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
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
7. Randomly throw three dice. The probability that the sum of the numbers on two of the dice is 7 is $\qquad$ . | 7. $\frac{5}{12}$.
Rolling three dice has a total of $6^{3}=216$ possible outcomes.
Notice that, $7=1+6=2+5=3+4$.
When rolling three dice, the number of outcomes where two dice show 1 and 6 is $6 \times 6-6=30$, which can be divided into
$$
\begin{array}{l}
(1,6, \times),(1, \times, 6),(6,1, \times), \\
(6, \times, 1)... | \frac{5}{12} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,491 |
8. Given points $A(1,-1), B(4,0), C(2,2)$, the plane region $D$ consists of all points $P(x, y)$ that satisfy
$$
\overrightarrow{A P}=\lambda \overrightarrow{A B}+\mu \overrightarrow{A C}(1<\lambda \leqslant a, 1<\mu \leqslant b)
$$
If the area of region $D$ is 8, then the minimum value of $a+b$ is . $\qquad$ | 8. 4 .
As shown in Figure 3, extend $A B$ to point $N$, and extend $A C$ to point $M$, such that
$$
|A N|=a|A B|,|A M|=b|A C| .
$$
Construct $\square A B E C$ and $\square A N G M$. Then quadrilateral $E H G F$ is a parallelogram.
From the conditions, the region $D$ composed of points $P(x, y)$ is the shaded area in... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,492 |
9. $A=\left[\frac{8}{9}\right]+\left[\frac{8^{2}}{9}\right]+\cdots+\left[\frac{8^{0114}}{9}\right]$ when divided by 63 leaves a remainder of $\qquad$ ( $[x]$ denotes the greatest integer not exceeding the real number $x$). | 9. 56 .
Notice that, for any positive integer $k$, $\frac{8^{2 k-1}}{9}$ and $\frac{8^{2 k}}{9}$ are not integers, and
$$
\frac{8^{2 k-1}}{9}+\frac{8^{2 k}}{9}=8^{2 k-1} \text {. }
$$
Therefore, for any positive integer $k$, we have
$$
\begin{array}{l}
{\left[\frac{8^{2 k-1}}{9}\right]+\left[\frac{8^{2 k}}{9}\right]=... | 56 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,493 |
13. As shown in Figure 1, in pentagon $A B C D E$, it is given that $B C \parallel A E, A B = B C + A E, \angle A B C = \angle C D E, M$ is the midpoint of $C E$, $O$ is the circumcenter of $\triangle B C D$, and $O M \perp M D$. Extend $D M$ to point $K$ such that $M K = M D$. Prove:
(1) $\angle B K C = \angle B D C$;... | 13. (1) Since $M$ is the midpoint of $K D$, and $O M \perp M D$, therefore, $O K=O D$.
Thus, point $K$ lies on the circumcircle of $\triangle B C D$.
Hence, $\angle B K C=\angle B D C$.
(2) As shown in Figure 4, extend $A E$ to point $T$, such that
$$
E T=B C \text {. }
$$
Connect $T B, T C, T D, T K, K E$.
From $A B=... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,494 |
14. Given
$$
f(x)=a \ln (x+1)+\frac{1}{x+1}+3 x-1 \text {. }
$$
(1) If $x \geqslant 0$, $f(x) \geqslant 0$ always holds, find the range of the real number $a$;
(2) Prove: For all positive integers $n$, we have
$$
\begin{array}{l}
\frac{2}{4 \times 1^{2}-1}+\frac{3}{4 \times 2^{2}-1}+\cdots+\frac{n+1}{4 \times n^{2}-1} ... | 14. (1) Notice,
$$
\begin{array}{c}
f^{\prime}(x)=\frac{a}{x+1}-\frac{1}{(x+1)^{2}}+3 \\
=\frac{3 x^{2}+(a+6) x+a+2}{(x+1)^{2}} .
\end{array}
$$
If $a \geqslant-2$, then for $x>0$, $f^{\prime}(x)>0$, at this time, $f(x)$ is an increasing function on the interval $[0,+\infty)$.
Therefore, when $x \geqslant 0$, $f(x) \g... | proof | Calculus | proof | Yes | Yes | cn_contest | false | 728,495 |
2. Construct squares $A B D E$, $C A F G$, and $B C H K$ outwardly on the sides of $\triangle A B C$. Prove that a triangle can be formed with sides $E F$, $G H$, and $K D$, and that its area is 3 times the area of $\triangle A B C$.
(2003, Beijing Middle School Mathematics Competition (Grade 8 Final)) | As shown in Figure 12, by translating $\triangle D B K$ to $\triangle E A P$, and using $\square D E P K$, $\square A P K B$, $\square A P H C$, $\square P H G F$, we can prove that $F P / / G H, F P = G H$. Using rotational transformation, we can prove:
$$
\begin{array}{l}
S_{\triangle D B K} = S_{\triangle G C H} \\
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,496 |
2. Calculate:
$$
\sin ^{3} 20^{\circ}-\sin ^{3} 100^{\circ}+\sin ^{3} 140^{\circ}=
$$ | 2. $-\frac{3 \sqrt{3}}{8}$.
From the triple angle formula, we get
$$
\begin{array}{l}
\sin ^{3} \alpha+\sin ^{3}\left(\alpha-120^{\circ}\right)+\sin ^{3}\left(\alpha+120^{\circ}\right) \\
=\frac{1}{4}\left[3 \sin \alpha-\sin 3 \alpha+3 \sin \left(\alpha-120^{\circ}\right)-\right. \\
\left.\quad \sin 3 \alpha+3 \sin \l... | -\frac{3 \sqrt{3}}{8} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,497 |
3. Let $a, b \in \mathbf{R}$, and $a+b=1$. Then
$$
f(a, b)=3 \sqrt{1+2 a^{2}}+2 \sqrt{40+9 b^{2}}
$$
the minimum value of $f(a, b)$ is $\qquad$ | 3. $5 \sqrt{11}$.
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
f(a, b) \\
=\sqrt{\left(\frac{81}{11}+\frac{18}{11}\right)\left(1+2 a^{2}\right)}+ \\
\sqrt{\left(\frac{40}{11}+\frac{4}{11}\right)\left(40+9 b^{2}\right)} \\
\geqslant\left(\frac{9}{\sqrt{11}}+\frac{6 a}{\sqrt{11}}\right)+\left(\frac{40}{... | 5 \sqrt{11} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,498 |
5. Given vectors $\boldsymbol{a}$ and $\boldsymbol{b}$ are two perpendicular unit vectors in a plane, and $(3 a-c) \cdot(4 b-c)=0$. Then the maximum value of $|c|$ is $\qquad$ . | 5.5.
Let $\overrightarrow{O A}=3 a, \overrightarrow{O B}=4 b, \overrightarrow{O C}=c$.
From the given, $\overrightarrow{A C} \perp \overrightarrow{B C}$.
Thus, point $C$ lies on the circle with $A B=5$ as its diameter, and this circle passes through the origin $O$.
Therefore, the maximum value of $|\overrightarrow{O C... | 5.5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,499 |
7. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=1, a_{n+1}=\frac{n a_{n}+2(n+1)^{2}}{n+2} \text {. }
$$
Then the general term formula of the sequence $\left\{a_{n}\right\}$ is . $\qquad$ | 7. $\frac{1}{6} n(n+1)(n+2)$.
From the given, we have
$$
\begin{array}{l}
(n+2) a_{n+1}=n a_{n}+2(n+1)^{2} \\
\Rightarrow(n+1)(n+2) a_{n+1} \\
\quad=n(n+1) a_{n}+2(n+1)^{3} .
\end{array}
$$
Let $b_{n}=n(n+1) a_{n}$. Then
$$
b_{1}=2, b_{n+1}=b_{n}+2(n+1)^{3} \text {. }
$$
Thus, $b_{n}=b_{1}+\sum_{i=1}^{n-1}\left(b_{i... | \frac{1}{6} n(n+1)(n+2) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,500 |
9. (16 points) Let real numbers $a, b, c, d$ satisfy
$$
\left\{\begin{array}{l}
a+b+c+d=3, \\
a^{2}+b^{2}+c^{2}+d^{2}=3, \\
a b c+b c d+c d a+d a b=1 .
\end{array}\right.
$$
Prove:
$$
a(1-a)^{3}=b(1-b)^{3}=c(1-c)^{3}=d(1-d)^{3} .
$$ | 9. According to the identity, we have
$$
\begin{array}{l}
a b+b c+c d+d a+a c+b d \\
=\frac{1}{2}\left[(a+b+c+d)^{2}-\left(a^{2}+b^{2}+c^{2}+d^{2}\right)\right] \\
=3 .
\end{array}
$$
Let $f(x)=x(1-x)^{3}$.
We only need to prove: $f(a)=f(b)=f(c)=f(d)$.
Notice that,
$$
\begin{aligned}
& (x-a)(x-b)(x-c)(x-d)-a b c d \\
... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 728,501 |
11. (20 points) Let $A$ and $B$ be the endpoints of the major axis of the ellipse
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)
$$
$P$ is any point on the ellipse, and a tangent line $CD$ is drawn through $P$, intersecting the tangents at points $A$ and $B$ at points $C$ and $D$, respectively. $N$ and $M$ are th... | 11. As shown in Figure 1.
From the problem, we have
$$
\begin{array}{l}
M(c, 0), N(-c, 0), A(a, 0), B(-a, 0), \\
a^{2}=b^{2}+c^{2} .
\end{array}
$$
Let $P(a \cos \alpha, b \sin \alpha)$.
Then the tangent line $l_{C D}: \frac{x \cos \alpha}{a}+\frac{y \sin \alpha}{b}=1$.
