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Question 1 In the right triangle $ABC$, $D$ is the foot of the perpendicular from $A$ to the hypotenuse $BC$, and the incenters of $\triangle ABD$ and $\triangle ACD$ are $I$ and $J$, respectively. The line $IJ$ intersects sides $AB$ and $AC$ at points $K$ and $L$. Prove that the area of $\triangle ABC$ is at least twi... | Prove that, as shown in Figure 1, extending $D I$ and $D J$ intersect the circumcircle of $\triangle A B D$ and the circumcircle of $\triangle A C D$ at points $M$ and $N$ respectively. Then $M$ and $N$ are the midpoints of arcs $\overparen{A B}$ and $\overparen{A C}$ respectively.
From $\angle N A C=\angle N D C=45^{\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,450 |
14. Given the parabola $y^{2}=a x(a>0)$ and the line $x=1$ enclose a closed figure with an area of $\frac{4}{3}$. Then, the coefficient of the $x^{-18}$ term in the expansion of $\left(x+\frac{a}{x}\right)^{20}$ is | 14. 20 .
According to the problem, we know $2 \int_{0}^{1} \sqrt{a x} \mathrm{~d} x=\frac{4}{3} \Rightarrow a=1$. Therefore, the term containing $x^{-18}$ is $\mathrm{C}_{20}^{19} x\left(\frac{1}{x}\right)^{19}$. | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,451 |
20. (12 points) The number of staff and the distribution of male and female staff in two departments, A and B, of a certain unit are shown in Table 1. Now, a stratified sampling method (using simple random sampling without replacement within strata) is adopted to select three staff members from the two departments for ... | 20. (1) The number of people to be drawn from departments A and B are $3 \times \frac{10}{15}=2$ and $3 \times \frac{5}{15}=1$, respectively.
(2) Let event $A$ be: at least one female is drawn from department A.
Then $P(A)=1-\frac{\mathrm{C}_{6}^{2}}{\mathrm{C}_{10}^{2}}=\frac{2}{3}$.
(3) The possible values of $\xi$ a... | \frac{9}{5} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,453 |
22. (12 points) Given the function
$$
f(x)=a x-1-\ln x(a \in \mathbf{R}) \text {. }
$$
(1) Discuss the number of extremum points of the function $f(x)$ within its domain;
(2) If the function $f(x)$ has an extremum at $x=1$, and for any $x \in(0,+\infty), f(x) \geqslant b x-2$ always holds, find the range of the real nu... | 22. (1) Notice that the domain of the function $f(x)$ is $(0,+\infty), f^{\prime}(x)=a-\frac{1}{x}$.
When $a \leqslant 0$, $f^{\prime}(x) < 0$, so $f(x)$ is monotonically decreasing on $(0,+\infty)$.
When $a > 0$, $f(x)$ is monotonically decreasing on the interval $\left(0, \frac{1}{a}\right)$ and monotonically incre... | proof | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 730,454 |
2. Given $\log _{\sqrt{7}}(5 a-3)=\log _{\sqrt{a^{2}+1}} 5$. Then the real number
$$
a=
$$
. $\qquad$ | 2. 2 .
Simplify the original equation to
$$
\log _{7}(5 a-3)=\log _{a^{2}+1} 5 \text {. }
$$
Since $f(x)=\log _{7}(5 x-3)$ is an increasing function for $x>\frac{3}{5}$, and $g(x)=\log _{5}\left(x^{2}+1\right)$ is also an increasing function, and $f(2)=$ $g(2)=1$, therefore, $a=2$. | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,455 |
3. Let $f(x)=x^{2}+a x+b$ have two real roots in the interval $[0,1]$. Then the range of $a^{2}-2 b$ is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 3. $[0,2]$.
Notice that, $f(x)=\left(x+\frac{a}{2}\right)^{2}+b-\frac{a^{2}}{4}$ has two real roots in the interval $[0,1]$.
According to the problem, $a$ and $b$ satisfy
$$
\begin{array}{l}
f(0)=b \geqslant 0, f(1)=a+b+1 \geqslant 0, \\
a^{2}-4 b \geqslant 0,0 \leqslant-\frac{a}{2} \leqslant 1 .
\end{array}
$$
In su... | [0,2] | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,456 |
5. Given two propositions:
Proposition $p$ : The function $f(x)=\log _{a} x(x>0)$ is monotonically increasing; Proposition $q$ : The function $g(x)=x^{2}+a x+1>0(x \in \mathbf{R})$. If $p \vee q$ is a true proposition, and $p \wedge q$ is a false proposition, then the range of real number $a$ is | 5. $(-2,1] \cup[2,+\infty)$.
It is easy to see that proposition $p$ holds if and only if $a>1$; proposition $q$ holds if and only if $-2<a<2$.
If $p \vee q$ is true, and $p \wedge q$ is false, then $a \in(-2,1] \cup[2,+\infty)$. | (-2,1] \cup[2,+\infty) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,457 |
6. Let $S$ be the set of all rational numbers in the interval $\left(0, \frac{5}{8}\right)$, for the fraction $\frac{q}{p} \in S, (p, q)=1$, define the function $f\left(\frac{q}{p}\right)=\frac{q+1}{p}$. Then the number of roots of $f(x)=\frac{2}{3}$ in the set $S$ is $\qquad$ | 6.5.
Since $f(x)=\frac{2}{3}$, let $q=2 m-1, p=3 m\left(m \in \mathbf{Z}_{+}\right)$.
Then, $0<\frac{2 m-1}{3 m}<\frac{5}{8} \Rightarrow \frac{1}{2}<m<8$. Upon verification, the number of roots of the equation is 5. | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,458 |
7. Given moving points $P$, $M$, and $N$ are on the $x$-axis, circle $\Gamma_{1}$: $(x-1)^{2}+(y-2)^{2}=1$, and circle $\Gamma_{2}$: $(x-3)^{2}+$ $(y-4)^{2}=3$, respectively. Then the minimum value of $|P M|+|P N|$ is $\qquad$ | 7. $2 \sqrt{10}-\sqrt{3}-1$
It is known that the center coordinates of circle $\Gamma_{1}$ are $(1,2)$, and the center coordinates of the circle $\Gamma_{2}$ symmetric to the $x$-axis are $(3,-4)$.
Then the minimum value of $|P M|+|P N|$ is
$$
\begin{array}{l}
\sqrt{(3-1)^{2}+(-4-2)^{2}}-1-\sqrt{3} \\
=2 \sqrt{10}-\sq... | 2 \sqrt{10}-\sqrt{3}-1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,459 |
9. Given $0<\lambda<1$. Planar vectors $\boldsymbol{a} 、 \boldsymbol{b} 、 \boldsymbol{c}$ satisfy
$$
|a|=1,|b|=2,|c|=3 \text {. }
$$
If $\boldsymbol{b} \cdot \boldsymbol{c}=0$, then the set of all values that $|\boldsymbol{a}-\lambda \boldsymbol{b}-(1-\lambda) \boldsymbol{c}|$ cannot take is $\qquad$ . | 9. $\left[\frac{6 \sqrt{13}}{13}-1,4\right]$.
Translate the vectors $\boldsymbol{b}$ and $\boldsymbol{c}$ to the origin $O$, and establish a Cartesian coordinate system with $b$ and $c$ as the positive directions of the $x$-axis and $y$-axis, respectively. Then, the coordinates of the point corresponding to the vector... | \left[\frac{6 \sqrt{13}}{13}-1,4\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,460 |
10. Given $f(x)=\left\{\begin{array}{cc}-2 x, & x<0 ; \\ x^{2}-1, & x \geqslant 0,\end{array}\right.$, the equation
$$
f(x)+2 \sqrt{1-x^{2}}+\left|f(x)-2 \sqrt{1-x^{2}}\right|-2 a x-4=0
$$
has three real roots $x_{1}<x_{2}<x_{3}$. If $x_{3}-x_{2}=2\left(x_{2}-x_{1}\right)$, then the real number $a=$ $\qquad$ | 10. $\frac{\sqrt{17}-3}{2}$.
Let $g(x)=2 \sqrt{1-x^{2}}(-1 \leqslant x \leqslant 1)$.
By $\max \{f(x), g(x)\}$
$$
=\frac{1}{2}(f(x)+g(x)+|f(x)-g(x)|),
$$
the equation can be transformed into
$$
\max \{f(x), g(x)\}=a x+2 .
$$
From $-2 x \geqslant 2 \sqrt{1-x^{2}} \Rightarrow x \leqslant-\frac{\sqrt{2}}{2}$.
Then $\ma... | \frac{\sqrt{17}-3}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,461 |
Question 3 Given a cyclic quadrilateral $ABCD$ with diagonals $AC$ and $BD$ intersecting at point $P$, the circumcircles of $\triangle ADP$ and $\triangle BCP$ intersect $AB$ at points $E$ and $F$ respectively. Let the incenters of $\triangle ADE$ and $\triangle BCF$ be $I$ and $J$ respectively, and the line $IJ$ inter... | Prove that, as shown in Figure 3, extending $E I$ intersects the circumcircle of $\triangle A D E$ at point $M$, then $M$ is the midpoint of arc $\overparen{A D}$; extending $F J$ intersects the circumcircle of $\triangle B C F$ at point $N$, then $N$ is the midpoint of arc $\overparen{B C}$. Let $I E$ and $J F$ inters... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,462 |
$$
\begin{array}{l}
\text { 1. Let } f(x)=x^{2}+a x+b \cos x \text {, and } \\
\{x \mid f(x)=0, x \in \mathbf{R}\} \\
=\{x \mid f(f(x))=0, x \in \mathbf{R}\} \neq \varnothing \text {. }
\end{array}
$$
Then the range of values for $a+b$ is | 1. $[0,4)$.
Let $x_{0} \in\{x \mid f(x)=0, x \in \mathbf{R}\}$.
Then $b=f(0)=f\left(f\left(x_{0}\right)\right)=0$.
Thus, $f(x)=x(x+a)$.
Therefore, $f(f(x))=x(x+a)\left(x^{2}+a x+a\right)$.
