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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class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
10. (20 points) In the Cartesian coordinate system $x O y$, it is known that the equation of an ellipse is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, $A_{1}$ and $A_{2}$ are the left and right vertices of the ellipse, respectively, $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse, respectively, a... | 10. Let $c=\sqrt{a^{2}-b^{2}}$. Then $A_{1}(-a, 0), A_{2}(a, 0), F_{1}(-c, 0), F_{2}(c, 0)$. Let $P\left(x_{0}, y_{0}\right), Q\left(x_{1}, y_{1}\right), R\left(x_{2}, y_{2}\right)$, where $\frac{x_{0}^{2}}{a^{2}}+\frac{y_{0}^{2}}{b^{2}}=1, y_{0} \neq 0$.
From $Q A_{1} \perp P A_{1}, Q A_{2} \perp P A_{2}$, we know
$\o... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,979 |
11. (20 points) Find all pairs of positive real numbers $(a, b)$ such that the function $f(x)=a x^{2}+b$ satisfies: for any real numbers $x, y$,
$$
f(x y)+f(x+y) \geqslant f(x) f(y) .
$$ | 11. From the problem, we have
$$
\begin{array}{l}
\left(a x^{2} y^{2}+b\right)+\left[a(x+y)^{2}+b\right] \\
\geqslant\left(a x^{2}+b\right)\left(a y^{2}+b\right) .
\end{array}
$$
First, find the necessary conditions that \(a\) and \(b\) must satisfy.
By setting \(y=0\) in equation (1), we get
$$
\begin{array}{l}
b+\le... | \{(a, b) \mid 0 < b \leq 1, 0 < a < 1, 2a + b \leq 2\} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 726,980 |
One, (40 points) As shown in Figure 1, $AB$ is a chord of circle $\Gamma$, and $P$ is a point on arc $\overparen{AB}$. Points $E$ and $F$ are on segment $AB$ such that $AE = EF = FB$. Lines $PE$ and $PF$ are extended to intersect circle $\Gamma$ at points $C$ and $D$, respectively. Prove:
$$
EF \cdot CD = AC \cdot BD \... | As shown in Figure 4, connect $A D, B C, C F, D E$. Let $d(A, l)$ denote the distance from point $A$ to line $l$. Since $A E=E F=F B$, we have
$$
\frac{B C \sin \angle B C E}{A C \sin \angle A C E}=\frac{d\left(B, l_{C P}\right)}{d\left(A, l_{C P}\right)}=\frac{B E}{A E}=2 \text {. }
$$
Similarly,
$$
\frac{A D \sin \an... | EF \cdot CD = AC \cdot BD | Geometry | proof | Yes | Yes | cn_contest | false | 726,981 |
Given positive integers $u, v$. The sequence $\left\{a_{n}\right\}$ is defined as follows: $a_{1}=u+v$, for integers $m \geqslant 1$,
$$
\begin{array}{l}
a_{2 m}=a_{m}+u, a_{2 m+1}=a_{m}+v . \\
\text { Let } S_{m}=a_{1}+a_{2}+\cdots+a_{m}(m=1,2, \cdots) .
\end{array}
$$
Prove: The sequence $\left\{S_{n}\right\}$ conta... | $$
\begin{array}{l}
S_{2^{n+1}-1} \\
= a_{1}+\left(a_{2}+a_{3}\right)+\left(a_{4}+a_{5}\right)+\cdots+ \\
\left(a_{2^{n+1}-2}+a_{2^{n+1}-1}\right) \\
= u+\dot{v}+\left(a_{1}+u+a_{1}+v\right)+ \\
\left(a_{2}+u+a_{2}+v\right)+\cdots+ \\
\left(a_{2 n-1}+u+a_{2 n-1}+v\right) \\
= 2^{n}(u+v)+2 S_{2^{n-1}} .
\end{array}
$$
... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 726,982 |
Three. (50 points) An exam consists of $m$ questions, with $n$ students participating, where $m, n \geqslant 2$ are given integers. The scoring rule for each question is: if exactly $x$ students fail to answer the question correctly, then each student who answers the question correctly gets $x$ points, and those who fa... | For $k=1,2, \cdots, m$, let the number of people who did not answer the $k$-th question correctly be $x_{k}$. Then the number of people who answered the $k$-th question correctly is $n-x_{k}$. According to the scoring rule, these $n-x_{k}$ people each score $x_{k}$ points on the $k$-th question.
Let the total score of ... | m(n-1) | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,983 |
Four, (50 points) Let $n, k\left(n<2^{k}\right)$ be integers greater than 1. Prove: There exist $2k$ integers not divisible by $n$, such that if they are arbitrarily divided into two groups, there will always be a group where the sum of some of its numbers is divisible by $n$.
| Four, first consider the case where $n$ is a power of 2.
Let $n=2^r (r \geqslant 1)$. Then $r<k$. Take 3 numbers $2^{r-1}$ and $2k-3$ ones. Clearly, these numbers are not divisible by $n$. Divide these $2k$ numbers into two groups arbitrarily, then there must be a group containing 2 numbers $2^{r-1}$, whose sum is $2^r... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 726,984 |
Example 6 Find the maximum number of elements in a set $S$ that satisfies the following conditions:
(1) Each element in set $S$ is a positive integer not exceeding 100;
(2) For any two distinct elements $a, b$ in set $S$, there exists an element $c$ in $S$ such that
$$
(a, c)=(b, c)=1 \text {; }
$$
(3) For any two dist... | Each positive integer can be expressed as
$$
n=2^{k_{1}} \times 3^{k_{2}} \times 5^{k_{3}} \times 7^{k_{4}} \times 11^{k_{5}} q,
$$
where $q$ is coprime with 2, 3, 5, 7, and 11, and $k_{1}, k_{2}, \cdots, k_{5}$ are non-negative integers.
Let $A=\left\{n \leqslant 100 \mid k_{1}, k_{2}, \cdots, k_{5}\right.$ have exac... | 72 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,985 |
1. Given $(1+\sqrt{3})^{n}=a_{n}+b_{n} \sqrt{3}$, where $a_{n}$ and $b_{n}$ are integers. Then $\lim _{n \rightarrow+\infty} \frac{a_{n}}{b_{n}}=$ $\qquad$
$(2011$, National High School Mathematics League Sichuan Preliminary Competition) | Hint: $\lim _{n \rightarrow+\infty} \frac{a_{n}}{b_{n}}$ $=\lim _{n \rightarrow+\infty} \sqrt{3} \times \frac{(1+\sqrt{3})^{n}+(1-\sqrt{3})^{n}}{(1+\sqrt{3})^{n}-(1-\sqrt{3})^{n}}=\sqrt{3}$. | \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,986 |
2. The first digit after the decimal point of $(\sqrt{2}+\sqrt{3})^{2010}$ is $\qquad$
$(2009$, National High School Mathematics League Jilin Province Preliminary) | Hint: The fractional part of $(\sqrt{2}+\sqrt{3})^{2010}$ is $1-(\sqrt{2}-\sqrt{3})^{2010}$.
Since $0.9<1-(\sqrt{2}-\sqrt{3})^{2010}<1$, the first digit after the decimal point of $(\sqrt{2}+\sqrt{3})^{2010}$ is 9. | 9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,987 |
3. Let $p, q \in \mathbf{Z}_{+}$, and $q \leqslant p^{2}$. Prove: For $n \in \mathbf{Z}_{+}$, there exists $N \in \mathbf{Z}_{+}$, such that
$$
\left(p-\sqrt{p^{2}-q}\right)^{n}=N-\sqrt{N^{2}-q^{n}},
$$
and $\square$
$$
\left(p+\sqrt{p^{2}-q}\right)^{n}=N+\sqrt{N^{2}-q^{n}} \text {. }
$$
(2012, Zhejiang Province High ... | Hint: $N=\frac{1}{2}\left[\left(p-\sqrt{p^{2}-q}\right)^{n}+\left(p+\sqrt{p^{2}-q}\right)^{n}\right]$ $\in \mathbf{Z}_{+}$. | proof | Algebra | proof | Yes | Yes | cn_contest | false | 726,988 |
4. Consider the sequence $x_{n}=(1+\sqrt{2}+\sqrt{3})^{n}$. Let
$$
x_{n}=a_{n}+b_{n} \sqrt{2}+c_{n} \sqrt{3}+d_{n} \sqrt{6} \text {, }
$$
where, $a_{n}, b_{n}, c_{n}, d_{n} \in \mathbf{Z}_{+}$.
Find $\lim _{n \rightarrow+\infty} \frac{b_{n}}{a_{n}}$, $\lim _{n \rightarrow+\infty} \frac{c_{n}}{a_{n}}$, and $\lim _{n \r... | The answer is:
$$
\lim _{n \rightarrow+\infty} \frac{b_{n}}{a_{n}}=\frac{\sqrt{2}}{2}, \lim _{n \rightarrow+\infty} \frac{c_{n}}{a_{n}}=\frac{\sqrt{3}}{3}, \lim _{n \rightarrow+\infty} \frac{d_{n}}{a_{n}}=\frac{\sqrt{6}}{6} .
$$ | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 726,989 |
1. Do there exist integers $a$, $b$, $c$ such that
$$
a^{2} b c+2, \quad a b^{2} c+2, \quad a b c^{2}+2
$$
are all perfect squares?
(Supplied by Li Qiusheng) | 1. Does not exist.
Assume it exists. If $a, b, c$ contain an even number (let's assume it is $a$), then $a^{2} b c + 2$ leaves a remainder of 2 when divided by 4, and thus cannot be a perfect square.