Let the tangent line $C D$ intersect the $y$-ax... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,502 |
Example 1 If $a, b$ are complex numbers, $a \neq 0, b \neq 0$, prove: $|a-b| \geqslant \frac{1}{2}(|a|+|b|)\left|\frac{a}{|a|}-\frac{b}{|b|}\right|$, equality holds if and only if $|a|=|b|$. | Proof: Let $a=r \mathrm{e}^{\mathrm{i} \theta}, b=-\rho \mathrm{e}^{\mathrm{i} \varphi}$, where,
$$
\begin{array}{l}
r=|a|>0, \rho=|b|>0 . \\
\text { Then } \frac{|a-b|}{|a|+|b|}=\left|\frac{r}{\rho+r} \mathrm{e}^{\mathrm{i} \theta}+\frac{\rho}{\rho+r} \mathrm{e}^{\mathrm{i} \varphi}\right|=\left|z_{1}\right|,
\end{arr... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 728,503 |
Example 2 Let $a, b, c$ be given complex numbers, and denote $|a+b|$ $=m,|a-b|=n$, with $m n \neq 0$. Prove:
$$
\max \{|a c+b|,|a+b c|\} \geqslant \frac{m n}{\sqrt{m^{2}+n^{2}}} \text {. }
$$ | $$
\begin{array}{l}
\max \{|a c+b|,|a+b c|\} \\
\geqslant \frac{|b||a c+b|+|a||b c+a|}{|b|+|a|} \\
\geqslant \frac{|b(a c+b)-a(b c+a)|}{|b|+|a|} \\
=\frac{|b-a||b+a|}{\sqrt{2\left(|b|^{2}+|a|^{2}\right)}} . \\
\text { Also, } m^{2}+n^{2}=|a+b|^{2}+|a-b|^{2} \\
=2\left(|a|^{2}+|b|^{2}\right),
\end{array}
$$
Therefore, ... | \max \{|a c+b|,|a+b c|\} \geqslant \frac{m n}{\sqrt{m^{2}+n^{2}}} | Inequalities | proof | Yes | Yes | cn_contest | false | 728,504 |
8. Let $\left(1+x+x^{2}\right)^{150}=\sum_{k=0}^{300} c_{k} x^{k}$, where $c_{0}$, $c_{1}, \cdots, c_{300}$ are constants. Then $\sum_{k=0}^{100} c_{3 k}=$ $\qquad$ . | $8.3^{149}$.
Notice that,
$$
\begin{array}{l}
\left(1+x+x^{2}\right)^{150}=\left(1+x+x^{2}\right)\left(1+x+x^{2}\right)^{149} \\
=\left(1+x+x^{2}\right) \sum_{k=0}^{298} d_{k} x^{k} .
\end{array}
$$
Let $x=1$. Then
$$
3^{150}=3 \sum_{k=0}^{100} c_{3 k} \Rightarrow \sum_{k=0}^{100} c_{3 k}=3^{149} .
$$ | 3^{149} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,505 |
10. (20 points) Through the focus $F_{0}$ of the parabola $\Gamma: x^{2}=4 y$, draw two different lines $l_{1}$ and $l_{2}$ with slopes $k_{1}$ and $k_{2}$ respectively, and $k_{1}+k_{2}=2$. Line $l_{1}$ intersects the parabola $\Gamma$ at points $A$ and $B$, and line $l_{2}$ intersects the parabola $\Gamma$ at points ... | 10. As shown in Figure 4, draw perpendiculars from points $M$ and $N$ to the directrix $y=-1$, with the feet of the perpendiculars being $E$ and $F$ respectively.
By the properties of the parabola, the directrix $y=-1$ is the common tangent of $\odot M$ and $\odot N$. Let the common chord of the two circles intersect ... | \frac{3 \sqrt{5}}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,506 |
One, (40 points) Given $a, b, c \in \mathbf{R}_{+}$, and
$$
4 a b c=a+b+c+1 \text{. }
$$
Prove: $\sum\left(a^{2}+a\right) \geqslant 2 \sum a b$, where “$\sum$” denotes the cyclic sum. | $$
\begin{array}{l}
\text{Notice,} \\
\sum \frac{1}{2 a+1}=1 \\
\Leftrightarrow \sum(2 a+1)(2 b+1)=\prod(2 a+ \\
\Leftrightarrow 4 \sum a b+4 \sum a+3 \\
\quad=8 a b c+4 \sum a b+2 \sum a+1 \\
\Leftrightarrow 4 a b c=a+b+c+1,
\end{array}
$$
where, “ ”” denotes the cyclic symmetric product.
$$
\begin{array}{l}
\text{Th... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 728,507 |
Four, (50 points) Prove: there exist infinitely many primes, such that for each of these primes $p$, there exists at least one $n \in \mathbf{Z}_{+}$, satisfying
$$
p \mid\left(2014^{2^{n}}+2014\right) .
$$ | Assume the conclusion does not hold, then we can set \( p_{1}, p_{2}, \cdots, p_{k} \) as all the prime numbers that divide at least one number of the form \( 2014^{2n} + 2014 \).
Consider \( k+1 \) numbers \( 2014^{2^{i}} + 2014 \) (for \( i=1,2, \cdots, k+1 \)). Since these numbers are finite, there exists a \( q \i... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 728,508 |
Example 2 Given $a, b \in \mathbf{R} \backslash\{0\}$, satisfying
$$
a^{2} b^{2}\left(a^{2} b^{2}+4\right)=2\left(a^{6}+b^{6}\right) \text {. }
$$
Prove: At least one of $a$ and $b$ is irrational.
(1999-2000 All-Russian High School Mathematics Olympiad) | $$
\begin{array}{l}
\left(a^{4}-2 b^{2}\right)\left(b^{4}-2 a^{2}\right)=0 \\
\Rightarrow\left(\frac{a^{2}}{b}\right)^{2}=2 \text { or }\left(\frac{b^{2}}{a}\right)^{2}=2 \\
\Rightarrow \frac{a^{2}}{b}= \pm \sqrt{2} \text { or } \frac{b^{2}}{a}= \pm \sqrt{2} .
\end{array}
$$
Therefore, the assertion in the problem is ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 728,509 |
Given the Fibonacci sequence $\left\{F_{n}\right\}$:
$$
F_{1}=F_{2}=1, F_{n+2}=F_{n+1}+F_{n} \text {. }
$$
Prove: For all $x \in \mathbf{R}, n \geqslant 2$, we have
$$
\sum_{k=1}^{n} F_{k}|x-k| \geqslant F_{n+2}+F_{n}-n-1 .
$$ | Prove that if $a_{i} \in \mathbf{R}_{+}(i=1,2, \cdots, n)$, the function
$$
f(x)=\sum_{i=1}^{n} a_{i}|x-i|
$$
reaches its minimum value at $x=i_{0}$, then $i_{0}$ is the largest positive integer satisfying
$$
a_{1}+a_{2}+\cdots+a_{i} \leqslant a_{i+1}+a_{i+2}+\cdots+a_{n}
$$
For example, if $a_{k}=k(k=1,2, \cdots, n)... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 728,511 |
For any positive integer $n$, the function $f(n)$ is the sum of the digits (i.e., the digital sum) of $n^{2}+3 n+1$ in decimal notation. Question: Does there exist an integer $n$ such that
$$
f(n)=2013 \text { or } 2014 \text { or } 2015 \text { ? }
$$ | When $3 \mid n$,
$$
f(n) \equiv n^{2}+3 n+1 \equiv 1(\bmod 3) ;
$$
When $3 \nmid n$,
$$
f(n) \equiv n^{2}+3 n+1 \equiv 2(\bmod 3) \text {. }
$$
Thus, $3 \nmid f(n)$.
Since 312013, there does not exist an integer $n$ such that
$$
f(n)=2013 \text {. }
$$
If there exists an integer $n$ such that $f(n)=2014$, then by $2... | 2015 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,512 |
Given $\triangle A B C$ with circumcenter and incenter $O$ and $I$ respectively, $A I, B I, C I$ intersect the circumcircles of $\triangle B I C, \triangle C I A, \triangle A I B$ at points $D, E, F$ different from $I$. Through $D, E, F$, draw perpendiculars to $B C, C A, A B$ as $l_{a}, l_{b}, l_{c}$ respectively.
Pr... | Proof (1) Let the intersections of $A I$, $B I$, and $C I$ with the circumcircle of $\triangle A B C$ be $L$, $M$, and $N$ respectively. Then $L$, $M$, and $N$ are the midpoints of the arcs $\overparen{B C}$, $\overparen{C A}$, and $\overparen{A B}$ of the circumcircle of $\triangle A B C$, and $L$, $M$, and $N$ are th... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,513 |
Example 7 Let $\alpha \in \mathbf{R}$, and $\cos \alpha \pi=\frac{1}{3}$. Prove: $\alpha$ is an irrational number.
(Note: $\alpha \pi$ is in radians.) | First, we prove: for all non-negative integers $k$, there exists an integer $\beta(k)$ not divisible by 3, such that
$$
\cos 3^{k} \alpha \pi=\frac{\beta(k)}{3^{3^{k}}} .
$$
When $k=0$, by the given condition, we can take $\beta(k)=1$, and equation (1) holds.
If for $k=m \geqslant 0$, equation (1) holds, then by the ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 728,514 |
In a convex quadrilateral $ABCD$, it is known that $\angle ABC = \angle CDA = 90^{\circ}$, $H$ is the foot of the perpendicular from point $A$ to $BD$, and points $S$ and $T$ are on sides $AB$ and $CD$ respectively, such that point $H$ is inside $\triangle SCT$, and
\[
\begin{array}{l}
\angle CHS - \angle CSB = 90^{\ci... | Proof As shown in Figure 2, extend $CB$ to point $X$ such that $BX = BC$; extend $CD$ to point $Y$ such that $DY = DC$, and connect $SX$ and $TY$.
It is easy to see that $KX = KC = KH = KS$,
$LY = LC = LH = LT$.
Thus, $C, H, S, X$ and $C, H, T, Y$ are concyclic, respectively.
