problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
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class | __index_level_0__ int64 0 742k |
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17. Let $N$ be a positive integer, and $5 | N$. Arrange one red ball and $N$ identical green balls randomly in a row. Let $P(N)$ be the probability that the number of green balls on one side of the red ball is at least $\frac{3}{5}$ of the total number of green balls. It is not hard to see that $P(5)=1$, and as $N$ app... | 17. A.
Let $n=\frac{N}{5}$. Arrange $N$ green balls in a row, dividing them into five equal parts. Insert a red ball randomly into one of the $N+1$ gaps, which gives $N+1$ different ways to insert the red ball. When the red ball is placed in the gaps within the middle part, it does not meet the condition that the gree... | A | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 729,479 |
19. As shown in Figure 4, in rectangle $A B C D$, $A B=6, B C$ $=3$, point $E$ is on side $B C$, point $F$ is on side $C E$, $B E=$ $E F=F C$. Connecting $A E$ and $A F$, they intersect $B D$ at points $P$ and $Q$ respectively. If $B P: P Q: Q D=r: s: t\left(r, s, t \in \mathbf{Z}_{+},(r, s, t)=1\right)$, then $r+s+t=(... | 19. E.
From $A D / / B C \Rightarrow \triangle A P D \backsim \triangle E P B$
$$
\Rightarrow \frac{r}{s+t}=\frac{B E}{A D}=\frac{1}{3} \text {. }
$$
Similarly, $\frac{r+s}{t}=\frac{B F}{A D}=\frac{2}{3}$.
$$
\begin{array}{l}
\text { Then }\left\{\begin{array}{l}
s+t=3 r, \\
3 r+3 s=2 t
\end{array} \Rightarrow t=4 s ... | E | Geometry | MCQ | Yes | Yes | cn_contest | false | 729,480 |
22. For a positive integer $n, 110 n^{3}$ has 110 different positive divisors, including 1 and $110 n^{3}$ itself. Then $81 n^{4}$ has ( ) different positive divisors.
(A) 110
(B) 191
( C) 261
(D) 325
(E) 425 | 22. D.
By the Fundamental Theorem of Arithmetic, let
$$
110 n^{3}=p_{1}^{n_{1}} p_{2}^{n_{2}} \cdots p_{k}^{n_{k}},
$$
where $n_{i} \in \mathbf{Z}_{+}, i=1,2, \cdots, k ; p_{1}, p_{2}, \cdots, p_{k}$ are prime factors, and $p_{1}<p_{2}<\cdots<p_{k}$.
By the Divisor Counting Theorem, we have
$$
\begin{array}{l}
\left(... | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 729,482 |
23. For any real numbers $a$, $b$, $c$, define an operation ※ with the following properties:
(1) $a ※(b ※ c)=(a ※ b) \cdot c$,
(2) $a ※ a=1$,
where “$\cdot$” denotes multiplication.
If the solution to the equation $2016 \times(6 ※ x)=100$ is $x=\frac{p}{q}(p, q$ are positive integers, $(p, q)=1)$, then the value of $p... | 23. A.
$$
\begin{array}{l}
\text { Given } 2016 ※(6 ※ x)=(2016 ※ 6) \cdot x \\
\Rightarrow(2016 ※ 6) \cdot 6=\frac{600}{x} \\
\Rightarrow 2016 ※(6 ※ 6)=\frac{600}{x} .
\end{array}
$$
Since $a ※ a=1$, we have,
$$
\begin{array}{l}
6 ※ 6=2016 ※ 2016=1 . \\
\text { Therefore } 2016 ※(2016 ※ 2016)=\frac{600}{x} \\
\Rightar... | 109 | Algebra | MCQ | Yes | Yes | cn_contest | false | 729,483 |
5. In a sequence of numbers $x_{1}, x_{2}, \cdots$, it is known that $x_{1}=1$, and for $k \geqslant 2$,
$$
x_{k}=x_{k-1}+1-4\left(\left[\frac{k-1}{4}\right]-\left[\frac{k-2}{4}\right]\right) \text {. }
$$
Then $x_{2010}=(\quad$.
(A) 1
(B) 2
(C) 3
( D) 4
(2010, National Junior High School Mathematics Competition) | Hint From the given, we have
$$
\begin{array}{l}
x_{1}=1, x_{2}=2, x_{3}=3, x_{4}=4, x_{5}=1, \\
x_{6}=2, x_{7}=3, x_{8}=4, \cdots \cdots
\end{array}
$$
Since $2010=4 \times 502+2$, therefore,
$$
x_{2010}=x_{2}=2 \text{. }
$$
So the answer is (B). | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 729,484 |
3. A physicist discovered 2015 atoms called "American Soap," each of which either has an electron or does not have an electron. The physicist cannot distinguish between the two. The only tool the physicist has is a diode, which can connect any American Soap A with any other American Soap B (this connection is direction... | 3. Cannot.
Let the American states be $U_{1}, U_{2}, \cdots, U_{m}(m=2015)$.
Consider the following pattern $M_{k}$: for some integer $k(0 \leqslant k \leqslant m)$, $U_{1}, U_{2}, \cdots, U_{k}$ all have electrons, and the other American states do not have electrons. Then, for any two different patterns, they have di... | Cannot | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 729,485 |
5. A tournament graph is a directed graph where any two distinct vertices form an unordered pair with a directed edge from one vertex to the other. A proper directed edge coloring is defined as assigning a color to each directed edge such that every pair of directed edges $\overrightarrow{u v}$ and $\overrightarrow{w v... | 5. For each positive integer $n, f(n)=\left\lceil\log _{2} n\right\rceil$, where $\lceil x\rceil$ denotes the smallest integer not less than the real number $x$.
First, use the second mathematical induction to prove: For each positive integer $n$, we have $f(n) \geqslant\left\lceil\log _{2} n\right\rceil$.
When $n=1$,... | f(n)=\left\lceil\log _{2} n\right\rceil | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,486 |
1. Given positive integers $a, b, c (a < b < c)$ form a geometric sequence, and
$$
\log _{2016} a+\log _{2016} b+\log _{2016} c=3 .
$$
Then the maximum value of $a+b+c$ is $\qquad$ | One, 1.4066 273.
From the given, we know
$$
\begin{array}{l}
b^{2}=a c, a b c=2016^{3} \\
\Rightarrow b=2016, a c=2016^{2} .
\end{array}
$$
Since $a$, $b$, and $c$ are positive integers, when $a=1$, $c=2016^{2}$, $a+b+c$ reaches its maximum value
$$
2016^{2}+2017=4066273 .
$$ | 4066273 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,487 |
2. The solution set of the equation about the real number $x$
$$
2^{1-2 \sin ^{2} x}=2+\log _{2}\left(1-\sin ^{2} x\right)
$$
is . $\qquad$ | 2. $\left\{x \mid x=k \pi\right.$ or $\left.\frac{k \pi}{2}+\frac{\pi}{4}, k \in \mathbf{Z}\right\}$.
Let $t=\cos 2 x$. Then $t \in(-1,1]$.
The original equation transforms into an equation about $t$:
$$
2^{t}=1+\log _{2}(1+t) \text {. }
$$
When $0<t+1>1+\log _{2}(1+t) \text {. }
$$
Therefore, the solution to the eq... | \left\{x \mid x=k \pi\right. \text{ or } \left.\frac{k \pi}{2}+\frac{\pi}{4}, k \in \mathbf{Z}\right\} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,488 |
3. The area of the closed figure enclosed by the curve $x^{2}+2 y^{2} \leqslant 4|y|$ is $\qquad$ | 3. $2 \sqrt{2} \pi$.
By the symmetry of the graph of the given curve equation, we only need to consider the case where $y \geqslant 0$.
When $y \geqslant 0$, the curve equation becomes
$$
\frac{x^{2}}{2}+(y-1)^{2} \leqslant 1 \text {, }
$$
which represents an elliptical surface. The area of this part is
$$
S=\pi \tim... | 2 \sqrt{2} \pi | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,489 |
4. For all complex numbers $z$ satisfying $z \neq \mathrm{i}$, we have $F(z) = \frac{z-\mathrm{i}}{z+\mathrm{i}}$. For all positive integers $n$, $z_{n}=F\left(z_{n-1}\right)$. If $z_{0}=2016+\mathrm{i}$, then $z_{2016}=$ $\qquad$ | 4. $2016+\mathrm{i}$.
From the problem, we know
$$
\begin{array}{l}
F(F(z))=\frac{\frac{z-\mathrm{i}}{z+\mathrm{i}}-\mathrm{i}}{\frac{z-\mathrm{i}}{z+\mathrm{i}}+\mathrm{i}}=\frac{z-\mathrm{i}-\mathrm{i} z+1}{z-\mathrm{i}+\mathrm{i} z-1} \\
=\frac{1-\mathrm{i}}{1+\mathrm{i}} \cdot \frac{z+1}{z-1}=-\mathrm{i} \cdot \fr... | 2016+\mathrm{i} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,490 |
5. Given that $P$ is a point on the edge $A B$ of the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, satisfying that the angle between line $A_{1} B$ and plane $B_{1} C P$ is $60^{\circ}$. Then the tangent value of the dihedral angle $A_{1}-B_{1} P-C$ is $\qquad$ . | 5. $-\sqrt{5}$.
Assume the edge length of the cube is $1, B P=x$, and let $A_{1} B$ intersect $B_{1} P$ at point $Q$.
It is easy to know that $Q B=\frac{\sqrt{2} x}{x+1}$.
Thus, the distance from point $B$ to the plane $B_{1} P C$ is
$h=Q B \sin 60^{\circ}=\frac{\sqrt{6} x}{2(x+1)}$.
Note that, in $\triangle B_{1} P C... | -\sqrt{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,491 |
Question 1 Given positive integers $m, n$ satisfying $m \geqslant n \geqslant 2$. Find the number of injections $f:\{1,2, \cdots, n\} \rightarrow\{1,2, \cdots, m\}$, such that there is a unique $i \in\{1,2, \cdots, n-1\}$, satisfying $f(i)>f(i+1)$. ${ }^{[1]}$
(2012, Romanian Mathematical Olympiad)
The original solutio... | Let the image set of the mapping $f$ be
$$
A=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\} .
$$
Then there are $\mathrm{C}_{m}^{n}$ such image sets.
