problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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Example 5 Given the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ has an eccentricity of $\frac{\sqrt{6}}{3}$, and the distance from one endpoint of the minor axis to the right focus is $\sqrt{3}$.
(1) Find the equation of the ellipse $C$;
(2) Suppose the line $l$ intersects the ellipse $C$ at points $A... | Solution: (1) Let the semi-focal distance of the ellipse be $c$. According to the problem, we have
$$
\left\{\begin{array}{l}
\frac{c}{a}=\frac{\sqrt{6}}{3}, \\
a=\sqrt{3} .
\end{array}\right.
$$
Solving this, we get $c=\sqrt{2}, b=1$.
Therefore, the equation of the ellipse is $\frac{x^{2}}{3}+y^{2}=1$.
(2) Let $A\left... | \frac{\sqrt{3}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,529 |
Three. (20 points) Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=1$,
$$
a_{n}=n^{2}\left[1+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\cdots+\frac{1}{(n-1)^{2}}\right]
$$
$(n=2,3, \cdots)$. Prove:
(1) $\frac{a_{n}+1}{a_{n+1}}=\frac{n^{2}}{(n+1)^{2}}(n=2,3, \cdots)$;
(2) $\left(1+\frac{1}{a_{1}}\right)\left(1+\frac{1}... | (1) From $\frac{a_{n}}{n^{2}}=1+\frac{1}{2^{2}}+\cdots+\frac{1}{(n-1)^{2}}$ $(n=2,3, \cdots)$, we get
$$
\begin{array}{l}
\frac{a_{n+1}}{(n+1)^{2}}=1+\frac{1}{2^{2}}+\cdots+\frac{1}{(n-1)^{2}}+\frac{1}{n^{2}} \\
=\frac{a_{n}}{n^{2}}+\frac{1}{n^{2}}=\frac{a_{n}+1}{n^{2}} . \\
\text { Hence } \frac{a_{n}+1}{a_{n+1}}=\fra... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 720,530 |
Four. (20 points) As shown in Figure 5, in the rectangular coordinate plane $x O y$, the four vertices of quadrilateral $A B C D$ are all on the hyperbola $y=\frac{1}{x}$. Among them, points $A$ and $B$ are in the first quadrant, and points $C$ and $D$ are in the second quadrant. The diagonal $A C$ passes through the o... | As shown in Figure 10, let \( A\left(a, \frac{1}{a}\right) \),
\[
\begin{array}{l}
B\left(b, \frac{1}{b}\right), \\
C\left(-a,-\frac{1}{a}\right), \\
D\left(d, \frac{1}{d}\right).
\end{array}
\]
Draw vertical lines through points \( B \) and \( D \) parallel to the \( y \)-axis, intersecting \( AC \) at points \( P \)... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,531 |
Five. (20 points) A function $f(x)$, if for any triangle, as long as its three side lengths $a, b, c$ are within the domain of $f(x)$, then $f(a), f(b), f(c)$ are also the side lengths of some triangle, is called a “triangle-preserving function”.
(1) If $g(x)$ is a periodic function defined on $\mathbf{R}$ with a range... | (1) Let $T>0$ be a period of the function $g(x)$. Since its range is $(0,+\infty)$, there exist $m > n > 0$ such that $g(m)=2, g(n)=1$.
Take a positive integer $\lambda > \frac{m-n}{T}$, then $m, \lambda T+n, \lambda T+n$ can be the three sides of a triangle. However, $g(m)=2, g(\lambda T+n)=1, g(\lambda T+n)=1$ canno... | \frac{5 \pi}{6} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,532 |
-、(50 points) As shown in Figure 6, in the convex quadrilateral $ABCD$, $M$ is the midpoint of side $AB$, and $MC=MD$. Perpendiculars are drawn from points $C$ and $D$ to sides $BC$ and $AD$, respectively, and the intersection of these two perpendiculars is point $P$. A perpendicular $PQ$ is drawn from point $P$ to $AB... | As shown in Figure 11, connect $PA$ and $PB$, and take the midpoints $E$ and $F$ of $PA$ and $PB$ respectively. Connect $EM$, $ED$, $FM$, and $FC$. Then quadrilateral $PEMF$ is a parallelogram. Thus, $\angle PEM = \angle PFM$. Since $ME = \frac{1}{2} BP = CF$, $MF = \frac{1}{2} AP = DE$,
$$
\begin{aligned}
MD & = MC \\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,533 |
II. (50 points) Let $a, b, c \in \mathbf{R}, m>0$. Prove:
$$
\left(a^{2}+m\right)\left(b^{2}+m\right)\left(c^{2}+m\right) \geqslant \frac{3 m^{2}}{4}(a+b+c)^{2} .
$$ | First, prove that if $\alpha, \beta \in\left[0, \frac{\pi}{2}\right)$, then
$$
\sin ^{2}(\alpha+\beta)+\cos ^{2} \alpha \cdot \cos ^{2} \beta \leqslant \frac{4}{3} \text {. }
$$
Notice that
$$
\begin{array}{l}
\sin ^{2}(\alpha+\beta)+\cos ^{2} \alpha \cdot \cos ^{2} \beta-\frac{4}{3} \\
=(\sin \alpha \cdot \cos \beta... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 720,534 |
Three. (50 points) Arrange the $n^2$ natural numbers $1, 2, \cdots, n^2$ (where $n \geqslant 2$) randomly in an $n \times n$ square grid. For any two numbers in the same row or the same column, compute the quotient of the larger number divided by the smaller number, resulting in $n^2(n-1)$ fractions. The smallest of th... | Three, first prove: for any permutation $A$, its characteristic value
$$
l(A)=\frac{n+1}{n} \text {. }
$$
If among the $n$ natural numbers $n^{2}-n+1, n^{2}-n+2$, $\cdots, n^{2}$, there are two in the same row or column, then
$$
l(A) \leqslant \frac{a}{b} \leqslant \frac{n^{2}}{n^{2}-n+1}b$.
If all these $n$ large num... | \frac{n+1}{n} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 720,535 |
From $\odot O$ outside point $P$, draw two tangents to $\odot O$, with $A$ and $B$ as the points of tangency. Then, from point $P$, draw two secants intersecting $\odot O$ at $C, D$ and $E, F$. Line $FC$ intersects $PA$ and $DA$ at $H$ and $G$ respectively. Prove that the intersection of $GB$ and $HE$ lies on $\odot O$... | Proof: As shown in Figure 2, let $G B$ intersect $\odot O$ at point $K_{1}$, and connect $A K_{1}$, $K_{1} F$, $A C$, $B C$, $B D$, $D F$. From
$$
\begin{array}{l}
\triangle G A K_{1} \sim \triangle G B D, \triangle G B C \sim \triangle G F K_{1}, \\
\triangle P B D \sim \triangle P C B, \triangle G A C \sim \triangle ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,536 |
Given a six-digit number $\overline{a b c d e f}$, it is known that $\overline{a b c d e f}$, $\bar{a}$, $\overline{b c}$, and $\overline{d e f}$ are all non-zero perfect squares. Find all six-digit numbers that satisfy these conditions. | Solution: Let $M=\overline{a b c d e f}=n^{2}, a=x^{2}, \overline{b c}=y^{2}$, $\overline{d e f}=z^{2}$, where $1 \leqslant x \leqslant 3, 4 \leqslant y \leqslant 9, 10 \leqslant z \leqslant$ 31.
Then $M=n^{2}=10^{5} x^{2}+10^{3} y^{2}+z^{2}$.
(1) If $x=1$, when $y=6$,
$$
\begin{array}{l}
M=136000+z^{2}, \\
368^{2}<136... | 136900,964324 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,537 |
As shown in Figure 3, from a point $P$ outside $\odot O$, draw a tangent to $\odot O$, with $A$ as the point of tangency. Draw a secant line $PD$ from point $P$, intersecting $\odot O$ at points $C$ and $D$ ($PC < PD$). Let $E$ be a point on $CD$, and $AE$ intersects $\odot O$ at another point $B$. The line $BC$ inters... | Prove: As shown in Figure 3, connect $C G, G A, A C, A D, D T$, and $T B$. Since line $H E F$ intersects $\triangle D B C$, by Menelaus' theorem, we have
$$
\frac{D H}{H B} \cdot \frac{B F}{F C} \cdot \frac{C E}{E D}=1.
$$
Since $\triangle F A B \sim \triangle F C A$, we have
$$
\begin{array}{l}
\frac{B F}{A F}=\frac{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,538 |
238 Test: Can 2007 be expressed in the form
$$
a_{1}^{x_{1}}+a_{2}^{x_{2}}+\cdots+a_{m}^{x_{n}}-b_{1}^{y_{1}}-b_{2}^{y_{2}}-\cdots-b_{n}^{y_{n}}
$$
where $m, n$ are both positive integers greater than 130 and less than 140 (allowing $m$ to equal $n$), $a_{1}, a_{2}, \cdots, a_{m}, b_{1}, b_{2}, \cdots, b_{n}$ are all ... | Solution: It can be done. The reasons are as follows:
Let $m=n=132$,
$$
\begin{array}{l}
a_{i}=\frac{16(i+2)^{2}+(2 i+3)^{2}}{(2 i+4)(2 i+3)}(i=1,2, \cdots, 125), \\
b_{i}=\frac{16(i+2)^{2}-(2 i+3)^{2}}{(2 i+4)(2 i+3)}(i=1,2, \cdots, 125), \\
a_{k}=\frac{(2 k+4)^{2}+(2 k+3)^{2}}{2(2 k+4)(2 k+3)}(k=126,127, \cdots, 132)... | 2007 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,539 |
Example $6 P$ is a point inside $\triangle A B C$, and $D, E, F$ are the feet of the perpendiculars from $P$ to the sides $B C, C A, A B$ respectively. Find all points $P$ that minimize $\frac{B C}{P D}+\frac{C A}{P E}+\frac{A B}{P F}$. | Solution: It is easy to know that
$$
S_{\triangle B C}=\frac{1}{2} B C \cdot P D+\frac{1}{2} C A \cdot P E+\frac{1}{2} A B \cdot P F \text {. }
$$
According to the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
\frac{B C}{P D}+\frac{C A}{P E}+\frac{A B}{P F} \\
=\frac{B C \cdot P D+C A \cdot P E+A B \cdot P F}... | P \text{ is the incenter of } \triangle ABC | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,540 |
Example 7 Find the real solutions of the system of equations:
$$
\left\{\begin{array}{l}
(x-2)^{2}+\left(y+\frac{3}{2}\right)^{2}+(z-6)^{2}=64, \\
(x+2)^{2}+\left(y-\frac{3}{2}\right)^{2}+(z+6)^{2}=25 .
