problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
II. (25 points) In $\triangle A B C$, $D, D^{\prime}$ are on $B C$, $E, E^{\prime}$ are on $C A$, $F, F^{\prime}$ are on $A B$, and lines $A D$ and $A D^{\prime}$, $B E$ and $B E^{\prime}$, $C F$ and $C F^{\prime}$ are symmetric with respect to the angle bisectors of $\angle A$, $\angle B$, $\angle C$, respectively. Pr... | $$
\begin{array}{l}
\frac{B D^{\prime}}{\sin \angle B A D^{\prime}} \\
= \frac{c}{\sin \angle A D^{\prime} B}, \\
\frac{D^{\prime} C}{\sin \angle C A D^{\prime}} \\
= \frac{b}{\sin \angle A D^{\prime} C}, \\
\therefore \frac{B D^{\prime}}{D^{\prime} C} \\
= \frac{c \sin \angle B A D^{\prime}}{b \sin \angle C A D^{\pri... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,895 |
Three, (25 points) There are both coins and banknotes for 1 fen, 2 fen, and 5 fen. To exchange one jiao into several 1 fen, 2 fen, and 5 fen coins or banknotes, how many different methods are there? | Three, suppose one dime can be exchanged for $x, y, z$ coins of 1 cent, 2 cents, and 5 cents respectively. Then we have
$$
x+2 y+5 z=10 \text {. }
$$
Solving this, we get $(x, y, z)=(0,0,2),(0,5,0),(1,2$,
$$
\begin{array}{l}
1),(2,4,0),(3,1,1)(4,3,0),(5,0,1),(6, \\
2,0),(8,1,0),(10,0,0) .
\end{array}
$$
Suppose $m$ c... | 134 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 707,896 |
1. The quadratic equation with complex coefficients $x^{2}+a x+b=0$ has exactly two pure imaginary roots, then ( )
(A) $a$ is zero, $b$ is a negative real number
(B) $a$ is zero, $b$ is a pure imaginary number
(C) $a$ is a pure imaginary number, $b$ is a real number
(D) $a$ is a pure imaginary number, $b$ is a pure ima... | ,- 1. (C).
Let the roots of the equation be $\alpha i, \beta i$, where $\alpha, \beta$ are non-zero real numbers. By Vieta's formulas, we know that $\alpha i + \beta i = -a$, so $a = -(\alpha + \beta) i$, and $(\alpha i)(\beta i) = b$, so $b = -\alpha \beta$. Therefore, it can only be (C). | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,897 |
2. $n$ is a positive integer, the number of integer solutions to $x^{2}+x y+y^{2}=n$ is ( ).
(A) a multiple of 4
(B) a multiple of 6
(C) a multiple of 2
(D) a multiple of 8 | 2. (B).
If $(x, y)$ is a solution, then $(x+y, -x)$, $(y, -x-y)$, $(-x, -y)$, $(-x-y, x)$, $(-y, x+y)$ are also solutions.
Since $x, y$ are not both zero, the above 6 solutions are pairwise distinct. | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 707,898 |
3. All real numbers $x$ that satisfy $2|x-| x+|x-1|||>|x+| x-|x+1|||$ are in ( ).
(A) $\left(\frac{1}{3}, 3\right)$
(B) $(3,+\infty)$
(C) $\left(-\infty, \frac{1}{3}\right)$
(D) $\left(-\infty, \frac{1}{3}\right) \cup(3,+\infty)$ | 3. (D).
$$
\begin{array}{l}
\text { Let } f(x)=-x+|x-1|, \\
g(x)=x-|x+1|, \\
f(x)=1+(x-1)+|x-1| \geqslant 1, \\
g(x)=(x+1)-|x+1|-1 \leqslant-1 .
\end{array}
$$
Given $2|x-1|>|x+1|$, squaring both sides, we get
$$
\begin{array}{l}
3 x^{2}-10 x+3>0, x \in\left(-\infty, \frac{1}{3}\right) \cup(3, \\
+\infty) .
\end{array... | D | Inequalities | MCQ | Yes | Yes | cn_contest | false | 707,899 |
4. $f(x)=x^{4}+a x^{3}+b x^{2}+c x+d$, here $a, b, c, d$ are real numbers. It is known that $f(1)=5, f(2)=10$, $f(3)=15$, then $f(8)+f(-4)$ is $(\quad)$.
(A) 2500
(B) uncertain
(C) 2540
(D) 860 | 4. (C).
$$
\begin{array}{c}
f(x)-5 x \\
=(x-1)(x-2)(x-3)(x+h), \\
f(8)-40=210(8+h), \\
f(-4)+20=-210(h-4), \\
\text { then, } f(8)+f(-4)=2540 .
\end{array}
$$ | 2540 | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,900 |
5. In the plane, let the graph of the function $f(x)$ be symmetric to the graph of $y=$ $-2 x^{2}$ with respect to the line $y=2 x+3$, then the analytical expression of $f(x)$ is ().
(A) $3 y+4 x+6=-\frac{2}{5}(4 y-3 x-12)^{2}$
(B) $3 y+4 x+6=-10(4 y-3 x-12)^{2}$
(C) $3 y+4 x=-\frac{2}{5}(4 y-3 x)^{2}$
(D) $3 y+4 x-6=-... | 5. (A).
Take a point $(x, y)$ on the graph of $f(x)$, and there is a corresponding point $\left(x^{*}, y^{*}\right)$ on $y=$ $-2 x^{2}$, such that these two points are symmetric with respect to the line $y=2 x+3$. Therefore, we have
$$
\begin{array}{l}
\frac{1}{2}\left(y+y^{*}\right)=x+x^{*}+3, \\
y^{*}-y=-\frac{1}{2}... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,901 |
6. $n \geqslant 3$, the equation $\frac{x_{1} x_{2} \cdots x_{n-1}}{x_{n}}+\frac{x_{1} x_{2} \cdots x_{n-2} x_{n}}{x_{n-1}}$ $+\cdots+\frac{x_{2} x_{3} \cdots x_{n}}{x_{1}}=n$ has the following number of ordered integer solutions:
(A) $n$ groups
(B) $2^{n-1}$ groups
(C) $2^{n}$ groups
(D) $2^{n+1}$ groups | 6. (B).
$x_{i} \neq 0(i=1,2, \cdots, n)$, from the original equation we have
$$
\begin{array}{l}
x_{1}^{2} x_{2}^{2} \cdots x_{n-1}^{2}+x_{1}^{2} x_{2}^{2} \cdots x_{n-2}^{2} x_{n}^{2} \cdots \cdots \\
+x_{2}^{2} x_{3}^{2} \cdots x_{n}^{2} \\
=n x_{1} x_{2} \cdots x_{n} .
\end{array}
$$
Then, $x_{1} x_{2} \cdots x_{n}... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,902 |
Example 12. Let $M, x, y$ be positive integers, and $\sqrt{M-\sqrt{28}}=\sqrt{x}-\sqrt{y}$. Then the value of $x+y+M$ is ( ).
(1994, Hope Cup Mathematics Competition) | Given $M-\sqrt{28}=(x+y)$ $-\sqrt{4 x y}$. Therefore, $x+y=M$, and $4 x y=28$. Since $x, y$ are positive integers, and it is specified that $x \geqslant y$, from $x y=7$, we get $x=7, y=1, M=8, x+y+M=16$. | 16 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,903 |
1. For any integers $x, y, z$, all real numbers $a, b, c$ that satisfy the equation $|a x+b y+c z|+|b x+c y+a z|+|c x+a y+b z|=| x|+| y|+| z|$ are $\qquad$. | $$
\begin{array}{r}
\text { = }, 1 .(a, b, c)=(1,0,0),(-1,0,0), \\
(0,1,0),(0,-1,0),(0,0,1),(0,0,-1)+
\end{array}
$$
Group.
$$
\begin{array}{l}
\text { Let } x=1, y=z=0, \text { then } \\
|a|+|b|+|c|=1 .
\end{array}
$$
$$
\text { Let } x=y=z=1 \text {, then }|a+b+c|=1 \text {. }
$$
Thus, $|a+b+c|=|a|+|b|+|c|, a, b, c... | (a, b, c)=(1,0,0),(-1,0,0),(0,1,0),(0,-1,0),(0,0,1),(0,0,-1) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,904 |
2. All positive rational triples $(x, y, z) (x \leqslant y \leqslant z)$ that make $x+y+z, \frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ and $x y z$ integers are $\qquad$ . | $$
\text { 2. }(x, y, z)=(1,1,1),(1,2,2),(2,3, }
$$
$6),(2,4,4),(3,3,3) 5$ groups.
$$
\text { Let } \begin{aligned}
f(u)= & (u-x)(u-y)(u-z) \\
= & u^{3}-(x+y+z) u^{2}+(x y+y z \\
& +z x) u-x y z .