When $a=0$, it clearly satisfies the condition;
When $a \neq 0$, since $0$ and $-a$ are not roots of $x^{2}+a x+a=0$, we have $a^{2... | [0,4) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,463 |
3. If the expansion of $(a+2 b)^{n}$ has three consecutive terms whose binomial coefficients form an arithmetic sequence, then the largest three-digit positive integer $n$ is $\qquad$ | 3. 959 .
Let the binomial coefficients $\mathrm{C}_{n}^{k-1}, \mathrm{C}_{n}^{k}, \mathrm{C}_{n}^{k+1}(1 \leqslant k \leqslant n-1)$ of three consecutive terms in the expansion of $(a+2 b)^{n}$ satisfy
$$
\begin{array}{l}
2 \mathrm{C}_{n}^{k}=\mathrm{C}_{n}^{k-1}+\mathrm{C}_{n}^{k+1} . \\
\text { Then } n^{2}-(4 k+1) ... | 959 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,465 |
4. Through a point $A$ on the line $x=-4$, draw a tangent line $l$ to the parabola $C$: $y^{2}=2 p x(p>0)$, with the point of tangency being $B(1,2)$. Line $l$ intersects the $x$-axis at point $D$. $P$ is a moving point on the parabola $C$ different from point $B$. $\overrightarrow{P E}=\lambda_{1} \overrightarrow{P A}... | 4. $y^{2}=\frac{8}{3}\left(x+\frac{1}{3}\right)\left(y \neq \frac{4}{3}\right)$.
It is easy to know that the parabola $y^{2}=4 x$, and the tangent line $l_{A B}: y=x+1$.
Then $D(-1,0), A(-4,-3)$
$$
\Rightarrow \overrightarrow{A D}=\frac{3}{2} \overrightarrow{D B} \text {. }
$$
Thus, $\overrightarrow{P D}=\frac{2}{5} ... | y^{2}=\frac{8}{3}\left(x+\frac{1}{3}\right)\left(y \neq \frac{4}{3}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,466 |
5. In the sequence $\left\{a_{n}\right\}$,
$$
a_{0}=a\left(a \in \mathbf{Z}_{+}\right), a_{n+1}=\frac{a_{n}^{2}}{a_{n}+1} \text {. }
$$
When $0 \leqslant n \leqslant \frac{a}{2}+1$, the greatest integer not exceeding $a_{n}$ is $\qquad$ | 5. $a-n$.
For any positive integer $n$, it is known by recursion that $a_{n}>0$.
By $a_{n}-a_{n+1}=\frac{a_{n}}{a_{n}+1}=1-\frac{1}{1+a_{n}} \in(0,1)$, we know the sequence $\left\{a_{n}\right\}$ is decreasing, and for any $n \in \mathbf{Z}_{+}$,
$a_{n}=a_{0}+\sum_{i=1}^{n}\left(a_{i}-a_{i-1}\right)>a-n$.
Thus, $a_{n-... | a-n | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,467 |
6. Given that the incircle of $\triangle A B C$ touches sides $A B$ and $A C$ at points $E$ and $F$ respectively, and $A D$ is the altitude from $A$ to side $B C$ of $\triangle A B C$, and $A E+A F=A D$. Then the range of $\sin \frac{A}{2}$ is | 6. $\left[\frac{3}{5}, \frac{\sqrt{2}}{2}\right)$.
Let $B C=a, A C=b, A B=c, A D=h$.
From the problem, we have $b+c=a+h$.
By the area formula of a triangle, we get $\sin A=\frac{a h}{b c}$.
By the cosine rule, we have
$$
\begin{array}{l}
\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}=\frac{(b+c)^{2}-a^{2}-2 b c}{2 b c} \\
=\f... | \left[\frac{3}{5}, \frac{\sqrt{2}}{2}\right) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,468 |
$$
\begin{array}{l}
\quad \mid\{(x, y) \mid x^{2}+y^{2} \equiv a(\bmod p), x, y \in\{0, \\
1, \cdots, p-1\}\} \mid .
\end{array}
$$ | 【Analysis】First, if $p \equiv 1(\bmod 4)$, then by Euler's criterion, we know $\left(\frac{-1}{p}\right)=1$, where $\left(\frac{d}{p}\right)$ is the Legendre symbol, defined as
$$
\left(\frac{d}{p}\right)=\left\{\begin{array}{ll}
1, & d \text { is a quadratic residue modulo } p; \\
-1, & d \text { is a quadratic non-re... | p+1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,469 |
7. Let the elements in set $T$ be integers between 1 and $2^{30}$, and their binary representations contain exactly two 1s. If a number is randomly selected from set $T$, then the probability that this number is divisible by 9 is $\qquad$ | 7. $\frac{5}{29}$.
Notice that, the elements in set $T$ are of the form $2^{a}+2^{b}$, where $0 \leqslant a<b \leqslant 29$.
Thus, $|T|=\mathrm{C}_{30}^{2}=435$.
By $9\left|\left(2^{a}+2^{b}\right) \Leftrightarrow 9\right| 2^{a}\left(2^{b-a}+1\right)$
$\Leftrightarrow 9 \mid\left(2^{b-a}+1\right)$
$\Leftrightarrow b-a... | \frac{5}{29} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,470 |
9. (16 points) Let $\{a_n\}$ be a sequence of positive integers with the sum of the first $n$ terms denoted as $S_n$. For any $n \in \mathbf{Z}_{+}$, we have $a_n < a_{n+1}$. Prove: There exists a unique $n_0 \in \mathbf{Z}_{+}$ such that
$$
\frac{S_{n_0+1}}{n_0} \in (a_{n_0+1}, a_{n_0+2}].
$$ | $$
\begin{array}{l}
\text { II. Let } f(n)=S_{n+1}-n a_{n+1} . \\
\text { Then } \frac{S_{n_{0}+1}}{n_{0}}>a_{n_{0}+1} \Leftrightarrow f\left(n_{0}\right)>0, \\
\frac{S_{n_{0}+1}}{n_{0}} \leqslant a_{n_{0}+2} \\
\Leftrightarrow S_{n_{0}+1}-n_{0} a_{n_{0}+2}=S_{n_{0}+2}-\left(n_{0}+1\right) a_{n_{0}+2} \leqslant 0 \\
\L... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 730,471 |
Example 1 As shown in Figure 4, in tetrahedron $A B C D$, $\triangle A B C$ is an equilateral triangle, $A D=B D=2, A D \perp$ $B D, A D \perp C D$. Then the distance from point $D$ to plane $A B C$ is [1] | (2016, National High School Mathematics League Jiangxi Province Preliminary)
【Analysis】According to the problem, we have
$$
A B=\sqrt{A D^{2}+B D^{2}}=2 \sqrt{2} \text {. }
$$
Then $B C=C A=A B=2 \sqrt{2}$,
$$
\begin{array}{l}
C D=\sqrt{A C^{2}-A D^{2}}=2, \\
A D=B D=C D=2 .
\end{array}
$$
Since $B C^{2}=B D^{2}+C D^... | \frac{2 \sqrt{3}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,473 |
Example 2 Given that the base of the regular tetrahedron $S-ABC$ is an equilateral triangle with a side length of 1, and the side edge length is 2. If a section through the line $AB$ divides the volume of the regular tetrahedron into two equal parts, then the cosine value of the plane angle of the dihedral angle formed... | 【Analysis】As shown in Figure 5, establish a spatial rectangular coordinate system. Let $O$ be the projection of point $S$ on the base, $O S=h$. Then
$$
\begin{array}{l}
A(0,0,0), \\
B(1,0,0), \\
C\left(\frac{1}{2}, \frac{\sqrt{3}}{2}, 0\right), \\
O\left(\frac{1}{2}, \frac{\sqrt{3}}{6}, 0\right), \\
S\left(\frac{1}{2},... | \frac{2 \sqrt{15}}{15} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,474 |
5. In a regular tetrahedron \( P-ABC \), \( AB=1, AP=2 \). A plane \( \alpha \) passing through \( AB \) bisects its volume. Then the cosine of the angle between edge \( PC \) and plane \( \alpha \) is \(\qquad\).
(2017, National High School Mathematics League) | Let the midpoints of $AB$ and $AC$ be $K$ and $M$, respectively. It is easy to prove that the plane $ABM$ is the plane $\alpha$.
By the median length formula, we know that $AM^2 = \frac{3}{2}$.
Thus, $KM = \sqrt{AM^2 + AK^2} = \sqrt{\frac{3}{2} - \left(\frac{1}{2}\right)^2} = \frac{\sqrt{5}}{2}$.
It is also easy to see... | \frac{3 \sqrt{5}}{10} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,475 |
Four, (50 points) Find all 2020-element rational number arrays (duplicate numbers are allowed in the array): If any one number is taken out, the remaining numbers can be evenly divided into three groups, and the product of the 673 numbers in each group is the same.
---
The translation maintains the original text's li... | (1) If the array contains 0, by the pigeonhole principle, we know there are at least four 0s; conversely, if there are at least four 0s, the condition is certainly satisfied.
(2) For any $a_{i} \neq 0 (i=1,2, \cdots, 2020)$, let $a_{i}=\frac{x_{i}}{y_{i}}\left(x_{i}, y_{i} \in \mathbf{Z}\right.$ and $\left.\left(x_{i},... | not found | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,476 |
Example 2 Let $\sum$ denote the cyclic sum. In an acute $\triangle ABC$, prove:
$$
\sum \cos A \geqslant 2 \sum \cos A \cdot \cos B .
$$ | Prove starting from a constant, to achieve the structural transformation of the relational expression.
Notice,
$$
\begin{array}{l}
2 \leqslant \frac{\sin A}{\sin B}+\frac{\sin B}{\sin A}, \\
\sin A \cdot \cos A+\sin B \cdot \cos B \\
=\frac{\sin 2 A+\sin 2 B}{2}=\sin (A+B) \cdot \cos (A-B) \\
=\sin C \cdot \cos (A-B) \... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 730,477 |
Example 3 Let $a_{1}, a_{2}, \cdots, a_{n}$ be positive numbers, satisfying for each $k \in\{1,2, \cdots, n\}$, we have $a_{1} a_{2} \cdots a_{k} \geqslant 1$. Prove:
$$
\sum_{k=1}^{n} \frac{k}{\prod_{i=1}^{k}\left(1+a_{j}\right)}<2 .
$$ | 【Analysis】For the proof of inequalities, sometimes it can also be converted to an equation to handle, that is, to hide the inequality sign. The principle is: if positive real numbers $A \leqslant B$, then there exists $\theta(0<\theta \leqslant 1)$, such that $A=B \theta$.
