problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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3. Let $a, b, c$ be distinct natural numbers, and $a b^{2} c^{3}=$ 675. Then the maximum value of $a+b+c$ is ( ).
(A) 50
(B) 79
(C) 76
(D) 78 | 3. B.
$$
\text { Sol } \begin{array}{l}
a b^{2} c^{3}=675=3^{3} 5^{2} \\
\quad=5^{2} \cdot 1^{2} \cdot 3^{3}=3^{3} \cdot 5^{2} \cdot 1^{3} \\
\quad=\left(5^{2} \cdot 3\right) \cdot 3^{2} \cdot 1^{3}=3 \cdot(3 \cdot 5)^{2} \cdot 1^{3} .
\end{array}
$$
$$
\text { Therefore } a+b+c=75+3+1=79 \text {. }
$$ | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 706,572 |
4. If $1+x+x^{2}+x^{3}+x^{4}=0$, then $1+x+x^{2}+x^{3}$ $+\cdots+x^{1993}+x^{1994}$ equals ( ).
(A) 0
(B) 1
(C) $x$
(D) 1994 | 4. A.
$$
\begin{array}{l}
\text { Solve } 1+x+x^{2}+x^{3}+x^{4}=0 . \\
1+x+x^{2}+x^{3}+\cdots+x^{1993}+x^{1991} \\
=\left(1+x+x^{2}+x^{3}+x^{4}\right) \\
+\left(x^{5}+x^{6}+x^{7}+x^{8}+x^{9}\right) \\
+\cdots+\left(x^{1990}+x^{1991}\right. \\
\left.+x^{1992}+x^{1993}+x^{1994}\right) \\
=\left(1+x+x^{2}+x^{3}+x^{4}\righ... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 706,573 |
5. If $\sqrt{3}$ is between $\frac{x+3}{x}$ and $\frac{x+4}{x+1}$, then the positive integer $x$ is ( ).
(A) 3
(B) 4
(C) 5
(D) 6 | 5. B.
Solution: From the given, we have $\frac{x+3}{x}<\sqrt{3}<\frac{x+4}{x+1}$ or $\frac{x+4}{x+1}<\sqrt{3}<\frac{x+3}{x}$.
(1) No solution. From (2), we solve to get
$$
\frac{4-\sqrt{3}}{\sqrt{3}-1}<x<\frac{3}{\sqrt{3}-1}. [x]=4.
$$ | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 706,574 |
Column 6. As shown in Figure 6, in $\triangle ABC$, $P$ is any point on side $BC$, $PE \parallel BA$, $PF \parallel CA$. If $S_{\triangle ABC}=1$, prove: $S_{\triangle BPF}$, $S_{\triangle PCE}$, and $S_{\triangle PEA}$ are at least one not less than $\frac{4}{9}$.
(1984, National High School League) | Proof: Let $\frac{B P}{P C}=m$, then $\frac{A F}{F B}=\frac{C E}{E A}=\frac{1}{m}$. According to the corollary, we can get
$$
\begin{aligned}
S_{\triangle B P Y} & =\frac{m}{(m+1)\left(\frac{1}{m}+1\right)} \cdot S_{\triangle A B C}=\frac{m^{2}}{(m+1)^{2}} . \\
S_{\triangle P C E} & =\frac{1}{(m+1)^{2}} . \\
S_{\square... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,575 |
6. As shown in the figure, a square is divided into 30 rectangles (not necessarily of the same shape) by 4 lines parallel to one pair of opposite sides and 5 lines parallel to the other pair of opposite sides, and the sum of the perimeters of these small rectangles is 30.
Then the area of the original square is ( ).
(A... | 6. C.
Solve $4 x+2(4+6) x=30$. ( $x$ is the side length of the original square) $x=\frac{15}{12}, x^{2}=\frac{225}{144}$. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,576 |
8. If the roots of the equation $4 x^{2}-2 m x+n=0$ with respect to $x$ are both greater than 0 and less than 1, then the values of the natural numbers $m, n$ are ( ).
(A) $m=1, \bar{n}=1$
(B) $m=1, n=2$
(C) $m=2, n=1$
(D) $m=2, n=2$ | 8. C.
Solution Let the two roots of the original equation be $x_{1}, x_{2}$. Then
$$
\begin{array}{l}
x_{1}+x_{2}=\frac{m}{2}, x_{1} x_{2}=\frac{n}{4} . \\
\left(x_{1}-1\right)\left(x_{2}-1\right) \\
=x_{1} x_{2}-\left(x_{1}+x_{2}\right)+1 \\
=\frac{n}{4}-\frac{m}{2}+1 . \\
\text { By }\left\{\begin{array} { l }
{ n ... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 706,578 |
1. If $a \cdot b \neq 1$, and $3 a^{3}+123456789 a+2=0$, $2 b^{2}+123456789 b+3=0$, then $\frac{a}{b}=$ $\qquad$ . | 二、1. $\frac{a}{b}=\frac{2}{3}$.
Solution Clearly, $a$ and $\frac{1}{b}$ are the two roots of the equation $3 x^{2}+123456789 x+2$ $=0$. By the relationship between roots and coefficients, we can get $a \cdot \frac{1}{b}=\frac{a}{b}=$ $\frac{2}{3}$. | \frac{2}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,579 |
2. $p, q$ are natural numbers, and there are $11 q-1 \geqslant 15 \psi, 10 p \geqslant 7 q$ $+1, \frac{7}{10}<\frac{q}{p}<\frac{11}{15}$. Then the smallest $q=$ $\qquad$ . | 2. $q=7$.
Solve $\left.\begin{array}{l}11 q-1 \geqslant 15 p, \\ 10 p \geqslant 7 q+1\end{array}\right\} \Rightarrow 11 q-1 \geqslant \frac{3}{2}(7 q+1) \Rightarrow q$
$\geqslant 5$. When $q=5,6$, check and find it does not meet the requirements, $q=7$ when, $p=5$. | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 706,580 |
3. In $\triangle A B C$,
$$
\angle A=100^{\circ} ; \angle B=50^{\circ} \text {. }
$$
$A H$ is the altitude from $A$ to side $B C$, and $B M$ is the median from $B$ to side $A C$.
Then $\angle M H C=$ | $\begin{array}{l}\text { 3. } \angle M H C=30^{\circ} \text {. } \\ \left.\text { Solution } \begin{array}{l}M \text { is the midpoint of } A C \text {, } \\ \angle A H C=90^{\circ}\end{array}\right\} \Rightarrow \\ A M=M C=M H \text {. } \\ \Rightarrow \angle M A H=\angle A H M=60^{\circ} \text {. } \\ \text { And } \... | 30^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,581 |
4. The vertices of an inscribed triangle in a circle divide the circumference into three arcs of lengths $3, 4, 5$. Then the area of this triangle is $\qquad$ . | $$
\text { 4. } S=\frac{9}{\pi^{2}}(\sqrt{3}+3) \text {. }
$$
Divide into three arcs of $90^{\circ}, 120^{\circ}$, and $150^{\circ}$.
$$
\begin{array}{l}
S_{\triangle A B C}=\frac{1}{2} r^{2}\left(\sin 90^{\circ}+\sin 120^{\circ}+\sin 150^{\circ}\right) \\
=\frac{1}{4} r^{2}(\sqrt{3}+3) \text {, } \\
2 \pi r=3+4+5=12,... | \frac{9}{\pi^{2}}(\sqrt{3}+3) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,582 |
II. (20 points) Let $P$ be any point inside the equilateral $\triangle ABC$. Construct the symmetric points $P_{1}, P_{2}, P_{3}$ of $P$ with respect to the three sides, and connect $P_{1} P_{2}, P_{2} P_{3}, P_{3} P_{1}$. Prove that $S_{\triangle P_{1} P_{2} P_{3}} \leqslant S_{\triangle A B C}$. | $$
\begin{array}{l}
P P_{1}=2 x, P P_{2}=2 y, \\
P P_{3}=2 z . \angle P_{1} P P_{2}= \\
\angle P_{2} P P_{3}=\angle P_{3} P P_{1} \\
=120^{\circ}. \\
S_{\triangle P_{1} P_{2} P_{3}} \\
=\frac{1}{2} \sin 120^{\circ}(4 x y \\
+4 y z+4 x z) \\
=\sqrt{3}(x y+y z+z x). \\
x^{2}+y^{2} \geqslant 2 x y, y^{2}+z^{2} \geqslant 2... | S_{\triangle P_{1} P_{2} P_{3}} \leqslant S_{\triangle A B C} | Geometry | proof | Yes | Yes | cn_contest | false | 706,584 |
Three, (20 points) If the three sides of a right-angled triangle are all integers, then, one of the legs of the right-angled triangle must be a multiple of 3. | Three, Solution Let the three sides of a right-angled triangle be $a, b, c$, and $a^{2}+b^{2}=c^{2}$.
If $a, b, c$ are not multiples of 3, then $a, b, c$ can be expressed as $3 k \pm 1$. Thus,
$$
(3 k \pm 1)^{2}=9 k^{2} \pm 6 k+1 .
$$
Therefore, $a^{2}+b^{2} \equiv 2(\bmod 3)$,
$$
c^{2} \equiv 1(\bmod 3) .
$$
This c... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 706,585 |
Example 7. On the sides of $\triangle ABC$, take points $D, E, F$ such that $BD \leqslant DC, CE \leqslant EA, AF \leqslant FB$ (Figure 1). If $S$ represents area, prove that $S_{\triangle DEF} \geqslant \frac{1}{4} S_{\triangle ABC}$.