If $a, b, c$ are all odd, then their remainders when divided by 4 can only be 1 or 3. By the pigeonhole principle, th... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,990 |
2. Let integer $n \geqslant 2$, and real numbers $x_{1}, x_{2}, \cdots, x_{n} \in$ $[0,1]$. Prove:
$$
\sum_{1 \leqslant k<l \leqslant n} k x_{k} x_{l} \leqslant \frac{n-1}{3} \sum_{k=1}^{n} x_{k} \text {. }
$$ | 2. Given $x_{1}, x_{2}, \cdots, x_{n} \in [0,1]$, we know $x_{i} x_{j} \leqslant x_{i}$. Therefore, $3 \sum_{1 \leqslant k<l \leqslant n} k x_{k} x_{1} = \sum_{1 \leqslant k<l \leqslant n} 3 k x_{k} x_{l} \leqslant \sum_{1 \leqslant k<l \leqslant n}^{1 \leqslant n} (k x_{k} + 2 k x_{l})$.
For $1 \leqslant k \leqslant n... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 726,991 |
3. In $\triangle A B C$, it is known that $B_{2}$ is the point symmetric to the excenter $B_{1}$ of side $A C$ with respect to the midpoint of $A C$, and $C_{2}$ is the point symmetric to the excenter $C_{1}$ of side $A B$ with respect to the midpoint of $A B$. The excircle of side $B C$ touches side $B C$ at point $D$... | 3. Auxiliary lines as shown in Figure 2. Let the excenter of side $B C$ be $A_{1}$.
By the properties of the excenter, $B_{1} 、 A 、 C_{1}, A_{1} 、 C 、 B_{1}, C_{1} 、 B$ 、 $A_{1}$ are collinear in threes, and $A_{1} A \perp B_{1} C_{1}$.
Take a point $P$ on the plane such that $\overrightarrow{C_{2} P}=\overrightarrow... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,992 |
4. Arrange $n(n \geqslant 2)$ coins in a row. If there is a coin with the head side up, then one can select one of them, and flip the consecutive odd number of coins (which can be one) starting from this coin from left to right simultaneously (flipping means turning a head-up coin to head-down, or a head-down coin to h... | 4. For each arrangement of the coins, define the corresponding sequence $c_{1} c_{2} \cdots c_{n}$: if the $i$-th coin from left to right is heads up, then $c_{i}=1$; otherwise, $c_{i}=0$.
Thus, the initial sequence is $\underbrace{11 \cdots 1}_{n \uparrow}$.
Each time, consider the first 1 from right to left, and star... | a_{n}=\left[\frac{2^{n+1}}{3}\right] | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,993 |
5. If a non-empty set $A \subseteq\{1,2, \cdots, n\}$ satisfies $|A|$ $\leqslant \min _{x \in A} x$, then $A$ is called a "good set of level $n$", where $|A|$ denotes the number of elements in set $A$, and $\min _{x \in A} x$ denotes the smallest element in set $A$. Let $a_{n}$ be the number of good sets of level $n$. ... | 5. Proof 1 Let $A$ be a good set of level $n$, and $|A| = k$. Then by the given condition, $\min _{x \in A} x \geqslant k$. Therefore,
$$
A \subseteq\{k, k+1, \cdots, n\} \text {. }
$$
Thus, the number of good sets of level $n$ with $|A|=k$ is $\mathrm{C}_{n-k+1}^{k}$.
Hence, $a_{n}=\sum_{k=1}^{\left[\frac{n+1}{2}\rig... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 726,994 |
6. As shown in Figure $1, P A$ and $P B$ are tangents to $\odot O$, point $C$ is on the minor arc $\overparen{A B}$ (different from points $A, B$), a line $l$ perpendicular to $P C$ is drawn through point $C$, intersecting the angle bisector of $\angle A O C$ at point $D$, and the angle bisector of $\angle B O C$ at po... | 6. Since point $C$ does not coincide with $A$, and $P C \perp C D$, $P A \perp A O$, it follows that $C D$ and $A O$ are not parallel. Therefore, let the line $l$ intersect $O A$ at point $M$. Then $P, C, A, M$ are concyclic (Figure 3 shows the case where point $M$ is on the extension of $O A$; if point $M$ coincides w... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 726,995 |
Example 7 Let $S=\{1,2, \cdots, 98\}$. Find the smallest positive integer $n$, such that in any $n$-element subset of $S$, one can always select 10 numbers, and no matter how these 10 numbers are divided into two groups, there is always one group in which there is a number that is coprime with the other four numbers, a... | The subset of 49 even numbers in set $S$ obviously does not satisfy the coprime condition, hence $n \geqslant 50$.
To prove that any 50-element subset $T$ of set $S$ contains 10 numbers that satisfy the condition, we use the following facts.
Lemma 1 If nine elements in $T$ have a common factor greater than 1, and the... | 50 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 726,996 |
7. Label the $n$ sides of a regular $n(n \geqslant 4)$-sided polygon in a clockwise direction with $1,2, \cdots, n$. Find all integers $n$ such that the $n$-sided polygon can be divided into $n-2$ triangular regions using $n-3$ non-intersecting diagonals inside the polygon, and each of these $n-3$ diagonals is labeled ... | 7. First, prove: if $n$ satisfies the condition, then $n \not \equiv 2(\bmod 4)$.
If a positive integer $n$ satisfies the condition, let the sum of the integers marked on the three sides of each triangle be $S$, and the sum of the integers marked on all $n-3$ diagonals be $m$. Summing the sums of the numbers marked on ... | n \neq 2(\bmod 4) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 726,997 |
8. Find all positive integers $a$ such that for any positive integer $n \geqslant 5$, we have $\left(2^{n}-n^{2}\right) \mid\left(a^{n}-n^{a}\right)$. | 8. Solution 1 If $2^{a}-a^{2} \neq 0$, then take a prime number
$$
p>\max \left\{\left|2^{a}-a^{2}\right|, 2\right\} \text {, and }(p, a)=1 \text {, }
$$
and take a positive integer $n=p(p-1)+2$.
Since $p \geqslant 3$, therefore, $n \geqslant 8>5$.
By Fermat's Little Theorem, we have
$$
\begin{array}{l}
2^{n}-n^{2} \e... | 2, 4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 726,998 |
1. Given the equation about $x$
$$
(1-2 k) x^{2}-2 \sqrt{k+1} x-1=0
$$
has real roots. Then the range of values for $k$ is ( ).
(A) $k \geqslant 2$
(B) $k \geqslant-1$, and $k \neq \frac{1}{2}$
(C) $-1 \leqslant k \leqslant 2$, and $k \neq \frac{1}{2}$
(D) $-1 \leqslant k \leqslant 2$ | - 1. D.
When $1-2 k=0 \Rightarrow k=\frac{1}{2}$, the original equation must have one real root.
When $k \neq \frac{1}{2}$, by $k+1 \geqslant 0$, and the discriminant of the original equation $\Delta=(-2 \sqrt{k+1})^{2}+4(1-2 k) \geqslant 0$, solving yields $-1 \leqslant k \leqslant 2$, and $k \neq \frac{1}{2}$.
In s... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 726,999 |
2. A mall deals in a certain commodity. Due to the current purchase price being $6.4 \%$ lower than the original purchase price, the profit margin has increased by 8 percentage points. Therefore, the current profit margin for dealing in this commodity is ( ).
(A) $23 \%$
(B) $25 \%$
(C) $27 \%$
(D) $29 \%$ | 2. B.
Let the original purchase price of this product be unit "1". Then the current purchase price is $(1-6.4\%)$. Let the original profit margin for dealing with this product be $x$. Then
$$
(1+x) \times 1=(1+x+8\%)(1-6.4\%) \text{. }
$$
Solving this, we get $1+x=1.17$, which means $x=0.17=17\%$. Therefore, the prof... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,000 |
3. As shown in Figure 1, in the right triangle $\triangle ABC$, $AC=3, BC=$ $4, D$ is a moving point on the hypotenuse $AB$, $DE \perp BC, DF \perp$ $AC$, with the feet of the perpendiculars being $E$ and $F$ respectively. When the length of segment $EF$ is minimized, $\cos \angle EFD=(\quad$.$) .$
(A) $\frac{3}{5}$
(B... | 3. A.
Connect $C D$. It is easy to know that quadrilateral $D E C F$ is a rectangle.
Therefore, $E F=C D$.
To make the length of line segment $E F$ the smallest, $C D \perp A B$ is required. By the Pythagorean theorem, $A B=5$.
Thus, $\cos \angle E F D=\cos \angle B C D$ $=\cos \angle B A C=\frac{3}{5}$. | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 727,001 |
4. Given the parabola $y=-x^{2}+x-m$ intersects the $x$-axis at two different points, and the sum of the distances from these two points to the origin does not exceed 5. Then the number of integer values of $m$ is ( ).
(A) 7
(B) 6
(C) 5
(D) 4 | 4. A.
Let the x-coordinates of the two distinct intersection points of the parabola with the x-axis be \( x_{1} \) and \( x_{2} \). Then
$$
\Delta=1-4 m>0 \Rightarrow m<\frac{1}{4} \text {, }
$$
and
$$
\begin{array}{l}
\left(\left|x_{1}\right|+\left|x_{2}\right|\right)^{2} \\
=\left(x_{1}+x_{2}\right)^{2}+2\left|x_{1... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,002 |
5. Given that $x$ is a real number, and
$$
|3 x-1|+|4 x-1|+\cdots+|17 x-1|
$$
is a definite constant. Then this constant is ( ).
(A) 75
(B) 15
(C) 10
(D) 5 | 5. D.
From the problem, we know that after removing the absolute value symbols, $x$ should be eliminated.
And $3+4+\cdots+12=13+14+\cdots+17$, so the original expression is
$$
\begin{aligned}
= & (1-3 x)+(1-4 x)+\cdots+(1-12 x)+ \\
& (13 x-1)+(14 x-1)+\cdots+(17 x-1) \\
= & 5 .
\end{aligned}
$$ | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,003 |
6. As shown in Figure 2, in the right trapezoid $A B C D$, $\angle B=$ $\angle C=90^{\circ}, A B=B C$, point $E$ is on side $B C$, and makes $\triangle A D E$ an equilateral triangle. Then the ratio of the area of $\triangle A D E$ to the area of trapezoid $A B C D$ is ( ).