Therefore, $\angle X = \angle SCB$
$\Righta... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,515 |
Question 1 As shown in Figure 1, in acute $\triangle ABC$, $AB > AC$, and $D$, $E$ are the midpoints of sides $AB$, $AC$ respectively. Let the circumcircle of $\triangle BCE$ intersect the circumcircle $\odot O$ of $\triangle ADE$ at point $P$, and the circumcircle of $\triangle BCD$ intersect $\odot O$ at point $Q$. P... | To prove that the known points $B, C, E, P$ and $C, B, D, Q$ are concyclic respectively.
$$
\begin{array}{l}
\text { Then } A P=A Q \Leftrightarrow \angle A P Q=\angle A Q P \\
\Leftrightarrow \angle A D Q=\angle A E P \\
\Leftrightarrow \angle B C Q=\angle C B P .
\end{array}
$$
For this, it is sufficient to prove th... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,516 |
Question 2 As shown in Figure 2, given an acute triangle $\triangle A B C$ with its circumcircle $\odot O$, a tangent line $l$ is drawn through point $A$ to $\odot O$, and $l$ intersects line $B C$ at point $D$. Point $E$ is on the extension of $D A$, and $F$ is a point on the minor arc $\overparen{B C}$. Line $E F$ in... | To prove the sufficiency.
Notice,
$$
\begin{array}{l}
E G=\frac{E Q \sin \angle G Q E}{\sin \angle E G Q}=\frac{E Q \sin \angle G Q E}{\sin \angle D B F}, \\
E F=\frac{P E \sin \angle F P E}{\sin \angle P F E}=\frac{P E \sin \angle F P E}{\sin \angle D C Q}, \\
D C=\frac{D Q \sin \angle G Q E}{\sin \angle D C Q}, \\
D ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,517 |
Question 3 As shown in Figure 3, in $\triangle ABC$, it is known that $\angle B$ and $\angle C$ are acute angles, $AD \perp BC$, $DE \perp AC$, and $M$ is the midpoint of $DE$. If $AM \perp BE$ at point $F$, prove: $\triangle ABC$ is an isosceles triangle.
(Tenth Northern Mathematical Olympiad Invitational Competition) | Prove as shown in Figure 3, connect $B M$.
Let $\angle A B C=\alpha$,
$$
\begin{array}{l}
\angle A C B=\theta \\
A B=x
\end{array}
$$
Then $\angle A D E$
$$
\begin{array}{l}
=\angle A C B=\theta \\
\angle D A E \\
=90^{\circ}-\theta \\
A D=x \sin \alpha \\
B D=x \cos \alpha \\
A E=A D \sin \theta=x \sin \alpha \cdot \... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,518 |
Question 2 Given a non-isosceles $\triangle ABC$, $N$ is the midpoint of the arc $\overparen{BAC}$ of its circumcircle, $M$ is a moving point on the angle bisector of $\angle BAC$, and $O_{1}$ and $O_{2}$ are the circumcenters of $\triangle ABM$ and $\triangle ACM$, respectively. Prove that $O_{1}$, $O_{2}$, $A$, and $... | Prove by connecting auxiliary lines as shown in Figure 1.
From the given, we have
$$
\angle B O_{1} M=2 \angle B A M=\angle B A C,
$$
and $O_{1} B=O_{1} M$.
Since $N$ is the midpoint of arc $\overparen{B A C}$, we know
$$
\angle B N C=\angle B A C, N B=N C \text {. }
$$
Therefore, $\triangle N B C \backsim \triangle ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,519 |
Question 3 In $\triangle ABC$, it is known that $\odot O$ is the circumcircle of $\triangle ABC$, $D$ is a point on side $BC$, $E$ is a point on arc $\overparen{BC}$, and satisfies $\angle BAD = \angle CAE$, $O_{1}$ and $O_{2}$ are the circumcenters of $\triangle ABD$ and $\triangle ACD$ respectively. Connect $EO$ and ... | To prove: As shown in Figure 2, draw auxiliary lines.
The original proposition is equivalent to: Let the circumcircle of $\triangle A O_{1} O_{2}$ intersect $\odot O$ at point $N$. Connect $N O$ and extend it to intersect $\odot O$ at point $E$. Then $\angle B A D = \angle C A E$.
In fact, by symmetry,
$$
\angle O_{2} ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,520 |
1. Given a real number $r \in(0,1)$. Prove: If $n$ complex numbers $z_{1}, z_{2}, \cdots, z_{n}$ satisfy
$$
\begin{array}{l}
\left|z_{k}-1\right| \leqslant r(k=1,2, \cdots, n), \\
\text { then } \quad\left|z_{1}+z_{2}+\cdots+z_{n}\right|\left|\frac{1}{z_{1}}+\frac{1}{z_{2}}+\cdots+\frac{1}{z_{n}}\right| \\
\geqslant n^... | 1. Let $z_{k}=x_{k}+y_{k} \mathrm{i}$, where $x_{k}, y_{k} \in \mathbf{R}, k=1,2, \cdots, n$.
First, prove:
$$
\frac{x_{k}^{2}}{x_{k}^{2}+y_{k}^{2}} \geqslant 1-r^{2} \quad (k=1,2, \cdots, n).
$$
Let $u=\frac{x_{k}^{2}}{x_{k}^{2}+y_{k}^{2}}$.
By $\left|x_{k}-1\right| \leqslant r0$.
Then $u>0$, and $y_{k}^{2}=\left(\fr... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 728,521 |
2. As shown in Figure 1, let $A, B, D, E, F, C$ be six points on a circle in that order, satisfying $A B=A C$. Line $A D$ intersects $B E$ at point $P$, line $A F$ intersects $C E$ at point $R$, line $B F$ intersects $C D$ at point $Q$, line $A D$ intersects $B F$ at point $S$, and line $A F$ intersects $C D$ at point ... | 2. As shown in Figure 2, connect $B C$, $R P$, and $D F$.
From $A B=A C$, we know $\angle A D C=\angle A F B$.
Therefore, $S$, $D$, $F$, and $T$ are concyclic.
Thus, $\angle Q S K=\angle T D F=\angle R A C$.
Combining $\angle S K Q=\angle A C E$, we get
$\triangle Q S K \backsim \triangle R A C$.
Similarly, $\triangle ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,522 |
3. Given an integer $n \geqslant 5$. Find the smallest integer $m$, such that there exist two sets of integers $A$ and $B$, simultaneously satisfying the following conditions:
(1) $|A|=n,|B|=m$, and $A \subseteq B$;
(2) For any two distinct elements $x, y$ in set $B$, $x+y \in B$ if and only if $x, y \in A$.
（Zhai Zhen... | 3. First, take an integer $N \geqslant 2 n$, and consider
$$
\begin{array}{l}
A=\{N+1, N+2, \cdots, N+n\}, \\
B=A \cup\{2 N+3,2 N+4, \cdots, 2 N+2 n-1\} .
\end{array}
$$
Condition (1) is clearly satisfied.
On one hand, for any $x, y \in A, x \neq y$, we have
$$
2 N+3 \leqslant x+y \leqslant 2 N+2 n-1 \text {. }
$$
Th... | 3n-3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,523 |
5. A certain meeting has 30 participants, each of whom knows at most five others; among any five people, at least two are not acquaintances. Find the largest positive integer $k$, such that in any group of 30 people satisfying the above conditions, there always exists a group of $k$ people, none of whom are acquaintanc... | 5, hence we have
$30-|X| \leqslant 5|X|$.
Therefore, $|X| \geqslant 5$.
If $|X|=5$, then by equation (1) the equality is achieved, which means the 25 edges are distributed among the five vertices in set $X$, i.e., the neighborhood of each vertex in set $X$ is a set of five points in set $V \backslash X$.
Since $|V \ba... | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,525 |
6. Let the infinite sequence of non-negative integers $a_{1}, a_{2}, \cdots$ satisfy: for any positive integers $m, n$, we have $\sum_{i=1}^{2 m} a_{i n} \leqslant m$. Prove: there exist positive integers $k, d$, such that
$$
\sum_{i=1}^{2 k} a_{i d}=k-2014
$$
(Zhai Zhenhua) | 6. Since $a_{1}, a_{2}, \cdots$ is a sequence of non-negative integers, taking $m=1$ in the condition, we know that for any positive integer $n$,
$$
0 \leqslant a_{n}+a_{2 n} \leqslant 1 \text {. }
$$
Thus, $a_{n} \in\{0,1\}$, and there are infinitely many $n$ such that $a_{n}=0$.
We now prove that there are infinitel... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 728,526 |
B. In the Cartesian coordinate system, there are points $A(-1,2)$, $B(3,2)$. If $C$ is a point on the coordinate axes, and $\triangle ABC$ is a right triangle, then the number of points $C$ that satisfy the condition is ( ) .
(A) 3
(B) 4
(C) 5
(D) 6 | B. C.
In $\triangle A B C$, if $\angle A$ or $\angle B$ is a right angle, then point $C(-1,0)$ or $C(3,0)$; if $\angle C$ is a right angle, then point $C$ is the point where the circle with $A B$ as its diameter is tangent to the $x$-axis, or one of the two intersection points with the $y$-axis. Therefore, there are 5... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 728,528 |
B. Use three types of regular polygon tiles with equal side lengths to pave the ground, with their vertices fitting together perfectly, leaving no gaps. If the number of sides of these three regular polygons are $x$, $y$, and $z$, then the value of $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ is $(\quad)$.
(A) $\frac{1}{3}$
(... | B. B.
The interior angles of these three regular polygons are respectively equal to
$$
\frac{x-2}{x} \times 180^{\circ}, \frac{y-2}{y} \times 180^{\circ}, \frac{z-2}{z} \times 180^{\circ} \text{. }
$$
Since the three tiles fit together perfectly without gaps, the sum of these three angles is $360^{\circ}$.
$$
\begin{... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 728,529 |
B. A fair cube die has the numbers $1,2,3,4,5,6$ on its six faces. The die is rolled four times, and the numbers obtained are $a, b, c, d$ respectively. The probability that one of the numbers $a$, $a+b$, $a+b+c$, $a+b+c+d$ equals 4 is ( ).