(1973, 34th Putnam Competition) | Proof: Let $\frac{A F}{F B}=m, \frac{B D}{D C}=n, \frac{C E}{E A}=l$, then from the given conditions, we have $0<m \leqslant 1,0<n \leqslant 1,0<l \leqslant 1$.
By the theorem, we have
$$
\begin{array}{l}
S_{\triangle D E F}=\frac{m n l+1}{(m+1)(n+1)(l+1)} \cdot S_{\triangle A B C} . \\
\because \quad 0<m, n, l \leqsla... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,586 |
Example 8. As shown in Figure 1, let the area $S$ of $\triangle ABC$ be 1. Try to find an internal point $D, E, F$ on each of the sides $BC, CA, AB$ respectively, such that the area $S'$ of $\triangle DEF$ satisfies $\frac{1}{4}<S'<\frac{1}{3}$. (1988, Jiangsu Junior High School Competition) | Let points $D, E, F$ be such that
$$
\frac{A F}{F B}=\frac{B D}{D C}=\frac{C E}{E A}=m .
$$
Then, by the theorem and the assumption $S=1$, we get
$$
S^{\prime}=\frac{m^{3}+1}{(m+1)^{3}}=\frac{m^{2}-m+1}{(m+1)^{2}} .
$$
From $\frac{1}{4}<\frac{m^{2}-m+1}{(m+1)^{2}}<\frac{1}{3}$, we solve to get
$$
\frac{1}{2}<m<2 \tex... | \frac{1}{2}<m<2 \text { and } m \neq 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,587 |
Example 9. In a convex hexagon $A B C D E F$, $A B \parallel E D, B C \parallel F E, C D \parallel A F$. Try to prove $S_{\triangle A C E}=S_{\triangle B D F}$. (1964, Hungarian Mathematical Olympiad) | Prove that extending line segments $A B, C D, E F$ to intersect at both ends: forming a $\triangle X Y Z$ (Figure 7).
Let $\frac{X A}{A Y}=m, \frac{Y C}{C Z}=n$,
$$
\begin{array}{l}
\frac{Z E}{E X}=t, \text { SU }^{\frac{Y}{F} \bar{Z}}=\frac{X A}{A Y}=\cdots \\
\frac{Y B}{B X}=\frac{Y C}{C Z}=n, \frac{Z U}{D Y}=\frac{Z... | S_{\triangle A C E}=S_{\triangle B D F} | Geometry | proof | Yes | Yes | cn_contest | false | 706,588 |
1. Given triangle $\triangle A B C$, the circle with $A B$ as diameter intersects the altitude $C C^{\prime}$ and its extension at $M: N$. Show that $A C$ is the diameter of the circle passing through $N, P, Q$.
（19th United States of America Mathematical Olympiad） | Analysis: Let $P Q$ and $M N$ intersect at point $K$, and connect $A P, A M$. To prove that $M, N, P, Q$ are concyclic, we need to prove that
$$
M K \cdot K N = P K \cdot K Q,
$$
which is equivalent to proving
$$
\left(M C^{\prime}-K C^{\prime}\right)\left(M C^{\prime}+K C^{\prime}\right) = \left(P B^{\prime}-K B^{\pri... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,590 |
2. $A, B, C$ are collinear points, point $O$ is outside the line, $O_{1}, O_{2}$, $O_{3}$ are the circumcenters of $\triangle O A B$, $\triangle O B C, \triangle O C A$ respectively. Prove that $O, O_{1}, O_{2}, O_{3}$ are concyclic. | Analysis: Draw all the auxiliary lines in the figure. It is easy to prove that $O_{1} O_{2}$ is the perpendicular bisector of $O B$, and $O_{1} O_{3}$ is the perpendicular bisector of $O A$. Observing $\triangle O B C$ and its circumcircle, we immediately get $\angle O O_{2} O_{1}=\frac{1}{2} \angle O O_{2} B=\angle O ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,591 |
Example 3. Find the integer solutions of $36 x+83 y=1$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | $$
\begin{array}{l}
\text{Since } (36,83)=1, \text{ by Theorem 1, there must be two integers } x, y \text{ that are solutions to the equation.} \\
\because 83=2 \times 36+11, 36=3 \times 11+3, \\
11=3 \times 3+2, \quad 3=2+1, \\
\therefore 1=3-2=3-(11-3 \times 3)=4 \times 3-11 \\
=4 \times(36-3 \times 11)-11 \\
=4 \tim... | x=30, y=-13 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 706,592 |
One, (20 points) Given $\triangle A B C, \angle A=60^{\circ}$, a circle with radius $R$ is tangent to $B C$ at $D$, to the extension of $A B$ at $E$, and to the extension of $A C$ at $F$ (this circle is the excircle of $B C$). Let $B C = a$. Prove: $R \leqslant \frac{2}{3} a$.
| One, as shown in the figure, construct the incircle $O_{1}$ of $\triangle A B C$, which touches $B C$ at $H$ and $A C$ at $G$.
Let $A B=c, A C=b$. Connect $O O_{1}, O_{1} G, O F$. Draw $O_{1} M \parallel F G$ intersecting $O F$ at $M$. Then $O_{1} M=F G, \angle O O_{1} M=\frac{A}{2}=30^{\circ}$.
By the tangent segmen... | R \leqslant \frac{2}{3} a | Geometry | proof | Yes | Yes | cn_contest | false | 706,594 |
$\because 、\left(20\right.$ points) For a quadratic equation with real coefficients $a x^{2}+2 b x$ $+ c=0$ having two real roots $x_{1}, x_{2}$. Let $d=\left|x_{1}-x_{2}\right|$. Find the range of $d$ when $a>b>c$ and $a+b+c=0$. | By Vieta's formulas, $x_{1}+x_{2}=-\frac{2 b}{a}, x_{1} x_{2}=\frac{c}{a}$.
$$
\begin{aligned}
d^{2} & =\left(x_{1}-x_{2}\right)^{2}=\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2} \\
& =\frac{4 b^{2}-4 a c}{a^{2}}=\frac{4 b^{2}+4 a(a+b)}{a^{2}} \\
& =4\left[\left(\frac{b}{a}\right)^{2}+\frac{b}{a}+1\right] .
\end{aligned}
... | \sqrt{3}<d<2 \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,595 |
Three, (20 points) Given that $p$ and $q$ are positive odd numbers, and $a, b, c, d$ are positive integers. Does there exist real numbers $x_{1}, x_{3}, y_{1}, y_{2}$, satisfying
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=p, \\
y_{1}+y_{2}=q, \\
x_{1}^{2}+y_{1}^{2}=a^{2}, \\
x_{1}^{2}+y_{2}^{2}=b^{2}, \\
x_{2}^{2}+y_{2}^{2... | Three, from (3)-(4), (4)-(5) we get
$$
y_{1}^{2}-y_{2}^{2}=a^{2}-b^{2}, x_{1}^{2}-x_{2}^{2}=b^{2}-c^{2}.
$$
Therefore, $\left(y_{1}-y_{2}\right) q=a^{2}-b^{2}, \left(x_{1}-x_{2}\right) p=b^{2}-c^{2}$.
$$
\begin{array}{l}
y_{1} p q - y_{2} p q = \left(a^{2}-b^{2}\right) p, \\
x_{1} p q - x_{2} p q = \left(b^{2}-c^{2}\r... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,596 |
1. Let $f_{1}(x)=\frac{x}{\sqrt{1+x^{2}}}$, for any natural number $n$, define $f_{x+1}(x)=f_{1}\left(f_{n}(x)\right)$. Then the analytical expression for $f_{1993}(x)$ is ).
(A) $\frac{1993 x}{\sqrt{1+x^{2}}}$
(B) $\frac{x}{\sqrt{1993+x^{2}}}$
(C) $\frac{x}{\sqrt{1+1993 x^{2}}}$
(D) $\frac{1993 x}{\sqrt{1+1993 x^{2}}}... | ,$- 1 . C$.
By induction, we can see that $f_{n}(x)=\frac{x}{\sqrt{1+n x^{2}}}$. When $n=1$, $f_{1}(x)=\frac{x}{\sqrt{1+x^{2}}}$. Assuming that when $n=k$, $f_{k}(x)=\frac{x}{\sqrt{1+k x^{2}}}$. Then, when $n=k+1$,
$$
\begin{aligned}
f_{k+1}(x) & =\frac{f_{k}(x)}{\sqrt{1+f_{k}^{2}(x)}}=\frac{\frac{x}{\sqrt{1+k x^{2}}}}... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 706,597 |
2. If $2^{x}=3^{y}=5^{z}>1$, then the order of $2 x, 3 y, 5 z$ from smallest to largest is ( ).
(A) $3 y<2 x<5 z$
(B) $5 z<2 x<3 y$
(C) $2 x<3 y<5 z$
(D) $5 z<3 y<\hat{i} x$ | 2. A.
