problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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class | __index_level_0__ int64 0 742k |
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9. Given $f(x)=\frac{2 x+3}{x-1}$. If the graph of $y=g(x)$ is symmetric to the graph of $y=f^{-1}(x+1)$ about the line $y=x$, then the value of $g(3)$ is
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 9. $\frac{7}{2}$.
From the given information, the inverse function of $y=f(x)$ is $f^{-1}(x)=\frac{x+3}{x-2}$. Therefore,
$$
f^{-1}(x+1)=\frac{x+4}{x-1} \text {. }
$$
Also, $g(x)$ is the inverse function of $f^{-1}(x+1)$, to find the value of $g(3)$, we need to solve the equation $3=\frac{x+4}{x-1}$, thus, $g(3)=\fra... | \frac{7}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,219 |
10. If the set $A=\{x \mid-2 \leqslant x \leqslant 5\}, B=\{x \mid m+1 \leqslant$ $x \leqslant 2 m-1\}$, and $A \cap B=B$, then the range of real number $m$ is $\qquad$. | $$
\text { 10. } m \leqslant 3 \text {. }
$$
From $A \cap B=B$, we know that $B$ is a subset of $A$.
When $B=\varnothing$, we have $m+1>2 m-1$, which gives $m<2$;
When $B \neq \varnothing$, we have
$$
\left\{\begin{array}{l}
-2 \leqslant m+1, \\
2 m-1 \leqslant 5, \\
m+1 \leqslant 2 m-1 .
\end{array}\right.
$$
Solvin... | m \leqslant 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,220 |
11. Given point $A(m, n)$ lies on the line $x+3 y=41$, where $0<n<m$. If the symmetric point of point $A$ with respect to the line $y=x$ is $B$, the symmetric point of point $B$ with respect to the $y$-axis is $C$, the symmetric point of point $C$ with respect to the $x$-axis is $D$, and the symmetric point of point $D... | 11. $(11,10)$.
By symmetry, we get $B(n, m)$, $C(-n, m)$, $D(-n, -m)$, and $E(n, -m)$.
The area of pentagon $ABCDE$ is
$$
\begin{array}{l}
S_{\text {pentagon } ABCDE}=2 m \times 2 n+\frac{1}{2} \times 2 m \times(m-n) \\
=m(m+3 n)=41 m,
\end{array}
$$
Thus, $41 m=451$. Therefore, $m=11, n=10$.
Hence, the coordinates o... | (11,10) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,221 |
12. Let $\left\{a_{n}\right\}$ be an arithmetic sequence with all terms being positive integers, an odd number of terms, a non-zero common difference, and the sum of all terms equal to 2004. Then the value of the second term $a_{2}$ is $\qquad$ | 12. 668 .
Let the first term and common difference of an arithmetic sequence be $a$ and $d$, and the number of terms be $n$. Then, $n a+\frac{1}{2} n(n-1) d=2004$, which means
$$
[2 a+(n-1) d] n=2004 \times 2=2^{3} \times 3 \times 167 \text {. }
$$
Since $n$ is odd and $d \neq 0$,
when $n=501$, $a+250 d=4$, which doe... | 668 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,222 |
13. The solution set of the inequality $a^{2}+2 a-\sin ^{2} x-2 a \cos x>$ 2 about $x$ is all real numbers. Find the range of real numbers for $a$. | Three, 13. Let $t=\cos x$, then the original inequality becomes
$$
t^{2}-2 a t+a^{2}+2 a-3>0, t \in[-1,1] \text {. }
$$
Thus, the problem is transformed into finding the minimum value of the function $f(t)=t^{2}-2 a t+a^{2}+2 a-3$ on $t \in[-1,1]$ and ensuring it is positive.
Since the function $f(t)=(t-a)^{2}+2 a-3$... | a > \sqrt{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 714,223 |
Example 4 As shown in Figure $5, \odot O_{1}$ and $\odot O_{2}$ are tangent to the three sides of $\triangle A B C$, with $E, F, G, H$ being the points of tangency. The extensions of $E G$ and $F H$ intersect at point $P$. Prove that line $P A$ is perpendicular to $B C$.
(1996, National High School Mathematics Competit... | Analysis: As shown in Figure 5, it is only necessary to prove that $O A$, $E G$, and $F H$ are concurrent, which means proving that the intersection point $P$ of $E G$ and $F H$ has an x-coordinate of 0. The equations of $E G$ and $F H$ depend on $\triangle A B C$, so we can introduce $\angle B$ and $\angle C$ as param... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,224 |
14. Given $A\left(x_{1}, y_{1}\right) 、 B\left(x_{2}, y_{2}\right)$ are two moving points on the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=$ $1(a>b>0)$, $O$ is the origin, and $O A \perp$ $O B$. Find the minimum length of segment $A B$. | 14. According to the problem, let's assume $A\left(r_{1} \cos \theta, r_{1} \sin \theta\right)$, $B\left(r_{2} \cos \left(\theta+\frac{\pi}{2}\right), r_{2} \sin \left(\theta+\frac{\pi}{2}\right)\right)$, then
$$
\begin{array}{l}
B\left(-r_{2} \sin \theta, r_{2} \cos \theta\right), A B^{2}=r_{1}^{2}+r_{2}^{2}, \\
\frac... | \frac{2 a b \sqrt{a^{2}+b^{2}}}{a^{2}+b^{2}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,225 |
15. Given the function $f(x)=x^{2}+x-2$.
(1) Try to find the analytical expression of the function $g(x)=\frac{|f(x)|-f(x)}{2}$;
(2) If $a>0$, when the line $y=a x+b$ intersects the curve $y=g(x)$ at three different points, try to determine the relationship between $a$ and $b$, and draw a diagram to represent the regio... | 15. (1) \( g(x) = \begin{cases}
0, & x \leqslant -2 \text{ or } x \geqslant 1, \\
-x^2 - x + 2, & -2 < x < 1
\end{cases} \)
When \( a > 0 \), the line \( y = ax + b \) intersects the curve \( y = g(x) \) at three different points if and only if the line intersects the curve in the interval \( -2 < x < 1 \). This re... | 2a < b < \frac{1}{4}(a+1)^2 + 2 \text{ and } 0 < a < 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,226 |
1. Let real numbers $a$, $b$, $c$ satisfy $c+b=3a^2-4a+6$, $c-b=a^2-4a+4$. Then the relationship between $a$, $b$, $c$ is ( ).
(A) $a<b \leqslant c$
(B) $b<a \leqslant c$
(C) $b<c \leqslant a$
(D) $c<a \leqslant b$ | $-1 . A$.
Since $b=\frac{(c+b)-(c-b)}{2}=a^{2}+1$, then $b-a=a^{2}-a+1=\left(a-\frac{1}{2}\right)^{2}+\frac{3}{4}>0$. Also, $c-b=a^{2}-4 a+4=(a-2)^{2} \geqslant 0$, thus $a<b \leqslant c$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,227 |
2. A store purchased a certain number of student-specific scientific calculators at the beginning of the school year for 880 yuan, and sold them at 50 yuan each, which were quickly sold out. It was found that students still needed 3 times as many calculators, so the store purchased the required calculators for 2580 yua... | 2.B.
Let the first purchase of calculators be $x$ units, then the second purchase of calculators is $3x$ units. According to the problem, we have
$$
\frac{880}{x}=\frac{2580}{3 x}+1 \text {. }
$$
Solving for $x$ gives $x=20$. Therefore, the total profit is
$$
(50 \times 76+4 \times 50 \times 0.9)-(880+2580)=520 \text ... | 520 | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,228 |
4. As shown in Figure 1, in $\triangle A B C$, the lengths of the three sides are $a$, $b$, and $c$ respectively. Point $I$ is the incenter, $I D \perp B C$ at point $D$, and $A H$ is the altitude. If $B D \cdot D C = I D \cdot A H$, then the relationship is ( ).
(A) $b^{2}+c^{2}=2 a^{2}$
(B) $b+c=2 a$
(C) $a^{2}+c^{2}... | 4.B.
Let $p=\frac{1}{2}(a+b+c)$. The area of $\triangle ABC$ is
$$
S=\sqrt{p(p-a)(p-b)(p-c)}.
$$
Since $B D \cdot D C=(p-b)(p-c)$,
$$
I D \cdot A H=\frac{S}{p} \cdot \frac{2 S}{a}=\frac{2(p-a)(p-b)(p-c)}{a},
$$
thus, $a=2(p-a)=b+c-a, b+c=2a$. | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,230 |
5. Given $x=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}, y=$ $\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}-\sqrt{n}}$ ( $n$ is a positive integer) is a solution to the equation $2 x^{2}+$ $207 x y+2 y^{2}=2003$. Then the value of $n$ is ( ).
(A) 7
(B) 8
(C) 9
(D) 10 | 5.A.
It is easy to get $x y=1, x+y=4 n+2$.