Proof For each $k$,
$\prod_{j=1}^{k}\left(1+a_... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 730,478 |
Example 6 Let $a, b, c, x, y, z \in \mathbf{R}_{+}$, satisfying:
$$
c y + b z = a, \quad a z + c x = b, \quad b x + a y = c.
$$
Try to find the minimum value of the function
$$
f(x, y, z) = \frac{x^{2}}{1+x} + \frac{y^{2}}{1+y} + \frac{z^{2}}{1+z}.
$$
[Analysis] On the surface, this problem is about an algebraic func... | From the conditions, we have
$$
\begin{array}{l}
b(a z+c x-b)+c(b x+a y-c)-a(c y+b z-a)=0 \\
\quad \Rightarrow 2 b c x+a^{2}-b^{2}-c^{2}=0 \\
\quad \Rightarrow x=\frac{b^{2}+c^{2}-a^{2}}{2 b c} .
\end{array}
$$
Similarly, \( y=\frac{a^{2}+c^{2}-b^{2}}{2 a c}, z=\frac{a^{2}+b^{2}-c^{2}}{2 a b} \).
Since \( a, b, c, x, ... | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,482 |
Proposition If the side lengths of the $n$-sided polygon $A_{1} A_{2} \cdots A_{n}(n \geqslant 3, n \in \mathbf{Z}_{+})$ are $a_{i}, \sum_{i=1}^{n} a_{i}=s$, then
(1) $\sum_{i=1}^{n} \frac{a_{i}}{s-2 a_{i}} \geqslant \frac{n}{n-2}$;
(2) $\sum_{i=1}^{n} \frac{s-2 a_{i}}{a_{i}} \geqslant n^{2}-2 n$. | Proof (1) From the condition, we have
$$
\begin{array}{l}
\sum_{i=1}^{n} \frac{a_{i}}{s-2 a_{i}}=\sum_{i=1}^{n} \frac{a_{i}^{2}}{a_{i} \sum_{j \neq i} a_{j}-a_{i}^{2}} \\
\geqslant \frac{\left(\sum_{i=1}^{n} a_{i}\right)^{2}}{M},
\end{array}
$$
where, \( M=\sum_{i=1}^{n}\left(a_{i} \sum_{j \neq i} a_{j}-a_{i}^{2}\righ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 730,483 |
Example 3 Suppose one edge of a tetrahedron is 6, and the other edges are all 5. Then the radius of the circumscribed sphere of this tetrahedron is $\qquad$ [3]
(2016, Hubei Provincial Preliminary of the National High School Mathematics League) | 【Analysis】Let
$$
P A=P B=B C=A B=A C=5, P C=6 \text {. }
$$
As shown in Figure 6, establish a spatial rectangular coordinate system.
Then $A(0,0,0), B(5,0,0), C\left(\frac{5}{2}, \frac{5 \sqrt{3}}{2}, 0\right)$,
$$
P A=P B \text {. }
$$
Let the projection of $P$ on the base $A B C$ be point $P^{\prime}$, and take the... | \frac{20 \sqrt{39}}{39} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,485 |
2. Let planar vectors $\boldsymbol{a} 、 \boldsymbol{b}$ satisfy $|\boldsymbol{a}+\boldsymbol{b}|=3$. Then the maximum value of $a \cdot b$ is $\qquad$ | 2. $\frac{9}{4}$.
Notice that,
$$
\boldsymbol{a} \cdot \boldsymbol{b}=\frac{1}{4}\left((\boldsymbol{a}+\boldsymbol{b})^{2}-(\boldsymbol{a}-\boldsymbol{b})^{2}\right) \leqslant \frac{9}{4},
$$
the equality holds if and only if $\boldsymbol{a}=\boldsymbol{b}$. Therefore, the maximum value of $\boldsymbol{a} \cdot \bold... | \frac{9}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,487 |
3. In the sequence $\left\{a_{n}\right\}$, for $1 \leqslant n \leqslant 5$, we have $a_{n}=n^{2}$, and for all positive integers $n$, we have
$$
a_{n+5}+a_{n+1}=a_{n+4}+a_{n} \text {. }
$$
Then $a_{2023}=$ . $\qquad$ | 3. 17 .
For all positive integers $n$, we have
$$
a_{n+5}+a_{n+1}=a_{n+4}+a_{n}=\cdots=a_{5}+a_{1}=26 \text {. }
$$
Then $a_{n}=26-a_{n+4}=26-\left(26-a_{n+8}\right)=a_{n+8}$, which means $\left\{a_{n}\right\}$ is a sequence with a period of 8.
Therefore, $a_{2023}=a_{7}=26-a_{3}=26-9=17$. | 17 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,488 |
5. In a certain social event, it was originally planned that each pair of people would shake hands exactly once, but four people each shook hands twice and then left. As a result, there were a total of 60 handshakes during the entire event. Then the number of people who initially participated in the event is $\qquad$ | 5. 15 .
Let the number of people participating in the activity be $n+4$, among which, the number of handshakes among the four people who quit is $x\left(0 \leqslant x \leqslant \mathrm{C}_{4}^{2}=6\right)$.
From the problem, we have $\mathrm{C}_{n}^{2}+4 \times 2=60+x$, which simplifies to $n(n-1)=104+2 x$.
Given $0 \... | 15 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,489 |
1. (16 points) Let positive numbers $x, y, z$ satisfy $x+y+z=1$. Prove: for any positive integer $n$, we have
$$
x^{n}+y^{n}+z^{n} \geqslant \frac{1}{3^{n-1}} .
$$ | $$
\begin{array}{l}
x^{n}+(n-1) \frac{1}{3^{n}} \geqslant n x \cdot \frac{1}{3^{n-1}}, \\
y^{n}+(n-1) \frac{1}{3^{n}} \geqslant n y \cdot \frac{1}{3^{n-1}}, \\
z^{n}+(n-1) \frac{1}{3^{n}} \geqslant n z \cdot \frac{1}{3^{n-1}} .
\end{array}
$$
Adding the three inequalities, we get
$$
x^{n}+y^{n}+z^{n}+3(n-1) \frac{1}{3... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 730,490 |
Example 4 As shown in Figure 7, in the quadrilateral pyramid $P-ABCD$, $PA \perp$ base $ABCD$, $BC=CD=2$, $AC=4$, $\angle ACB=\angle ACD=\frac{\pi}{3}$, $F$ is the midpoint of $PC$, and $AF \perp PB$.
Find (1) the length of $PA$;
(2) the sine value of the dihedral angle $B-AF-D$.
(2016, Gansu Provincial Preliminary of ... | 【Analysis】(1) As shown in Figure 8, connect $B D$, intersecting $A C$ at point $O$.
Given $B C=C D=2$, we know that $\triangle B C D$ is an isosceles triangle.
Since $A C$ bisects $\angle B C D$, it follows that $A C \perp B D$.
Taking $O$ as the origin, the directions of $\overrightarrow{O B}$, $\overrightarrow{O C}$,... | \frac{\sqrt{63}}{8} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,491 |
2. (20 points) Let the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ have an eccentricity of $\frac{1}{2}$, and the upper and lower endpoints of the ellipse's minor axis be $A$ and $B$, respectively. A circle centered at $A$ with the semi-major axis length of the ellipse as its radius intersects the ellips... | 2. (1) From the problem, we have
$$
\left\{\begin{array}{l}
a^{2}=b^{2}+c^{2}, \\
e=\frac{c}{a}=\frac{1}{2}
\end{array} \Rightarrow \left\{\begin{array}{l}
a=2c, \\
b=\sqrt{3}c.
\end{array}\right.\right.
$$
Thus, point $A(0, \sqrt{3}c)$, and the semi-major axis of the ellipse is $2c$.
Therefore, the equation of the ci... | \frac{x^{2}}{4}+\frac{y^{2}}{3}=1, (4,0) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,492 |
3. (20 points) There are nine players participating in a table tennis singles tournament, with each pair of players playing at most one match. It is known that a total of 28 matches were played. Prove: There must be four players who have all played against each other. | 3. Let points $v_{1}, v_{2}, \cdots, v_{9}$ represent nine players participating in a table tennis singles tournament. Construct the graph $G=(V, E)$: the vertex set $V=\left\{v_{1}, v_{2}, \cdots, v_{9}\right\}$, and $v_{i}$ is connected to $v_{j}$ (denoted as $\left.v_{i} v_{j} \in E\right)$ if and only if $v_{i}$ an... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 730,493 |
2. Given $P$ is the center of the upper base $\triangle A_{1} B_{1} C_{1}$ of a regular triangular prism $A B C-A_{1} B_{1} C_{1}$, construct a plane $B C D \perp A P$, intersecting the edge $A A_{1}$ at point $D$. If $A A_{1}=2 A B=2$, then the volume of the tetrahedron $D-A B C$ is ( ).
(A) $\frac{\sqrt{3}}{48}$
(B) ... | 2. A.
As shown in Figure 3, let the plane $A A_{1} P$ intersect the edges $B C$ and $B_{1} C_{1}$ at points $E$ and $E_{1}$, respectively.
In the rectangle $A E E_{1} A_{1}$, $A A_{1}=2$,
$A E=A_{1} E_{1}=\frac{\sqrt{3}}{2}$,
$$
A_{1} P=\frac{\sqrt{3}}{3} \text {. }
$$
From $\frac{A A_{1}}{A E}=\frac{A P}{E D}$, we g... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 730,494 |
Example 5 Given a quadrilateral pyramid $P-ABCD$, the base $ABCD$ is a rhombus with side length 2, $\angle ABC=60^{\circ}, E$ is the midpoint of $AB$, $PA \perp$ plane $ABCD$, and the sine of the angle formed by $PC$ and plane $PAD$ is $\frac{\sqrt{6}}{4}$.
(1) Find a point $F$ on edge $PD$ such that $AF \parallel$ pla... | 【Analysis】(1) Let the intersection of $A C$ and $B D$ be $O$. Establish a spatial rectangular coordinate system with $B D$ as the $x$-axis, $C A$ as the $y$-axis, and a line through point $O$ perpendicular to plane $A B C D$ as the $z$-axis, as shown in Figure 9.