Let $2^{x}=3^{y}=5^{z}=k$, then $k>1$, and
$$
2 x=\frac{2}{\lg 2} \lg k, 3 y=\frac{3}{\lg 3} \lg k, 5 z=\frac{5}{\lg 5} \lg k \text {. }
$$
Since $2^{3}1, \lg k>0$, therefore, $3 y<2 x<5 z$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 706,598 |
4. A frustum of a cone has an upper base radius of $1 \mathrm{~cm}$, a lower base radius of $2 \mathrm{~cm}$, and a generatrix $A B$ of $4 \mathrm{~cm}$. A certain ant starts from point $A$ on the circumference of the lower base, and crawls around the lateral surface of the frustum (i.e., it must intersect every genera... | 4. A.
Let the radii of the upper and lower bases of the frustum be $r^{\prime}, r$, and the central angle of the sector ring in the lateral development of the frustum be $\theta$. It is easy to get
$$
\begin{aligned}
\theta & =\frac{r-r^{\prime}}{A B} \cdot 2 \pi \\
& =\frac{2-1}{4} \cdot 2 \pi \\
& =\frac{\pi}{2} \\
... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,600 |
5. If the complex conjugate of the complex number $z$ is $\bar{z}$, and $|z|=1$, and $A$ $=(-1,0)$ and $B=(0,-1)$ are fixed points, then the function $f(z)$ $=|(z+1)(\bar{z}-i)|$ takes its maximum value when the figure formed by points $Z, A, B$ on the complex plane is ( ).
(A) Isosceles triangle
(B) Right triangle
(C)... | 5.C.
Let $z=\cos \theta+i \sin \theta$, then
$$
\begin{aligned}
f(z) & =|(z+1)(\bar{z}-i)| \\
& =|(\cos \theta+1+i \sin \theta)[\cos \theta-i(\sin \theta+1)]| \\
& =|1+\sin \theta+\cos \theta+i(1+\sin \theta+\cos \theta)| \\
& =\sqrt{2\left[1+\sqrt{2} \sin \left(\theta+\frac{\pi}{4}\right)\right]^{2}} .
\end{aligned}
... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,601 |
4. Find the integer solutions of $11 x+15 y=7$ | Solution 1: Transform the equation, we get
$$
x=\frac{7-15 y}{11} \text{. }
$$
Since $x$ is an integer, $7-15y$ must be a multiple of 11. By observation, we find that $x_{0}=2, y_{0}=-1$ is a set of integer solutions to this equation.
Therefore, by Theorem 3, the integer solutions to the equation $11 x+15 y=7$ are $\... | \left\{\begin{array}{l}x=2-15 t, \\ y=-1+11 t .\end{array}\right.} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 706,603 |
1. The points $(x, y)$ that satisfy the inequality $\log _{x} y \geqslant \log _{\frac{x}{y}}(x y)$ | $$
\text { II. 1. }\{(x, y) \mid x>1
$$
and $y>x\} \cup\{(x, y) \mid 0<x<1$ and $y<x\} \\
\begin{aligned}
& \Leftrightarrow \{(x, y) \mid x>1 \text { and } y>x\} \cup\{(x, y) \mid 0<x<1 \text { and } y<x\} \\
& \Leftrightarrow >1 .
\end{aligned}
$$
Therefore, the required set is $\{(x, y) \mid x>1$ and $y>x\} \cup$ $... | \{(x, y) \mid x>1 \text { and } y>x\} \cup \{(x, y) \mid 0<x<1 \text { and } y<x\} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 706,604 |
*2. A sphere and a cylinder are on the same plane, and they have a common inscribed sphere. Let the volume of the sphere be $V_{1}$, the volume of the cylinder be $V_{2}$, and $V_{1}=k V_{3}$. Then the minimum value of $k$ is | 2. $\frac{4}{3}$.
Take their common axial section, as shown in the figure, then $r=r_{2}$, $r=r_{1} \tan \theta, h=r_{1} \tan 2 \theta$.
If $V_{1}=k V_{2}$, then we have
$$
\begin{array}{l}
\frac{1}{3} \pi r_{1}^{2} \cdot r_{1} \tan 2 \theta \\
=2 \pi r_{1}^{3} \cdot k \tan^{3} \theta .
\end{array}
$$
Simplifying, we... | \frac{4}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,605 |
4. As shown in the figure, given the ellipse $\frac{x^{2}}{2} + y^{2}=1, P A \perp A B, C B \perp$ $A B$, and $D A=3 \sqrt{2}, C B$ $=\sqrt{2}$. Point $P$ moves on the arc $\overparen{A B}$. Then the minimum value of the area of $\triangle P C D$ is $\qquad$ | 4. $4-\sqrt{6}$.
The equation of the line $CD$ is: $x+y-2 \sqrt{2}=0$.
Let the parametric coordinates of the moving point $P$ on the ellipse be $P(\sqrt{2} \cos \theta, \sin \theta), \theta \in [0,2 \pi)$.
The distance from $P$ to $CD$ is
$$
\begin{aligned}
d & =\frac{|\sqrt{2} \cos \theta+\sin \theta-2 \sqrt{2}|}{\sq... | 4-\sqrt{6} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,607 |
* 5. For the quartic equation $x^{4}-20 x^{3} + k x^{2}-400 x+384=0$, the product of two of its four roots is 24, then the value of $k$ is $\qquad$. | 5. 140 .
Let the four roots of the equation be $x_{1}, x_{2}, x_{3}, x_{4}$, then by Vieta's formulas, we have
$$
\left\{\begin{array}{l}
x_{1}+x_{2}+x_{3}+x_{4}=20, \\
x_{1} x_{2}+x_{1} x_{3}+x_{1} x_{4}+x_{2} x_{3}+x_{2} x_{4}+x_{3} x_{4}=k \text { (2) } \\
x_{1} x_{2} x_{3}+x_{1} x_{2} x_{4}+x_{1} x_{3} x_{4}+x_{2}... | 140 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,608 |
* 6. The sum of four positive numbers is 4, and the sum of their squares is 8. Then, the maximum value of the largest number among these four numbers is | $6.1+\sqrt{3}$.
Let $a \geqslant b \geqslant c \geqslant d>0$.
Given $a+b+c+d=4, a^{2}+b^{2}+c^{2}+d^{2}=8$.
Then
$$
\begin{array}{l}
b+c+d=4-a, \\
b^{2}+c^{2}+d^{2}=8-a^{2} .
\end{array}
$$
By the Cauchy-Schwarz inequality
$$
3\left(b^{2}+c^{2}+d^{2}\right) \geqslant(b+c+d)^{2},
$$
i.e., $3\left(8-a^{2}\right) \geqs... | 1+\sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,609 |
$*$ Three. (20 points) $a_{1}, a_{2}, \cdots, a_{n}$ are distinct natural numbers. Prove:
$$
\begin{array}{l}
\left(a_{1}{ }^{7}+a_{2}{ }^{7}+\cdots+a_{n}{ }^{7}\right)+\left(a_{1}{ }^{5}+a_{2}{ }^{5}+\cdots+a_{n}^{5}\right) \\
\geqslant 2\left(a_{1}{ }^{3}+a_{2}{ }^{3}+\cdots+a_{n}{ }^{3}\right)^{2} .
\end{array}
$$ | For $n$ using induction.
When $n=1$, the left side $=a_{1}^{7}+a_{1}^{5} \geqslant 2 \sqrt{a_{1}^{5} \cdot a_{1}^{7}}$ $=2 \cdot\left(a_{1}^{3}\right)^{2}=$ the right side.
Assume the inequality holds for $n=k$.
Then for $n=k+1$, without loss of generality, assume
$$
\begin{array}{l}
a_{1}<a_{2}<\cdots<a_{k}<a_{k+1}, \... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 706,610 |
Four. (20 points) Let $P, M$ be points on the sides $BC, CD$ of square $ABCD$, respectively. $PM$ is tangent to the circle with radius $AB$. Segments $PA$ and $MA$ intersect the diagonal $BD$ at $Q$ and $N$, respectively. Prove that the pentagon $PQNMC$ is cyclic. | As shown in the figure, since $P A$ bisects $\angle T A D$ and $M A$ bisects $\angle T A B$, we have $\angle M A P = 45^{\circ} = \angle Q D P$. Therefore, $\angle D P Q = \angle Q N A = \angle Q N C$. Hence, points $P, Q, N, C$ are concyclic. Similarly, points $M, N, Q, C$ are concyclic. Therefore, points $P, Q, N, M,... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,611 |
*Five, (20 points) 100 matchboxes, numbered 1 to 100. We can ask whether the total number of matches in any 15 boxes is odd or even. What is the minimum number of questions needed to determine the parity (odd or even) of the number of matches in box 1? | Let $a_{i}$ represent the number of matches in the $i$-th box.