Since $2003=2 x^{2}+207 x y+2 y^{2}=2(x+y)^{2}+203$ $=2(4 n+2)^{2}+203$, therefore, $(4 n+2)^{2}=900, n=7$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,231 |
6. As shown in Figure $2, \odot 0$ has two diameters $A B \perp$ $C D$, with its area being $S$. Point $P$ is on $\overparen{B D}$, $P A$ intersects $O D$ at point $E$, and $P C$ intersects $O B$ at point $F$. The area of quadrilateral $A C F E$ is ( ).
(A) $\frac{S}{2}$
(B) $\frac{S}{3}$
(C) $\frac{S}{\pi}$
(D) $\frac... | 6.C.
Connect $P B$, then $P F$ bisects $\angle A P B$. Since $\triangle A O E \sim \triangle A P B$, we have $\frac{A O}{O E}=\frac{A P}{P B}=\frac{A F}{F B}$.
Let the radius of $\odot O$ be $r$, then $S=\pi r^{2}$.
Let $O E=x$, $O F=y$. Substituting, we get
$$
\begin{array}{l}
\frac{r}{x}=\frac{r+y}{r-y} \Rightarrow ... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,232 |
1. Real numbers $x, y, z$ satisfy $x=y+\sqrt{2}, 2 x y+$ $2 \sqrt{2} z^{2}+1=0$. Then the value of $x+y+z$ is $\qquad$ | $=1.0$.
Since $x=y+\sqrt{2}$, we have,
$$
\begin{array}{l}
(x-y)^{2}=2, (x+y)^{2}-4 x y=2, \\
2 x y=\frac{1}{2}(x+y)^{2}-1 .
\end{array}
$$
Substituting this into $2 x y+2 \sqrt{2} z^{2}+1=0$ yields
$$
\frac{1}{2}(x+y)^{2}+2 \sqrt{2} z^{2}=0 \text {. }
$$ | 1.0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,233 |
Example 5 Two circles $\odot O_{1}$ and $\odot O_{2}$ have radii
$$
R \text{ and } r, O_{1} O_{2}=\sqrt{R^{2}+r^{2}-r \sqrt{4 R^{2}+r^{2}}}, R \geqslant
$$
$\sqrt{2} r . A$ is a point on the larger circle $\odot O_{1}$, and lines $A B$ and $A C$ are tangent to the smaller circle $\odot O_{2}$
at points $B$ and $C$, res... | Analysis: As shown in Figure 6, let $\angle O_{2} A x=\alpha, \angle O_{2} A D=\beta, A O_{2}=m$, then
$$
\begin{array}{l}
\cos \alpha=\frac{R^{2}+m^{2}-\left(R^{2}+r^{2}-r \sqrt{4 R^{2}+r^{2}}\right)}{2 R m} \\
=\frac{m^{2}-r^{2}+r \sqrt{4 R^{2}+r^{2}}}{2 R m}, \\
\sin \beta=\frac{r}{m} .
\end{array}
$$
Since $(R-r)^... | r^{2} | Geometry | proof | Yes | Yes | cn_contest | false | 714,235 |
3. Points $A(-4,0)$ and $B(2,0)$ are two fixed points on the $x O y$ plane, and $C$ is a moving point on the graph of $y=-\frac{1}{2} x+2$. How many right triangles $\triangle A B C$ can be drawn that satisfy the above conditions? | 3. 4 .
As shown in Figure 6, draw perpendiculars from $A$ and $B$ to the $x$-axis, intersecting the line $y=-\frac{1}{2} x + 2$ at points $C_{1}$ and $C_{2}$, respectively; with $AB$ as the diameter, draw a semicircle intersecting the line
$$
y=-\frac{1}{2} x + 2 \text { at }
$$
points $C_{3}$ and $C_{4}$. Then,
$\t... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,236 |
4. Given that $a$, $b$, $c$, and $d$ are all prime numbers, and satisfy 10 $<c<d<20$, also $c$ and $a$ differ by a larger prime number, $d^{2}-c^{2}=a^{3} b(a+b)$. Then the value of $a b(c+d)$ is $\qquad$. | 4. 180.
Since $a, b, c, d$ are all prime numbers, and $10<c<d<20$, $c, d$ can only be 11, 13, 17, or 19, and $c \neq 19$. Also, $c-a$ is a larger prime number (not even), so $a=2$. By taking $c=11, 13, 17$, $c-a$ are 9, 11, 15, respectively. Only when $c=13, c-a=11$ does it meet the condition. Substituting $c=13, a=2$... | 180 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,237 |
One, (20 points) A woodworking factory produces two types of chairs, A and B. Each chair must go through two processes, woodworking and painting, to be completed. The daily working hours of the woodworkers and painters and the time required to make (paint) one chair are shown in Table 1. It is known that the factory ea... | Let's do $x$ chairs of type A and $y$ chairs of type B each day, with daily profit being $M$, then we have
$$
\left\{\begin{array}{l}
x+2 y \leqslant 8, \\
3 x+y \leqslant 9, \\
M=2 x+3 y, \\
x \geqslant 0, y \geqslant 0 .
\end{array}\right.
$$
From Figure 7, it is clear that,
$$
\text { when } x=2, y=3
$$
$M$ is max... | 13 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,238 |
Three. (25 points) If the natural number $n$ has $m(m \geqslant 2)$ positive odd divisors (including the natural number 1), prove that $n$ can be expressed as the sum of natural numbers in $m-1$ ways.
| Three, let $n$ have $m$ positive odd divisors:
$$
\begin{array}{l}
1,2 k_{1}+1,2 k_{2}+1, \cdots, 2 k_{m-1}+1 . \\
\text { Let } n=\left(2 k_{1}+1\right) a_{4}, i=1,2, \cdots, m-1 .
\end{array}
$$
When $a_{i} \leqslant k_{i}$,
$$
\begin{array}{l}
n=\left(2 k_{i}+1\right) a_{i} \\
=\underbrace{\left(k_{i}+1-a_{i}\right... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 714,240 |
1. There are 4 cards (as shown in Figure 1), with one side of each card written with an English letter, and the other side written with a number. The rule is: when one side of the card is the letter $R$, its other side must be the number 2. Your task is: to check which of the following 4 cards need to be flipped to ens... | $-、 1 . C$.
Thought 1: Turn over each card one by one, imagining the various possibilities on the back, which can lead to the conclusion that (b) and (c) do not need to be flipped. To aid understanding, imagine that “the red light is on” represents “one side is written with the English letter $R$”, and “pedestrians do ... | C | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 714,241 |
2. On a basketball court, there are 5 people practicing. The tactic is for Player A to start the game (the 1st pass), and after 6 passes (with each person having an equal chance of passing and receiving the ball), it returns to Player A, who then attempts a 3-point shot. The number of different passing methods is ( ).
... | 2.D.
Let the positions of the ball correspond to points $A_{0}, A_{1}, \cdots, A_{5}$. If there is a passing relationship between two positions, a straight line is drawn between the corresponding two points. Since it finally returns to player A, the passing relationship corresponds to a hexagon $A_{0} A_{1} \cdots A_{... | D | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 714,242 |
3. Given two sets of complex numbers
$$
\begin{array}{l}
M=\left\{z \mid z=\cos \alpha+\left(4-\cos ^{2} \alpha\right) \mathrm{i}, \alpha \in \mathbf{R}\right\}, \\
N=\{z \mid z=\cos \beta+(\lambda+\sin \beta) \mathrm{i}, \beta \in \mathbf{R}\} .
\end{array}
$$
When $M \cap N \neq \varnothing$, the range of the real n... | 3. B.
From $M \cap N \neq \varnothing$, we know that the two sets must have common elements. Let $\cos \alpha+\left(4-\cos ^{2} \alpha\right) \mathrm{i}=\cos \beta+(\lambda+\sin \beta) \mathrm{i}$, then we have $\left\{\begin{array}{l}\cos \alpha=\cos \beta, \\ 4-\cos ^{2} \alpha=\lambda+\sin \beta .\end{array}\right.... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,243 |
4. $P$ is a point on the central conic section $\frac{x^{2}}{m}+\frac{y^{2}}{n}=1(m>$ $|n|>0)$, and the lines connecting $P$ to the two foci $F_{1}$ and $F_{2}$ are perpendicular to each other. Then the area of $\mathrm{Rt} \triangle P F_{1} F_{2}$ is ( ).
(A) $|n|$
(B) $m$
(C) $n^{2}$
(D) cannot be determined, depends... | 4.A.
When $n>0$, the curve is an ellipse, with the semi-major axis $a=\sqrt{m}$, the semi-minor axis $b=\sqrt{n}$, and the semi-focal distance $c=\sqrt{a^{2}-b^{2}}=\sqrt{m-n}$. According to the problem, we have
$$
\begin{array}{l}
P F_{1}+P F_{2}=2 \sqrt{m}, \\
P F_{1}^{2}+P F_{2}^{2}=4(m-n) .
\end{array}
$$
(1) ${ }... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,244 |
5. In a non-coplanar quadrilateral $A B C D$, the sum of the four interior angles is denoted as $\theta=\angle A B C+\angle B C D+\angle C D A+$ $\angle D A B$. Then $(\quad)$.
(A) $\theta>360^{\circ}$
(B) $\theta=360^{\circ}$
(C) $\theta<360^{\circ}$
(D) Cannot be determined, depends on the shape | 5.C.