(A) $\sqrt{3}-1$
(B) $\frac{\sqrt{3}}{2}$
(C)... | 6. D.
As shown in Figure 6, extend trapezoid $ABCD$ to form square $ABCF$, with side length 1, and let $CD = CE = x$. Then,
$$
BE = FD = 1 - x.
$$
By the Pythagorean theorem, we have $2x^2 = 1 + (1 - x)^2$.
Solving for $x$ gives $x = \sqrt{3} - 1$.
$$
\text{Therefore, } \frac{S_{\triangle ADE}}{S_{\text{trapezoid } A... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 727,004 |
1. The number of positive integers $n$ such that $n+1$ divides $n^{2012}+2012$ is $\qquad$ . | From the problem, we know
$$
\begin{array}{l}
n^{2012}+2012 \equiv(-1)^{2012}+2012 \\
=2013 \equiv 0(\bmod n+1) .
\end{array}
$$
Since $2013=3 \times 11 \times 61$, $n$ has $2^{3}-1=7$ solutions. | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,005 |
1. Let the function $y=f(x)$ be continuous and non-negative on $[a, b]$, and the sum $\sum_{a<t \leqslant b}[f(t)]$ (where $t$ is an integer in $[a, b]$) represents the number of integer points in the plane region $a<x \leqslant b, 0<y \leqslant f(x)$. Prove: If $(p, q)=1$, then the number of integer points in the rect... | Prompt: As shown in Figure 2, calculate the number of elements in
$$
\begin{aligned}
M= & \left\{p x-q y \mid x=1,2, \cdots, \frac{q-1}{2} ;\right. \\
& \left.y=1,2, \cdots, \frac{p-1}{2}\right\}
\end{aligned}
$$
in two ways.
On one hand, when $x$ runs through $1,2, \cdots, \frac{q-1}{2}$ and $y$ runs through $1, 2, \c... | \frac{p-1}{2} \cdot \frac{q-1}{2} | Number Theory | proof | Yes | Yes | cn_contest | false | 727,007 |
3. As shown in Figure 4, points $A$ and $D$ are on the semicircle with diameter $BC$, and $D$ is the midpoint of arc $\overparen{AC}$. $AC$ intersects $BD$ at point $E$. If $AE=3, CD=2\sqrt{5}$, then $AB=$ $\qquad$ ,$BC=$ $\qquad$ | 3. 6, 10.
As shown in Figure 7, extend $BA$ and $CD$ to intersect at point $F$.
It is easy to prove that Rt $\triangle CDE \sim$ Rt $\triangle CAF$.
Thus, $\frac{CE}{CF} = \frac{CD}{AC}$
$$
\begin{array}{l}
\Rightarrow \frac{CE}{4 \sqrt{5}} = \frac{2 \sqrt{5}}{CE + 3} \\
\Rightarrow CE = 5 .
\end{array}
$$
By the Py... | 6, 10 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,008 |
4. Given that $P(-2,3)$ is a point on the graph of the inverse proportion function $y=\frac{k}{x}$, $Q$ is a moving point on the branch of the hyperbola in the fourth quadrant. A line is drawn through point $Q$ such that it intersects the hyperbola $y=\frac{k}{x}$ at only one point, and intersects the $x$-axis and $y$-... | 4. 48.
It is known that $k=-6, y=-\frac{6}{x}, A(-4,0), B(0,6)$.
Let the line passing through point $Q$ be $y=a x+b$.
Then $\left\{\begin{array}{l}y=a x+b, \\ y=-\frac{6}{x}\end{array}\right.$ has only one solution.
Eliminating and rearranging gives $a x^{2}+b x+6=0$.
By $\Delta=b^{2}-24 a=0 \Rightarrow b^{2}=24 a$.
T... | 48 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,009 |
One. (20 points) As shown in Figure 5, $P$ is a point outside circle $\odot O$, $PA$ is tangent to $\odot O$ at point $A$, and $PBC$ is a secant of $\odot O$. $AD \perp PO$ at point $D$. If $PB=4$, $CD=5$, $BC=6$, find the length of $BD$. | Connect $O A, O B, O C$. Then $O B=O C$. It is easy to see that
$$
\begin{array}{l}
P A \perp O A, \\
P A^{2}=P B \cdot P C=P D \cdot P O \\
\Rightarrow \frac{P O}{P C}=\frac{P B}{P D} .
\end{array}
$$
Since $\angle B P D$ is a common angle, we have
$$
\begin{array}{l}
\triangle P O B \sim \triangle P C D . \\
\text {... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,010 |
II. (25 points) Does there exist a prime number $p$ such that the equation
$$
x^{2}-4 p(x-p)-5 p-1=0
$$
has two integer roots? If it exists, find all possible values of $p$ and the roots of the equation; if not, explain why. | There exists a prime number $p$ that satisfies the condition. The reason is as follows.
The original equation is equivalent to $(x-2 p)^{2}=5 p+1$. According to the problem, $5 p+1$ must be a perfect square. Let $5 p+1=n^{2}$.
Note that, $p \geqslant 2$. Then $n \geqslant 4$, and $n$ is an integer. Therefore, $5 p=n^{... | p=3 \text{ or } 7, \text{ with roots } (2, 10) \text{ and } (8, 20) \text{ respectively} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,011 |
Three. (25 points) Let $k$, $x$, $y$ be positive integers, and the quotients when $k$ is divided by $x^{2}$ and $y^{2}$ are $n$ and $n+148$, respectively.
(1) If $(x, y)=1$, prove: $x^{2}-y^{2}$ is coprime with $x^{2}$ and $y^{2}$;
(2) If $(x, y)=1$, find the value of $k$;
(3) If $(x, y)=4$, find the value of $k$. | (1) Let $\left(x^{2}-y^{2}, x^{2}\right)=t$. Then $x^{2}-y^{2}=t p, x^{2}=t q\left(p, q \in \mathbf{N}_{+}\right)$.
Thus, $y^{2}=t(q-p) \Rightarrow t \mid y^{2}$.
Since $(x, y)=1$, it follows that $\left(x^{2}, y^{2}\right)=1$.
Therefore, $t=1$.
Hence, $x^{2}-y^{2}$ and $x^{2}$ are coprime.
Similarly, $x^{2}-y^{2}$ and... | 2^{8} \times 3^{4} \times 19^{2} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,012 |
2. Given vectors $\boldsymbol{a} 、 \boldsymbol{b}$ satisfy $|\boldsymbol{a}|=1,|\boldsymbol{b}|=\sqrt{3}$, and $(3 a-2 b) \perp a$. Then the angle between $a 、 b$ is $\qquad$ | 2. $\frac{\pi}{6}$.
Let the required angle be $\theta$. Then
$$
0=(3 a-2 b) \cdot a=3|a|^{2}-2|a||b| \cos \theta \text {. }
$$
Combining $|a|=1,|b|=\sqrt{3}$, we have
$$
\cos \theta=\frac{3}{2} \cdot \frac{|a|}{|b|}=\frac{\sqrt{3}}{2} \Rightarrow \theta=\frac{\pi}{6} .
$$ | \frac{\pi}{6} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,014 |
3. For each positive integer $n$, let the tangent line to the curve $y=x^{n+1}$ at the point $(1,1)$ intersect the $x$-axis at a point with abscissa $x_{n}$. Let $a_{n}=\lg x_{n}$. Then
$$
a_{1}+a_{2}+\cdots+a_{99}=
$$
$\qquad$ | 3. -2 .
Given $y=x^{n+1}$, we know $y^{\prime}=(n+1) x^{n}$.
Then according to the condition, we have
$$
\begin{array}{l}
\frac{0-1}{x_{n}-1}=y^{\prime}(1)=n+1 \Rightarrow x_{n}=\frac{n}{n+1} . \\
\text { Therefore, } a_{1}+a_{2}+\cdots+a_{99}=\lg \left(x_{1} x_{2} \cdots x_{99}\right) \\
=\lg \left(\frac{1}{2} \times... | -2 | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 727,015 |
4. In the range of real numbers, the solution set of the equation
$$
\left(x^{2}-x-1\right)^{x^{3}+2 x^{2}-9 x-18}=1
$$
is $\qquad$ . | 4. $\{0, \pm 1, \pm 2, \pm 3\}$.
Discuss three possible cases.
(1) $x^{2}-x-1=1$, solving gives $x=-1,2$.
(2) $x^{2}-x-1=-1$, solving gives $x=0,1$, in this case, the exponent $x^{3}+2 x^{2}-9 x-18$ is even, so $x=0,1$ are both solutions to the original equation.
(3) $x^{3}+2 x^{2}-9 x-18=0$, and $x^{2}-x-1 \neq 0$, s... | \{0, \pm 1, \pm 2, \pm 3\} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,016 |
5. In the Cartesian coordinate system, line $l$ passes through one of the foci of the hyperbola $x^{2}-y^{2}=1$ and intersects the hyperbola at points $A$ and $B$. If the circle with $A B$ as its diameter is tangent to the $y$-axis, then $|A B|=$ $\qquad$ . | $5.2+2 \sqrt{2}$.
It is known that the two foci of the hyperbola $x^{2}-y^{2}=1$ are $( \pm \sqrt{2}, 0)$.
Without loss of generality, assume the line $l$ passes through the right focus $(\sqrt{2}, 0)$.
(1) When $A B$ is not perpendicular to the $x$-axis, let the equation of the line $l$ be $y=k(x-\sqrt{2})$. Substitu... | 2+2 \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,017 |
4. Let $S=\{1,2, \cdots, 50\}$. Find the smallest positive integer $k$, such that in any $k$-element subset of $S$, there exist two distinct numbers $a$ and $b$ satisfying $(a+b) \mid a b$.
$(1996$, China Mathematical Olympiad) | Let the greatest common divisor of $a$ and $b$ be $d$, and $a = a_{1}d$, $b = b_{1}d$. Then $\left(a_{1}, b_{1}\right) = 1$.