The first time, boxes 1 to 15 are taken, so the parity of $\sum_{k=1}^{15} a_{k}$ is known;
The second time, boxes 2 to 8 and 16 to 23 are taken, so the parity of $\sum_{k=2}^{8} a_{k}+\sum_{k=16}^{23} a_{k}$ is known;
The third time, boxes 9 to 23 are ta... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 706,612 |
一、(35 Points)
In the right figure, $\triangle A B C$, $\triangle B C D$, $\triangle C D E$ are all equilateral triangles, and line segment $F G \parallel B A$. Connect $D G$ and $E F$ to intersect at $O$, and connect $C O$ and extend it to intersect the extension of $A B$ at $P$. Prove that $C P = E F$. | $$
\left.\begin{array}{ll}
CF = CG, \\
CE = CD, \\
\angle ECF = \angle GCD = 120^{\circ}
\end{array}\right\} \Rightarrow
\triangle FCE \cong \triangle GCD \Rightarrow \angle CEF = \angle CDG \Rightarrow O, C, E,
D \text{ are concyclic} \Rightarrow \angle OCD = \angle OED \Rightarrow \angle BCP = 60^{\circ} -
\angle OC... | CP = EF | Geometry | proof | Yes | Yes | cn_contest | false | 706,613 |
Example 5. Find all integer solutions to the equation $\frac{x+y}{x^{2}-x y+y^{2}}=\frac{3}{7}$. (1986, 12th All-Russian Mathematics Competition) | The original equation can be transformed into
$$
3 x^{2}-(3 y+7) x+3 y^{2}-7 y=0 \text {. }
$$
Since $x$ is an integer, the original equation should have real roots. Therefore,
$$
\Delta=(3 y+7)^{2}-4 \times 3 \times\left(3 y^{2}-7 y\right) \geqslant 0 .
$$
Solving this, we get $\frac{21-14 \sqrt{3}}{9} \leqslant y \... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,614 |
*Two. (35 points) Suppose $a_{1}$, $a_{2}, \cdots, a_{10}$ and $b_{1}, b_{2}, \cdots, b_{10}$
are sequences composed of unequal complex numbers. It is known that for $i=1,2$, $\cdots, 10$,
$$
\left(a_{1}+b_{i}\right)\left(a_{2}+b_{i}\right) \cdots\left(a_{10}+b_{i}\right)=100 .
$$
Prove that for any $j=1,2, \cdots, 10... | Let
$$
F(x)=\left(a_{1}+x\right)\left(a_{2}+x\right) \cdots\left(a_{10}+x\right)-100 .
$$
Then, by the given conditions, $b_{1}, b_{2}, \cdots, b_{10}$ are the 10 distinct complex roots of the polynomial $F(x)$. Since $F(x)$ is a 10th-degree polynomial in $x$ with the coefficient of $x^{10}$ being 1, it follows from t... | -100 | Algebra | proof | Yes | Yes | cn_contest | false | 706,615 |
* Three, (35 points) Prove that among any 28 different positive integers between 104 and 208 (inclusive), there must be two numbers that are not coprime (i.e., their greatest common divisor is greater than 1). | Let $E_{2}$ denote the set of all multiples of 2 between 104 and 208, and similarly define $E_{3}, E_{5}, E_{7}$. It is easy to calculate:
$$
\begin{array}{l}
\left|E_{2}\right|=\left[\frac{208}{2}\right]-\left[\frac{103}{2}\right] \\
=104-51=53, \\
\left|E_{3}\right|=\left[\frac{208}{3}\right]-\left[\frac{103}{3}\righ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 706,616 |
On 6. Person A and Person B simultaneously solve the radical equation $\sqrt{x+a}+$ $\sqrt{x+b}=7$. When copying, A mistakenly copies it as $\sqrt{x-a}+\sqrt{x+b}=7$, and ends up solving one of its roots as 12. B mistakenly copies it as $\sqrt{x+a}+\sqrt{x+d}$ $=7$, and ends up solving one of its roots as 13. It is kno... | Given that one root of $\sqrt{x-a}+\sqrt{x+b}=7$ is 12, we have
$$
\sqrt{12-a}+\sqrt{12+b}=7.
$$
Also, given that one root of $\sqrt{x+a}+\sqrt{x+d}=7$ is 13, we have
$$
\sqrt{13+a}+\sqrt{13+d}=7.
$$
According to the definition of arithmetic roots, we have $12-a \geqslant 0, 13+a \geqslant 0$.
Thus, $0 \leqslant 12-a... | a=12, b=37; a=3, b=4; a=-4, b=-3; a=-13, b=-8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,617 |
朋 7. The number of distinct integer solutions to $x^{2}-y^{2}=1983$ is $\qquad$ | Right $1988=( \pm 2) \times( \pm 2) \times 7 \times 71$.
The original equation is equivalent to
$$
(x+y)(x-y)=( \pm 2) \times( \pm 2) \times 7 \times 71 .
$$
The integer solutions of the equation $x^{2}-y^{2}=1988$ should be determined by the following system of equations:
$$
\begin{array}{l}
\left\{\begin{array} { l ... | not found | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 706,618 |
Example 8. There is a rectangular prism with length, width, and height of $m, n, r(m \leqslant n \leqslant r)$, respectively. After its surface is painted red, it is cut into cubes with an edge length of 1. It is known that the sum of the number of cubes without any red faces and the number of cubes with two red faces,... | According to the problem, the number of unit cubes without red faces is $(m-2)(n-2)(r-2)$, the number of unit cubes with one red face is $2[(m-2)(n-2)+(m-2)(r-2)+(n-2)(r-2)]$, and the number of unit cubes with two red faces is $4[(m-2)+(n-2)+(r-2)]$. Setting up the equation, we get
$$
\begin{array}{l}
\quad(m-2)(n-2)(r... | (1) m=5, n=7, r=663; (2) m=5, n=5, r=1981; (3) m=3, n=3, r=1981; (4) m=1, n=3, r=1987; (5) m=1, n=7, r=399 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 706,619 |
Example 9. Prove that the equation $6 x+3 y=17$ has no integer solutions. | Proof by contradiction. Transform the equation $6 x+3 y=17$ into
$$
y=\frac{-6 x+17}{3} \text {. }
$$
If the equation has integer solutions, then $y$ must be expressible as the sum of an integer and an integer expression, i.e.,
$$
\begin{aligned}
y & =\frac{-6 x+(15+2)}{3}=\frac{(-6 x+15)+2}{3} \\
& =(-2 x+5)+\frac{2}... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 706,620 |
Example 10. Find the integer solutions of the equation $x^{2}=y^{2}+1990$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 1990 is an even number, and the square of an odd number remains odd, the square of an even number remains even, and $x^{2}=y^{2}+1990$, so $x, y$ have the same parity.
When $x, y$ are both odd, let $x=2 m+1, y=2 n+1(m, n$ be integers).
Since $1990=4 \times 497+2$, the original equation can be transformed into
$$
(2 m+... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 706,621 |
Example 3. In trapezoid $ABCD$, $AB // DC$, $AB > CD$, $K$, $M$ are on $AD$, $BC$ respectively, $\angle DAM = \angle CBK$.
Prove: $\angle DMA = \angle CKB$.
(2nd Zu Chongzhi Cup Junior High School Competition) | Analysis: It is evident that points $A, B, M$, and $K$ are concyclic. Connecting $K M$, we have
$$
\begin{array}{l}
\angle D A B=\angle C M K . \\
\because \angle D A B+\angle A D C \\
=180^{\circ}, \\
\therefore \angle C M K+\angle K D C \\
=180^{\circ} .
\end{array}
$$
Thus, points $C, D, K, M$ are concyclic $\Right... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,624 |
4. A truck is traveling at a constant speed on the highway. Initially, the number on the milestone is $\overline{A B}$. After 1 hour, the number on the milestone is $\overline{B A}$. After another hour, the number on the milestone is $\overline{A 0 B}$. Find the numbers seen each time and the speed of the truck. | $$
\begin{array}{c}
\text { (Hint: }(10 B+A)-(10 A+B)=(100 A+B) \\
-(10 B+A), A=1, B=6 . \overline{A B}=16, \overline{B A}=61, \overline{A 0 B}=
\end{array}
$$
106. Speed is 45 kilometers/hour. | 16, 61, 106, 45 \text{ kilometers/hour} | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 706,626 |
Example 1. Prove: (1) $H(m, n)$ is not an integer;
(2) If $H(m, n)$ is written as a reduced fraction $\frac{q}{p},(p, q)=1$, then $p$ is even, and $q$ is odd ${ }^{[1]}$. | Proof. Obviously, if (2) holds, then (1) holds, so we only need to prove (2).