This means that the sum of the interior angles of a non-coplanar quadrilateral is less than $360^{\circ}$. The proof involves the properties of trihedral angles.
As shown in Figure 7, in quadrilateral $ABCD$, point $D$ is not in the plane $ABC$, and $AC$ is connected. By the property of trihedral angles that the... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,245 |
Example 1 As shown in Figure 1, in hexagon $A B C D E F$, $A B \cong$ $E D$, $A F \cong C D$, $B C \xlongequal[=]{=}$ $F E$, diagonal $F D \perp$ $B D$. It is known that $F D=24 \mathrm{~cm}$, $B D=18 \mathrm{~cm}$. What is the area of hexagon $A B C D E F$ in square centimeters? | Analysis: At first glance, this problem seems intractable, but upon closer inspection, we notice three sets of parallel and equal line segments, as well as two perpendicular diagonals with known lengths. This leads to the idea of translating the shapes to form a rectangle.
Solution: As shown in Figure 2, translate
$\tr... | 432 \mathrm{~cm}^{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,246 |
Example 2 Two line segments $A B$ and $C D$, both of length 1, intersect at point $O$, and $\angle B O D=60^{\circ}$. Prove that: $A C+B D \geqslant 1$.
| Analysis: To prove $A C+B D \geqslant 1$, it is easy to think of concentrating $A C$, $B D$, and a line segment of length 1 into a triangle, and then proving it using the triangle inequality.
To keep the length of $A C$ unchanged and the angle of $60^{\circ}$ unchanged, we can translate $A C$ to $B B_{1}$ (as shown in... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,247 |
Example 4 Let $f(x)=3x+2$. Prove: there exists a positive integer $m$, such that $f^{(100)}(m)$ is divisible by 1988, where $f^{(k)}(x)$ denotes $\underbrace{f(f(\cdots f}_{k \uparrow \uparrow}(x) \cdots))$.
(1988, China National Team Selection Exam) | Explanation: First, derive the expression for $f^{(100)}(m)$ using the knowledge of fixed points, then analyze the numerical characteristics given in the problem to conclude that $\left(3^{n}, 1988\right) = 1$ (where $n$ is a natural number).
From the problem, we know that the fixed point of the iteration is $x = -1$. ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 714,248 |
9. Let $n$ be a positive integer not exceeding 2003. If there is an angle $\theta$ such that $(\sin \theta+\mathrm{i} \cos \theta)^{n}=\sin n \theta+\mathrm{i} \cos n \theta$ holds. Then the total number of such $n$ is $\qquad$. | 9. 501 .
$$
\begin{array}{l}
(\sin \theta + i \cos \theta)^{n} = [i(\cos \theta - i \sin \theta)]^{n} \\
= i^{n}[\cos (-\theta) + i \sin (-\theta)]^{n} \\
= i^{n}(\cos n \theta - i \sin n \theta) = i^{n-1}(\sin n \theta + i \cos n \theta).
\end{array}
$$
If \( i^{n-1}(\sin n \theta + i \cos n \theta) = \sin n \theta +... | 501 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,249 |
10. Among three-digit numbers, if the digit in the tens place is smaller than the digits in the hundreds and units places, the number is called a concave number, such as 504, 746, etc., which are all concave numbers. Therefore, among three-digit numbers with no repeated digits, the number of concave numbers is $\qquad$... | 10. 240 .
When the tens digit is 0, there are $9 \times 8$ concave numbers that meet the condition, when the tens digit is 1, there are $8 \times 7$ concave numbers that meet the condition, .....
When the tens digit is 7, there are $2 \times 1$ concave numbers that meet the condition, in total there are $72+56+42+30+2... | 240 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 714,250 |
11. Given $\boldsymbol{a}=(\cos \alpha, \sin \alpha), \boldsymbol{b}=(\cos \beta, \sin \beta), \boldsymbol{a}$ and $\boldsymbol{b}$ satisfy the relation $|k \boldsymbol{a}+\boldsymbol{b}|=\sqrt{3}|\boldsymbol{a} - k \boldsymbol{b}|$, where $k>$ 0. Then the minimum value of $\boldsymbol{a} \cdot \boldsymbol{b}$ is $\qqu... | 11. $\frac{1}{2}$.
From $|k \boldsymbol{a}+\boldsymbol{b}|^{2}=(\sqrt{3}|\boldsymbol{a}-k \boldsymbol{b}|)^{2}$ we get
$8 k \boldsymbol{a} \cdot \boldsymbol{b}=\left(3-k^{2}\right) \boldsymbol{a}^{2}+\left(3 k^{2}-1\right) \boldsymbol{b}^{2}$.
Therefore, $\boldsymbol{a} \cdot \boldsymbol{b}=\frac{\left(3-k^{2}\right) ... | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,251 |
12. Given that $x$, $y$, and $z$ are all positive integers. Then the equation $x+y+z=$ 15 has solutions. | 12. 91 .
Express 15 as 15 ones, i.e., $1+1+\cdots+1=15$, where any 2 of the 14 plus signs are chosen, and the 1s separated by these two plus signs are combined into a single number to form a solution of the equation. Therefore, the number of solutions is $\mathrm{C}_{14}^{2}=91$. | 91 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 714,252 |
14. Given $A(\sqrt{5}, 0)$ and points $P_{1}, P_{2}, \cdots, P_{n}$ on the curve $\frac{x^{2}}{4}-y^{2}=1(2 \leqslant x \leqslant 2 \sqrt{5}, y \geqslant 0)$. Does there exist an $n$ such that $\left|P_{1} A\right|,\left|P_{2} A\right|, \cdots,\left|P_{n} A\right|$ form an arithmetic sequence with a common difference $... | 14. Since $d>0$, the sequence is increasing, so $\left|P_{\mid} A\right|$ is the smallest, and $\left|P_{n} A\right|$ is the largest.
As shown in Figure 5, the curve $\frac{x^{2}}{4}$
$$
-y^{2}=1(2 \leqslant x \leqslant
$$
$2 \sqrt{5}, y \geqslant 0)$ is part of a hyperbola,
with $A(\sqrt{5}, 0)$ being
its right focus,... | n=3,4,5, \cdots, 14 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,253 |
15. City $A$ has 4 suburban counties $(B, C, D, E)$, as shown in Figure 2. There are 5 colors available. How many different coloring methods are there such that no two adjacent areas share the same color, and each area is painted with only one color? | 15. The coloring method that meets the requirements must use at least three colors, so it can be divided into three categories:
(1) Using five colors, there are $P_{5}^{s}=120$ methods.
(2) Using four colors. The number of ways to choose four colors is $\mathrm{C}_{5}^{4}$. Among them, choosing one color to paint $A$ h... | 420 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 714,254 |
16. As shown in Figure 3, the circumcenter of $\triangle ABC$ is O. Using the sides of $\triangle ABC$ as axes of symmetry, three symmetric points of O, $O_{1}$, $O_{2}$, and $O_{3}$, are obtained. Now, all points are erased except for $O_{1}$, $O_{2}$, and $O_{3}$. Try to reconstruct $\triangle ABC$ based on these thr... | 16. (1) As shown in Figure 6, connect $OA$, $OC$, $O_{2}A$, and $O_{2}C$. These four line segments are equal, making $OAO_{2}C$ a rhombus. Therefore,
Similarly, $\mathrm{OAO}_{3}B$ is also
Thus, $O_{3}B \perp O_{2}C$. Consequently, $\mathrm{BCO}_{2}\mathrm{O}_{3}$ is a parallelogram, so $\mathrm{BC} \parallel \mathrm{O... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,255 |
1. Prove: For any positive integer $n$, there exists an $n$-digit number whose digits are all odd and which is divisible by $5^{n}$. | 1. Prove by mathematical induction.
When $n=1$, the conclusion is obviously true. Assume $N=\overline{a_{1} a_{2} \cdots a_{n}}$ can be divided by $5^{n}$, and each digit is odd. Consider the following numbers:
$$
\begin{array}{l}
N_{1}=\overline{1 a_{1} a_{2} \cdots a_{n}}=1 \times 10^{n}+5^{n} M=5^{n}\left(1 \times 2... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 714,256 |
2. A convex polygon $P$ in the plane is divided into smaller convex polygons by all its diagonals. If all sides and diagonals of polygon $P$ have rational lengths, prove: The side lengths of all the smaller polygons formed by the division are also rational.
untranslated text remains the same as requested. | 2. Let $P=A_{1} A_{2} \cdots A_{n}(n \geqslant 3)$. When $n=3$, the conclusion is obviously true, so we only need to consider the case when $n \geqslant 4$. To prove the conclusion, we first prove the following lemma.
Lemma: Suppose the sides and diagonals of a convex quadrilateral $A B C D$ are all rational numbers, ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,257 |
Example: In $5 m$ boxes, each contains a certain number of balls. Each time, add one ball to each of $n(n<m)$ boxes. Prove: When $(m, n)=1$, regardless of the initial distribution, it is always possible to make the number of balls in each box equal by adding balls in the above manner a finite number of times.