(A) $\frac{33}{1296}$
(B) $\frac{334}{1296}$
(C) $\frac{343}{1296}$
(D) $\frac{... | B. C.
If $a=4$, the probability is $\frac{1}{6}$;
From $4=1+3=2+2=3+1$, we know that the probability of $a+b=4$ is $\frac{3}{6^{2}}$;
From $4=1+1+2=1+2+1=2+1+1$, we know that the probability of $a+b+c=4$ is $\frac{3}{6^{3}}$;
If $a+b+c+d=4$, then it can only be $a=b=c=d=1$, with a probability of $\frac{1}{6^{4}}$.
... | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 728,530 |
6. A. The minimum value of the algebraic expression $\sqrt{x^{2}+4}+\sqrt{(12-x)^{2}+9}$ is | $=、 6$. A. 13.
This problem can be transformed into finding the minimum value of the sum of distances from a point $(x, 0)$ on the $x$-axis to the points $(0,2)$ and $(12,3)$ in a Cartesian coordinate system.
Since the symmetric point of $(0,2)$ with respect to the $x$-axis is $(0,-2)$, the length of the line segment ... | 13 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,531 |
B. As shown in Figure 2, in the square $A B C D$ with side length 1, $E$ and $F$ are points on $B C$ and $C D$ respectively, and $\triangle A E F$ is an equilateral triangle. Then the area of $\triangle A E F$ is $\qquad$. | B. $2 \sqrt{3}-3$.
From the given information, we have
$$
\triangle A B E \cong \triangle A D F \Rightarrow B E=D F, E C=F C \text {. }
$$
Let $E C=F C=a$. Then
$$
\begin{array}{l}
A E=E F=\sqrt{2} a, B E=1-a . \\
\text { Hence } A E=\sqrt{A B^{2}+B E^{2}} \\
=\sqrt{1+(1-a)^{2}}=\sqrt{2} a \\
\Rightarrow a^{2}+2 a-2=... | 2 \sqrt{3}-3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,532 |
9. A. Given two inverse proportion functions $y=\frac{k}{x}$ and $y=\frac{1}{x}$, their graphs in the first quadrant are shown in Figure 4. Point $P$ is on the graph of $y=\frac{k}{x}$, $PC \perp x$-axis, with the foot of the perpendicular at $C$, and intersects the graph of $y=\frac{1}{x}$ at point $A$; $PD \perp y$-a... | 9. A. (1)(2)(4).
Let $P\left(a, \frac{k}{a}\right)$ be any point on the graph of the function $y=\frac{k}{x}$ in the first quadrant. Then $A\left(a, \frac{1}{a}\right), B\left(\frac{a}{k}, \frac{k}{a}\right)$.
Thus, $S_{\text {quadrilateral } O A P B}=S_{\text {rectangle } O C P D}-\left(S_{\triangle A O C}+S_{\triang... | (1)(2)(4) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,533 |
11. A. Given the equation in terms of $x$
$$
\frac{x^{2}+k x+3}{x-1}=3 x+k
$$ | Three, 11. A. The original equation can be transformed into
$$
2 x^{2}-3 x-(k+3)=0 \text {. }
$$
When $\Delta=9+8(k+3)=0$, $k=-\frac{33}{8}$, at this time, $x_{1}=x_{2}=\frac{3}{4}$ satisfies the condition;
If $x=1$ is a root of equation (1), then $2 \times 1^{2}-3 \times 1-(k+3)=0 \Rightarrow k=-4$. At this time, the... | k=-\frac{33}{8} \text{ or } k=-4 \text{ or } k \geqslant -3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,534 |
B. Originally planned to spend 1500 yuan to buy $x$ pieces of product A and $y$ pieces of product B, but the price of product A increased by 1.5 yuan per piece, and the price of product B increased by 1 yuan per piece. Despite buying 10 fewer pieces of product A than originally planned, the total amount is still 29 yua... | B. (1) Let the original planned unit prices for goods A and B be $a$ yuan and $b$ yuan, respectively. Then the original plan is
$$
a x + b y = 1500 \text{.}
$$
When the unit price of good A increases by 1.5 yuan, the unit price of good B increases by 1 yuan, and the quantity of good A decreases by 10 units, we get
$$
... | x = 76, y = 55 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,535 |
12. A. As shown in Figure 5, in rectangle $A B C D$, $A B=a$, $B C=b$, and $a<b$. A line through the center $O$ of rectangle $A B C D$ intersects segments $B C$ and $D A$ at points $E$ and $F$, respectively. The quadrilateral $E C D F$ is folded along $E F$ to the plane of quadrilateral $B E F A$, such that point $C$ c... | 12. A. (1) It is easy to know that quadrilateral $A E C F$ is a rhombus.
Since $\angle A O F=\angle A D C=90^{\circ}$, we have
$\triangle A O F \backsim \triangle A D C \Rightarrow \frac{A F}{A C}=\frac{A O}{A D}$.
Given $A O=\frac{A C}{2}=\frac{\sqrt{a^{2}+b^{2}}}{2}, A D=B C=b$, we know $A F=\frac{A C \cdot A O}{A D... | \frac{11}{8} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,536 |
B. Given that $m$, $n$, and $t$ are positive integers, and satisfy $t(m^2 - n^2) + m - n^2 - n = 0$.
Prove: (1) $m - n$ is a perfect square;
(2) For any positive integer $t$, there exists a set of positive integers $m$, $n$ that satisfy the above equation. | B. (1) Since $m$, $n$, and $t$ are all positive integers, and
$$
t(m+n)(m-n)+(m-n)=n^{2} \text {, }
$$
it follows that $m>n$.
Let $m=n+k$ (where $k$ is a positive integer). Then
$$
n^{2}-2 k t n-t k^{2}-k=0 \text {. }
$$
Since $n$ is a positive integer, its discriminant
$$
\Delta=4 k[t(t+1) k+1]
$$
must be a perfect... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 728,537 |
7. Given that $C$ is a moving point on the line segment $A B$ of length 1, point $D$ is such that $D B \perp A B$, and $B D=A C$, connect $A D$, and construct $C E \perp A D$, with the foot of the perpendicular being $E$. Let $A C=x, B E=$ $y$. Then the function expression of $y$ in terms of $x$ is $\qquad$, and the mi... | 7. $y=\sqrt{\frac{2 x^{2}-2 x+1}{x^{2}+1}}(0<x \leqslant 1), \frac{\sqrt{5}-1}{2}$.
From the problem, we know
$$
A E=\frac{x}{\sqrt{1+x^{2}}}, \cos A=\frac{1}{\sqrt{1+x^{2}}} .
$$
In $\triangle A B E$, by the cosine rule we have
$$
\begin{array}{l}
B E^{2}=A E^{2}+A B^{2}-2 A E \cdot A B \cos A \\
\Rightarrow y=\sqrt... | \frac{\sqrt{5}-1}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,542 |
8. In the Cartesian coordinate system $x O y$, let point $P$ be in the first quadrant, and points $A$ and $B$ be on $O x$ and $O y$ respectively, satisfying $P A = P B = 2$. Then the maximum value of the area of the convex quadrilateral $P A O B$ is $\qquad$ | 8. $2+2 \sqrt{2}$.
It is easy to see that when quadrilateral $O B P A$ is a right trapezoid, where $\angle A=45^{\circ}$ or $\angle B=45^{\circ}$, its area reaches the maximum value of $2+2 \sqrt{2}$. | 2+2 \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,543 |
11. Let real numbers $x_{1}, x_{2}, \cdots, x_{2014}$ satisfy
$$
\left|x_{1}\right|=99,\left|x_{n}\right|=\left|x_{n-1}+1\right| \text {, }
$$
where, $n=2,3, \cdots, 2014$. Find the minimum value of $x_{1}+x_{2}+\cdots+x_{2014}$. | 11. From the given, we have
$$
x_{n}^{2}=x_{n-1}^{2}+2 x_{n}+1(n=2,3, \cdots, 2014) \text {. }
$$
Adding the above 2013 equations, we get
$$
\begin{array}{l}
x_{2014}^{2}=x_{1}^{2}+2\left(x_{1}+x_{2}+\cdots+x_{2013}\right)+2013 \\
\Rightarrow 2\left(x_{1}+x_{2}+\cdots+x_{2014}\right) \\
\quad=x_{2014}^{2}+2 x_{2014}-2... | -5907 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,544 |
2. Given $\sin \alpha+\cos \alpha=\frac{\sqrt{2}}{2}$. Then $\sin ^{4} \alpha+\cos ^{4} \alpha=$ $\qquad$ | 2. $\frac{7}{8}$.
Squaring the given equation yields
$$
\begin{array}{l}
1+2 \sin \alpha \cdot \cos \alpha=\frac{1}{2} \\
\Rightarrow \sin \alpha \cdot \cos \alpha=-\frac{1}{4} . \\
\text { Therefore, } \sin ^{4} \alpha+\cos ^{4} \alpha \\
=\left(\sin ^{2} \alpha+\cos ^{2} \alpha\right)^{2}-2(\sin \alpha \cdot \cos \a... | \frac{7}{8} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,545 |
5. Given a regular tetrahedron $D-ABC$ with a base edge length of 1 and a side edge length of 2. A plane is drawn through point $A$ intersecting the side edges $BD$ and $CD$ at points $E$ and $F$ respectively. When the perimeter of $\triangle AEF$ is minimized, the area of $\triangle AEF$ is $\qquad$ | 5. $\frac{3 \sqrt{55}}{64}$.
As shown in Figure 2, the tetrahedron is cut along the edge $D A$ and unfolded into a pentagon.