\end{aligned}
$$
$f(u)$ is a polynomial with integer coefficients. Given this, the roots must all be integers, so $x, y, z$... | (1,1,1),(1,2,2),(2,3,6),(2,4,4),(3,3,3) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 707,905 |
3. The sum of the upper base radius $r$ and the lower base radius $R (R > r)$ of a frustum is 6 times the slant height $l$, and the areas of the upper base, lateral surface, and lower base form a geometric sequence. The height of this frustum is $20 - 3r^2$, then the maximum volume of the frustum is $\qquad$ | $$
\begin{array}{l}
\text { 3. } \frac{500}{9} \pi(2+\sqrt{3}) . \\
r+R=6 l,(\pi(r+R) l)^{2}=\pi r^{2} \cdot \pi R^{2}, \\
l=\frac{R r}{R+r},(r+R)^{2}=6 R r .
\end{array}
$$
Given $R>r$, then $R=(2+\sqrt{3}) r$.
$$
\begin{array}{l}
V=\frac{1}{3} \pi\left(20-3 r^{2}\right)\left(\pi^{2}-1-R r+r^{2}\right) \\
=\frac{1}{3... | \frac{500}{9} \pi(2+\sqrt{3}) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,906 |
4. Let $f(x)=\frac{x^{4}+k x^{2}+1}{x^{4}+x^{2}+1}(x \in R)$, for any three real numbers $a, b, c$, it is known that there exists a triangle with side lengths $f(a), f(b), f(c)$, then all real numbers $k$ that satisfy the above condition are $\qquad$ . | 4. $\left(-\frac{1}{2}, 4\right)$.
From $\frac{x^{2}}{x^{4}+x^{2}+1} \leqslant \frac{1}{3}$, then
$$
f(x)=1+\frac{(k-1) x^{2}}{x^{4}+x^{2}+1} \text {. }
$$
If $k \geqslant 1$, then $1 \leqslant f(x) \leqslant \frac{1}{3}(k+2)$;
If $k \leqslant 1, \frac{1}{3}(k+2) \leqslant f(x) \leqslant 1$.
According to the problem,... | \left(-\frac{1}{2}, 4\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,907 |
5. Let $f(z)=3 z\left(\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right)$, where $z$ is a complex number. Let $A$ represent the point $f(1+\sqrt{3} i)$, $B$ represent the point $f(0)$, and $C$ represent the point $f(4 i)$. Then $\angle A B C=$ $\qquad$ | 5. $\frac{\pi}{6}$.
Let $f_{1}(z)=3 z$,
$f_{2}(z)=z\left(\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right)$.
$f(z)=f_{1} f_{2}(z) \cdot f_{1}(z)$ is a similarity transformation, $f_{2}(z)$ is a rotation transformation, then $f(z)$ is a transformation that preserves angles. Let $A^{*}$ represent the point $1+\sqrt{3} i$, ... | \frac{\pi}{6} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,908 |
6. $a, b$ are positive real numbers, $x_{0}=a, x_{1}=b, x_{n+1}=$ $\frac{1}{2}\left(x_{n-1}+\frac{1}{x_{n}}\right)$, where $x \in N$. If this sequence is a periodic sequence, then $a, b$ must satisfy the condition | $6 . a b=1$.
$x_{n} x_{n+1}=\frac{1}{2}\left(x_{n-1} x_{n}+1\right)$, let $y_{n}=x_{n} x_{n+1}$, $y_{n}=\frac{1}{2}\left(y_{n-1}+1\right)$. If $x_{n}$ is periodic, then $y_{n}$ is also periodic. If $y_{n-1}>1$, then $y_{n-1}>y_{n}>1$. Therefore, for $y_{n}$ to be periodic, it must be that $y_{n}=1$ for any non-negative... | a b=1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,909 |
一、(Given) $f(1)=2$, for positive integer $n, f(n+1)=\rho(n)+1$. Prove: when positive integer $n \geqslant 2$, we have $1-\frac{1}{2^{2^{n-1}}}<\frac{1}{f(1)}+\frac{1}{f(2)}$ $+\cdots+\frac{1}{f(n)}<1-\frac{1}{n^{n}}$ | $$
\begin{array}{l}
\text {Prove } f(n-1)-1 \\
=f(n)(f(n)-1) .
\end{array}
$$
Since $f(1)=2$, from the above equation, we know that $f(n)$ is monotonically increasing.
$$
\begin{array}{l}
\frac{1}{f(n+1)-1}=\frac{1}{f(n)(f(n)-1)} \\
=\frac{1}{f(n)-1}-\frac{1}{f(n)} .
\end{array}
$$
Thus, we have
$$
\frac{1}{f(n)}=\fr... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 707,910 |
II. (6 points) Let $A^{*}$ be the symmetric point of point $A$ with respect to line $BC$, $B^{*}$ be the symmetric point of point $B$ with respect to line $CA$, and $C^{*}$ be the symmetric point of point $C$ with respect to line $AB$. Find the necessary and sufficient condition for $\triangle ABC$ such that $\triangle... | Let $B C=a, C A=b, A B=c$.
In $\triangle A B^{*} C^{*}$,
$A B^{\cdot}=c, A C^{\cdot}=b$.
$\angle B^{*} A C^{\cdot}=|2 \pi-3 \angle A|$.
Using the cosine rule, we have
$B^{\cdot} C^{\cdot 2}=b^{2}+c^{2}-$
$2 b c \cos 3 A$.
Similarly,
$C^{\cdot} A^{\cdot 2}=a^{2}+c^{2}-$
$2 a c \cos 3 B$.
$A^{*} B^{\circ 2}=a^{2}+b^{2}-2... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,911 |
Three, (Full marks 35 points) Does there exist a positive integer $n>1$, such that $1,2,3, \cdots, n^{2}$ can be placed in an $n \times n$ grid, so that the product of each row is the same? Prove your conclusion. | Three, Solution: No empty cells. Proof by contradiction. If there exists such a square grid where the product of numbers in each row is $p$, then the total product of all numbers is $\dot{p}^{3}$, leading to
\[ 1 \cdot 2 \cdot 3 \cdot \cdots \cdot n^{2} = p^{n}. \]
Since $n > 1$, $p$ must be even. Let $m$ be the expone... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 707,912 |
Four, (Full marks 35 points) $a_{1}, a_{2}, \cdots, a_{n}$ is a sequence, for each $k, 1 \leqslant k \leqslant n, a_{k} \in\{1,0\}$. If $a_{k}, a_{k+1}$ are different, write $b_{k}=1$; if $a_{k}, a_{k+1}$ are the same, write $b_{k}=0$. Thus, a new sequence $b_{1}, b_{2}$, $\cdots, b_{n-1}$ is obtained, where $b_{k} \in... | Let $x_{n}$ denote the maximum value of the number in question. When $n=1$, $x_{n}=1$; for $n=2$, $x_{n}=2$; for $n=3$, $x_{n}=4$, because $a_{1}, a_{2}, a_{3}$ have 6 possible permutations: $0,0,0 ; 1,1,1 ; 1,0,1 ; 1,1,0 ; 0,1,1 ; 0, 1,0$; in the 2nd, 3rd, and 4th cases, $x_{3}=4$, and in the other cases, it is less t... | x_{n}=\left[\frac{n(n+1)+1}{3}\right] | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 707,913 |
Example 1. Given $n$ circles with centers $O_{i}$ $(i=1,2,3, \cdots, n)$. These $n$ circles are simultaneously internally tangent to a large semicircle at point $C$; and also tangent to the diameter of the large semicircle at point $D_{1}$. Prove: The $n$ lines $C, D$ intersect at one point. | Consider the diameter of a semicircle as $A B$, with $O$ as the center. To prove that $n$ lines $C_{i} D_{i}$ intersect at a single point, let's first take one of these lines, $C_{1} D_{1}$, and assume that $C_{1} D_{1}$ intersects $\odot O$ at $M$. Clearly, $O, O_{1}, C_{1}$ are collinear. Connect $O_{1} D_{1}, O M$. ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,914 |
Given 31. $\odot O_{1}$ is the circumcircle of isosceles $\triangle A B C$, $\odot O$ is a circle with the base $B C$ as a chord, $\odot O_{2}$ is internally tangent to $\odot O$, and is tangent to $A B$ and $A C$ at points $P$ and $Q$ respectively, and point $I$ is the incenter of $\triangle A B C$.