Then $A(0,1,0), B(-\sqrt{3}, 0,0), C(0,-1,0)$,
$D(\sqrt{... | \frac{4 \sqrt{31}}{31} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,495 |
9. Given $z \in \mathbf{C}$. If the equation in terms of $x$
$$
4 x^{2}-8 z x+4 \mathrm{i}+3=0
$$
has real roots. Then the minimum value of $|z|$ is $\qquad$ | 9. 1 .
Let $z=a+b \mathrm{i}(a, b \in \mathbf{R}), x=x_{0}$ be the real root of the given equation. Then
$$
\begin{array}{l}
4 x_{0}^{2}-8(a+b \mathrm{i}) x_{0}+4 \mathrm{i}+3=0 \\
\Rightarrow\left\{\begin{array}{l}
4 x_{0}^{2}-8 a x_{0}+3=0, \\
-8 b x_{0}+4=0 .
\end{array}\right.
\end{array}
$$
Eliminating $x_{0}$ a... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,496 |
10. Let $S_{n}=n!\left(\sum_{k=1}^{n} \frac{k}{(k+1)!}-1\right)$. Then
$$
S_{2017}=
$$
$\qquad$ . | 10. $-\frac{1}{2018}$.
From $\frac{k}{(k+1)!}=\frac{1}{k!}-\frac{1}{(k+1)!}$, we get
$$
\begin{array}{l}
S_{n}=n!\left(\sum_{k=1}^{n}\left(\frac{1}{k!}-\frac{1}{(k+1)!}\right)-1\right) \\
=n!\left(-\frac{1}{(n+1)!}\right)=-\frac{1}{n+1} .
\end{array}
$$
Therefore, $S_{2017}=-\frac{1}{2018}$. | -\frac{1}{2018} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,497 |
12. Let the function $f(x)$ be a differentiable function defined on the interval $(-\infty, 0)$, with its derivative being $f^{\prime}(x)$, and $2 f(x) + x f^{\prime}(x) > x^{2}$. Then
$$
(x+2017)^{2} f(x+2017)-f(-1)>0
$$
The solution set is $\qquad$ | 12. $(-\infty,-2018)$.
Transform the original inequality into
$$
(x+2017)^{2} f(x+2017)>(-1)^{2} f(-1) \text {. }
$$
Construct $F(x)=x^{2} f(x)$, so that equation (1) becomes
$$
F(x+2017)>F(-1) \text {. }
$$
Given the condition $2 f(x)+x f^{\prime}(x)>x^{2}$, multiply both sides by $x (x<0)$, we get
$$
F^{\prime}(x)... | (-\infty,-2018) | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 730,498 |
14. (16 points) Given the sequence $\left\{a_{n}\right\}$ satisfies:
$$
a_{1}=2, a_{n+1}=-\frac{\left(S_{n}-1\right)^{2}}{S_{n}}\left(n \in \mathbf{Z}_{+}\right) \text {, }
$$
where $S_{n}$ is the sum of the first $n$ terms of $\left\{a_{n}\right\}$.
(1) Prove: $\left\{\frac{1}{S_{n}-1}\right\}$ is an arithmetic seque... | 14. (1) When $n \geqslant 1$, from the condition we get
$$
\begin{array}{l}
S_{n+1}-S_{n}=-\frac{\left(S_{n}-1\right)^{2}}{S_{n}} \\
\Rightarrow S_{n+1}-1=\frac{S_{n}-1}{S_{n}} .
\end{array}
$$
Thus, $\frac{1}{S_{n+1}-1}-\frac{1}{S_{n}-1}=\frac{S_{n}}{S_{n}-1}-\frac{1}{S_{n}-1}=1$.
Also, $\frac{1}{S_{1}-1}=\frac{1}{2-... | 3 | Algebra | proof | Yes | Yes | cn_contest | false | 730,499 |
2. If the function
$$
f(x)=\left(x^{2}-1\right)\left(x^{2}+a x+b\right)
$$
satisfies $f(x)=f(4-x)$ for any $x \in \mathbf{R}$, then the minimum value of $f(x)$ is $\qquad$ . | 2. -16 .
Notice that, $f(1)=f(-1)=0$.
Also, $f(x)=f(4-x)$, so $f(3)=f(5)=0$.
Therefore, $f(x)=(x^{2}-1)(x-3)(x-5)$ $=(x^{2}-4x+3)(x^{2}-4x-5)$.
Let $t=x^{2}-4x+4 \geqslant 0$, then $f(x)=(t-1)(t-9)=(t-5)^{2}-16$.
Thus, the minimum value of $f(x)$ is -16. | -16 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,500 |
Example 6 In a regular tetrahedron $ABCD$, $E$ and $F$ are on edges $AB$ and $AC$ respectively, satisfying $BE=3, EF=4$, and $EF$ is parallel to plane $BCD$. Then the area of $\triangle DEF$ is $\qquad$
(2017, National High School Mathematics Joint Competition (B Volume)) | 【Analysis】According to property (6), establish a rectangular coordinate system as shown in Figure 10.
Since the edge length of the regular tetrahedron is 7, we have
$A(0,0,0), B(7,0,0), C\left(\frac{7}{2}, \frac{7 \sqrt{3}}{2}, 0\right)$,
$D\left(\frac{7}{2}, \frac{7 \sqrt{3}}{6}, \frac{7 \sqrt{6}}{3}\right), E(4,0,0),... | 2 \sqrt{33} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,501 |
3. As shown in Figure 1, in the regular triangular prism $A B C-A_{1} B_{1} C_{1}$, $D$ and $E$ are points on the side edges $B B_{1}$ and $C C_{1}$, respectively, with $E C=B C=2 B D$. Then the size of the dihedral angle formed by the section $A D E$ and the base $A B C$ is | $3.45^{\circ}$.
Let $B C=2$. Then the area of $\triangle A B C$ is $\sqrt{3}$. Since $E C=B C=2 B D$, we have,
$$
\begin{array}{l}
E C=2, B D=1 \\
\Rightarrow A E=2 \sqrt{2}, A D=D E=\sqrt{5} \\
\Rightarrow S_{\triangle A D E}=\sqrt{6} \\
\Rightarrow \cos \theta=\frac{\sqrt{3}}{\sqrt{6}}=\frac{\sqrt{2}}{2} \Rightarrow ... | 45^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,502 |
5. Let $x, y$ be real numbers. Then the maximum value of $\frac{2 x+\sqrt{2} y}{2 x^{4}+4 y^{4}+9}$ is . $\qquad$ | 5. $\frac{1}{4}$.
$$
\begin{array}{l}
\text { Given } 2 x^{4}+2+2+2 \geqslant 8 x, \\
4 y^{4}+1+1+1 \geqslant 4 \sqrt{2} y, \\
2 x^{4}+4 y^{4}+9 \geqslant 8 x+4 \sqrt{2} y \\
\Rightarrow \frac{2 x+\sqrt{2} y}{2 x^{4}+4 y^{4}+9} \leqslant \frac{1}{4} .
\end{array}
$$
When $x=1, y=\frac{\sqrt{2}}{2}$, the maximum value ... | \frac{1}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,503 |
$$
\begin{array}{l}
\text { 6. Let } a_{n}=1+2+\cdots+n\left(n \in \mathbf{Z}_{+}\right) , \\
S_{m}=a_{1}+a_{2}+\cdots+a_{m}(m=1,2, \cdots) \text {. }
\end{array}
$$
Then among $S_{1}, S_{2}, \cdots, S_{2017}$, the numbers that are divisible by 2 but not by 4 are $\qquad$ in number.
$$ | 6. 252 .
Notice that, $S_{m}=\frac{m(m+1)(m+2)}{6}$.
Thus $S_{m} \equiv 2(\bmod 4)$
$$
\begin{array}{l}
\Leftrightarrow m(m+1)(m+2) \equiv 4(\bmod 8) \\
\Leftrightarrow m \equiv 3(\bmod 8) .
\end{array}
$$
Therefore, among $S_{1}, S_{2}, \cdots, S_{2017}$, the numbers that are divisible by 2 but not by 4 are $\left[\... | 252 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,504 |
7. In the Cartesian coordinate system $x O y$, $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $x^{2}-\frac{y^{2}}{b^{2}}=1(b>0)$, respectively. A tangent line is drawn from point $F_{1}$ to the circle $x^{2}+y^{2}=1$, intersecting the left and right branches of the hyperbola at points $A$ and $B$, res... | $7.1+\sqrt{3}$.
From the known information and the definition of a hyperbola, we get
$$
\begin{array}{l}
A F_{1}=B F_{1}-A B=B F_{1}-B F_{2}=2, \\
A F_{2}=2+A F_{1}=4 .
\end{array}
$$
As shown in Figure 3, let $A B$ be tangent to the circle at point $T$. | 1+\sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,505 |
一、(40 points) As shown in Figure 2, in the inscribed pentagon $ABCDE$ of $\odot O$, $AD$ and $BE$ intersect at point $F$, the extension of $CF$ intersects $\odot O$ at point $P$, and $AB \cdot CD = BC \cdot ED$. Prove:
$$
OP \perp AE.
$$ | Connect $P A$ and $P E$.
Since points $A, B, D, E$ are on $\odot O$, we have,
$$
\begin{array}{l}
\triangle A B F \backsim \triangle E D F \\
\Rightarrow \frac{A B}{E D}=\frac{F B}{F D} .
\end{array}
$$
Similarly, $\frac{P E}{B C}=\frac{P F}{B F}$,
$$
\frac{D C}{P A}=\frac{D F}{P F} \text {. }
$$
(1) $\times$ (2) $\ti... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,507 |
Three, (50 points) On a plane, there are $2 n(n>1, n \in \mathbf{N})$ points, with no three points collinear. Line segments are drawn between any two points, and any $n^{2}+1$ of these line segments are colored red. Prove: There are at least $n$ triangles with all three sides red. | Three, first prove: there must exist a red triangle (a triangle with all three sides red is called a red triangle, hereinafter the same).
Let the red line segments drawn from vertex $A$ be the most.