Without loss of generality, let $A=\{1,2, \cdots, n\}$.
Next, we calculate the number of one-to-one mappings $f$:
$$
\{1,2, \cdots, n\} \rightarrow\{1,2, \cdots, n\}
$$
suc... | \mathrm{C}_{m}^{n}\left(2^{n}-n-1\right) | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,493 |
Question 2 Given positive integers $m, n$ satisfying $m \geqslant n \geqslant 3$. Find the number of injections $f:\{1,2, \cdots, n\} \rightarrow\{1,2, \cdots, m\}$, such that there is a unique $i \in\{1,2, \cdots, n-2\}$, satisfying
$$
f(i)>f(i+1)>f(i+2) \text {. }
$$
For other $j \in\{1,2, \cdots, n-1\} \backslash\{... | Let the image set of the mapping $f$ be
$$
A=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\} .
$$
Then there are $\mathrm{C}_{m}^{n}$ such image sets.
Without loss of generality, let $A=\{1,2, \cdots, n\}$.
We now calculate the number of one-to-one mappings $f$ :
$$
\{1,2, \cdots, n\} \rightarrow\{1,2, \cdots, n\}
$$
such... | \mathrm{C}_{m}^{n}\left[(n-4) 2^{n-1}+n+2\right] | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,494 |
Question 1 In the acute triangle $\triangle ABC$, $AB < AC$, $O$ and $G$ are the circumcenter and centroid of $\triangle ABC$ respectively, and $D$ is the midpoint of side $BC$. Point $E$ lies on the circle with diameter $BC$, satisfying $AE \perp BC$, and $A$ and $E$ are on the same side of $BC$. Extend $EG$ to inters... | Let $B C=a, C A=b, A B=c, \angle B A C=\alpha$,
$\angle A B C=\beta, \angle B C A=\gamma$.
It is easy to know $K D=D L \triangleq x$. Then
$B K=L C \triangleq_{t, x+t}=\frac{a}{2}$,
$M K=t \tan \beta, N L=t \tan \gamma$.
In Figure 1, let $A E \perp B C$ at point $T$. Then $A T / /$
$O D$, and the orthocenter $H$ of $\t... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,495 |
Question 2 As shown in Figure 2, in the acute triangle $\triangle ABC$, $AB > AC$, $O$ is the circumcenter, and $D$ is the midpoint of side $BC$. A circle is constructed with $AD$ as its diameter, intersecting sides $AB$ and $AC$ at points $E$ and $F$, respectively. A line $DM \parallel AO$ is drawn through point $D$, ... | Prove as shown in Figure 2, construct the altitudes $C H$ and $B G$ on sides $A B$ and $A C$ of $\triangle A B C$. Connect $H G$, $D E$, and $D F$. Let $N$ be the midpoint of side $H G$.
From $D$ being the midpoint of side $B C$, we know
$$
D H=\frac{1}{2} B C=D G \text {. }
$$
Then $D N \perp H G$.
Draw the tangent ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,496 |
Question 3 In $\triangle B C F$, $\angle B$ is a right angle. On the line $C F$, take point $A$ such that $F A=F B$, and $F$ is between points $A$ and $C$; take point $D$ such that $D A=D C$, and $A C$ is the angle bisector of $\angle D A B$; take point $E$ such that $E A=E D$, and $A D$ is the angle bisector of $\angl... | Proof as shown in Figure 3.
Given that
$$
\begin{array}{l}
\angle E A D=\angle E D A=\angle D A C \\
=\angle B A F=\angle A B F=\alpha .
\end{array}
$$
Then $E D \parallel A C$.
Also, $A M \parallel E X$, so $E, D, X$ are collinear.
Since $D A=D C$, we have
$$
\begin{array}{l}
\angle A E D=\angle A D C=180^{\circ}-2 \... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,497 |
Find all integer-coefficient polynomials $P(x)$ such that there exists an infinite integer sequence $\left\{a_{n}\right\}$, where any two terms are distinct, and satisfies
$$
P\left(a_{1}\right)=0, P\left(a_{k+1}\right)=a_{k}(k=1,2, \cdots) .
$$
(2016, Peking University Mathematical Science Summer Camp Preliminary Cont... | First, we prove a lemma.
Lemma If $F(x)$ is a polynomial with integer coefficients, and $u, v$ are integers, then
$$
(u-v) \mid (F(u)-F(v)).
$$
Proof Let
$$
F(x)=b_{n} x^{n}+b_{n-1} x^{n-1}+\cdots+b_{1} x+b_{0}.
$$
Then $F(u)-F(v)$
$$
\begin{array}{l}
=b_{n}\left(u^{n}-v^{n}\right)+b_{n-1}\left(u^{n-1}-v^{n-1}\right)... | P(x)=x-C | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,498 |
Find all complex-coefficient polynomials $P(x)$ such that there exists an infinite sequence of integers $\left\{a_{n}\right\}$, all distinct, and satisfying
$$
P\left(a_{1}\right)=0, P\left(a_{k+1}\right)=a_{k}(k=1,2, \cdots) .
$$ | 【Analysis】Let $P(x)=\sum_{j=0}^{n}\left(\alpha_{j}+\beta_{j} \mathrm{i}\right) x^{j}$, where $\alpha_{j}, \beta_{j} \in \mathbf{R}(j=0,1, \cdots, n), \alpha_{n} \neq 0$.
Then by the given condition, for any $k \in \mathbf{Z}_{+}, P\left(a_{k+1}\right)=$ $a_{k} \in \mathbf{R}$, i.e., $\square$
$$
\sum_{j=0}^{n} \beta_{... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,499 |
493 Given the sequence $\left\{S_{n}\right\}$ satisfies
$$
S_{n}=\sum_{i=1}^{n} i!\left(n \in \mathbf{Z}_{+}\right) .
$$ | Prove: For any given prime $p$, there exist infinitely many positive integers $a$ such that $v_{p}\left(S_{a+1}\right) \geqslant v_{p}\left(S_{a}\right)$, where $v_{p}(d)$ denotes the power of $p$ in $d$. | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 729,500 |
Given a positive sequence $\left\{a_{n}\right\}$, let the set $A=\left\{k \in \mathbf{Z}_{+} \mid k=1\right.$ or $k \geqslant 2$, and $\left.a_{k}>\frac{1}{2} a_{k-1}\right\}$. If $\sum_{k \in A} a_{k}$ converges, prove: $\sum_{k=1}^{+\infty} a_{k}$ converges. | Prove that for any positive integer $k$, if the set $A$ does not contain any number greater than $k$, denote $f(k)=+\infty$; if the set $A$ contains numbers greater than $k$, denote
$$
\begin{array}{l}
f(k)=\min \{m \mid m>k, m \in A\} . \\
\text { Define } S_{k}=\sum_{i=k}^{f(k)-1} a_{i} .
\end{array}
$$
We need to p... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 729,501 |
Example 3 Proof: If the distances between the opposite edges of a tetrahedron are $d_{1}, d_{2}, d_{3}$, then the volume of the tetrahedron
$$
V \geqslant \frac{1}{3} d_{1} d_{2} d_{3} .
$$
(48th All-Soviet Union Mathematical Competition) | Prove that constructing a parallelepiped such that the edges of the original tetrahedron are the diagonals of the parallelepiped's faces. Thus, the distances between the opposite faces are equal to the distances between the three pairs of opposite edges of the tetrahedron, and it is easy to see that the volume of the p... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,502 |
Let $k$ be a positive integer. Prove: there exists a positive real number $c_{k}$, such that for any positive integer $n$ and any distinct positive integers $a_{1}, a_{2}, \cdots, a_{n}$, we have
$$
\left[a_{1}, a_{2}, \cdots, a_{n}\right] \geqslant c_{k} n^{k}
$$ | Let $d(n)$ denote the number of positive divisors of the positive integer $n$. Let $\left[a_{1}, a_{2}, \cdots, a_{n}\right]=m$. Then $a_{1}, a_{2}, \cdots, a_{n}$ are all positive divisors of $m$. Therefore, $d(m) \geqslant n$. Thus, it suffices to prove that for any positive integer $m$,
$$
\frac{d^{k}(m)}{m} \leqsla... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 729,503 |
In $\triangle A B C$, point $D$ is on side $B C$, and the incircles of $\triangle A B D$ and $\triangle A D C$ are equal. $\odot M$ is the excircle of $\triangle A B C$ internal to $\angle B$, with radius $m$; $\odot N$ is the excircle of $\triangle A B C$ internal to $\angle C$, with radius $n$. Prove: $A D$ is the ge... | Proof As shown in Figure 2, let $B C=a, C A=b, A B=c$, $p$ be the semiperimeter of $\triangle A B C$, and $\odot O_{1}, \odot O_{2}$ be the incircles of $\triangle A B D, \triangle A D C$ with radii $r_{1}, r_{2}$, respectively.
Extend $B O_{1}, C O_{2}$ to intersect at point $O$, and draw $O G \perp B C$, $O_{1} E \p... | A D^{2}=m n | Geometry | proof | Yes | Yes | cn_contest | false | 729,504 |
492 Select five subsets $A_{1}, A_{2}, \cdots, A_{5}$ from the set $\{1,2, \cdots, 1000\}$ such that $\left|A_{i}\right|=500(i=1,2$, $\cdots, 5)$. Find the maximum value of the number of common elements in any three of these subsets. | Let $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{5}$ be five 1000-dimensional vectors, and let $\alpha_{i}(j)$ denote the $j$-th element of $\alpha_{i}$. Then $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{5}$ satisfy
$$
\alpha_{i}(j)=\left\{\begin{array}{l}
1, j \in A_{i} ; \\
0, j \notin A_{i} .
\end{array}\right.
$$
Let the ... | 50 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,505 |
Example 4 If three lines $a$, $b$, and $c$ in space are pairwise skew lines, then the number of lines that intersect with $a$, $b$, and $c$ is ( ).