Prove: $\angle P I... | $$
\begin{array}{c}
\text{Given:} \\
\odot O \text{ and } \odot O_{2} \text{ are tangent at point } E, \\
A O_{1} \text{ extended intersects } \odot O_{1} \text{ at point } D, \text{ clearly } A, O, I, O_{1}, O_{2}, D, E \text{ are collinear.} \\
\text{Connect } B D, B E, B I, O_{2} P, P E. \\
\because \angle A B I=\an... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,915 |
32. As shown in the figure, in pentagon $A B C D E$, $\angle B A C=\angle C A D=\angle D A E$, $\angle A C B=\angle A D C=\angle A E D=90^{\circ}$, $F$ is the midpoint of $C D$. Prove that $A F, B D$, and $C E$ are concurrent. | Proof: Let $A F$ and $B D$ intersect at point $H$, connect $C H$, $H E$, and extend $A D$, $B C$ to intersect at point $G$. Then $C G = C B$. It is easy to prove that
$$
\begin{array}{l}
\triangle C D G \sim \triangle A D C . \\
\therefore \frac{C G}{A C}=\frac{G D}{C D}, \frac{2 C G}{A C}=\frac{G D}{\frac{1}{2} C D},
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,916 |
31. In $\triangle ABC$, prove that:
$$
\sin A + 2(\sin B + \sin C) \leqslant \frac{\sqrt[4]{108}}{2}(\sqrt{3} + 1).
$$
Determine the conditions under which equality holds. | Proof: Let $t \neq 0$ be a real number, then
$$
\begin{array}{l}
\left\lfloor\sin A+2(\sin B+\sin C)\right]^{2} \\
=\left[\sin A+4 \sin \frac{B+C}{2} \cos \frac{B-C}{2}\right]^{2} \\
=\left[2 \cos \frac{A}{2}\left(\sin \frac{A}{2}+2 \cos \frac{B-C}{2}\right)\right]^{2} \\
\leqslant\left[2 \cos \frac{A}{2}\left(\sin \fr... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 707,917 |
Example 2. Find the smallest positive integer $n$, such that in any 9-vertex and $n$-edge graph with edges colored in two colors, there must exist a monochromatic triangle.
(33rd IMO, Problem 3) | Since $K_{2}$ can have one edge removed and be colored with 0 colors to ensure it contains no monochromatic triangles, and $K_{5}$ can be colored with 2 colors to ensure it contains no monochromatic triangles. $K$ can be divided into 4 $K_{2}$'s and 1 $K_{1}$ with no common vertices, so in $K$, 4 edges can be removed a... | 33 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 707,919 |
Example 2. $\odot A$ and $\odot B$ are equal and intersect. Draw two externally tangent circles $\odot C$ and $\odot D$, such that they are both internally tangent to $\odot A$ and externally tangent to $\odot B$. Let the common internal tangent of $\odot C$ and $\odot D$ be $i$. Clearly, there can be infinitely many p... | Let $\odot A$ and $\odot B$ have radii $R$, and $\odot C$ and $\odot D$ have radii $c$ and $d$ respectively. Connect $A C$, $B C$, $M C$; $A D$, $B D$, $M D$.
Using the median formula in a triangle, we have
$$
\begin{array}{l}
A C^{2}+B C^{2}=2 \cdot C M^{2}+2 \cdot A M^{2}, \\
A D^{2}+B D^{2}=2 \cdot D M^{2}+2 \cdot ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,921 |
Example 3. Given that $\odot O$ is separated from line $l$, and there are $n$ points $P_{i}(i=1,2, \cdots, n)$ on $l$. From $P_{i}$, two tangents to $\odot O$ are drawn, with points of tangency $A_{i}, B_{i}$. Try to prove: the $n$ lines $A_{i} B_{i}$ intersect at one point. | Analyze: Draw $O H \perp l$, denote the radius of $\odot O$ as $R$, and $O H=h$. First, consider the line $A_{1} B_{1}$ obtained from point $P_{1}$.
Connect $O P_{1}, O A_{1}, O B_{1}$, $H B_{1}$, and let $A_{1} B_{1}$ intersect $O H$ at point $K$.
It is easy to prove that $O, A_{1}, P_{1}, B_{1}$ and $O, P_{1}, H, B_... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,922 |
Example 4. $A B$ is the diameter of $\odot O$, and through a fixed point $D$ on $O B$, $C D \perp A B$ intersects $\odot O$ at $C$. On $\odot O$, there are any two sets of points $M_{i}, N_{i}(i=1,2, \cdots, n)$, satisfying $\angle C D M_{i}=\angle C D N_{i}$. Try to prove: $n$ lines $M_{i} N_{i}$ concur. | First, consider $M_{1}, N_{1}$. Clearly, $\angle A D M_{1} = \angle B D N_{1}$. Extend $N_{1} D$ to intersect $\odot O$ at $K$, and connect $O M_{1}$, $O N_{1}$, $O K$. It is easy to prove that
$$
\begin{array}{l}
\triangle O M_{1} D \simeq \triangle O K D \\
\Rightarrow \angle O M_{1} D = \angle O K D = \angle O K N_{... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,923 |
Example 5. In trapezoid $ABCD$, the ratio of the two bases $DC: AB = 1: 3$. Extend the non-parallel sides $AD, BC$ to meet at point $P$. Draw $n$ line segments $X_{i} Y_{i} (i=1,2, \cdots, n)$ such that $P$ is the midpoint of $X_{i} Y_{i}$. Take the midpoints $M_{i}$ of $AX_{i}$ and the midpoints $N_{i}$ of $BY_{i}$. P... | Let $M_{1} N_{1}$ intersect $A C$ at point $Q$. We will study the properties of point $Q$.
It is easy to see that point $C$ is the centroid of $\triangle B X_{1} Y_{1}$, i.e., $X_{1}, C, Y_{1} \equiv$ are collinear. Draw $X_{1} K / / A C$ intersecting line $M_{1} N_{1}$ at $K$. Clearly,
$\triangle M_{1} A Q \cong \tri... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,924 |
Example 6. Through the midpoints of the sides of a cyclic convex quadrilateral, perpendiculars are drawn to the opposite sides. Prove that the four perpendiculars are concurrent.
| Analyzing quadrilateral $ABCD$ as a convex cyclic quadrilateral inscribed in $\odot O$, $M, N$ are the midpoints of $AB, CD$; $E, F$ are the midpoints of $BC, DA$. $EE' \perp DA$ at $E'$, $FF' \perp BC$ at $F'$, and they intersect at point $P$.
It is easy to know that $OE \perp BC, OF \perp DA$, thus quadrilateral $PEO... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,925 |
Example 8. Given $\triangle A B C$ and $n$ lines $l,(i=1,2$, $\cdots, n)$. If each line divides the perimeter of $\triangle A B C$ into two parts in the same ratio as the area of $\triangle A B C$ is divided into two parts. Prove: the $n$ lines must be concurrent. | Let line $l_{1}$ intersect $AB, AC$ at $E, F$, respectively. According to the problem, we have
$$
\begin{array}{l}
\frac{AE + AF}{BE + CF + BC} \\
= \frac{S_{\triangle AEF}}{S_{EBCF}}.
\end{array}
$$
Construct the angle bisector of $\angle BAC$ intersecting $l_{1}$ at $I$. Let the distances from $I$ to $AB$, $AC$ be $... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,926 |
Example 9. $\odot O$ is the incircle of $\triangle A B C$, and the points of tangency on $B C, C A$, and $A B$ are $D, E, F$ respectively. The ray $D O$ intersects $E F$ at $A^{\prime}$, and similarly we get $B^{\prime}, C^{\prime}$. Prove that the lines $A A^{\prime}$, $B B^{\prime}$, $C C^{\prime}$ are concurrent. | Analyzing the connection $A^{\prime} B, A^{\prime} C$. It is easy to see that $B, D, O, F$ and $C, D, O, E$ are respectively concyclic, thus we have
$$
\begin{array}{c}
\angle A^{\prime} O F= \\
\angle B, \angle A^{\prime} O E= \\
\angle C . \\
\because \frac{A^{\prime} F}{\sin \angle A^{\prime} O F}=\frac{O A^{\prime}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,927 |
Example 3. When $x=\frac{1+\sqrt{1994}}{2}$, the value of the polynomial
$$
\left(4 x^{3}-1997 x-1994\right)^{2001}
$$
is ( ).
(A) 1
(B) -1
(C) $2^{2001}$
(D) $-2^{2001}$ | Analyzing direct substitution, it is certainly not the best strategy. From the given, we have $2 x-1=\sqrt{19} 94$. Squaring and rearranging gives
$$
4 x^{2}-4 x-1993=0 \text {. }
$$
And
$$
\begin{array}{l}
4 x^{3}-1997 x-1994 \\
=(x-1)\left(4 x^{2}-4 x-1993\right)-1 \\
=-1 .
\end{array}
$$
Therefore, the desired res... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,928 |
Example 10. On the arc $\overparen{B C}$ of the circumcircle of $\triangle A B C$, take $n$ points $P_{i}(i=1,2, \cdots, n)$. Construct the symmetric points $M_{i}$ of $P_{i}$ with respect to $A B$, and the symmetric points $N_{i}$ of $P_{i}$ with respect to $A C$. Prove that the $n$ lines $M_{i} N_{i}$ are concurrent. | Analyzing first,
consider the line $M_{1} N_{1}$
intersecting $\triangle A B C$'s
altitude $A D \mp H$
at point. The feet of the perpendiculars $E, F$
are the midpoints of $P_{1} M_{1}, P_{1} N_{1}$; draw $P_{1} G \perp B C$ at $G$, clearly $E, G, F$ are collinear (Simson line).