\end{array}\right.
$$ | Solving: By adding and subtracting the two given equations, we get
$$
\left\{\begin{array}{l}
x^{2}+y^{2}+z^{2}=\left(\frac{3}{2}\right)^{2}, \\
-8 x+6 y-24 z=39 .
\end{array}\right.
$$
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
39=-8 x+6 y-24 z \\
\leqslant \sqrt{(-8)^{2}+6^{2}+(-24)^{2}} \cdot \sq... | \left(-\frac{6}{13}, \frac{9}{26},-\frac{18}{13}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,541 |
Example 8 Given that $a, b$ are positive numbers, $n$ is a positive integer, and $\frac{\sin ^{4} \theta}{a}+\frac{\cos ^{4} \theta}{b}=\frac{1}{a+b}$.
Prove: $\frac{\sin ^{2 n} \theta}{a^{n-1}}+\frac{\cos ^{2 n} \theta}{b^{n-1}}=\frac{1}{(a+b)^{n-1}}$. | Proof: By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
1=(a+b)\left(\frac{\sin ^{4} \theta}{a}+\frac{\cos ^{4} \theta}{b}\right) \\
\geqslant\left(\sqrt{a} \cdot \frac{\sin ^{2} \theta}{\sqrt{a}}+\sqrt{b} \cdot \frac{\cos ^{2} \theta}{\sqrt{b}}\right)^{2} \\
=\left(\sin ^{2} \theta+\cos ^{2} \theta\right)... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 720,542 |
Example 1 Given that for any integer $x$, the trinomial $a x^{2}+$ $b x+c$ is a perfect square. Prove: there must be
$$
a x^{2}+b x+c=(d x+e)^{2} \text {. }
$$ | Let $f(x)=a x^{2}+b x+c$.
We need to prove: $a, b, c$ are integers, and $b^{2}=4 a c$.
It is easy to see that $c=f(0)$ is a perfect square, and
$2 b=f(1)-f(-1)$,
$2 a=f(1)+f(-1)-2 c$
are both integers.
If $b$ is not an integer, then $2 b$ is odd.
Let $2 b=2 n+1$.
Thus, $4 b \equiv 2(\bmod 4)$.
Also, $c \equiv 0$ or $1(... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 720,543 |
Example 2 Let $d$ be a positive integer not equal to $2, 5, 13$. Prove: in the set $\{2, 5, 13, d\}$, there can be found two distinct elements $a, b$ such that $ab-1$ is not a perfect square. | To prove: It is sufficient to prove that among $2d-1$, $5d-1$, and $13d-1$, there must be one that is not a perfect square.
Proof by contradiction.
Suppose all three numbers are perfect squares. Since the sum of two numbers
$$
(2d-1) + (5d-1) + (13d-1) = 20d - 3
$$
is a number of the form $4N + 1$, hence among these th... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 720,544 |
Example 1 Let the incircle and circumcircle of acute triangle $\triangle ABC$ be $\omega$ and $\Omega$, respectively, with circumradius $R$. Circle $\omega_{A}$ is internally tangent to $\Omega$ at point $A$ and externally tangent to $\omega$; circle $\Omega_{A}$ is internally tangent to $\Omega$ at point $A$ and inter... | Proof: As shown in Figure 1, let the inradius of $\triangle ABC$ be $r$, the incenter be $I$, and the circumcenter be $O$. The radii of circles $\omega_{A}$ and $\Omega_{A}$ are $u$ and $v$, respectively.
Then $A P_{A}=u, P_{A} O=R-u, I P_{A}=r+u$.
Notice that $P_{A}$ is a point on the side $OA$ of $\triangle A O I$. ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,545 |
Example 2 Let the semi-perimeter of $\triangle A B C$ be $p$, the inradius be $r$, and the distances from the incenter to the vertices $A, B, C$ be $l_{A}, l_{B}, l_{C}$. Prove: $\frac{3}{4}+\frac{r}{l_{A}}+\frac{r}{l_{B}}+\frac{r}{l_{C}} \leqslant \frac{p^{2}}{12 r^{2}}$. | Proof: Let the three sides of $\triangle ABC$ be $a, b, c$, and the area be $S$. Notice that
$$
\begin{array}{l}
\frac{r}{l_{A}}=\sin \frac{A}{2}=\sqrt{\frac{(p-b)(p-c)}{b c}} \\
\leqslant \frac{(p-b)+(p-c)}{2 \sqrt{b c}}=\frac{a}{2 \sqrt{b c}} \\
=\frac{a \sqrt{b c}}{2 b c} \leqslant \frac{a^{2}(b+c)}{4 a b c} . \\
\t... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 720,546 |
Example 3 In $\triangle A B C$, the angle bisector of $\angle A$ intersects $B C$ at point $D$, and the projections of $D$ on $A B$ and $A C$ are $E$ and $F$ respectively. Let the length of $E F$ be $l_{A}$, and similarly define $l_{B}$ and $l_{C}$. If the perimeter of $\triangle A B C$ is $l$, prove: $\frac{l_{A} l_{B... | Proof: Let the three sides of $\triangle ABC$ be $a, b, c$, the area be $S$, the semi-perimeter be $p$, and the lengths of the three angle bisectors be $t_{a}, t_{b}, t_{c}$.
Since $A, E, D, F$ are concyclic, and $AD$ is the diameter, we have
$l_{A}=t_{a} \sin A=\frac{2}{b+c} \sqrt{b c p(p-a)} \sin A \leqslant \sqrt{p... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 720,547 |
Example 4 As shown in Figure 2, given a convex quadrilateral $ABCD, \angle B + \angle D < 180^{\circ}, P$ is a moving point on the plane. Let
$$
\begin{aligned}
f(P)= & PA \cdot BC + \\
& PD \cdot CA + \\
& PC \cdot AB.
\end{aligned}
$$
(1) Prove that when $f(P)$ reaches its minimum value, points $P, A, B, C$ are concy... | (1) Proof: In quadrilateral $PABC$ (including concave and degenerate quadrilaterals), by the generalized Ptolemy's theorem, we have
$PA \cdot BC + PC \cdot AB \geqslant PB \cdot AC$.
Thus, $f(P) \geqslant PB \cdot AC + PD \cdot CA$
$= AC(BP + PD) \geqslant AC \cdot BD$.
Equality holds if and only if $P$ is the intersec... | \sqrt{10} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,548 |
Question 1 As shown in Figure 1, given a convex quadrilateral $ABCD$, $\angle B + \angle D < 180^{\circ}$, and $P$ is a moving point on the plane. Let
$$
\begin{array}{l}
f(P) \\
= PA \cdot BC + \\
PD \cdot CA + \\
PC \cdot AB.
\end{array}
$$
(1) Prove:
When $f(P)$ reaches its minimum value, points $P$, $A$, $B$, and ... | (1) Proof: As shown in Figure 1, construct \(\triangle A P B^{\prime} \backsim \triangle C P B\). Then \(\angle B P B^{\prime} = \angle C P A\),
\[
\frac{A P}{C P} = \frac{B^{\prime} P}{B P}, \quad \frac{A B^{\prime}}{B C} = \frac{P A}{P C}.
\]
Thus, \(\triangle B^{\prime} P B \backsim \triangle A P C\).
Therefore, \(... | \sqrt{10} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,550 |
The third question: Let $a_{k}>0(k=1,2, \cdots, 2008)$. Prove: When and only when $\sum_{k=1}^{2008} a_{k}>1$, there exists a sequence $\left\{x_{n}\right\}$ satisfying the following conditions:
(1) $0=x_{0}<x_{n}<x_{n+1}(n=1,2, \cdots)$;
(2) $\lim _{n \rightarrow \infty} x_{n}$ exists;
(3) $x_{n}-x_{n-1}=\sum_{k=1}^{2... | Prove: Let $y_{n}=x_{n}-x_{n-1}$.
Then the original problem is equivalent to the following problem:
Let $a_{k}>0(k=1,2, \cdots, 2008)$. Prove: When and only when $\sum_{k=1}^{2008} a_{k}>1$, there exists a sequence $\left\{y_{n}\right\}$ satisfying:
(i) $y_{n}>0(n=1,2, \cdots)$;
(ii) $\lim _{n \rightarrow \infty}\left(... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 720,552 |
2. As shown in Figure 1, quadrilateral $ABCD$ is a rectangular piece of paper. Fold $AD$ and $BC$ so that points $A$ and $B$ coincide at point $P$ on side $CD$, then flatten to get creases $EF$ and $GH$. If $PE=8 \text{ cm}$, $PG=6 \text{ cm}$, $EG=10 \text{ cm}$, then the area of the rectangular piece of paper $ABCD$ ... | 2. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,554 |
3. Simplify $\frac{\sqrt{2}+\sqrt{6}}{\sqrt{2+\sqrt{3}}}$, the result is $(\quad)$.
(A) 1
(B) $\sqrt{3}$
(C) 2
(D) 4 | 3. C
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,555 |
4. Point $P$ on the plane of $\triangle A B C$ such that the areas of $\triangle A B P$, $\triangle B C P$, and $\triangle A C P$ are equal. The number of such points $P$ is ( ).