Substituting into $(a+b) \mid ab$, we get
$$
\left(a_{1} + b_{1}\right) \mid a_{1} b_{1} d \Rightarrow \left(a_{1} + b_{1}\right) \mid d.
$$
Let $d = k\left(a_{1} + b_{1}\right)$, we obtain all ... | 39 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,018 |
6. A racing robot has the following characteristics:
(1) The step length can be manually set to 0.1 meters, 0.2 meters, $\cdots$, 1.8 meters, or 1.9 meters:
(2) After the start command, the robot immediately takes the first step of the set length, and each step is completed instantaneously;
(3) When the step length is ... | 6.48.6 seconds.
Let $\lceil x$ denote the smallest integer not less than the real number $x$.
Let the step length be $a$ meters, $a \in\{0.1,0.2, \cdots, 1.9\}$. The robot takes $\left[\frac{50}{a}\right]$ steps to complete exactly 50 meters, and the number of intervals required is $\left\lceil\frac{50}{a}\right\rceil... | 48.6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,019 |
7. Xiao Zhang, Xiao Li, Xiao Hua, and Xiao Ming are playing a game of taking turns to roll a standard die. If one person rolls the smallest number, and no one ties with him, he is declared the winner; if more than one person rolls the smallest number, those who did not roll the smallest number are eliminated first, and... | 7. $\frac{175}{864}$.
Consider the four possible scenarios in the first round.
(1) Xiao Zhang wins. The probability of this is
$$
P_{1}=\left(\frac{3}{6}\right)^{3}=\frac{1}{8} \text {. }
$$
(2) Xiao Zhang ties with one other person. The probability of this is
$$
\mathrm{C}_{3}^{1} \times \frac{1}{6} \times\left(\frac... | \frac{175}{864} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,020 |
8. A grain silo can be roughly regarded as a frustum of a cone, with the radius of the upper base being 3 meters, the radius of the lower base being 6 meters, and the height being $\sqrt{7}$ meters. A well-fed mouse is exercising and plans to start from point $A$ on the circumference of the lower base, run around the s... | 8. $8 \sqrt{3}+\frac{10 \pi}{3}$.
Unfolding the lateral surface of the frustum, we get a part of an annulus, as shown in Figure 2. In the unfolded diagram, arcs $\overparen{A D}$ and $\overparen{E F}$ correspond to the lower and upper bases of the frustum, respectively, and point $A(D)$ is the starting point of the mo... | 8 \sqrt{3}+\frac{10 \pi}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,021 |
9. (16 points) Given a positive integer $n(n \geqslant 3)$. It is known that $a_{1}<a_{2}<\cdots<a_{n}$, and $(k+1) a_{k} \in A(k=1,2$, $\left.\cdots, n-1, A=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}\right)$. Find all ordered tuples $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ that satisfy the conditions. | 9. First, $a_{1} \geqslant 0$. Otherwise, if $a_{1}<0$, then $2 a_{1}<a_{1}$. Therefore, it must be that
$$
a_{2}=2 a_{1}, a_{3}=3 a_{2}, \cdots, a_{n}=n a_{n-1} \text {. }
$$
Let $a_{1}=c$, then for $k \in\{2,3, \cdots, n\}$, we have
$$
a_{k}=k a_{k-1}=k(k-1) a_{k-2}=\cdots=k!a_{1}=k!c \text {. }
$$
(2) If $a_{1}=0$,... | (c, 2!c, \cdots, n!c) \text { or }\left(0, \frac{2!d}{2}, \frac{3!d}{2}, \cdots, \frac{n!d}{2}\right) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,022 |
10. (20 points) Given real numbers $x, y$ satisfy
$$
2^{x}+2^{y}=4^{x}+4^{y} \text {. }
$$
Try to find the range of values for $U=8^{x}+8^{y}$. | 10. Let $a=2^{x}, b=2^{y}$. Then the given condition can be rewritten as
$$
a+b=a^{2}+b^{2} \quad (a, b>0).
$$
Completing the square, we get
$$
\left(a-\frac{1}{2}\right)^{2}+\left(b-\frac{1}{2}\right)^{2}=\left(\frac{\sqrt{2}}{2}\right)^{2}.
$$
Observing the image of $(a, b)$ satisfying equation (1) in the rectangul... | (1,2] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,023 |
11. (20 points) Let the ellipse $C$ be:
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{20}=1(a>2 \sqrt{5})
$$
with its left focus at $F$, and point $P(1,1)$. It is known that there exists a line $l$ passing through point $P$ intersecting the ellipse at points $A$ and $B$, with $M$ being the midpoint of $A B$, such that $|F M|$ i... | 11. By the median length formula, we have
$|F M|^{2}=\frac{1}{2}\left(|F A|^{2}+|F B|^{2}\right)-\frac{1}{4}|A B|^{2}$.
Also, $|F M|^{2}=|F A||F B|$, so
$|A B|^{2}=2(|F A|-|F B|)^{2}$.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$, and
$c=\sqrt{a^{2}-20}$.
Then $|F A|-|F B|=\frac{c}{a} x_{1}-\frac{c}{a} ... | 7 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,024 |
一、(40 Points) As shown in Figure 1, in the acute triangle $\triangle ABC$, points $D$, $E$, and $F$ are on sides $BC$, $CA$, and $AB$ respectively, satisfying $DB = DF$ and $DC = DE$. Let $H$ be the orthocenter of $\triangle ABC$. Prove that points $A$, $E$, $H$, and $F$ are concyclic. | Extend $B H$ and $C H$ to intersect $A C$ and $A B$ at points $K$ and $L$, respectively. Draw $D M \perp A C$ at point $M$ and $D N \perp A B$ at point $N$, and connect $H E$ and $H F$.
It is easy to see that $M$ and $N$ are the midpoints of $C E$ and $B F$, respectively, and $D M \parallel B K, D N \parallel C L$.
$$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,025 |
II. (40 points) Given that $a$, $b$, and $c$ are positive integers, and $a(a+b)(a+bc)$ is a power of a prime. Prove that $1+\frac{bc}{a}$ must be a positive integer power of $1+\frac{b}{2}$. | Let $a(a+b)(a+bc)$ be a power of a prime $p$. Note that, $1 \leqslant a < a+b \leqslant a+bc$.
Thus, we can set
$$
a=p^{k}, a+b=p^{k+m}, a+bc=p^{k+n},
$$
where $k \in \mathbf{N}, m, n \in \mathbf{N}_{+}, m \leqslant n$.
Therefore, $b=p^{k}(p^{m}-1), bc=p^{k}(p^{n}-1)$.
Thus, $c=\frac{p^{n}-1}{p^{m}-1}$, which means $(... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 727,026 |
Three. (50 points) Remove a $2 \times 2$ small square from any corner of a $3 \times 3$ grid, and the remaining shape is called a "corner shape". Now, place some non-overlapping corner shapes in a $9 \times 9$ grid, with the requirement that the boundaries of the corner shapes coincide with the boundaries or grid lines... | Three, first, $k_{\max }<6$. This is because, if 6 corner shapes are placed in the manner shown in Figure 4, it is impossible to place another complete corner shape on this grid.
Now, place 5 corner shapes in any manner.
Next, we will prove that it is still possible to place another complete corner shape.
Consider th... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,027 |
Four. (50 points) Let
$$
f(x)=(x-x_{1})(x-x_{2}) \cdots(x-x_{n}),
$$
where $x_{1}, x_{2}, \cdots, x_{n} \in[-1,1]$.
Prove: For any $a \in(-1,0), b \in(0,1)$, we have $\min \{|f(a)|,|f(b)|\}<1$. | Four, using proof by contradiction. Assume there exist $a \in(-1,0)$, $b \in(0,1)$, such that
$$
\begin{array}{l}
|f(a)| \geqslant 1,|f(b)| \geqslant 1 . \\
\text { Let } g(x)=\sum_{k=1}^{n}\left|x-x_{k}\right| .
\end{array}
$$
Since $g(x)$ is (non-strictly) convex on $\mathbf{R}$, we have
$$
\left\{\begin{array}{l}
g... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 727,028 |
Example 1 Given real numbers $a, b, c, d$ satisfy: for any real number $x$,
$a \cos x + b \cos 2x + c \cos 3x + d \cos 4x \leqslant 1$.
Find the maximum value of $a + b - c + d$ and the values of the real numbers $a, b, c, d$ at this time.
(Ninth China Southeast Mathematical Olympiad) | Let
$$
\begin{array}{l}
f(x)=a \cos x+b \cos 2 x+c \cos 3 x+d \cos 4 x . \\
\text { By } f(0)=a+b+c+d, \\
f(\pi)=-a+b-c+d, \\
f\left(\frac{\pi}{3}\right)=\frac{a}{2}-\frac{b}{2}-c-\frac{d}{2},
\end{array}
$$
then \(a+b-c+d\)
$$
=f(0)+\frac{2}{3} f(\pi)+\frac{4}{3} f\left(\frac{\pi}{3}\right) \leqslant 3 .
$$
Equality... | 3 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 727,029 |
In rhombus $A B C D$, $\angle A=60^{\circ}, E$ is a point on segment $C D$, extend $A E$, and intersect the extension of $B C$ at point $F$, connect $F D$, extend $B E$, and intersect $D F$ at point $G$. If $B G=a$, try to express the area of quadrilateral $B C G D$ in terms of $a$. | Solve As shown in Figure 2, draw $D H \perp B E$ at point $H$, $C K \perp$ $B G$ at point $K$, and take point $M$ on $B G$ such that $D M=D G$.
It is easy to know that $\triangle A B D$ is an equilateral triangle.