Extend $A D$ to intersect the circle at... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,929 |
Example 2. The four vertices of a regular tetrahedron $ABCD$ lie on a sphere with radius $R$. Find the length of $AB$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
(Note: The note itself is not part of the translation, it's... | Consider a regular tetrahedron $ABCD$ formed by the diagonals on the faces of a cube inscribed in a sphere with radius $R$. Clearly, the vertices are on the sphere. Let $AB=a$, then the edge length of the cube is $\frac{\sqrt{2}}{2} a$.
$$
\begin{array}{l}
\because(2 R)^{2}=3 \cdot\left(\frac{\sqrt{2}}{2} a\right)^{2},... | null | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,931 |
Example 3. Connect the midpoints of the sides of an acute triangle with side lengths $2 a, 2 b, 2 c$, forming four triangles, which is the unfolded diagram of a tetrahedron. Find the volume of this tetrahedron. | Given the figure, let the tetrahedron be \(A_{1} B C_{1} D\). According to the problem, each face of the tetrahedron is an isosceles triangle, and the three pairs of opposite edges are equal in length. Therefore, it can be extended to form a rectangular prism \(A B C D A_{1} B_{1} C_{1} D_{1}\). At this point, each sid... | \frac{\sqrt{2}}{12}\sqrt{\left(b^{2} + c^{2} - a^{2}\right)\left(a^{2} + c^{2} - b^{2}\right)\left(a^{2} + b^{2} - c^{2}\right)} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,932 |
Example 4. Prove that if the distances between the opposite edges of a tetrahedron are $h_{1}, h_{2}$, and $h_{3}$, then the volume of the tetrahedron is not less than $\frac{1}{3} h_{1} h_{2} h_{3}$. | Proof: Let the tetrahedron be \(ABCD\), with the distances between \(AB\) and \(CD\), \(AD\) and \(BC\), \(AC\) and \(BD\) being \(h_1\), \(h_2\), and \(h_3\) respectively. Extend the tetrahedron to form a parallelepiped \(AD_1BC_1-A_1DB_1C\) (as shown in the figure), such that each edge of the tetrahedron is a diagona... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,933 |
1. Given a regular tetrahedron \(ABCD\) with the midpoints of the opposite edges \(AB, CD\) being \(M, N\) respectively. Find the angle between \(MN\) and \(BC\). | (Hint: Complete the regular tetrahedron into a cube, it is easy to see that the formed angle is $45^{\circ}$ ) | 45^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,934 |
2. In the tetrahedron $S-ABC$, there are three pairs of opposite edges that are equal and are sequentially $2 \sqrt{5}, \sqrt{13}, 2 \sqrt{2}$. Find the volume of the tetrahedron. | $\left(\right.$ Hint: $\because(\sqrt{13})^{2}+(2 \sqrt{2})^{2}>(2$ $\sqrt{5})^{2}, \therefore$ the tetrahedron $S-A B C$ can be completed into a rectangular parallelepiped. It is easy to get $V=\frac{5 \sqrt{30}}{12}$.) | \frac{5 \sqrt{30}}{12} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,935 |
3. In tetrahedron $ABCD$, $AC \perp BC$, $AD \perp BD$. Prove: The cosine of the angle between line $AC$ and $BD$ is less than $\frac{CD}{AB}$. | (提示: Complement the tetrahedron $A B C D$ to form a parallelepiped $A C B E-D C^{\prime} B^{\prime} E^{\prime}$, it is easy to see that $A, C, B, E, D$ are all on the surface of a sphere with the intersection point $O$ of $A B$ and $C E$ as the center. Also, the circumcircle of $\triangle B D E$ is a small circle of sp... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,936 |
In some materials about the "Pigeonhole Principle", the following question is often used:
Put the natural numbers from 1 to 10 in a circle, prove that there exist three adjacent numbers whose sum is not less than 17. | Proof As shown in the figure, let $a_{1}$, $a_{2}$, $a_{3}$, ..., $a_{9}$, $a_{10}$ represent ten natural numbers not exceeding 10, arranged in a circle. The three adjacent numbers form the groups $\left(a_{1}, a_{2}, a_{3}\right)$, $\left(a_{2}, a_{3}, a_{4}\right)$, $\left(a_{3}, a_{4}, a_{5}\right)$, ..., $\left(a_{... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 707,937 |
Example 4. If $a>1$.
$$
\begin{array}{l}
p=\sqrt{1993 a}-\sqrt{1993 a-1}, \\
q=\sqrt{1993 a-1}-\sqrt{1993 a}, \\
r=\sqrt{1993 a}-\sqrt{1993 a+1}, \\
s=\sqrt{1993 a+1}-\sqrt{1993 a},
\end{array}
$$
Among $p, q, r, s$, the one with the smallest value is ( ).
(A) $p$
(B) $q$
(C) $r$
(D) $s$ | Given $\because a>1$,
$$
\therefore p>0, s>0, q<0, r<0 \text {. }
$$
Therefore, we only need to determine the sizes of $\boldsymbol{y}, \boldsymbol{\%}$.
$$
\begin{array}{l}
\text { For } q_{2}=\frac{1}{\sqrt{1993 a-1}+\sqrt{1993 a}}, \\
r=\frac{-1}{\sqrt{1993 a}+\sqrt{1953 a+1}} . \\
\text { Also } 0<\sqrt{1993 a-1}+... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 707,939 |
Let $R, r, r_{a}, r_{b}, r_{c}$ be the circumradius, inradius, and exradii of $\triangle A B C$, respectively, then
$$
\frac{r_{a}}{r_{b} r_{c}}+\frac{r_{b}}{r_{c} r_{a}}+\frac{r_{c}}{r_{a} r_{b}} \geqslant \frac{2}{r}-\frac{2}{R} .
$$ | Prove as shown in the figure, we have
$$
\begin{array}{l}
r \operatorname{ctg} \frac{B}{2} + r = \operatorname{tg} \frac{C}{2} \\
= \alpha = r_{a} \operatorname{cog} \frac{\pi - B}{2} + r_{a} \operatorname{ctg} \frac{\pi - C}{2}. \\
\therefore r_{a} = \frac{r}{\operatorname{tg} \frac{B}{2} \operatorname{tg} \frac{C}{2}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 707,940 |
Proposition In an acute triangle $\triangle ABC$, $AD, BE, CF$ are altitudes, and $H$ is the orthocenter. Let $h=\sqrt{H D \cdot H A}$. Then $H D^{2}+H E^{2}+H F^{2} \geqslant \frac{3}{2} h^{2}$. | Proof As shown in the figure, it is easy to prove by applying the concyclic four points:
$$
\begin{array}{l}
H D \cdot H A \\
=H E \cdot H B \\
=H F \cdot H C \\
=h^{2} .
\end{array}
$$
Since $P H=h$, and $P H \perp$ plane $A B C$, then $B C \perp P D$.
$$
\begin{array}{l}
\text { Also } P H^{2}=h^{2}=H D \cdot H A, \... | H D^{2}+H E^{2}+H F^{2} \geqslant \frac{3}{2} \cdot h^{2} | Inequalities | proof | Yes | Yes | cn_contest | false | 707,942 |
Proposition Let the three segments remaining after removing $\triangle A B C$ from its circumcircle be $h_{1}, h_{2}, h_{3}$, and let the inradius and circumradius of $\triangle A B C$ be $r$ and $R$, respectively. Then
$$
2 r \leqslant 2 \sqrt[3]{h_{1} h_{2} h_{3}} \leqslant R .
$$ | Let $h_{1}$ be the altitude of the triangle $BC$. It is easy to know from the left figure below that
$$
h_{1}=R-R \cos A=2 R \sin ^{2} \frac{A}{2} ;
$$
From the right figure below, we have
$$
h_{1}=R+R \cos \left(180^{\circ}-A\right)=2 R \sin ^{2} \frac{A}{2} \text {, }
$$
Thus,
$$
h_{1} h_{2} h_{3}=(2 R)^{3}\left(\s... | 2 r \leqslant 2 \sqrt[3]{h_{1} h_{2} h_{3}} \leqslant R | Inequalities | proof | Yes | Yes | cn_contest | false | 707,944 |
1. Let $A, B, C, D$ be four distinct points on a straight line in that order. Two circles are drawn with diameters $A C$ and $B D$, respectively, intersecting at points $X$ and $Y$. The line $X Y$ intersects $B C$ at $Z$. If $P$ is a point on the line $X Y$ different from $Z$, the line $C P$ intersects the circle with ... | Analyzing the typical method to prove three lines are concurrent is: select one of the lines, examine the intersection of this line with the other two, and utilize known properties of concurrency, such as the three medians of a triangle intersecting at a point, etc.
In this problem, we should consider two cases: point... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,945 |
2. Let $a, b, c$ be positive real numbers and satisfy $abc=1$. Prove:
$$
\begin{array}{l}
\frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)} \\
\geqslant \frac{3}{2} .
