problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
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values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
10. (14 points) As shown in Figure 1, in the right triangular prism $A B C-A_{1} B_{1} C_{1}$, it is known that $\angle B A C=90^{\circ}, A B=a, A C=2, A A_{1}=1$, and point $D$ is on edge $B_{1} C_{1}$, with $B_{1} D: D C_{1}=1: 3$.
(1) Prove: $B D \perp A_{1} C$;
(2) For what value of $a$ is the dihedral angle $B-A_{... | 10. Draw $D E / / A_{1} B_{1}$ intersecting $A_{1} C_{1}$ at point $E$. Then $D E \perp A_{1} C_{1}$.
From the right triangular prism $A B C-A_{1} B_{1} C_{1}$, we know that plane $A_{1} B_{1} C_{1} \perp$ plane $A_{1} C$.
Therefore, $D E \perp$ plane $A_{1} C$.
Connect $A E$. Then $A E$ is the projection of $B D$ on p... | a=\frac{2 \sqrt{3}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,517 |
11. (14 points) Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
\begin{array}{l}
a_{1}=2, a_{2}=3, \\
2 a_{n+1}=3 a_{n}-a_{n-1}(n \geqslant 2) .
\end{array}
$$
Find (1) the general term formula $a_{n}$ of the sequence $\left\{a_{n}\right\}$;
(2) all positive integer values of $m$ that satisfy the inequality $\f... | 11. (1) From $2 a_{n+1}=3 a_{n}-a_{n-1}$, we know
$$
2\left(a_{n+1}-a_{n}\right)=a_{n}-a_{n-1} \text {. }
$$
Thus, the sequence $\left\{a_{n}-a_{n-1}\right\}$ is a geometric sequence with the first term $a_{2}-a_{1}=1$ and the common ratio $\frac{1}{2}$.
Therefore, $a_{n}-a_{n-1}=\left(\frac{1}{2}\right)^{n-2}$.
Summi... | (m, n)=(1,1),(2,1),(3,2) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,518 |
12. (14 points) In an ellipse, the chord intercepted by the ellipse on a line passing through a focus and perpendicular to the major axis is called the "latus rectum" of the ellipse. As shown in Figure 2, given the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ with foci $F_{1}$ and $F_{2}$, its eccentricit... | 12. (1) From $\frac{c}{a}=\frac{1}{2}$, we get $a=2c$.
Given that the length of the latus rectum is 3, substituting $x=c$ into the ellipse equation yields $y^{2}=\frac{b^{4}}{a^{2}} \Rightarrow \frac{2 b^{2}}{a}=3$.
Solving this, we get $a=2, b=\sqrt{3}$.
Thus, the equation of the ellipse is $\frac{x^{2}}{4}+\frac{y^... | \left(\frac{11}{8}, 0\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,519 |
13. (15 points) Given the function
$$
f(x)=\left(a-\frac{1}{2}\right) \mathrm{e}^{2 x}+x(a \in \mathbf{R}) \text {. }
$$
(1) If $f(x)$ is monotonically increasing on the interval $(-\infty, 0)$, find the range of real number $a$;
(2) If on the interval $(0,+\infty)$, the graph of the function $f(x)$ is always below the... | 13. (1) Since $f(x)$ is monotonically increasing on the interval $(-\infty, 0)$, we know that on the interval $(-\infty, 0)$,
$$
f^{\prime}(x)=(2 a-1) \mathrm{e}^{2 x}+1 \geqslant 0 .
$$
Thus, $1-2 a \leqslant \frac{1}{\mathrm{e}^{2 x}}$.
When $x \in(-\infty, 0)$, $\frac{1}{\mathrm{e}^{2 x}}>1$, so
$$
\begin{array}{l}... | a \in\left[-\frac{1}{2}, \frac{1}{2}\right] | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 727,520 |
14. (15 points) Let
$$
A=x^{4}+2 x^{3}-x^{2}-5 x+34 \text {. }
$$
Find the integer values of \( x \) for which \( A \) is a perfect square. | 14. Notice that,
$$
A=\left(x^{2}+x-1\right)^{2}-3(x-11) \text {. }
$$
So, when $x=11$, $A=131^{2}$ is a perfect square.
Next, we prove: there are no other integer $x$ that satisfy the condition.
(1) When $x>11$, we have $A>0$, thus, $A>\left(x^{2}+x-2\right)^{2}$.
Therefore, $\left(x^{2}+x-2\right)^{2}<A<\left(x^{2}+... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,521 |
$$
\begin{array}{l}
A=\{x|| x-2 \mid<a\}, \\
B=\left\{x \mid x^{2}-2 x-3<0\right\} .
\end{array}
$$
If $B \subseteq A$, then the range of real number $a$ is $\qquad$ | $$
-1 . a \geqslant 3 \text {. }
$$
From the problem, we know that $A=\{x \mid 2-a<x<2+a\}$,
$$
B=\{x \mid-1<x<3\} \text {. }
$$
Since $B \subseteq A$, then $2-a \leqslant-1$, and $3 \leqslant 2+a$. Therefore, $a \geqslant 3$. | a \geqslant 3 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 727,522 |
4. Given the function $f(x)$ for any $x \in \mathbf{R}$, it holds that $f(x+2)=\frac{f(x)-1}{f(x)+1}$, and $f(1)=-2$. Then $f(2013)=$ $\qquad$ | 4. $\frac{1}{2}$.
Notice that,
$$
\begin{array}{l}
f(x+4)=f(x+2+2)=\frac{f(x+2)-1}{f(x+2)+1} \\
=\frac{\frac{f(x)-1}{f(x)+1}-1}{\frac{f(x)-1}{f(x)+1}+1}=-\frac{1}{f(x)} .
\end{array}
$$
Thus, $f(x+8)=f(x)$.
Then $f(2013)=f(251 \times 8+5)=f(5)$
$$
=-\frac{1}{f(1)}=\frac{1}{2} \text {. }
$$ | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,525 |
5. Seven balls of different colors are placed into three boxes numbered 1, 2, and 3. It is known that the number of balls in each box is not less than its number. The number of different ways to place the balls is $\qquad$ | 5.455.
(1) If the number of balls placed in boxes 1, 2, and 3 are 2, 2, and 3 respectively, then the number of different ways to place them is $\mathrm{C}_{7}^{2} \mathrm{C}_{5}^{2}=210$;
(2) If the number of balls placed in boxes 1, 2, and 3 are 1, 3, and 3 respectively, then the number of different ways to place them... | 455 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,526 |
Example 1 Let $m \in \mathbf{Z}_{+}$. If $\left(2^{m+1}+1\right) \mid\left(3^{2^{m}}+1\right)$, prove: $2^{m+1}+1$ is a prime. ${ }^{[2]}$
(2003, Korean Mathematical Olympiad) | Proof According to the condition, we have
$$
3^{2^{m}}=-1\left(\bmod 2^{m+1}+1\right) \text {. }
$$
Therefore, $3^{m+1} \equiv 1\left(\bmod 2^{m+1}+1\right)$.
Let the order of 3 modulo $2^{m+1}+1$ be $\lambda$.
Then $\lambda \mid 2^{m+1}$, and $\lambda \mid \varphi\left(2^{m+1}+1\right)$.
If $\lambda=2^{k}(k \leqslant... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 727,527 |
Example 2 Find all integers $m, n$ such that
$$
m n \mid \left(3^{m}+1\right), m n \mid \left(3^{n}+1\right) \text {. }
$$
(2005, Korean Mathematical Olympiad) | Solve by discussing two cases.
(1) One of $m$ and $n$ is 1.
Without loss of generality, let $m=1$. Then $mn=n, 3^{m}+1=4$.
Thus, $n \mid 4 \Rightarrow n=1,2$ or 4.
Upon verification, we find that $(m, n)=(1,1),(1,2)$ satisfy the given conditions.
Symmetrically, the solutions to the equation in this case are
$$
(m, n)... | (m, n)=(1,1),(1,2),(2,1) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,528 |
Example 3 Let $0<\alpha, \beta, \gamma<\frac{\pi}{2}$, and $\sin ^{3} \alpha+\sin ^{3} \beta+$ $\sin ^{3} \gamma=1$. Prove:
$$
\tan ^{2} \alpha+\tan ^{2} \beta+\tan ^{2} \gamma \geqslant \frac{3 \sqrt{3}}{2} .
$$
(2nd China Southeast Mathematical Olympiad) | 【Analysis】To deduce the target expression $\tan ^{2} \alpha+\tan ^{2} \beta+\tan ^{2} \gamma$ from the condition
$$
\sin ^{3} \alpha+\sin ^{3} \beta+\sin ^{3} \gamma=1
$$
or to deduce the condition from the target expression, is often difficult. In such cases, one can often use the Cauchy-Schwarz inequality to forcibly... | \tan ^{2} \alpha+\tan ^{2} \beta+\tan ^{2} \gamma \geqslant \frac{3 \sqrt{3}}{2} | Inequalities | proof | Yes | Yes | cn_contest | false | 727,529 |
Given $AB$ is the diameter of $\odot O$, the tangents at points $A$ and $B$ are $l_a$ and $l_b$ respectively, $C$ is any point on the circumference, $BC$ intersects line $l_a$ at point $K$, $M$ is the midpoint of arc $\overparen{CAB}$, the tangent at $M$ intersects line $l_b$ at point $T$, line $KT$ intersects $\odot O... | Prove as shown in Figure 1, connect $A C$, $M C$, and $M B$.
Since $M$ is the midpoint of arc $\overparen{C A B}$, therefore,
$M T / / B C$.
Then $\frac{M T}{M S}=\frac{H K}{H S}$.
Also, $S_{\triangle \triangle B T}=S_{\triangle K B T}=S_{\triangle K B M}$,
then $A B \cdot B T$
$$
\begin{array}{l}
=H M \cdot K B \sin ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,530 |
380 The four planes in space are mutually parallel, and the distance between each adjacent pair of planes is $h$. A regular tetrahedron has its vertices on the four planes. Find the edge length of the regular tetrahedron. | Solve As shown in Figure 2, the vertices of the regular tetrahedron $ABCD$ lie on four parallel planes $\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}$.
Let $AD$ intersect planes $\alpha_{2}, \alpha_{3}$ at points $E, F$ respectively, and $AC$ intersect plane $\alpha_{2}$ at point $H$. Connect $BE, EH, HB$.