(Adapted ... | Explanation: Starting from extreme cases, examine the boxes with the most and the least balls.
Since $(m, n)=1$, by Bézout's theorem, there exist $u$, $v \in \mathbf{Z}_{\text{+}}$ such that $u n = v m + 1 = v(m-1) + v + 1$.
This equation indicates that by adding balls to the boxes $u$ times continuously, it is possi... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 714,258 |
4. A circle passes through the vertices $A, B$ of $\triangle A B C$, intersecting the line segments $A C, B C$ at points $D, E$ respectively. The line $B A$ and $E D$ intersect at point $F$, and the line $B D$ and $C F$ intersect at point $M$. Prove: $M F=M C \Leftrightarrow M B \cdot M D=M C^{2}$. | 4. As shown in Figure 3, extend $DM$ to $G$ such that $FG \parallel CD$, then $MF = MC \Leftrightarrow$ quadrilateral $CDFG$ is a parallelogram $\Leftrightarrow FD \parallel CG \Leftrightarrow \angle GCD = \angle FDA$, i.e., $\angle CGF + \angle FDA = 180^{\circ}$.
Furthermore, points $A, B, E, D$ lie on the circumfere... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,260 |
5. Let $a, b, c$ be positive real numbers. Prove:
$$
\frac{(2 a+b+c)^{2}}{2 a^{2}+(b+c)^{2}}+\frac{(2 b+c+a)^{2}}{2 b^{2}+(c+a)^{2}}+\frac{(2 c+a+b)^{2}}{2 c^{2}+(a+b)^{2}} \leqslant 8 \text {. }
$$ | 5. Multiplying $a, b, c$ by a suitable factor can reduce the original problem to the case of $a+b+c=3(a, b, c>0)$. Therefore, we only need to prove
$$
\begin{array}{c}
\frac{(a+3)^{2}}{2 a^{2}+(3-a)^{2}}+\frac{(b+3)^{2}}{2 b^{2}+(3-b)^{2}}+\frac{(c+3)^{2}}{2 c^{2}+(3-c)^{2}} \leqslant 8 . \\
\text { Let } f(x)=\frac{(x... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 714,261 |
6. On the 6 vertices of a regular hexagon, there are 6 non-negative integers, and their sum is 2003. An operation is allowed: choose a point, and replace the number at that point with the absolute difference of the numbers at the two adjacent points. Prove: a series of operations can be performed so that the number at ... | 6. Use $A_{F E}^{B} C_{D}$ to represent a certain state during the operation process, where $A, B, C, D, E, F$ are the numbers written at 6 points. ${ }^{A}{ }_{F}^{B} C_{D}(\bmod 2)$ represents the numbers written modulo $2$, $s$ represents the sum of all numbers in a certain state, and $M$ represents the maximum valu... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 714,262 |
2. The lengths of two sides of a triangle are 2 and 9, respectively, then the length of the third side is ( ).
(A) 4
(B) 6
(C) 8
(D) 12
(E) 14 | 2.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 | 714,264 |
5. The figure with the largest area among the following is ( ).
(A) a circle with a radius of 3
(B) a square with a side length of 5
(C) a rectangle with length and width of 3 and 9
(D) a right triangle with side lengths of 6, 8, and 10
(E) an equilateral triangle with a side length of 7 | 5.A.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | null | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,267 |
6. The interval that $\sqrt{87654321}$ lies in is ( ).
(A) $[900,1000]$
(B) $[9000,10000]$
(C) $[90000,100000]$
(D) $[2000,3000]$
(E) $[20000,30000]$ | $\begin{array}{l}\text { 6. B. } \\ 81000000<87654321<100000000 \text {. }\end{array}$ | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 714,268 |
Example 6 Given that $p, q$ are positive integers, and $(p, q) = 1$. If an integer $n$ can be written in the form $n = px + qy$, where $x, y$ are non-negative integers, then this number is called a good number; otherwise, it is called a bad number.
(1) Prove: There must exist an integer $c$, such that among the two int... | Explanation: From the given condition $(p, q)=1$, we think of applying Bézout's theorem.
(1) By the theorem, there exist integers $u, v$ such that $n = pu + qv$. Thus, for any integer $t$, we have
$$
n = p(u - tq) + q(v + tp).
$$
Therefore, $n$ can be expressed as
$$
n = px + qy,
$$
where $x, y \in \mathbf{Z}$, $0 \l... | not found | Number Theory | proof | Yes | Yes | cn_contest | false | 714,269 |
7. The number of distinct real roots of the equation $\left(x^{2}+4 x-2\right)^{2}=\left(5 x^{2}-1\right)^{2}$ is ( ).
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4 | 7.D.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,270 |
10. Casey, Bob, David are members of the same science team. Now they need to answer a true/false question. The three answer independently, with Casey and Bob each having an $80 \%$ probability of answering correctly, and David having a $50 \%$ probability of answering correctly. The team is considered to have answered ... | 10.E.
Let $A=0.8$, then the probability that all three people answer correctly is $A \times$ $A \times 0.5=0.5 A^{2}$; the probability that Casey and Bob answer correctly while David answers incorrectly is $A \times A \times 0.5=0.5 A^{2}$; the probability that Casey and David answer correctly while Bob answers incorr... | E | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 714,273 |
11. For any real number $a$, define $[a]$ as the greatest integer less than or equal to $a$, for example, $[4.9]=4,[5]=5$. If real numbers $x, y$ satisfy $[\sqrt{x}]=9,[\sqrt{y}]=12$, then the maximum possible value of $[x+y]$ is ( ).
(A) 225
(B) 242
(C) 256
(D) 268
(E) 270 | $\begin{array}{l}\text { 11. D. } \\ \text { Since }[x]=9,[y]=12 \text {, then } x<100, y<169 .[x+y] \\ <269 \text {. Also, when } x=99.5, y=168.5 \text {, the conditions are satisfied. Thus }[x+y] \\ =268 \text {. }\end{array}$ | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,274 |
12. Given that $a$ and $b$ are integers. If $a^{2}-b^{2}=2003$, then the value of $a^{2}+b^{2}$ is ( ).
(A) 2006005
(B) 2005004
(C) 2004003
(D) 2003002
(E) 2002001 | 12.A.
An intuitive method is that it is easy to see the last digit of $a^{2}+b^{2}$ can only be 5, therefore, it can only be $(\mathrm{A})$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,275 |
13. As shown in Figure 3, two perpendicular lines intersect at the center of a circle with a radius of 1, dividing the circle into four parts. Another smaller circle is tangent to one of these parts. The radius of the smaller circle is ( ).
(A) $\frac{1}{3}$
(B) $\frac{2}{5}$
(C) $\sqrt{2}-1$
(D) $\frac{1}{2}$
(E) $2-\... | 13. C.
Let the radius of the small circle be $x$, then we have $x^{2}+x^{2}=(1-x)^{2}$, which simplifies to $x^{2}+2 x-1=0$.
Solving for $x$ gives $x= \pm \sqrt{2}-1$.
Since $x>0$, we have $x=\sqrt{2}-1$. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,276 |
14. If $f(x)=(x-1)+(x-2)^{2}+\cdots+(x-10)^{10}$, then the sum of the 10 roots of $f(x)$ is ( ).
(A) 55
(B) 99
(C) 100
(D) 110
(E) 120 | 14.B.
The sum of the roots of an $n$-degree polynomial is equal to the opposite of the quotient of the coefficient of $x^{n-1}$ divided by the coefficient of $x^{n}$. In the original expression, the coefficient of $x^{10}$ is 1. Therefore, the sum of the roots is the opposite of the coefficient of $x^{9}$. In $(x-1)^{9... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,277 |
15. Given $x, y \neq 0$, and $x, y$ satisfy $xy = \frac{x}{y} = x - y$. Then the value of $x + y$ is ( ).
(A) $-\frac{3}{2}$
(B) $-\frac{1}{2}$
(C) 0
(D) $\frac{1}{2}$
(E) $\frac{3}{2}$ | 15.A.
From the left equation, we know $y=\frac{1}{y}$, then $y^{2}=1$. If $y=1$, the ancient equation becomes $x=x-1$, which is clearly not true. Therefore, $y=-1$. Thus, the right equation becomes $-x=x+1$. That is, $x=-\frac{1}{2}$. Therefore, $x+y=-\frac{3}{2}$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,278 |
16. Given that $a$ and $b$ are the two positive roots of the equation $\log _{3 x} 3+\log _{77} 3 x=-\frac{4}{3}$. Then the value of $a+b$ is ().
(A) $\frac{4}{27}$
(B) $\frac{10}{27}$
(C) $\frac{4}{81}$.
(D) $\frac{10}{81}$
(E) $\frac{28}{81}$ | 16.D.