\end{array}
$$ | Analyzing this problem, which seems easy at first glance, should not be taken lightly. Among the 412 contestants, as many as 300 stumbled on this problem, scoring 0 points. This is often due to the difficulty in converting the given condition \(abc = 1\) into other forms, such as converting it to
\[
a b \cdot b c \cdot... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 707,946 |
5. Let $A B C D E F$ be a hexagon, satisfying $A B=$ $B C=C D, D E=E F=F A, \angle B C D=\angle E F A$ $=60^{\circ}$. Let $G$ and $H$ be two points inside this hexagon such that $\angle A G B=\angle D H E=120^{\circ}$. Prove:
$$
A G+G B+G H+D H+H E=C F .
$$ | Analyzing the given convex hexagon, it can be divided into two equilateral triangles and a quadrilateral. Noting that quadrilateral $A B D E$ is symmetrical about line $B E$, the problem can be easily solved.
Proof 1: Taking line $B E$ as the axis of symmetry, construct the axisymmetric points $C^{\prime}$ and $F^{\pr... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,949 |
Example 5. Simplify $\frac{\sqrt{5}+3 \sqrt{3}+4 \sqrt{2}}{(\sqrt{5}+\sqrt{2})(\sqrt{2}+\sqrt{3})}$. (1994, Tianjin Mathematical Competition) | Analyzing, if we rationalize the denominator in the conventional way, it would be very complicated. Observing the characteristics of the numerator, we get
the original expression
$$
\begin{array}{l}
=\frac{(\sqrt{5}+\sqrt{2})+3(\sqrt{2}+\sqrt{3})}{(\sqrt{5}+\sqrt{2})(\sqrt{2}+\sqrt{3})} \\
=\frac{1}{\sqrt{2}+\sqrt{3}}+... | \sqrt{3}+\sqrt{5}-2 \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,950 |
57.04. There are 60 students. It is known that from any 10 of these students, 3 can be found who are in the same class. Is it true that among these 60 students, one can definitely find: (a) 15, (b) 16 students in the same class? | E\%. 84. (a) It is definitely possible to find 15 students with the same surname.
If it is not possible to find 15 students with the same surname, then if the students with surnames Jia, Yi, Bing, and Ding each have at least two students, then by selecting two students from each of these four surnames, and then selecti... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 707,955 |
57. 05. A factory reduced the volume of its product by $51 \%$ over two years, and the percentage reduction was the same each year. Find the percentage reduction each year. | 57.05. Let the annual reduction percentage be $x$, then we have
$$
(1-x)^{2}=0.51 \text {. }
$$ | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,956 |
57.07. Now there are many red, yellow, and green cubes of size $1 \times 1$ $\times 1$. Can they be used to form a $3 \times 3 \times 3$ cube such that each $3 \times 1 \times 1$ column contains all 3 colors? | 57. 07. It is possible. Just assemble the top, middle, and bottom 3 layers of the $3 \times 3 \times 3$ cube (they are all $3 \times 3 \times 1$) in the following way (see the figure below). | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 707,958 |
57.08. A school organized a total of 20 astronomical observation activities, with exactly 5 students participating in each activity, and no two students have participated in more than one activity together. Prove that there are at least 20 different students who participated in these activities. | 57. 08. Assuming there are no more than 19 different students
participating in these activities. Since a total of 20 activities were held, with 5 people participating in each, there were a total of 100 participants. Since $100 > 19 \times 5$, there must be at least one student (let's call him A) who participated in at... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 707,959 |
Example 6. Let $M=\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+$ $\cdots+\frac{1}{\sqrt{1993}+\sqrt{1994}}, N=1-2+3-4+\cdots$ $+1993-1994$. Then the value of $\frac{N}{(M+1)^{2}}$ is ( ). | $$
\begin{array}{l}
M=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\cdots \\
+\sqrt{1994}-\sqrt{1993} \\
= \sqrt{1994}-1, \\
N=-1 \times \frac{1994}{2}=-997 . \\
\text { Therefore } \frac{N}{(M+1)^{2}}=-\frac{1}{2} .
\end{array}
$$ | -\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,961 |
Number, they mistook them for a 6-digit number, and this 6-digit number happens to be 7 times the original product. Try to find the original two numbers and the mistaken 6-digit number.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result di... | 57. 10. 143,143143 .
Let these two 3-digit numbers be $x$ and $y$. According to the problem, we have:
$$
7 x y=10^{3} x+y \text {, }
$$
Thus, $y=\frac{10^{3} x}{7 x-1}$.
Therefore, $10^{3} x$ should be a multiple of $7 x-1$. We have
$$
10^{3} x=143(7 x-1)+143-x .
$$
Hence, $143- x=k(7 x-1)$, where $k$ is an integer.... | null | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 707,962 |
57. 11. In $\triangle A B C$, draw the angle bisectors of $\angle A$ and $\angle C$. From point $B$, draw perpendiculars to these two angle bisectors, with the feet of the perpendiculars being points $P$ and $Q$. Prove: $P Q / / A C$. | 57. 11. As shown in Figure 3, let $A P$ and $C Q$ intersect at $D$, then $B D$ is the angle bisector of $\angle B$.
Let $\angle A=2 \alpha$, $\angle B=2 \beta$, $\angle C=2 \gamma$, then
$$
\alpha+\beta+\gamma=90^{\circ} \text{. }
$$
Since $\angle B P^{2} D=\angle B Q D=90^{\circ}$, points $B, P, D, Q$ are concyclic.... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,963 |
57. 18. Two circles intersect at points $A$ and $B$. From point $A$, draw tangents to both circles, intersecting the circles at points $M$ and $N$. Lines $B M$ and $B N$ intersect the circles again at points $P$ and $Q$ (with $P$ on $B M$ and $Q$ on $B N$). Prove that segments $M P$ and $N Q$ are equal. | 57. 18. As shown in Figure 6, connect $A P$ and $A Q$. To prove $M P = N Q$, it suffices to prove
$$
\begin{array}{l}
\triangle A P M \cong \triangle A N Q . \\
\because \angle A P B = \angle A N B, \\
\angle A Q B = \angle A M B,
\end{array}
$$
$\therefore \triangle A P M \subset \triangle A N Q$, and $P M$ and $N Q$ ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,970 |
The last digit is not 0, after removing one digit (but not the first digit) from it, it becomes a smaller number that is an integer multiple of the resulting number (i.e., the original number is an integer multiple of the resulting number -- translator's note).
Translate the above text into English, please retain ... | 57. 19. Let the original number be $b$ times the later number, i.e.,
$$
\begin{array}{l}
\overline{a_{k} \cdots a_{l+1}} \cdot 10^{t}+a_{t} \cdot 10^{t-1}+\overline{a_{i-1} \cdots a_{1}} \\
=b\left(\overline{a_{k} \cdots a_{l+1}} \cdot 10^{t-1}+\overline{a_{t-1} \cdots a_{1}}\right) .
\end{array}
$$
It is easy to see ... | 180625 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 707,971 |
7. Simplify $\frac{\sqrt{5-\sqrt{24}}}{\sqrt{10}+3-\sqrt{6}-\sqrt{15}}$. (1994, Tianjin Mathematical Competition) | $$
\begin{array}{l}
\frac{\sqrt{5-2 \sqrt{6}}}{\sqrt{3}(\sqrt{3}-\sqrt{2})-\sqrt{5}(\sqrt{3}-\sqrt{2})} \\
=\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3}-\sqrt{2})(\sqrt{3}-\sqrt{5})} \\
=\frac{1}{\sqrt{3}-\sqrt{5}}=-\frac{1}{2}(\sqrt{3}+\sqrt{5}) .
\end{array}
$$
Solve the original expression:
$$
\begin{array}{l}
\frac{\sqrt{5-... | -\frac{1}{2}(\sqrt{3}+\sqrt{5}) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,972 |
57. 20. Jia arranges some cardboard pieces on a $10 \times 10$ grid paper as follows: one $1 \times 4$ piece, two $1 \times 3$ pieces, three $1 \times 2$ pieces, and four $1 \times 1$ pieces. After placing them, the pieces cannot be placed along the edges of the grid paper (the pieces must be placed within the rectangl... | 57. 20. First answer $(b)$. From figure (1), it can be seen that when nine smaller pieces of paper are placed in the box, it may not be possible to fit a tenth $(1 \times 4)$ cardboard piece.