For conve... | \sqrt{10}h | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,531 |
381 A chess piece starts from one corner of an $8 \times 8$ chessboard, and moving one square in the row direction or column direction is called a step. After several steps, it passes through every square without repetition and returns to the starting point.
(1) Prove: the number of steps in the row direction is differ... | (1) Construct an $8 \times 8$ grid, and connect every adjacent point with a line segment to form a graph. Assume the distance between adjacent points is 1, and call a square with an area of 1 and vertices on the grid points a "cell."
The original problem is equivalent to: Prove that in every Hamiltonian cycle of the g... | 4 | Combinatorics | proof | Yes | Yes | cn_contest | false | 727,532 |
382 cm tall, Xiao Wang's daily sleep time is uniformly distributed between 6.5 hours and 8 hours.
(1) If a day's sleep time is guaranteed to be 7 hours, then record 1, otherwise record 0. Convert the seven numbers recorded in a week into a binary number and then to a decimal number. Find the probability that the result... | (1) Let the number recorded on the $i$-th day be $b_{i}(i=1,2, \cdots, 7)$. Clearly,
$$
P\left(b_{i}=1\right)=\frac{2}{3}, P\left(b_{i}=0\right)=\frac{1}{3} \text {. }
$$
The binary number $b_{1} b_{2} b_{3} b_{4} b_{5} b_{6} b_{7}$ converted to a decimal number exceeds 32 if and only if: $b_{1}=1$ or $b_{1}=0$, $b_{2... | \frac{1942}{2187}, \frac{1448}{2187}, \frac{5}{6} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,533 |
Example 4 Let $a_{i} \in \mathbf{R}_{+}(i=1,2, \cdots, n)$. Prove:
$$
\begin{array}{l}
\frac{1}{a_{1}}+\frac{2}{a_{1}+a_{2}}+\cdots+\frac{n}{a_{1}+a_{2}+\cdots+a_{n}} \\
<2\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}\right) .
\end{array}
$$ | Prove that the general term on the left is $\frac{k}{\sum_{i=1}^{k} a_{i}}$, and extract it using the Cauchy-Schwarz inequality:
For each $k$,
$$
\begin{array}{l}
{\left[\frac{k(k+1)}{2}\right]^{2}=\left(\sum_{i=1}^{k} i\right)^{2}=\left(\sum_{i=1}^{k} \frac{i}{\sqrt{a_{i}}} \cdot \sqrt{a_{i}}\right)^{2}} \\
\leqslant\... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 727,534 |
Example 5 Consider the transformation $T$: For an ordered triplet $X=(x, y, z)$ of three positive integers, the transformation $T$ changes it into a new ordered triplet $T(X)=X_{1}=\left(x_{1}, y_{1}, z_{1}\right)$, where,
$$
\begin{array}{l}
\left(x_{1}, y_{1}, z_{1}\right) \\
=\left\{\begin{array}{ll}
(x+2 z, z, y-x-... | Proof First, it is easy to verify that the transformation $T$ keeps the value of $x^{2}+4 y z$ unchanged. For example, if we apply this transformation according to case (1), then
$$
\begin{array}{l}
x_{1}^{2}+4 y_{1} z_{1} \\
=(x+2 z)^{2}+4 z(y-x-z) \\
=x^{2}+4 y z .
\end{array}
$$
The other two cases can also be veri... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 727,535 |
In 1993, the 2nd problem of the Junior Group Autumn Competition of the World Cities Mathematics Competition was:
As shown in Figure 1, given that the square $P Q R S$ inside the square $A B C D$ satisfies that the line segments $A P, B Q, C R, D S$ do not intersect each other. Prove:
$$
\begin{array}{l}
S_{\text {quad... | Prove by first considering a special case.
When the centers of square $ABCD$ and square $PQRS$ coincide, as shown in Figure 3.
By symmetry, it is easy to see that
$S_{\text{quadrilateral } ABQP} = S_{\text{quadrilateral } BCRQ} = S_{\text{quadrilateral } CDSR} = S_{\text{quadrilateral } DAPS}$.
Clearly, the conclusion ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,536 |
Proposition As shown in Figure 5, inside a regular $2n (n \geqslant 2)$-sided polygon $A_{1} A_{2} \cdots A_{2 n}$, a smaller regular $2n$-sided polygon $B_{1} B_{2} \cdots B_{2 n}$ is placed such that the line segments $A_{1} B_{1}, A_{2} B_{2}, \cdots, A_{2 n} B_{2 n}$ do not intersect each other. Let the area of the... | Let $A$ and $B$ represent the regular $2n$-sided polygons $A_{1} A_{2} \cdots A_{2 n}$ and $B_{1} B_{2} \cdots B_{2 n}$, respectively.
When the centers of figures $A$ and $B$ coincide, it is easy to see that
$S_{1}=S_{2}=\cdots=S_{2 n}$.
The proposition is clearly true.
When the centers of figures $A$ and $B$ d... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,537 |
Example 1 The terms of the sequence $\left\{a_{n}\right\}$ are all positive, and for any $n \in \mathbf{Z}_{+}$, it satisfies
$a_{n+1}=a_{n}+c a_{n}^{2}$ (constant $\left.c>0\right)$.
Prove: (1) For any positive number $M$, there exists $N \in \mathbf{Z}_{+}$, such that when $n>N$, we have $a_{n}>M$; | Prove (1) For any $n \in \mathbf{Z}_{+}$, given $a_{n}>0, c>0$, we know $a_{n+1}=a_{n}+c a_{n}^{2}>a_{n}$,
and $a_{n+1}-a_{n}=a_{n}-a_{n-1}+c a_{n}^{2}-c a_{n-1}^{2}$
$$
>a_{n}-a_{n-1}>\cdots>a_{2}-a_{1} \text {. }
$$
Then $a_{n}=a_{n}-a_{n-1}+a_{n-1}-a_{n-2}+\cdots+a_{2}-a_{1}+a_{1}$
$$
>(n-1)\left(a_{2}-a_{1}\right)... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 727,539 |
(2) If $b_{n}=\frac{1}{1+c a_{n}}, S_{n}$ is the sum of the first $n$ terms of the sequence $\left\{b_{n}\right\}$, then for any $d>0$, there exists $N \in \mathbf{Z}_{+}$, such that when $n>N$, we have
$$
0<\left|S_{n}-\frac{1}{c a_{1}}\right|<d \text {. }
$$
(2013, Joint Autonomous Admission Examination of Tsinghua U... | (2) From $a_{n+1}=a_{n}+c a_{n}^{2}$, we know
$$
\begin{array}{l}
\frac{1}{1+c a_{n}}=\frac{1}{c} \cdot \frac{1}{\frac{1}{c}+a_{n}}=\frac{1}{c}\left(\frac{1}{a_{n}}-\frac{1}{a_{n+1}}\right) \\
\Rightarrow S_{n}=\sum_{i=1}^{n} b_{i}=\frac{1}{c a_{1}}-\frac{1}{c a_{n+1}} \\
\Rightarrow\left|S_{n}-\frac{1}{c a_{1}}\right|... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 727,540 |
Example 3 Find all positive integers $n$ such that $n^{2} \mid\left(2^{n}+1\right)$.
(31st IMO) | First, we prove a lemma using mathematical induction.
Lemma If $3 \|(a+1)$, then for any $k \in \mathbf{N}$, $3^{k+1} \|\left(a^{3^{k}}+1\right)$.
Proof: When $k=0$, according to the condition, the conclusion is obviously true.
Assume that when $k=t$, $3^{t+1} \|\left(a^{3^{k}}+1\right)$.
Then when $k=t+1$,
$$
a^{3^{t+... | n=1,3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,543 |
Example 5 In the sequence $\left\{a_{n}\right\}$,
$$
a_{0}=2007, a_{n+1}=\frac{a_{n}^{2}}{a_{n}+1}(n \in \mathbf{N}) \text {. }
$$
Prove: When $0 \leqslant n \leqslant 1004$, we have
$$
\left[a_{n}\right]=2007-n \text {, }
$$
where $[x]$ denotes the greatest integer not exceeding the real number $x$. ${ }^{[5]}$
(3rd... | Prove the general problem first.
Let $a_{0}>0, a_{n+1}=\frac{a_{n}^{2}}{a_{n}+1}$. Prove:
$$
\left[a_{n}\right]=a_{0}-n\left(0 \leqslant n \leqslant \frac{1}{2}\left(a_{0}+2\right)\right) \text {. }
$$
Proof: For any positive integer $n$, by the recurrence relation, we know $a_{n}>0$.
By $a_{n}-a_{n+1}=a_{n}-\frac{a_{... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 727,545 |
1. The sequence satisfies $a_{0}=\frac{1}{4}$, and for natural number $n$, $a_{n+1}=a_{n}^{2}+a_{n}$.
Then the integer part of $\sum_{n=0}^{201} \frac{1}{a_{n}+1}$ is $\qquad$.
(2011, National High School Mathematics League Gansu Province Preliminary) | Given: $\frac{1}{a_{n}+1}=\frac{1}{a_{n}}-\frac{1}{a_{n+1}}$
$$
\Rightarrow \sum_{n=0}^{2011} \frac{1}{a_{n}+1}=\frac{1}{a_{0}}-\frac{1}{a_{2012}}=4-\frac{1}{a_{2012}} \text {. }
$$
Obviously, the sequence $\left\{a_{n}\right\}$ is monotonically increasing.
$$
\begin{array}{l}
\text { Also, } a_{1}=\frac{5}{16}, a_{2}... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,546 |
2. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{0}=\frac{1}{2}, a_{n+1}=a_{n}+\frac{1}{2006} a_{n}^{2}(n \in \mathbf{N}) \text {. }
$$
Prove: $1-\frac{1}{2008}<a_{2006}<1$.
(2nd Northern Mathematical Olympiad Invitational Competition) | $\begin{array}{l}\text { Hint: From } \frac{1}{2006+a_{n}}=\frac{1}{a_{n}}-\frac{1}{a_{n+1}} \\ \Rightarrow \frac{1}{a_{0}}-\frac{1}{a_{n}}=\sum_{i=0}^{n-1} \frac{1}{a_{i}+2006} \\ \Rightarrow 2-\frac{1}{a_{2006}}=\sum_{i=0}^{2005} \frac{1}{a_{i}+2006}\sum_{i=0}^{2005} \frac{1}{1+2006}=\frac{2006}{2007} \\ \Rightarrow ... | a_{2006}>\frac{2007}{2008} | Algebra | proof | Yes | Yes | cn_contest | false | 727,547 |
Note: All line segments mentioned in the lemma and its proof are directed line segments.