Let $t=1+\log _{3} x$. Transform the original equation to contain only logarithms with base 3, we get
$$
\begin{array}{l}
\frac{\log _{3} 3}{\log _{3} 3+\log _{3} x}+\frac{\log _{3} 3+\log _{3} x}{\log _{3} 27}=-\frac{4}{3} \\
\Leftrightarrow \frac{1}{t}+\frac{t}{3}=-\frac{4}{3} \Leftrightarrow t^{2}+4 t+3=0 \\
... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,279 |
Example 7 Let $n, k$ be positive integers, $(k, n)=1$, and $0<k<n$. Let $M=\{1,2, \cdots, n-1\}$. Now, each $i$ in the set $M$ is painted blue or white, and satisfies:
(a) $i$ and $n-i$ must be the same color;
(b) When $i \neq k$, $i$ and $|k-i|$ must be the same color.
Prove: All numbers are the same color.
(26th IMO) | Explanation: Since $(k, n)=1$, using Bézout's identity, we can categorically prove that all $i \in M$ are the same color as $k$. During the discussion, note that from (b) we can derive two conclusions: one is that the integer multiples of $k$ in $M$ are the same color as $k$; the second is that when $i > k$, $i$ and $i... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 714,280 |
17. The equation $\cos 15 \theta=\cos 3 \theta$ has $(\quad)$ roots in $\left[0,180^{\circ}\right]$.
(A) 11
(B) 12
(C) 13
(D) 14
(E) 15 | 17. C.
Let $x=3 \theta$. Then the problem becomes finding how many $x \in[0,3 \pi]$, such that $\cos 5 x=\cos x$. This equation holds if and only if there exists an integer $k$, such that $5 x=2 k \pi \pm x$, i.e.,
$$
x=\frac{k \pi}{2} \text { or } x=\frac{k \pi}{3} \text {. }
$$
When $x$ is an integer multiple of $\... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,281 |
18. A circle with an area of 20 has its center at the fixed point as shown in Figure 4. $\triangle A B C$ is inscribed in this circle, and $S_{\triangle B B C}=8$, $\angle \alpha, \angle \beta, \angle \gamma$ are as shown in the figure. Then the value of $\sin \alpha+\sin \beta+\sin \gamma$ is ( ).
(A) $\frac{4 \pi}{5}... | 18. A.
Let the center of the circle be $O$, and the radius be $r$. From the given conditions, we easily get
$$
\begin{array}{l}
S_{\triangle A B C}=\frac{1}{2} \cdot r^{2} \sin \alpha+\frac{1}{2} r^{2} \sin \beta+\frac{1}{2} r^{2} \sin \gamma=8 \\
\Rightarrow \sin \alpha+\sin \beta+\sin \gamma=\frac{16}{r^{2}} .
\end{... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,282 |
19. Given positive real numbers $a$, $b$, $c$ satisfy the system of equations
$$
\left\{\begin{array}{l}
a+b^{2}+2 a c=29, \\
b+c^{2}+2 a b=18, \\
c+a^{2}+2 b c=25 .
\end{array}\right.
$$
Then the value of $a+b+c$ is ( )
(A)9
(B) 10
(C) 11
(D) 7
(E) 8 | 19.E.
Let $x=a+b+c$. By adding the three equations, we get $x+x^{2}=72$, which simplifies to $(x-8)(x+9)=0$. Since $a, b, c$ are all positive numbers, then $x>0$, hence $a+b+c=8$.
Note that $a=4, b=1, c=3$ satisfies this system of equations. | E | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,283 |
20. $\left(1-\frac{1}{2^{2}}\right)\left(1-\frac{1}{3^{2}}\right)\left(1-\frac{1}{4^{2}}\right) \cdots\left(1-\frac{1}{2003^{2}}\right)$'s value is ( ).
(A) $\frac{2005}{4006}$
(B) $\frac{1001}{2003}$
(C) $\frac{1}{2}$
(D) $\frac{1002}{2003}$
(E) $\frac{2007}{4006}$ | 20.D.
$$
\begin{array}{l}
\text { Original expression }=\frac{1 \times 3}{2^{2}} \times \frac{2 \times 4}{3^{2}} \times \cdots \times \frac{2001 \times 2003}{2002^{2}} \times \frac{2002 \times 2004}{2003^{2}} \\
=\frac{2004}{2 \times 2003}=\frac{1002}{2003} .
\end{array}
$$ | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,284 |
21. Given that $f(x)$ is a real function, and $x \neq 0$, and $f(x)$ satisfies $f\left(\frac{1}{x}\right)+\frac{1}{x} f(-x)=2 x$. Then the value of $f(2)$ is $(\quad$.
(A) 2.5
(B) 3
(C) 3.5
(D) 4
(E) 4.5 | 21.E.
Let $u=f(2), v=f\left(-\frac{1}{2}\right)$. Set $x=\frac{1}{2}, x=$
-2, to get $u+2 v=1, v-\frac{1}{2} u=-4$. Solving these, we get $u=\frac{9}{2}$. Therefore, $f(2)=u=\frac{9}{2}$. The function $f(x)=\frac{x^{3}+1}{x}$ satisfies these conditions. | E | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,285 |
23. Given that $x$, $y$, and $z$ are positive integers less than 10, and satisfy $(100 x+10 y+z)^{2}=(x+y+z)^{5}$. Then the value of $x^{2}+y^{2}+z^{2}$ is ( ).
(A) 21
(B) 23
(C) 29
(D) 33
(E) 37 | 23.C.
Let $N=(100 x+10 y+z)^{2}=(x+y+z)^{5}$. Since $N$ is a power of 2 and also a power of 5, it must be a power of 10. Also, because $N$ is the square of a three-digit number and
$$
\begin{array}{l}
2^{10}=\left(2^{5}\right)^{2}=32^{2}, 3^{10}=\left(3^{5}\right)^{2}=243^{2}, \\
4^{10}=\left(4^{5}\right)^{2}=1024^{2}... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 714,287 |
$24 . N$ is the smallest number among those which leave a remainder of 2 when divided by 5, a remainder of 3 when divided by 7, and a remainder of 4 when divided by 9. Then the sum of the digits of $N$ is ( ).
(A) 4
(B) 8
(C) 13
(D) 22
(E) 40 | 24.C.
Since $N$ leaves a remainder of 2 when divided by 5, we can set $N=5q+2$ ($q$ is an integer), then $2N+1=10q+5$. Therefore, 5 divides $2N+1$. Similarly, if $2N+1$ is divisible by 7 and 9, then $2N+1$ is divisible by $5 \times 7 \times 9 = 315$. Thus, $2N+1 \geqslant 315$, which means $N \geqslant 157$. Since 157... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 714,288 |
25. Among the positive integers $n$ between $1 \sim 100$, the number of $n$ for which $n^{2}+4$ and $n+3$ have a common divisor greater than 1 is ( ).
(A) 0
(B) 3
(C) 5
(D) 7
(E) 11 | 25.D.
Assume $n^{2}+4$ and $n+3$ can both be divided by $d$. Since $n^{2}+4-(n+3)(n-3)=13$, if $n^{2}+4$ and $n+3$ have a common divisor greater than 1, it must be 13. Therefore, there must exist an integer $k$ such that $n=13 k+10$ for $n+3$ to be divisible by 13. At this time,
$$
\begin{array}{l}
(13 k+10)^{2}+4=169... | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 714,289 |
26. As shown in Figure $5, A B$ and $C D$ are two perpendicular chords of the same circle, intersecting at point $E$. Given that $A E=12, D E=4$, and $C E=6$. Then the area of the circle is ( ).
(A) $50 \pi$
(B) $45 \pi$
(C) $40 \pi$
(D) $35 \pi$
(E) $30 \pi$ | 26. A.
$A E \times B E=C E \times D E$, from the given we have $B E=\frac{6 \times 4}{12}=2$.
Take the midpoint $F$ of $A B$, then $O F \perp A B$, take the midpoint $G$ of $D C$, then $O G \perp D C, A F=7, D G=5$, so $O F=G E=1$. Therefore, $O A^{2}=7^{2}+1=50$. Hence, the area of the circle is $50 \pi$. | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,290 |
Example 8 Determine all pairs of positive integers $(n, p)$, such that: $p$ is a prime, $n \leqslant 2 p$, and $(p-1)^{n}+1$ is divisible by $n^{p-1}$.
(IMO 40th) | Explanation: First find a breakthrough in solving the problem from $n$, then use Fermat's Little Theorem to solve it.
Clearly, $(1, p)$ and (2,2) satisfy the problem.
Now consider the case where $n \geqslant 2, p \geqslant 3$.
Since $(p-1)^{n}+1$ is odd, $n$ must also be odd. Therefore, $n<2 p$. Let $q$ be any prime fa... | (2,2),(3,3) \text{ and } (1, p) | Number Theory | proof | Yes | Yes | cn_contest | false | 714,291 |
27. The number of positive integers $n$ that make $3^{n}+81$ a perfect square is ( ).
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4 | 27. B.
When $n \leqslant 4$, it is easy to see that $3^{n}+81$ is not a perfect square. Therefore, assume $n=k+4$, where $k$ is a positive integer. Then
$$
3^{k}+81=81\left(3^{k}+1\right) \text {. }
$$
Since 81 is a perfect square, there must exist a positive integer $x$ such that
$$
3^{k}+1=x^{2} \Leftrightarrow(x-1... | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 714,292 |
28. As shown in Figure 6, a, b, c, $d, e, f, g, h, i$ represent one of the numbers $1,2,3,4,5,6,7,8,9$ respectively. By adding 3 numbers in 9 circles, we get 9 sums. If these 9 sums are equal, then the value of $a+d+g$ is ( ).