(a) According to the arrangement in figure 7(2), a $10 \times 10$ grid paper is divided into nine rectangular areas. When placin... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 707,973 |
1. Write $20 \uparrow 1.1$ and $20 \uparrow 1.11$ on a piece of paper, cross out some numbers so that the sum of the remaining numbers is 19.93, please explain how to do it. | 1. If the paper is left with $n$ 1.11s, then the last digit of $n$ should be 3, because $n \leqslant 20$. So $n=3$ or $n=13$. From this, it can be deduced that 7 1.11s and 15 1.1s should be crossed out. | 7 \text{ 1.11s and 15 1.1s should be crossed out} | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 707,974 |
3. During the New Year festival, the school bought a chocolate bar for each student. It is known that if the chocolate bars are packed 20 to a box, then 5 more boxes are used compared to when they are packed 24 to a box. How many students are there? | 3. If each tray contains 20 pieces and requires $x$ boxes per week, then, $20 x=24(x-5)$. Solving for $x$ gives $x=30$. Therefore, the number of students is $20 \times 30=600$. | 600 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,976 |
The size of the acute angle formed between them.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: It seems there might be a typo or an unclear term in the original text ("锐命" doesn't make sense in Chinese and is transla... | 4. The minute hand points to "9", and the angle it forms with the vertical upward line is $90^{\circ}$, while the hour hand is between "11" and "12", and the angle it forms with the vertical upward line is $\frac{1}{4} \cdot \frac{360^{\circ}}{12}=7.5^{\circ}$. Therefore, the required angle is $90^{\circ}$
$$
-7^{\circ... | null | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,977 |
5. In a certain month, there are more Mondays than Tuesdays, and more Sundays than Saturdays. What day of the week is the 5th of this month? Could this month be December? | 5. From the 1st to the 28th, these four weeks have four Mondays, four Tuesdays, ... four Sundays. From the given conditions, we know that the 29th is a Tuesday, the 30th is a Wednesday, and this month does not have a 31st. Therefore, the 1st of this month is a Sunday, and the 5th is a Thursday.
This month cannot be Dec... | Thursday | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 707,978 |
2. Given a ratio $a: b=c: d$.
(1) Prove: $a: b=(a+c):(b+d)$;
(2) The ratio $a: b=(a c):(b d)$ is also correct. Is this conclusion true? | 2. (1) From $a d=b c$, we get $a(b+d)=b(a+c)$.
Therefore, $a: b=(a+c):(b+d)$.
(2) If $a: b=c{ }^{2} d \neq 1$, then,
$a: b \neq(a c):(b d)$.
For example, $2: 1=4: 2$, but $2: 1 \neq 8: 2$. So $a: b=(a c):(b d)$ is not true. | proof | Algebra | proof | Yes | Yes | cn_contest | false | 707,980 |
3. On the diagonal $BD$ of square $ABCD$, take two points $E$ and $F$, such that the extension of $AE$ intersects side $BC$ at point $M$, and the extension of $AF$ intersects side $CD$ at point $N$, with $CM = CN$. If $BE = 3$, $EF = 4$, what is the length of the diagonal of this square? | 3. From the condition, we get
$\triangle A^{\prime} B M I \triangle \triangle A C M$.
$\therefore \angle B E=\angle D A E$.
Also, $A B=A D$,
$\angle A B E=\angle A D F$,
$\therefore \triangle A B E \simeq \triangle A D F$.
Therefore, $D F=B E=3$.
$\therefore A C=B D=10$. | 10 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,981 |
4. Two commercial vendors buy goods at the same price. After a week, the first vendor's price decreases by $10 \%$, and after another week, the price increases by $20 \%$, while the second vendor's price increases by $10 \%$ after two weeks. Which vendor's price is lower after two weeks? | 4. Let the original price of the goods be $x$, then the price of the first vendor after two weeks is $x \cdot \frac{90}{100} \cdot \frac{120}{100}=1.08 x$. While the price of the second vendor is $x \cdot \frac{110}{100}=1.1 x$. Therefore, the price of the first vendor is lower. $(x>0)$. | 1.08x < 1.1x | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,982 |
Example 8. Calculate $(\sqrt{5}+\sqrt{6}+\sqrt{7})(\sqrt{5}$
$$
\begin{array}{l}
+\sqrt{6}-\sqrt{7})(\sqrt{5}-\sqrt{6}+\sqrt{7}) \\
\cdot(-\sqrt{5}+\sqrt{6}+\sqrt{7}) .
\end{array}
$$ | Analyze the repeated application of the difference of squares formula to get the result as 104. | 104 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,983 |
1. Do the lines $x+y-1=0, 2x-5y+1=0$ and $4x-3y-1=0$ pass through the same point? | 1. A system of three equations with two unknowns:
$$
\left\{\begin{array}{l}
x+y-1=0, \\
2 x-5 y+1=0 . \\
4 x-3 y-1=0 .
\end{array}\right.
$$
Solving (1) and (2) gives $x=\frac{4}{7}, y=\frac{3}{7}$. Substituting into (3), the equation holds,
Therefore, the three lines intersect at the same point $\left(\frac{4}{7}, ... | \left(\frac{4}{7}, \frac{3}{7}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,985 |
3. In acute triangle $\triangle A B C$, the altitudes $A A_{1}$ and $C C_{1}$ intersect at point $O$. Prove that if $O A=O C$, then $\triangle A B C$ is an isosceles triangle. | 3. From $O A=O C, \angle C_{1} O A=\angle A_{1} O C$, we get
$$
\begin{array}{l}
\text { Rt } \triangle A O C_{1} \mathscr{2 R t} \triangle C O A_{1} . \\
\therefore O C_{1}=O A_{1}, \\
A A_{1}=C C_{1} .
\end{array}
$$
Thus, we have
$$
\triangle A B A_{1} \simeq \triangle C B C_{1} .
$$
Therefore, $A B=B C$. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 707,987 |
5. In a certain month, there are three Sundays that fall on even dates. What day of the week is the 5th of this month? | 5. In the 28 days from the 1st to the 28th, there are four Sundays, two of which fall on prime dates, so there must be another Sunday on the 30th, which means the first Sunday is the 2nd, and the 5th is a Wednesday. | Wednesday | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 707,989 |
3. Among the one million natural numbers from 1 to 1000000, are there more numbers that can be divided by 11 but not by 13, or more numbers that can be divided by 13 but not by 11? | 3. Among the one million natural numbers from 1 to $10^{6}$, the numbers that can be divided by 11 but not by 13 are more than the numbers that can be divided by 13 but not by 11.
In fact, let the number of such numbers be $A$ and $B$, respectively, and the number of numbers from 1 to $10^{6}$ that can be divided by b... | A > B | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 707,992 |
4. Draw a triangle on a grid paper with its vertices at the intersections of the grid lines. How can you use a ruler to find the intersection of the medians of this triangle?
| 4. Let the vertices of the given triangle be $A, B, C$. Then, as long as we find points $A_{1}, B_{1}$ such that quadrilaterals $A B A_{1} C$ and $B C B_{1} A$ are both parallelograms, the intersection point of lines $A A_{1}$ and $B B_{1}$ is the centroid of $\triangle A B C$. | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 707,993 |
Example 9. If $a, b, c$ are pairwise distinct rational numbers. Prove that $\sqrt{\frac{1}{(a-b)^{2}}+\frac{1}{(b-c)^{2}}+\frac{1}{(c-a)^{2}}}$ is a rational number. (1991, Beijing Mathematical Competition) | To prove that $w=(a-b)^{2}(b-c)^{2}+(b-c)^{2}(c-a)^{2}+(c-a)^{2}(a-b)^{2}$ is a perfect square, we need to show that:
$$
\begin{aligned}
w & =[(a-b)(b-c)+(b-c)(c-a) \\
& +(c-a)(a-b)]^{2}-2(a-b)(b-c) \\
& \cdot(c-a)(a-b+b-c+c-a) \\
& =[(a-b)(b-c)+(b-c)(c-a) \\
& +(c-a)(a-b)]^{2} .
\end{aligned}
$$
Rational numbers. | proof | Algebra | proof | Yes | Yes | cn_contest | false | 707,994 |
5. Divide a square with a side length of 1993 into unit squares using lines parallel to its sides. In the game played on this square: two people take turns moving a chip. On each move, one can move the chip up any number of squares or to the right any number of squares (but not both up and to the right at the same time... | 5. The second player to move can win. The winning strategy is: no matter where the first player moves the code, he just needs to move the equal code to the square on the diagonal from the bottom left to the top right. | proof | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 707,995 |
1. On the coordinate plane, plot the set of points $L(x, y)$ that are solutions to the equation $\frac{x^{2}+x}{y^{2}+y}=1$. | 1. The lines $y=x$ and $y=-x-1$ excluding the points with coordinates (0, 0), (0, -1), (-1, 0), (-1, -1), form the set of points as shown in the figure. | y=x \text{ and } y=-x-1 \text{ excluding the points (0, 0), (0, -1), (-1, 0), (-1, -1)} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,996 |
4. For positive $x$, we have $x^{2}+\frac{1}{x^{2}}=7$. Prove that $x^{5}+\frac{1}{x^{5}}$ is a number, and calculate the result. | 4. From the equation $\left(x+\frac{1}{x}\right)^{2}=x^{2}+\frac{1}{x^{2}}+2$ and $x+\frac{1}{x}=3$.
We get
$$
\begin{array}{l}
27:=\left(x+\frac{1}{x}\right)^{3} \\
=x^{3}+3 x^{2} \cdot \frac{1}{x}+3 x \cdot \frac{1}{x^{2}}+\frac{1}{x^{3}} \\
=x^{3}+\frac{1}{x^{3}}+3\left(x+\frac{1}{x}\right) .