As shown in Figure 1, on line $l_{1}$, there are two fixed points $A$ and $D$ and a moving point $P$. On line $l_{2}$, there are two fixed points $B$ and $C$ and a moving point $Q$ such that $\frac{A P}{P D}=\frac{B Q}{Q C}$. Addi... | Proof of Lemma 1: As shown in Figure 2, let points $E$ and $F$ be on $AB$ and $CD$ respectively, such that $\frac{AE}{EB} = \frac{DF}{FC} = u$, and let points $I$ and $J$ be on $PE$ and $PF$ respectively, such that $\frac{PE}{EI} = \frac{PF}{FJ} = u$.
Then $AP \parallel BI, DP \parallel CJ$. Hence $BI \parallel CJ$.
A... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,549 |
Example 1 Given that the midpoints of sides $BC, CD, DE,$ and $EA$ of pentagon $ABCDE$ are $P, Q, R, S$ respectively, and the midpoints of $PR$ and $QS$ are $M$ and $N$. Prove: $MN \parallel AB, MN=\frac{1}{4} AB$. | From Lemma 1, we know that points $R$, $N$, and $L$ are collinear, and $N$ is the midpoint of $RL$.
In $\triangle ABC$ and $\triangle PRL$, by the properties of the midline, we have
$$LP \parallel AB, \quad LP = \frac{1}{2} AB; \quad MN \parallel LP, \quad MN = \frac{1}{2} LP.$$
Thus, $MN \parallel AB$, and $MN = \frac... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,551 |
Example 2 Given $\triangle A B C$, construct $\triangle A B D \backsim \triangle A C E$ outside the triangle with sides $A B$ and $A C$. Let $M$ and $N$ be the midpoints of $B C$ and $D E$ respectively, and let $B E$ and $C D$ intersect at point $F$. Prove: $A F / / M N$. | Proof As shown in Figure 5, let $AC$ intersect $BD$ and $AB$ intersect $CE$ at points $X$, $Y$ respectively.
$$
\begin{array}{l}
\text { From } \triangle A B D \backsim \triangle A C E, \angle B A X=\angle C A Y \\
\Rightarrow \triangle A B X \backsim \triangle A C Y, \triangle A D X \backsim \triangle A E Y \\
\Righta... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,552 |
Example 3 Given $\triangle A B C$, construct $\triangle A B D \backsim \triangle A C E$ outside the triangle with $A B$ and $A C$ as the hypotenuses, such that $\angle A D B=\angle A E C=90^{\circ}$. Let $B E$ and $C D$ intersect at point $F$. Prove: $A F \perp D E$.
(23rd IMO Shortlist) | Proof As shown in Figure 6, let the midpoints of $BC$, $DE$, $AB$, and $AC$ be $M$, $N$, $P$, and $Q$ respectively, and denote $\angle BAD = \angle CAE = \theta$.
Given $\angle ADB = \angle AEC = 90^{\circ}$, we have
$$
\begin{array}{l}
DP = \frac{1}{2} AB = MQ, EQ = \frac{1}{2} AC = MP, \\
\angle MPD = \angle MPB + \a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,553 |
Example 4 Determine all pairs of positive integers $(n, p)$ such that: $p$ is a prime, $n \leqslant 2 p$, and $n^{p-1} \mid\left[(p-1)^{n}+1\right]$.
(40th IMO) | (1) When $n=1$, obviously, any prime $p$ satisfies the condition.
(2) When $n=2$, since $n^{p-1} \mid\left[(p-1)^{n}+1\right]$, thus, $(p-1)^{n}+1$ is even.
Therefore, $p$ is even, and can only be $p=2$.
Upon verification, $(n, p)=(2,2)$ satisfies the condition.
(3) When $n \geqslant 3$, obviously, $p \geqslant 3$.
At ... | (n, p)=(2,2),(3,3),(1, p)(p \text{ is any prime}) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,554 |
Example 4 As shown in Figure 7, given a convex hexagon $A B C D E F$ with the midpoints of its six sides $A B, B C, C D, D E, E F, F A$ being $G, H, I, J, K, L$ respectively. If $A B=B C, D E=E F, \triangle A B C \backsim \triangle F E D, A F \parallel C D$, prove: $G J, H K, I L$ are concurrent. | Proof As shown in Figure 7, let $AC$ and $DF$ intersect at point $S$.
Given $AF \parallel CD$ and $\frac{AL}{LF}=\frac{CI}{ID}=1$, we know that points $S, L, I$ are collinear.
Let the midpoints of $HJ, BE, GK$ be $P, Q, R$ respectively.
By Lemma 1, we know that $I, P, Q$ and $L, R, Q$ are collinear respectively, and $P... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,555 |
Example 5 As shown in Figure 8, given points $D, E$ are on the sides $AB, AC$ of $\triangle ABC$, respectively, and $M, N$ are the midpoints of $BC, DE$, respectively. $MD$ intersects $NC$ at point $F$, the circumcenters of $\triangle ABC$ and $\triangle ADE$ are $O_{1}$ and $O_{2}$, respectively. Point $P$ is on line ... | Proof: Let the circumcircle $\odot O_{1}$ of $\triangle ABC$ intersect $AP$ at a point $P'$ different from $A$, and let the projections of $P'$ onto $AB$ and $AC$ be $Q'$ and $R'$, respectively. By homothety, we have $Q'R' \parallel QR$.
Let the circumcircle $\odot O_{2}$ of $\triangle ADE$ intersect $\odot O_{1}$ at ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,556 |
In $\triangle A B C$, $\sum \sin A \leqslant \frac{3 \sqrt{3}}{2}$ is a basic trigonometric inequality. Below, we use it to prove a three-variable inequality problem. Given positive numbers $a, b, c$ satisfying $\sum a=1$. Prove:
$$
\sum \frac{1}{b c+a+\frac{1}{a}} \leqslant \frac{27}{31},
$$
where “$\sum$” denotes th... | Prove that let $a=y z, b=z x, c=x y(x, y, z>0)$. Then
$$
\sum x y=1 \text {. }
$$
Let $x=\cot A, y=\cot B, z=\cot C$, where $\angle A$, $\angle B$, $\angle C \in\left(0, \frac{\pi}{2}\right)$.
Then from equation (1) we get
$$
\begin{array}{l}
\cot C=z=\frac{1-x y}{x+y}=\frac{1-\cot A \cdot \cot B}{\cot A+\cot B} \\
=-... | \frac{27}{31} | Inequalities | proof | Yes | Yes | cn_contest | false | 727,557 |
1. As shown in Figure 1, let the circumcenter of acute $\triangle ABC$ be $O$, and the projection of point $A$ on side $BC$ be $H_{A}$. The extension of $AO$ intersects the circumcircle of $\triangle BOC$ at point $A'$. The projections of point $A'$ on lines $AB$ and $AC$ are $D$ and $E$, respectively. The circumcenter... | 1. As shown in Figure 2, let $T$ be the reflection of point $A$ over $BC$, and let $A'$ be the projection of $A$ onto side $BC$, and $F$ be the projection of $T$ onto line $AC$.
From $AC = CT$, we know $\angle TCM = 2 \angle TAM$.
$$
\begin{aligned}
& \text{Also, } \angle TAM = \frac{\pi}{2} - \angle ACB = \angle OAB, ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,558 |
2. Let $A_{1} A_{2} \cdots A_{101}$ be a regular 101-gon. Each vertex is colored either red or blue. Let $N$ be the number of obtuse triangles that satisfy the following conditions: the three vertices of the triangle are vertices of the 101-gon, the two acute vertices have the same color, and the color of the obtuse ve... | 2. Let $x_{i}=0$ or 1 represent $A_{i}$ being red or blue, respectively. For an obtuse triangle $\triangle A_{i-a} A_{i} A_{i+b}$, where vertex $A_{i}$ is the obtuse angle vertex, i.e., $a+b \leqslant 50$.
The coloring of the three vertices satisfies the condition if and only if
$$
\left(x_{i}-x_{i-a}\right)\left(x_{i}... | 32175 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,559 |
3. Prove: The indeterminate equation
$$
\begin{array}{l}
(x+1)(x+2) \cdots(x+2014) \\
=(y+1)(y+2) \cdots(y+4028)
\end{array}
$$
has no positive integer solutions $(x, y)$.
(Supplied by Li Weiguo) | 3. For $n=2^{k} m$ (where $k$ is a non-negative integer, $m$ is an odd number), let $v(n)=2^{k}$.
Proof by contradiction.
Assume $(x, y)$ is a positive integer solution to the original equation.
Let $v(x+i)=\max _{1 \leq j2^{1007} .
\end{array}
$$
Also, $\prod_{j=1}^{2014}(x+j)=\prod_{j=1}^{4088}(y+j)$ is a multiple o... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 727,560 |
4. Let $k (k>3)$ be a given odd number. Prove: there exist infinitely many positive odd numbers $n$, such that there are two positive integers $d_{1}, d_{2}$, satisfying $d_{1} \left| \frac{n^{2}+1}{2}, d_{2} \right| \frac{n^{2}+1}{2}$, and $d_{1}+d_{2}=n+k$.
（Supplied by Hongbing Yu） | 4. Consider the indeterminate equation
$$
\left[(k-2)^{2}+1\right] x y=(x+y-k)^{2}+1 \text {. }
$$
We only need to prove: Equation (1) has infinitely many positive odd solutions $(x, y)$.
Obviously, $(1,1)$ is a set of positive odd solutions.
Let $\left(x_{1}, y_{1}\right)=(1,1)$.