(A) 8
(B) 4
(C) 3
(D) 1 | 4. B
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,556 |
5. In a Cartesian coordinate system, let $A(-1,-2)$, $B(4,-1)$, $C(m, 0)$, and $D(n, n)$ be the four vertices of a quadrilateral. When the perimeter of quadrilateral $ABCD$ is minimized, the value of $\frac{m}{n}$ is ( ).
(A) -2
(B) -1
(C) $-\frac{1}{2}$
(D) 1 | 5.A
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,557 |
Example 4 Given 1996 positive integers $a_{1}, a_{2}, \cdots, a_{1996}$ satisfy that the sum of any two numbers can be divided by their difference. Let $n=a_{1} a_{2} \cdots a_{1996}$. Prove that:
$$
n, n+a_{1}, n+a_{2}, \cdots, n+a_{1996}
$$
these 1997 numbers also satisfy the above condition. | Proof: Let $x, y$ be any two numbers from $n, n+a_{1}, n+a_{2}, \cdots, n+a_{1996}$.
(1) If one of $x, y$ is $n$, then without loss of generality, let $x=n, y=n+a_{i}(1 \leqslant i \leqslant 1996)$. In this case,
$x+y=2 n+a_{i}, x-y=-a_{i}$.
Since $a_{i} \mid n$, it follows that $(x-y) \mid (x+y)$.
Thus, the conclusion... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 720,560 |
5. In the convex pentagon $A B C D E$, $\angle B A E +$ $\angle A E D=270^{\circ}, \angle B C D=90^{\circ}, A B=3, B C=12$, $C D=5, D E=4, A E=8$. Then the area of pentagon $A B C D E$ is equal to . $\qquad$ | 5.55 .2
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,563 |
Three, (10 points) Given
$$
\frac{x}{y+z+u}=\frac{y}{z+u+x}=\frac{z}{u+x+y}=\frac{u}{x+y+z} \text {. }
$$
Find the value of $\frac{x+y}{z+u}+\frac{y+z}{u+x}+\frac{z+u}{x+y}+\frac{u+x}{y+z}$. | $$
\begin{array}{l}
\frac{x+y+z+u}{y+z+u}=\frac{x+y+z+u}{z+u+x} \\
=\frac{x+y+z+u}{u+x+y}=\frac{x+y+z+u}{x+y+z} .
\end{array}
$$
(1) If the numerator $x+y+z+u \neq 0$, then from the denominators we get $x=y=z=u$. At this time,
$$
\begin{array}{l}
\frac{x+y}{z+u}+\frac{y+z}{u+x}+\frac{z+u}{x+y}+\frac{u+x}{y+z} \\
=1+1+1... | 4 \text{ or } -4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,564 |
Four. (15 points) In hexagon $A B C D E F$, $A B / /$ $D E, B C / / E F, C D / / F A, A B+D E=B C+E F$, $A_{1} D_{1}=B_{1} E_{1}, A_{1}, B_{1}, D_{1}, E_{1}$ are the midpoints of sides $A B$, $B C, D E, E F$ respectively. Prove: $\angle C D E=\angle A F E$.
---
The translation is provided as requested, maintaining th... | As shown in Figure 3, construct $\square A B P F$, connect $D P$, and take the midpoint $M$ of $D P$. Then quadrilateral $B C D P$ is a trapezoid. Connect $B_{1} M$ and $E_{1} M$. By the midline theorem of trapezoids, we have
$$
\begin{array}{l}
B_{1} M / / C D / / B P / / A F, \\
M E_{1} / / D E / / F P / / A B,
\end{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,565 |
Five. (15 points) Prove:
(1) The remainder of a natural number squared when divided by 7 can only be $0, 1, 4, 2$;
(2) For any positive integer $n$,
$$
[\sqrt{n(n+2)(n+4)(n+6)}]
$$
is not divisible by 7 (where $[x]$ denotes the greatest integer less than or equal to the real number $x$). | (1) Let the natural number
$$
m=7 q+r(r=0,1, \cdots, 6) \text {. }
$$
Then $m^{2}=(7 q+r)^{2}=49 q^{2}+14 q r+r^{2}$. Since $r^{2}$ can only take $0,1,4,9,16,25,36$, the remainders when divided by 7 correspond to $0,1,4,2,2,4,1$. Therefore, the remainder when a natural number's square is divided by 7 can only be $0,1,... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 720,566 |
1. The minimum value of the function $f(x)=\frac{5-4 x+x^{2}}{2-x}$ on $(-\infty, 2)$ is ( ).
(A) 0
(B) 1
(C) 2
(D) 3 | - 1. C.
When $x<2$. Therefore,
$$
f(x)=\frac{1}{2-x}+(2-x) \geqslant 2,
$$
with equality holding if and only if $\frac{1}{2-x}=2-x$. This equation has a solution $x=1 \in(-\infty, 2)$. Therefore, the minimum value of $f(x)$ on $(-\infty, 2)$ is 2. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,567 |
2. Let $A=[-2,4), B=\left\{x \mid x^{2}-a x-4 \leqslant 0\right\}$. If $B \subseteq A$, then the range of real number $a$ is ( ).
(A) $[-1,2)$
(B) $[-1,2]$
(C) $[0,3]$
(D) $[0,3)$ | 2.D.
Since $x^{2}-a x-4=0$ has two real roots, we have $x_{1}=\frac{a}{2}-\sqrt{4+\frac{a^{2}}{4}}, x_{2}=\frac{a}{2}+\sqrt{4+\frac{a^{2}}{4}}$. Therefore, $B \subseteq A$ is equivalent to $x_{1} \geqslant-2$ and $x_{2}<4$, which means $\frac{a}{2}-\sqrt{4+\frac{a^{2}}{4}} \geqslant-2$ and $\frac{a}{2}+\sqrt{4+\frac{a... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,568 |
3. Two players, A and B, are playing a table tennis match, agreeing that the winner of each game gets 1 point and the loser gets 0 points. The match stops when one player is 2 points ahead or after 6 games have been played. Suppose the probability of A winning each game is $\frac{2}{3}$, and the probability of B winnin... | 3. B.
According to the problem, all possible values of $\xi$ are $2, 4, 6$.
Let each two games be a round. Then the probability that the match stops at the end of this round is $\left(\frac{2}{3}\right)^{2}+\left(\frac{1}{3}\right)^{2}=\frac{5}{9}$.
If the match continues after this round, then A and B must each have... | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 720,569 |
4. If the surface areas of three cubes with integer edge lengths (unit: $\mathrm{cm}$) sum up to $564 \mathrm{~cm}^{2}$, then the sum of the volumes of these three cubes is ( ).
(A) $764 \mathrm{~cm}^{3}$ or $586 \mathrm{~cm}^{3}$
(B) $764 \mathrm{~cm}^{3}$
(C) $586 \mathrm{~cm}^{3}$ or $564 \mathrm{~cm}^{3}$
(D) $586 ... | 4.A.
Let the edge lengths of the three cubes be $a$, $b$, and $c$. Then $6\left(a^{2}+b^{2}+c^{2}\right)=564 \Rightarrow a^{2}+b^{2}+c^{2}=94$.
Assume without loss of generality that $1 \leqslant a \leqslant b \leqslant c \leqslant 31$.
Thus, $6 \leqslant c < 10$. Therefore, $c$ can only be $9, 8, 7, 6$.
If $c=9$, the... | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 720,570 |
Example 5 Let $p$ be an odd prime, and
$$
1+\frac{1}{2}+\cdots+\frac{1}{p-2}+\frac{1}{p-1}=\frac{a}{b}\left(a 、 b \in \mathbf{N}_{+}\right) \text {. }
$$
Prove: $p \mid a$. | Prove: Let $p-1=2k$. Then
$$
\begin{array}{l}
1+\frac{1}{p-1}=\frac{p}{1 \times(p-1)}, \\
\frac{1}{2}+\frac{1}{p-2}=\frac{p}{2(p-2)}, \\
\cdots \cdots \\
\frac{1}{k}+\frac{1}{p-k}=\frac{p}{k(p-k)}. \\
\text { Hence } \frac{a}{b}=1+\frac{1}{2}+\cdots+\frac{1}{p-2}+\frac{1}{p-1} \\
=\frac{p}{1 \times(p-1)}+\frac{p}{2(p-2... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 720,571 |
5. The number of rational solutions $(x, y, z)$ for the system of equations $\left\{\begin{array}{l}x+y+z=0, \\ x y z+z=0, \\ x y+y z+x z+y=0\end{array}\right.$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | 5.B.
If $z=0$, then $\left\{\begin{array}{l}x+y=0, \\ x y+y=0 .\end{array}\right.$
Solving, we get $(x, y)=(0,0),(-1,1)$.
If $z \neq 0$, from $x y z+z=0$, we get $x y=-1$.
From $x+y+z=0$, we get $z=-x-y$.
Substituting equation (2) into $x y+y z+x z+y=0$, we get $x^{2}+y^{2}+x y-y=0$.
From equation (1), we get $x=-\fra... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,572 |
6. Let $\triangle A B C$ have $\angle A, \angle B, \angle C$ opposite sides $a, b, c$ in a geometric progression. Then the range of $\frac{\sin A \cdot \cot C+\cos A}{\sin B \cdot \cot C+\cos B}$ is ( ).