The red line segments drawn from point $A$ are $A B_{1}, A B_{2}, \cdots, A B_{k}$, then $k \geqslant n+1$.
If there ex... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 730,508 |
1. In the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, the size of the dihedral angle $B-A_{1} C-D$ is $\qquad$
(2016, National High School Mathematics League Fujian Province Preliminary)
Hint Establish a spatial rectangular coordinate system as shown in Figure 11. | Let the edge length of the cube be 1. Then
$$
\begin{array}{l}
D(0,0,0), A(0,1,0), B(1,1,0), \\
C(1,0,0), A_{1}(0,1,1) .
\end{array}
$$
By property (7), the equation of plane $B A_{1} C$ is
$$
x+z-1=0 \text {, }
$$
and the equation of plane $A_{1} C D$ is $y-z=0$.
The cosine of the angle between the normal vectors of... | 120^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,509 |
Four. (50 points) Let $n$ be a positive integer,
$$
1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}=\frac{a_{n}}{b_{n}},
$$
where $a_{n}, b_{n}$ are coprime positive integers. For a prime $p$, let the set
$$
S_{p}=\left\{n\left|n \in \mathbf{Z}_{+}, p\right| a_{n}\right\} \text {. }
$$
Prove: For every prime $p \geqslan... | Lemma: Let $p \geqslant 5$ be a prime number, and $k$ be a non-negative integer. Let $\sum_{i=1}^{p-1} \frac{1}{k p+i}=\frac{t_{k}}{s_{k}}$ (where $t_{k}$ and $s_{k}$ are coprime positive integers). Then $p^{2} \mid t_{k}$.
Proof: Note that,
$$
\begin{array}{l}
\frac{t_{k}}{s_{k}}=\frac{1}{2} \sum_{i=1}^{p-1}\left(\fr... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 730,510 |
2. Given a positive geometric sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{6}+a_{5}+a_{4}-a_{3}-a_{2}-a_{1}=49 \text {. }
$$
Then the minimum value of $a_{9}+a_{8}+a_{7}$ is $\qquad$ | 2. 196.
Let the common ratio be $q$. From the condition, we have
$$
\left(q^{3}-1\right)\left(a_{3}+a_{2}+a_{1}\right)=49.
$$
Clearly, $q^{3}-1>0$.
Then $a_{3}+a_{2}+a_{1}=\frac{49}{q^{3}-1}$.
Thus, $a_{9}+a_{8}+a_{7}=q^{6}\left(a_{3}+a_{2}+a_{1}\right)$
$$
\begin{array}{l}
=\frac{49 q^{6}}{q^{3}-1}=49\left(\sqrt{q^{... | 196 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,511 |
3. Let the function be
$$
f(x)=x^{3}+a x^{2}+b x+c \quad (x \in \mathbf{R}),
$$
where $a, b, c$ are distinct non-zero integers, and
$$
f(a)=a^{3}, f(b)=b^{3} \text {. }
$$
Then $a+b+c=$ $\qquad$ | 3. 18 .
Let $g(x)=f(x)-x^{3}=a x^{2}+b x+c$.
From the problem, we have $g(a)=g(b)=0$.
Thus, $g(x)=a(x-a)(x-b)$
$$
\begin{array}{l}
\Rightarrow b=-a(a+b), c=a^{2} b \\
\Rightarrow b=-\frac{a^{2}}{a+1}=1-a-\frac{1}{a+1} .
\end{array}
$$
Since $b$ is an integer, we have $a+1= \pm 1$.
Also, $a \neq 0$, so $a=-2, b=4, c=1... | 18 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,512 |
4. Let the function be
$$
f(x)=\left(\frac{1}{2}\right)^{x}+\left(\frac{2}{3}\right)^{x}+\left(\frac{5}{6}\right)^{x}(x \in[0,+\infty)) \text {. }
$$
Then the number of integer points on the graph of the function is $\qquad$ | 4.3.
It is known that the function $f(x)$ is monotonically decreasing on the interval $[0,+\infty)$, and
$$
f(0)=3, f(1)=2, f(3)=1 .
$$
When $x>3$, we have
$$
0<f(x)<f(3)=1 \text {. }
$$
Therefore, the number of integer points on the graph of the function $y=f(x)(x \in[0,+\infty))$ is 3. | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,513 |
6. As shown in Figure 1, in the right triangular prism $A B C-A_{1} B_{1} C_{1}$, $\angle A C B=90^{\circ}, B C=C C_{1}$ $=2, A C=4 \sqrt{2}, P$ is a point on $B C_{1}$. Then the minimum value of $C P+P A_{1}$ is $\qquad$ . | 6. $2 \sqrt{13}$.
From the problem, we know that $\triangle C C_{1} B$ is an isosceles right triangle.
Also, $A_{1} C_{1} \perp$ plane $B C_{1}$, so $\angle A_{1} C_{1} B=90^{\circ}$.
Unfolding the dihedral angle $A_{1}-B C_{1}-C$ along $B C_{1}$ into a plane figure, we get quadrilateral $A_{1} C_{1} C B$,
as shown in... | 2 \sqrt{13} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,514 |
8. Six people are playing a coin-tossing game in a circle (the coin is fair), with each person tossing the coin once. The rule is: those who get the coin landing tails up have to perform a show, while those who get heads up do not have to perform. What is the probability that no two performers are adjacent? $\qquad$ | 8. $\frac{9}{32}$.
There are a total of $2^{6}=64$ possible scenarios, with the number of performers being at most three.
It is easy to see that the number of scenarios where the number of performers is exactly $0$, $1$, or $3$ and meets the conditions are $1$, $6$, and $2$, respectively. The number of scenarios wher... | \frac{9}{32} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,515 |
9. Given that a line passing through the focus $F$ of the parabola $y^{2}=4 x$ intersects the parabola at points $M$ and $N$, and $E(m, 0)$ is a point on the $x$-axis. The extensions of $M E$ and $N E$ intersect the parabola at points $P$ and $Q$ respectively. If the slopes $k_{1}$ and $k_{2}$ of $M N$ and $P Q$ satisf... | 9.3.
When $M P$ is not perpendicular to the $x$-axis, let $l_{\text {MР }}: y=k(x-m)$.
Substitute into $y^{2}=4 x$ to get
$$
x^{2}-\left(\frac{4}{k^{2}}+2 m\right) x+m^{2}=0 \text {. }
$$
Then $x_{M} x_{P}=m^{2} \Rightarrow y_{M} y_{P}=-4 m$ $\Rightarrow y_{P}=\frac{-4 m}{y_{M}}$.
When $M P \perp x$-axis, the conclus... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,516 |
2. Let the height of the regular triangular prism $A B C-A_{1} B_{1} C_{1}$ be 2, and the side length of the base be 1. The centroid of the upper base $\triangle A_{1} B_{1} C_{1}$ is $P$. A plane $B C D \perp A P$ is made through the lower base edge $B C$, intersecting the edge $A A_{1}$ at point $D$. Then the area of... | ```
As shown in Figure 12, establish a spatial rectangular coordinate system.
\[
\begin{array}{c}
\text { Let } A D=h \text {. Then } \\
A(0,0,0), \\
B(1,0,0), \\
C\left(\frac{1}{2},-\frac{\sqrt{3}}{2}, 0\right), \\
A_{1}(0,0,2), \\
B_{1}(1,0,2), \\
C_{1}\left(\frac{1}{2},-\frac{\sqrt{3}}{2}, 2\right), \\
P\left(\frac{... | \frac{\sqrt{13}}{8} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,517 |
10. Arrange all positive integers that are coprime with 70 in ascending order. The 2017th term of this sequence is $\qquad$ . | 10. 5881.
It is easy to know that the number of positive integers not exceeding 70 and coprime with 70 is
$$
35-(7+5)+1=24 \text{.}
$$
Let the sequence of all positive integers coprime with 70, arranged in ascending order, be $\left\{a_{n}\right\}$. Then
$$
\begin{array}{l}
a_{1}=1, a_{2}=3, a_{3}=9, \cdots, a_{24}=6... | 5881 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,518 |
1. Given that $a$, $b$, and $c$ are three non-zero real numbers, and $x^{2}-1$ is a factor of the polynomial $x^{3}+a x^{2}+b x+c$. Then the value of $\frac{a b+3 a}{c}$ is ( ).
(A) -2
(B) -1
(C) 1
(D) 2 | - 1. A.
From the fact that $\pm 1$ are roots of the given polynomial, we have
$$
\begin{array}{l}
\left\{\begin{array} { l }
{ 1 + a + b + c = 0 , } \\
{ - 1 + a - b + c = 0 }
\end{array} \Rightarrow \left\{\begin{array}{l}
a=-c, \\
b=-1
\end{array}\right.\right. \\
\Rightarrow \frac{a b+3 a}{c}=-2 .
\end{array}
$$ | -2 | Algebra | MCQ | Yes | Yes | cn_contest | false | 730,519 |
2. In the Cartesian coordinate system, a moving line segment $AB$ of length 3 is sliding on the $x$-axis, with point $A$ to the left of point $B$, and $C(4,4), D(0,1)$. Then the minimum perimeter of quadrilateral $ABCD$ is $(\quad)$.
(A) $7+3 \sqrt{3}$
(B) 13
(C) $8+\sqrt{26}$
(D) $8+3 \sqrt{3}$ | 2. C.
As shown in Figure 2, the point $D$ is symmetric to the $x$-axis at point $D_{1}(0,-1)$. Translating point $D_{1}$ 3 units to the right yields point $D_{2}(3,-1)$. Connecting $C D_{2}$, we find that quadrilateral $A B D_{2} D_{1}$ is a parallelogram,
$$
\begin{array}{l}
B D_{2}=A D_{1}=A D, \\
A D+B C=B D_{2}+B ... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 730,520 |
5. Given 2023 identical-looking coins, among which there are two counterfeit coins of the same weight, and the remaining 2021 coins are genuine and of the same weight. The counterfeit coins weigh differently from the genuine coins. Now, you only want to know whether a counterfeit coin is heavier or lighter than a genui... | 5. B.
Take out a coin and place it elsewhere, then divide the remaining coins into two groups $A$ and $B$ (each group has 1011 coins), and let $W(N)$ represent the total weight of the coins in group $N$.