Assume $\left(x_{i}, y_{i}\right)\lef... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 727,561 |
5. Let $n (n>1)$ be a given integer. Find the largest constant $\lambda(n)$, such that for any $n$ non-zero complex numbers $z_{1}, z_{2}, \cdots, z_{n}$, we have
$$
\sum_{k=1}^{n}\left|z_{k}\right|^{2} \geqslant \lambda(n) \min _{1 \leqslant k \leqslant n}\left\{\left|z_{k+1}-z_{k}\right|^{2}\right\}\left(z_{n+1}=z_{1... | ```
\begin{array}{l}
\text { 5. Let } \quad \lambda_{0}(n)=\left\{\begin{array}{ll}
\frac{n}{4}, & \text { when } n \text { is even; } \\
\frac{n}{4 \cos ^{2} \frac{\pi}{2 n}}, & \text { when } n \text { is odd. }
\end{array}\right. \\
\text { Next, we prove: } \lambda_{0}(n) \text { is the desired maximum constant val... | \lambda_{0}(n) = \left\{\begin{array}{ll}
\frac{n}{4}, & \text { when } n \text { is even; } \\
\frac{n}{4 \cos^{2} \frac{\pi}{2n}}, & \text { when } n \text { is odd. }
\end{array}\right.} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 727,562 |
6. For an integer $k>1$, let $f(k)$ denote the number of ways to decompose $k$ into a product of integers greater than 1 (the order of the factors in the product does not matter, e.g., $f(12)=4$, because 12 has the following four decompositions: $12, 2 \times 6, 3 \times 4, 2 \times 2 \times 3$). If $n$ is an integer g... | 6. Let $P(n)$ denote the largest prime factor of $n$, and define
$P(1)=f(1)=1$.
First, prove two lemmas.
Lemma 1 For any positive integer $n$ and prime $p, p \mid n$, we have
$$
f(n) \leqslant \sum_{d \left\lvert\, \frac{n}{p}\right.} f(d) .
$$
Proof of Lemma 1 For convenience, we will refer to a valid decomposition o... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 727,563 |
1. If on the interval $[2,3]$, the function $f(x)=x^{2}+b x+c$ and $g(x)=x+\frac{6}{x}$ take the same minimum value at the same point, then the maximum value of the function $f(x)$ on $[2,3]$ is | $$
-1.15-4 \sqrt{6} \text {. }
$$
Notice that, $g(x)=x+\frac{6}{x} \geqslant 2 \sqrt{6}$, when $x=\sqrt{6}$ $\in[2,3]$, the equality holds, i.e., $g(x)$ reaches its minimum value $2 \sqrt{6}$ at $x=\sqrt{6}$.
Thus, by the problem statement, we have
$$
f(x)=(x-\sqrt{6})^{2}+2 \sqrt{6} \text {. }
$$
Also, $3-\sqrt{6}>\... | 15-4 \sqrt{6} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,564 |
Example 5 Find all integers $n>1$ such that there exists a unique integer $a$ satisfying $0 \leqslant a<n!$, and $n! \mid \left(a^{n}+1\right)$.
(46th IMO Shortlist) | (1) When $a=2$, only $a=1$ satisfies the condition.
Therefore, $n=2$ satisfies the problem's condition.
(2) When $n>2$, and $n$ is even, obviously, $4 \mid n!$.
And $a^{n}+1 \equiv\left(a^{2}\right)^{\frac{n}{2}}+1 \equiv 1$ or $2(\bmod 4)$, then $4 \nmid\left(a^{n}+1\right)$.
At this time, $n$ has no solution.
(3) W... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,565 |
2. If $a, b, c, d$ are integers, and
$$
a \lg 2+b \lg 3+c \lg 5+d \lg 7=2013 \text {, }
$$
then the ordered tuple $(a, b, c, d)=$ $\qquad$ | 2. $(2013,0,2013,0)$.
From the problem, we have
$$
2^{a} \times 3^{b} \times 5^{c} \times 7^{d}=10^{2013}=2^{2013} \times 5^{2013} \text {. }
$$
By the Fundamental Theorem of Arithmetic, we know
$$
(a, b, c, d)=(2013,0,2013,0) .
$$ | (2013,0,2013,0) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,566 |
3. Given the function
$$
y=\sqrt{\left(x^{2}-2\right)^{2}+(x-5)^{2}}+\sqrt{\left(x^{2}-3\right)^{2}+x^{2}} \text {. }
$$
Then the minimum value of the function is $\qquad$ | 3. $\sqrt{26}$.
The function $y$ can be regarded as the sum of the distances from point $P\left(x, x^{2}\right)$ on the parabola $y=x^{2}$ to points $M(5,2)$ and $N(0,3)$.
From the graph, it is easy to see that when point $P$ is the intersection of segment $M N$ and the parabola, $y$ is minimized, at which time,
$$
y... | \sqrt{26} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,567 |
4. Given the line segment $x+y=9(x \geqslant 0, y \geqslant 0)$ intersects the $y$-axis, the graph of the exponential function $y=a^{x}$, the graph of the logarithmic function $y=\log _{a} x$, and the $x$-axis at points $A$, $B$, $C$, and $D$ respectively, where $a>0, a \neq 1$. If the middle two points exactly trisect... | 4. $\sqrt[3]{6}$ or $\sqrt[6]{3}$.
Obviously, $A(0,9), D(9,0)$.
When $A B=\frac{1}{3} A D$, point $B(3,6)$.
Thus, $6=a^{3} \Rightarrow a=\sqrt[3]{6}$.
When $A B=\frac{2}{3} A D$, point $B(6,3)$.
Thus, $3=a^{6} \Rightarrow a=\sqrt[6]{3}$. | \sqrt[3]{6} \text{ or } \sqrt[6]{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,568 |
5. As shown in Figure 1, given
Ellipse $C: \frac{x^{2}}{25}+y^{2}=1$ and $\odot O: x^{2}+y^{2}=1$, the maximum radius of a circle contained within the region inside ellipse $C$ and outside $\odot O$ (including the boundaries) is $\qquad$ | 5. $\frac{23}{25}$.
To maximize the radius $r$ of the circle, whose center lies on the $x$-axis and is externally tangent to $\odot O$, and intersects the ellipse $C$ at only one point in the first quadrant.
By symmetry, let the equation of the desired circle $M$ be
$$
\begin{array}{l}
(x-r-1)^{2}+y^{2}=r^{2} \\
\Righ... | \frac{23}{25} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,569 |
6. The integer solutions $(m, n)=$ $\qquad$ for the equation $\frac{1}{m}+\frac{1}{n}-\frac{1}{m n^{2}}=\frac{3}{4}$. | 6. $(3,2)$.
Obviously, $m \neq 0, n \neq 0$.
Transform the given equation to
$$
m=\frac{4(n+1)(n-1)}{n(3 n-4)} \text {. }
$$
Since $(n, n+1)=1=(n, n-1)$, it follows that $n \mid 4$.
Thus, $n= \pm 1, \pm 2, \pm 4$.
Upon inspection, only when $n=2$, $m$ is a non-zero integer 3. | (3,2) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,570 |
7. A bag contains 6 red balls and 8 white balls. Any five balls are placed in box $A$, and the remaining nine balls are placed in box $B$. The probability that the sum of the number of white balls in box $A$ and the number of red balls in box $B$ is not a prime number is $\qquad$ (answer with a number). | 7. $\frac{213}{1001}$.
Let the number of white balls in box $A$ be $n (0 \leqslant n \leqslant 5)$, then the number of red balls in box $A$ is $5-n$; the number of white balls in box $B$ is $8-n$, and the number of red balls is $n+1$. Therefore, the sum of the number of white balls in box $A$ and the number of red bal... | \frac{213}{1001} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,571 |
8. If one element is deleted from the set $\{1!, 2!, \cdots, 100!\}$, and the product of the remaining elements is exactly a perfect square, then the deleted element is $\qquad$ . | 8. 50 !.
Let $A=1!\times 2!\times \cdots \times 100!$.
Since $(2 k)!=(2 k) \cdot(2 k-1)!$, we have
$$
\begin{aligned}
A & =(1!)^{2} \times 2 \times(3!)^{2} \times 4 \times \cdots \times(99!)^{2} \times 100 \\
& =(1!\times 3!\times \cdots \times 99!)^{2}(2 \times 4 \times 6 \times \cdots \times 100) \\
& =(1!\times 3!\... | 50! | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,572 |
9. (12 points) For the sequence of positive numbers $\left\{a_{n}\right\}$, the sum of the first $n$ terms is $b_{n}$, and the product of the first $n$ terms of the sequence $\left\{b_{n}\right\}$ is $c_{n}$, and it is given that $b_{n}+2 c_{n}=1$ $\left(n \in \mathbf{Z}_{+}\right)$. Find the number in the sequence $\l... | 9. Since $a_{1}=b_{1}=c_{1}$, therefore, in $b_{n}+2 c_{n}=$ $\left(n \in \mathbf{Z}_{+}\right)$, let $n=1$, we get
$$
3 a_{1}=1 \Rightarrow a_{1}=\frac{1}{3} \text {. }
$$
When $n \geqslant 2$, from $b_{n}=\frac{c_{n}}{c_{n-1}}$ and $b_{n}+2 c_{n}=1$, we get $\frac{c_{n}}{c_{n-1}}+2 c_{n}=1 \Rightarrow \frac{1}{c_{n}... | 2024 \frac{3}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,573 |
10. (12 points) Given a positive number $p$ and the parabola $C: y^{2}=2 p x$ $(p>0)$, $A\left(\frac{p}{6}, 0\right)$ is a point on the axis of symmetry of the parabola $C$, $O$ is the vertex of the parabola $C$, and $M$ is any point on the parabola $C$. Find the maximum value of $\frac{|O M|}{|A M|}$. | 10. Let point $M(x, y)$. Then $x \geqslant 0$, and $y^{2}=2 p x$.
$$
\begin{array}{l}
\text { Hence }\left(\frac{|O M|}{|A M|}\right)^{2}=\frac{x^{2}+y^{2}}{\left(x-\frac{p}{6}\right)^{2}+y^{2}} \\
=\frac{x^{2}+2 p x}{x^{2}-\frac{p}{3} x+\frac{p^{2}}{36}+2 p x} \\
=\frac{x^{2}+2 p x}{x^{2}+\frac{5 p}{3} x+\frac{p^{2}}{... | \frac{3 \sqrt{2}}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,574 |
11. (18 points) Given
$$
k(a b+b c+c a)>5\left(a^{2}+b^{2}+c^{2}\right) .
$$
(1) If there exist positive numbers $a, b, c$ such that inequality (1) holds, prove: $k>5$;
(2) If there exist positive numbers $a, b, c$ such that inequality (1) holds, and any set of positive numbers $a, b, c$ that satisfy inequality (1) are... | 11. (1) Since $a^{2}+b^{2}+c^{2} \geqslant a b+b c+c a$, therefore, from inequality (1) we know
$$
k(a b+b c+c a)>5(a b+b c+c a) .
$$
Noting that $a, b, c$ are positive numbers.