(A) 15
(B) 16
(C) 18
(D) 19
(E) 21 | 28.C.
Let this common sum be $s$. Adding these 9 sums, we get
$$
\begin{array}{l}
9 s=1+2+\cdots+9+2(a+b+\cdots+i) \\
=3 \times 45=9 \times 15 .
\end{array}
$$
Thus, $s=15$. The sum of the numbers in the 6 circles is
$$
\begin{array}{l}
90=6 s \\
=1+2+4+5+7+8+(a+b+\cdots+i)+a+d+g \\
=27+45+a+d+g=72+a+d+g .
\end{array... | C | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 714,293 |
$29 . n$ is the smallest positive integer such that $5^{n}+n$ is divisible by 31. Then the value of $n$ is ( ).
(A) 23
(B) 30
(C) 51
(D) 68
(E) 88 | 29. B.
Let $n=3 q+r, r \in\{0,1,2\}$. Since
$$
5^{3}-1=(5-1)\left(5^{2}+5+1\right)=4 \times 31 \text {, }
$$
there exists an integer $k$ such that
$$
\begin{array}{l}
5^{n}-5^{r}=5^{3 q+r}-5^{r}=5^{r}\left(5^{3 q}-1\right) \\
=5^{r}\left(5^{3}-1\right)\left(5^{3 q-3}+5^{3 q-6}+\cdots+5^{3}+1\right)=31 k .
\end{array}... | B | Number Theory | MCQ | Yes | Yes | cn_contest | false | 714,294 |
30. As shown in Figure 7, extend the three sides of $\triangle ABC$ such that $BD=\frac{1}{2} AB$, $CE=\frac{1}{2} BC$, and $AF=\frac{1}{2} CA$. Then the ratio of the area of $\triangle DEF$ to the area of $\triangle ABC$ is ( ).
(A) $3: 1$
(B) $13: 4$
(C) $7:2$
(D) $14: 15$
(E) $10: 3$ | 30.B.
As shown in Figure 8, draw the altitude from point $C$ to side $AB$ of $\triangle ABC$, with the foot of the altitude being point $X$. Draw the altitude from point $F$ to side $AD$ of $\triangle ADF$, with the foot of the altitude being point $Y$. Then,
$$
\begin{array}{l}
\triangle A X C \backsim \triangle A Y ... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,295 |
1. Given point $A(2,1)$, determine point $P$ on the coordinate axes such that $\triangle A O P$ is an isosceles triangle. Then the number of points that satisfy this condition is $(\quad)$.
(A) 4
(B) 6
(C) 8
(D) More than 8 | $-、 1 . C$.
(1) With $A$ as the center and $O A$ as the radius, draw $\odot A$ intersecting the $x$-axis and $y$-axis at $P_{1}$ and $P_{2}$, then we have $P_{1}(4,0), P_{2}(0,2)$.
(2) With $O$ as the center and $O A$ as the radius, draw $\odot O$ intersecting the $x$-axis and $y$-axis at $P_{3}$, $P_{4}$, $P_{5}$, and... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,296 |
2. As shown in Figure 1, point $D$ is a point on side $B C$ of $\triangle A B C$. If $\angle C A D=$
$\angle D A B=60^{\circ}, A C$
$=3, A B=6$, then
the length of $A D$ is ( ).
(A) $\frac{3}{2}$
(B) 2
(C) $\frac{4}{3}$
(D) None of the above answers is correct | 2. B.
Extend $B A$ to $E$, such that $A E=A C$. Since $\angle C 1 B=120^{\circ}$, then $\angle C A E=60^{\circ}$. Therefore, $\triangle A C E$ is an equilateral triangle. Since $C E=A C=3$, and $A D / / C E$, we have $\frac{A D}{C E}=\frac{B A}{B E}$, so $A D=\frac{3 \times 6}{9}=2$. | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,297 |
3. As shown in Figure 2, quadrilaterals $A B C D, A_{1} B_{1} B A, \cdots$, $A_{100} B_{100} B_{99} A_{99}$ are all small squares with side length 1. Given $\angle A C B=\alpha, \angle A_{1} C B_{1}=\alpha_{1}, \cdots, \angle A_{100} C B_{100}$ $=\alpha_{100}$. Then $\tan \alpha \cdot \tan \alpha_{1}+\tan \alpha_{1} \c... | 3. D.
$$
\begin{array}{l}
\tan \alpha \cdot \tan \alpha_{1}+\tan \alpha_{1} \cdot \tan \alpha_{2}+\cdots+\tan \alpha_{99} \cdot \tan \alpha_{100} \\
=\frac{A B}{B C} \cdot \frac{A_{1} B_{1}}{B_{1} C}+\frac{A_{1} B_{1}}{B_{1} C} \cdot \frac{A_{2} B_{2}}{B_{2} C}+\cdots+\frac{A_{99} B_{99}}{B_{99} C} \cdot \frac{A_{100} ... | \frac{100}{101} | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,298 |
4. The following are 4 propositions:
(1) Given $45^{\circ}<\alpha<90^{\circ}, \sin \alpha \cdot \cos \alpha=\frac{1}{4}$, then $\cos \alpha-\sin \alpha=\frac{\sqrt{2}}{2}$;
(2) Given that the diagonal $A C$ of quadrilateral $A B C D$ bisects $\angle A$ and $\angle C$, and $\angle A=\angle C$, then this quadrilateral is... | 4.C.
Since $\cos \alpha - \sin \alpha = -\frac{\sqrt{2}}{2}$, (1) is incorrect.
(2) is correct.
If point $P$ is on the arc $AD$ or arc $BC$ between $AB$ and $CD$, then $\angle APD + \angle BPC = 180^{\circ}$; if point $P$ is on arc $AB$ or arc $CD$, then $\angle APD = \angle BPC$. Hence (3) is incorrect.
Since $1 - a... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,299 |
5. Given that $\sin \alpha, \cos \alpha$ are the roots of the equation $x^{2}-\sqrt{m} x+$ $\frac{1}{2}=0$. Then the triangle with side lengths $\sin \alpha, \cos \alpha, \frac{1}{2} m$ is ( ).
(A) scalene triangle
(B) isosceles triangle
(C) isosceles right triangle
(D) none of the above answers is correct | 5.C.
From the given, we have $\sin \alpha + \cos \alpha = \sqrt{m}, \sin \alpha \cdot \cos \alpha = \frac{1}{2}$. Solving this, we get $m=2$.
Also, $\Delta = (-\sqrt{m})^2 - 4 \times \frac{1}{2} = m - 2 = 0$, so $\sin \alpha = \cos \alpha$.
Since $\sin^2 \alpha + \cos^2 \alpha = 1 = \left(\frac{1}{2} m\right)^2$, the ... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,300 |
6. Given $a+b+c=3, a^{2}+b^{2}+c^{2}=3$. Then the value of $a^{2003}+b^{2003}+c^{2003}$ is ( ).
(A) 0
(B) 3
(C) 6
(D) $3^{2003}$ | 6.B.
$$
\begin{array}{l}
(a-1)^{2}+(b-1)^{2}+(c-1)^{2} \\
=\left(a^{2}+b^{2}+c^{2}\right)-2(a+b+c)+3=0,
\end{array}
$$
Then $a=1, b=1, c=1$. Therefore, $a^{2 \alpha B}+b^{2 \alpha B}+c^{2 \alpha B}=3$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,301 |
2. Given $p>0, q>0$, and satisfy $2 p+\sqrt{p q}-$ $q+\sqrt{p}+\sqrt{q}=0$. Then $(2 \sqrt{p}-\sqrt{q}+2)^{3}=$ $\qquad$ | 2. 1 .
The original expression can be transformed into
$$
2(\sqrt{p})^{2}+\sqrt{p q}-(\sqrt{q})^{2}+(\sqrt{p}+\sqrt{q})=0 \text {, }
$$
which is $(\sqrt{p}+\sqrt{q})(2 \sqrt{p}-\sqrt{q}+1)=0$.
Since $p>0, q>0$, then $\sqrt{p}+\sqrt{q}>0$. Therefore, $2 \sqrt{p}-\sqrt{q}+1=0$, which means $2 \sqrt{p}-\sqrt{q}+2=1$.
Th... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,304 |
3. Given that the three sides of a triangle $a, b, c$ are unequal integers, and satisfy
$$
a b c+a b+b c+a c+a+b+c=119 .
$$
Then this triangle is a $\qquad$ triangle. | 3. Right Angle.
The original equation can be transformed into $(a+1)(b+1)(c+1)=120$.
From the given information, $a+1$, $b+1$, and $c+1$ are also positive integers.