(A) $(0,+\infty)$
(B) $\left(0, \frac{\sqrt{5}+1}{2}\right)$
(C) $\left(\frac{\sqrt{5}-1}{2}, \frac{\sqrt{5}+1}{2}\r... | 6. C.
Let the common ratio of $a, b, c$ be $q$. Then $b=a q, c=a q^{2}$.
$$
\begin{array}{l}
\text { and } \frac{\sin A \cdot \cot C+\cos A}{\sin B \cdot \cot C+\cos B} \\
= \frac{\sin A \cdot \cos C+\cos A \cdot \sin C}{\sin B \cdot \cos C+\cos B \cdot \sin C} \\
=\frac{\sin (A+C)}{\sin (B+C)}=\frac{\sin (\pi-B)}{\s... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,573 |
7. Let $f(x)=a x+b(a, b$ be real numbers),
$$
\begin{array}{l}
f_{1}(x)=f(x), \\
f_{n+1}(x)=f\left(f_{n}(x)\right)(n=1,2, \cdots) .
\end{array}
$$
If $f_{7}(x)=128 x+381$, then $a+b=$ | Ni.7.5.
From the problem, we know
$$
\begin{array}{l}
f_{n}(x)=a^{n} x+\left(a^{n-1}+a^{n-2}+\cdots+a+1\right) b \\
=a^{n} x+\frac{a^{n}-1}{a-1} \cdot b .
\end{array}
$$
From $f_{7}(x)=128 x+381$, we get
$$
a^{7}=128, \frac{a^{7}-1}{a-1} \cdot b=381 \text {. }
$$
Therefore, $a=2, b=3, a+b=5$. | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,574 |
8. Let $f(x)=\cos 2x-2a(1+\cos x)$ have a minimum value of $-\frac{1}{2}$. Then $a=$ $\qquad$ . | $$
\begin{array}{l}
\text { 8. }-2+\sqrt{3} \\
f(x)=2 \cos ^{2} x-1-2 a-2 a \cos x \\
=2\left(\cos x-\frac{a}{2}\right)^{2}-\frac{1}{2} a^{2}-2 a-1
\end{array}
$$
(1) When $a>2$, $f(x)$ reaches its minimum value $1-4a$ when $\cos x=1$;
(2) When $a>2$ or $a<-2$, the minimum value of $f(x)$ cannot be $-\frac{1}{2}$, then... | -2+\sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,575 |
9. Distribute 24 volunteer slots to 3 schools, such that each school gets at least one slot and the number of slots each school receives is different. The number of ways to do this is $\qquad$.
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 | 9.222 .
Using 4 sticks to represent the gaps between three schools, and * to represent quotas. For example,
$$
\text { | * * * * | } * \cdots * \mid * * \text { | }
$$
indicates that the first, second, and third schools have 4, 18, and 2 quotas, respectively.
If each “*” and each “|” is considered a position, and si... | null | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 720,576 |
10. Let the sequence $\left\{a_{n}\right\}$ have the sum of the first $n$ terms $S_{n}$ satisfying
$$
S_{n}+a_{n}=\frac{n-1}{n(n+1)}(n=1,2, \cdots) \text {. }
$$
Then the general term $a_{n}=$ | 10. $\frac{1}{2^{n}}-\frac{1}{n(n+1)}$.
Notice
$$
\begin{array}{l}
a_{n+1}=S_{n+1}-S_{n} \\
=\frac{n}{(n+1)(n+2)}-a_{n+1}-\frac{n-1}{n(n+1)}+a_{n},
\end{array}
$$
which means
$$
\begin{array}{l}
2 a_{n+1}=\frac{n+2-2}{(n+1)(n+2)}-\frac{1}{n+1}+\frac{1}{n(n+1)}+a_{n} \\
=\frac{-2}{(n+1)(n+2)}+a_{n}+\frac{1}{n(n+1)} .
... | a_{n}=\frac{1}{2^{n}}-\frac{1}{n(n+1)} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,577 |
11. Let $f(x)$ be a function defined on $\mathbf{R}$. If $f(0)=2008$, and for any $x \in \mathbf{R}$, it satisfies
$$
\begin{array}{l}
f(x+2)-f(x) \leqslant 3 \times 2^{x}, \\
f(x+6)-f(x) \geqslant 63 \times 2^{x},
\end{array}
$$
then $f(2008)=$ | $$
11.2^{2008}+2007
$$
From the given conditions, we know
$$
\begin{array}{l}
f(x+2)-f(x) \\
=-(f(x+4)-f(x+2))- \\
(f(x+6)-f(x+4))+ \\
(f(x+6)-f(x)) \\
\geqslant-3 \times 2^{x+2}-3 \times 2^{x+4}+63 \times 2^{x}=3 \times 2^{x} .
\end{array}
$$
Therefore, \( f(x+2)-f(x)=3 \times 2^{x} \). Hence,
$$
\begin{aligned}
f(... | 2^{2008}+2007 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,578 |
12. A small ball with a radius of 1 can move freely in all directions inside a regular tetrahedron container with an internal edge length of $4 \sqrt{6}$. Then the area of the container's inner wall that the ball can never touch is $\qquad$ | $12.72 \sqrt{3}$.
As shown in Figure 3, consider the situation when the small ball is squeezed into a corner.
Let the radius of the small ball be $r$, and construct the plane $A_{1} B_{1} C_{1} /$ plane $A B C$, tangent to the small ball at point $D$. Then the center of the small ball $O$ is the center of the regular t... | 72 \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,579 |
13. Given the function $f(x)=|\sin x|$ intersects the line $y=k x(k>0)$ at exactly three points, and the maximum value of the x-coordinates of the intersection points is $\alpha$. Prove:
$$
\frac{\cos \alpha}{\sin \alpha+\sin 3 \alpha}=\frac{1+\alpha^{2}}{4 \alpha} \text {. }
$$ | Three, 13. The graph of $f(x)$ intersects the line $y=k x(k>0)$ at three points as shown in Figure 5, and is tangent to it in $\left(\pi, \frac{3 \pi}{2}\right)$, with the point of tangency being $A(\alpha,-\sin \alpha), \alpha \in\left(\pi, \frac{3 \pi}{2}\right)$.
Since $f^{\prime}(x)=-\cos x, x \in\left(\pi, \frac{3... | \frac{1+\alpha^{2}}{4 \alpha} | Calculus | proof | Yes | Yes | cn_contest | false | 720,580 |
14. Solve the inequality
\[
\begin{array}{l}
\log _{2}\left(x^{12}+3 x^{10}+5 x^{8}+3 x^{6}+1\right) \\
<1+\log _{2}\left(x^{4}+1\right) .
\end{array}
\] | 14. Solution 1: From
$$
1+\log _{2}\left(x^{4}+1\right)=\log _{2}\left(2 x^{4}+2\right),
$$
and $\log _{2} y$ is an increasing function on $(0,+\infty)$, the original inequality is equivalent to
$$
x^{12}+3 x^{10}+5 x^{8}+3 x^{6}+1x^{6}+3 x^{4}+3 x^{2}+1+2 x^{2}+2 \\
=\left(x^{2}+1\right)^{3}+2\left(x^{2}+1\right), \\... | \left(-\sqrt{\frac{\sqrt{5}-1}{2}}, \sqrt{\frac{\sqrt{5}-1}{2}}\right) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 720,581 |
1. Let $x$, $y$, $z$ be integers, and $16 \mid (3x + y - 7z)$, $16 \mid (6x - 4y + z)$. Prove: $16 \mid (4x - 2y - z)$. | (Hint: From $2(3 x+y-7 z)+5(6 x-4 y+$ $z)=9(4 x-2 y-z)$, we get $16 \mid 9(4 x-2 y-z)$. Also, $(16,9)=1$, so $16 \mid (4 x-2 y-z))$. | 16 \mid (4 x-2 y-z) | Number Theory | proof | Yes | Yes | cn_contest | false | 720,582 |
15. As shown in Figure $1, P$ is a moving point on the parabola $y^{2}=2 x$, points $B$ and $C$ are on the $y$-axis, and the circle $(x-1)^{2}+y^{2}=1$ is inscribed in $\triangle P B C$. Find the minimum value of the area of $\triangle P B C$. | 15. Let \( P\left(x_{0}, y_{0}\right) \), \( B(0, b) \), and \( C(0, c) \), and assume \( b > c \).
The equation of line \( PB \) is \( y - b = \frac{y_{0} - b}{x_{0}} x \).
Simplifying, we get \( \left(y_{0} - b\right) x - x_{0} y + x_{0} b = 0 \).
The distance from the circle center \((1,0)\) to \( PB \) is 1, so
\[ ... | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,583 |
Three. (50 points) Let $a_{k}>0(k=1,2, \cdots, 2008)$. Prove: There exists a sequence $\left\{x_{n}\right\}$ satisfying the following conditions if and only if $\sum_{k=1}^{2008} a_{k}>1$:
(1) $0=x_{0}<x_{n}<x_{n+1}(n=1,2, \cdots)$;
(2) $\lim _{n \rightarrow \infty} x_{n}$ exists;
$$
\begin{array}{l}
\quad \text { (3) ... | Three, Necessity: Suppose there exists $\left\{x_{n}\right\}$ satisfying (1), (2), and (3).
Notice that the equation in (3) can be transformed into
$$
x_{n}-x_{n-1}=\sum_{k=1}^{20 B} a_{k}\left(x_{n+k}-x_{n+k-1}\right)\left(n \in \mathbf{N}_{+}\right) \text {, }
$$
where $x_{0}=0$.