For the first weighing, compare $W(A)$ with $W(B)$. Assume $W(A) \geqslant W(B)$, then divide group $A$ into three... | B | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 730,521 |
6. Given that the incircle $\odot O$ of quadrilateral $ABCD$ touches $AB$, $BC$, $CD$, and $DA$ at points $E$, $F$, $G$, and $H$ respectively, and $AF$, $DF$ intersect $EG$ at points $M$ and $N$. If $BF=CF=5$, $EG=6$, then the length of $MN$ is ( ).
(A) 3
(B) 4
(C) 5
(D) an indeterminate value | 6. A.
As shown in Figure 3, connect $EF$, $GF$, and $HF$. Draw a line through point $A$ parallel to $EG$, intersecting lines $EF$ and $FH$ at points $P$ and $Q$, respectively. $HF$ intersects $EG$ at point $T$. | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 730,522 |
3. In two regular tetrahedrons $A-OBC$ and $D-OBC$ with their bases coinciding, $M$ and $N$ are the centroids of $\triangle ADC$ and $\triangle BDC$ respectively. Let $\overrightarrow{OA}=\boldsymbol{a}, \overrightarrow{OB}=\boldsymbol{b}, \overrightarrow{OC}=\boldsymbol{c}$. If point $P$ satisfies $\overrightarrow{OP}... | Take $O$ as the origin and the line $O B$ as the $x$-axis, establishing a spatial rectangular coordinate system as shown in Figure 13.
Let $B(1,0,0)$. Then
$$
\begin{array}{l}
C\left(\frac{1}{2}, \frac{\sqrt{3}}{2}, 0\right), \\
A\left(\frac{1}{2}, \frac{\sqrt{3}}{6}, \frac{\sqrt{6}}{3}\right), \\
D\left(\frac{1}{2}, \... | 439 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,523 |
1. Given the sum of $n$ positive integers is 2017. Then the maximum value of the product of these $n$ positive integers is $\qquad$ . | $$
\text { II. } 1.2^{2} \times 3^{671} \text {. }
$$
Since when $1 \times m4$, $2(m-2)>2+(m-2)$.
Therefore, when a $m$ is replaced by a 2 and a $m-2$, the product increases.
Also, $4+4=2+3+3$, but $4 \times 4<2 \times 3 \times 3$, so the $n$ numbers can only be 2 and 3, where the number of 2s does not exceed 2, and ... | 2^{2} \times 3^{671} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,524 |
2. As shown in Figure 1, in the Cartesian coordinate system $x O y$, $A(1,3), B\left(8 \frac{1}{3}, 1 \frac{2}{3}\right)$, $C\left(7 \frac{1}{3}, 4 \frac{2}{3}\right)$, $O A$ intersects the extension of $B C$ at point $D$, $M$ and $N$ are points on line segments $O D$ and $B D$ respectively, and $O M=M N=B N$. Then the... | 2. $\frac{5 \sqrt{10}}{3}$.
As shown in Figure 4, the line segment $B D$ is translated downward by $1 \frac{2}{3}$ units to the line segment $B_{1} D_{1}$, where $B_{1}\left(8 \frac{1}{3}, 0\right)$. The point $O$ is translated to the right to point $P\left(1 \frac{2}{3}, 0\right)$, and the point $B_{1}$ is translated... | \frac{5 \sqrt{10}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,525 |
4. Given the equation about $x$: $k x^{2}+\frac{|x|}{x+6}=0$ has four distinct real solutions. Then the range of the real number $k$ is $\qquad$ . | 4. $k<-\frac{1}{9}$.
Obviously, $k \neq 0, x=0$ is a known solution of the equation.
When $x \neq 0$, the condition becomes $|x|(x+6)=-\frac{1}{k}$ having
three distinct non-zero real solutions.
From the graph of $f(x)=|x|(x+6)$ (as shown in Figure 6), we get $0<-\frac{1}{k}<f(-3) \Rightarrow 0<-\frac{1}{k}<9$
$\Right... | k<-\frac{1}{9} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,526 |
One, (20 points) Let $a, b$ be real numbers, and the equation with respect to $x$
$$
\frac{x}{x-1}+\frac{x-1}{x}=\frac{a+b x}{x^{2}-x}
$$
has no real roots. Find the value of the expression $8 a+4 b+|8 a+4 b-5|$. | One, the original equation can be transformed into
$$
2 x^{2}-(b+2) x+(1-a)=0 \text {. }
$$
Thus, $\Delta=(b+2)^{2}+8 a-8$.
(1) When $\Delta>0$, equation (1) has two distinct real roots.
Since the original equation has no solution, it follows that the two distinct real roots of equation (1) are 0 and 1, i.e.,
$$
\beg... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,527 |
2. In $\triangle A B C$, the sides opposite to $\angle A, \angle B, \angle C$ are $a, b, c$ respectively, and $G$ is the centroid of $\triangle A B C$. If
$$
a \overrightarrow{G A}+b \overrightarrow{G B}+\frac{\sqrt{3}}{3} c \overrightarrow{G C}=\mathbf{0} \text {, }
$$
then $\angle A=$ . $\qquad$ | 2. $30^{\circ}$.
Notice that,
$$
\overrightarrow{G A}+\overrightarrow{G B}+\overrightarrow{G C}=\mathbf{0} \Rightarrow \overrightarrow{G A}=-\overrightarrow{G B}-\overrightarrow{G C} \text {. }
$$
Then $a(-\overrightarrow{G B}-\overrightarrow{G C})+b \overrightarrow{G B}+\frac{\sqrt{3}}{3} c \overrightarrow{G C}=0$
$... | 30^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,528 |
3. There are $n$ ellipses centered at the origin and symmetric with respect to the coordinate axes
the directrices of which are all $x=1$. If the eccentricity of the $k$-th $(k=1,2, \cdots, n)$ ellipse is $e_{k}=2^{-k}$, then the sum of the major axes of these $n$ ellipses is $\qquad$ . | 3. $2-2^{1-n}$.
Let the semi-major axis of the $k$-th ellipse be $a_{k}$, and the semi-focal distance be $c_{k}$. According to the problem,
$$
\begin{array}{l}
\frac{a_{k}^{2}}{c_{k}}=1, e_{k}=\frac{c_{k}}{a_{k}}=2^{-k} \\
\Rightarrow a_{k}=2^{-k} \\
\Rightarrow \sum_{k=1}^{n} a_{k}=\sum_{k=1}^{n} 2^{-k}=1-2^{-n} .
\e... | 2-2^{1-n} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,529 |
5. For the positive integer $n$, define $a_{n}$ as the unit digit of $n^{(n+1)^{n-1}}$. Then $\sum_{n=1}^{2018} a_{n}=$ $\qquad$ . | 5. 5857 .
When $n \equiv 0,1,5,6(\bmod 10)$, $a_{n} \equiv n^{(n+1)^{n+2}} \equiv n(\bmod 10)$; when $n \equiv 2,4,8(\bmod 10)$, $(n+1)^{n+2} \equiv 1(\bmod 4)$, then $a_{n} \equiv n^{(n+1)^{n+2}}=n^{4 k+1} \equiv n(\bmod 10)$; when $n \equiv 3,7,9(\bmod 10)$, $(n+1)^{n+2} \equiv 0(\bmod 4)$, then $a_{n} \equiv n^{(n+... | 5857 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,530 |
Example 1 If the equation concerning $x$
$$
\begin{array}{l}
x^{4}-16 x^{3}+(81-2 a) x^{2}+(16 a-142) x+ \\
a^{2}-21 a+68=0
\end{array}
$$
has all integer roots, find the value of $a$ and solve this equation.
(2009, National Junior High School Mathematics League (Jiangxi Division)) | Consider the original equation as a quadratic equation in $a$, and rearrange it to get
$$
\begin{array}{l}
a^{2}-\left(2 x^{2}-16 x+21\right) a+ \\
\left(x^{4}-16 x^{3}+81 x^{2}-142 x+68\right)=0 .
\end{array}
$$
Notice that, $21=4+17,68=4 \times 17$.
Decompose the "constant term" of equation (1) to get
$$
\begin{arra... | a=-4, x=2,3,4,7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,531 |
6. A five-digit number $\overline{a b c d e}$ satisfies:
$$
ac>d, dd, b>e \text {. }
$$
For example, 34 201, 49 412. If the digits of the number change in a pattern similar to the monotonicity of a sine function over one period, then the five-digit number is said to follow the "sine rule." The number of five-digit num... | 6. 2892.
From the problem, we know that $b$ and $d$ are the maximum and minimum numbers, respectively, and $2 \leqslant b-d \leqslant 9$.
Let $b-d=k$, at this point, there are $10-k$ ways to choose $(b, d)$, and $a$, $c$, and $e$ each have $k-1$ ways to be chosen, i.e., $(a, c, e)$ has $(k-1)^{3}$ combinations. There... | 2892 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,532 |
8. Planar Region
$$
\begin{aligned}
S= & \left\{(x, y) \left\lvert\, x 、 y \in\left[0, \frac{\pi}{2}\right]\right.,\right. \\
& \left.\sin ^{2} x+\sin x \cdot \sin y+\sin ^{2} y \leqslant \frac{3}{4}\right\}
\end{aligned}
$$
The area of the region is | 8. $\frac{\pi^{2}}{6}$.
$$
\begin{aligned}
\text { Given } & 2\left(\sin ^{2} x-\sin x \cdot \sin y+\sin ^{2} y\right) \\
= & 2-2 \cos (x+y) \cdot \cos (x-y)+ \\
& \cos (x+y)-\cos (x-y) \\
= & \frac{3}{2}-2\left(\cos (x+y)+\frac{1}{2}\right)\left(\cos (x-y)-\frac{1}{2}\right) \\
\leqslant & \frac{3}{2} \\
\Rightarrow &... | \frac{\pi^{2}}{6} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,533 |
9. (16 points) If the function
$$
f(x)=256 x^{9}-576 x^{7}+432 x^{5}-120 x^{3}+9 x \text {, }
$$
find the range of the function $f(x)$ for $x \in[-1,1]$. | II. 9. Note that,
$$
\begin{array}{l}
\cos 3 \theta=\cos 2 \theta \cdot \cos \theta-\sin 2 \theta \cdot \sin \theta \\
=\left(2 \cos ^{2} \theta-1\right) \cos \theta-2\left(1-\cos ^{2} \theta\right) \cos \theta \\
=4 \cos ^{3} \theta-3 \cos \theta .
