A|=2$. | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,340 |
3. In the tetrahedron $S-ABC$, $BC=3, AC=4$, the distance from point $C$ to the midpoint of $AB$ is $\frac{5}{2}$, and the dihedral angles between the lateral faces $SAB$, $SBC$, and $SAC$ and the base are all $60^{\circ}$. Then the volume of the tetrahedron is $\qquad$ | 3. $2 \sqrt{3}$.
Since the angles between the three lateral faces and the base are equal, the projection of vertex $S$ onto the base is the incenter of $\triangle A B C$.
From the given information, $\angle A C B=90^{\circ}$, and $A B=5$. Therefore, the inradius $r$ of $\triangle A B C$ is $1$. Thus, the distance $h$... | 2 \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,341 |
4. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=a_{2}=3, a_{n+2}=3 a_{n+1}-2 a_{n}-1 \text {, }
$$
where, $n \in \mathbf{N}_{+}, S_{n}$ is the sum of the first $n$ terms of $\left\{a_{n}\right\}$. Then the maximum value of $S_{n}$ is $\qquad$ . | 4. 8 .
From the given information, we have
$$
\begin{array}{l}
a_{n+2}-a_{n+1}-1=2\left(a_{n+1}-a_{n}-1\right) \\
=2^{n}\left(a_{2}-a_{1}-1\right)=-2^{n} \\
\Rightarrow a_{n}-a_{n-1}=1-2^{n-2} \\
\Rightarrow a_{n}=n+3-2^{n-1}\left(n \in \mathbf{N}_{+}\right)
\end{array}
$$
Therefore, $a_{3}=2$.
Also, when $n \geqslan... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,342 |
5. If the hyperbola
$$
\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>0, b>0)
$$
has a point with the abscissa $\frac{3}{2} a$ whose distance to the right focus is greater than its distance to the left directrix, then the range of the eccentricity $e$ of the hyperbola is $\qquad$. | 5. $(2,+\infty)$.
Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola, respectively. Then
$$
\begin{array}{l}
\left|P F_{2}\right|>\frac{3 a}{2}+\frac{a^{2}}{c} \\
\Rightarrow e\left(\frac{3 a}{2}-\frac{a}{c}\right)^{2}>\frac{3 a}{2}+\frac{a^{2}}{c} \\
\Rightarrow e\left(\frac{3}{2}-\frac{1}{e}\right)>... | (2,+\infty) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,343 |
6. From the set $\{1,2, \cdots, 2011\}$, any two different numbers $a, b$ are selected such that $a+b=n$ (where $n$ is some positive integer) with a probability of $\frac{1}{2011}$. Then the minimum value of $ab$ is | 6.2010.
Let the number of ways such that $a+b=n$ be $k$. Then
$$
\frac{k}{\mathrm{C}_{2011}^{2}}=\frac{1}{2011} \Rightarrow k=1005 \text {. }
$$
Consider $a b$ to be as small as possible, and the number of ways such that $a+b=n$ is 1005.
Take $n=2011$. Then
$$
1+2010=2+2009=\cdots=1005+1006 \text {. }
$$
At this poi... | 2010 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 724,344 |
7. Given real numbers $a, b, c$ satisfy $a+b+c=1, a^{2}+b^{2}+c^{2}=3$. Then the maximum value of $a b c$ is $\qquad$ . | 7. $\frac{5}{27}$.
Notice that among $a^{2}, b^{2}, c^{2}$, there is always one that does not exceed 1, let's assume $c^{2} \leqslant 1$. Then $c \in[-1,1]$.
$$
\begin{array}{l}
\text { Also } a b c=\frac{1}{2}\left[(a+b)^{2}-\left(a^{2}+b^{2}\right)\right] c \\
=\frac{1}{2}\left[(1-c)^{2}-\left(3-c^{2}\right)\right] ... | \frac{5}{27} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,345 |
3. All the diagonals drawn from one vertex of a convex $n$-sided polygon divide this convex $n$-sided polygon into $m$ small triangles. If $m$ is equal to $\frac{4}{9}$ of the number of diagonals of this convex $n$-sided polygon, then the sum of the interior angles of this $n$-sided polygon is $\qquad$ . | Answer: $720^{\circ}$. | 720^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 724,346 |
8. In an acute triangle $\triangle A B C$, the lengths of the sides opposite to $\angle A, \angle B, \angle C$ are $a, b, c$ respectively. If
$$
\frac{b^{2}}{a c} \geqslant \frac{\cos ^{2} B}{\cos A \cdot \cos C},
$$
then the range of $\angle B$ is $\qquad$ . | 8. $\left[\frac{\pi}{3}, \frac{\pi}{2}\right)$.