Summing the above equation from the ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 720,586 |
3. Using "Kepler's Third Law: the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit," we can calculate the time it takes for a planet in the solar system to complete one orbit around the Sun (i.e., its orbital period). If the average distance of Earth from... | 3.
$$
687.19 ; 1.881
$$ | 687.19 ; 1.881 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,589 |
5. The line $y=x$ intersects the graph of the function $y=\cot x$ in the intervals $(-\pi, 0)$ and $(0, \pi)$ at points $A$ and $B$, respectively. The length of the line segment $AB$ is $\qquad$ (rounded to 0.01). | 5.2.43
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
However, since the provided text "5.2.43" does not contain any non-English words or phrases, it remains unchanged in the translation:
5.2.43 | 5.24 | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 720,591 |
2. Let $u, v \in \mathbf{Z}$, and $9 \mid\left(u^{2}+u v+v^{2}\right)$. Prove: $3 \mid u$ and $3 \mid v$.
| (Notice that $u^{2}+u v+v^{2}=(u-v)^{2}+$ $3 u v$, it is easy to know that 3 divides $(u-v)^{2}$. Since 3 is a prime number, then $3 \mid(u-v)$. Therefore, $9 \mid(u-v)^{2}$. It is easy to know that $9 \mid 3 u v$, which means $3 \mid u v$. Also, $3 \mid\left(u^{2}+u v+v^{2}\right)$, so, 3 divides [( $u^{2}+u v$ $+v^{2... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 720,593 |
7. A closed cylinder with a base diameter of 4 and a height of 18, filled with spheres of radius 1, can hold at most $\qquad$.
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Note: The translation provided above already meets the request, but if you meant to have the note included in the translation as well, it would look like t... | $7.24$ | not found | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 720,594 |
II. (20 points) Find the positive integer $n$ such that
$$
\left[\log _{3} 1\right]+\left[\log _{3} 2\right]+\cdots+\left[\log _{3} n\right]=2007 \text {, }
$$
where $[x]$ denotes the greatest integer not exceeding the real number $x$. | $$
\begin{array}{l}
{\left[\log _{3} 1\right]+\left[\log _{3} 2\right]=0,} \\
{\left[\log _{3} 3\right]+\left[\log _{3} 4\right]+\cdots+\left[\log _{3} 8\right]} \\
=6 \times 1=6, \\
{\left[\log _{3} 9\right]+\left[\log _{3} 10\right]+\cdots+\left[\log _{3} 26\right]} \\
=18 \times 2=36, \\
{\left[\log _{3} 27\right]+\... | 473 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,596 |
Three. (20 points) As shown in Figure 1, square $A B C D$ is inscribed in an ellipse, and squares $E F G H$ and $I J H K$ have vertices $E$, $F$, and $I$ on the ellipse, vertices $K$, $H$, and $G$ on side $A B$, and vertex $J$ on side $H E$. It is known that the area ratio of square $A B C D$ to square $E F G H$ is $4:... | Three, let the equation of the ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, and the side length of the square $A B C D$ be $2 \mathrm{~m}$. Then
$A(m, m), E\left(2 m, \frac{m}{2}\right)$.
Thus, $\left\{\begin{array}{l}\frac{m^{2}}{a^{2}}+\frac{m^{2}}{b^{2}}=1, \\ \frac{4 m^{2}}{a^{2}}+\frac{m^{2}}{4 b^{2}}=1... | 0.144 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,597 |
Four, (20 points) We know that $\frac{49}{98}$ simplifies to $\frac{1}{2}$, but by the method $\frac{49}{98}$, it also results in $\frac{1}{2}$. Find all proper fractions where both the numerator and the denominator are two-digit positive integers in decimal, the unit digit of the numerator is the same as the tens digi... | Let the true fraction $\frac{a b}{-}$ have the given property. Then $b c$
$\overline{a b}6$ when, $0<2 a-12<3 a+2$, at this time, $c$
is not an integer;
When $a=6$, $c=6$, contradiction;
When $a<6$, $12-2 a \geqslant 3 a+2$, so, $a \leqslant$ 2, and when $a=1$, $c=4$, at this time, the true fraction that satisfies the... | \frac{16}{64}, \frac{26}{65}, \frac{19}{95}, \frac{49}{98} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,598 |
One, (20 points) Find the solution to the following mixed system of equations in $x, y$
$$
\left\{\begin{array}{l}
\frac{x^{2}+3}{2}=x-\cos (x+y), \\
100 \mathrm{e} x \leqslant y \leqslant \frac{100 \pi}{x},
\end{array}\right.
$$
where, $\mathrm{e} \approx 2.71828, \pi \approx 3.14159$. | From the first equation, we have
$$
(x-1)^{2}+2=-2 \cos (x+y) \text {. }
$$
Since $(x-1)^{2}+2 \geqslant 2,-2 \cos (x+y) \leqslant 2$,
it follows that $x=1$, and $\cos (x+y)=-1$.
From $\cos (1+y)=-1$, we get
$1+y=(2 k+1) \pi(k \in \mathbf{Z})$.
Thus, $x=1, y=(2 k+1) \pi-1(k \in \mathbf{Z})$.
Substituting (1) into the ... | \left\{\begin{array}{l}
x=1, \\
y=(2 k+1) \pi-1(k=43,44, \cdots, 49) .
\end{array}\right.
} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,599 |
II. (20 points) Find all positive decimal integers $n$ that satisfy the following conditions: $n$ is an $a$-digit number, and $a^{a}=n$.
| II. Since $1^{1}=1$, therefore, $n=1$ satisfies the condition.
When $a=2,3,4,5,6,7$, because
$$
\begin{array}{l}
2^{2}=4,3^{3}=27,4^{4}=256,5^{5}=3125, \\
6^{6}=46656,7^{7}=823543,
\end{array}
$$
thus, $n$ cannot be a two-digit number, three-digit number, $\cdots$, seven-digit number.
And when $a=8,9$,
$$
8^{8}=167772... | 1,16777216,387420489 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,600 |
Three. (20 points) Arrange the numbers $1, 2, \cdots, 2007$ in any order to get 2007! different sequences. Does there exist 4 sequences:
$$
\begin{array}{l}
a_{1}, a_{2}, \cdots, a_{2007} ; b_{1}, b_{2}, \cdots, b_{2007} ; \\
c_{1}, c_{2}, \cdots, c_{2007} ; d_{1}, d_{2}, \cdots, d_{2007}
\end{array}
$$
such that $a_{... | $$
\begin{array}{l}
a_{1} b_{1}+a_{2} b_{2}+\cdots+a_{2007} b_{2007} \\
\leqslant 1 \times 1+2 \times 2+\cdots+2007 \times 2007 \\
=\frac{2007 \times 2008 \times 4015}{6}=2696779140, \\
c_{1} d_{1}+c_{2} d_{2}+\cdots+c_{2007} d_{2007} \\
\geqslant 1 \times 2007+2 \times 2006+\cdots+2007 \times 1 \\
=\sum_{k=1}^{2007} k... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 720,601 |
10.2. In an $n \times n$ grid, the $n$ columns from left to right are called the 1st column, 2nd column, $\cdots, n$th column. Fill the grid with the $n$ positive integers $1 \sim n$ such that each row and each column contains $n$ different numbers. If the number in a cell is greater than the column number of that cell... | 10.2. First, find the total number of good squares.
In the 1st column, there are $n-1$ (all except the one filled with number 1 are good), in the 2nd column, there are $n-2$ (all except the ones filled with numbers 1 and 2 are good), $\cdots \cdots$ the $n$-th column has no good squares. Therefore, the total number of... | n = 2k + 1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 720,602 |
10.3. $O$ is the center of circle $\omega$, which is tangent to the two sides of $\angle BAC$ at points $B$ and $C$. $Q$ is a point inside $\angle BAC$, and a point $P$ on segment $AQ$ satisfies $AQ \perp OP$. Line $OP$ intersects the circumcircle $\omega_{1}$ of $\triangle BPQ$ and the circumcircle $\omega_{2}$ of $\t... | 10.3. Let $\triangle B P Q, \triangle C P Q$ intersect the rays $A B, A C$ at points $D, E$ respectively.
By the power of a point theorem, we have
$$
A B \cdot A D=A P \cdot A Q, A C \cdot A E=A P \cdot A Q \text {. }
$$
Thus, $A B \cdot A D=A C \cdot A E$.
Given $A B=A C$, we get $A D=A E$.
Therefore, $\triangle A D ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,603 |
3. A positive integer $a$, if its digits are rearranged to form a new positive integer $b$. If $a=3b$, then $a$ is called a "Hope Number". Prove: If $a$ and $b$ are both Hope Numbers, then $729|ab$.
| (For any positive integer $n$, let $S(n)$ denote the sum of the digits of $n$. Let $a$ be any hopeful number. First, prove that $27 \mid a$. According to the definition of a hopeful number, there exists a positive integer $p$ such that $a=3 p$, and $S(a)=S(p)$. From $a=3 p$, we know $3 \mid a$, so $3 \mid S(a)$, which ... | 729 \mid ab | Number Theory | proof | Yes | Yes | cn_contest | false | 720,604 |
10.4. The sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ satisfy $a_{1}=1, b_{1}=$ $2, a_{n+1}=\frac{1+a_{n}+a_{n} b_{n}}{b_{n}}, b_{n+1}=\frac{1+b_{n}+a_{n} b_{n}}{a_{n}}$. Prove: $a_{2008}<5$. | 10.4. First prove by induction: For any positive integer $n$, we have $\frac{1}{a_{n}+1}-\frac{1}{b_{n}+1}=\frac{1}{6}$.
When $n=1$, it is clear that $\frac{1}{2}-\frac{1}{3}=\frac{1}{6}$.