(A) 0
(B) 2
(C) More than 1 but finite
(D) Infinite
(1997, National High School Mathematics Competition) | Regardless of the positional relationship between lines $a$, $b$, and $c$, a parallelepiped $A B C D-A_{1} B_{1} C_{1} D_{1}$ can always be constructed such that $A B$ lies on line $a$, $B_{1} C_{1}$ lies on line $b$, and $D_{1} D$ lies on line $c$, as shown in Figure 4.
Take a point $M$ on the extension of $D D_{1}$.... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 729,506 |
Example 7 A rectangular wooden board (neglecting thickness) is placed on a horizontal desktop, denoted as rectangle $A B C D$. Now the board is rotated upward around side $A B$ by an angle $\alpha$, and then rotated upward around side $A D$ by an angle $\beta$ and fixed, where $\alpha, \beta$ are both acute angles. If ... | Construct the rectangular prism $A B C D-A_{1} B_{1} C_{1} D_{1}$, and connect $B D, A D_{1}$, as shown in Figure 9.
Assume the plane of the rectangular wooden board after the first rotation is the plane $A B C_{1} D_{1}$, and the plane after the second rotation is the plane $A B_{1} D_{1}$.
In the plane $B C C_{1} B... | m g \sqrt{1-\cos ^{2} \alpha \cdot \cos ^{2} \beta} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,508 |
Given that $a$, $b$, and $c$ are three distinct real numbers. If in the quadratic equations
$$
\begin{array}{l}
x^{2}+a x+b=0, \\
x^{2}+b x+c=0, \\
x^{2}+c x+a=0
\end{array}
$$
any two of these equations have exactly one common root, find the value of $a^{2}+$ $b^{2}+c^{2}$. | Let the equations (1) and (3) have only one common root \( x_{1} \), equations (1) and (2) have only one common root \( x_{2} \), and equations (2) and (3) have only one common root \( x_{3} \). Therefore, the roots of equation (1) are \( x_{1} \) and \( x_{2} \), the roots of equation (2) are \( x_{2} \) and \( x_{3} ... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,509 |
Example 1 Given a positive integer $n$. There are $2n$ cards stacked, numbered from $1$ to $2n$ from top to bottom. Perform the following operation: each time, extract all the cards in even positions from top to bottom, maintaining their order and placing them at the bottom of the stack. For example, when $n=3$, the in... | 【Analysis】First, we mathematize the practical problem.
Let the $k$-th card from the top become the $f(k)$-th card from the top after one operation. The original problem is to prove that there exists a positive integer $r$ not exceeding $2n-2$ such that $f_{r}(k)=k$ for all $k (1 \leqslant k \leqslant 2n)$ after $r$ ite... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 729,510 |
Example 2: A rope of length 2009 is operated as follows: first, it is divided into two ropes of positive integer lengths, and the lengths of the two ropes are recorded, then the above operation is repeated on one of the ropes, ... until 2009 ropes of length 1 are obtained. If the lengths of the two ropes obtained in a ... | 【Analysis】(1) It is easy to know that a rope of length 2 can only be divided into two segments of length 1, meaning the last operation on the rope must not be a good operation. A rope of length 2009 can be divided into 2009 segments of length 1 after exactly 2008 operations, so the number of good operations is no more ... | 7 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,511 |
Example 4 If the irreducible fraction $\frac{n}{m}$ satisfies
$$
m n \leqslant 2009\left(m, n \in \mathbf{Z}_{+}\right),
$$
then $\frac{n}{m}$ is called a "Bull Fraction". Now, all Bull Fractions are arranged in increasing order to form a sequence $\frac{n_{1}}{m_{1}}, \frac{n_{2}}{m_{2}}, \frac{n_{3}}{m_{3}}, \cdots$... | Prove that for any positive integer $n$, the subsequence of the Farey sequence with denominators not exceeding $n$ is denoted as $T_{n}$. When $n=1$, the sequence
$$
T_{1}=\left(\frac{1}{1}, \frac{2}{1}, \frac{3}{1}, \cdots, \frac{2009}{1}\right)
$$
obviously satisfies the condition.
We proceed by induction on $n$.
Fr... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 729,512 |
1. There are $n\left(n \in \mathbf{Z}_{+}\right)$ cities arranged from left to right on a highway. Each city has a left bulldozer (placed on the left side of the city and facing left) and a right bulldozer (placed on the right side of the city and facing right). These $2 n$ bulldozers are all of different sizes. Each t... | 1. Let the $n$ cities from left to right be $T_{1}, T_{2}$, $\cdots, T_{n}$. If city $T_{i}$ can overrun city $T_{j}$, then $T_{i}$ can also overrun the cities between $T_{i}$ and $T_{j}$.
Prove the conclusion by the second mathematical induction on $n$.
When $n=1$, it is trivial.
Assume the conclusion holds for all po... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 729,513 |
3. For a finite set $A$ consisting of positive integers, if $A$ is divided into two non-empty disjoint subsets $A_{1}$ and $A_{2}$, and the least common multiple (LCM) of the elements in $A_{1}$ equals the greatest common divisor (GCD) of the elements in $A_{2}$, then such a partition is called "good". Find the minimum... | 3. 3024.
Let $A=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}\left(a_{1}<a_{2}<\cdots<a_{n}\right)$.
For any non-empty finite set of positive integers $B$, let $\operatorname{lcm} B$ and $\operatorname{gcd} B$ denote the least common multiple and greatest common divisor of the elements in $B$, respectively.
Consider any... | 3024 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 729,515 |
4. Let $n$ be a positive integer, and two players $A$ and $B$ play a game where they take turns choosing a positive integer $k (k \leqslant n)$. The rules of the game are as follows:
（1）Neither player can choose a number that has already been chosen by either player in any previous round.
（2）Neither player can choose a... | 4. When $n=1,2,4,6$, the game is a draw, and in other cases, player $B$ wins.
First, prove that when $n \neq 1,2,4,6$, player $B$ wins.
To achieve this, provide a strategy to ensure that after player $A$ chooses a number, player $B$ can always choose a number and ensure the game does not end in a draw.
Next, prove an i... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,516 |
Example 3 A secondary school has an odd number of classes, and each class has an odd number of students. A student is selected from each class to form the school committee. Try to prove that the following two statements are equivalent:
(1) The number of ways the school committee can contain an odd number of boys is gre... | 【Analysis】Suppose there are $2 k+1$ classes, denoted as $X_{1}, X_{2}$, $\cdots, X_{2 k+1}$. A composition method of the school committee is
$A=\left\{x_{1}, x_{2}, \cdots, x_{2 k+1}\right\}\left(x_{i} \in X_{i}\right)$.
When $x_{i}$ is a boy, let $f\left(x_{i}\right)=-1$;
When $x_{i}$ is a girl, let $f\left(x_{i}\righ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 729,519 |
1. Let $a_{1}, a_{2}, \cdots, a_{n}$ be positive real numbers, and
$$
a_{1}+a_{2}+\cdots+a_{n}=n \text {. }
$$
Prove: $\sum_{m=1}^{n} \frac{a_{m}}{\prod_{1}^{m}\left(1+a_{k}\right)} \leqslant 1-\frac{1}{2^{n}}$. | 1. For any positive integer $m(2 \leqslant m \leqslant n)$, we have
$$
\begin{array}{l}
\frac{a_{m}}{\prod_{k=1}^{m}\left(1+a_{k}\right)}=\frac{1+a_{m}}{\prod_{k=1}^{m}\left(1+a_{k}\right)}-\frac{1}{\prod_{k=1}^{m}\left(1+a_{k}\right)} \\
=\frac{1}{\prod_{k=1}^{m-1}\left(1+a_{k}\right)}-\frac{1}{\prod_{k=1}^{m}\left(1+... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 729,520 |
5. Let $\theta_{i} \in\left(0, \frac{\pi}{2}\right)(i=1,2, \cdots, n)$. Prove:
$$
\begin{array}{l}
\left(\sum_{i=1}^{n} \tan \theta_{i}\right)\left(\sum_{i=1}^{n} \cot \theta_{i}\right) \\
\geqslant\left(\sum_{i=1}^{n} \sin \theta_{i}\right)^{2}+\left(\sum_{i=1}^{n} \cos \theta_{i}\right)^{2}
\end{array}
$$ | 5. By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
\left(\sum_{i=1}^{n} \tan \theta_{i}\right)\left(\sum_{i=1}^{n} \cot \theta_{i}\right) \\
\geqslant\left(\sum_{i=1}^{n} \sqrt{\tan \theta_{i}} \sqrt{\cot \theta_{i}}\right)^{2}=n^{2}, \\
\left(\sum_{i=1}^{n} \sin \theta_{i}\right)^{2}+\left(\sum_{i=1}^{n}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 729,523 |
2. As shown in Figure 3, $\triangle A B C$ is inscribed in $\odot O$, and the angle bisector of $\angle A B C$ intersects $\odot O$ at point $D$. Two tangents $P B$ and $P C$ to $\odot O$ from points $B$ and $C$ respectively meet at point $P$. Connect $P D$, intersecting $A C$ at point $E$ and $\odot O$ at point $F$. L... | 2. As shown in Figure 6, let $B D$ intersect $A C$ at point $X$, extend $B D$ to intersect the extension of $P C$ at point $Y$, draw $P Z$ parallel to $B D$, intersecting the extension of $A C$ at point $Z$, and connect $M E$ and $D C$.
Figure 6
By Menelaus' Theorem, we have
$\frac{X D}{D Y} \cdot \frac{Y P}{P C} \cdot... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,525 |
3. Let $m$ be an integer greater than 1, and the sequence $\left\{a_{n}\right\}$ is defined as follows:
$$
\begin{array}{l}
a_{0}=m, a_{1}=\varphi(m), \\
a_{2}=\varphi^{(2)}(m)=\varphi(\varphi(m)), \cdots, \\
a_{n}=\varphi^{(n)}(m)=\varphi\left(\varphi^{(n-1)}(m)\right),
\end{array}
$$
where $\varphi(m)$ is the Euler'... | 3. It is known that if $a \mid b$, then $\varphi(a) \mid \varphi(b)$.
Moreover, when $m>2$, $\varphi(m)$ is even.
Therefore, it is only necessary to find integers $m$ such that $a_{1} \mid a_{0}$, which implies $a_{k+1} \mid a_{k}$.
If $m$ is an odd number greater than 2, it is impossible for $a_{1} \mid m$.