By the cosine rule, we have
$$
\frac{b^{2}}{a c} \geqslant \frac{\left(\frac{a^{2}+c^{2}-b^{2}}{2 a c}\right)^{2}}{\frac{b^{2}+c^{2}-a^{2}}{2 b c} \cdot \frac{a^{2}+b^{2}-c^{2}}{2 a b}} \text {. }
$$
Simplifying it, we get
$$
b^{2} \geqslant \frac{a^{4}+c^{4}}{a^{2}+c^{... | \left[\frac{\pi}{3}, \frac{\pi}{2}\right) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 724,347 |
9. (16 points) Let the graph $C_{1}$ of the function $f(x)=\ln x$ intersect with the graph $C_{2}$ of the function $g(x)=\frac{1}{2} a x^{2}+b x(a \neq 0)$ at points $P$ and $Q$. Draw a vertical line through the midpoint of segment $PQ$ that intersects the graphs $C_{1}$ and $C_{2}$ at points $M$ and $N$, respectively.... | Let $P\left(x_{1}, y_{1}\right)$ and $Q\left(x_{2}, y_{2}\right)\left(01\right)$.
Let $r(t)=\ln t-\frac{2(t-1)}{1+t}$. Then
$$
r^{\prime}(t)=\frac{1}{t}-\frac{4}{(t+1)^{2}}=\frac{(t-1)^{2}}{t(t+1)^{2}} \text {. }
$$
Since when $t>1$, $r^{\prime}(t)>0$, so $r(t)$ is monotonically increasing on $[1,+\infty)$.
Thus, $r(t... | proof | Calculus | proof | Yes | Yes | cn_contest | false | 724,348 |
10. (20 points) Given $f(x)=\frac{2 x}{x+1}$. The sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=2$, and
$$
a_{n}=\frac{1}{2} f\left(a_{n-1}\right)\left(n \in \mathbf{N}_{+}, n \geqslant 2\right) \text {, }
$$
(1) Find the general term formula for the sequence $\left\{a_{n}\right\}$;
(2) Prove: For all positive integ... | 10. (1) From the given, we have
$$
\begin{array}{l}
a_{n}=\frac{a_{n-1}}{a_{n-1}+1} \\
\Rightarrow \frac{1}{a_{n}}=1+\frac{1}{a_{n-1}}=\cdots=(n-1)+\frac{1}{a_{1}} \\
\quad=\frac{2 n-1}{2} \\
\Rightarrow a_{n}=\frac{2}{2 n-1} .
\end{array}
$$
When $n=1$, $a_{1}=2$, which also fits the above formula.
Thus, $a_{n}=\frac... | a_{n}=\frac{2}{2 n-1}\left(n \in \mathbf{N}_{+}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,349 |
11. (20 points) Let the parabola $C: y^{2}=2 p x(p>0)$ pass through the point $M(3,-6)$, with its directrix being $l$ and focus being $F$. For any point $N$ on the directrix $l$ of the parabola, and any chord $A B$ of the parabola passing through the focus $F$, let $k_{1} 、 k_{2} 、 k$ be the slopes of $N A 、 N B 、 N F$... | 11. From the point $M(3,-6)$ on the parabola $C$, we get $p=6$. Therefore, the equation of the parabola $C$ is $y^{2}=12 x$.
Since the point $F(3,0)$, we set $l_{A B}: x=t y+3$.
Let points $A\left(x_{1}, y_{1}\right)$, $B\left(x_{2}, y_{2}\right)$, and $N(-3, m)$.
From $\left\{\begin{array}{l}x=t y+3, \\ y^{2}=12 x\end... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 724,350 |
一、(40 points) As shown in Figure 1, let $A D, B E, C F$ be the altitudes of $\triangle A B C$, intersecting at point $H$, and let $M$ be the midpoint of side $B C$. The circumcircle $\Gamma_{1}$ of $\triangle B M E$ and the circumcircle $\Gamma_{2}$ of $\triangle C M F$ intersect at points $M$ and $N$. The line $A N$ i... | One, the auxiliary line as shown in Figure 2.
Since points $B, C, E, F$ are concyclic, we have
$H B \cdot H E = H C \cdot H F$.
This indicates that point $H$ lies on the radical axis of circles $\Gamma_{1}$ and $\Gamma_{2}$. Therefore, points $M, H, N$ are collinear. Thus,
$\angle F N H = \angle F N M = \angle F C M = ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,351 |
Let $\left\{p_{i} \mid i \in \mathbf{N}_{+}\right\}$ be the sequence of all prime numbers arranged in ascending order, and let $a_{n}=\sum_{i=1}^{n} i^{2} p_{i}(n \in \mathbf{N}_{+})$. Prove: for any positive integer $n$, the closed interval $\left[a_{n}, a_{n+1}\right]$ contains at least $n+1$ perfect squares. | Second, $a_{1}=2, a_{2}=14,\left[a_{1}, a_{2}\right]$ contains two perfect squares 4,9.