Thus, $a b+b c+c a>0$.
Hence $k>5$.
(2) From (1) we know $k>5$.
Since $k$ is an integer, then $k \geqslant 6$.
Let $a=1, b=1, c=2$. It is ea... | 6 | Inequalities | proof | Yes | Yes | cn_contest | false | 727,575 |
1. Prove: For any $n>1\left(n \in \mathbf{Z}_{+}\right)$, we have $n \nmid\left(2^{n}-1\right)$.
(33rd Annual William Lowell Putnam Mathematical Competition; 1992, Friendship Cup International Mathematical Competition) | Prompt: If there exists $n>1\left(n \in \mathbf{Z}_{+}\right)$, such that $n \mid\left(2^{n}-1\right)$.
Since $2^{n}-1$ is an odd number, $n$ must be odd. Let $p$ be the smallest prime factor of $n$. Then $2^{n} \equiv 1(\bmod p)$. By Fermat's Little Theorem, we know $2^{p-1} \equiv 1(\bmod p)$. Let the order of 2 modu... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 727,576 |
12. (18 points) As shown in Figure 2, given a cube $A B C D E F G H$ with edge length 1, let $P$ be the set of its eight vertices. Define a sequence of $2 n+1$ ordered vertices $\left(A_{0} A_{1} \cdots A_{2 n}\right)$ such that point $A_{0}$ coincides with $A$, and for each $i \in\{0,1, \cdots, 2 n-1\}, A_{i+1}$ is an... | 12. (1) Establish a spatial rectangular coordinate system with $A B$, $A D$, and $A E$ as the $x$, $y$, and $z$ axes, respectively. Then
$$
A(0,0,0), B(1,0,0), \cdots, H(0,1,1) \text {. }
$$
If $A_{i+1}$ and $A_{i}$ are adjacent vertices, then $A_{i+1}$ only changes one of the coordinates of $A_{i}$ from 0 to 1, or fr... | \frac{9^{n}-1}{4} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,577 |
1. Given the universal set $U=\mathbf{R}$, the sets
$$
\begin{array}{l}
N=\left\{x \left\lvert\,\left(\frac{1}{2}\right)^{x} \leqslant 1\right.\right\}, \\
M=\left\{x \mid x^{2}-6 x+8 \leqslant 0\right\} .
\end{array}
$$
Then the set represented by the shaded area in Figure 1 is ( ).
(A) $\{x \mid x \leqslant 0\}$
(B)... | 1. D.
From the problem, we know that the solution sets of sets $N$ and $M$ are $N=\{x \mid x \geqslant 0\}, M=\{x \mid 2 \leqslant x \leqslant 4\}$.
Therefore, the required solution set is $\{x \mid 0 \leqslant x \leqslant 4\}$. | D | Inequalities | MCQ | Yes | Yes | cn_contest | false | 727,578 |
2. Given that \( i \) is the imaginary unit. Then
$$
\mathrm{i}+\mathrm{i}^{2}+\mathrm{i}^{3}+\mathrm{i}^{4}+\cdots+\mathrm{i}^{2013}=(\quad) \text {. }
$$
(A) \(\mathrm{i}\)
(B) \(-\mathrm{i}\)
(C) 0
(D) 1 | 2. A.
Notice that $\mathrm{i}+\mathrm{i}^{2}+\mathrm{i}^{3}+\mathrm{i}^{4}=0$.
Therefore, $\mathrm{i}+\mathrm{i}^{2}+\mathrm{i}^{3}+\mathrm{i}^{4}+\cdots+\mathrm{i}^{2013}$ $=\left(\mathrm{i}+\mathrm{i}^{2}+\mathrm{i}^{3}+\mathrm{i}^{4}\right) \times 503+\mathrm{i}=\mathrm{i}$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,579 |
4. Given that the line $l$ passes through the focus of the parabola $C: x^{2}=4 y$ and is perpendicular to the $y$-axis. Then the area of the figure enclosed by the line $l$ and the parabola $C$ is ( ).
(A) $\frac{4}{3}$
(B) 2
(C) $\frac{8}{3}$
(D) $\frac{16 \sqrt{2}}{3}$ | 4. C.
From the problem, we know the coordinates of the focus are $(0,1)$. Therefore, by symmetry, the area of the required figure is $S=2 \int_{0}^{1} \sqrt{4 y} \, \mathrm{d} y=4 \times\left.\frac{2}{3} y^{\frac{3}{2}}\right|_{0} ^{1}=\frac{8}{3}$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,581 |
5. Divide four students, A, B, C, and D, into three different classes, with each class having at least one student, and students A and B cannot be in the same class. The number of different ways to do this is ( ).
(A) 24
(B) 30
(C) 36
(D) 81 | 5. B.
First, choose two classes out of three to arrange A and B, which can be done in $\mathrm{C}_{3}^{2} \mathrm{~A}_{2}^{2}$ ways. Then arrange C and D: if both C and D are in the class without A or B, there is 1 way to choose; if one is in the class without A or B, and the other is in a class with A or B, there are... | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 727,582 |
7. The program flowchart is shown in Figure 3. Given that the real number $x \in [1, 9]$. Then the probability that the output $x$ is not less than 55 is ( ).
(A) $\frac{1}{3}$
(B) $\frac{2}{3}$
(C) $\frac{3}{8}$
(D) $\frac{5}{8}$ | 7. C.
It is easy to know that when $x=6$, the output result is $x=55$. Therefore, the required probability is the case when $x \geqslant 6$. Note that, this problem fits a geometric distribution. Thus, $p=\frac{3}{8}$. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,584 |
8. Given a function $f(x)$ defined on $\mathbf{R}$, which is increasing on $(-\infty, 2)$, and the graph of $f(x+2)$ is symmetric about the $y$-axis. Then ( ).
(A) $f(-1)f(3)$
(C) $f(-1)=f(3)$
(D) $f(0)=f(3)$ | 8. A.
From the problem, we know that $f(x)$ is symmetric about $x=2$.
Also, the function $f(x)$ is monotonically increasing on $(-\infty, 2)$, hence
$$
\begin{array}{l}
f(-1)=f(5)<f(3), \\
f(0)=f(4)<f(3) .
\end{array}
$$ | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,585 |
9. Simplify $\frac{\sin 4 \alpha}{4 \sin ^{2}\left(\frac{\pi}{4}+\alpha\right) \cdot \tan \left(\frac{\pi}{4}-\alpha\right)}=(\quad)$.
(A) $\cos 2 \alpha$
(B) $\sin 2 \alpha$
(C) $\cos \alpha$
(D) $\sin \alpha$ | 9. B.
$$
\begin{array}{l}
\text { Original expression }=\frac{2 \sin 2 \alpha \cdot \cos 2 \alpha}{4 \sin ^{2}\left(\frac{\pi}{4}+\alpha\right) \cdot \tan \left[\frac{\pi}{2}-\left(\frac{\pi}{4}+\alpha\right)\right]} \\
=\frac{2 \sin 2 \alpha \cdot \cos 2 \alpha}{4 \sin ^{2}\left(\frac{\pi}{4}+\alpha\right) \cdot \cot ... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,586 |
3. Find all pairs of prime numbers $(p, q)$ such that $p q \mid\left(p^{p}+q^{q}+1\right)$.
(2007, Korean Mathematical Olympiad) | If $p=q$, then $p^{2} \mid\left(2 p^{p}+1\right)$. This is impossible. By symmetry, we may assume $p<q$.
From the condition, we have
$$
p q \left\lvert\,\left(p^{p}+q^{q}+1\right) \Leftrightarrow\left\{\begin{array}{l}
p \mid\left(q^{q}+1\right), \\
q \mid\left(p^{p}+1\right) .
\end{array}\right.\right.
$$
We discuss ... | (2,5),(5,2) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,587 |
10. Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola
$$
\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>0, b>0)
$$
respectively. If there exists a point $P$ on the right branch of the hyperbola such that
$$
\left(\overrightarrow{O P}+\overrightarrow{O F_{2}}\right) \cdot \overrightarrow{F_{2} P}=0 \text... | 10. B.
From $\left(\overrightarrow{O P}+\overrightarrow{O F_{2}}\right) \cdot \overrightarrow{F_{2} P}=0$, we know
$$
\begin{array}{l}
\left(\overrightarrow{O P}+\overrightarrow{O F_{2}}\right) \cdot\left(\overrightarrow{O P}-\overrightarrow{O F_{2}}\right)=0 \\
\Rightarrow(\overrightarrow{O P})^{2}-\left(\overrightar... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 727,588 |
11. In the right triangular prism $A_{1} B_{1} C_{1}-A B C$, it is known that $\angle B A C=\frac{\pi}{2}, A B=A C=A A_{1}=1, G$ and $E$ are the midpoints of $A_{1} B_{1}$ and $C C_{1}$, respectively, and $D$ and $F$ are moving points on segments $A C$ and $A B$ (excluding the endpoints). If $G D \perp E F$, then the r... | 11. A.
Establish a rectangular coordinate system, then
$$
\begin{array}{l}
F(m, 0,0)(0<m<1), D(0, n, 0)(0<n<1), \\
G\left(\frac{1}{2}, 0,1\right), E\left(0,1, \frac{1}{2}\right) . \\
\text { Hence } \overrightarrow{G D}=\left(-\frac{1}{2}, n,-1\right), \overrightarrow{E F}=\left(m,-1,-\frac{1}{2}\right) .
\end{array}
... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 727,589 |
13. Given variables $x, y$ satisfy the constraint conditions
$$
\left\{\begin{array}{l}
x-y+2 \leqslant 0, \\
x \geqslant 1, \\
x+y-7 \leqslant 0 .
\end{array}\right.
$$
Then the range of $\frac{y}{x}$ is $\qquad$ | 13. $\left[\frac{9}{5}, 6\right]$.