Let $m=2^... | 1944 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 729,526 |
5. Given $a_{1}=2$,
$$
a_{n+1}=\frac{2^{n+1} a_{n}}{\left(n+\frac{1}{2}\right) a_{n}+2^{n}}\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$
(1) Find the general term formula for the sequence $\left\{a_{n}\right\}$;
(2) Let $b_{n}=\frac{n^{3}+2 n^{2}+2 n+2}{n(n+1)\left(n^{2}+1\right) a_{n}}$, find the sum of the first $n... | 5. (1) From the given information,
$$
\frac{\frac{2^{n+1}}{a_{n+1}}}{\left(n+\frac{1}{2}\right)+\frac{2^{n}}{a_{n}}}=1
$$
Let $c_{n}=\frac{2^{n}}{a_{n}}$. Then $c_{n+1}=c_{n}+n+\frac{1}{2}$.
Thus, $c_{2}-c_{1}=1+\frac{1}{2}, c_{3}-c_{2}=2+\frac{1}{2}, \cdots$, $c_{n}-c_{n-1}=n-1+\frac{1}{2}$.
Adding all these equatio... | 2-\frac{(n+2)^{2}}{2^{n+1}(n+1)} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,528 |
8. Given the set
$$
I=\left\{\left(x_{1}, x_{2}, x_{3}, x_{4}\right) \mid x_{i} \in\{1,2, \cdots, 11\}\right\} \text {, }
$$
$A$ is a subset of $I$ and satisfies: for any
$$
\left(x_{1}, x_{2}, x_{3}, x_{4}\right) 、\left(y_{1}, y_{2}, y_{3}, y_{4}\right) \in A \text {, }
$$
there exist $i 、 j(1 \leqslant i<j \leqslant... | 8. First, consider the set satisfying $x_{1}+x_{2}+x_{3}+x_{4}=24$
$$
\begin{aligned}
A= & \left\{\left(x_{1}, x_{2}, x_{3}, x_{4}\right) \mid x_{i} \in\{1,2, \cdots, 11\},\right. \\
& i=1,2,3,4\},
\end{aligned}
$$
The number of its elements is
$$
\begin{array}{l}
\mathrm{C}_{23}^{3}-4\left(\mathrm{C}_{2}^{2}+\mathrm{... | 891 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,529 |
1. Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=1, a_{2}=\frac{1}{2}$, and for any integer $n \geqslant 2$, we have
$$
n(n+1) a_{n+1} a_{n}+n a_{n} a_{n-1}=(n+1)^{2} a_{n+1} a_{n-1} \text {. }
$$
(1) Find the general term of the sequence $\left\{a_{n}\right\}$;
(2) For any integer $n>2$, prove:
$\frac{2}{... | 1. (1) The given equation is equivalent to
$$
\begin{array}{l}
n a_{n} a_{n-1}=(n+1) a_{n+1}\left[(n+1) a_{n-1}-n a_{n}\right] \\
\Leftrightarrow \frac{1}{(n+1) a_{n+1}}-\frac{1}{a_{n}}=\frac{1}{n a_{n}}-\frac{1}{a_{n-1}} .
\end{array}
$$
According to equation (1), and noting that $a_{1}=1, a_{2}=\frac{1}{2}$, we have... | a_{n}=\frac{1}{n!} | Algebra | proof | Yes | Yes | cn_contest | false | 729,530 |
2. As shown in Figure $1, P A B$ and $P C D$ are two secants of $\odot O$, $A D$ intersects $B C$ at point $Q, T$ is a point on segment $B Q$, segment $P T$ intersects $\odot O$ at point $K$, line $Q K$ intersects segment $P A$ at point $S$. Prove: If $S T \parallel P Q$, then points $B, S, K, T$ are concyclic. | 2. Connect $A K$.
$$
\begin{array}{l}
\text { Given } S T / / P Q \\
\Rightarrow \frac{S K}{K Q}=\frac{T K}{K P} .
\end{array}
$$
Notice, $\frac{S K}{K Q}=\frac{S_{\triangle A S K}}{S_{\triangle A K Q}}=\frac{A S \sin \angle S A K}{A Q \sin \angle K A Q}$.
Since $\angle S A K=\angle T C K, \angle K A Q=\angle P C K$, ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,531 |
4. Four points on a plane, if the areas of the triangles formed by any three of these points are four distinct positive integers, and the distances between any two points are six distinct positive integers, then the convex hull of these four points is called a "Lotus Diagram".
(1) Provide a specific example of a Lotus ... | 4. (1) The two complete quadrilaterals $ABCD$ shown in Figures 3 and 4 are formed by piecing together four right-angled triangles with integer side lengths.
Since the side lengths and areas of each right-angled triangle are integers, the six side lengths of the quadrilaterals formed are distinct positive integers, and... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,532 |
7. As shown in Figure $2, \triangle A B C$ has an incircle $\odot I$ that touches sides $B C$, $C A$, and $A B$ at points $D$, $E$, and $F$, respectively. Lines $B I$, $C I$, and $D I$ intersect $E F$ at points $M$, $N$, and $K$, respectively. Line $B N$ intersects $C M$ at point $P$, and line $A K$ intersects $B C$ at... | 6. From the conditions, we have
$$
\begin{array}{l}
a(n)=\mathrm{C}_{n}^{0}+\mathrm{C}_{n}^{3}+\cdots+\mathrm{C}_{n}^{3 k}+\cdots, \\
b(n)=\mathrm{C}_{n}^{0}+\mathrm{C}_{n}^{6}+\cdots+\mathrm{C}_{n}^{6 k}+\cdots .
\end{array}
$$
Thus, for any positive integer $n$, we have
$$
\sum_{k=1}^{3} z_{k}^{n}=\left\{\begin{arra... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,533 |
8. Let the sequence of positive integers $\left\{a_{n}\right\}$ satisfy: for any integer $n \geqslant 2016$, we have
$$
n^{2} \mid\left(a_{1}+a_{2}+\cdots+a_{n}\right) \text { and } a_{n} \leqslant(n+2016)^{2} \text {. }
$$
Let $b_{n}=a_{n+1}-a_{n}$. Prove: the sequence $\left\{b_{n}\right\}$ is constant from some ter... | 7. Draw a line through point $K$ parallel to $BC$, intersecting $AB$ and $AC$ at points $R$ and $S$ respectively. Then $IK \perp RS$.
Also, $IF \perp AB$, so points $I, K, F, R$ are concyclic.
Similarly, points $I, K, S, E$ are concyclic.
Thus, $\angle IRK = \angle IFK = \angle IEK = \angle ISK$
$\Rightarrow KR = KS$.
... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 729,534 |
2. Let $n$ be a positive integer. Given $n$ positive numbers $x_{1}, x_{2}$, $\cdots, x_{n}$ whose product is 1. Prove:
$$
\sum_{i=1}^{n}\left(x_{i} \sqrt{x_{1}^{2}+x_{2}^{2}+\cdots+x_{i}^{2}}\right) \geqslant \frac{(n+1) \sqrt{n}}{2} .
$$ | 2. From the conditions, we have
$$
\begin{array}{l}
\sum_{i=1}^{n}\left(x_{i} \sqrt{x_{1}^{2}+x_{2}^{2}+\cdots+x_{i}^{2}}\right) \\
\geqslant \sum_{i=1}^{n}\left(x_{i} \cdot \frac{x_{1}+x_{2}+\cdots+x_{i}}{\sqrt{i}}\right)=\sum_{i=1}^{n} \sum_{j=1}^{i} \frac{x_{i} x_{j}}{\sqrt{i}} \\
\geqslant \frac{1}{\sqrt{n}} \sum_{... | \frac{(n+1) \sqrt{n}}{2} | Inequalities | proof | Yes | Yes | cn_contest | false | 729,536 |
4. A substitute teacher leads a group of students on an outing. The homeroom teacher informs the substitute that there are two mischievous students who sometimes lie (but the substitute does not know which two students they are), and the other students always tell the truth. They get lost in the forest and eventually g... | 4. Let $A, B, C, D$ represent four paths. In the first 20 minutes, the teacher explores path $A$. If the camp is found, the direction of the camp is successfully determined. Assuming the teacher does not find the camp, the camp must be on one of the paths $B, C, D$.
(1) In the first 20 minutes, the teacher sends three ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 729,537 |
5. Let the constant $\alpha$ satisfy $0<\alpha<1$. Prove that:
(1) There exists a constant $C(\alpha)>0$, such that for all $x \geqslant 0$, we have
$$
\ln (1+x) \leqslant C(\alpha) x^{\alpha} ;
$$
(2) For any two non-zero complex numbers $z_{1} 、 z_{2}$, we have
$$
|\ln | \frac{z_{1}}{z_{2}}|| \leqslant C(\alpha)\left... | 5. (1) Let $f(x)=C(\alpha) x^{\alpha}-\ln (1+x)$.
Then for $x>0$,
$$
\begin{array}{c}
f^{\prime}(x)=\alpha C(\alpha) x^{\alpha-1}-\frac{1}{1+x} \\
=\frac{\alpha C(\alpha) x^{\alpha}(1+x)-x}{x(1+x)} .
\end{array}
$$
Notice that, $0<\alpha \leqslant 1<1+\alpha$.
Then $x^{\alpha}(1+x)=x^{\alpha}+x^{1+\alpha}$
$$
\geqsla... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 729,538 |
1. The NBA regular season involves 30 teams. Question: Can the 30 teams be divided into Eastern and Western Conferences, and the number of games between any two teams be appropriately designated, so that each team plays exactly 82 games and the number of games between Eastern and Western Conference teams is exactly hal... | 1. Not possible.
Let $x$, $y$, $z$ represent the number of games played between eastern teams, between western teams, and between eastern and western teams, respectively, and let $k$ be the number of teams in the eastern conference. Then,
$$
z=x+y, 82 k=2 x+z, x+y+z=\frac{30 \times 82}{2} \text {. }
$$
Thus, $82 k-2 ... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,539 |
2. In a cyclic quadrilateral $ABCD$, the diagonals $AC$ and $BD$ intersect at point $P$. Point $Q$ on segment $BC$ satisfies $PQ \perp AC$. Prove: The line connecting the circumcenters of $\triangle APD$ and $\triangle BQD$ is parallel to $AD$. | 2. Take a point $T$ on the line $Q P$ such that $D T \perp D A$. Since $\angle A P T=90^{\circ}=\angle A D T$, it follows that points $A$, $P$, $D$, and $T$ are concyclic.