From $A D / / B F \Rightarrow \triangle A D E \backsim \triangle F C E$ $\Rightarrow \frac{F C}{C E}=\frac... | \frac{\sqrt{3}}{4} a^{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,030 |
Given 352 As shown in Figure $3, \odot O_{1}$ is internally tangent to $\odot O_{2}$ at point $P$, points $A, B$ are on $\odot O_{1}$, lines $A C, B D$ are tangent to $\odot O_{2}$ at points $C, D$ respectively, and line $A B$ intersects $C D$ at point $E$. Prove:
(1) $\frac{A C}{B D}=\frac{A P}{B P}$;
(2) $P E$ bisect... | Proof (1) As shown in Figure 4, let $P A$ and $P B$ intersect $\odot O_{2}$ at points $M$ and $N$, respectively, and connect $M N$.
It is easy to see that $M N \parallel A B$. Therefore, $\frac{A M}{B N}=\frac{A P}{B P}$.
By the secant-tangent theorem, we have
$$
\frac{A C^{2}}{B D^{2}}=\frac{A M \cdot A P}{B N \cdot B... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,031 |
Given $a_{1}, a_{2}, \cdots, a_{n}>0$. Prove:
$$
\sum_{k=1}^{n} \frac{k^{2}}{a_{1}^{2}+a_{2}^{2}+\cdots+a_{k}^{2}} \leqslant\left(\sum_{k=1}^{n} \frac{1}{a_{k}}\right)^{2} .
$$ | Prove that when $n=1$, the equality of the inequality holds.
When $n \geqslant 2$, by the arithmetic-geometric mean inequality, we have $\frac{4}{a_{1}^{2}+a_{2}^{2}} \leqslant \frac{2}{a_{1} a_{2}}$.
In general,
$$
\begin{array}{l}
\frac{(2 m+1)^{2}}{a_{1}^{2}+a_{2}^{2}+\cdots+a_{2 m+1}^{2}} \\
\leqslant \frac{2 m+1}{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 727,032 |
A building has $2^{10}$ floors, the $i$-th floor has $i$ $\left(i=1,2, \cdots, 2^{10}\right)$ windows, and the top floor has $2^{9}+2013$ windows open. From a distance, all the windows form an inverted Pascal's triangle, and more specifically, if two adjacent windows on the same floor are both open (or both closed), th... | Lemma $a_{1,1}=\left(a_{n, 1}\right)^{\mathrm{C}_{n-1}^{0}}\left(a_{n, 2}\right)^{\mathrm{C}_{n-1}^{1}} \cdots\left(a_{n, n}\right)^{\mathrm{C}_{n-1}^{n-1}}$.
Proof When $n=1$, the conclusion is obviously true.
Assume that when $n=k-1$, the conclusion holds.
Then when $n=k$, by the induction hypothesis, we have
$$
\beg... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 727,033 |
Example 1 As shown in Figure 1, in quadrilateral $A B C D$, $A C$ and $B D$ are diagonals, $\triangle A B C$ is an equilateral triangle, $\angle A D C=30^{\circ}, A D=3, B D=$ 5. Then the length of $C D$ is ( ). ${ }^{[1]}$
(A) $3 \sqrt{2}$
(B) 4
(C) $2 \sqrt{5}$
(D) 4.5 | Solve As shown in Figure 2, rotate $C D$ clockwise around point $C$ by $60^{\circ}$ to get $C E$, and connect $D E$ and $A E$. Then $\triangle C D E$ is an equilateral triangle.
Since $A C=B C$,
$$
\begin{aligned}
& \angle B C D \\
= & \angle B C A+\angle A C D \\
= & \angle D C E+\angle A C D=\angle A C E,
\end{aligne... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 727,034 |
Example 2 In an isosceles right $\triangle ABC$, it is known that $\angle ACB=$ $90^{\circ}, P$ is a point inside $\triangle ABC$, such that $PA=11, PB=7, PC=$ 6. Then the length of side $AC$ is $\qquad$ [2] | As shown in Figure 3, $\triangle P B C$ is rotated $90^{\circ}$ clockwise around point $C$, making point $B$ coincide with $A$, and point $P$ moves to position $P^{\prime}$, connecting $P P^{\prime}$.
Then $\triangle C P P^{\prime}$ is an isosceles right triangle. Thus,
$$
P P^{\prime}=\sqrt{P C^{2}+P^{\prime} C^{2}}=... | \sqrt{85+42 \sqrt{2}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,035 |
Example 3 Given positive integers $a_{1}, a_{2}, \cdots, a_{n}$ are all composite numbers, and they are pairwise coprime. Prove:
$$
\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}<\frac{1}{2} \text {. }
$$ | Proof: Let the smallest prime factor of $a_{k}$ be $p_{k}(k=1,2, \cdots, n)$. Since $a_{k}$ is not a prime, we have $a_{k} \geqslant p_{k}^{2}$.
Notice that $a_{1}, a_{2}, \cdots, a_{n}$ are pairwise coprime.
Thus, $p_{1}, p_{2}, \cdots, p_{n}$ are all distinct.
$$
\begin{array}{l}
\text { Hence } \sum_{k=1}^{n} \frac{... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 727,036 |
5. Given that line $A B$ intersects the parabola $y^{2}=4 x$ at points $A$ and $B$, $M$ is the midpoint of $A B$, and $C$ is a moving point on the parabola. If point $C_{0}$ satisfies
$$
\overrightarrow{C_{0} A} \cdot \overrightarrow{C_{0} B}=\min \{\overrightarrow{C A} \cdot \overrightarrow{C B}\} \text {, }
$$
then ... | 5. B.
Notice,
$$
\begin{array}{l}
\overrightarrow{C A} \cdot \overrightarrow{C B}=(\overrightarrow{C M}-\overrightarrow{A M}) \cdot(\overrightarrow{C M}-\overrightarrow{B M}) \\
=|\overrightarrow{C M}|^{2}-\overrightarrow{C M} \cdot(\overrightarrow{A M}+\overrightarrow{B M})+\overrightarrow{A M} \cdot \overrightarrow{... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 727,037 |
6. If the three-digit number $\overline{b b c}$ is divisible by 7, and $a$, $b$, $c$ form an arithmetic sequence with a non-zero common difference, then the number of such integers is ( ).
(A) 4
(B) 6
(C) 7
(D) 8 | 6. D.
Let the three-digit number that satisfies the condition be
$$
\overline{(b-d) b(b+d)}=111 b-99 d \text {, }
$$
where, $0 \leqslant b<9,-9<d<9, d \neq 0$.
$$
\begin{array}{l}
\text { By } 7 \mid(111 b-99 d) \Rightarrow 7 \text { divides } (b+d) \\
\Rightarrow(b, d)=(1,-1),(2,-2),(3,-3), \\
(4,3),(4,-4),(5,2),(6... | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 727,038 |
9. Let the function $f(x)=x(x-1)^{2}(x-2)^{3}(x-3)^{4}$. Then the point of local maximum of the function $y=f(x)$ is ( ).
(A) $x=0$
(B) $x=1$
(C) $x=2$
(D) $x=3$ | 9. B.
From the graph, we know that $x=1$ is a point of maximum value for the function, $x=3$ is a point of minimum value, and $x=0,2$ are not points of extremum. | B | Calculus | MCQ | Yes | Yes | cn_contest | false | 727,041 |
10. Given that $f(x)$, $g(x)$, and $h(x)$ are all linear functions. If for real numbers $x$ the following holds:
$$
\begin{array}{l}
|f(x)|-|g(x)|+h(x) \\
=\left\{\begin{array}{ll}
-1, & x<-1 ; \\
3 x+2, & -1 \leqslant x<0 ; \\
-2 x+2, & x \geqslant 0,
\end{array}\right.
\end{array}
$$
then $h(x)=(\quad)$.
(A) $x-\fra... | 10. C.
$$
h(x)=\frac{-2 x+2+(-1)}{2}=-x+\frac{1}{2} .
$$ | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,042 |
11. If $\tan x \cdot \tan y=2, \sin x \cdot \sin y=\frac{1}{3}$, then $x-y$ $=$ | $$
\text { II, } 11.2 k \pi \pm \frac{\pi}{3} \text {. }
$$
Given $\tan x \cdot \tan y=2, \sin x \cdot \sin y=\frac{1}{3}$
$$
\begin{array}{l}
\Rightarrow \cos x \cdot \cos y=\frac{1}{6} \\
\Rightarrow \cos (x-y)=\frac{1}{2} \\
\Rightarrow x-y=2 k \pi \pm \frac{\pi}{3} .
\end{array}
$$ | x-y=2 k \pi \pm \frac{\pi}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,043 |
12. Let $f(x)=x^{2}-(k+1) x+2$. If for $x>0$, $f(x)$ is always greater than zero, then the range of values for $k$ is $\qquad$. | 12. $(-\infty, 2 \sqrt{2}-1)$.
From $x^{2}-(k+1) x+2>0 \Rightarrow k+1<x+\frac{2}{x}$.
Also, $x+\frac{2}{x} \geqslant 2 \sqrt{2}$, equality holds when $x=\sqrt{2}$, at this point, $k<2 \sqrt{2}-1$. | (-\infty, 2 \sqrt{2}-1) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,044 |
13. Given the sequence $\{\sqrt[n]{n}\}\left(n \in \mathbf{Z}_{+}\right)$. Then the value of the largest term in the sequence is $\qquad$ . | 13. $\sqrt[3]{3}$.
Let $f(x)=x^{\frac{1}{x}}=\mathrm{e}^{\frac{1}{x} \ln x}$. Then $f^{\prime}(x)=\frac{x^{\frac{1}{x}}}{x^{2}}(1-\ln x)$.
Solving, we find that $x=\mathrm{e}$ is a point of maximum.
Thus, the largest term in the sequence is the third term, which is $\sqrt[3]{3}$. | \sqrt[3]{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,045 |
14. If $x, y \in \mathbf{R}$ satisfy
$$
2 x-2 x^{2} y^{2}-2 y\left(x+x^{2}\right)-x^{2}=5 \text {, }
$$
then $(x, y)=$ $\qquad$ | 14. $\left(3,-\frac{2}{3}\right)$.