When points $A, E, F, A_{1}$ are collinear, the perimeter of the section $\triangle A E F$ is minimized. At this time, the isosceles $\triangle D A A_{1}$ and $\triangle D B C$ have the same an... | \frac{3 \sqrt{55}}{64} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,546 |
8. Consider each permutation of $1,2, \cdots, 8$ as an eight-digit number. Then the number of eight-digit numbers that are multiples of 11 is $\qquad$ | 8. 4608 .
For each such eight-digit number $\overline{a_{1} a_{2} \cdots a_{8}}$, let
$$
A=\left\{a_{1}, a_{3}, a_{5}, a_{7}\right\}, B=\left\{a_{2}, a_{4}, a_{6}, a_{8}\right\} .
$$
Let $S(A)$ and $S(B)$ denote their digit sums, and assume $S(A) \geqslant S(B)$.
Then $S(A)+S(B)=36$.
Thus, $S(A)$ and $S(B)$ have the ... | 4608 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,547 |
9. (20 points) Let \( a, b, c \) be positive numbers. Prove:
\[
\begin{array}{l}
\sqrt{a^{2}+a b+b^{2}}+\sqrt{a^{2}+a c+c^{2}} \\
\geqslant 4 \sqrt{\left(\frac{a b}{a+b}\right)^{2}+\left(\frac{a b}{a+b}\right)\left(\frac{a c}{a+c}\right)+\left(\frac{a c}{a+c}\right)^{2}} .
\end{array}
\] | $$
\begin{array}{l}
(a+b)^{2} \geqslant 4 a b, (a+c)^{2} \geqslant 4 a c . \\
\text { Then } \frac{a b}{a+b} \leqslant \frac{a+b}{4}, \frac{a c}{a+c} \leqslant \frac{a+c}{4} .
\end{array}
$$
Thus, it suffices to prove
$$
\begin{array}{l}
\sqrt{a^{2}+a b+b^{2}}+\sqrt{a^{2}+a c+c^{2}} \\
\geqslant \sqrt{(a+b)^{2}+(a+b)(... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 728,548 |
10. (22 points) Let the ellipse $\frac{y^{2}}{a^{2}}+\frac{x^{2}}{b^{2}}=1(a>b>0)$ and the parabola $x^{2}=2 p y(p>0)$ have a common focus $F$, and let $P Q$ be a common tangent line, with $P$ and $Q$ being the points of tangency. Prove:
$$
P F \perp Q F .
$$ | 10. Let $P\left(x_{1}, y_{1}\right)$ be on the parabola, and $Q\left(x_{2}, y_{2}\right)$ be on the ellipse, with the focus $F\left(0, \frac{p}{2}\right)$.
Thus, the equation of the tangent line to the parabola is
$$
x_{1} x=p\left(y+y_{1}\right),
$$
and the equation of the tangent line to the ellipse is
$$
\frac{y_{2... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,549 |
$$
\begin{array}{l}
\text { 2. Let } z=\sum_{k=1}^{n} z_{k}^{2}, \\
z_{k}=x_{k}+\mathrm{i} y_{k}\left(x_{k} \backslash y_{k} \in \mathbf{R}, k=1,2, \cdots, n\right) \text {, }
\end{array}
$$
$p$ is the real part of the square root of $z$. Prove:
$$
|p| \leqslant \sum_{k=1}^{n}\left|x_{k}\right| .
$$ | Let $p+q \mathrm{i}(p, q \in \mathbf{R})$ be the square root of $z$. Then $p^{2}-q^{2}=\sum_{k=1}^{n}\left(x_{k}^{2}-y_{k}^{2}\right), p q=\sum_{k=1}^{n} x_{k} y_{k}$.
If $|p|>\sum_{k=1}^{n}\left|x_{k}\right|$, then
$$
\begin{array}{l}
p^{2}>\sum_{k=1}^{n} x_{k}^{2} \Rightarrow q^{2}>\sum_{k=1}^{n} y_{k}^{2} . \\
\text... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 728,550 |
11. (22 points) As shown in Figure $1, C$ is the midpoint of the semicircle arc, $P$ is a point on the extension of the diameter $B A$, a tangent line $P D$ is drawn from point $P$ to the semicircle $\odot O$, with $D$ as the point of tangency, the angle bisector of $\angle D P B$ intersects $A C$ and $B C$ at points $... | 11. As shown in Figure 3, connect $O D, D E, B D$.
From $O D \perp P D, C O \perp P O$, we know
$$
\begin{array}{l}
\angle D P B=\angle D O C=2 \angle D A C \\
=2 \angle D B C=2 \angle D P F \\
\Rightarrow \angle D A C=\angle D P E, \angle D P F=\angle D B F \\
\Rightarrow P, A, E, D \text { and } P, B, F, D \text { a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,551 |
1. Given natural numbers $a, b, c$ whose sum is $S$, satisfying $a+b=1014, c-b=497, a>b$. Then the maximum value of $S$ is ( ).
(A) 2014
(B) 2015
(C) 2016
(D) 2017 | -1. D.
From the given, we have $a \geqslant b+1$.
Then $1014=a+b \geqslant 2 b+1$
$$
\begin{array}{l}
\Rightarrow b \leqslant 506.5 \Rightarrow b \leqslant 506 . \\
\text { Also } S=(a+b)+(c-b)+b \\
=1014+497+b=1511+b \\
\leqslant 1511+506=2017,
\end{array}
$$
Therefore, the maximum value of $S$ is 2017. | 2017 | Algebra | MCQ | Yes | Yes | cn_contest | false | 728,552 |
2. As shown in Figure 1, given that $E$ and $F$ are the midpoints of sides $AD$ and $CD$ of a square $ABCD$ with side length 6, $AF$ and $BE$ intersect at point $P$. Then the length of $PC$ is ( ).
(A) 6
(B) 6.5
(C) 5
(D) 5.5 | 2. A.
As shown in Figure 2, extend $A F$ and $B C$ to intersect at point $G$. It is easy to prove
Rt $\triangle A B E$
$\cong \operatorname{Rt} \triangle D A F$
$\cong \mathrm{Rt} \triangle C G F$.
Thus, $B C=A D=C G, \angle A P B=90^{\circ}$. Therefore, $P C=\frac{1}{2} B G=B C=6$. | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 728,553 |
5. Several years ago, the age of the grandfather, who is over seventy, was 10 times the age of the grandson. Several years later, it will be 8 times the age of the grandson. If both the grandfather and the grandson are celebrating their birthdays today, then their ages are ( ) years.
(A) 71,8
(B) $72 、 9$
(C) $73 、 9$
... | 5. A.
Let the current ages of the grandfather and grandson be $x$ years and $y$ years, respectively. The grandfather's age was 10 times the grandson's age $a$ years ago, and will be 8 times the grandson's age $b$ years from now. From the problem, we have:
$$
\left\{\begin{array}{l}
x-a=10(y-a), \\
x+b=8(y+b) .
\end{ar... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 728,554 |
1. Given a set of data consisting of seven positive integers, the only mode is 6, and the median is 4. Then the minimum value of the sum of these seven positive integers is $\qquad$ | 2, 1.26.
Arrange these seven positive integers in ascending order, it is clear that the fourth number is 4. If 6 appears twice, then the other four numbers are $1, 2, 3, 5$, their sum is smaller, being 27; if 6 appears three times, the other three numbers are $1, 1, 2$, their sum is 26.
Therefore, the minimum sum of th... | 26 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,555 |
3. In quadrilateral $A B C D$, $\angle A B D=70^{\circ}$, $\angle C A D=20^{\circ}$, $\angle B A C=48^{\circ}$, $\angle C B D=40^{\circ}$. Then $\angle A C D=$ $\qquad$ | 3. $22^{\circ}$.
As shown in Figure 5, construct the circumcircle of $\triangle A B C$, and let it intersect the extension of $B D$ at point $E$. Connect $A E$.
It is easy to find that $\angle A B C=110^{\circ}$,
$$
\begin{array}{l}
\angle A E B=\angle A C B \\
=22^{\circ}, \\
\angle A B C+\angle A B D \\
=180^{\circ}... | 22^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,556 |
4. Given the parabola $y=a x^{2}+b x+c$ passing through points $A(-1,0)$ and $B(2,0)$, for any non-zero real number $a$, the parabola never passes through point $P\left(x_{0}+1,2 x_{0}^{2}-2\right)$. Then the coordinates of point $P$ are $\qquad$ . | 4. $(-1,6)$.
Notice that, the parabola $y=a(x+1)(x-2)$.
Since point $P$ is not on the parabola, we know
$$
2 x_{0}^{2}-2 \neq a\left(x_{0}+2\right)\left(x_{0}-1\right) \text {. }
$$
Since the above equation holds for any non-zero real number $a$, we have
$$
\left(x_{0}+2\right)\left(x_{0}-1\right)=0 \text {, and } 2-... | (-1,6) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,557 |
II. (25 points) On the arc $\overparen{A B}$ of a sector $O A B$ with a central angle of a right angle, there is a moving point $P (P$ does not coincide with $A, B$), and $P Q$ is tangent to the arc $\overparen{A B}$ and intersects the line $O A$ at point $Q$. Identify the locus of the incenter $I$ of $\triangle O P Q$... | II. The trajectory of the incenter $I$ is the line segment $AB$ (excluding points $A$ and $B$).
In fact, as shown in Figure 6, construct the angle bisector $PM$ of $\angle OPQ$, intersecting $AB$ at point $M$, and connect $OM$ and $AP$.
Since $PQ$ is tangent to arc $\overparen{AB}$, we know $\angle OPQ = 90^{\circ}$.
... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,558 |
2. Given
$$
\left(2 x+\sqrt{4 x^{2}+1}\right)\left(\sqrt{y^{2}+4}-2\right) \geqslant y>0 \text {. }
$$
Then the minimum value of $x+y$ is $\qquad$. | 2. 2 .