Thus, the circumcenter of $\triangle A P D$ lies on the perpendicular bisector $l$ of $D T$.
Given that $A$, $B$, $C$, and $D$ are concyclic,
$$
\a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,540 |
3. Given that $f(x)$ is a cubic polynomial. If for a set of three distinct real numbers $(a, b, c)$ we have
$$
f(a)=b, f(b)=c, f(c)=a,
$$
then the set of real numbers is called “cyclic”.
Assume there exist eight cyclic sets $\left(a_{i}, b_{i}, c_{i}\right)(1 \leqslant i \leqslant 8)$, containing 24 distinct real numb... | 3. Suppose the eight sums $a_{i}+b_{i}+c_{i}$ take at most two different values. Then, four of these sums must take the same value. Without loss of generality, let
$$
\begin{array}{l}
s=a_{i}+b_{i}+c_{i}(1 \leqslant i \leqslant 4) . \\
\text { Let } g(x)=x+f(x)+f(f(x)) .
\end{array}
$$
Then $g(x)$ is a ninth-degree po... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 729,541 |
Example 6 Consider the following action as one operation: cutting a rope of positive integer length into three ropes of positive integer length, with the longest rope being exactly one. Given a rope of length $a^{b}$ (where $a$ and $b$ are both integers greater than 1), two players, A and B, take turns performing the f... | 【Analysis】Obviously, if $1 \leqslant n \leqslant 3$, player A must fail; if $4 \leqslant n \leqslant 7=3+2+2$, then player A has a winning strategy; if $8 \leqslant n \leqslant 9$, then the length of the longest rope after player A's operation must satisfy $4 \leqslant m \leqslant 7$, so player B has a winning strategy... | a=2, b=3 \text{ or } a \text{ is a power of 3} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 729,542 |
1. Given real numbers $a, b$ satisfy
$$
\left(a-\sqrt{a^{2}-1}\right)\left(b-\sqrt{b^{2}-1}\right)=1 \text {. }
$$
Statements: (1) $a=b$; (2) $a+b=0$; (3) $a b=1$;
(4) $a b=-1$. Among these, the correct number of statements is ( ).
(A) 1
(B) 2
(C) 3
(D) 0 | - 1. B.
Given that multiplying both sides of the equation by $a+\sqrt{a^{2}-1}$ yields
$$
b-\sqrt{b^{2}-1}=a+\sqrt{a^{2}-1} \text {. }
$$
Similarly, $b+\sqrt{b^{2}-1}=a-\sqrt{a^{2}-1}$.
Adding (1) and (2) gives $a=b$.
Thus $\left(a-\sqrt{a^{2}-1}\right)^{2}=1 \Rightarrow a= \pm 1$
$$
\Rightarrow a b=a^{2}=1
$$ | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 729,543 |
2. Let $A D$ be the angle bisector of $\angle B A C$, $A B=8, A C$ $=10, A D=6, E$ be a point on $A C$, $A E=2, M$ be the midpoint of $A E$, $N$ be the midpoint of $B C$. Then $M N=$ ( ).
(A) 5
(B) $\frac{26}{5}$
(C) $\frac{27}{5}$
(D) $\frac{11}{2}$ | 2. C.
As shown in Figure 4, connect $B E$, take the midpoint $P$ of $B E$, and connect $P M \backslash P N$.
$$
\begin{array}{l}
\text { Then } P M \Perp \frac{1}{2} A B, \\
P N \Perp \frac{1}{2} C E .
\end{array}
$$
Therefore, $A B=C E=8 \Rightarrow P M=P N$
$$
\begin{array}{l}
\Rightarrow \angle P M N=\angle P N M=... | \frac{27}{5} | Geometry | MCQ | Yes | Yes | cn_contest | false | 729,544 |
5. As shown in Figure $1, M$ is the midpoint of side $CD$ of the inscribed quadrilateral $ABCD$ in the semicircle $\odot O$ with diameter $AB$, $MN \perp AB$ at point $N$, $AB=10, AD=AN=$ 3. Then $BC=(\quad)$.
(A) 5
(B) 6
(C) 7
(D) 8 | 5. C.
As shown in Figure 5, draw perpendiculars from points $C$ and $D$ to $AB$, with the feet of the perpendiculars being $E$ and $F$, respectively. Connect $AC$ and $BD$.
Let $AF = x$. Then
$$
\begin{array}{c}
E N = F N = 3 - x, \\
B E = B N - E N = (10 - 3) - (3 - x) = 4 + x .
\end{array}
$$
In right triangles $\t... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 729,545 |
1. Given that $a$ and $b$ are integers, $\frac{127}{a}-\frac{16}{b}=1$. Then the maximum value of $b$ is $\qquad$ . | $$
\text { Two, 1.2 } 016 .
$$
The original equation is transformed into $b=\frac{16 a}{127-a}=\frac{127 \times 16}{127-a}-16$. When and only when $127-a=1$, $b$ reaches its maximum value, at this point, $b=127 \times 16-16=2016$. | 2016 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 729,546 |
2. Given a set of data $x_{1}, x_{2}, \cdots, x_{6}$ with the variance
$$
S^{2}=\frac{1}{6}\left(x_{1}^{2}+x_{2}^{2}+\cdots+x_{6}^{2}-24\right) \text {. }
$$
then the mean of the data $x_{1}+1, x_{2}+1, \cdots, x_{6}+1$ is | 2. -1 or 3. From $S^{2}=\frac{1}{6}\left(x_{1}^{2}+x_{2}^{2}+\cdots+x_{6}^{2}-6 \bar{x}^{2}\right)$, we know $6 \bar{x}^{2}=24, \bar{x}= \pm 2$.
Therefore, the required average of the data is $\bar{x}+1=-1$ or 3. | -1 \text{ or } 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,547 |
3. In rectangle $A B C D$, $A B=10$ cm, $B C=$ 20 cm, a moving point $M$ starts from point $B$ along side $A B$ towards endpoint $A$, with a speed of 1 cm per second, and a moving point $N$ starts from point $C$ along side $B C$ towards point $B$, with a speed of 1 cm per second. Then, by the 10th second, the distance ... | 3. $5 \sqrt{2}$ cm.
As shown in Figure 7, take the midpoints $E$ and $F$ of $AD$ and $BC$ respectively, and connect $EF$, $EM$, $EN$, $EP$, and $PF$.
It is easy to prove that quadrilateral $ABFE$ is a square.
Then $\triangle MAE \cong \triangle NFE$
$\Rightarrow \triangle EMN$ is an isosceles right triangle
$\Rightarr... | 5 \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,548 |
One, (20 points) Given that when $-1 \leqslant x \leqslant 0$, the function
$$
y=\left|x^{2}+(m-1) x+\left(m^{2}-3 m+1\right)\right|
$$
increases as $x$ increases. Find the range of real values for $m$.
| $$
\text { I. Let } f(x)=x^{2}+(m-1) x+\left(m^{2}-3 m+1\right) \text {. }
$$
Then $\Delta=(m-1)^{2}-4\left(m^{2}-3 m+1\right)$
$$
=-(3 m-1)(m-3) \text {, }
$$
The axis of symmetry of $f(x)$ is $x=\frac{1-m}{2}$.
When $\Delta \leqslant 0$, that is, $m \leqslant \frac{1}{3}$ or $m \geqslant 3$,
$$
\frac{1-m}{2} \leqsla... | m=1 \text{ or } m \geqslant 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,549 |
1. Given non-zero vectors $\boldsymbol{a} 、 \boldsymbol{b}$ with an angle of $120^{\circ}$ between them. If vector $\boldsymbol{a}-\boldsymbol{b}$ is perpendicular to $\boldsymbol{a}+2 \boldsymbol{b}$, then $\left|\frac{2 a-b}{2 a+b}\right|=$ | $-1 . \frac{\sqrt{10+\sqrt{33}}}{3}$.
Let $|\boldsymbol{a}|=x,|\boldsymbol{b}|=y$.
According to the problem, $(4 x-y)^{2}=33 y^{2}$.
Also, $\left|\frac{2 a-b}{2 a+b}\right|=\sqrt{\frac{(2 a-b)^{2}}{(2 a+b)^{2}}}$
$=\sqrt{\frac{(4 x+y)^{2}+3 y^{2}}{(4 x-y)^{2}+3 y^{2}}}=\frac{\sqrt{10+\sqrt{33}}}{3}$. | \frac{\sqrt{10+\sqrt{33}}}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,551 |
2. The system of equations
$$
\left\{\begin{array}{l}
\frac{9}{2(x+y)}=\frac{1}{x}+\frac{1}{y}, \\
\sqrt{x^{2}-2}=\sqrt{3-y^{2}}
\end{array}\right.
$$
has the real solutions | 2. $(2,1),(-2,-1)$.
$$
\begin{array}{l}
\text { Given } \frac{9}{2(x+y)}=\frac{1}{x}+\frac{1}{y} \\
\Rightarrow 2(x+y)^{2}=9 x y \\
\Rightarrow x=2 y \text { or } y=2 x . \\
\text { Also } \sqrt{x^{2}-2}=\sqrt{3-y^{2}} \Rightarrow x^{2}+y^{2}=5 .
\end{array}
$$
$$
\text { It is easy to see that } y= \pm 1 \text { or } ... | (2,1),(-2,-1) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,552 |
5. Given $P$ as a moving point on the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, draw the perpendiculars from the foci $F_{1}$ and $F_{2}$ to the external angle bisector of $\angle F_{1} P F_{2}$. Then the locus of the feet of these perpendiculars is $\qquad$ . | 5. $x^{2}+y^{2}=a^{2}$.