Assume $a < c$; the triangle can still exist. When $a+1$ takes the values $1, 2, 3$, $b$ and $c$ take positive integers that satisfy the conditions, but t... | a=3, b=4, c=5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,305 |
4. The graph of the quadratic function $y=$ $a x^{2}+b x+c$ is shown in Figure 3. Also, $k=$ $\frac{b(2 a+b)(a+b+c)}{c(2 a-b)(a-b+c)}$, then the sign of $k$ is $\qquad$ | $4 . k>0$.
From Figure 3, we can see that the intersection point with the $y$-axis $(0, c)$ is below the $x$-axis, so $c<0$;
When $x=-1$, $y=a-b+c<0$. | k>0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,306 |
One, (20 points) A research study group from Class 1, Grade 3 of Yuhong Middle School conducted a survey on students' lunch time at the school canteen. It was found that within a unit of time, the number of people buying lunch at each window and the number of people choosing to eat outside due to unwillingness to wait ... | Let each window sell lunch to $x$ people per minute, and $y$ people go out to eat per minute, with the total number of students being $z$ people, and assume that at least $n$ windows need to be open simultaneously. According to the problem, we have
$$
\left\{\begin{array}{l}
45 x=z-45 y, \\
2 \times 30 x=z-30 y, \\
20 ... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,307 |
II. (25 points) Given that $\odot O_{1}$ and $\odot O_{2}$ are externally tangent at point $P, A P$ and $B P$ are diameters of $\odot O_{1}$ and $\odot O_{2}$ respectively, the external common tangent $C D$ touches $\odot O_{1}$ at point $C$ and $\odot O_{2}$ at point $D, B E$ is parallel to $C D$ and intersects $C A$ ... | As shown in Figure 6, extend $AC$ and $BD$ to intersect at $Q$, and connect $PC$, $PD$, and $PQ$. Since $CD$ is the external common tangent and $FP$ is the internal common tangent, we have
$$
FP = FC = FD.
$$
Thus, $\angle CPD = 90^{\circ}$. Since $AP$ and $BP$ are diameters, $\angle QCP$ and $\angle QDP$ are both $90... | FG = \frac{R_{2}}{R_{1}} \sqrt{R_{1}R_{2}} | Geometry | proof | Yes | Yes | cn_contest | false | 714,308 |
Three. (25 points) Given the system of equations about $x$ and $y$:
$$
\left\{\begin{array}{l}
k x^{2} - a y + k - 3 b c = 0, \\
y = (k-3) x
\end{array}\right.
$$
has only one integer solution. Here, $k, a, b, c$ are all integers, and $a > 0$, $a, b, c$ satisfy $a^{2} - a - b c = -1$, $b + c = 2$. Find:
(1) The value ... | (1) From $b+c=2, bc=a^{2}-a+1$ we know that $b, c$ are the two real roots of the equation $x^{2}-2x+a^{2}-a+1=0$, then we have
$$
\Delta=(-2)^{2}-4\left(a^{2}-a+1\right) \geqslant 0 \text {. }
$$
Therefore, $a^{2}-a \leqslant 0$.
Since $a>0$, then $a \leqslant 1$. Also, $a$ is a positive integer, so, $a=1$.
(2) When $... | a=1, k=-1, (x, y)=(2, -8); k=0, (x, y)=(1, -3); k=2, (x, y)=(-1, 1); k=3, (x, y)=(0, 0) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,309 |
1. Given $\omega$ is a positive real number, $f(x)=4 \sin \frac{\omega x}{2} \cos \frac{\omega x}{2}$ is increasing on $\left[-\frac{\pi}{3}, \frac{\pi}{4}\right]$. Then the range of $\omega$ is ( ).
(A) $\left(0, \frac{2}{3}\right]$
(B) $\left(0, \frac{3}{2}\right]$
(C) $\left[\frac{3}{2},+\infty\right)$
(D) $[1,+\inf... | $-1 . B$.
$f(x)=2 \sin \omega x, f(x)$ has a monotonically increasing interval $\left[-\frac{\pi}{2 \omega}, \frac{\pi}{2 \omega}\right]$. Given that $\left[-\frac{\pi}{3}, \frac{\pi}{4}\right] \subseteq\left[-\frac{\pi}{2 \omega}, \frac{\pi}{2 \omega}\right]$. Therefore, $\frac{\pi}{2 \omega} \geqslant \frac{\pi}{3}$.... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,310 |
2. Given the curve $y^{2}=a x$ and its curve symmetrical about the point $(1,1)$ intersect at two different points. If the line passing through these two points has an inclination angle of $45^{\circ}$, then the value of $a$ is ( ).
(A) 1
(B) $\frac{\sqrt{2}}{2}$
(C) 2
(D) $\pm 2$ | 2.C.
Let $y^{2}=a x$ be the curve symmetric to point $(1,1)$. For any point $(x, y)$ on this curve, $\left(x^{\prime}, y^{\prime}\right)$ is the symmetric point of $(x, y)$ about point $(1,1)$. Then,
$$
y^{\prime 2}=a x^{\prime} \text {. }
$$
Also, $x+x^{\prime}=2$ and $y+y^{\prime}=2$, so $x^{\prime}=2-x$ and $y^{\p... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,311 |
3. In a transparent sealed cubic container, there is exactly half the volume of the container filled with water. When the cubic container is rotated arbitrarily, the shape of the water surface in the container can be:
(1) triangle, (2) rhombus, (3) rectangle, (4) square, (5) regular hexagon.
The correct conclusion is ... | 3.C.
The problem is transformed into determining the cross-sectional shape of a cube, and this cross-section divides the cube's volume equally. Clearly, the sections that meet the requirements are (3) and (4). The cross-section is a rhombus, as shown in Figure 1; the cross-section is a regular hexagon, as shown in Fig... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 714,312 |
2. The sequence $f_{n+1}(x)=f_{1}\left(f_{n}(x)\right)$, where $f_{1}(x)=2 x+$ $1, n \in \mathbf{N}$. Prove: for any $n \in\{11,12,13, \cdots\}$, there must exist a unique $m_{0} \in\{0,1, \cdots, 1993\}$, such that $1995 \mid f_{n}\left(m_{0}\right)$. | (Given: $f_{n}(m)=2^{n}(m+1)-1$. From $\left(2^{n}, 1995\right)=1$, we know that there exists $1 \leqslant u_{0} \leqslant 1994, u_{0} \in \mathbf{Z}$, such that $2^{n} u_{0}+1995 k_{1}$ $=1, k_{1} \in \mathbf{Z}$. Therefore, 1995 divides $\left(2^{n} u_{0}-1\right)$. Taking $m_{0}=u_{0}-1$, then $1995 \mid f_{n}\left(... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 714,313 |
5. Given the planar region $D$ consists of all points on the boundary and inside of $\triangle A B C$, and $A(1,5)$, $B(1,1)$, $C(4,3)$. For the function $w=m x+y(m>0)$, if there are infinitely many points $(x, y) \in D$ such that $w$ attains its maximum value, then the value of $m$ is ( ).
(A) $\frac{3}{2}$
(B) $\frac... | 5.B.
From $w=m x+y$ we get
$$
y=-m x+w \text {. }
$$
Equation (1) represents a family of parallel lines with a slope of $-m$, and $w$ is the intercept on the $y$-axis. According to the problem, when $y=-m x+u$ coincides with line $A C$, there are infinitely many points that make $u$ reach its maximum value. Therefore... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,315 |
6. If the equation $\frac{4\left(x^{2}-y^{2}-\frac{1}{2}\right)^{2}}{49}+\frac{x^{2} y^{2}}{3}=1$ represents a non-circular quadratic curve, then it represents ( ).
(A) Ellipse
(B) Hyperbola
(C) Parabola with focus at $(7,0)$
(D) Parabola with focus at $(\sqrt{3}, 0)$ | 6. A.
In the equation, replacing $x$ with $-x$ leaves the equation unchanged, indicating that the $y$-axis is the axis of symmetry; replacing $y$ with $-y$ also leaves the equation unchanged, indicating that the $x$-axis is also an axis of symmetry. From this, we can infer that it represents an ellipse or a hyperbola,... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 714,316 |
1. Given $\alpha, \beta \in\left(0, \frac{\pi}{2}\right)$, and
$$
\sin \beta=2 \cos (\alpha+\beta) \cdot \sin \alpha\left(\alpha+\beta \neq \frac{\pi}{2}\right) \text {. }
$$
then the maximum value of $\tan \beta$ is | $$
\begin{array}{l}
\text { II, 1. } \frac{\sqrt{3}}{3} . \\
\sin [(\alpha+\beta)-\alpha]=2 \cos (\alpha+\beta) \cdot \sin \alpha, \\
\sin (\alpha+\beta) \cdot \cos \alpha-\cos (\alpha+\beta) \cdot \sin \alpha \\
=2 \cos (\alpha+\beta) \cdot \sin \alpha, \\
\sin (\alpha+\beta) \cdot \cos \alpha=3 \cos (\alpha+\beta) \c... | \frac{\sqrt{3}}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,317 |
2. In a tetrahedron $ABCD$ with volume 12, points $E$, $F$, and $G$ are on edges $AB$, $BC$, and $AD$ respectively, such that $AE = 2EB$, $BF = FC$, and $AG = 2GD$. A plane through points $E$, $F$, and $G$ intersects the tetrahedron in a section $EFGH$, and the distance from point $C$ to this section is 1. The area of ... | 2. 7 .