If $k$ can be divided by $2^{a}$ but not by $2^{a+1}$ $(\alpha \in \mathbb{N})$, then $\alpha$ is called the "parity exponent" of $k$. $2^{*}, 3 \cdot 2^{*}, 5 \cdot 2^{*}, 7 \cdot 2^{\infty}$, $\cdots$ are numbers with a parity exponent of ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 706,628 |
Example 2. If $H(m, n)=\frac{q}{p}(p, q \in N)$, and $m+n$ is an odd prime, then $(m+n) \mid q$.
| Prove that since $m+n$ is an odd prime, $n-m+1$ is even. Therefore, $H(m, n)$ has an even number of terms.
$$
\begin{array}{l}
\text { Let } \frac{1}{m n}+\frac{1}{(m+1)(n-1)}+\cdots \\
\frac{m+n-1}{2} \cdot \frac{1}{2}+\frac{n+1}{2}=\frac{s}{r}, \\
(r, s)=1, r, s \in N \text {. } \\
\end{array}
$$
It is easy to see t... | 1979 \mid q | Number Theory | proof | Yes | Yes | cn_contest | false | 706,629 |
4. Prove:
$$
\frac{2(n-m+1)}{m+n}<H(m, n)<\frac{m+n}{2 m n}(n-m+1) .
$$ | Prove that it is easy to know
\[ H(m, n)=\frac{1}{2} \sum_{k=0}^{n-n}\left(\frac{1}{m+k}+\frac{1}{n-k}\right) \]
\[
\begin{array}{l}
=\frac{m+n}{2} \sum_{k=0}^{n-m} \frac{1}{(m+k)(n-k)} . \\
\text { and } (m+k)(n-k) \leqslant\left(\frac{m+k+n-k}{2}\right)^{2} \\
=\frac{(m+n)^{2}}{4}, \\
k(n-m-k) \geqslant 0 \Rightarrow... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 706,631 |
Example 6. Prove that any positive rational number is the sum of finitely many distinct terms of the harmonic series $1+\frac{1}{2}+$ $\frac{1}{3}+\cdots+\frac{1}{n}+\cdots$ [2]. | Proof: Let $\frac{A}{B} (A, B \in \mathbb{N})$ be a positive rational number. It is easy to find a natural number $m_{0}$ such that $\frac{A}{B} > \frac{1}{m_{0}}$. Since the harmonic series $H(m_{0}, n)$ diverges as $n \rightarrow \infty$, there exists a natural number $n_{0}$ such that $H(m_{0}, n_{0}) > 1$.
By (2),... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 706,633 |
The number $\sum_{j=1}^{\infty} \frac{1}{j(j+1)}$ of terms closely related to 1, and find out how many ways it can be represented? (The American Mathematical Monthly) Vol. 56, No. 872
Express the above text in English, please retain the original text's line breaks and format, and output the translation result directly... | Prove $\because \frac{1}{j(j+1)}=\frac{1}{j}-\frac{1}{j+1}$,
$$
\therefore \sum_{j=0}^{b-1} \frac{1}{j(j+1)}=\frac{1}{a}-\frac{1}{b} \text {. }
$$
Thus, the problem is equivalent to finding positive integers $a, b$, such that for a fixed integer $m>1$ we have
$$
\frac{1}{a}-\frac{1}{b}=\frac{1}{m} .
$$
From (1), the ... | P(m)=\frac{1}{2}\left[d\left(m^{2}\right)-1\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,634 |
㤡 4. $\odot O$ passes through the vertices $A, C$ of $\triangle ABC$, and intersects $AB, BC$ at $K, N$ (where $K$ and $N$ are distinct). The circumcircle of $\triangle ABC$ and the circumcircle of $\triangle BKN$ intersect at $B$ and $M$. Prove that $\angle BMO = 90^{\circ}$. (26th IMO, Problem 5) | Analysis: This international math competition problem has made many contestants hesitate. In fact, as long as the known conditions and characteristics of the figure are utilized, with the help of "four points on a circle," the problem is not difficult to solve.
Connect $O C, O K, M C, M K$, and extend $B M$ to $G$. It... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,635 |
Example 8. From the harmonic series $1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}+\cdots$, remove all terms where the denominator's decimal representation contains the digit 9. The resulting series is necessarily bounded $[1]$. | Prove that in the interval $\left[10^{m}, 10^{m+1}-1\right]$, natural numbers all have $m$ digits, except that the first digit cannot be 0 or 9, while the other digits can be any of the 9 digits $0,1,2$, $\cdots, 8$. Therefore, the number of natural numbers in this interval that do not contain the digit 9 is $8 \cdot 9... | 80 | Number Theory | proof | Yes | Yes | cn_contest | false | 706,636 |
Example 1. Outside $\triangle ABC$, construct three isosceles triangles $BCA'$, $CAB'$, and $ABC'$ with the same base angles, as shown in Figure 1. Prove that $AA'$, $BB'$, and $CC'$ are concurrent.
This is a problem from the 4th (1989) National Training Team Selection Contest for the Mathematical Olympiad, Question 4... | Let $A A^{\prime}$ intersect $B C$ at $D$. Then we have
$$
\begin{array}{l}
\frac{B D}{D C}=\frac{S_{\triangle A B A}}{S_{\triangle A A^{\prime} C}} \\
=\frac{A B \cdot A^{\prime} B \cdot \sin (B+0)}{A C \cdot A^{\prime} C \cdot \sin (C+0)} \\
=\frac{c \cdot \sin (B+0)}{b \sin (C+0)} .
\end{array}
$$
Now, let $B B^{\p... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,637 |
Example 2. On the sides of the acute-angled $\triangle ABC$, construct three similar acute-angled triangles outwardly: $\triangle AC_{1}B, \triangle BA_{1}C$, and $\triangle CB_{1}A$ (simultaneously
$$
\begin{array}{l}
\angle AB_{1}C = \angle ABC_{1} = \\
\angle A_{1}BC, \angle BA_{1}C = \\
\left.\angle BAC = \angle B_... | Let $A A^{\prime}, B B^{\prime}$, and $C C^{\prime}$ intersect $B C, C A$, and $A B$ at $D, E$, and $F$, respectively. Let the internal angles of the three similar triangles be $\alpha$, $\beta$, and $\gamma$, as shown in Figure 2. Then we have
$$
\begin{array}{l}
\frac{B D}{D C}=\frac{S_{\triangle A B A_{1}}}{S_{\tria... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,638 |
Example 3. There does not exist a quadrilateral $P Q R S$ inscribed in $\triangle A B C$, as shown in Figure 3, such that $S_{1}=S_{2}=S_{3}=S_{4}$. Prove this.
This is a problem from [3], the original proof involves setting the midpoints of each side as $A^{\prime}, B^{\prime}$, and $C^{\prime}$, first proving the di... | Assume that $P Q R S$ divides $\triangle A B C$ into four equal parts, then we have
$$
\begin{aligned}
& \frac{S_{1}}{S_{\triangle A B C}} \\
= & \frac{A P \cdot A Q}{A B \cdot A C} \\
= & \frac{1}{4} .
\end{aligned}
$$
That is,
$$
\begin{array}{l}
A P \cdot A Q \\
=\frac{1}{4} b c .
\end{array}
$$
Similarly, $B P \c... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,639 |
Example 5. The side length of square $ABCD$ is $1$, and there are points $P, Q$ on $AB, AD$ respectively. If the perimeter of $\triangle APQ$ is 2, find $\angle PCQ$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | This is a problem from the 1st (1987) Mathematical Winter Camp for middle school students in our country, and in the 4th (1991) "Zu Chongzhi Cup" Mathematical Competition, it was also used as a contest problem ${ }^{[5]}$. The usual proof method involves the use of congruent shapes. In fact, points $P$ and $Q$ in this ... | 45^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,641 |
Example 6. Let the three sides of $\triangle A B C$ be $a, b, c$ with opposite angles $A, B, C$, respectively, and satisfy the condition $a \cos A+b \cos B=c \cos C$. Then $\triangle A B C$ must be ( ).
(A) an equilateral triangle
(B) a right triangle with $a$ as the hypotenuse
(C) a right triangle with $b$ as the hypo... | Answer the multiple-choice question, just get it right, no need to argue. Therefore, it is valuable to be quick and accurate. From the condition
$$
a \cos A + r \cos B = c \cos C
$$
it is easy to see that $A(a), [B(b)]$ and $C(c)$ cannot be interchanged, so we know that conclusion (A) is incorrect.
Since the above eq... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,642 |
Example 7. In the acute-angled $\triangle ABC$, $AC=1$, $AB=c$, and the circumradius $R \leqslant 1$ of $\triangle ABC$. Prove that $\cos A < c \leqslant \cos A + \sqrt{3} \sin A$.
This is a problem from the 1984 Tianjin Junior High School Mathematics Competition. Books [7], [8], and [9] all start by applying the cosi... | Draw $C D \perp A B$ at $D$, then it is easy to know
$$
c=a \cos B+b \cos A \text {. }
$$
And $a=2 R \sin A, b=1$,
$$
\therefore \quad c=2 R \sin A \cos B+\cos A \text {. }
$$
Obviously, $\cos A<c \leqslant 2 \sin A \cos B+\cos A$.
Since the condition $R \leqslant 1$ is a crucial factor for the problem, further explo... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,643 |
Let $D$ be a point inside an acute triangle $\triangle A B C$ such that
$$
\begin{array}{l}
\angle A D B=\angle A C B+90^{\circ}, \\
A C \cdot B D=A D \cdot B C .
\end{array}
$$
(1) Calculate the ratio $\frac{A B \cdot C D}{A C \cdot B D}$. The key related to the perpendicular tangents at point $C$ is to find the relat... | Solve (1) Construct
the circumcircle of $\triangle A B C$, intersecting the extensions of $A D$ and $B D$ at $E$ and $F$ respectively, and connect $B E$, $C E$, $A F$, and $C F$, as shown in the figure.
$$
\begin{aligned}
& \because \angle A D B \\
= & \angle A C B+90^{\circ}, \\
\text { and } & \angle A D B \\
= & \an... | \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,644 |
Title: A horizontal line $m$ passes through the center of $\odot O$, line $l \perp m$, $l$ intersects $m$ at $M$, and point $M$ is to the right of the center. Line $l$ has three distinct points $A, B, C$ outside the diagram, all located above line $m$, with point $A$ being the farthest from $M$ and point $C$ being the ... | Proof 1 transforms the expression to be proved into determining the sign of $I$. By taking the difference,
$$
\begin{aligned}
I= & A B \cdot C R + B C \cdot A P - A C \cdot B Q \\
= & (a-b) \sqrt{c^{2}+d} + (b-c) \sqrt{a^{2}+d} - (a-c) \\
& \cdot \sqrt{b^{2}+d} \\
= & (b-c)\left(\sqrt{a^{2}+d}-\sqrt{b^{2}+d}\right) - (... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,645 |
Example 5. Quadrilateral $ABCD$ is inscribed in a circle, the incenters of $\triangle BCD, \triangle ACD$, $\triangle ABD, \triangle ABC$ are denoted as $I_{A}, I_{B}, I_{C}, I_{D}$ respectively. Prove that $I_{A} I_{B} I_{C} I_{D}$ is a rectangle.