\end{array}
$$
Therefore, $\cos 9 \theta=4 \cos ^{3} 3 \theta-3 \cos ... | [-1,1] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,534 |
11. (20 points) Let the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ satisfy:
$$
a_{1}=3, b_{1}=1
$$
and for any $n \in \mathbf{Z}_{+}$, we have
$$
\left\{\begin{array}{l}
a_{n+1}=a_{n}+b_{n}+\sqrt{a_{n}^{2}-a_{n} b_{n}+b_{n}^{2}} \\
b_{n+1}=a_{n}+b_{n}-\sqrt{a_{n}^{2}-a_{n} b_{n}+b_{n}^{2}}
\end{array}... | 11. (1) From the problem, we have
$$
\begin{array}{l}
a_{n+1}+b_{n+1}=2\left(a_{n}+b_{n}\right), \\
a_{n+1} b_{n+1}=3 a_{n} b_{n} . \\
\text { Also, } a_{1}+b_{1}=4, a_{1} b_{1}=3 \text {, then } \\
a_{n}+b_{n}=\left(a_{1}+b_{1}\right) 2^{n-1}=2^{n+1}, \\
a_{n} b_{n}=a_{1} b_{1} 3^{n-1}=3^{n} .
\end{array}
$$
Notice t... | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,535 |
One, (40 points) Let $S$ be a set of positive integers with the property: for any $x \in S$, the arithmetic mean of the remaining numbers in $S$ after removing $x$ is a positive integer, and it satisfies $1 \in S$, 2016 is the largest element in $S$. Find the maximum value of $|S|$. | Let the elements of set $S$ be
$$
1=x_{1}<x_{2}<\cdots<x_{n}=2016 \text {. }
$$
Then for $1 \leqslant j \leqslant n$, we have
$$
y_{j}=\frac{\sum_{i=1}^{n} x_{i}-x_{j}}{n-1} \in \mathbf{Z}_{+} \text {. }
$$
Thus, for $1 \leqslant i<j \leqslant n$, we have
$$
\begin{array}{l}
y_{i}-y_{j}=\frac{x_{j}-x_{i}}{n-1} \in \m... | 32 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,536 |
Sure, here is the translated text:
```
(40 points) Let real numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy
$$
0<a_{1}<a_{2}<\cdots<a_{n} \text {. }
$$
Prove: $\left(\sum_{i=1}^{n} \frac{1}{1+a_{i}}\right)^{2} \leqslant \frac{1}{a_{1}}+\sum_{i=1}^{n-1} \frac{1}{a_{i+1}-a_{i}}$.
``` | By the Cauchy inequality, the left side of equation (1)
$$
\begin{aligned}
\leqslant & \left(\frac{1}{a_{1}}+\sum_{i=1}^{n-1} \frac{1}{a_{i+1}-a_{i}}\right) . \\
& \left(\frac{a_{1}}{\left(1+a_{1}\right)^{2}}+\sum_{i=1}^{n-1} \frac{a_{i+1}-a_{i}}{\left(1+a_{i+1}\right)^{2}}\right) .
\end{aligned}
$$
Notice that, $\fra... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 730,537 |
Conclusion 1 In a semicircle with diameter $A B$, there is an inscribed circle that is tangent to the semicircle arc and the diameter $A B$ at point $T$. Then the radius $r$ of the inscribed circle is $r=\frac{A T \cdot T B}{A B}$. | Conclusion 1 Proof: As shown in Figure 1, the midpoint of $AB$ is $O$, and the center of the inscribed circle of the semicircle is $O_{1}$. Connect $O O_{1}$ and $O_{1} T$.
Then $O_{1} T=r$.
Let $A T=x, T B=y$.
Without loss of generality, assume $x \geqslant y$. Then
$$
O A=O B=\frac{1}{2}(x+y), O T=\frac{1}{2}(x-y) \t... | r=\frac{A T \cdot T B}{A B} | Geometry | proof | Yes | Yes | cn_contest | false | 730,538 |
Inference 1 As shown in Figure $1, \odot O_{1}$ is the incircle of the semicircle, tangent to the diameter at point $T$, and tangent to the line $C D$ perpendicular to $A B$. $C D$ intersects $A B$ at point $D$, and intersects the semicircle arc at point $C$, then
(1) $A C=A T$;
(2) $C T$ bisects $\angle B C D$. | Proof of Inference 1 (1) Let $r$ be the radius of $\odot O_{1}$, and let $A T=x, T B=y, A D=a, D B=b$.
When $x \geqslant y$, $x=a+r, y=b-r$.
By $r=\frac{x y}{x+y}=\frac{(a+r)(b-r)}{a+b}$
$\Rightarrow r^{2}+2 a r-a b=0$
$\Rightarrow r=\sqrt{a(a+b)}-a$
$\Rightarrow A T=A D+D T=\sqrt{a(a+b)}$.
By the projection theorem,
$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,539 |
14. In 1993, American mathematician F. Smarandache proposed many number theory problems, attracting the attention of scholars both at home and abroad. One of these is the famous Smarandache function. The Smarandache function of a positive integer \( n \) is defined as
\[
S(n)=\min \left\{m \left| m \in \mathbf{Z}_{+}, ... | 14. (1) It is easy to know that $16=2^{4}$. Then $S(16)=6$.
From $2016=2^{5} \times 3^{2} \times 7$, we know
$S(2016)=\max \left\{S\left(2^{5}\right), S\left(3^{2}\right), S(7)\right\}$.
Also, $S(7)=7, S\left(3^{2}\right)=6, S\left(2^{5}\right)=8$, so $S(2016)=8$.
(2) From $S(n)=7$, we know $n \mid 7!$.
Thus, $n \leq... | 5040 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 730,541 |
15. As shown in Figure 1, points $A$ and $A^{\prime}$ are on the $x$-axis and are symmetric with respect to the $y$-axis. A line passing through point $A^{\prime}$ and perpendicular to the $x$-axis intersects the parabola $y^{2}=2 x$ at points $B$ and $C$. Point $D$ is a moving point on segment $A B$, and point $E$ is ... | 15. (1) Let $A\left(-2 a^{2}, 0\right), A^{\prime}\left(2 a^{2}, 0\right)$. Then $B\left(2 a^{2}, 2 a\right), C\left(2 a^{2},-2 a\right)$.
Let $D\left(x_{1}, y_{1}\right), \overrightarrow{A D}=\lambda \overrightarrow{A B}$. Then $\overrightarrow{C E}=\lambda \overrightarrow{C A}$.
Thus, $\left(x_{1}+2 a^{2}, y_{1}\righ... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,542 |
2. The function $f(x)=\log _{a} \frac{x+1}{x-1}$ has a range of $(1,+\infty)$ on the interval $(1, a-2)$. Then $a=$ $\qquad$ . | 2. $2+\sqrt{3}$.
From $13$.
Since $\frac{x+1}{x-1}=1+\frac{2}{x-1}$ is decreasing on the interval $(1,+\infty)$, therefore, $f(x)$ is decreasing on the interval $(1,+\infty)$.
Hence $f(a-2)=1$
$\Rightarrow a^{2}-4 a+1=0$
$\Rightarrow a=2 \pm \sqrt{3}$ (discard the smaller root). | 2+\sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,544 |
4. In a chess tournament, $n$ players participate in a round-robin competition. After players A and B each played two games, they withdrew from the competition due to certain reasons. It is known that a total of 81 games were ultimately played. Then $n=$ $\qquad$ | 4. 15 .
If there is no match between A and B, then
$$
\mathrm{C}_{n-2}^{2}+2 \times 2=81 \text{, }
$$
this equation has no positive integer solution.
If A and B have a match, then
$$
\mathrm{C}_{n-2}^{2}+3=81 \Rightarrow n=15 .
$$ | 15 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,545 |
6. Let the right focus of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>b>0)$ be $F$, the eccentricity be $e$, and the line passing through point $F$ with a slope of 1 intersect the two asymptotes of the hyperbola at points $A$ and $B$. If the midpoint of $A B$ is $M,|F M|=c$, then $e=$ $\qquad$ | 6. $\sqrt[4]{2}$.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right), M\left(x_{0}, y_{0}\right)$. Then
$$
\frac{\left(x_{1}+x_{2}\right)\left(x_{1}-x_{2}\right)}{a^{2}}-\frac{\left(y_{1}+y_{2}\right)\left(y_{1}-y_{2}\right)}{b^{2}}=0 \text {. }
$$
Substituting the coordinates of the midpoint $M$ and the slo... | \sqrt[4]{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 730,546 |
Example 5 Let $H$ be the orthocenter of acute $\triangle A B C$, and let $A P, A Q$ be the tangents from point $A$ to the circle with diameter $B C$, touching the circle at points $P, Q$ respectively. Prove that $P, H, Q$ are collinear.
$(1996, \mathrm{CMO})$ | Proof As shown in Figure 8, extend $A P$ and $A Q$ to intersect line $B C$ at points $L$ and $N$. Let line $A H$ intersect side $B C$ at point $D$.
Since $H$ is the orthocenter, we know $A D \perp B C$.
By Conclusion 3, we know that points $A, P, D, Q$ are concyclic, and
$\angle A P D + \angle A Q D = 180^{\circ}$.
Let... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,547 |
9. (16 points) Let the domain of $f(x)$ be $[0,+\infty)$, and $f\left(\lg \left(1+\tan ^{2} x\right)\right)=\cos 2 x$. Solve the inequality $f\left(x^{2}-1\right) \geqslant 1$.
| Let $t=\lg \left(1+\tan ^{2} x\right)$. Then $t \geqslant 0$,
$$
\begin{array}{l}
10^{t}=1+\tan ^{2} x=\frac{1}{\cos ^{2} x} \Rightarrow \cos ^{2} x=10^{-t} \\
\Rightarrow \cos 2 x=2 \cos ^{2} x-1=2 \times 10^{-t}-1 .