According to the problem, the intersection points of the desired region are
$$
\left(\frac{5}{2}, \frac{9}{2}\right),(1,6)
$$ | \left[\frac{9}{5}, 6\right] | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 727,591 |
14. Define $f(x)$ as an odd function on $\mathbf{R}$, and when $x \geqslant 0$, $f(x)=x^{2}$. If for any $x \in[a, a+2]$, $f(x+a) \geqslant 2 f(x)$, then the range of the real number $a$ is $\qquad$ | 14. $[\sqrt{2},+\infty)$.
From the problem, we know that for any real numbers $x, y$ satisfying the domain, $f(x y)=f(x) f(y)$.
Given that $f(x)$ is an odd function on $\mathbf{R}$, and when $x \geqslant 0$, $f(x)$ is monotonically increasing. Therefore,
$$
\begin{array}{l}
f(x+a) \geqslant 2 f(x) \\
\geqslant f(\sqrt... | [\sqrt{2},+\infty) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,592 |
15. Given $O A=1, O B=3, \overrightarrow{O A} \cdot \overrightarrow{O B}=0$, point $C$ is inside $\angle A O B$, and $\angle A O C=30^{\circ}$, set
$$
\overrightarrow{O C}=m \overrightarrow{O A}+n \overrightarrow{O B}(m 、 n \in \mathbf{R}) \text {. }
$$
Then $\frac{m}{n}=$ $\qquad$ | 15. $3 \sqrt{3}$.
Establish a Cartesian coordinate system. Without loss of generality, let
$$
\begin{array}{l}
A(1,0), B(0,3), C(\sqrt{3} y, y) . \\
\text { Then } \overrightarrow{O C}=(\sqrt{3} y, y)=m(1,0)+n(0,3) . \\
\text { Therefore }\left\{\begin{array}{l}
\sqrt{3} y=m, \\
y=3 n
\end{array} \Rightarrow \frac{m}{... | 3 \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,593 |
16. Given that one side of the square $A B C D$ lies on the line $y=2 x-17$, and the other two vertices are on the parabola $y=x^{2}$. Then the minimum value of the area of the square is $\qquad$ . | 16. 80 .
From the problem, we know that the distance from a point $\left(x, x^{2}\right)$ on the parabola to the line $l$ is
$$
d=\frac{\left|2 x-x^{2}-17\right|}{\sqrt{5}}>0 .
$$
When $x=1$, $d$ reaches its minimum value $\frac{16}{\sqrt{5}}$.
To minimize the area of the square, and by the symmetry of the square, th... | 80 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,594 |
17. (10 points) Given the function
$$
f(x)=\frac{\sqrt{3}}{2} \sin 2 x-\cos ^{2} x-\frac{1}{2}(x \in \mathbf{R}) \text {. }
$$
(1) When $x \in\left[-\frac{\pi}{12}, \frac{5 \pi}{12}\right]$, find the minimum and maximum values of the function $f(x)$;
(2) Let the internal angles $\angle A, \angle B, \angle C$ of $\trian... | $$
\begin{array}{l}
f(x)=\frac{\sqrt{3}}{2} \sin 2 x-\frac{1+\cos 2 x}{2}-\frac{1}{2} \\
=\frac{\sqrt{3}}{2} \sin 2 x-\frac{1}{2} \cos 2 x-1 \\
=\sin \left(2 x-\frac{\pi}{6}\right)-1
\end{array}
$$
$$
\begin{array}{l}
\text { (1) From }-\frac{\pi}{12} \leqslant x \leqslant \frac{5 \pi}{12} \\
\Rightarrow-\frac{\pi}{3} ... | a=1, b=2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,595 |
18. (12 points) As shown in Figure 4, in the equilateral $\triangle ABC$, points $D$ and $E$ are on sides $AC$ and $AB$ respectively, and $AD = \frac{1}{3} AC$, $AE = \frac{2}{3} AB$, $BD$ intersects $CE$ at point $F$.
(1) Prove that points $A$, $E$, $F$, and $D$ are concyclic;
(2) If the side length of the equilateral... | 18. (1) From $A E=\frac{2}{3} A B$, we know $B E=\frac{1}{3} A B$.
In the equilateral $\triangle A B C$, from $A D=\frac{1}{3} A C \Rightarrow A D=B E$.
Also, $A B=B C, \angle B A D=\angle C B E$, then $\triangle B A D \cong \triangle C B E \Rightarrow \angle A D B=\angle B E C$.
Therefore, $\angle A D F+\angle A E F=... | \frac{2}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,596 |
19. (12 points) A bag contains 2 white balls and $n$ red balls $\left(n \geqslant 2\right.$, and $\left.n \in \mathbf{Z}_{+}\right)$, each time two balls are drawn from the bag (the two balls are returned to the bag after each draw). If the two balls drawn are of the same color, it is considered a win; otherwise, it is... | 19. (1) When selecting two balls from $n+2$ balls, there are $\mathrm{C}_{n+2}^{2}$ ways to choose, among which the ways to select two balls of the same color are $\mathrm{C}_{n}^{2}+\mathrm{C}_{2}^{2}$.
Thus, the probability of winning in one draw is
$$
p_{1}=\frac{\mathrm{C}_{n}^{2}+\mathrm{C}_{2}^{2}}{\mathrm{C}_{n+... | n=2 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,597 |
4. Find all pairs of prime numbers $(p, q)$ such that $p q \mid\left(5^{p}+5^{q}\right)$.
(2009, China Mathematical Olympiad) | (1) $\min \{p, q\}=2$.
By symmetry, we may assume $p=2$. Then $2 q \mid\left(25+5^{q}\right)$.
By Fermat's Little Theorem, we know $5^{q} \equiv 5(\bmod q)$.
Thus, $25+5^{q} \equiv 30 \equiv 0(\bmod q)$
$\Rightarrow q \mid 30 \Rightarrow q=2,3$ or 5.
Upon verification, $q=3,5$ satisfy the given conditions.
Therefore, ... | (2,3),(2,5),(3,2),(5,2),(5,5),(5,313),(313,5) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,598 |
20. (12 points) The sequence $\left\{a_{n}\right\}$ has the sum of the first $n$ terms as $S_{n}$, satisfying
$$
a_{1}=1,3 t S_{n}-(2 t+3) S_{n-1}=3 t \text {, }
$$
where $t>0, n \geqslant 2$, and $n \in \mathbf{Z}$.
(1) Prove that the sequence $\left\{a_{n}\right\}$ is a geometric sequence;
(2) Let the common ratio o... | 20. (1) When $n \geqslant 2$,
$$
\begin{array}{l}
3 t S_{n}-(2 t+3) S_{n-1}=3 t \\
3 t S_{n+1}-(2 t+3) S_{n}=3 t
\end{array}
$$
Subtracting the above two equations yields
$$
\begin{array}{l}
3 t a_{n+1}-(2 t+3) a_{n}=0 \\
\Rightarrow \frac{a_{n+1}}{a_{n}}=\frac{2 t+3}{3 t}(n \geqslant 2) .
\end{array}
$$
Given $a_{1}... | -\frac{20}{9} | Algebra | proof | Yes | Yes | cn_contest | false | 727,599 |
21. (12 points) Given points $F_{1}(-1,0), F_{2}(1,0)$,
$\odot F_{2}:(x-1)^{2}+y^{2}=1$,
a moving circle on the right side of the $y$-axis is tangent to the $y$-axis and externally tangent to $\odot F_{2}$. The locus of the center of this moving circle is curve $C$, and curve $E$ is an ellipse with foci at $F_{1}$ and ... | 21. (1) Let the center of the moving circle be $D(x, y) (x>0)$.
Since $\odot D$ is tangent to the $y$-axis on its right side and is externally tangent to $\odot F_{2}$, we have $\left|D F_{2}\right|-x=1$.
Thus, $\sqrt{(x-1)^{2}+y^{2}}=x+1$.
Therefore, the equation of curve $C$ is $y^{2}=4 x (x>0)$.
(2) Since curve $E$... | -\frac{\sqrt{6}}{8}<k<\frac{\sqrt{6}}{8}, k \neq 0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,600 |
22. (12 points) Let
$$
f(x)=\frac{\ln (1+x)}{x}(x>0) \text {. }
$$
(1) Determine the monotonicity of the function $f(x)$.
(2) Does there exist a real number $a$ such that for all $x \in(0,+\infty)$,
$$
\ln (1+x)<a x ?
$$
If it exists, find the range of $a$; if not, explain why.
(3) Prove: $\left(1+\frac{1}{n}\right)^{... | 22. From the given, we have
$$
f^{\prime}(x)=\frac{\frac{x}{1+x}-\ln (1+x)}{x^{2}} \text {. }
$$
Let $g(x)=\frac{x}{1+x}-\ln (1+x)(x \geqslant 0)$. Then
$$
g^{\prime}(x)=\frac{1+x-x}{(1+x)^{2}}-\frac{1}{1+x} \leqslant 0 \text {. }
$$
Thus, the function $g(x)$ is decreasing on $[0,+\infty)$.
Therefore, $g(x)=\frac{x}{... | proof | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 727,601 |
1. Given arrays $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ and $\left(b_{1}, b_{2}, \cdots, b_{n}\right)$ are both permutations of $1,2, \cdots, n$. Then
$$
a_{1} b_{1}+a_{2} b_{2}+\cdots+a_{n} b_{n}
$$
the maximum value is | $$
-1 . \frac{n(n+1)(2 n+1)}{6} \text {. }
$$
Notice that, $1<2<\cdots<n$.
By the rearrangement inequality, we know that when $a_{i}=b_{i}=i(i=1,2, \cdots, n)$, the value of the sum $a_{1} b_{1}+a_{2} b_{2}+\cdots+a_{n} b_{n}$ is the maximum. At this time,
$$
\begin{array}{l}
a_{1} b_{1}+a_{2} b_{2}+\cdots+a_{n} b_{n}... | \frac{n(n+1)(2 n+1)}{6} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,602 |
2. If a space diagonal of a rectangular prism forms angles $\alpha, \beta, \gamma$ with the three edges starting from the same vertex, then
$$
\frac{\cos \beta \cdot \cos \gamma}{\cos \alpha}+\frac{\cos \gamma \cdot \cos \alpha}{\cos \beta}+\frac{\cos \alpha \cdot \cos \beta}{\cos \gamma}
$$
the minimum value is $\qqu... | 2. $\sqrt{3}$.