(1st National Training Selection Test for Mathematical Olympiad) | Analysis: Connect $A I_{C}$, $A I_{D}$, $B I_{C}$, $B I_{D}$, and $D I_{B}$.
It is easy to see that $\angle A I_{C} B=90^{\circ}$
$$
\begin{array}{l}
+\frac{1}{2} \cdot \angle A D B=: 90^{\circ}+ \\
\frac{1}{2} \angle A C B=\angle A I_{D} B \Rightarrow
\end{array}
$$
$A, B, I_{D}, I_{C}$ are concyclic.
Similarly, $A, D... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,646 |
Theorem: Let the side lengths and areas of $\triangle A_{1} B_{1} C_{1}$ and $\triangle A_{2} B_{2} C_{2}$ be $a_{1}, b_{1}, c_{1}, a_{2}, b_{2}, c_{2}$ and $\triangle_{1}, \triangle_{2}$, respectively. Let $s_{i}=a_{i}^{2}+b_{i}^{2}+c_{i}^{2}$, $i=1,2$, and $H=a_{1}^{2}\left(-a_{2}^{2}+b_{2}^{2}+c_{2}^{2}\right)+b_{1}... | Let $\Sigma a^{4}=a^{4}+b^{4}+c^{4}$, etc. Then, by Heron's formula:
$$
8 \triangle_{1}^{2}=\frac{1}{2} s_{1}^{2}-\Sigma a_{1}^{4}, 8 \triangle_{2}^{2}=\frac{1}{2} s_{2}^{2}-\Sigma a_{2}^{4},
$$
we have $8 s_{1}^{2} \triangle_{2}^{2}+8 s_{2}^{2} \triangle_{1}^{2}$
$$
\begin{array}{l}
=\left(\frac{1}{2} s_{1}^{2} s_{2}... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,647 |
If $A B C D$ is a square, $\angle E A F=45^{\circ}$. Prove that $S_{\triangle A E F}=2 S_{\triangle A P Q} \cdot(1990$, Four J Problem)
Generalization 1 If $A B C D$ is a square. If we set $\angle E A F=\alpha$, $\angle D A F=\theta$, then
$$
\frac{S_{\triangle A P Q}}{S_{\triangle A E F}}=\frac{\cos \theta \sin (\alp... | Given $A B=b, A D=B C=a$, then
$$
\triangle A Q B \backsim \triangle F Q D \text {. }
$$
Thus, $\frac{A Q}{A F}=\frac{b}{b+D F}$.
Similarly, $\frac{A P}{A F}=\frac{a}{a+B E}$.
Also, $D F=a \operatorname{tg} \theta, B E=b \operatorname{ctg}(\alpha+\theta)$,
Therefore,
$$
\begin{array}{l}
\frac{A Q}{A F} \cdot \frac{A P... | \frac{\sin 2 \varphi \cos \theta \sin (\alpha+\theta)}{2 \sin (\varphi+\theta) \sin (\alpha+\theta+\varphi)} | Geometry | proof | Yes | Yes | cn_contest | false | 706,648 |
2. The four-digit number $\overline{a a b b}$ is a perfect square. Then $\overline{a a b b}=(\quad)$.
(A) 7744
(B) 6655
(C) 8833
(D) 4477 | 2. A
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 706,650 |
3. In an $n$-sided polygon, the sum of $(n-1)$ interior angles is $8940^{\circ}$. Then $n$ equals ( ).
(A) 60
(B) 51
(C) 52
(D) 53 | 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 | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,651 |
4. As shown in the figure, along the side of a square with a side length of 90 meters. In the direction of $A \rightarrow B \rightarrow C$ $\rightarrow D \rightarrow A \cdots$ . Person A starts from $A$ at a speed of 65 meters/min, and Person B starts from $B$ at a speed of 72 meters/min. When B catches up with A for t... | 4. B
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 | 706,652 |
6. Given that the difference in the circumferences of two circles is $1 \mathrm{~cm}$. Then the difference in the radii of these two circles is ( ).
(A) uncertain
(B) $\frac{1}{n} \mathrm{~cm}$
(C) $\sqrt{3} \mathrm{~cm}$
(D) $\frac{1}{2 \pi} \mathrm{cm}$ | 6. D
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 | 706,654 |
8. Among the following expressions, the largest result is ( ).
(A) $4^{\frac{1}{4}}-3^{\frac{1}{3}}$
(B) $7-5 \sqrt{2}$
(C) $5 \sqrt{2}-7$
(D) $2+\sqrt{3}-(2-\sqrt{3})^{-}$ | 8. 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 | 706,656 |
Example 6. The center of square $ABCD$ is $O$, and its area is $1989 \mathrm{~cm}^{2} . P$ is a point inside the square, and $\angle O P B=45^{\circ}, P A : P B=5 : 14$. Then $P B=$ $\qquad$ . (1989, National Junior High School League) | Analysis: The answer is $P B=42 \mathrm{~cm}$. How do we get it?
Connect $O A, O B$. It is easy to know that $O, P, A, B$ are concyclic, so $\angle A P B=\angle A O B=90^{\circ}$.
Therefore, $P A^{2}+P B^{2}=A B^{2}=1989$. Since $P A: P B=5: 14$, we can find $P B$. | 42 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,657 |
1. Given that the side length of square $A B C D$ is $1 \mathrm{~cm}$. Then the area of the common part of the two circles with diameters $A B, B C$ is | 1. $\frac{1}{8}(\pi-2) \mathrm{cm}^{2}$ | \frac{1}{8}(\pi-2) \mathrm{cm}^{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,658 |
$+1 \mid=k$ has four distinct branches. Find the range of $k$.
The above text translated into English, please retain the original text's line breaks and format, and output the translation result directly. | Three, first $k \geqslant 0$. Secondly, consider two cases:
(1) When $x^{2}-2 \sqrt{3} x+1=k$,
$$
\begin{array}{l}
\Delta=12-4(1-k)>0, \\
\therefore k>-2 .
\end{array}
$$
$$
\begin{array}{l}
\text { (2) When } x^{2}-2 \sqrt{3} x+1=-k \text {, } \\
\quad \Delta=12-4(1+k)>0, \\
\therefore k<2 .
\end{array}
$$
Combining ... | 0 \leqslant k < 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,662 |
Four. (20 points) As shown in the figure, given that $Q$ is the intersection of the diagonals of the cyclic quadrilateral $ABCD$, $PB$ and $PD$ are tangents to the circle, and $P$ lies on the line $AC$.
Prove:
1. $\frac{QA}{QC} = \frac{AB \cdot AD}{CB \cdot CD}$.
2. $\frac{QA}{QC} = \frac{PA}{PC}$. | 1.
$$
\begin{aligned}
\frac{Q A}{Q C} & =\frac{Q A^{2}}{Q C \cdot Q A} \\
& =\frac{Q A^{2}}{Q B \cdot Q D} \\
& =\frac{Q A}{Q B} \cdot \frac{Q A}{Q D} \\
& =\frac{D A}{C B} \cdot \frac{B A}{C D} \\
& =\frac{A B \cdot A D}{C B \cdot C D}
\end{aligned}
$$
2.
$$
\begin{array}{l}
\because \triangle P C D \subset \triangle ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,663 |
Five. (20 points) Fill in the right table with $1,2,3,4,5,6$ respectively, so that in each row, the number on the left is less than the number on the right, and in each column, the number on top is less than the number below. How many ways are there to fill the table? Provide an analysis process. | Five, as shown in the right figure, from the known information we can get: $a$ is the smallest, $f$ is the largest, so $a=1$, $f=6$. From $bd$, we need to discuss the following two cases:
(1) When $bd$, then $b=3, d=2, c=4$ or
5.
In this case, there are the following two ways to fill:
\begin{tabular}{|l|l|l|}
\hline 1 ... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 706,664 |
1. Let $x<-1$, simplify $2-|2-| x-2||$ the result is ).
(A) $2-x$
(B) $2+x$
(C) $-2+x$
(D) $-2-x$ | $-、 1$. When $x<-1$, the original expression $=2-|2+x-2|=2$ $-|x|=2+x$. Choose (B) | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 706,665 |
2. Simplify the result of $x \sqrt{-\frac{1}{x}}$ is ( ).
(A) $\sqrt{x}$
(B) $-\sqrt{x}$
(C) $\sqrt{-x}$
(D) $-\sqrt{-x}$ | 2. $\because x<0, \therefore$ original expression $=x \sqrt{-\frac{x}{x^{2}}}=\frac{x}{|x|} \sqrt{-x}$ $=-\frac{x}{x} \sqrt{-x}=-\sqrt{-x}$. Choose (D). | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 706,666 |
3. $[x],[y],[z]$ respectively represent the greatest integers not exceeding $x, y, z$. If $[x]=5,\left[y^{\prime}\right]=-3,[z]=-1$, then the number of possible values for $[x-y-z]$ is ( ).