If we consider the equation in the problem as a quadratic equation in $x$, then
$$
\begin{array}{l}
\Delta=4(y-1)^{2}-20\left(2 y^{2}+2 y+1\right) \geqslant 0 \\
\Rightarrow(3 y+2)^{2} \leqslant 0 \\
\Rightarrow y=-\frac{2}{3}, x=3 .
\end{array}
$$ | \left(3,-\frac{2}{3}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,046 |
Example 4 Let $\sigma(n)$ denote the sum of all positive divisors of the positive integer $n$. Prove:
(1) $\frac{\sigma(n)}{n} \leqslant \log _{2} 2 n$;
(2) for any given positive integer $k$, there exist infinitely many $n$ such that $\frac{\sigma(n)}{n}>k .{ }^{[3]}$ | Proof (1) It is easy to see that
$$
\begin{array}{l}
\sigma(n) \leqslant \sum_{j=1}^{n} \frac{n}{j} \\
=n\left[1+\left(\frac{1}{2}+\frac{1}{3}\right)+\cdots+\left(\frac{1}{2^{\left[\log _{2} n\right]-1}}+\cdots+\right.\right. \\
\left.\left.\quad \frac{1}{2^{\left[\log _{2} n\right]}-1}\right)+\left(\frac{1}{2^{\left[\... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 727,047 |
15. Let line $l$ intersect the curve $y=x^{3}+x+1$ at three distinct points $A$, $B$, and $C$, and $|A B|=|B C|=\sqrt{5}$. Then the equation of line $l$ is $\qquad$. | 15. $y=2 x+1$.
Notice that, the curve is symmetric about the point $(0,1)$.
Let the equation of the line be $y=k x+1$, and point $A(x, y)$. Then
$$
\left.\begin{array}{l}
\left\{\begin{array}{l}
y=k x+1, \\
y=x^{3}+x+1, \\
\sqrt{x^{2}+(y-1)^{2}}=\sqrt{5}
\end{array}\right. \\
\Rightarrow(k-2)\left(k^{2}+k+3\right)=0
\... | y=2 x+1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,048 |
16. If $a>0, b>0$, then
$$
\min \left\{\max \left\{a, b, \frac{1}{a^{2}}+\frac{1}{b^{2}}\right\}\right\}=
$$
$\qquad$ | 16. $\sqrt[3]{2}$.
$$
\begin{array}{l}
\text { Let } \max \left\{a, b, \frac{1}{a^{2}}+\frac{1}{b^{2}}\right\}=m \text {. Then } \\
a \leqslant m, b \leqslant m, \frac{1}{a^{2}}+\frac{1}{b^{2}} \leqslant m \\
\Rightarrow m \geqslant \frac{2}{m^{2}} \Rightarrow m \geqslant \sqrt[3]{2} \\
\Rightarrow \min \left\{\max \le... | \sqrt[3]{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 727,049 |
17. A moving point moves on the integer points in the first quadrant of the Cartesian coordinate system (including the integer points on the $x$-axis and $y$-axis of the first quadrant), with the movement rules being $(m, n) \rightarrow(m+1, n+1)$ or $(m, n) \rightarrow$ $(m+1, n-1)$. If the moving point starts from th... | $\begin{array}{l}\text { 17. } 9 \\ C_{6}^{2}-C_{6}^{1}=9\end{array}$
The translation is as follows:
$\begin{array}{l}\text { 17. } 9 \\ C_{6}^{2}-C_{6}^{1}=9\end{array}$
Note: The original text is already in a mathematical format, which is universal and does not require translation. However, if you intended to have... | 9 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,050 |
18. Given the parabola $y^{2}=4 x$, a line passing through point $K$ on the $x$-axis intersects the parabola at points $P$ and $Q$. Prove: There exists a unique point $K$ such that $\frac{1}{|P K|^{2}}+\frac{1}{|K Q|^{2}}$ is a constant, and determine the coordinates of point $K$. | Three, 18. Let point $K(a, 0)$. The equation of the line passing through $K$ is $y=k(x-a)$, intersecting the parabola at points $P\left(x_{1}, y_{1}\right), Q\left(x_{2}, y_{2}\right)$.
Substituting the line equation into the parabola equation, we get
$$
\begin{array}{l}
k^{2} x^{2}-2\left(a k^{2}+2\right) x+a^{2} k^{2... | K(2,0) | Algebra | proof | Yes | Yes | cn_contest | false | 727,051 |
19. Let the quadratic function be
$$
f(x)=a x^{2}+(2 b+1) x-a-2(a, b \in \mathbf{R}, a \neq 0)
$$
have at least one root in $[3,4]$. Find the minimum value of $a^{2}+b^{2}$. | 19. Regarding the given equation as a linear equation in $a$ and $b$:
$$
\left(x^{2}-1\right) a+2 x b+x-2=0 \text {. }
$$
From the fact that the distance from a point $(a, b)$ on the line to the origin is greater than or equal to the distance from the origin to the line, we get:
$$
\sqrt{a^{2}+b^{2}} \geqslant \frac{|... | \frac{1}{100} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,052 |
20. Let $x(x \in \mathbf{N})$ satisfy
$$
\left(\frac{1+x}{x}\right)^{2013}<\frac{2014}{2013} \text {. }
$$
Given that the sequence $a_{1}, a_{2}, \cdots, a_{2013}$ is an arithmetic sequence with a common difference of $x^{2013}$ and the first term $(x+1)^{2} x^{2012}-1$; the sequence $b_{1}, b_{2}, \cdots, b_{2013}$ i... | 20. From the problem, we know
$$
\begin{array}{l}
a_{i}=(x+1)^{2} x^{2012}-1+(i-1) x^{2013}, \\
b_{i}=(x+1) x^{2013}\left(\frac{1+x}{x}\right)^{i-1}=(x+1)^{i} x^{2014-i} .
\end{array}
$$
Thus, \( b_{i+1}-b_{i}=x^{2013}\left(\frac{1+x}{x}\right)^{i} \).
First, we use mathematical induction to prove:
$$
a_{i}-b_{i} \geq... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 727,053 |
21. Let $a, b, c \in \mathbf{R}_{+}$, satisfying $a b + b c + c a \geqslant 3$. Prove:
$$
\sum a^{5} + \sum a^{3}(b^{2} + c^{2}) \geqslant 9,
$$
where, “$\sum$” denotes the cyclic sum. | 21. The original proposition is equivalent to
$$
\left(a^{3}+b^{3}+c^{3}\right)\left(a^{2}+b^{2}+c^{2}\right) \geqslant 9 \text {. }
$$
Also, $\left(a^{3}+b^{3}+c^{3}\right)^{2} \geqslant 9\left(\frac{a^{2}+b^{2}+c^{2}}{3}\right)^{3}$, so it suffices to prove that $a^{2}+b^{2}+c^{2} \geqslant 3$.
Using the given condi... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 727,054 |
22. Choose several different numbers from $0,1, \cdots, 10$ to fill each circle in the diagram, which is called a "filling method". If the absolute values of the differences between the numbers in the circles connected by each line segment are all different, then such a filling method is called a "perfect filling metho... | 22. The perfect filling of Figure 3 is not unique, as shown in Figure 5.
For Figure 4, there is no perfect filling. This is because Figure 4 has a total of 10 lines, so the absolute values of the differences between the numbers at the ends of each line are exactly \(1, 2, \cdots, 10\), and their sum
\[
\begin{aligned}... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,055 |
1. Given $x_{1}, x_{2}, \cdots, x_{2013} \in[-1,1]$. Then the minimum value of $x_{1} x_{2}+x_{2} x_{3}+\cdots+x_{2012} x_{2013}+x_{2013} x_{1}$ is ( ).
(A) -2013
(B) -2012
(C) -2011
(D) -2010 | - 1. C.
If 2012 variables are fixed, the original expression becomes a linear function of one variable. Note that, when the variable takes 1 or -1, the function reaches its maximum value, hence $x_{1}, x_{2}, \cdots, x_{2013} \in\{-1,1\}$.
By trying, at least one of $x_{1} x_{2}, x_{2} x_{3}, \cdots, x_{2013} x_{1}$ ... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,056 |
2. As shown in Figure 1, the side length of the equilateral $\triangle ABC$ is $6$, and $D$, $E$ are points on sides $BC$, $AC$ respectively, such that $CD = AE$. Let $BE$ and $AD$ intersect at point $F$, and connect $CF$. Draw $EG \parallel CF$ intersecting $AD$ at point $G$. If $EF = 1$, then $AG$ $=(\quad)$.
(A) $\f... | 2. C.
Notice,
$$
\begin{array}{l}
\triangle A B E \cong \triangle A C D \Rightarrow \angle E F D=120^{\circ} \\
\Rightarrow E, F, D, C \text { are concyclic } \\
\Rightarrow \angle F D E=\angle F C E=\angle G E A \\
\Rightarrow \triangle A E G \backsim \triangle A D E \\
\Rightarrow A E^{2}=A G \cdot A D . \\
\text {... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 727,057 |
Example 5 Let $n$ be an integer greater than 1, and all positive divisors be $d_{1}, d_{2}, \cdots, d_{k}$, where $1=d_{1}<d_{2}<\cdots<d_{k}=n$, and let $D=d_{1} d_{2}+d_{2} d_{3}+\cdots+d_{k-1} d_{k}$.
(1) Prove: $D<n^{2}$;
(2) Determine all $n$ such that $D \mid n^{2}$.
(4) | (1) Proof Note that, all positive divisors of $n$ from smallest to largest can also be written as $\frac{n}{d_{k}}, \frac{n}{d_{k-1}}, \cdots, \frac{n}{d_{1}}$, hence
$$
\begin{array}{l}
D=d_{k} d_{k-1}+d_{k-1} d_{k-2}+\cdots+d_{2} d_{1}=n^{2} \sum_{i=1}^{k-1} \frac{1}{d_{i} d_{i+1}} \\
\leqslant n^{2} \sum_{i=1}^{k-1}... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 727,058 |
3. Given $x, y>0, x^{2}+y^{2}=24$, $(\sqrt{x}+\sqrt{y})^{4}+(\sqrt{x}-\sqrt{y})^{4}=180$. Then $x y=(\quad)$.
(A) 8
(B) 9
(C) 10
(D) 11 | 3. D.