Next, prove that when $n \geqslant 2$, we have
$$
\begin{array}{l}
\sqrt{a_{n+1}}-\sqrt{a_{n}} \geqslant n+1 . \\
\text { Equation (1) } \Leftrightarrow a_{n+1} \geqslant\left(\sqrt{a_{n}}+n+1\right)^{2} \\
\Leftrightarrow a_{n+1}-a... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 724,352 |
Three. (50 points) Let $n$ be a given positive integer greater than 1. For any $d_{1}, d_{2}, \cdots, d_{n}>1$, when
$$
\sum_{1 \leqslant i<j \leqslant n} d_{i} d_{j}=(n-1) \sum_{i=1}^{n} d_{i}
$$
find the maximum value of
$$
S=\sum_{i=1}^{n}\left[d_{i} \sum_{j=1}^{n} d_{j}-n d_{i}+n(n-2)\right]^{-1}
$$ | $$
\text { Three, } S_{\max }=\frac{1}{n} \text {. }
$$
First, let $d_{1}=d_{2}=\cdots=d_{n}=2$. Clearly, $\sum_{1 \leqslant i0(i=1,2, \cdots, n)$, so,
$$
\sum_{i=1}^{n} a_{i} \geqslant n \sqrt{\frac{\sum_{1 \leq i<j \leq n} a_{i} a_{j}}{C_{n}^{2}}}=n \text {. }
$$
Then, $S=\sum_{i=1}^{n}\left[d_{i} \sum_{j=1}^{n} d_... | \frac{1}{n} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,353 |
Four, (50 points) Each point in the plane is colored with one of $n$ colors, while satisfying:
(1) Each color has infinitely many points, and they are not all on the same line;
(2) There is at least one line on which all points are exactly two colors.
Find the minimum value of $n$ such that there exist four points of ... | Given $n \geqslant 4$.
If $n=4$, take a fixed circle $\odot O$ and three points $A, B, C$ on it. Color the arc $\overparen{A B}$ (including point $A$ but not $B$), the arc $\overparen{B C}$ (including point $B$ but not $C$), and the arc $\overparen{C A}$ (including point $C$ but not $A$) with colors 1, 2, and 3, respec... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 724,354 |
In $\triangle B E F$, $\angle B E F=90^{\circ}$, $B E=E F$, quadrilateral $A B C D$ is a square, connect $D F$, $G$ is the midpoint of $D F$, connect $E G$ and $C G$.
Prove:
$$
\begin{array}{c}
E G=C G \\
E G \perp C G .
\end{array}
$$ | Prove as shown in Figure 2, extend $C G$ to point $M$, such that $M G = C G$, connect $M F, M E, E C$, extend $M F$ to intersect the extension of $C B$ at point $N$, and $M N$ intersects $B E$ at point $H$.
Since $F G = D G, M G = C G$, we have,
From the fact that quadrilateral $A B C D$ is a square, we know
$\angle D ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 724,355 |
Given $a$, $b$, $x$ are positive integers, and $a \neq b$, $\frac{1}{x}=\frac{1}{a^{2}}+\frac{1}{b^{2}}$. Try to find the minimum value of $x$.
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The above text translated into English, preserving the original text's line breaks and format, is as follows:
Given $a$, $b$, $x$ are positive integers, and $a \neq b$, ... | Solve: It is easy to know that $x=\frac{a^{2} b^{2}}{a^{2}+b^{2}}$.
Let $d=(a, b), a=d a_{0}, b=d b_{0},\left(a_{0}, b_{0}\right)=1$.
Then $x=\frac{d^{2} a_{0}^{2} b_{0}^{2}}{a_{0}^{2}+b_{0}^{2}} \in \mathbf{N}_{+}$.
By $\left(a_{0}^{2}+b_{0}^{2}, a_{0}^{2} b_{0}^{2}\right)=1$, we know $\left(a_{0}^{2}+b_{0}^{2}\right)... | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 724,356 |
Example 1 On a circle, $n$ points $A_{1}, A_{2}, \cdots, A_{n}$ are arranged in sequence. Each point is arbitrarily colored white or black. Prove: Among the $n$ arcs $\overparen{A_{1} A_{2}}$, $\overparen{A_{2} A_{3}}, \cdots, \overparen{A_{n} A_{1}}$ connecting adjacent points, the number of arcs with endpoints of dif... | Prove that an arc with endpoints of different (same) colors is called a bichromatic (monochromatic) arc. Use numbers to represent colors, with white denoted as 1 and black as -1. Each point $A_{k}(k=1,2, \cdots, n)$ uniquely corresponds to a number $a_{k}$, where $a_{k}=1$ or -1.
Clearly, the arc $\overparen{A}_{k} A_{... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 724,357 |
Example 2 Divide the square $A B C D$ into $n^{2}$ equal small squares ($n$ is a positive integer), color the opposite vertices $A, C$ red, and $B, D$ blue, and color their intersection points in either red or blue. Prove: the number of small squares with exactly three vertices of the same color must be even. | Proof: Let's denote red as 1 and blue as -1; and number the small squares as $1, 2, \cdots, n^{2}$, and let the product of the numbers at the four vertices of the $i$-th small square be $A_{i}$. If exactly three vertices are the same color, then $A_{i}=-1$; otherwise, $A_{i}=1$.
In the product $A_{1} A_{2} \cdots A_{n... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 724,358 |
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