(A) 3
(B) 4
(C) 5
(D) 6 | 3. $\because 5 \leqslant x<6,-3 \leqslant y<-2,-1 \leqslant z<0$, there is $2<$ $-y \leqslant 3,0<-z \leqslant 1, \therefore 7<x-y-z<10$. Therefore, the values that $\{x-y$ $-z]$ can take are $7,8,9$. Hence, the answer is (A). | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 706,667 |
Inverting 7. Given a square with side length 1. Try to find the largest and the smallest area of an inscribed equilateral triangle within this square, and calculate these two areas (You must prove your argument).
(1978, National High School League) | Analysis: Let $\triangle E F G$ be
an inscribed equilateral triangle
in square $A B C D$. Since the three vertices of an
equilateral triangle must lie on at
least three sides of the square, we
can assume that points $F$ and $G$
lie on a pair of opposite sides of
the square.
Construct the altitude $E K$ of $\triangle E ... | 2 \sqrt{3}-3 \text{ and } \frac{\sqrt{3}}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,668 |
4. In $\triangle A B C$, the side lengths $a, b, c$ satisfy $(a+b+c)(a+b-c)=a b$, then the angle opposite to side $A B$ is ( ).
(A) $30^{\circ}$
(B) $60^{\circ}$
(C) $120^{\circ}$
(D) $150^{\circ}$ | 4. $(a+b+c)(a+b-c)=ab \Rightarrow a^2+b^2-c^2=-ab$ $\Rightarrow \frac{a^2+b^2-c^2}{2ab}=-\frac{1}{2}$, by the cosine rule, $\cos \angle C=-\frac{1}{2}$, $\therefore \angle C=120^{\circ}$. Choose $(C)$. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,669 |
f. Among the four students Zhang, Wang, Ben, and Zhao, one of them did a good deed outside of school and received praise. After being questioned, Zhang said: "It was Li who did it," Wang said: "It was Zhang who did it," Li said: "What Wang said is not true," and Zhao said: "It wasn't me who did it." After investigation... | 6. If Zhang did the good deed, then Wang and Zhao are correct,
If Wang did the good deed, then Li and Zhao are correct,
If Li did the good deed, then Zhang, Li, and Zhao are correct,
If Zhao did the good deed, then Li is correct,
$\because$ Only one person is correct, $\therefore$ the good deed was done by Zhao. Hence,... | D | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 706,671 |
1. For any real number $x$, the quadratic trinomial $x^{2}+3 m x+m^{2}-$ $m+\frac{1}{4}$ is a perfect square, then $m=$ $\qquad$. | 二、1. $\Delta=9 m^{2}-4\left(m^{2}-m+\frac{1}{4}\right)=0$. Solving it yields $m=\frac{1}{5}$ or -1. | \frac{1}{5} \text{ or } -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,672 |
$\begin{array}{l}\text { 3. Let } x>\frac{1}{4} \text {, simplify } \sqrt{x+\frac{1}{2}+\frac{1}{2} \sqrt{4 x+1}} \\ -\sqrt{x+\frac{1}{2}-\frac{1}{2} \sqrt{4 x+1}}=\end{array}$ | $\begin{array}{l}\text { 3. } \because \sqrt{x+\frac{1}{2} \pm \frac{1}{2} \sqrt{4 x+1}} \\ =\sqrt{\frac{2 x+1 \pm \sqrt{4 x+1}}{2}} \\ =\frac{1}{2} \sqrt{4 x+2 \pm 2 \sqrt{4 x+1}} \\ =\frac{1}{2} \sqrt{(\sqrt{4 x+1} \pm 1)^{2}} \\ =\frac{1}{2}|\sqrt{4 x+1} \pm 1|, \\ \therefore \text { when } x>\frac{1}{4} \text {, th... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,674 |
$4.1992^{1002}$ when divided by 1993 leaves a remainder of | $$
\begin{aligned}
4 \cdot 1992^{1002} & =(1993-1)^{2 \cdot 2 \cdot 2 \cdot 3 \cdot 83} \\
& =\left(1993^{2}-2 \cdot 1993+1\right)^{2 \cdot 2 \cdot 3 \cdot 83}
\end{aligned}
$$
Continuing the calculation in a similar manner, the first term that appears is 1993, and the last term is 1. Therefore, the remainder when $19... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 706,675 |
5. The area of convex quadrilateral $A B C D$ is 1, and its diagonals intersect at $P$. The centroids of $\triangle P A B, \triangle P B C, \triangle P C D, \triangle P D A$ are $M_{1}, M_{2}, M_{3}, M_{4}$, respectively. Then the area of quadrilateral $M_{1} M_{2} M_{3} M_{4}$ is | 5. Let the midpoints of $P B$ and $P D$ be $K$ and $H$ respectively. Connect $A K$, $A U$, $C K$, $C H$, then $M_{1}, M_{2}, M_{3}, M_{4}$ are the trisection points of $A K$, $C K$, $C H$, $A H$ respectively.
$$
\begin{aligned}
& \because S_{A B C D}=1, \therefore S_{A K C H}=\frac{1}{2} . \\
\because & \frac{A M_{1}}{... | \frac{2}{9} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,676 |
One, (14 points) Solve the system of equations
$$
\left\{\begin{array}{l}
x+y+\frac{9}{x}+\frac{4}{y}=10, \\
\left(x^{2}+9\right)\left(y^{2}+4\right)=24 x y .
\end{array}\right.
$$ | $$
\left\{\begin{array}{l}
x+\frac{9}{x}+y+\frac{4}{y}=10, \\
\left(x+\frac{9}{x}\right)\left(y+\frac{4}{y}\right)=24 .
\end{array}\right.
$$
Thus, $x+\frac{9}{x}$ and $y+\frac{4}{y}$ can be seen as the two roots of the equation $t^{2}-10 t+24=0$.
By $(t-4)(t-6)=0$, we get $t_{1}=4, t_{2}=6$.
We have (I) $\left\{\beg... | \left\{\begin{array}{l}x=3, \\ y=2 .\end{array}\right.} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,678 |
Example 8. $NS$ is the diameter of $\odot O$, chord $AB \perp NS$ at $M$, $P$ is any point on $\overparen{ANB}$ other than $N$, $PS$ intersects $AB$ at $R$, and the extension of $PM$ intersects $\odot O$ at $Q$. Prove: $RS > MQ$.
(1991, Jiangsu Province Junior High School Competition) | Connect $N P, N Q$, $N R, N R$ extended to meet $\odot O$ at $Q^{\prime}$. Connect $M Q^{\prime}, S Q^{\prime}$.
It is easy to prove that $N, M, R, P$ are concyclic, thus, $\angle S N Q^{\prime}=$ $\angle M N R=\angle M P R=$ $\angle S P Q=\angle S N Q$.
According to the axial symmetry property of the circle, $Q$ and... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,679 |
II. (16 points) Find the maximum and minimum values of the quadratic function $y=x^{2}+m x+m$ in the interval -3 $\leqslant x \leqslant-1$. | $$
\begin{array}{l}
\text { 2. } y=\left(x+\frac{m}{2}\right)^{2}+ \\
\frac{4 m-m^{2}}{4},
\end{array}
$$
The graph of the given function moves left and right and up and down as $m$ takes different values.
When $-2 \leqslant-\frac{m}{2} \leqslant 6$, $x=-\frac{m}{2}$ is to the left of $-3 \leqslant x \leqslant$ -1, th... | y_{\text{min}}=9-2m, y_{\text{max}}=1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,680 |
Three, (16 points) Divide each side of an equilateral triangle with side length $a$ into 1993 equal parts, then connect the corresponding points on each side with parallel lines, resulting in several smaller equilateral triangles. Find the sum of the areas of the incircles of all the smaller equilateral triangles.
| $$
\begin{array}{l}
\text { Three, the number of small equilateral triangles is } \\
1+3+5+\cdots+(2 \times 1993 \\
-1)=1993^{2} .
\end{array}
$$
The height of each small equilateral triangle is
$$
\frac{\sqrt{3}}{2} a \cdot \frac{1}{1993} \text {, }
$$
The radius of the incircle of each small equilateral triangle i... | \frac{\pi a^{2}}{12} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,681 |
Four, (16 points) Given that the altitudes $A D, B E$ of $\triangle A B C$ intersect at $H$, and the circumcircles of $\triangle A B C$ and $\triangle A B H$ are $\odot O$ and $\odot O_{1}$ respectively. Prove: The radii of $\odot O$ and $\odot O_{1}$ are equal. | Draw diameters $A P, A Q$ of $\odot O$ and $\odot O_{1}$ through $A$, and connect $P B, Q B, \because \angle A B P=90^{\circ}=\angle A B Q, \therefore P, B, Q$ are collinear.