Notice,
$$
\begin{array}{l}
x^{2}+y^{2} \\
=(\sqrt{x}+\sqrt{y})^{4}-2(2 x \sqrt{x y}+3 x y+2 y \sqrt{x y}), \\
x^{2}+y^{2} \\
=(\sqrt{x}-\sqrt{y})^{4}+2(2 x \sqrt{x y}-3 x y+2 y \sqrt{x y}) .
\end{array}
$$
(1) + (2) gives
$$
\begin{array}{l}
2\left(x^{2}+y^{2}\right)=(\sqrt{x}+\sqrt{y})^{4}+(\sqrt{x}-\sqrt{y})^... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,059 |
4. As shown in Figure 2, the incircle of $\triangle A B C$ touches sides $A B$, $A C$, and $B C$ at points $E$, $F$, and $G$ respectively. If the product of the lengths of the three sides of $\triangle A B C$ is 64, then the maximum value of $A E \cdot B G \cdot C F$ is ( ).
(A) 6
(B) 8
(C) 10
(D) 16 | 4. B.
Let $BC=a$, $AC=b$, $AB=c$. Then $abc=64$, $AE \cdot BG \cdot CF=\frac{a+b-c}{2} \cdot \frac{a+c-b}{2} \cdot \frac{b+c-a}{2}$.
Thus, we only need to prove:
$$
(a+b-c)(a+c-b)(b+c-a) \leqslant abc \text {. }
$$
Let $a+b-c=x$, $a+c-b=y$, $b+c-a=z$.
Then equation (1) becomes $xyz \leqslant \frac{x+y}{2} \cdot \frac... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 727,060 |
5. As shown in Figure 3, in a $4 \times 4$ dot array, the probability of forming a triangle by randomly selecting three points is ( ).
(A) $\frac{129}{140}$
(B) $\frac{67}{70}$
(C) $\frac{27}{28}$
(D) $\frac{41}{42}$ | 5. A.
The number of ways to choose any three points (disregarding order) is $\mathrm{C}_{16}^{3}=560$, and the number of ways that do not form a triangle (i.e., the three points are collinear) is 44. Therefore,
$$
P=\frac{560-44}{560}=\frac{129}{140} .
$$ | A | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 727,061 |
6. If $n$ is a positive integer and satisfies
$$
(n-1)^{2} \mid\left(n^{2013}-1\right) \text {, }
$$
then the number of possible values of $n$ is ( ).
(A) 2
(B) 4
(C) 6
(D) 8 | 6. D.
Notice,
$$
\begin{array}{l}
(n-1)^{2} \mid\left(n^{2013}-1\right) \\
\Rightarrow(n-1) \mid\left(n^{2012}+n^{2011}+\cdots+n^{2}+n+1\right) \\
\Rightarrow(n-1) \mid\left[\left(n^{2012}-1\right)+\left(n^{2011}-1\right)+\cdots+(n-1)+2013\right] \\
\Rightarrow(n-1) \mid 2013 .
\end{array}
$$
Since $2013=3 \times 11 ... | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 727,062 |
1. As shown in Figure $4, \triangle A B C$ has an incircle $\odot O_{1}$ that touches side $B C$ at point $D, \odot O_{2}$ is the excircle of $\triangle A B C$ inside $\angle A$. If $O_{1} B=6, O_{1} C=3, O_{1} D=2$, then $O_{1} O_{2}=$ $\qquad$ | $$
\begin{aligned}
& \angle O_{1} B O_{2}=\angle O_{1} C O_{2}=90^{\circ} \\
\Rightarrow & O_{1} 、 C 、 O_{2} 、 B \text { are concyclic } \\
\Rightarrow & \angle O_{1} C D=\angle O_{1} O_{2} B \\
\Rightarrow & \triangle O_{1} D C \backsim \triangle O_{1} B O_{2} \\
\Rightarrow & O_{1} O_{2}=\frac{O_{1} B \cdot O_{1} C}{... | 9 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,063 |
3. If the quadratic function $y=a x^{2}+(1-4 a) x-2$ intersects the reciprocal function $y=\frac{8}{x}(x>0)$ at only one point, then this intersection point is $\qquad$ | 3. $(4,2)$.
The quadratic function passes through the point $(4,2)$ (regardless of the value of $a$), and $y=\frac{8}{x}$ always passes through the point $(4,2)$. | (4,2) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,065 |
4. If a number, from the highest digit to the lowest digit, does not decrease at each digit, it is called a "positive number" (such as $12$, $22$, $566$, $1448$, $123456789$, etc.); if a number, from the highest digit to the lowest digit, does not increase at each digit, it is called a "negative number" (such as $21$, ... | 4.525.
According to the definition of a wavy number, we discuss in two cases:
(1) The hundreds, tens, and units digits are in the pattern "small-large-small": there are
$$
2 \times 1+3 \times 2+\cdots+9 \times 8=240 \text { (numbers). }
$$
(2) The hundreds, tens, and units digits are in the pattern "large-small-large"... | 525 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,066 |
One. (20 points) Equations
$$
x^{2}+a x+b=0 \text { and } x^{2}+b x+a=0
$$
have a common root, and let the other two roots be $x_{1} 、 x_{2}$; Equations
$$
x^{2}-c x+d=0 \text { and } x^{2}-d x+c=0
$$
have a common root, and let the other two roots be $x_{3} 、 x_{4}$. Find the range of $x_{1} x_{2} x_{3} x_{4}$ $(a, ... | Let the common root of the first two equations be $x_{0}$. Then
$$
\left\{\begin{array}{l}
x_{0}^{2} + a x_{0} + b = 0, \\
x_{0}^{2} + b x_{0} + a = 0 .
\end{array}\right.
$$
Subtracting the two equations gives $(a-b) x_{0} = a-b \Rightarrow x_{0} = 1$.
By Vieta's formulas, the other two roots are
$$
x_{1} = \frac{b}{... | 0 < x_{1} x_{2} x_{3} x_{4} < \frac{1}{16} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,067 |
II. (25 points) As shown in Figure 5, given that the circumcenter of $\triangle ABC$ is $O$, and $I$, $G$ are the incenter and centroid of $\triangle ABC$ respectively. If $IG \parallel BC$, prove: $OI \perp AI$. | II. As shown in Figure 6, extend $A I$ to intersect $B C$ at point $E$ and $\odot O$ at point $D$. Let the inradius be $r$.
From $I G / / B C$, we get $\frac{A I}{I E}=2$.
Then $\frac{S_{\triangle A B}+S_{\triangle A I C}}{S_{\triangle B I C}}=2 \Rightarrow \frac{0.5(A B+A C) r}{0.5 B C \cdot r}=2$
$$
\Rightarrow A B+A... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,068 |
Example 6 Let $a_{1}, a_{2}, \cdots, a_{k}$ be positive integers,
$$
1 \leqslant a_{1}<a_{2}<\cdots<a_{k} \leqslant n \text {, }
$$
and for any $i, j \in\{1,2, \cdots, k\}$, we have $\left[a_{i}, a_{j}\right] \leqslant n$. Prove: $k<2 \sqrt{n}$. | Prove that taking $m=\sqrt{n}$.
Let $a_{1}, a_{2}, \cdots, a_{k}$ be the numbers not greater than $m$, which are $a_{1}, a_{2}, \cdots, a_{t}$. Then
$$
t \leqslant[m] \leqslant m .
$$
Now consider $a_{i}(t+1 \leqslant i \leqslant k)$.
For any $i \in\{t+1, t+2, \cdots, k-1\}$, we have
$$
\begin{array}{l}
n \geqslant\le... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 727,069 |
Three. (25 points) Given the quadratic function $y=a x^{2}+b x+c$ intersects the $x$-axis at points $A$ and $B$, and its vertex is $C$.
(1) Prove: If $\triangle A B C$ is an isosceles right triangle, then
$$
\Delta=b^{2}-4 a c=4 \text {; }
$$
(2) The quadratic function $y=x^{2}-(2 a+2) x+a^{2}+5 a+3$ intersects the $x$... | (1) It is only necessary to prove: $|A B|=\frac{\sqrt{\Delta}}{|a|}$, the distance from vertex $C$ to the $x$-axis is $\left|\frac{\Delta}{4 a}\right|$.
Given that $\triangle A B C$ is an isosceles right triangle
$$
\Rightarrow \frac{\sqrt{\Delta}}{|a|}=2\left|\frac{\Delta}{4 a}\right| \Rightarrow \frac{\Delta}{a^{2}}=... | a=-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,070 |
2. In the quadrilateral pyramid $P-ABCD$, it is known that quadrilateral $ABCD$ is a rectangle, and $AB=4, BC=3, PA=PB=PC=PD=5$, $AC$ intersects with $BD$ at point $O$, and $M$ is the midpoint of side $PC$. Then the angle between $OM$ and the plane $PBC$ is | 2. $\arcsin \frac{4 \sqrt{273}}{91}$.
Take the midpoint $E$ of side $BC$, then $OE \perp BC$, $PE \perp BC$.
Thus, $BC \perp$ plane $POE$.
Draw $OF \perp PE$ through point $O$, then $OF \perp$ plane $PBC$. Therefore, $\angle OMF$ is the angle we need.
Since $OM$ is the median of the hypotenuse of right $\triangle POC$... | \arcsin \frac{4 \sqrt{273}}{91} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,072 |
3. Among $m$ students, it is known that in any group of three, two of them know each other, and in any group of four, two of them do not know each other. Then the maximum value of $m$ is $\qquad$ | 3. 8 .
When $m=8$, the requirement is satisfied.
It only needs to prove: $m \leqslant 8$.
First, prove that the following two scenarios are impossible.