Assume without loss of generality that $F_{1} M$ is perpendicular to the external angle bisector $P M$ of $\angle F_{1} P F_{2}$ at point $M$, and intersects the extension of $P F_{2}$ at point $N$. Then $P N=P F_{1}$. Therefore, $F_{2} N=2 a$, and $O M$ is the midline of $\triangle N F_{1} F_{... | x^{2}+y^{2}=a^{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,553 |
6. Given that $P$ is a point on a sphere $O$ with radius $r$, and through $P$ three mutually perpendicular moving chords $P A$, $P B$, and $P C$ are drawn. If the maximum distance from point $P$ to the plane $A B C$ is 1, then $r=$ $\qquad$ | 6. $\frac{3}{2}$.
Let $P A=x, P B=y, P C=z$. Then
$$
\begin{array}{l}
A B^{2}=x^{2}+y^{2}, B C^{2}=y^{2}+z^{2}, C A^{2}=z^{2}+x^{2}, \\
S_{\triangle A B C}=\frac{1}{2} A B \cdot B C \sin \angle A B C \\
=\frac{1}{2} A B \cdot B C \sqrt{1-\cos ^{2} \angle A B C} \\
=\frac{1}{2} \sqrt{x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}... | \frac{3}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,554 |
7. In a lottery with 100000000 tickets, each ticket number consists of eight digits. A ticket number is called "lucky" if and only if the sum of its first four digits equals the sum of its last four digits. Then the sum of all lucky ticket numbers, when divided by 101, leaves a remainder of $\qquad$ | 7.0.
If the eight-digit number $x=\overline{a b c d e f g h}$ is lucky, then $y=99999999-x$ is also lucky, and $x \neq y$.
$$
\begin{array}{l}
\text { and } x+y=99999999=9999 \times 10001 \\
=99 \times 101 \times 10001,
\end{array}
$$
Therefore, the sum of all lucky numbers must be a multiple of 101. | 0 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 729,555 |
Example 1 If a natural number $K$ can be expressed as
$$
\begin{aligned}
K & =V(a, b, c) \\
& =a^{3}+b^{3}+c^{3}-3 a b c \quad (a, b, c \in \mathbf{N})
\end{aligned}
$$
then $K$ is called a "Water Cube Number".
Now, arrange all different Water Cube Numbers in ascending order to form a sequence $\left\{K_{n}\right\}$, ... | 【Analysis】Let $M=\{3 x \mid x \in \mathbf{N},(3, x)=1\}$.
Lemma A natural number $k$ is a water cube number if and only if $k \in \mathbf{N} \backslash M$.
Proof Note that,
$V(a, b, c)=a^{3}+b^{3}+c^{3}-3 a b c$
$=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)$
$=(a+b+c)^{3}-3(a+b+c)(a b+b c+c a)$.
Then when $3 \mid... | 2614032 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 729,556 |
493 Given the sequence $\left\{S_{n}\right\}$ satisfies
$$
S_{n}=\sum_{i=1}^{n} i!\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$
Prove: For any given prime $p$, there exist infinitely many positive integers $a$ such that $v_{p}\left(S_{a+1}\right) \geqslant v_{p}\left(S_{a}\right)$, where $v_{p}(d)$ denotes the power... | Prove the boundedness of $\left\{v_{p}\left(S_{n}\right)\right\}$.
(1) If $\left\{v_{p}\left(S_{n}\right)\right\}$ is bounded, then there exists $c \in \mathbf{Z}_{+}$,
thus $v_{p}\left(S_{t p}-S_{t p-1}\right)=v_{p}((t p)!)$
$$
\geqslant c \geqslant v_{p}\left(S_{t p-1}\right) \text {, }
$$
where $t$ is any integer n... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 729,558 |
Example 3 The sequence $\left\{a_{n}\right\}$:
$1,1,2,1,1,2,3,1,1,2,1,1,2,3,4, \cdots$, is called a "fractal sequence", and its construction method is as follows:
First, give $a_{1}=1$, then copy this item 1 and add its successor number 2, to get $a_{2}=1, a_{3}=2$;
Then copy all the previous items $1, 1, 2$, and add... | Solving the construction method of the sequence $\left\{a_{n}\right\}$, we easily know
$$
a_{1}=1, a_{3}=2, a_{7}=3, a_{15}=4, \cdots \cdots
$$
Generally, $a_{2^{n-1}}=n$, which means the number $n$ first appears at the $2^{n}-1$ term, and if
$$
m=2^{n}-1+k\left(1 \leqslant k \leqslant 2^{n}-1\right),
$$
then $a_{m}=... | 3950 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,560 |
Example 1 Let $P(x)$ be a polynomial of degree $2n$, such that
$$
\begin{array}{l}
P(0)=P(2)=\cdots=P(2 n)=0, \\
P(1)=P(3)=\cdots=P(2 n-1)=2, \\
P(2 n+1)=-6 .
\end{array}
$$
Determine $n$ and $P(x)$. | From Theorem 3 (2), we get
$$
\begin{array}{l}
\Delta^{2 n+1} P(0)=\sum_{k=0}^{2 n+1}(-1)^{k} \mathrm{C}_{2 n+1}^{k} P(2 n+1-k) \\
=0+\sum_{j=1}^{n} \mathrm{C}_{2 n+1}^{2 j}+(-6) \\
=2 \times \frac{1}{2}\left(\sum_{j=1}^{n} \mathrm{C}_{2 n+1}^{2 j}+\sum_{j=1}^{n} \mathrm{C}_{2 n+1}^{2 j-1}\right) \\
=(-6)+2 \times \fra... | P(x)=-2 x^{2}+4 x | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,561 |
Example 2 Let $a \geqslant 3, g(x)$ be a polynomial with real coefficients, $\operatorname{deg} g(x)=n$. Prove:
$$
\max _{0 \leqslant i \leqslant n+1}\left|a^{i}-g(i)\right| \geqslant 1 .
$$ | Let $f(x)=a^{x}-g(x)$.
By Theorem 3 (2), we have
$$
\begin{array}{l}
\Delta^{n+1} a^{x}=(E-I)^{n+1} a^{x} \\
=\sum_{i=0}^{n+1}(-1)^{n+1-i} \mathrm{C}_{n+1}^{i} E^{i} a^{x} \\
=\sum_{i=0}^{n+1}(-1)^{n+1-i} \mathrm{C}_{n+1}^{i} a^{x+i} \\
=a^{x} \sum_{i=0}^{n+1}(-1)^{n+1-i} \mathrm{C}_{n+1}^{i} a^{i} \\
=a^{x}(a-1)^{n+1}... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 729,562 |
In $\triangle A B C$, $X 、 Y$ are two points on the line $B C$ $(X 、 B 、 C 、 Y$ are arranged in sequence $)$, such that
$$
B X \cdot A C=C Y \cdot A B \text {. }
$$ | Let the circumcenters of $\triangle A C X$ and $\triangle A B Y$ be $O_{1}$ and $O_{2}$, respectively, and let the line $O_{1} O_{2}$ intersect $A B$ and $A C$ at points $U$ and $V$. Prove that $\triangle A U V$ is an isosceles triangle.
The standard solution method one is concise and clear, the key is to prove that t... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,563 |
Question 1 In the acute triangle $\triangle ABC$, $AB > BC$, $P$ and $Q$ are the midpoints of the minor arc $\overparen{AC}$ and the major arc $\overparen{ABC}$ of its circumcircle $\odot O$, respectively. A perpendicular line is drawn from point $Q$ to segment $AB$, with the foot of the perpendicular being $M$. Prove:... | Prove that, as shown in Figure 1, extend $C M$ to intersect $\odot O$ at point $R$, connect $R P$, intersecting $A B$ at point $T$, and connect $Q B$, $Q T$. Let the midpoint of line segment $B P$ be $S$.
It suffices to prove that $B$, $C$, $S$, and $M$ are concyclic.
From the given, we have
$$
\angle Q R T=\angle Q M ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,564 |
Question 2 If all vertices of a polygon are colored the same color, then the polygon is called "monochromatic". Given that every point in a plane is colored either red or blue. Prove: either there exists a monochromatic equilateral triangle with side length 2, or there exists a monochromatic equilateral triangle with s... | Proof Assume there does not exist such a polygon.
Take any point on the plane, let's say it is a red point $A$, and draw circles with this point as the center and radii of 1 and 2, respectively, and divide each circle into six equal parts, as shown in Figure 2, and mark them as $1 \sim 6, 7 \sim 12$.
If $1 \sim 6$ are... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 729,565 |
Question 3 Let $a, b, b^{\prime}, c, m, q$ be positive integers, where $m>1, q>1, \left|b-b^{\prime}\right| \geqslant a$. It is known that there exists a positive integer $M$ such that
$$
S_{q}(a n+b) \equiv S_{q}\left(a n+b^{\prime}\right)+c(\bmod m)
$$
holds for all integers $n \geqslant M$. Prove: Equation (1) hold... | Prove that taking $n=q^{k}+t(t \in\{1,2, \cdots, M-1\})$, $k$ sufficiently large, such that $(M-1) a+b$ and $(M-1) a+b^{\prime}$ have fewer than $k-1$ digits in base $q$, i.e.,
$$
q^{k}>\max \left\{a(M-1)+b, a(M-1)+b^{\prime}\right\} .
$$
It is easy to see that in this case $n \geqslant M$.
Then $S_{q}\left(a\left(q^{... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 729,566 |
3. Given $\triangle A B C$ satisfies $\angle C=90^{\circ}$, point $C$ is the projection of $A B$ onto $H$, and $D$ is a point inside $\triangle C B H$ such that $C H$ bisects segment $A D$. Line $B D$ intersects $C H$ at point $P$, $\Gamma$ is a semicircle with $B D$ as its diameter, and it intersects $B C$ at a point ... | 3. As shown in Figure 2, let the projection of point $D$ on $AB$ be $K$.
Then $A H = H K$.
Since $P H \parallel D K$, we have
$$
\frac{P D}{P B} = \frac{H K}{H B} = \frac{A H}{H B}.
$$
Let the projection of point $Q$ on $DB$ be $L$.
Since $P Q$ is tangent to the semicircle $\Gamma$ at point $Q$, and $\angle D Q B =$
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,567 |
Example 3 Let $n$ be a positive integer,
$$
\begin{aligned}
S= & \{(x, y, z) \mid x, y, z \in\{0,1, \cdots, n\}, \\
& x+y+z>0\}
\end{aligned}
$$
is a set of $(n+1)^{3}-1$ points in three-dimensional space. Try to find the minimum number of planes whose union contains $S$ but does not contain $(0,0,0)$. ${ }^{[4]}$ | Suppose there exist $m$ planes
$$
a_{i} x+b_{i} y+c_{i} z-d_{i}=0\left(1 \leqslant i \leqslant m, d_{i} \neq 0\right)
$$
satisfying the conditions.