$\because A, B, P, C$ are concyclic,
$$
\therefore \angle P=\angle C \text {. }
$$
$\because A, H, B, Q$ are concyclic,
$$
\therefore \angle Q=\a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,682 |
Five. (16 points) A certain middle school originally had several classrooms, each with an equal number of desks, totaling 539 desks. This year, the school added 9 new classrooms in a newly built teaching building, increasing the total number of desks to 1080. At this point, the number of desks in each classroom remaine... | Five, let there be $x$ classrooms now, then the original number of classrooms is $x-9$. According to the problem, we have
$\frac{1080}{x}>\frac{539}{x-9}$, which means $\frac{1080}{x}-\frac{539}{x-9}>0$,
or $\frac{1080 x-9720-539 x}{x(x-9)}>0, x>\frac{9720}{541}$.
Thus, $x>17$.
Also, since $\frac{539}{x-9}$ is a natura... | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,683 |
Example 1.40 A centipede with 40 legs and a dragon with 3 heads are in the same cage, with a total of 26 heads and 298 legs. If the centipede has 1 head, then the dragon has $\qquad$ legs. (1988, Shanghai Junior High School Competition)
If the centipede has $x$ individuals, and the dragon has $y$ individuals, and each ... | From (1), we get $x=26-3 y$.
Since $0 \leqslant x \leqslant 26$, and $x$ takes integer values, we have $0 \leqslant y \leqslant 8 \frac{2}{3}$, and $y$ takes integer values, so $0 \leqslant y \leqslant 8$.
Substituting (3) into (2), we get
$$
\begin{array}{l}
n y+40(26-3 y)=298, \\
n y=120 y-742 . \\
n=\frac{120 y-742}... | 14 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 706,690 |
4. As shown in the figure, the radius of semicircle $O$ is 1, $A C \perp A B$ at $A, B D \perp A B$ at $B$, and $A C=1, B D=3, P$ is any point on the semicircle, then the maximum value of the area of the closed figure $A B D P C$ is | 4. $2+\sqrt{2}$ | 2+\sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 706,696 |
One. (20 points) Let the quadratic function $y=-x^{2}+(m-2) x$ $+3(m+1)$ have its graph intersect the $x$-axis at points $A, B$ ($A$ is to the left of $B$), and the $y$-axis at point $C$. The product of the lengths of segments $AO$ and $OB$ equals 6 ($O$ is the origin). Connect $AC, BC$. Find the value of $\sin C$. | According to the problem, we have $A O \cdot O B=6$.
Let $A\left(x_{1}, 0\right), B\left(x_{2}, 0\right)$, then $A O=\left|x_{1}\right|, O B=\left|x_{2}\right|$.
Thus, $\left|x_{1}\right| \cdot\left|x_{2}\right|=6$, which means $\left|x_{1} \cdot x_{2}\right|=6$.
We consider the following two cases:
(1) $x_{1} \cdot x_... | \frac{\sqrt{2}}{10} \text{ or } \frac{\sqrt{2}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,697 |
II. (20 points) As shown in the figure, in $\triangle ABC$, $AB=AC$, $O$ is the incenter of $\triangle ABC$. Connect $AO$ and extend it to intersect the circumcircle of $\triangle ABC$ at point $D$. Draw a line through $O$ parallel to $BC$, intersecting $AB$ and $AC$ at $E$ and $F$ respectively. Prove that $AB$ and $AC... | $$
\begin{array}{c}
\text{Connect } O C, D C, F O^{\prime}. \text{ Let the circumcircle of } \triangle E F D \text{ be } \odot O^{\prime}. \\
\because A B=A C, \therefore A D \text{ is the diameter of the circumcircle, and the incenter of } \triangle A B C \text{ and the circumcenter of } \triangle E F D \text{ are bot... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 706,698 |
Three. (20 points) Let $a, b$ be integers. How many solutions does the system of equations
$$
\left\{\begin{array}{l}
{[x]+2 y=a,} \\
{[y]+2 x=b}
\end{array}\right.
$$
(here $[x]$ denotes the greatest integer less than or equal to $x$) have? | Obviously, if $2x$ and $2y$ are integers, then $x$ and $y$ are either integers or differ from some integer by $\frac{1}{2}$. We will discuss several cases below:
(1) Suppose $x, y$ are both integers, let $x=[x], y=[y]$, then from the system of equations we can obtain
$$
x=\frac{2b-a}{3}, \quad y=\frac{2a-b}{3}.
$$
Th... | 2 \text{ or } 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,699 |
1. Given a convex pentagon $A B C D E, \angle A B C=\angle A E D=$ $90^{\circ}, \angle D A C=30^{\circ}, \angle B A E=70^{\circ}, F$ is the midpoint of side $C D$, and $F B=F E$. Then $\angle B A C$ equals ( ).
(A) $10^{\circ}$
(B) $20^{\circ}$
(C) $30^{\circ}$
(D) $15^{\circ}$ | $-1 . \mathrm{B}$
Take the midpoints $P, Q$ of $A C, A D$, and connect $P B, P F$, $Q E, Q F$, then $P B=P A$ $=F Q, Q E=Q A=$ $F P$. Furthermore, since $F B=F E$, we have $\triangle F P B \cong \triangle F Q E$. Thus, $\angle 17 \angle 3=\angle 2+\angle 4$. Also, $\angle 1=\angle 2=\angle C A D$, so $\angle 3=\angle 4... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,700 |
Side 2. 100 yuan was used to buy exactly 100 pens of three types, where each gold pen costs 10 yuan, each silver pen costs 3 yuan, and each ballpoint pen costs 0.5 yuan. How many of each type of pen were bought? (1984, Wuhu Junior High School Competition) | Let $x, y, z$ be the number of gold pens, iridium pens, and ballpoint pens purchased, respectively. According to the problem, we have
$$
\left\{\begin{array}{l}
x+y+z=100, \\
10 x+3 y+\frac{1}{2} z=100 .
\end{array}\right.
$$
(2) $\times 2-$ (1), we get
$$
19 x+5 y=100 .
$$
Since $x, y$ must be positive integers, from... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 706,701 |
2. Given that the cubic polynomial $x^{3}-a x^{2}-1003 x$ $+b$ can be divided by $x^{2}-999 x+1994$. Then the value of $b-6 a$ is ( ).
(A) 6
(B) 7
(C) 8
(D) 9 | 2. $\mathbf{A}$
Since $x^{2}-999 x+1994=(x-2)(x-997)$, when $x=2$ and $x=997$, by the assumption, if $x^{3}-a x^{2}-1003 x$ $+b$ can be divided by $x^{2}-999 x+1994$, then we have
$$
\left\{\begin{array}{l}
2^{3}-a \times 2^{2}-1003 \times 2+b=0, \\
997^{3}-a \times 997^{2}-1003 \times 997+b=0 .
\end{array}\right.
$$
... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 706,702 |
3. Let the two legs of a right triangle be $p$ and $q$, and the hypotenuse be $r$, where $p, q, r$ are positive integers, and $p$ is a prime number. Then the remainder when $2(p+q+1)$ is divided by 3 could be ( ).
(A) 0 or 1
(B) 0 or?
(C)1 or?
(D)C,1,2, that hole can
Translate the above text into English, please ret... | 3. A
$p^{2}=r^{2}-q^{2}=(r-q)(r+q)$, since $p$ is a prime, then $r-q=1, r+q=p^{2}$. Thus, $p^{2}-1=2 q$. Therefore, $2(p+q+1)=2 p+2 q+2=2 p+p^{2}+1=(p+1)^{2}$. Hence, the remainder when divided by 3 can only be 0 or 1. | null | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 706,703 |
4. Given that $a$ and 6 satisfy
$$
5 a^{2}+2 b^{2}-1=6 a b+4 a-2 b \text{. }
$$
Then the value of $(a-b)^{1004}$ is ( ).
(A) 0
(B) 1
(C) 2
(D) Cannot be determined | 4. A
Given the equation is transformed into $5 a^{2}-(6 b+4) a+\left(2 b^{2}+2 b+1\right)$ $=0$, this is a quadratic equation in terms of $a$. Since it has real roots, then $\Delta=(6 b+$ $4)^{2}-20\left(2 b^{2}+2 b+1\right) \geqslant 0$. Solving this, we get $(b-1)^{2} \leqslant 0$. Therefore, $b$ $=1$. Subsequently,... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 706,704 |
5. In $\triangle A B C$, points $E$ and $F$ on side $B C$ trisect $B C$, $B M$ is the median of side $A C$, and $A E, A F$ intersect $B M$ at points $G, H$ respectively. Then $B G: G H: H M$ equals ( ).
(A) $4: 3 \div 2$
(B) $4: 2: 3$
(C) $5 \pm 3: 2$
(D) $3: 3: 2$ | 5. C
As shown in the figure, draw $C H^{\prime} / / A H, C G^{\prime} / / A G$, intersecting the extension of $B M$ at $H^{\prime}, G^{\prime}$, then $G H = G^{\prime} H^{\prime}, H M = H^{\prime} M$. For convenience, let $B G = x, G H = y$, $H M = z$, then $\frac{x}{2 y + 2 z} = \frac{B E}{E C} = \frac{1}{2}$. Theref... | 5: 3: 2 | Geometry | MCQ | Yes | Yes | cn_contest | false | 706,705 |
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