(1) If a student $A$ knows at least 6 people, by Ramsey's theorem, among these 6 people, there exist 3 people who either all know each other or all do not know each ot... | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,073 |
4. Let $n>0, n=2^{k} m$ (where $m$ is odd). Then the greatest common divisor of $\mathrm{C}_{2 n}^{1}, \mathrm{C}_{2 n}^{3}$, $\cdots, \mathrm{C}_{2 n}^{2 n-1}$ is $\qquad$ | 4. $2^{k+1}$.
First, $2^{k+1} \| \mathrm{C}_{2 n}^{1}$.
It is easy to prove that $2^{k+1} \| \mathrm{C}_{2 n}^{i} (i=3,5, \cdots, 2 n-1)$. This is because $i \mathrm{C}_{2 n}^{i}=2 n \mathrm{C}_{2 n-1}^{i-1}, (i, 2 n)=1$.
Therefore, the greatest common divisor of $\mathrm{C}_{2 n}^{1}, \mathrm{C}_{2 n}^{3}, \cdots, \m... | 2^{k+1} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,074 |
5. If $n$ is a positive integer greater than 1, then
$$
\begin{array}{l}
\cos \frac{2 \pi}{n}+\cos \frac{4 \pi}{n}+\cos \frac{6 \pi}{n}+\cdots+\cos \frac{2 n \pi}{n} \\
=
\end{array}
$$ | 5. 0 .
$$
\sum_{k=1}^{n} \cos \frac{2 k \pi}{n}=\operatorname{Re} \sum_{k=1}^{n} \mathrm{e}^{\frac{2 k \pi i}{n}}=0 .
$$ | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,075 |
6. Given positive real numbers $x, y, z$ satisfy
$$
x y z + x y + y z + z x + x + y + z = 3 \text{. }
$$
Then the maximum value of $u = x y z (x + y + z)$ is $\qquad$ | 6. $3(\sqrt[3]{4}-1)^{4}$.
When $x=y=z=\sqrt[3]{4}-1$, $u=3(\sqrt[3]{4}-1)^{4}$.
It is only necessary to prove: $u \leqslant 3(\sqrt[3]{4}-1)^{4}$.
Notice,
$$
\begin{array}{l}
27 u=27 x y z(x+y+z) \\
\leqslant(x+y+z)^{3}(x+y+z)=(x+y+z)^{4} .
\end{array}
$$
Thus, $x+y+z \geqslant \sqrt[4]{27 u}$.
Also, $x y z+\frac{(\... | 3(\sqrt[3]{4}-1)^{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,076 |
7. Let the 10 complex roots of the equation $x^{10}+(13 x-1)^{10}=0$ be $x_{1}, x_{2}, \cdots, x_{10}$. Then
$$
\frac{1}{x_{1} \overline{x_{1}}}+\frac{1}{x_{2} \overline{x_{2}}}+\cdots+\frac{1}{x_{5} \overline{x_{5}}}=
$$
$\qquad$ | 7. 850 .
Let $\varepsilon=\cos \frac{\pi}{10}+\mathrm{i} \sin \frac{\pi}{10}$. Then $\varepsilon^{10}=-1$.
Given that the 10 complex roots of the equation $x^{10}+(13 x-1)^{10}=0$ are $x_{1}, x_{2}, \cdots, x_{10}$, we can assume them to be $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$, $\overline{x_{1}}, \overline{x_{2}}, \ove... | 850 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,077 |
9. (16 points) Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
\begin{array}{l}
a_{1}=2, \\
\sum_{k=1}^{n} \frac{a_{k}}{k(k+1)(k+2)}=\frac{3 n^{2}+5 n}{4\left(n^{2}+3 n+2\right)}\left(n \in \mathbf{Z}_{+}\right) .
\end{array}
$$
Prove: $\log _{\sqrt{3}} \frac{\sqrt{1+a_{n}}}{2}<\sum_{k=1}^{n} \frac{1}{a_{k}}<\f... | 9. Let $S_{n}=\sum_{k=1}^{n} \frac{a_{k}}{k(k+1)(k+2)}$. Then
$$
S_{1}=\frac{a_{1}}{1 \times 2 \times 3}=\frac{3 \times 1^{2}+5 \times 1}{4\left(1^{2}+3 \times 1+2\right)} \Rightarrow a_{1}=2 \text {. }
$$
When $n \geqslant 2$,
$$
\begin{array}{l}
\frac{a_{n}}{n(n+1)(n+2)}=S_{n}-S_{n-1} \\
=\frac{3 n^{2}+5 n}{4\left(n... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 727,079 |
Example 7 Given an integer $n \geqslant 2$, for any coprime positive integers $a_{1}, a_{2}, \cdots, a_{n}$, let
$$
A=a_{1}+a_{2}+\cdots+a_{n} .
$$
Let the greatest common divisor of $A$ and $a_{i}(i=1,2, \cdots, n)$ be $d_{i}$; the greatest common divisor of the remaining $n-1$ numbers after removing $a_{i}$ from $a_... | Consider $D_{1}=\left(a_{2}, a_{3}, \cdots, a_{n}\right)$ and $d_{2}=\left(a_{2}, A\right)=\left(a_{2}, a_{1}+a_{2}+\cdots+a_{n}\right)$. Let $\left(D_{1}, d_{2}\right)=d$. Then $d\left|a_{2}, d\right| a_{3}, \cdots, d\left|a_{n}, d\right| A$. Hence $d \mid a_{1}$. Therefore, $d \mid\left(a_{1}, a_{2}, \cdots, a_{n}\ri... | (n-1)^{n} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,080 |
10. (20 points) Let the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ have left and right foci $F_{1}$ and $F_{2}$, respectively. A line $l$ passing through point $F_{1}$ with an inclination angle $\theta$ intersects the ellipse $C$ at points $A$ and $B$. If $\cos \theta=\frac{3}{5}$, $3 \overrightarrow... | 10. (1) Let the projections of points $A$ and $B$ on the left directrix of the ellipse be points $A_{1}$ and $B_{1}$, respectively.
Given $3 \overrightarrow{F_{1} A}=5 \overrightarrow{B F_{1}}$, let $\left|A F_{1}\right|=5 t$ and $\left|B F_{1}\right|=3 t$ (where $t>0$).
Draw $B T \perp A A_{1}$ at point $T$.
Then $|T ... | \left(\frac{x-169}{\frac{144}{5}-x}\right)^{2}=\frac{338}{5} x+y^{2}-\frac{169 \times 119}{25} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,081 |
11. (20 points) Let $x_{i} \in \mathbf{R}_{+}(i=1,2, \cdots, n, n \geqslant 2)$, $m \in \mathbf{N}, m \geqslant 2$, and $\sum_{i=1}^{n} x_{i}=T$, where $k$, $a$, $T$, and $m$ are given positive real numbers, and let $W=\sum_{i=1}^{n} \sqrt[m]{k x_{i}+a}$. Find the range of $W$. | 11. Let $y_{i}=\left(k x_{i}+a\right)^{\frac{1}{m}}(i=1,2, \cdots, n)$. Then
$$
y_{i}^{m}=k x_{i}+a \Rightarrow \sum_{i=1}^{n} y_{i}^{m}=k T+n a \text {. }
$$
By Jensen's inequality, we have
$$
\begin{array}{l}
\frac{\sum_{i=1}^{n} y_{i}^{m}}{n} \geqslant\left(\frac{\sum_{i=1}^{n} y_{i}}{n}\right)^{m}=\left(\frac{W}{n... | \left[\sqrt[m]{k T+a}+(n-1) \sqrt[m]{a}, n \sqrt[m]{\frac{k T+n a}{n}}\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,082 |
In Rt $\triangle A B C$, it is known that $C D$ is the altitude on the hypotenuse $A B$, $I 、 I_{1} 、 I_{2}$ are the incenters of $\triangle A B C 、 \triangle A D C$ 、 $\triangle B D C$ respectively, $I E \perp A B$ at point $E$, line $A I$ intersects $B C 、 B I$ intersects $A C 、 M N$ intersects $C D$ at points $N 、 M... | Because $A X=\frac{1}{2}(A D+A C-C D)$,
$A E=\frac{1}{2}(A B+A C-B C)$,
therefore, $X E=\frac{1}{2}(D B+D C-B C)=D Y=I_{2} Y$.
Thus, $I X=X D=E Y$.
Also, $\angle I_{2} Y E=\angle I_{1} X E=90^{\circ}$, then
$\triangle I_{2} Y E \cong \triangle I_{1} X E$.
Hence $I_{1} E=I_{2} E$,
$\angle I_{1} E X+\angle ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,083 |
(40 points) If $t (t \in \mathbf{R})$ is a root of some polynomial with integer coefficients, then $t$ is called an "algebraic number." Otherwise, $t$ is called a "transcendental number." Prove:
(1) The union of countably many countable sets is countable;
(2) There exist transcendental numbers. | (1) Let $I_{1}, I_{2}, \cdots, I_{n}, \cdots$ be a countable set (note that, the countable sets mentioned in the problem are countable. Therefore, these sets can be numbered by natural numbers).
Let $I_{i}=\left\{x_{1}^{i}, x_{2}^{i}, \cdots\right\}, I=I_{1} \cup I_{2} \cup \cdots$.
Map $x_{j}^{i}$ to $\frac{i}{j}$ (wh... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 727,084 |
Four, (50 points) 11 interest classes, several students participate (can participate repeatedly), and each interest class has the same number of students (full, unknown number). It is known that any nine interest classes include all students, while any eight interest classes do not include all students. Find the minimu... | Let the set of students in 11 interest classes be $A_{1}$, $A_{2}, \cdots, A_{11}$.
By the problem, we know $\left|A_{1}\right|=\left|A_{2}\right|=\cdots=\left|A_{11}\right|=x$.
Let $T=A_{1} \cup A_{2} \cup \cdots \cup A_{11}$.
By the problem, we know the union of any nine sets is $T$, and the union of any eight sets i... | 165 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,086 |
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