\end{array}
$$
Then $f(t)=2 \times 10^{-t}-1$ is decreasing on the interval $[0,+\infty)$.
$$
\begin{... | \{-1,1\} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 730,548 |
Three. (50 points) On a line $l$, there are $n+1$ points numbered $1, 2, \cdots, n+1$ arranged from left to right. In the plane above $l$, connect these points with $n$ continuous curves, satisfying:
(1) Any curve connects only two different points, and at most one curve connects any two points;
(2) If point $i$ is con... | Three, first prove a lemma using mathematical induction.
Lemma: Among $n+1$ points, at most $n$ lines can be connected.
Proof: When $n=0,1$, the conclusion is obviously true.
Assume that when $n \leqslant i(i \geqslant 1)$, the conclusion holds.
Suppose when $n=i+1$, at least $i+2$ lines can be connected.
First, explai... | A_{n}=\frac{1}{n+1} \mathrm{C}_{2 n}^{n} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 730,549 |
For $x_{1}, x_{2}, \cdots, x_{n}$ being real numbers. For $1 \leqslant k \leqslant n$, let $\sigma_{k}=\sum_{1 \leqslant i_{1}<i_{2}<\cdots<i_{k} \leqslant n} x_{i_{1}} x_{i_{2}} \cdots x_{i_{k}}$. Prove:
$$
\begin{array}{l}
\prod_{k=1}^{n}\left(x_{k}^{2}+1\right) \\
\geqslant 2\left|\sum_{k=0}^{\left[\frac{n}{2}\right... | Let $f(x)=\prod_{k=1}^{n}\left(x+x_{k}\right)$.
By definition, we know $f(x)=\sum_{k=0}^{n} \sigma_{n-k} x^{k}$.
$$
\begin{array}{l}
\text { Then } \prod_{k=1}^{n}\left(x_{k}^{2}+1\right) \\
=\left(\prod_{k=1}^{n}\left(x_{k}+\mathrm{i}\right)\right)\left(\prod_{k=1}^{n}\left(x_{k}-\mathrm{i}\right)\right) \\
=f(\mathrm... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 730,550 |
In a convex quadrilateral $ABCD$, $AB + CD = BC + DA$. If points $P, Q, M, N$ are on sides $DC$, $AB$, $AD$, $BC$ respectively, and
$$
\frac{DP}{PC} = \frac{AQ}{QB} = \frac{DA}{BC}, \frac{AM}{MD} = \frac{BN}{NC} = \frac{AB}{CD},
$$
prove: $\frac{MN^2}{PQ^2} = \frac{AB \cdot CD}{BC \cdot DA}$. | Let $A B=a, B C=b, C D=c, D A=d$. The four interior angles are $\alpha, \beta, \gamma, \delta$.
Then $a+c=b+d$.
As shown in Figure 1, connect $P M, P N, Q M, Q N$, and let $P Q$ intersect $M N$ at point $O$.
$$
\begin{array}{l}
\text { By } \frac{A Q}{Q B}=\frac{D A}{B C}=\frac{d}{b} \Rightarrow \frac{A Q}{A B}=\frac{d... | \frac{AB \cdot CD}{BC \cdot DA} | Geometry | proof | Yes | Yes | cn_contest | false | 730,551 |
Given 575 As shown in Figure $2, \triangle A B C$ has a circumcircle $\odot O$. A tangent line to $\odot O$ is drawn through point $A$, intersecting $B C$ at point $D$. $P$ is any point on $O D$, and $P M \perp A B$ at point $M$, $P N \perp$ $A C$ at point $N$. Let the orthocenter of $\triangle A M N$ be $H$. Prove:
$$... | Prove that the orthocenter $H^{\prime}$ of $\triangle P M N$ lies on $B C$. As shown in Figure 2, let $L$ be the midpoint of side $B C$, $A I$ be the diameter of $\odot O$, and $I B, I C$ intersect $O D$ at points $J, K$ respectively.
From $A, O, L, D$ being concyclic,
$\Rightarrow \angle A O D=\angle A L D$.
Also, $\a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,552 |
Given 576 As shown in Figure $4, \odot O_{1}$ and $\odot O_{2}$ are separate, and both are tangent to $\odot O$. The chord $AB$ of $\odot O$ is the external common tangent of $\odot O_{1}$ and $\odot O_{2}$. The two internal common tangents $CD$ and $EF$ of $\odot O_{1}$ and $\odot O_{2}$ intersect $AB$ at points $C$ a... | Prove that as shown in Figure 4, extend $FE$ and $DC$ to intersect $\odot O$ at points $T$ and $S$, respectively, and connect $SA$, $SB$, $TA$, and $TB$.
Let $\odot O_{1}$ and $\odot O_{2}$ be tangent to $AB$ at points $K$ and $L$, to $SD$ at points $U$ and $V$, and to $FT$ at points $Y$ and $X$.
For $\odot O_{1}$ an... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,553 |
Title As shown in Figure 1, in $\triangle A B C$, $A B>A C$, the incircle $\odot I$ touches the three sides at points
$D, E, F, M$ is the midpoint of
side $B C$,
$A H \perp B C$ at point $H$,
the angle bisector of $\angle B A C$ $A I$ intersects
lines $D E, D F$ at
points $K, L$. Prove:
$M, L, H, K$ are concyclic. ${ }... | After carefully interpreting this problem, the author believes that its geometric configuration has richer connotations. For this reason, the author further explores this problem.
First, let's look at two variations.
Variation 1 As shown in Figure 2, in $\triangle ABC$, $AC > AB$, the incircle $\odot I$ touches the thr... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,554 |
Variation 3 As shown in Figure 4, in $\triangle A B C$, $A B>A C$,
$D, E, F$ are the midpoints of the three sides, $A H \perp B C$ at point $H$, the angle bisector of $\angle B A C$ intersects lines $D E$ and $D F$ at points $L$ and $K$ respectively. Prove: $D, L, H, K$ are concyclic. | Proof of Variant 3 As shown in Figure 4, connect $B K$ and $H K$.
By the given conditions,
$$
\begin{array}{l}
F D=A E, \\
\angle F A L=\angle E L A=\angle D L K, \\
\angle E A L=\angle F K A .
\end{array}
$$
Since $A K$ is the angle bisector of $\angle B A C$, we have
$$
\begin{array}{l}
\angle F A L=\angle E L A=\an... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,555 |
Conclusion 2 The semicircular arc with diameter $A B$ is tangent to the side $P K$ of $\angle A P K$ at point $E$, the diameter $A B$ lies on the side $A P$, $C$ is the midpoint of the semicircular arc, the angle bisector of $\angle A P K$ intersects $C A$ and $C B$ at points $G$ and $H$ respectively, then $P, A, G, E$... | For the proof of Conclusion 2, as shown in Figure 2, let the midpoint of $AB$ be $O$, and the line $OC$ intersects $PK$ at point $K$.
By $K O \perp O P, O E \perp P K$
$$
\begin{aligned}
\Rightarrow & 2 \angle E P G=\angle E P O=\angle E O C \\
& =2 \angle E B C=2 \angle E A C \\
\Rightarrow & \angle E P G=\angle E A G... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,561 |
Conclusion As shown in Figure 1, circles $\Gamma_{1}$ and $\Gamma_{2}$ intersect at points $A$ and $X$. Through point $A$, lines $AB$ and $AC$ are drawn, intersecting circle $\Gamma_{1}$ at points $E$ and $F$, and circle $\Gamma_{2}$ at points $B$ and $C$. Then
(1) $\triangle X E B \backsim \triangle X F C$,
$\triangle... | This conclusion generally has two uses in problem-solving.
(1) Using cyclic quadrilaterals to prove similarity.
If the problem has already given circles $\Gamma_{1} 、 \Gamma_{2}$, then
$\triangle X E B \backsim \triangle X F C$,
$\triangle X E F \backsim \triangle X B C$.
(2) Using similarity to prove cyclic quadrilat... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,564 |
Example 1 Given that the circumcircle of $\triangle ABC$ is $\Gamma$, the incenter is $I$, $M$ is the midpoint of $BC$, and $D$, $E$, $F$ are on sides $BC$, $CA$, $AB$ respectively, such that $ID \perp BC$, $IE \perp AI$, $IF \perp AI$. If the circumcircle of $\triangle AEF$ intersects circle $\Gamma$ at a point $X$ di... | Proof As shown in Figure 2, let $A M$ intersect the circle $\Gamma$ at point $H$, and extend $H D$ to intersect the circle $\Gamma$ at point $X^{\prime}$.
It suffices to prove that point $X$ coincides with $X^{\prime}$.
Thus, the problem is reduced to conclusion (2).
Hence, it suffices to prove: $\frac{X^{\prime} C}{X^... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,565 |
Example 2 As shown in Figure 3, let $H$ be the orthocenter of acute $\triangle A B C$, and $D$ be the midpoint of side $B C$. A line through point $H$ intersects sides $A B$ and $A C$ at points $F$ and $E$ respectively, such that $A E=A F$. Ray $D H$ intersects the circumcircle of $\triangle A B C$ at point $P$. Prove:... | Proof As shown in Figure 3, extend $H D$ to intersect the circumcircle of $\triangle A B C$ at point $Q$, and connect the auxiliary lines as shown in the figure.
Since $D$ is the midpoint of side $B C$, we have
$\frac{S_{\triangle P B Q}}{S_{\triangle P C Q}}=\frac{\frac{1}{2} B P \cdot B Q \sin \angle P B Q}{\frac{1}{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,566 |
Example 3 As shown in Figure $4, \odot O$ is the circumcircle of $\triangle ABC$, $AD$ bisects $\angle BAC$, intersecting $\odot O$ at point $D, H$ is the orthocenter of $\triangle ABC$, $CE \perp AB$ at point $E, BF \perp AC$ at point $F$, the circumcircle $\odot P$ of $\triangle AFE$ intersects $\odot O$ at point $G,... | $$
\begin{array}{l}
\text { Given } \angle E B H=\angle F C H, \angle B E C=\angle B F C=90^{\circ} \\
\Rightarrow \triangle B E H \sim \triangle C F H \Rightarrow \frac{B H}{C H}=\frac{B E}{C F} . \\
\text { Given } \angle A E G=\angle A F G, \angle E B G=\angle F C G \\
\Rightarrow \triangle B E G \sim \triangle C F ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 730,567 |
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