Let $y=\sum \frac{\cos \beta \cdot \cos \gamma}{\cos \alpha}$, where “$\sum$” denotes the cyclic sum.
Since $00$.
Also, $\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1$, then
$$
\begin{aligned}
y^{2} & =\sum \frac{\cos ^{2} \beta \cdot \cos ^{2} \gamma}{\cos ^{2} \alpha}+2 \sum \cos ^{2} \alpha \\
... | \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,603 |
3. Let $x, y$ be real numbers. Then
$$
f(x, y)=x^{2}+x y+y^{2}-x-y
$$
the minimum value is $\qquad$ . | 3. $-\frac{1}{3}$.
Let $u=x+y, v=x-y$. Then $x=\frac{u+v}{2}, y=\frac{u-v}{2}$.
Thus $x^{2}+x y+y^{2}-x-y$ $=\left(\frac{u+v}{2}\right)^{2}+\frac{u+v}{2} \cdot \frac{u-v}{2}+\left(\frac{u-v}{2}\right)^{2}-u$ $=\frac{3 u^{2}-4 u+v^{2}}{4}$ $=\frac{3\left(u-\frac{2}{3}\right)^{2}}{4}+\frac{v^{2}}{4}-\frac{1}{3} \geqslan... | -\frac{1}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,604 |
4. In $\triangle A B C$, it is known that the lengths of the three sides are $A B=\sqrt{7}$,
$$
\begin{aligned}
B C= & \sqrt{3}, C A=\sqrt{2} \text {. Then } \\
& \overrightarrow{A B} \cdot \overrightarrow{B C}+\sqrt{2} \overrightarrow{B C} \cdot \overrightarrow{C A}+\sqrt{3} \overrightarrow{C A} \cdot \overrightarrow{... | 4. $\sqrt{2}-3 \sqrt{3}-4$.
Let $A B=c, B C=a, C A=b$. Then
$$
\begin{array}{l}
\overrightarrow{A B} \cdot \overrightarrow{B C}=-a c \cos B=\frac{1}{2}\left(b^{2}-a^{2}-c^{2}\right), \\
\overrightarrow{B C} \cdot \overrightarrow{C A}=\frac{1}{2}\left(c^{2}-a^{2}-b^{2}\right), \\
\overrightarrow{C A} \cdot \overrightar... | \sqrt{2}-3 \sqrt{3}-4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,605 |
5. Given $a, b, c \in \mathbf{R}_{+}$, and $a+3b+c=9$. Then the minimum value of $a+b^{2}+c^{3}$ is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 5. $\frac{243-8 \sqrt{3}}{36}$.
Let $\alpha, \beta>0$. Then $b^{2}+\alpha^{2} \geqslant 2 \alpha b, c^{3}+\beta^{3}+\beta^{3} \geqslant 3 \beta^{2} c$.
Thus $a+b^{2}+c^{3} \geqslant a+2 \alpha b+3 \beta^{2} c-\alpha^{2}-2 \beta^{3}$.
By the given conditions, take $\alpha=\frac{3}{2}, \beta=\frac{\sqrt{3}}{3}$.
Then $a... | \frac{243-8 \sqrt{3}}{36} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,606 |
6. From the set of positive integers not exceeding 2013, $\{1,2, \cdots, 2013\}$, any two positive integers are chosen. Then the probability that they are exactly the solutions to the equation $x^{3}+y^{3}=x^{2} y^{2}$ is | 6. $\frac{1}{2013^{2}}$.
First, study the positive integer solutions of $x^{3}+y^{3}=x^{2} y^{2}$.
Let $(x, y)$ be a solution of the equation satisfying $x \geqslant y$. Then
$$
\begin{array}{l}
x^{2} \mid y^{3} \Rightarrow y^{3} \geqslant x^{2} \\
\Rightarrow 4 y^{3} \geqslant 4 x^{2} .
\end{array}
$$
Since $x \geqs... | \frac{1}{2013^{2}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,607 |
7. In the Cartesian coordinate system $x O y$, point $A(-1, -1)$, let $B$ and $C$ be two different points on the curve $x y=1(x>0)$, and $\triangle A B C$ is an equilateral triangle. Then the area of its circumcircle is $\qquad$ | $7.8 \pi$.
Since points $B$ and $C$ are symmetric with respect to the altitude through point $A$ on side $BC$, and point $A$ lies on the axis of symmetry of the curve $xy=1 (x>0)$, the slope of line $BC$ must be -1.
Let the line $l_{BC}: y=-x+b (b>0)$.
Substituting into $xy=1$, we get $x^2 - bx + 1 = 0$.
Thus, $\Delta ... | 8\pi | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,608 |
Example 1 Let $a, b, c, x, y, z$ be positive numbers, satisfying:
$$
a=b z+c y, b=c x+a z, c=a y+b x \text {. }
$$
Prove: A triangle with side lengths $a, b, c$ can be formed and it is an acute triangle. | To prove that the problem can be solved by appropriately using three 0s:
$$
\begin{array}{l}
0=a-b z-c y, \\
0=b-c x-a z, \\
0=c-a y-b x,
\end{array}
$$
From \(a(a-b z-c y)\)
$$
=b(b-c x-a z)+c(c-a y-b x),
$$
we get \(x=\frac{b^{2}+c^{2}-a^{2}}{2 b c}\).
Similarly, \(y=\frac{a^{2}+c^{2}-b^{2}}{2 a c}\), \(z=\frac{a^{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,609 |
8. As shown in Figure 1, the base $ABCD$ of the quadrilateral pyramid $F-ABCD$ is a rhombus, with diagonals $AC=2$, $BD=\sqrt{2}$, and $AE$, $CF$ are both perpendicular to the plane $ABCD$, $AE=1$, $CF=2$. Then the volume of the common part of the quadrilateral pyramids $E-ABCD$ and $F-ABCD$ is $\qquad$ | 8. $\frac{2 \sqrt{2}}{9}$.
As shown in Figure 2, let the line $A F$ intersect the line $C E$ at point $H$. Then the common part of the quadrilateral pyramid $E-A B C D$ and the quadrilateral pyramid $F-A B C D$ is the quadrilateral pyramid $H-A B C D$.
Draw $H P \perp$ plane $A B C D$, with $P$ being the foot of the p... | \frac{2 \sqrt{2}}{9} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,610 |
9. (16 points) Prove: The equation
$$
3 x^{3}+9(1+\sqrt{3}) x^{2}+18(1+\sqrt{3}) x+12+10 \sqrt{3}=0
$$
has a unique real root. | II. 9. The original equation can be transformed into
$$
x^{3}+3(1+\sqrt{3}) x^{2}+6(1+\sqrt{3}) x+4+\frac{10 \sqrt{3}}{3}=0 \text {. }
$$
Let $x=y-1$. Then
$$
\begin{array}{l}
y^{3}+3 \sqrt{3} y^{2}+3 y+\frac{\sqrt{3}}{3}=0 \\
\Rightarrow \sqrt{3} y^{3}+9 y^{2}+3 \sqrt{3} y+1=0 .
\end{array}
$$
Let $y=-\frac{1}{z}$. ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 727,611 |
10. (20 points) Given $\angle A, \angle B, \angle C$ are the three interior angles of $\triangle ABC$, and the vector
$$
\boldsymbol{\alpha}=\left(\cos \frac{A-B}{2}, \sqrt{3} \sin \frac{A+B}{2}\right),|\boldsymbol{\alpha}|=\sqrt{2}.
$$
If when $\angle C$ is maximized, there exists a moving point $M$ such that $|\over... | 10. Since $|\alpha|=\sqrt{2}$, we have
$$
\begin{array}{l}
\cos ^{2} \frac{A-B}{2}+3 \sin ^{2} \frac{A+B}{2} \\
=2+\frac{1}{2} \cos (A-B)-\frac{3}{2} \cos (A+B)=2 \\
\Leftrightarrow \cos (A-B)=3 \cos (A+B) \\
\Leftrightarrow 2 \sin A \cdot \sin B=\cos A \cdot \cos B \\
\Leftrightarrow \tan A \cdot \tan B=\frac{1}{2} . ... | \frac{2 \sqrt{3}+\sqrt{2}}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,612 |
11. (20 points) In the sequence $\left\{x_{n}\right\}$,
$$
x_{n}=p^{n}+q^{n}, x_{1}=1, x_{3}=4 \text {. }
$$
(1) Prove: $x_{n+2}=x_{n+1}+x_{n}$;
(2) Determine the last digit of $x_{2011}$, but no proof is required. | 11. (1) Notice,
$$
\begin{array}{l}
p+q=1, p^{3}+q^{3}=4, \\
p^{3}+q^{3}=(p+q)\left(p^{2}-p q+q^{2}\right) .
\end{array}
$$
Thus, $p^{2}-p q+q^{2}=4, p^{2}+2 p q+q^{2}=1$.
$$
\begin{array}{l}
\text { Then } x_{2}=p^{2}+q^{2}=3 \\
\Rightarrow p^{2}+(1-p)^{2}=3 \\
\Rightarrow p^{2}=p+1, q^{2}=q+1 \text {. }
\end{array}
... | 6 | Algebra | proof | Yes | Yes | cn_contest | false | 727,613 |
12. (20 points) Given the parabola $C: x^{2}=2 y$ and the line $l$: $y=k x-1$ have no common points, $P$ is a moving point on the line $l$, and two tangent lines are drawn from $P$ to the parabola $C$, with $A$ and $B$ being the points of tangency.
(1) Prove: the line $A B$ always passes through a fixed point $Q$;
(2) ... | 12. (1) Let point $A\left(x_{1}, y_{1}\right)$. Then $y_{1}=\frac{1}{2} x_{1}^{2}$.
From $y=\frac{1}{2} x^{2}$, we get $y^{\prime}=x$. Therefore, $\left.y^{\prime}\right|_{x=x_{1}}=x_{1}$.
Thus, the equation of the tangent line to the parabola $C$ at point $A$ is
$y-y_{1}=x_{1}\left(x-x_{1}\right) \Rightarrow y=x_{1} ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 727,614 |
1. In the complex number range, the equation
$$
x^{2}+p x+1=0(p \in \mathbf{R})
$$
has two roots $\alpha, \beta$. If $|\alpha-\beta|=1$, then $p=$ $\qquad$ | 1. $\pm \sqrt{3}$ or $\pm \sqrt{5}$.