Assume the equation holds for $n$. Then
$$
\begin{array}{l}
\frac{1}{a_{n+1}+1}-\frac{1}{b_{n+1}+1} \\
=\frac{b_{n}}{1+a_{n}+b_{n}+... | a_{n}<5 | Algebra | proof | Yes | Yes | cn_contest | false | 720,605 |
10.5. Find all triples of real numbers $(x, y, z)$ that satisfy:
$$
\begin{array}{l}
1+x^{4} \leqslant 2(y-z)^{2}, \\
1+y^{4} \leqslant 2(z-x)^{2}, \\
1+z^{4} \leqslant 2(x-y)^{2} .
\end{array}
$$ | 10.5. The ternary real number groups that satisfy the conditions are
$$
\begin{array}{l}
(-1,0,1),(-1,1,0),(0,-1,1), \\
(0,1,-1),(1,-1,0),(1,0,-1) .
\end{array}
$$
By symmetry, without loss of generality, assume $x \geqslant y \geqslant z$. Then
$$
2 x^{2} \leqslant 1+x^{4} \leqslant 2(y-z)^{2} \text {, }
$$
which im... | (-1,0,1),(-1,1,0),(0,-1,1),(0,1,-1),(1,-1,0),(1,0,-1) | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 720,606 |
10.6. Given that $A A_{1}$ and $C C_{1}$ are the altitudes of a non-isosceles acute triangle $\triangle A B C$, $H$ and $O$ are the orthocenter and circumcenter of $\triangle A B C$, respectively, and $B_{0}$ is the midpoint of side $A C$. The lines $B O$ and $B H$ intersect side $A C$ and $A_{1} C_{1}$ at points $P$ a... | 10.6. Let $O_{1}$ be the midpoint of $B H$.
Since $\angle B A_{1} H=\angle B C_{1} H=90^{\circ}$, $O_{1}$ is the circumcenter of quadrilateral $A_{1} B C_{1} H$.
Also, since Rt $\triangle B A A_{1} \backsim \mathrm{Rt} \triangle B C C_{1}$, we have $\frac{A B}{A_{1} B}=\frac{C B}{C_{1} B}$.
Thus, $\triangle A_{1} B C_... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,607 |
10.7. Find all integers $n$ $(n>1)$: there exist not all equal positive integers $b_{1}, b_{2}, \cdots$, $b_{n}$, such that for any given positive integer $k$, there exist positive integers $a, b (a, b>1)$ satisfying $\prod_{i=1}^{n}\left(b_{i}+k\right)=a^{b}$. | 10.7. $n>1$ satisfies the condition if and only if $n$ is a composite number.
When $n=r s(r>1, s>1)$ is a composite number, let
$b_{1}=b_{2}=\cdots=b_{r}=1$,
$b_{r+1}=b_{r+2}=\cdots=b_{n}=2$.
Then for any positive integer $k$,
$$
\left(b_{1}+k\right)\left(b_{2}+k\right) \cdots\left(b_{n}+k\right)
$$
is a positive inte... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,608 |
10.8. In the plane, there are a finite number of rectangles with sides parallel to the coordinate axes. It is known that for any two of these rectangles, there exists a vertical or horizontal line that intersects both of them. Prove: There exists a horizontal and a vertical line such that each rectangle intersects at l... | 10.8. First, prove a lemma.
Lemma: In the plane, there are a finite number of rectangles with edges parallel to the coordinate axes, satisfying: for any two rectangles among them, there exists a vertical line intersecting both of them. Then there exists a vertical line intersecting all these finite rectangles.
Proof ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,609 |
1. The number of prime solutions to the equation $x y + 41 z = 2009$ is ( ) groups.
(A) 4
(B) 6
(C) 8
(D) 10 | - 1.A.
From the original equation, we get
$$
x y=2009-41 z=41(49-z) \text {. }
$$
Since $x, y$ are prime numbers, then $x=41$ or $y=41$.
(1) When $x=41$, $y=49-z$, it follows that $y$ or $z$ must be even, hence $(y, z)=(2,47),(47,2)$, at this point, the original equation has two sets of prime solutions;
(2) When $y=4... | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 720,610 |
2. There are $n$ points on a straight line, and the number of line segments on this line is denoted as $a_{n}$; there are $n$ lines in a plane, and these $n$ lines divide the plane into at most $b_{n}$ parts and at least $c_{n}$ parts. The relationship between $a_{n}$ and $b_{n}-c_{n}$ is ( ).
(A) $a_{n}>b_{n}-c_{n}$
(... | 2. B.
Notice that
$$
\begin{array}{l}
a_{n}=1+2+\cdots+(n-1)=\frac{n(n-1)}{2}, \\
b_{n}=1+(1+2+\cdots+n)=\frac{n^{2}+n+2}{2}, \\
c_{n}=n+1 .
\end{array}
$$
Thus, $b_{n}-c_{n}=\frac{n^{2}+n+2}{2}-(n+1)=a_{n}$. | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 720,611 |
3. A middle school selects student council members from students in grades 7 to 12, and at least one of the following conditions must be met:
(1) At most 1 person from grade 7;
(2) At most 2 people from grade 8;
(3) At most 3 people from grade 9;
(4) At most 4 people from grade 10;
(5) At most 5 people from grade 11;
(... | 3.C.
Consider the opposite.
If none of the six conditions are met, then at least 2 students from Grade 7; at least 3 students from Grade 8; … at least 7 students from Grade 12 must be selected. In this case, the total number of students selected is at least
$$
2+3+4+5+6+7=27 \text { students. }
$$
Therefore, at most ... | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 720,612 |
4. In the Cartesian coordinate system, $P$ is a moving point on the parabola $y=x^{2}$, point $M\left(0, \frac{1}{4}\right), N\left(0,-\frac{1}{4}\right)$, and a line $l$ is drawn through $N$ parallel to the $x$-axis. Then the positional relationship between the circle $\odot P$ with center $P$ and radius $PM$ and the ... | 4. B.
As shown in Figure 1, draw $PA \perp y$-axis at point $A$, and $PB \perp l$ at point $B$. Let $P\left(a, a^{2}\right)$. Then
$$
\begin{array}{l}
PA=|a|, \\
PB=\left|a^{2}-\left(-\frac{1}{4}\right)\right| \\
=a^{2}+\frac{1}{4} .
\end{array}
$$
In the right triangle $\triangle PAM$, $AM=\left|a^{2}-\frac{1}{4}\ri... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,613 |
5. If real numbers $x, y, a, b$ satisfy $x^{2}+y^{2}=1$, $a^{2}+b^{2}=25$, then the maximum value of $a x+b y$ is ( ).
(A) 3
(B) 4
(C) 5
(D) 6 | 5. C.
Notice that
$$
\begin{array}{l}
0 \leqslant(a-5 x)^{2}+(b-5 y)^{2} \\
=\left(a^{2}+b^{2}\right)+25\left(x^{2}+y^{2}\right)-10(a x+b y) \\
=50-10(a x+b y) .
\end{array}
$$
Then $a x+b y \leqslant 5$.
When $a x+b y=5$, $a=5 x, b=5 y(x, y$ are any real numbers).
Therefore, the maximum value of $a x+b y$ is 5. | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,614 |
4. Let $a, b, c, d$ be distinct integers, and let the integer $x$ satisfy $(x-a)(x-b)(x-c)(x-d)=9$. Prove that 4 । $(a+b+c+d)$.
Note: The symbol "।" in the original text is typically used in some languages to denote division. In this context, it means "4 divides $(a+b+c+d)$". | (Tip: Since $x-a$, $x-b$, $x-c$, $x-d$ are distinct divisors of 9, $x-a$, $x-b$, $x-c$, $x-d$ are a permutation of $1$, $-1$, $3$, $-3$. Therefore,
$$
\begin{array}{l}
(x-a)+(x-b)+(x-c)+(x-d) \\
=1+(-1)+3+(-3)=0,
\end{array}
$$
which means $a+b+c+d=4x$.) | a+b+c+d=4x | Number Theory | proof | Yes | Yes | cn_contest | false | 720,615 |
6. In $\odot O$, chord $CD$ is perpendicular to diameter $AB$ at point $E$, chord $DF$ bisects $BC$ and intersects $BC$ at point $G$, chord $AF$ intersects $OC$ at point $H$. Then the value of $\frac{GH}{AB}$ is ( ).
(A) $\frac{1}{3}$
(B) $\frac{1}{4}$
(C) $\frac{1}{5}$
(D) $\frac{2}{5}$ | 6. B.
As shown in Figure 2, connect $C F$.
By the perpendicular diameter theorem, we have
$$
\overparen{A C}=\overparen{A D} \text {, }
$$
Thus, $\angle B=\angle A F D$.
Since $O C=O B$, we get
$$
\angle B=\angle O C B \text {. }
$$
Therefore, $\angle A F D=\angle O C B$.
Hence, $C, H, G, F$ are concyclic.
Thus, $\a... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,616 |
1. In a convex quadrilateral $A B C D$, $A D=B C, \angle A$ $=\angle C=100^{\circ}$. Then the shape of quadrilateral $A B C D$ is $\qquad$ | 2. Parallelogram.
As shown in Figure 3, draw $D E \perp$
$A B$ at $E$, and $B F \perp C D$ at
$F$. Then
$\angle E A D=\angle F C B$.
Also, $A D=B C$, so
$\triangle E A D \cong \triangle F C B$.
Therefore, $D E=B F, A E=C F$.
Connect $B D$.
Then Rt $\triangle B E D \cong$ Rt $\triangle D F B$.
Thus, $B E=D F$.
(2) - (1)... | Parallelogram | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,617 |
2. From $1,2, \cdots, 2008$, at least $\qquad$ even numbers must be taken to ensure that there are definitely two even numbers whose sum is 2012. | 2.504.