Construct the $m$-degree polynomial
$$
f(x, y, z)=\prod_{i=1}^{m}\left(a_{i} x+b_{i} y+c_{i} z-d_{i}\right) \text {. }
$$
By the problem, when $(x, y, z) \in S$,
$$
f(x, ... | 3n | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,568 |
5. Given $\triangle A B C$ satisfies $C A \neq C B$, the midpoints of sides $A B$, $A C$, and $B C$ are $D$, $F$, and $G$ respectively. Circle $\Gamma$ passes through point $C$, and is tangent to $A B$ at point $D$, and intersects segments $A F$ and $B G$ at points $H$ and $I$ respectively. Point $H^{\prime}$ is symmet... | 5. Let's assume $C A>C B$.
Since $H^{\prime} 、 I^{\prime}$ lie on segments $C F 、 C G$ respectively, point $M$ is outside $\triangle A B C$, as shown in Figure 5.
By $D$ being the midpoint of $A B$ and the Power of a Point theorem, we have
$C H^{\prime} \cdot C A=A H \cdot A C=A D^{2}$
$=B D^{2}=B I \cdot B C=C I^{\pr... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,569 |
2. Let the complex numbers $z, w$ satisfy
$$
|z|=3,(z+\bar{w})(\bar{z}-w)=7+4 \mathrm{i} \text{, }
$$
where $\mathrm{i}$ is the imaginary unit, and $\bar{z}, \bar{w}$ are the conjugates of $z, w$ respectively. Then the modulus of $(z+2 \bar{w})(\bar{z}-2 w)$ is $\qquad$ | 2. $\sqrt{65}$.
From the properties of operations, we know
$$
\begin{array}{l}
7+4 \mathrm{i}=(z+\bar{w})(\bar{z}-w) \\
=|z|^{2}-|w|^{2}-(z w-\overline{z w}) .
\end{array}
$$
$$
\begin{array}{l}
\text { Since }|z|^{2}, |w|^{2} \in \mathbf{R}, \operatorname{Re}(z w-\overline{z w})=0 \\
\Rightarrow|z|^{2}-|w|^{2}=7, z w... | \sqrt{65} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,571 |
3. Let positive real numbers $u, v, w$ be not equal to 1. If
$$
\log _{u} v w+\log _{v} w=5, \log _{v} u+\log _{w} v=3,
$$
then the value of $\log _{w} u$ is | 3. $\frac{4}{5}$.
Let $\log _{u} v=a, \log _{v} w=b$. Then
$$
\begin{array}{l}
\log _{v} u=\frac{1}{a}, \log _{w} v=\frac{1}{b}, \\
\log _{u} v w=\log _{u} v+\left(\log _{u} v\right)\left(\log _{v} w\right)=a+a b .
\end{array}
$$
Therefore, $a+a b+b=5, \frac{1}{a}+\frac{1}{b}=3$
$$
\Rightarrow a b=\frac{5}{4} \text {... | \frac{4}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,572 |
6. Let the function be
$$
f(x)=\sin ^{4} \frac{k x}{10}+\cos ^{4} \frac{k x}{10}\left(k \in \mathbf{Z}_{+}\right) .
$$
If for any real number $a$, we have
$$
\{f(x) \mid a<x<a+1\}=\{f(x) \mid x \in \mathbf{R}\} \text {, }
$$
then the minimum value of $k$ is $\qquad$ | 6. 16 .
From the given conditions, we have
$$
\begin{array}{l}
f(x)=\left(\sin ^{2} \frac{k x}{10}+\cos ^{2} \frac{k x}{10}\right)^{2}-2 \sin ^{2} \frac{k x}{10} \cdot \cos ^{2} \frac{k x}{10} \\
=1-\frac{1}{2} \sin ^{2} \frac{k x}{5}=\frac{1}{4} \cos \frac{2 k x}{5}+\frac{3}{4},
\end{array}
$$
The function $f(x)$ re... | 16 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,573 |
7. Given the hyperbola $C: x^{2}-\frac{y^{2}}{3}=1$, with left and right foci $F_{1}$ and $F_{2}$ respectively. A line is drawn through point $F_{2}$ intersecting the right branch of the hyperbola $C$ at points $P$ and $Q$, such that $\angle F_{1} P Q=90^{\circ}$. Then the inradius of $\triangle F_{1} P Q$ is | 7. $\sqrt{7}-1$.
As shown in Figure 4.
By the properties of the hyperbola, we have
$$
\begin{array}{l}
F_{1} F_{2}=2 \sqrt{1+3}=4, \\
P F_{1}-P F_{2}=Q F_{1}-Q F_{2}=2 . \\
\text { Given } \angle F_{1} P Q=90^{\circ} \\
\Rightarrow P F_{1}^{2}+P F_{2}^{2}=F_{1} F_{2}^{2} \\
\Rightarrow P F_{1}+P F_{2} \\
\quad=\sqrt{2... | \sqrt{7}-1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,574 |
Example 4 Find all positive integers $n$, such that there exists a polynomial $f(x)$ with real coefficients, satisfying the following two properties:
$(1)$ For any integer $k$, the number $f(k)$ is an integer if and only if $k$ is not divisible by $n$;
(2) The degree of the polynomial $f(k)$ is less than $n$.
[5] | Proof of a lemma first.
Lemma If $n$ has two different prime factors, then $\operatorname{gcd}\left(\mathrm{C}_{n}^{1}, \mathrm{C}_{n}^{2}, \cdots, \mathrm{C}_{n}^{n-1}\right)=1$.
Proof by contradiction.
If there exists a prime $p$, such that
$p \mid \operatorname{gcd}\left(\mathrm{C}_{n}^{1}, \mathrm{C}_{n}^{2}, \cdot... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,575 |
8. Let $a_{1}, a_{2}, a_{3}, a_{4}$ be four distinct numbers from $1, 2, \cdots, 100$, satisfying
$$
\begin{array}{l}
\left(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}\right)\left(a_{2}^{2}+a_{3}^{2}+a_{4}^{2}\right) \\
=\left(a_{1} a_{2}+a_{2} a_{3}+a_{3} a_{4}\right)^{2} .
\end{array}
$$
Then the number of such ordered quadruples... | 8. 40 .
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
\left(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}\right)\left(a_{2}^{2}+a_{3}^{2}+a_{4}^{2}\right) \\
\geqslant\left(a_{1} a_{2}+a_{2} a_{3}+a_{3} a_{4}\right)^{2},
\end{array}
$$
Equality holds if and only if
$$
\frac{a_{1}}{a_{2}}=\frac{a_{2}}{a_{3}}=\frac{a_{... | 40 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,576 |
9. (16 points) In $\triangle A B C$,
$$
\overrightarrow{A B} \cdot \overrightarrow{A C}+2 \overrightarrow{B A} \cdot \overrightarrow{B C}=3 \overrightarrow{C A} \cdot \overrightarrow{C B} \text{. }
$$
Find the maximum value of $\sin C$. | 9. From the definition of the dot product and the cosine rule, we have
$$
\overrightarrow{A B} \cdot \overrightarrow{A C}=c b \cos A=\frac{b^{2}+c^{2}-a^{2}}{2} \text {. }
$$
Similarly, $\overrightarrow{B A} \cdot \overrightarrow{B C}=\frac{a^{2}+c^{2}-b^{2}}{2}$,
$$
\overrightarrow{C A} \cdot \overrightarrow{C B}=\fr... | \frac{\sqrt{7}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,577 |
One. (40 points) Let real numbers $a_{1}, a_{2}, \cdots, a_{2016}$ satisfy
$$
\begin{aligned}
9 a_{i} & >11 a_{i+1}^{2}(i=1,2, \cdots, 2015) \text {. Find } \\
& \left(a_{1}-a_{2}^{2}\right)\left(a_{2}-a_{3}^{2}\right) \cdots\left(a_{2015}-a_{2016}^{2}\right)\left(a_{2016}-a_{1}^{2}\right)
\end{aligned}
$$
the maximum... | $$
\text { I. Let } P=\prod_{i=1}^{2016}\left(a_{i}-a_{i+1}^{2}\right), a_{2017}=a_{1} \text {. }
$$
From the given, for $i=1,2, \cdots, 2015$, we have
$$
a_{i}-a_{i+1}^{2}>\frac{11}{9} a_{i+1}^{2}-a_{i+1}^{2}>0 \text {. }
$$
If $a_{2016}-a_{1}^{2} \leqslant 0$, then $P \leqslant 0$.
Now consider the case where $a_{2... | \frac{1}{4^{2016}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,578 |
Three. (50 points) Given ten points in space, where no four points lie on the same plane. Connect some of the points with line segments. If the resulting figure contains no triangles and no spatial quadrilaterals, determine the maximum number of line segments that can be drawn. | Three, using these ten points as vertices and the connected line segments as edges, we get a simple graph $G$ of order 10.
The following proves: The number of edges in graph $G$ does not exceed 15.
Let the vertices of graph $G$ be $v_{1}, v_{2}, \cdots, v_{10}$, with a total of $k$ edges, and use $\operatorname{deg}\le... | 15 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,579 |
1. Let integer $n \geqslant 3$. Place $n^{2}$ cards labeled $1,2, \cdots, n^{2}$ into $n$ boxes, with each box containing $n$ cards. The following operation is allowed: each time, select two boxes and take two cards from each, then place them into the other box. Prove: regardless of the initial placement, it is always ... | 1. First, prove: It is possible to swap any two cards between any two boxes with two operations, while keeping the positions of the other cards unchanged.
Assume one of the boxes contains three cards $A, X, Y$ (ignoring the other cards for now), and another box contains three cards $B, Z, W$ (also ignoring the other c... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 729,581 |
2. Let $\triangle ABC$ have side lengths $BC=a$, $AC=b$, $AB=c$, and let $\Gamma$ be the circumcircle of $\triangle ABC$.