\end{array}
$$
Thus, ... | 123 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 707,999 |
5. In a rectangle that is 19 units wide and 93 units long (with the longer side horizontal), use lines parallel to the sides to divide it into unit squares. Place a token in the unit square at the bottom-left corner of the original rectangle. Two students play a game where they take turns moving the token. On each turn... | 5. The student who moves the token first can win. His winning strategy is:
Start from the top right corner and draw a $19 \times$
19 square's diagonal (as shown in the figure).
The first student places the token at the endpoint of the
diagonal, then, no matter how the opponent moves, he always moves the token to the sq... | not found | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 708,000 |
4. Given $\cos \alpha+\cos \beta=a, \sin \alpha+\sin \beta=b, a^{2}+b^{2} \neq$ 0. Write the expression for $\cos (\alpha+\beta)$ in terms of $a$ and $b$. | $$
\begin{array}{l}
\cos \alpha+\cos \beta=2 \cos \frac{\alpha+\beta}{2} \cdot \cos \frac{\alpha-\beta}{2}=a, \\
\sin \alpha+\sin \beta=2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}=b,
\end{array}
$$
We can get $\operatorname{tg} \frac{\alpha+\beta}{2}=\frac{b}{a}$.
$$
\begin{array}{l}
\because \cos x=\fra... | \frac{a^{2}-b^{2}}{a^{2}+b^{2}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,004 |
16. Calculate $(\sqrt{1300}]+[\sqrt{1901}]+\cdots$ $+[\sqrt{1993}]=(\quad)$.
(1993, Hope Cup Mathematics Competition) | $$
\begin{array}{l}
\text { Analysis } \because 43<\sqrt{1900}, \sqrt{1901}, \cdots, \\
\sqrt{1935}<44,44 \leqslant \sqrt{1936}, \sqrt{1937}, \cdots, \\
\sqrt{1993}<45, \\
\therefore[\sqrt{1900}],[\sqrt{1901}], \cdots,[\sqrt{1935}] \text { are all } 43, \text { a total of 36, } [\sqrt{1936}],[\sqrt{1937}], \cdots, [\sq... | 4100 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 708,005 |
5. On a square wooden board with $9 \times 9$ squares,
some squares are painted black, while the rest are painted white. We define that three small squares forming a "corner" (i.e., the four arrangements shown in the figure) can simultaneously be repainted to the opposite color (black to white, white to black). Prove:... | 5. According to the given conditions, using the help of "angles", the four small squares in a $2 \times 2$ square can be transformed three times (as shown in the figure). After such transformations, the color of one square in the $2 \times 2$ square changes. Using this method, all the small squares in the original boar... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 708,006 |
1. A perfect square number has its first two digits as 19 and the last two digits as 95. Then such a perfect square number ( ).
(A) does not exist
(B) has only one
(C) has two
(卫) 葆个.以.亡.
Note: The last option (卫) 葆个.以.亡. seems to be in Chinese and does not have a clear English translation. It might be a typo or an er... | -1 (A).
Obviously, the unit digit of the arithmetic square root of this square number is 5. Let's assume it is $10a+5$, then $(10a+5)^2=$ $100a^2+100a+25$, which means the last two digits of this square number can only be 25. | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 708,007 |
3. As shown in the figure, in $\triangle A B C$, $D$ is a point on $A B$, and $D C=$ $\frac{A D \cdot B C}{A C}$. Then ().
(A) $\triangle A D C \sim \triangle A C B$
(B) $\triangle B D C \sim \triangle B C A$
(C) $\triangle A D C \sim \triangle C D B$
(D) There are no similar triangles in the figure | 3. (A).
From the problem, we know $\frac{D C}{A D}=\frac{B C}{A C}$,
$$
\text { and } \frac{D C}{A D}=\frac{\sin A}{\sin \angle A C D}, \frac{B C}{A C}=\frac{\sin A}{\sin B} \text {, }
$$
then $\sin \angle A C D=\sin B$.
But $\angle B+\angle A C D$
$$
<\angle B+\angle A C B<180^{\circ} \text {, }
$$
then $\angle A C... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 708,009 |
5. Given the quadratic inequality $a x^{2}+b x+c \geqslant 0$ has the solution set $-2 \leqslant x \leqslant 3$. Then the solution set of the inequality $c x^{2}+b x+a \leqslant 0$ is ( ).
(A) $x \geqslant 3$ or $x \leqslant-2$
(B) $x \leqslant-\frac{1}{2}$ or $x \geqslant \frac{1}{3}$
(C) $-\frac{1}{2} \leqslant x \le... | 5. (B).
From the problem, we have $a<0$, and
$$
\begin{array}{l}
a x^{2}+b x+c \\
=a(x+2)(x-3)=a x^{2}-a x-6 a,
\end{array}
$$
then $c x^{2}+b x+a$
$$
\begin{aligned}
= & -6 a x^{2}-a x+a \\
& =-a(3 x-1)(2 x+1) \leqslant 0
\end{aligned}
$$
Thus, the solution set is $x \leqslant-\frac{1}{2}$ or $x \geqslant \frac{1}{... | B | Inequalities | MCQ | Yes | Yes | cn_contest | false | 708,011 |
6. In square $ABCD$, the distances from two vertices to line $l$ are equal and are twice the distances from the other two vertices to line $l$. How many such lines $l$ are there?
(A) 4
(B) 8
(C) 12
(D) 16 | 6. (C).
As shown in the figure, $A E=\frac{1}{2} E D, B F$ $=\frac{1}{2} F C$, then there are 8 such lines $l_{1}$. Also, the distance from $B, D$ to $l_{2}$ is twice the distance from $A, C$ to $l_{2}$, there are 4 such lines $l$, making a total of 12 lines. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 708,012 |
1. Given $(a x+n)^{4}=a x^{4}+b x^{3}+c x^{2}+d x$
+16 . Then $b+c+d=$ $\qquad$ | $=、 1.64$ or $\cdots 16$.
From the problem, we have $c^{\prime}=a, n^{\prime}=16$, then
$$
a=1, n= \pm 2 \text {. }
$$
When $a=2$,
$$
b+c+d=(1+2)^{4}-1-16=64 \text {; }
$$
When $n=-2$,
$$
b+c+d=(1-2)^{4}-1-16=-16 .
$$ | 64 \text{ or } -16 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,013 |
3. As shown in the left figure, it is a regular pentagram, and $S_{\triangle A F J}$ $=10$. Then the area of the entire pentagram is $\qquad$
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 3. $50+10 \sqrt{5}$.
Connect $G J, G I$. By the property of the golden ratio, we get $\frac{F G}{A F}=$ $\frac{\sqrt{5}-1}{2}=\frac{S_{\triangle G F J}}{S_{\triangle A F J}}$, then $S_{\triangle G F J}:=5(\sqrt{5}-1)$, $S_{\triangle G H I}=5(\sqrt{5}-1)$, and $\frac{S_{\triangle G I I}}{S_{\triangle G J A}}=\frac{J I}... | 50+10 \sqrt{5} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 708,015 |
1. Find all such positive integers $a$ such that the equation $a x^{2}+2(2 a-1) x+4(a-3)=0$ has at least one real root. (3rd Zu Chongzhi Cup Junior High School Competition) | Analysis 1 Since the equation has integer solutions, then its discriminant $\Delta=4(2 a-1)^{2}-16 a(a-3)=4(8 a+1)$ is a perfect square, thus $8 a+1$ is a perfect square.
Since $8 a+1$ is odd. Therefore, we can set $8 a+1$ $=(2 m+1)^{2}\left(m\right.$ is a positive integer), then $a=\frac{1}{2} m(m+$ 1).
Hence $x_{1,2... | a=1,3,6,10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,016 |
Example 1. Given that $x, y, z$ are all in $(0,1)$. Prove:
$$
x(1-y)+y(1-z)+z(1-x)<1 .
$$ | $$
\begin{array}{l}
\because x, y, z \text{ are all in } (0,1), \text{ we think of taking points } D, E, F \text{ on the sides of an equilateral } \triangle ABC \text{ with side length 1, such that } AD=x, BE=y, CF=z. \text{ Then } x(1-y), y(1-z), z(1-x) \text{ are related to the areas of the triangles, as shown in the... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 708,018 |
Four, (Full marks 25 points) In the positive arithmetic sequence $\left\{a_{n}\right\}, a_{1}, a_{2}, \cdots, a_{3 n}$ with common difference $d>0$, any $n+2$ numbers are taken. Try to prove that there must exist two numbers $a_{i}, a_{j}$ satisfying the inequality
$$
1<\frac{\left|a_{i}-a_{j}\right|}{n d}<2 .
$$ | In the extracted $n+2$ numbers, let $a_{i}$ be the maximum, $a_{l} \leqslant a_{s n}$. By adding $a_{m}-a_{i}$ to each number, this processing does not change the beauty value of any two numbers, so we can always assume that the extracted $n+2$ numbers include $a_{2 n}$, and let $a_{t}=a_{m}$.