From $1,2, \cdots, 2008$ select two even numbers, the sum of which is 2012, there are a total of 501 pairs, that is
$$
4+2008,6+2006, \cdots, 1004+1008 \text {. }
$$
Since the sum of 2 or 1006 and any one of the even numbers above does not equal 2012, therefore, at least $501+2+$ $1=504$ even numbers must be t... | 504 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 720,618 |
3. In the acute triangle $\triangle A B C$, $\angle A=60^{\circ}, A D \perp B C$ at point $D, B D=1, C D=2$. Then the area of $\triangle A B C$ is $\qquad$ | 3. $\frac{3(\sqrt{3}+\sqrt{11})}{4}$.
As shown in Figure 4, draw $B E \perp$
$A C$ at point $E$, intersecting $A D$ at
point $F$. Then
$$
\angle 1=\angle 2 \text {. }
$$
From $\triangle A F E \backsim$
$\triangle B C E$
$$
\begin{aligned}
\Rightarrow & \frac{A F}{B C}=\frac{A E}{B E} \\
\Rightarrow A F & =\frac{A E}{... | \frac{3(\sqrt{3}+\sqrt{11})}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,619 |
4. If the inequality $(x+y)\left(\frac{1}{x}+\frac{a}{y}\right) \geqslant 16$ holds for any positive real numbers $x, y$, then the minimum value of the real number $a$ is $\qquad$
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly... | 4.9.
From the problem, transform and rearrange the original inequality to
$$
y^{2}-(15-a) x y+a x^{2} \geqslant 0 \text {. }
$$
Consider equation (1) as a quadratic inequality in $y$. Since equation (1) holds for all real numbers $y$, we have
$$
\Delta=[-(15-a) x]^{2}-4 a x^{2} \leqslant 0 \text {, }
$$
which simpli... | null | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 720,620 |
One. (20 points) Two cars, A and B, start from point A at the same time, driving in the same direction on a straight line. Car A can carry at most $a \mathrm{~L}$ of gasoline, and Car B can carry at most $b \mathrm{~L}$ of gasoline $(a \geqslant b$ and both are the maximum capacities of the fuel tanks). They cannot ref... | One car travels from $A$ to $B$, and another car travels from $A$ to $B$. Car B stays at $B$, first lending some gasoline to Car A to fill its tank. Car A travels the farthest to $C$ using half of the gasoline in the tank, and then returns to $B$ using the other half of the gasoline in the tank. At this point, Car B le... | 6(2a+b) | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 720,621 |
II. (25 points) Given in a convex quadrilateral $ABCD$, $AB=CD$, $G$ and $H$ are the midpoints of $BD$ and $AC$ respectively. Extend $BA$ to point $E$ and extend $DC$ to point $F$ such that $AE=CF$. Prove that $GH$ bisects $EF$.
| II. As shown in the figure,
5, let the line $G H$
intersect $E F$ at point $P$,
connect $A F$, and take
the midpoints of
$E F$, $A F$, and $B C$
as $P^{\prime}$, $Q$, and $R$, respectively, and connect
$G R$, $H R$, $P^{\prime} Q$,
$H Q$, $H P^{\prime}$. It is easy to see
Figure 5
$G R \Perp \frac{1}{2} C D$,
Encounter... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,622 |
Three, (25 points) "If $a, b, c$ are positive real numbers, then $\sqrt[3]{a b c} \leqslant \frac{a+b+c}{3}$, where the equality holds if and only if $a=b=c$." Using the above conclusion, find the maximum value of the function $y=2 x(4-x)(3-2 x)\left(0<x<\frac{3}{2}\right)$. | $$
\begin{array}{l}
\text { Given the known inequality } a b c \leqslant\left(\frac{a+b+c}{3}\right)^{3}. \\
\text { To find the maximum value of } y, \text { it is necessary that the sum of the three factors of } y \text { is a constant and the three factors are equal when } x \text { exists. } \\
\text { Clearly, the... | \frac{200}{27} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 720,623 |
1. Let $P$ be a moving point in the plane of $\triangle A B C$. Then the point $P$ that minimizes $\boldsymbol{P A} \cdot \boldsymbol{P B}+\boldsymbol{P B} \cdot \boldsymbol{P C}+\boldsymbol{P C} \cdot \boldsymbol{P A}$ is the ( ) of $\triangle A B C$.
(A) Circumcenter
(B) Incenter
(C) Centroid
(D) Orthocenter | - 1.C.
Notice
$$
\begin{array}{l}
P A \cdot P B+P B \cdot P C+P C \cdot P A \\
=P A \cdot(P A+A B)+(P A+A B) \text {. } \\
(P A+A C)+(P A+A C) \cdot P A \\
=3 P A^{2}+2(A B+A C) \cdot P A+A B \cdot A C \\
=3\left(P A+\frac{A B+A C}{3}\right)^{2}-\frac{(A B+A C)^{2}}{3}+ \\
A B \cdot A C \text {. } \\
\end{array}
$$
W... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,624 |
2. As shown in Figure 1, in rectangle $A B C D$, $A B=1, B C$ $=m, O$ is the center of the rectangle, $P O \perp$ plane $A B C D, P O=n$, and there exists a unique point $E$ on side $B C$ such that $P E \perp D E$. If the angle between plane $P D E$ and plane $A B C D$ is $60^{\circ}$, then $(m, n)$ is ( ).
(A) $\left(... | 2. B.
As shown in Figure 7, let $C E = x$, and draw $O F \perp B C$ at $F$. Then
$$
\begin{array}{l}
O F = \frac{1}{2}, \\
E F = \frac{m}{2} - x .
\end{array}
$$
Thus, $P E^{2} = P O^{2} + O F^{2} + E F^{2}$
$$
\begin{array}{l}
= n^{2} + \frac{1}{4} + \left(\frac{m}{2} - x\right)^{2} \\
= x^{2} - m x + \frac{m^{2}}{4... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,625 |
Example 4 Given the inequality $(x+y)\left(\frac{1}{x}+\frac{a}{y}\right) \geqslant 9$ holds for any positive real numbers $x, y$. Then the minimum value of the positive real number $a$ is ( ).
(A) 2
(B) 4
(C) 6
(D) 8 | Solution: By Cauchy-Schwarz inequality, we can find
$$
\begin{array}{l}
(x+y)\left(\frac{1}{x}+\frac{a}{y}\right) \geqslant\left(\sqrt{x} \cdot \frac{1}{\sqrt{x}}+\sqrt{y} \cdot \frac{\sqrt{a}}{\sqrt{y}}\right)^{2} \\
=(1+\sqrt{a})^{2} . \\
\text { When } x=1, y=\sqrt{a} \text {, } (x+y)\left(\frac{1}{x}+\frac{a}{y}\ri... | B | Inequalities | MCQ | Yes | Yes | cn_contest | false | 720,626 |
3. Given real numbers $a, b, c$ satisfy $abc=1$. Then the numbers $2a-\frac{1}{b}, 2b-\frac{1}{c}, 2c-\frac{1}{a}$, at most ( ) of these numbers can be greater than 1.
(A) 3
(B) 2
(C) 1
(D) 0 | 3. B.
Assume $2a-\frac{1}{b}$, $2b-\frac{1}{c}$, $2c-\frac{1}{a}$ are all greater than 1.
From $abc=1$, we know that at least one of $a$, $b$, $c$ is positive. Without loss of generality, let $a>0$.
Then $2c>2c-\frac{1}{a}>1 \Rightarrow c>0$.
Thus, $2b>2b-\frac{1}{c}>1 \Rightarrow b>0$.
Therefore, $a$, $b$, $c$ are al... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,627 |
4. As shown in Figure 2, given the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the line $l$ is tangent to the ellipse at point $P$, and intersects the positive $x$-axis and the positive $y$-axis at points $A$ and $B$, respectively. If $|P A|=b$, then $|P B|$ is ( ).
(A) $a$
(B) $\sqrt{a b}$
(C) $\sqr... | 4.A.
It is known that the point of tangency $P\left(x_{0}, y_{0}\right)$ is in the first quadrant. Then the equation of the tangent line $l$ is $\frac{x_{0} x}{a^{2}}+\frac{y_{0} y}{b^{2}}=1$.
By setting $y=0, x=0$, we get $A\left(\frac{a^{2}}{x_{0}}, 0\right), B\left(0, \frac{b^{2}}{y_{0}}\right)$.
From $|P A|=b$, we... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 720,628 |
5. Given the function $y=f(x)\left(x, y \in \mathbf{N}_{+}\right)$ satisfies:
(1) For any $a, b \in \mathbf{N}_{+}, a \neq b$, it holds that $a f(a)+b f(b)>a f(b)+b f(a)$;
(2) For any $n \in \mathbf{N}_{+}$, it holds that $f(f(n))=3 n$. Then the value of $f(5)+f(12)$ is $(\quad$.
(A) 17
(B) 21
(C) 25
(D) 29 | 5.D.
For any $n \in \mathbf{N}_{+}$, from (1) we get
$$
\begin{array}{l}
(n+1) f(n+1)+n f(n) \\
>(n+1) f(n)+n f(n+1),
\end{array}
$$
which means $f(n+1)>f(n)$.
Therefore, $f(x)$ is a monotonically increasing function on $\mathbf{N}_{+}$.