Notice,
$$
\begin{array}{l}
\left(2 x+\sqrt{4 x^{2}+1}\right)\left(\sqrt{1+\frac{4}{y^{2}}}-\frac{2}{y}\right) \geqslant 1 \\
\Rightarrow 2 x+\sqrt{4 x^{2}+1} \\
\geqslant \frac{1}{\sqrt{1+\frac{4}{y^{2}}}-\frac{2}{y}}=\sqrt{1+\frac{4}{y^{2}}}+\frac{2}{y} .
\end{array}
$$
When $x=\frac{1}{y}$,
$$
\left(2 x+\sq... | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 728,559 |
$$
\begin{array}{l}
\text { 4. Given } f(x)=\frac{x^{2}}{2 x+1}, f_{1}(x)=f(x), \\
f_{n}(x)=\underbrace{f(\cdots f}_{n \uparrow}(x) \cdots) .
\end{array}
$$
Then $f_{6}(x)=$ | 4. $\frac{1}{\left(1+\frac{1}{x}\right)^{64}-1}$.
Let $a_{n}=f_{n}(x), a_{1}=\frac{x^{2}}{2 x+1}$, and
$$
a_{n}=\frac{a_{n-1}^{2}}{2 a_{n-1}+1} \text {. }
$$
Then $\frac{1}{a_{n}}=\frac{1}{a_{n-1}^{2}}+\frac{2}{a_{n-1}}$
$$
\begin{array}{l}
\Rightarrow \frac{1}{a_{n}}+1=\left(\frac{1}{a_{n-1}}+1\right)^{2}=\cdots=\le... | \frac{1}{\left(1+\frac{1}{x}\right)^{64}-1} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,560 |
5. In the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, it is known that the edge length is 1, point $E$ is on $A_{1} D_{1}$, point $F$ is on $C D$, $A_{1} E$ $=2 E D_{1}, D F=2 F C$. Then the volume of the tetrahedron $B-F E C_{1}$ is $\qquad$ . | 5. $\frac{5}{27}$.
As shown in Figure 2, draw $F F_{1} \perp C_{1} D_{1}$, connect $B_{1} F_{1}$, which intersects $E C_{1}$ at point $K$.
It is easy to see that $B_{1} F_{1} \perp E C_{1}$, and $E C_{1} \perp$ plane $B F K$.
Since $B F$ and $E C_{1}$ are skew perpendicular lines with a distance of 1, and $B F=E C_{1}... | \frac{5}{27} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,561 |
Example 1 Let $S$ be a finite set of points in the plane, no three of which are collinear. For each convex polygon $P$ with vertices in the set $S$, let $a(P)$ be the number of vertices of $P$, and $b(P)$ be the number of points in $S$ that are outside the polygon $P$. Prove that for any real number $x (0 < x < 1)$, we... | Prove that for a set $S$ of points, if each point is randomly colored, with the probability of coloring a point black being $x$ and the probability of coloring it white being $1-x$.
For any convex polygon $P$, let event $E_{P}$ denote the event that all points on the boundary of the polygon $P$ are colored black and a... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 728,562 |
Example 2 Let $n \in \mathbf{Z}_{+}$, real numbers $a_{1}, a_{2}, \cdots, a_{n} \in (0,1)$, for any subset $S \subseteq \{1,2, \cdots, n\}$, define the set function
$$
f(S)=\left(\prod_{i \in S} a_{i}\right)\left[\prod_{j \notin S}\left(1-a_{j}\right)\right] .
$$
If $\sum_{|S| \text { is odd }} f(S)=\frac{1}{2}$, prov... | Prove that $X=\sum_{|S| \text { is odd }} f(S)$.
Consider $n$ coins, each with a probability of landing heads up as $a_{1}, a_{2}, \cdots, a_{n}$. Then $X$ represents the probability that an odd number of coins land heads up.
Obviously, if there exists an $a_{k}=\frac{1}{2}$, then $X=\frac{1}{2}$. Therefore, $X-\frac{... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 728,563 |
1. There are $n$ students who participated in the 2014 China Mathematical Olympiad Winter Camp, each with an admission card. Now, all the admission cards are collected and shuffled before being randomly distributed back to each student. Let $S$ denote the number of students who receive their own admission card. Prove: ... | Hint: Decompose $S$ into indicator random variables for each student. | 1 | Combinatorics | proof | Yes | Yes | cn_contest | false | 728,564 |
Given the sequence $\left\{a_{n}\right\}$ satisfies
$a_{1}=1, a_{2 n}=\left\{\begin{array}{ll}a_{n}, & n \text { is even; } \\ 2 a_{n}, & n \text { is odd, }\end{array}\right.$
$a_{2 n+1}=\left\{\begin{array}{ll}2 a_{n}+1, & n \text { is even; } \\ a_{n}, & n \text { is odd. }\end{array}\right.$
Find the number of posi... | Solve: From the given, we have
$$
\begin{array}{l}
a_{2014}=2 a_{1007}=2 a_{503}=2 a_{251}=2 a_{125} \\
=2\left(2 a_{62}+1\right)=2\left(4 a_{31}+1\right) \\
=2\left(4 a_{15}+1\right)=2\left(4 a_{7}+1\right) \\
=2\left(4 a_{3}+1\right)=2\left(4 a_{1}+1\right)=10 .
\end{array}
$$
In fact, $a_{n}$ can be obtained throug... | 320 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,567 |
Given 424 As shown in Figure $1, A$ is a point inside $\triangle B C D$, satisfying $B C=B A, \angle B C D=\angle C A E$, and the circumcircle of $\triangle E A D$ intersects side $C D$ at point $F$. Prove: $E C=E F$.
Translate the above text into English, please retain the original text's line breaks and format, and ... | Prove as shown in Figure 1, construct the circumcircle of $\triangle C A E$ intersecting $B C$ and $B A$ at points $H$ and $I$ respectively.
Since $B C = B A$, quadrilateral $A C H I$ is an isosceles trapezoid.
Thus, $\frac{B H}{B C} = \frac{B I}{B A}$.
Since $H, C, A, E$ are concyclic,
$$
\begin{array}{l}
\angle B H E... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,568 |
2. $A_{1}, A_{2}, \cdots, A_{s}$ are some subsets of $\{1,2, \cdots, M\}$. Assume that $A_{1}, A_{2}, \cdots, A_{s}$ do not contain each other (i.e., none is a subset of another), and let $a_{i}=\left|A_{i}\right|$. Prove:
$$
\sum_{i=1}^{s} \frac{1}{\mathrm{C}_{M}^{a_{i}}} \leqslant 1 .
$$
(1981, St. Petersburg Mathema... | Let $\sigma=\left(x_{1}, x_{2}, \cdots, x_{M}\right)$ be a random permutation of $\{1,2, \cdots, M\}$, and let the set of all such permutations be denoted by the space $\Omega . E_{i}$ represents the event that the elements of $A_{i}$ appear in the first $a_{i}$ positions of $\sigma$, i.e., $\left\{x_{1}, x_{2}, \cdots... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 728,569 |
Example 2 Label the four vertices of a tetrahedron as $1, 2, 3, 4$. Now divide it into several smaller tetrahedra in such a way that any two of these smaller tetrahedra either have no common points, or share only one common point, or share only one common edge (not part of an edge), or share only one common base (not p... | Proof: There exists a small tetrahedron whose four vertices are labeled exactly as $1, 2, 3, 4$.
Proof: Suppose in this subdivision, the number of small triangles with vertices labeled $1, 2, 3$ that lie entirely on the base triangle (123) of the original tetrahedron is $m$, and the number that lie inside the tetrahed... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,573 |
Example 3 Can the two 1s, two 2s, two 3s, ..., two 2013s be arranged in a row so that the two 1s are separated by one number, the two 2s are separated by two numbers, ..., and the two 2013s are separated by 2013 numbers? Prove your conclusion.
Arrange the two 1s, two 2s, two 3s, ..., two 2013s in a row so that the two... | In general, consider the following problem: For a positive integer $n$, can two 1s, two 2s, ..., two $n$s be arranged in a row such that the two 1s are separated by one number, the two 2s are separated by two numbers, ..., and the two $n$s are separated by $n$ numbers?
To explore this, let's conduct some trials:
When $... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 728,574 |
(1) Draw an accurate diagram, clarify the given, to be proved, and the order of generation of each element;
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 【Thought Analysis】(1) It is not easy to draw an accurate diagram according to the problem, the difficulty lies in how to use the two given equations to determine the positions of points $H$ and $I$.
First, analyze the geometric properties of point $H$.
From $\angle C E H - \angle C H D = 90^{\circ}$, it is not difficul... | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,576 |
Example 3 In an $n \times n$ grid, the numbers $1,2, \cdots, n$ each appear $n$ times. Prove that there must exist a row or a column containing at least $[\sqrt{n}]$ different numbers, where $[x]$ denotes the greatest integer not exceeding the real number $x$.
(2004, USA Math Summer Camp) | To prove that randomly selecting a row or a column, there are a total of $2n$ ways to do so. Let the random variable $X$ represent the number of different numbers in the selected row or column.
Let $I_{i}$ be the indicator random variable, i.e., if $i$ appears in the selected row or column, then $I_{i}=1$, otherwise, ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 728,578 |
Example 2 Given $P$ is a moving point on a fixed circle $\odot O$, a circle $\Gamma$ is constructed with $P$ as the center and a radius smaller than that of $\odot O$, intersecting $\odot O$ at points $T$ and $Q$. Let $TR$ be a diameter of circle $\Gamma$. Construct circles with $R$ and $P$ as centers and $RQ$ as the r... | 【Analysis】(1) Draw the accurate figure 4, understand the given, to be proved, and the order of generation of each element.