If the equation has real roots, then these two real roots are $\frac{ \pm \sqrt{5}+1}{2}$ and $\frac{ \pm \sqrt{5}-1}{2}$, at this time, $p= \pm \sqrt{5}$;
If the equation has no real roots, then these two complex roots are conjugate complex numbers, which are $\frac{ \pm \sqrt{3}+... | \pm \sqrt{3} \text{ or } \pm \sqrt{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,615 |
2. Given that for any real number $x$ we have $a \cos x + b \cos 2x \geqslant -1$.
Then the maximum value of $a + b$ is $\qquad$ | 2. 2 .
Let $x=\frac{2 \pi}{3}$, then $a+b \leqslant 2$.
When $a=\frac{4}{3}, b=\frac{2}{3}$,
$$
\begin{array}{l}
a \cos x+b \cos 2 x=\frac{4}{3} \cos x+\frac{2}{3} \cos 2 x \\
=\frac{4}{3} \cos ^{2} x+\frac{4}{3} \cos x-\frac{2}{3} \\
=\frac{4}{3}\left(\cos x+\frac{1}{2}\right)^{2}-1 \\
\geqslant-1 .
\end{array}
$$ | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 727,616 |
3. Given the sequence $\left\{a_{n}\right\}_{n \geqslant 1}$ satisfies
$$
a_{n+2}=a_{n+1}-a_{n} \text {. }
$$
If the sum of the first 1000 terms of the sequence is 1000, then the sum of the first 2014 terms is $\qquad$ . | 3. 1000 .
From $a_{n+2}=a_{n+1}-a_{n}$, we get
$$
a_{n+3}=a_{n+2}-a_{n+1}=\left(a_{n+1}-a_{n}\right)-a_{n+1}=-a_{n} \text {. }
$$
Therefore, for any positive integer $n$ we have
$$
a_{n}+a_{n+1}+a_{n+2}+a_{n+3}+a_{n+4}+a_{n+5}=0 \text {. }
$$
Also, $2014=1000(\bmod 6)$, so
$$
S_{2014}=S_{1000}=1000 \text {. }
$$ | 1000 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,617 |
4. Given the function
$$
f(x)=x^{3}-6 x^{2}+17 x-5 \text {, }
$$
real numbers $a, b$ satisfy $f(a)=3, f(b)=23$. Then $a+b=$ $\qquad$ | 4.4 .
Notice,
$$
\begin{array}{l}
f(x)=x^{3}-6 x^{2}+17 x-5 \\
=(x-2)^{3}+5(x-2)+13 \text {. } \\
\text { Let } g(y)=y^{3}+5 y .
\end{array}
$$
Then $g(y)$ is an odd function and monotonically increasing.
And $f(a)=(a-2)^{3}+5(a-2)+13=3$, so, $g(a-2)=-10$.
Also, $f(b)=(b-2)^{3}+5(b-2)+13=23$, then $g(b-2)=10$.
Thus $... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,618 |
5. The axial section $SAB$ of the circular cone is an equilateral triangle with a side length of 2, $O$ is the center of the base, $M$ is the midpoint of $SO$, and the moving point $P$ is within the base of the cone (including the circumference). If $AM \perp MP$, then the length of the trajectory formed by point $P$ i... | 5. $\frac{\sqrt{7}}{2}$.
Establish a rectangular coordinate system in space. Let
$$
\begin{array}{l}
A(0,-1,0), B(0,1,0), S(0,0, \sqrt{3}), \\
M\left(0,0, \frac{\sqrt{3}}{2}\right), P(x, y, 0) .
\end{array}
$$
Thus, $\overrightarrow{A M}=\left(0,1, \frac{\sqrt{3}}{2}\right), \overrightarrow{M P}=\left(x, y,-\frac{\sq... | \frac{\sqrt{7}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,619 |
Example 2 In $\triangle A B C$, prove:
$$
\cot A+\cot B+\cot C \geqslant \frac{1}{3}\left(\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2}\right) .
$$ | Let $x=\tan \frac{A}{2}, y=\tan \frac{B}{2}, z=\tan \frac{C}{2}$.
Then $x y+y z+z x=1$.
It suffices to prove
$$
\begin{array}{l}
\frac{1-x^{2}}{2 x}+\frac{1-y^{2}}{2 y}+\frac{1-z^{2}}{2 z} \geqslant \frac{1}{3}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) \\
\Leftarrow \frac{1}{6}\left(\frac{1}{x}+\frac{1}{y}+\frac{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 727,620 |
6. In the Cartesian coordinate system, the "rectilinear distance" between points $P\left(x_{1}, y_{1}\right)$ and $Q\left(x_{2}, y_{2}\right)$ is defined as
$$
d(P, Q)=\left|x_{1}-x_{2}\right|+\left|y_{1}-y_{2}\right| \text {. }
$$
If point $C(x, y)$ has equal rectilinear distances to $A(1,3)$ and $B(6,9)$, where the ... | 6. $5(\sqrt{2}+1)$.
From the problem, we have
$$
|x-1|+|y-3|=|x-6|+|y-9| \text {. }
$$
When $y \geqslant 9, y \leqslant 3$, equation (1) transforms into
$$
\begin{array}{l}
|x-1|+6=|x-6|, \\
|x-1|=6+|x-6| .
\end{array}
$$
Both have no solutions.
When $3 \leqslant y \leqslant 9$, equation (1) transforms into
$$
2 y-1... | 5(\sqrt{2}+1) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,621 |
7. The positive integer solutions of the equation $x+y^{2}+(x, y)^{3}=x y(x, y)$ are $\qquad$ groups $((x, y)$ represents the greatest common divisor of integers $x, y)$.
| 7.4.
Let $(x, y)=d$. Then $d^{2} \mid x$.
Let $x=a d^{2}, y=b d$. Then $(a d, b)=1$.
Thus, the original equation becomes
$$
a+b^{2}+d=a b d^{2} \text {. }
$$
Therefore, $b \mid(a+d) \Rightarrow b \leqslant a+d$.
Then $a+b^{2}+d=a b d^{2}$
$$
\begin{array}{l}
=(a+d) b+(a+d) b+b\left(a d^{2}-2 a-2 d\right) \\
\geqslant... | 4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,622 |
8. A middle school has 35 lights on each floor. To save electricity while ensuring the lighting needs of the corridors, the following requirements must be met:
(1) Two adjacent lights cannot be on at the same time;
(2) Any three consecutive lights cannot be off at the same time.
If you were to design different lighting... | 8. 31572 .
Notice that, $a_{n+3}=a_{n}+a_{n+1}, a_{3}=4, a_{4}=5$. Therefore, $a_{35}=31572$. | 31572 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,623 |
9. (16 points) Given the polynomial
$$
\begin{aligned}
f(x)= & 2 a x^{4}+(2-7 a) x^{3}+(5 a-3) x^{2}+ \\
& (11-7 a) x+3 a-5,
\end{aligned}
$$
where $a$ is a real number. Prove: For any real number $a$, the equation $f(x)=0$ always has the same real root. | $$
\begin{array}{l}
f(x)=a\left(2 x^{4}-7 x^{3}+5 x^{2}-7 x+3\right)+ \\
\quad\left(2 x^{3}-3 x^{2}+11 x-5\right) \\
=(2 x-1)\left[a\left(x^{3}-3 x^{2}+x-3\right)+\left(x^{2}-x+5\right)\right] \\
=(2 x-1)\left[a(x-3)\left(x^{2}+1\right)+\left(x^{2}-x+5\right)\right] .
\end{array}
$$
Thus, for any real number $a$, the ... | x=0.5 | Algebra | proof | Yes | Yes | cn_contest | false | 727,624 |
10. (20 points) Given $P\left(2 a^{2}, 4 a\right)$ (where $a$ is a non-zero constant) is a point on the parabola $y^{2}=8 x$. Two lines with complementary inclination angles are drawn through $P$ and intersect the parabola at two other points $A$ and $B$.
(1) Prove that the slope of line $A B$ is a constant;
(2) If poi... | 10. (1) Let points $A\left(2 t_{1}^{2}, 4 t_{1}\right)$ and $B\left(2 t_{2}^{2}, 4 t_{2}\right)$.
Given that the slopes of lines $PA$ and $PB$ are complementary,
$$
\begin{array}{l}
\Rightarrow k_{P A}+k_{P B}=0 \\
\Rightarrow \frac{4 t_{1}-4 a}{2 t_{1}^{2}-2 a^{2}}+\frac{4 t_{2}-4 a}{2 t_{2}^{2}-2 a^{2}}=0 \\
\Righta... | 24 a^{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,625 |
11. (20 points) Given positive real numbers $a, b$ satisfy $\frac{1}{a^{2}}+\frac{4}{b^{2}} \leqslant 1, a^{2}+2 b^{2} \leqslant 15$.
Find the range of $a+b$. | 11. From $a^{2}+2 b^{2} \leqslant 15 \Rightarrow a^{2} \leqslant 15-2 b^{2}$.
From $\frac{1}{a^{2}}+\frac{4}{b^{2}} \leqslant 1 \Rightarrow \frac{1}{15-2 b^{2}}+\frac{4}{b^{2}} \leqslant 1$
$\Rightarrow b^{4}-11 b^{2}+30 \leqslant 0 \Rightarrow 5 \leqslant b^{2} \leqslant 6$.
Also, $\frac{1}{a^{2}}+\frac{4}{b^{2}} \le... | [\sqrt{3}+\sqrt{6}, 2 \sqrt{5}] | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 727,626 |
一、(40 points) As shown in Figure 1, $\triangle ABC$ and $\triangle PQR$ satisfy the following conditions: $A$ and $P$ are the midpoints of segments $QR$ and $BC$ respectively, and $QR$ and $BC$ are the angle bisectors of $\angle BAC$ and $\angle QPR$ respectively. Prove:
$$
AB + AC = PQ + PR
$$ | As shown in Figure 2, let $X$ be the intersection of the perpendicular bisectors of segments $BC$ and $RQ$, and let $Q', R'$ be the reflections of points $Q, R$ over the line $XP$, and $B', C'$ be the reflections of points $B, C$ over the line $XA$.
Then $\angle Q'PB = \angle QPC = \angle RPC$
$\Rightarrow Q', P, R$ ar... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,627 |
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