If among the extracted $... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 708,019 |
In the figure, for the scalene $\triangle ABC$, the three sides satisfy the relationship $BC = \frac{1}{2}(AB + AC)$. $O$ and $I$ are the circumcenter and incenter of $\triangle ABC$, respectively. The external angle bisector of $\angle BAC$ intersects the circumcircle $\odot O$ at point $E$.
Prove: $OI = \frac{1}{2} ... | Extend $A I$ to intersect $\overparen{B C}$ at point $D$, connect $D E$, then by the given, $D$ is the midpoint of $\mathscr{B C}$, and $D F$ is a diameter.
Thus, $D E \perp B C$ (by the perpendicular diameter theorem), let the foot of the perpendicular be $H$, then $H B=H C=\frac{1}{2} \because C$.
Now connect $I B,... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 708,020 |
Initial 34. Solve the equation in the set of real numbers
$$
\sqrt[3]{(x+14)(x-16)}+\sqrt[5]{(x+15)(17-x)}=1 .
$$ | Let $a=\sqrt[5]{(x+14)(x-16)}$,
$$
b=\sqrt[5]{(x+15)(17-x)} \text {. }
$$
Then we have
$$
\left\{\begin{array}{l}
a+b=1 \\
a^{5}+b^{5}=31
\end{array}\right.
$$
(2) $\div$ (1) gives $a^{4}+b^{4}-a^{3} b-a b^{3}+a^{2} b^{2}=31$.
From (1), squaring and rearranging gives $a^{2}+b^{2}=1 - 2ab$.
From (4), squaring and rearr... | x=1 \pm 4 \sqrt{14}, x=1 \pm \sqrt{257} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,021 |
33. Given that $h(x), k(x)$ are real-valued functions defined on $R$, and satisfy
$$
\begin{array}{l}
h(x) \cdot h(y)=\frac{k(x+y)-k(x-y)}{2}, \\
\quad k(x) \cdot k(y)=\frac{k(x+y) 2+k(x-y)}{2}. \\
x, y \in R .
\end{array}
$$
Find all possible values of $h(0)$ and $k(0)$. | Let $x=y=0$ in (1) and (2), we get
$$
h^{2}(0)=0 \text { and } k^{2}(0)=k(0) .
$$
Thus, $h(0)=0, k(0)=0$ or $k(0)=1$.
But $k(0)=0$ cannot hold. Otherwise, by setting $x=y$ in (1) and (2), we get
$$
\begin{array}{l}
h^{2}(x)=\frac{k(2 x)-k(0)}{2}=\frac{k(2 x)}{2}, \\
k^{2}(x)=\frac{k(2 x)+k(0)}{2}=\frac{k(2 x)}{2} .
\e... | h(0)=0, k(0)=1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,022 |
For any $a>0, b>0, c>0$, we have
$$
\sqrt{a^{2}-a b+b^{2}}+\sqrt{b^{2}-b c+c^{2}} \geqslant \sqrt{a^{2}+a c+c^{2}},
$$
with equality when $\frac{1}{b}=\frac{1}{a}+\frac{1}{c}$. | Proof $\because \sqrt{a^{2}+b^{2}-a b}$
$$
\begin{array}{c}
=\sqrt{a^{2}+b^{2}-2 a b \cos 60^{\circ}}, \\
\sqrt{b^{2}+c^{2}-5}=v^{2}+a^{2}-2 \dot{u} c \cos 60^{\circ}, \\
\sqrt{a^{2}+c^{2}+a c}= \\
\sqrt{a^{2}+c^{2}-2 a c \cos 120^{\circ}},
\end{array}
$$
$\therefore$ we can construct $\triangle A B D$, $\triangle B D ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 708,026 |
6. Prove the identity
$$
\operatorname{ctg} 10^{\circ}-4 \cos 10^{\circ}=\sqrt{3}
$$ | To prove that in order to construct $\operatorname{ctg} 10^{\circ}$ and $\sqrt{3}$, we can construct Rt $\triangle A C B$ and Rt $\triangle A D E$ such that $\angle B A C = 30^{\circ}, B C = E D = 1, \angle E A D = 10^{\circ}$. Then $A C = \sqrt{3}, A B = 2, A D = \operatorname{ctg} 10^{\circ}$. If we can prove that $C... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 708,028 |
Prove that $\cos \frac{\pi}{7}-\cos \frac{2 \pi}{7}+\cos \frac{3 \pi}{7}=$ $\frac{1}{2} \cdot($ 5th IMO $)$ | To prove that because the directed line segment $\overrightarrow{A B}$ has a projection angle on axis $l$, the sum of the projections of a closed curve on one axis equals zero, and $\frac{2 \pi}{7}$ is exactly one exterior angle of a regular heptagon, we can construct the following vector model. Construct a regular hep... | \frac{1}{2} | Algebra | proof | Yes | Yes | cn_contest | false | 708,029 |
1. In a regular $n$-sided pyramid, the range of the dihedral angle $\alpha$ formed by two adjacent lateral faces is ( ).
(A) $\left(\frac{n-2}{n} \pi, \pi\right)$
(B) $\left(\frac{n-1}{n} \pi, \pi\right)$
(C) $\left(0, \frac{\pi}{2}\right)$
(D) $\left(\frac{n-2}{n} \pi, \frac{n-1}{n} \pi\right)$ | Solving for the apex of a regular $n$-sided pyramid lies on a line passing through the center of the base and perpendicular to the base (as shown in the figure), hence there are two extreme positions:
The apex tends to the infinitely far point: $\alpha \rightarrow \alpha_{1}=\frac{n-2}{n} \pi$ (the base angle of a reg... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 708,030 |
Example 2. Given a directed line segment $P Q$ with the coordinates of the starting point $P$ and the end point $Q$ being $(-1,1)$ and $(2,2)$ respectively. If the line $l_{1} x+m y+m=0$ intersects the extension of $P Q$, then the range of values for $m$ is $\qquad$ | Let $k_{l}$ denote the solution for the line $l$, then
$$
k_{i}=-\frac{1}{m} \text {. }
$$
One of the extreme positions of this slope is the line $l_{1}$ passing through $Q$. Substituting the coordinates of $Q$ $(2,2)$ into the equation of $l_{1}$, we get
$$
2+2 m_{i}+m_{1}=
$$
0,
$$
\begin{aligned}
\therefore k_{l_{1... | -3<m<-\frac{2}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 708,031 |
Example 3. Given $0<b<1,0<a<\frac{\pi}{4}$. Then the size relationship of the following three numbers $x=(\sin \alpha)^{\log _{6} \sin }, y=(\cos \alpha)^{\log _{b} \cos }, z=(\sin \alpha)^{\log _{b} \cos \alpha}$ is ( ).
(A) $x<z<y$
(B) $y<x<x$
(C) $z<x<y$
(D) $x<y<z$ | Solve by taking the extreme first. Let $a \rightarrow 0$, then $y \rightarrow 1, z \rightarrow 0$. That is, $y$ is greater than $z$, so we can rule out (B) and (D).
$$
\text { Also, } \frac{x}{z}=(\sin \alpha)^{\log _{6} \sin \alpha-\log _{b} \cos \alpha}=(\sin \alpha)^{\log _{b} 48} \text {. }
$$
Now, take a special ... | A | Inequalities | MCQ | Yes | Yes | cn_contest | false | 708,032 |
Example 3. Given positive integers $a, b, c$ satisfying $a>b>c$, and $\quad\left\{\begin{array}{l}36-6(a+b+c)+(a b+b c+c a)=0, \\ 81-9(a+b+c)+(a b+b c+c a)=0 .\end{array}\right.$ Find the values of $a, b, c$. | The problem statement indicates that 6 and 9 are the roots of the quadratic equation in $x$:
$$x^{2}-(a+b+c) x+(a b+b c+c a)=0.$$
By Vieta's formulas, we get
$$
\left\{\begin{array}{l}
a+b+c=6+9=15 . \\
a b+b c+i a=6 \times 9=54 .
\end{array}\right.
$$
Given $a>b>1$, we have $1.5=a+b+c>3 c$, which implies $c<5$.
$\th... | a=10, b=4, c=1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 708,033 |
gode 4. In the Cartesian coordinate system, the curve represented by
$$
\frac{|x+y|}{2 a}+\frac{|x-y|}{2 b}=1
$$
$(a, b$ are two distinct positive numbers) is ( ).
(A) Triangle
(B) Square
(C) Non-square Rectangle
(D) Non-square Rhombus | If we remove the absolute value in equation (1), we need to consider four cases, which is rather complicated. By using symmetry, we can avoid the hassle of classification.
$\because$ In (1), interchanging $x$ and $y$ does not change the equation,
$\therefore$ it is symmetric about the line $y=2$;
$\therefore$ it is sym... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 708,034 |
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