(2) Hope to eliminate the circles as much as possible. Each point is determined by two conditions. Clearly, $M$ is the circumcenter of $\triangle P R N$, points $N$ and $Q$ are on $\odot P$, and $R... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,580 |
$\begin{array}{l}\text { 2. If } \frac{y}{x}+\frac{x}{z}=a, \frac{z}{y}+\frac{y}{x}=b, \frac{x}{z}+\frac{z}{y}=c, \\ \text { then }(b+c-a)(c+a-b)(a+b-c)=\end{array}$ | 2.8.
Notice that,
$$
\begin{array}{l}
b+c-a=\left(\frac{z}{y}+\frac{y}{x}\right)+\left(\frac{x}{z}+\frac{z}{y}\right)-\left(\frac{y}{x}+\frac{x}{z}\right) \\
=\frac{2 z}{y} .
\end{array}
$$
Similarly, $c+a-b=\frac{2 x}{z}, a+b-c=\frac{2 y}{x}$.
Therefore, $(b+c-a)(c+a-b)(a+b-c)$ $=\frac{2 z}{y} \cdot \frac{2 x}{z} \c... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,582 |
3. Given quadrilateral $A B C D$ is an isosceles trapezoid, $A B / / C D, A B=6, C D=16 . \triangle A C E$ is a right triangle, $\angle A E C=90^{\circ}, C E=B C=A D$. Then the length of $A E$ is . $\qquad$ | 3. $4 \sqrt{6}$.
As shown in Figure 2, draw $A F \perp C D$ at point $F$. Then
$$
D F=5, C F=11 \text {. }
$$
Let $C E=B C$
$$
=A D=x \text {. }
$$
Thus, $A C^{2}=x^{2}+A E^{2}$, $A C^{2}=A F^{2}+11^{2}=x^{2}-5^{2}+11^{2}$.
Therefore, $x^{2}+A E^{2}=x^{2}-5^{2}+11^{2}$
$$
\Rightarrow A E^{2}=96 \Rightarrow A E=4 \sq... | 4 \sqrt{6} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,583 |
5. In $\triangle A B C$, it is known that $\angle A B C=44^{\circ}, D$ is a point on side $B C$, satisfying $D C=2 A B, \angle B A D=$ $24^{\circ}$. Then the size of $\angle A C B$ is $\qquad$ | $5.22^{\circ}$.
It is easy to know that $\angle A D C=44^{\circ}+24^{\circ}=68^{\circ}$.
As shown in Figure 3, draw ray
$A K$, such that $\angle D A K$
$=68^{\circ}$, and let it intersect $D C$
at point $K$.
Then $A K=D K$. Since $\angle A K D=180^{\circ}-2 \times 68^{\circ}=\angle B$, we have $A K=A B$.
Therefore, $... | 22^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,584 |
6. In the Cartesian coordinate plane $x O y$, the area of the region determined by the system of inequalities
$$
\left\{\begin{array}{l}
|x| \leqslant 2, \\
|y| \leqslant 2, \\
|| x|-| y|| \leqslant 1
\end{array}\right.
$$
is $\qquad$ | 6. 12 .
Obviously, the region is symmetric with respect to both the $x$-axis and the $y$-axis. Therefore, we can assume $x \geqslant 0, y \geqslant 0$.
From $\left\{\begin{array}{l}0 \leqslant x \leqslant 2, \\ 0 \leqslant y \leqslant 2, \\ -1 \leqslant x-y \leqslant 1\end{array}\right.$, draw the part of the region ... | 12 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 728,585 |
Example 4 Let $S=\left\{0,1, \cdots, N^{2}-1\right\}, A$ be an $N$-element subset of $S$. Prove: there exists an $N$-element subset $B$ of $S$ such that the set
$$
A+B=\{a+b \mid a \in A, b \in B\}
$$
contains at least half the number of elements of $S$ when considered modulo $N^{2}$.
(40th IMO Shortlist) | Prove that if we select $N$ elements from the set of $N^{2}$ residue classes with equal probability and independence, the set of these elements is denoted as $B$ (since the selection is independent, the probability of each number being selected each time is $\frac{1}{N^{2}}$, although some elements may be selected repe... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 728,586 |
7. Find all positive real numbers $a$ such that the equation
$$
a^{2} x^{2}+a x+1-13 a^{2}=0
$$
has two integer roots. $\qquad$ . | $7.1, \frac{1}{3}, \frac{1}{4}$.
Let the two integer roots be $\alpha, \beta$. Then, from $\alpha+\beta=-\frac{1}{a}$ being an integer, we know that the positive number $a=\frac{1}{n}$ (where $n$ is a positive integer).
Thus, the original equation can be written as
$$
x^{2}+n x+n^{2}-13=0 \text {. }
$$
By $\Delta=n^{2... | 1, \frac{1}{3}, \frac{1}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,587 |
8. Let the positive divisors of $2014^{2}$ be $d_{1}, d_{2}, \cdots$, $d_{k}$. Then
$$
\frac{1}{d_{1}+2014}+\frac{1}{d_{2}+2014}+\cdots+\frac{1}{d_{k}+2014}
$$
$=$ . $\qquad$ | 8. $\frac{27}{4028}$.
Notice that, $2014^{2}=2^{2} \times 19^{2} \times 53^{2}$.
Thus, $2014^{2}$ has $(2+1)^{3}=27$ positive divisors, i.e., $k=27$.
Let $1=d_{1}<d_{2}<\cdots<d_{27}=2014^{2}$.
Therefore, $d_{i} d_{28-i}=2014^{2}(i=1,2, \cdots, 27)$.
Then $\frac{1}{d_{i}+2014}+\frac{1}{d_{28-i}+2014}$
$$
\begin{array}... | \frac{27}{4028} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,588 |
9. (15 points) Solve the equation for $x$
$$
\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}=(a+1) \sqrt{\frac{x}{x+\sqrt{x}}} .
$$ | From the given, we have
$$
\left\{\begin{array}{l}
x \geqslant 0, \\
x-\sqrt{x} \geqslant 0, \\
x+\sqrt{x} \neq 0
\end{array} \Rightarrow x \geqslant 1 .\right.
$$
The original equation is equivalent to
$$
\begin{array}{l}
x+\sqrt{x}-\sqrt{x^{2}-x}=(a+1) \sqrt{x} \\
\Leftrightarrow \sqrt{x}+1-\sqrt{x-1}=a+1 \\
\Leftri... | x=\left(\frac{a^{2}+1}{2 a}\right)^{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,589 |
11. (20 points) Given a square $A B C D$ with side length $a$ that contains $n$ circles, each with an area no greater than 1, and any line parallel to the sides of the square $A B C D$ intersects at most one circle. Prove: the sum of the areas of these $n$ circles is less than $a$. | 11. Let the diameter of the $i$-th circle $(i=1,2, \cdots, n)$ be $d_{i}$. Then the area of the circle $S_{i}=\frac{\pi}{4} d_{i}^{2}$.
By the problem, we know $S_{i} \leqslant 1$.
If $d_{i}>1$, then $S_{i} \leqslant 1<d_{i}$;
If $0<d_{i} \leqslant 1$, then $S_{i}=\frac{\pi}{4} d_{i}^{2} \leqslant \frac{\pi}{4} d_{i}<d... | S_{1}+S_{2}+\cdots+S_{n}<a | Geometry | proof | Yes | Yes | cn_contest | false | 728,590 |
12. (20 points) Prove: (1) It is possible to divide all positive integers into three groups $A_{1} 、 A_{2} 、 A_{3}$, such that for every integer $n \geqslant 15$, in each of $A_{1} 、 A_{2} 、 A_{3}$, there are two different numbers whose sum is $n$.
(2) If all positive integers are arbitrarily divided into four groups $... | 12. (1) Divide all positive integers into the following three groups:
$A_{1}=\{1,2,3,3 \times 3+3,3 \times 4+3, \cdots\}$,
$A_{2}=\{4,5,6,3 \times 3+2,3 \times 4+2, \cdots\}$,
$A_{3}=\{7,8,9,3 \times 3+1,3 \times 4+1, \cdots\}$.
The following proof shows that for every integer $n \geqslant 15$, in each of $A_{1}$, $A_{... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 728,591 |
2. Let the solution set of $x^{2}-5 x+6 \leqslant 0$ be $P$, and the solution set of $x^{2}-2 a x+a^{2}-1 \leqslant 0$ be $Q$. Given that $P \subset Q$. Then the range of real number $a$ is $\qquad$ | $2.2 \leqslant a \leqslant 3$.
From the problem, we know
$$
\begin{array}{l}
P=\{x \mid 2 \leqslant x \leqslant 3\}, \\
Q=\{x \mid a-1 \leqslant x \leqslant a+1\} .
\end{array}
$$
Since $P \subset Q$, therefore, $a-1 \leqslant 2$, and $3 \leqslant a+1$. Solving this, we get $2 \leqslant a \leqslant 3$. | 2 \leqslant a \leqslant 3 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 728,592 |
3. Given that Figure 1 is the graph of an even function $f(x)$, and Figure 2 is the graph of an odd function $g(x)$.
Let the number of real roots of the equations $f(f(x))=0, f(g(x))=0$,
$$
g(g(x))=0, g(f(x))=0
$$
be $a, b, c, d$ respectively. Then
$$
a+b+c+d=
$$
$\qquad$ | 3. 30 .
From the graph, we know that the range of the function $y=f(x)$ is $[-1,1]$, and the range of $y=g(x)$ is $[-2,2]$.
Notice that, the roots of the equation $f(x)=0$ are $0, x_{1}, x_{2}$, with $\left|x_{1}\right|=\left|x_{2}\right| \in(1,2)$;
The roots of the equation $g(x)=0$ are $0, x_{3}, x_{4}$, with $\le... | 30 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,593 |
Example 5 There are 1600 deputies in the Duma, forming 16000 committees, each with 80 members. Prove: It is possible to find two committees, each of which has at least 4 common members.