For any $n \in \mathbf{N}_{+}$, from (2) we get
$$
f(3 n)=f(f(f(n)))=3 f(n) \tex... | 29 | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,629 |
1. As shown in Table 1, in the use of computers, codes are arranged according to certain rules, and they are infinite from left to right and from top to bottom. In Table 1, the code 100 appears $\qquad$ times.
\begin{tabular}{|c|c|c|c|c|c|}
\hline 1 & 1 & 1 & 1 & 1 & $\cdots$ \\
\hline 1 & 2 & 3 & 4 & 5 & $\cdots$ \\
\... | From Table 1, we know that the $m$-th row is an arithmetic sequence with the first term 1 and common difference $m-1$. Therefore, the $n$-th number in the $m$-th row is
$$
\begin{array}{l}
a_{m n}=1+(n-1)(m-1) . \\
\text { Let } a_{m n}=100 \text {. Then } \\
(m-1)(n-1)=99=3^{2} \times 11 \text {. }
\end{array}
$$
The... | 6 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,631 |
2. Given $f(x)=2^{x}(x \in \mathbf{R})$ can be expressed as the sum of an odd function $g(x)$ and an even function $h(x)$. If the inequality $a g(x)+h(2 x) \geqslant 0$ holds for all $x \in[1,2]$, then the range of the real number $a$ is . $\qquad$ | 2. $\left[-\frac{17}{6},+\infty\right)$.
Obviously, $f(x)=g(x)+h(x)$,
$$
f(-x)=-g(x)+h(x) \text {. }
$$
Then $g(x)=\frac{f(x)-f(-x)}{2}=\frac{2^{x}-2^{-x}}{2}$,
$h(x)=\frac{f(x)+f(-x)}{2}=\frac{2^{x}+2^{-x}}{2}$.
Let $t=2^{x}-2^{-x}\left(\frac{3}{2} \leqslant t \leqslant \frac{15}{4}\right)$. Then $2^{2 x}+2^{-2 x}=t^... | -\frac{17}{6} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,632 |
3. As shown in Figure 3, in the Cartesian coordinate system $x O y$, $A B$ is the diameter of the semicircle $\odot O$: $x^{2}+y^{2}=1(y \geqslant 0)$, $C$ is any point on the semicircle $\odot O$, and $A C$ is extended to point $P$ such that $C P=C B$. When point $C$ moves from $B$ to $A$, the length of the trajectory... | 3. $\sqrt{2} \pi$.
As shown in Figure 8, let the intersection point of the semicircle $\odot O$ with the $y$-axis be $M$, and connect $M P$, $M B$, and $M C$. Then
$$
\begin{array}{l}
\angle B C M \\
= \angle B C A+ \\
\angle A C M \\
= \angle B C A+ \\
\angle A B M \\
= 90^{\circ}+45^{\circ}=135^{\circ} .
\end{array... | \sqrt{2} \pi | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,633 |
4. Let $\alpha, \beta, \gamma \in \mathbf{R}$. Then
$$
M=\sin ^{2}(\alpha-\beta)+\sin ^{2}(\beta-\gamma)+\sin ^{2}(\gamma-\alpha)
$$
The maximum value of $M$ is $\qquad$ . | 4. $\frac{9}{4}$.
$$
\begin{aligned}
M= & \sin ^{2}(\alpha-\beta)+\sin ^{2}(\beta-\gamma)+\sin ^{2}(\gamma-\alpha) \\
= & \frac{1-\cos 2(\alpha-\beta)}{2}+\frac{1-\cos 2(\beta-\gamma)}{2}+ \\
& \frac{1-\cos 2(\gamma-\alpha)}{2} \\
= & \frac{3}{2}-\frac{1}{2}[\cos 2(\alpha-\beta)+ \\
& \cos 2(\beta-\gamma)+\cos 2(\gamma... | \frac{9}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,634 |
5. As shown in Figure 4. In the right prism $A B C-A_{1} B_{1} C_{1}$, $\angle A C B=90^{\circ}, A C$ $=6, B C=C C_{1}=\sqrt{2}, P$ is a moving point on the line segment $B_{1} C$. Then the minimum value of $A P+P C_{1}$ is | $5.5 \sqrt{2}$.
As shown in Figure 9, connect $A B_{1}$, and unfold $\triangle C_{1} B_{1} C$ along $B_{1} C$ onto the plane $A B_{1} C$. Then the minimum value of $A P + P C_{1}$ is the length of line segment $A C_{1}$.
In $\triangle A C C_{1}$, $A C=6, C C_{1}=\sqrt{2}$, $\angle A C C_{1}=\angle A C B_{1} + \angle B_... | 5 \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 720,635 |
6. A positive integer $n$, if it does not contain any perfect square factor greater than 1, is called a "simple number". Then, among the 1000 numbers $1,2, \cdots, 1000$, the number of simple numbers is $\qquad$. | 6.508.
Notice that $1000<33^{2}$. If a number in the set $M=\{1,2, \cdots, 1000 \}$ is not a simple number, then it must contain one or more factors from the set
$$
A=\left\{2^{2}, 3^{2}, 5^{2}, 7^{2}, 11^{2}, 13^{2}, 17^{2}, 19^{2}, 23^{2}, 29^{2}, 31^{2}\right\}.
$$
Let $P(a)$ denote the number of multiples of $a^{... | 508 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 720,636 |
1. The number of rational numbers $x$ that satisfy the equation
$$
|x-1|-2|x-2|+3|x-3|=4
$$
is ( ) .
(A) 1
(B) 2
(C) 3
(D) infinitely many | -.1.D.
Points $1,2,3$ divide the number line into four intervals:
$$
(-\infty, 1],(1,2],(2,3],(3,+\infty) \text {. }
$$
Under these conditions, removing the absolute value symbol yields the solutions $1 \leqslant x \leqslant 2, x=5$, which include infinitely many rational numbers. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,637 |
2. If $2 x^{2}-6 y^{2}+x y+k x+6$ can be factored into the product of two linear factors, then the number of integer values of $k$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | 2. B.
Using the discriminant or the method of undetermined coefficients, we get $k= \pm 7$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 720,638 |
One. (20 points) Let the constant $a \geqslant 0$. Solve the equation $x^{4}+6 x^{3}+(9-2 a) x^{2}-6 a x+a^{2}-4=0$ for $x$.
| $$
\begin{array}{l}
a^{2}-\left(2 x^{2}+6 x\right) a+x^{4}+6 x^{3}+9 x^{2}-4=0 \\
\Rightarrow a^{2}-\left(2 x^{2}+6 x\right) a+\left(x^{2}+3 x\right)^{2}-4=0 \\
\Rightarrow a^{2}-\left[\left(x^{2}+3 x+2\right)+\left(x^{2}+3 x-2\right)\right] a+ \\
\quad\left(x^{2}+3 x+2\right)\left(x^{2}+3 x-2\right)=0 \\
\Rightarrow a... | x_{1,2}=\frac{-3 \pm \sqrt{4 a+1}}{2}, x_{3,4}=\frac{-3 \pm \sqrt{4 a+17}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 720,639 |
9. (40 points) Among nine visually identical gold coins, one weighs $a$, seven weigh $b$, and the last one weighs $c$, and $a < b < c$. Using a balance scale, find the minimum number of weighings required to identify the coin weighing $a$ and the coin weighing $c$.
To solve this problem, we need to devise a strategy t... | 9. Solution 1: First, label the nine coins as
$A, B, C, D, E, F, G, H, I$.
Weigh $(A, B, C, D)$ against $(E, F, G, H)$. If they balance, then $a+c=2b$. If they do not balance, assume without loss of generality that $(A, B, C, D) > (E, F, G, H)$.
For the second weighing, weigh $(A, B)$ against $(C, D)$ and $(E, F)$ aga... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 720,641 |
10. (40 points) Given that $n$ is a positive integer, such that
$$
1+n+\frac{n(n-1)}{2}+\frac{n(n-1)(n-2)}{6}
$$
$=2^{k}$ ( $k$ is a positive integer).
Find the sum of all possible values of $n$. | 10. Since
$$
\begin{array}{l}
1+n+\frac{n(n-1)}{2}+\frac{n(n-1)(n-2)}{6} \\
=\frac{(n+1)\left(n^{2}-n+6\right)}{6},
\end{array}
$$
Therefore, $n+1$ is a power of 2 or 3 times a power of 2.
(1) If $n+1=2^{m}\left(m \in \mathbf{N}_{+}\right)$, then
$$
n^{2}-n+6=2^{2 m}-3 \times 2^{m}+8
$$
is 3 times a power of 2.
When ... | 36 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 720,642 |
1. Given that when the value of $x$ is $2$, $m_{1}$, $m_{2}$, the polynomial $a x^{2}+b x+c$ has values $0$, $p_{1}$, $p_{2}$ respectively. If $a>b>c$, and $p_{1} p_{2}-c p_{1}+a p_{2}-a c=0$, can we guarantee that: when the value of $x$ is $m_{1}+5$, $m_{2}+5$, at least one of the values of the polynomial is positive?... | 1. From the given information,
$$
a x^{2}+b x+c=a(x-2)\left(x-\frac{c}{2 a}\right),
$$
and $4 a+2 b+c=0$.
Also, from $a>b>c$ we get $a>0, c<0$, and $2>\frac{c}{2 a} \text {. }
$$
From the given information, $c=-4 a-2 b>-6 a$, then
$$
\frac{c}{a}>-6, \frac{c}{2 a}>-3, m_{i}+5>2 \text {. }
$$
When $x=m_{i}+5$, the val... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 720,643 |
2. In $\triangle A B C$, $\angle A=75^{\circ}, \angle B=35^{\circ}$, and $D$ is a point on side $B C$ such that $B D=2 C D$. Prove:
$$
A D^{2}=(A C+B D)(A C-C D) \text {. }
$$ | 2. From the given, $\angle C=70^{\circ}$.
Extend $B C$ to $E$, such that $A C=C E$. Connect $A E$. Then
$$
\begin{array}{l}
\angle C E A=\angle C A E=\frac{1}{2} \angle A C B \\
=35^{\circ}=\angle A B C .
\end{array}
$$
Therefore, $\triangle C A E \backsim \triangle A E B$.
Thus, $A E^{2}=A C \cdot B E$, which means
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 720,644 |
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