(1) If on the arc $\overparen{BC}$ of circle $\Gamma$ (excluding point $A$) there is a unique point $P$ (where $P$ does not coincide with $B$ or $C$) such that $PA=PB+PC$, find the necessary and suf... | 2. (1) If the conditions stated in the problem hold, let $P$ be the unique point mentioned. By Ptolemy's theorem, we have
$$
a P A=b P B+c P C .
$$
Combining $P A=P B+P C$, we get
$$
(b-a) P B+(c-a) P C=0 \text {. }
$$
If $b=a$, it is clear that $c=a$. In this case, $\triangle A B C$ is an equilateral triangle.
By P... | b \neq a, c \neq a, \frac{c(c-a)}{b(a-b)}>0, \angle BAC < 60^{\circ} | Geometry | proof | Yes | Yes | cn_contest | false | 729,582 |
3. Let integers $m$ and $n$ be coprime, and both greater than 1. Prove: There exist positive integers $a$, $b$, $c$, satisfying
$$
m^{a}=1+n^{b} c \text {, and }(c, n)=1 \text {. }
$$ | 3. For a non-zero integer $t$ and a prime $p$, define $v_{p}(t)$ as the power of $p$ in $t$, i.e., the largest non-negative integer $\alpha$ such that $p^{\alpha} \mid t$.
First, prove a lemma.
Lemma If $d$ is an integer greater than 1, $s$ is a positive integer, $p$ is a prime, and $v_{p}(d-1)=u \geqslant 2$, then
$$
... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 729,583 |
3. Several pairwise non-overlapping isosceles right triangles with leg lengths of 1 are placed on a $100 \times 100$ grid paper. It is known that the hypotenuse of any right triangle is the diagonal of some unit square; each side of a unit square is the leg of a unique right triangle. A unit square whose diagonals are ... | 3. For a general $2 n \times 2 n$ grid paper, the maximum number of empty cells is $n(n-1)$.
In fact, the $2 n \times 2 n$ grid paper is enclosed by $2 n+1$ horizontal lines and $2 n+1$ vertical lines:
$$
\begin{array}{l}
\{(x, y) \mid x=k, 0 \leqslant k \leqslant 2 n, k \in \mathbf{Z}\}, \\
\{(x, y) \mid y=k, 0 \leqs... | 2450 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 729,584 |
4. Space is divided into a union of tetrahedra and octahedra by the set of planes
$$
A=\{x \pm y \pm z=n, n \in \mathbf{Z}\}
$$
Suppose a rational point $\left(x_{0}, y_{0}, z_{0}\right)$ does not belong to any plane in $A$. Prove: there exists a positive integer $k$, such that the point $\left(k x_{0}, k y_{0}, k z_{... | 4. Let $[x]$ denote the greatest integer not exceeding the real number $x$, and $\{x\}=x-[x]$.
First, prove a lemma.
Lemma If the rational numbers $a, b, c, a+b+c$ are all non-integers, then there exists a positive integer $k$ such that $ka, kb, kc$ are all non-integers and
$$
12 \text {. }
$$
Let $k$ be the smallest ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,585 |
7. Let $a, b, c, d$ be positive real numbers whose sum is 3. Prove:
$$
\frac{1}{a^{3}}+\frac{1}{b^{3}}+\frac{1}{c^{3}}+\frac{1}{d^{3}} \leqslant \frac{1}{(a b c d)^{3}} .
$$ | 7. It suffices to prove its equivalent proposition
$$
a^{3} b^{3} c^{3}+a^{3} b^{3} d^{3}+a^{3} c^{3} d^{3}+b^{3} c^{3} d^{3} \leqslant 1 \text {. }
$$
Without loss of generality, assume $a \geqslant b \geqslant c \geqslant d$.
By the AM-GM inequality, we have
$$
\begin{array}{l}
a b(c+d) \leqslant\left[\frac{a+b+(c+d... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 729,586 |
8. Let $M_{A}, M_{B}, M_{C}$ be the midpoints of sides $BC, CA, AB$ of $\triangle ABC$, respectively, and let $M$ be the centroid of $\triangle ABC$. Circle $\Gamma_{A}$ passes through the midpoint of $AM$ and is tangent to $BC$ at point $M_{A}$. Similarly define circles $\Gamma_{B}, \Gamma_{C}$. Prove that circles $\G... | 8. Let the midpoints of $AM$, $BM$, and $CM$ be $K_{A}$, $K_{B}$, and $K_{C}$, respectively. The circumcircle of $\triangle K_{A} M_{B} M_{C}$ intersects the circumcircle of $\triangle K_{B} M_{A} M_{C}$ at point $X$. Without loss of generality, assume that points $K_{A}$ and $X$ are on the same side of line $M_{B} M_{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 729,587 |
1. If there exists $x_{0} \in[0,1]$, such that
$$
2^{x_{0}}\left(3 x_{0}+a\right)<1 \text {, }
$$
then the range of real number $a$ is $\qquad$ | $$
\text { I, 1. } a<1 \text {. }
$$
Notice that,
$$
2^{x_{0}}\left(3 x_{0}+a\right)<1 \Leftrightarrow a<\left(\frac{1}{2}\right)^{x_{0}}-3 x_{0} .
$$
Given that $f(x)=\left(\frac{1}{2}\right)^{x}-3 x$ is decreasing on $x \in[0,1]$, we know that $a<f(x)_{\max }=1$. | a<1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 729,588 |
2. Given the sets
$$
A=\left\{n^{2}+1 \mid n \in \mathbf{Z}_{+}\right\}, B=\left\{n^{3}+1 \mid n \in \mathbf{Z}_{+}\right\} \text {. }
$$
Arrange all elements in $A \cap B$ in ascending order to form the sequence $a_{1}, a_{2}, \cdots$. Then the units digit of $a_{99}$ is | 2. 2 .
From the given, we know that $A \cap B=\left\{n^{6}+1 \mid n \in \mathbf{Z}_{+}\right\}$.
Therefore, $a_{99}=99^{6}+1$.
Thus, its unit digit is 2. | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 729,589 |
3. If the positive real numbers $x \backslash y$ satisfy $y=2016 x$, and $x^{y}=y^{x}$, then $\log _{2016} x+\log _{2016} y=$ $\qquad$ | 3. $\frac{2017}{2015}$.
Notice that,
$$
x^{y}=y^{x} \Leftrightarrow x \log _{2016} y=y \log _{2016} x \text{. }
$$
By $y=2016 x$
$$
\begin{array}{l}
\Rightarrow 2016 x \log _{2016} x=x \log _{2016} 2016 x \\
\Rightarrow \log _{2016} x=\frac{1}{2015} \\
\Rightarrow \log _{2016} x+\log _{2016} y \\
\quad=2 \log _{2016}... | \frac{2017}{2015} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,590 |
4. If acute angles $\alpha$ , $\beta$ satisfy
$$
\sin \alpha=\cos (\alpha+\beta) \cdot \sin \beta,
$$
then the maximum value of $\tan \alpha$ is | $\begin{array}{l}\text { 4. } \frac{\sqrt{2}}{4} \text {. } \\ \text { Given } \sin \alpha=\cos (\alpha+\beta) \cdot \sin \beta \\ =\sin (\alpha+\beta-\beta) \\ =\sin (\alpha+\beta) \cdot \cos \beta-\cos (\alpha+\beta) \cdot \sin \beta \\ \Rightarrow \tan (\alpha+\beta)=2 \tan \beta \\ \Rightarrow \tan \alpha=\frac{\ta... | \frac{\sqrt{2}}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,591 |
6. Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola $C: \frac{x^{2}}{4}-\frac{y^{2}}{5}=1$, respectively. Point $P$ is on the right branch of the hyperbola $C$, and the excenter of $\triangle P F_{1} F_{2}$ opposite to $\angle P F_{1} F_{2}$ is $I$. The line $P I$ intersects the $x$-axis at point $Q$... | 6. 4 .
Since $I F_{1}$ is the angle bisector of $\angle P F_{1} Q$, we have
$$
\frac{|P Q|}{|P I|}=1+\frac{\left|F_{1} Q\right|}{\left|F_{1} P\right|} \text {. }
$$
Let $P\left(x_{0}, y_{0}\right)$. Then, $\left|P F_{1}\right|=\frac{3}{2} x_{0}+2$.
By the optical property of the hyperbola, the tangent line to the hyp... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,592 |
7. In the tetrahedron $S-ABC$,
$$
SA=SB=SC=\sqrt{21}, BC=6 \text{.}
$$
If the projection of point $A$ onto the plane containing the side $SBC$ is exactly the orthocenter of $\triangle SBC$, then the volume of the inscribed sphere of the tetrahedron $S-ABC$ is | 7. $\frac{4 \pi}{3}$.
Since the projection of point $A$ on the plane containing side $SBC$ is exactly the orthocenter of $\triangle SBC$, we have
$$
S A \perp B C, S B \perp A C, S C \perp A B .
$$
Thus, the projection of point $S$ on the plane $ABC$ is the orthocenter of $\triangle ABC$.
Given $\sqrt{S A}=\sqrt{S B... | \frac{4 \pi}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 729,593 |
II. Answer Questions (Total 56 points)
9. (16 points) Given the sequence $\left\{a_{n}\right\}$ satisfies:
$$
\begin{array}{l}
a_{1}=1, a_{2}=2, a_{3}=4, \\
a_{n}=a_{n-1}+a_{n-2}-a_{n-3}+1(n \geqslant 4) .
\end{array}
$$
(1) Find the general term formula of the sequence $\left\{a_{n}\right\}$;
(2) Prove: $\frac{1}{a_{1... | (1) For $n \geqslant 4$, summing up we get
$$
a_{n}-a_{n-2}=a_{3}-a_{1}+n-3=n \text {. }
$$
When $n=2 m\left(m \in \mathbf{Z}_{+}\right)$,
$$
\begin{aligned}
a_{n} & =a_{2}+\sum_{k=1}^{m-1}\left(a_{2 k+2}-a_{2 k}\right) \\
& =2+\sum_{k=1}^{m-1}(2 k+2)=\frac{1}{4} n(n+2) ;
\end{aligned}
$$
When $n=2 m+1\left(m \in \ma... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 729,594 |
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