(1995-1996, All-Russian Mathematical Olympiad) | To prove that by randomly selecting a pair of committees, there are $\mathrm{C}_{16000}^{2}$ possible pairs, let the random variable $X$ represent the number of common members in this pair of committees. Let $I_{i}$ be the indicator random variable, i.e., if $i$ is a common member of this pair of committees, $I_{i}=1$;... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 728,594 |
5. Let the set $P=\{1,2, \cdots, 2014\}, A \cong P$. If any two numbers in set $A$ have a difference that is not a multiple of 99, and the sum of any two numbers is also not a multiple of 99, then the set $A$ can contain at most $\qquad$ elements. | 5.50.
Let the set
$$
B_{i}=\{99 \times 1+i, 99 \times 2+i, \cdots, 99 \times 20+i\} \text {, }
$$
where, $i=0,1, \cdots, 34$;
$$
B_{j}=\{99 \times 1+j, 99 \times 2+j, \cdots, 99 \times 19+j\},
$$
where, $j=35,36, \cdots, 98$.
Take any $a, b \in A$.
Since the difference between any two numbers in set $A$ is not a mul... | 50 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,595 |
2. Let condition $p$ :
real numbers $m, n$ satisfy $\left\{\begin{array}{l}2<m+n<4, \\ 0<m n<3 ;\end{array}\right.$
condition $q$ :
real numbers $m, n$ satisfy $\left\{\begin{array}{l}0<m<1, \\ 2<n<3 .\end{array}\right.$ Then the correct option is ( ).
(A) $p$ is a sufficient but not necessary condition for $q$
(B) $p... | 2. B.
From condition $q$, we know
$$
2<m+n<4,0<m n<3 \text {. }
$$
Therefore, $p$ is a necessary condition for $q$.
Notice that, when $m=1, n=2$, condition $p$ is satisfied, but condition $q$ is not.
Thus, $p$ is not a sufficient condition for $q$.
In conclusion, $p$ is a necessary but not sufficient condition for $q... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 728,597 |
4. Given the vector $\boldsymbol{a}=(1,1)$. Then the vector $\boldsymbol{b}=$ $\left(\frac{1-\sqrt{3}}{2}, \frac{1+\sqrt{3}}{2}\right)$ is obtained by rotating vector $a$ through $(\quad)$.
(A) clockwise rotation by $60^{\circ}$
(B) clockwise rotation by $120^{\circ}$
(C) counterclockwise rotation by $60^{\circ}$
(D) c... | 4. C.
Let the angle between the two vectors be $\theta$. Then
$$
\cos \theta=\frac{\left(\frac{1-\sqrt{3}}{2}, \frac{1+\sqrt{3}}{2}\right) \cdot(1,1)}{2}=\frac{1}{2} \text {. }
$$
Given $\theta \in\left[0^{\circ}, 180^{\circ}\right]$, thus, $\theta=60^{\circ}$.
Notice that, $\frac{1-\sqrt{3}}{2}<0$.
Therefore, vector... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 728,598 |
5. As shown in Figure 1, in a triangular prism, it is known that the plane determined by points $A$, the midpoint $M$ of $B B_{1}$, and the midpoint $N$ of $B_{1} C_{1}$ divides the triangular prism into two parts with different volumes. The ratio of the volume of the smaller part to the volume of the original triangul... | 5. B.
As shown in Figure 4, connect $NM$, intersecting the extensions of $CB$ and $CC_{1}$ at points $D$ and $E$, respectively, and connect $AD$.
According to the problem, point $E$ is also in the plane determined by points $A$, $M$, $N$, and $D$.
Let the plane $AMN$ intersect $A_{1}C_{1}$ at point $F$. Let the base... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 728,599 |
Example: Given $6 n$ vectors $\boldsymbol{v}_{1}, v_{2}, \cdots, v_{n} \in \mathbf{R}^{n}$, each of length 1, i.e.,
$$
\left|\boldsymbol{v}_{i}\right|=\sqrt{v_{1 i}^{2}+v_{2 i}^{2}+\cdots+v_{n i}^{2}}=1 \text {. }
$$
Prove: There exist $\varepsilon_{1}, \varepsilon_{2}, \cdots, \varepsilon_{n} \in\{1,-1\}$, such that
... | Prove that if $\varepsilon_{1}, \varepsilon_{2}, \cdots, \varepsilon_{n}$ are independently and uniformly drawn from $\{1,-1\}$, and the random variable
$$
X=\left|\varepsilon_{1} v_{1}+\varepsilon_{2} v_{2}+\cdots+\varepsilon_{n} v_{n}\right|^{2} .
$$
Expanding it, we get
$$
\begin{aligned}
X= & \left(\varepsilon_{1}... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 728,600 |
7. Given
$$
f(x)=x^{2}-53 x+196+\left|x^{2}-53 x+196\right| \text {. }
$$
Then $f(20)+f(14)=$ $\qquad$ . | $$
\begin{array}{l}
\text { II.7.0. } \\
\text { Let } g(x)=x^{2}-53 x+196 \\
=(x-4)(x-49)
\end{array}
$$
Then when $x<4$ or $x>49$, $g(x)>0$;
when $4 \leqslant x \leqslant 49$, $g(x) \leqslant 0$.
Thus, when $x=4,5, \cdots, 49$,
$$
f(x)=g(x)+|g(x)|=g(x)-g(x)=0 \text {. }
$$
Therefore, $f(20)+f(14)=0$. | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,601 |
8. Given $0<x<\frac{\pi}{2}, \sin x-\cos x=\frac{\pi}{4}$. If $\tan x+\frac{1}{\tan x}$ can be expressed in the form $\frac{a}{b-\pi^{c}}$ ($a$, $b$, $c$ are positive integers), then $a+b+c=$ $\qquad$ . | 8. 50 .
Squaring both sides of $\sin x-\cos x=\frac{\pi}{4}$ and rearranging yields $\sin x \cdot \cos x=\frac{16-\pi^{2}}{32}$.
Therefore, $\tan x+\frac{1}{\tan x}=\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}$
$$
=\frac{1}{\sin x \cdot \cos x}=\frac{32}{16-\pi^{2}} \text {. }
$$
Thus, $a=32, b=16, c=2$.
Hence, $a+b+... | 50 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,602 |
10. Given that an infinite arithmetic sequence contains 3 terms (arranged in order but not necessarily consecutive): $13, 25, 41$. Then 2013 $\qquad$ (fill in “is” “is not” or “cannot be determined”) a term in the sequence. | 10. Yes.
Let the arithmetic sequence that satisfies the conditions be $\left\{a_{n}\right\}$, with a common difference of $d$.
Without loss of generality, let
$$
a_{1}=13, a_{m}=25, a_{n}=41(n>m>1) \text {. }
$$
Then $d=\frac{25-13}{m-1}=\frac{41-13}{n-1}$.
Thus, $(m-1) d=12,(n-1) d=28$.
Notice that,
$$
\begin{array}... | is | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,603 |
11. Randomly select a three-digit number $I$. Then the probability that the number $I$ contains the factor 5 is $\qquad$ . | 11. $\frac{1}{5}$.
From the problem, we know that this is a classical probability model, because all the events in the experiment are three-digit numbers, totaling 999 $-99=900$ numbers. The events that meet the condition are numbers $I$ that contain the factor 5, i.e., numbers $I$ that are multiples of 5. Among these... | \frac{1}{5} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,604 |
12. Given real numbers $x, y$ satisfy
$$
\left\{\begin{array}{l}
x-y \leqslant 0, \\
x+y-5 \geqslant 0, \\
y-3 \leqslant 0 .
\end{array}\right.
$$
If the inequality $a\left(x^{2}+y^{2}\right) \leqslant(x+y)^{2}$ always holds, then the maximum value of the real number $a$ is $\qquad$ | 12. $\frac{25}{13}$.
From the problem, we know that $a \leqslant \frac{(x+y)^{2}}{x^{2}+y^{2}}$.
Let $f(x, y)=\frac{(x+y)^{2}}{x^{2}+y^{2}}$. Then $a \leqslant f(x, y)_{\text {min }}$.
And $f(x, y)=1+\frac{2}{\frac{x}{y}+\frac{y}{x}}$, the inequality represents a planar region as shown in Figure 5.
Considering that $... | \frac{25}{13} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 728,605 |
13. (16 points) As shown in Figure 2, given that $O$ is a point inside $\triangle ABC$ and satisfies
$$
\begin{array}{l}
\angle B A O \\
=\angle C A O \\
=\angle C B O \\
=\angle A C O .
\end{array}
$$
Investigate the relationship that the three sides of $\triangle ABC$ satisfy, and prove your conclusion. | Three, 13. The three sides of $\triangle A B C$ satisfy the relationship:
$$
B C^{2}=A B \cdot A C \text {. }
$$
The proof is given below.
As shown in Figure 6, extend $A O$,
intersecting the circumcircle of $\triangle B O C$
at point $D$, and connect $B D$ and $C D$.
$$
\begin{array}{l}
\text { Let } \angle B A O \\... | B C^{2}=A B \cdot A C | Geometry | proof | Yes | Yes | cn_contest | false | 728,606 |
14. (16 points) The number of tourists received by a certain tourist area each month approximately follows a periodic pattern, and thus, the number of people working in tourism services in the $n$-th month, $f(n)$, can be approximately described by the function
$$
f(n)=100[A \cos (\omega n+\alpha)+k]
$$
where $n$ is a... | 14. According to the rules (i), (ii), (iii), we know that the function is a periodic function with a period of 12.
Thus,
$$
T=\frac{2 \pi}{\omega}=12 \Rightarrow \omega=\frac{\pi}{6} \text {. }
$$
Since the number of people is the highest in August, we have
$$
\frac{\pi}{6} \times 8+\alpha=2 k \pi(k \in \mathbf{Z}) \t... | n=6,7,8,9,10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,607 |
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