As shown in Figure 3, it is easy to know that $G E / /$ $D B$. Since $D B / /$ plane $E F H G$, then $H F$ // BD. Therefore, $H$ is also the midpoint of $D C$.
Let the distances from points $A, B, C, D$ to the section $E F H G$ be $h_{a}, h_{b}, h_{c}, h_{d}$, respectively. We have
$$
\begin{array}{l}
h_{b}=h_... | 7 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 714,318 |
3. One focus of a hyperbola is $F(0,4)$, the length of the real axis is 6, and point $A(0,3)$ lies on this hyperbola. Then the maximum value of the eccentricity of this hyperbola is $\qquad$ | 3. $\frac{4}{3}$.
Let the center of the hyperbola be $O^{\prime}(x, y)$, and the other focus be $F^{\prime}(x_{0}, y_{0})$. Since $O^{\prime}$ is the midpoint of $F F^{\prime}$, then $x_{0}=2 x, y_{0}=2 y-4$.
Since $A(0,3)$ is on the hyperbola, then
$||A F^{\prime}|-| A F||=2 a=6$,
i.e., $\left|\sqrt{(2 x)^{2}+(2 y-7... | \frac{4}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,319 |
4. Let $f(x)=a x^{2}+b x+c$ be a quadratic trinomial with integer coefficients. If integers $m, n$ satisfy $f(m)-f(n)=1$. Then $|m-n|=$ $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 4. 1 .
$$
\begin{array}{l}
f(m)-f(n)=a\left(m^{2}-n^{2}\right)+b(m-n) \\
=(m-n)[a(m+n)+b],
\end{array}
$$
then $(m-n)(a m+a n+b)=1$.
Since $m-n$ and $a m+a n+b$ are both integers, it must be that
$$
|m-n|=1 \text{. }
$$ | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 714,320 |
5. A non-empty set $S$ satisfies:
(1) $S \subseteq\{1,2, \cdots, 2 n+1\}, n \in \mathbf{N}_{+}$;
(2) If $a \in S$, then $(2 n+2-a) \in S$.
Then, the number of non-empty sets $S$ that satisfy (1) and (2) is $\qquad$ | 5. $2^{n+1}-1$.
Pair the natural numbers $1,2, \cdots, 2 n+1$ into $n+1$ pairs:
$$
\{1,2 n+1\},\{2,2 n\}, \cdots,\{n, n+2\},\{n+1\} \text {. }
$$
The elements of $S$ are chosen from the above $n+1$ pairs, and there are
$$
C_{n+1}^{1}+C_{n+1}^{2}+\cdots+C_{n+1}^{n+1}=2^{n+1}-1
$$
ways to choose. | 2^{n+1}-1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 714,321 |
6. The sequence $\left\{a_{n}\right\}$ is a monotonically increasing sequence, and $a_{n}=$ $2^{n-1}-3 a_{n-1}, n=1,2, \cdots$. Then the first term $a_{0}$ is equal to $\qquad$. | 6. $\frac{1}{5}$.
From the given, we have $a_{n}-\frac{2^{n}}{5}=-3\left(a_{n-1}-\frac{2^{n-1}}{5}\right)$, hence $a_{n}-\frac{2^{n}}{5}=(-3)^{n}\left(a_{0}-\frac{1}{5}\right)$.
$a_{n}-a_{n-1}=\frac{2^{n-1}}{5}\left[1+4(-1)^{n-1}\left(\frac{3}{2}\right)^{n-1}\left(1-5 a_{0}\right)\right]$.
If $1-5 a_{0}>0$, then for s... | \frac{1}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,322 |
Three, (20 points) For the complex numbers $z_{1}$ and $z_{2}$, the following three conditions are given:
(1) $\frac{z_{1}-\overline{z_{1}}}{z_{2}-\overline{z_{2}}}=0$;
(2) $\overline{z_{2}}+6=\frac{2}{z_{2}+6}$;
(3) $z_{1} \cdot z_{2}^{2}+z_{2}+2=0$.
Does there exist complex numbers $z_{1}$ and $z_{2}$ that simultaneo... | Three, assume that $z_{1}$ and $z_{2}$ simultaneously satisfy the three conditions given in the problem. From (1), we get $z_{1} \in \mathbf{R}, z_{2}$ is an imaginary number. From (2), we get
$$
\left|z_{2}+6\right|^{2}=2 \text {. }
$$
Let $z_{1}=a \in \mathbf{R}$. From (3), we get $a z_{2}^{2}+z_{2}+2=0$.
Since $z_{... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 714,323 |
3. Let $m>n>1$. It is known that there are some balls in $2^{m}-1$ boxes. If one ball is added to $2^{n}-1$ of these boxes, it is called one operation. Prove: when $(m, n)=1$, regardless of how many balls are already in the $2^{m}-1$ boxes, after a finite number of operations, it is always possible to make the number o... | (Given: $(m, n)=1$, there exist positive integers $a, b$, such that $n a$ $m b=1$. Let $x=2^{m}-1, y=2^{n}-1$, then $2^{\mathrm{m}}=(x+1)^{b}$, $2^{a n}=(y+1)^{a}$. Thus, $(y+1)^{a}=2(x+1)^{b}$. By the binomial theorem expansion, we know that there exist positive integers $M, N$, such that $y M=x N+1$, i.e., $\left(2^{... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 714,324 |
Four. (20 points) Given a right triangle inscribed in the parabola $y=x^{2}$. Prove: the length of the hypotenuse of the right triangle is not less than 2. | Let the endpoints of the hypotenuse be \( A\left(x_{1}, x_{1}^{2}\right) \) and \( B\left(x_{2}, x_{2}^{2}\right) \), and the right-angle vertex be \( C\left(x_{0}, x_{0}^{2}\right) \). The slopes of the lines \( AC \) and \( BC \) are respectively
\[
k_{A C}=\frac{x_{1}^{2}-x_{0}^{2}}{x_{1}-x_{0}}=x_{1}+x_{0}, \quad k... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,325 |
Five, (20 points) Prove: There exist infinitely many positive irrational number pairs \((a, b)\), such that \(a^{b}\) is a rational number. | Let $a=\sqrt{4 n^{2}+1}, b=\log _{\sqrt{4 n^{2}+1}} 2, n \in \mathbf{N}$ and $n>1$.
Obviously, $a$ is a positive irrational number, and it can be proven that $b$ is also a positive irrational number, while $a^{b}=2$ is a rational number. Since there are infinitely many $n \in \mathbf{N}$ and $n>1$ that satisfy the cond... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 714,326 |
One, (50 points) Prove: If the ratio of two sides of a triangle is a constant (not equal to 1), then, if and only if this triangle is a right triangle with these two sides as the legs, the angle between the median and the altitude to the third side is minimized. | As shown in Figure 4, in $\triangle ABC$, $\frac{CA}{CB}=k$ (a constant $k \neq 1$), and $E$ is the midpoint of side $AB$. The original problem is to prove that $\angle CEB$ is maximized if and only if $\angle ACB=90^{\circ}$.
Let points $M$ and $N$ be the internal and external division points of $AB$ in the ratio $k$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 714,327 |
ニ、(50 points) $f(n)$ is defined on the set of positive integers, and satisfies $f(1)=2, f(n+1)=(f(n))^{2}-f(n)+$ $1, n=1,2, \cdots$. Prove that for all integers $n>1$, we have
$$
1-\frac{1}{2^{2^{n-1}}}<\sum_{i=1}^{n} \frac{1}{f(i)}<1-\frac{1}{2^{2^{2}}} \text {. }
$$ | From the problem, it is clear that $f(n) \geqslant 2$.
Transform $f(n+1)=(f(n))^{2}-f(n)+1$ into $f(n+1)-1=f(n)[f(n)-1]$,
then $\frac{1}{f(n+1)-1}=\frac{1}{f(n)-1}-\frac{1}{f(n)}$.
Thus, $\sum_{i=1}^{n} \frac{1}{f(i)}=\sum_{i=1}^{n}\left(\frac{1}{f(i)-1}-\frac{1}{f(i+1)-1}\right)$
$=\frac{1}{f(1)-1}-\frac{1}{f(n+1)-1}=... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 714,328 |
Three. (50 points) A finite set $S$ of integer points in the plane is called a bi-neighbor set if for each point $(p, q)$ in $S$, exactly two of the points $(p+1, q)$, $(p, q+1)$, $(p-1, q)$, and $(p, q-1)$ are in $S$. For which positive integers $n$ does there exist a bi-neighbor set that contains exactly $n$ integer ... | Three, first prove the conclusion:
A double-neighbor set $S$ contains exactly $n$ integer points, then $n$ must be even. Two integer points with one coordinate the same and the other differing by 1 are called adjacent integer points. Connect the adjacent integer points in $S$ with line segments. Since each integer poin... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 714,329 |
Given that $ABCD$ is a unit square, $E$ and $F$ are the midpoints of sides $CD$ and $BC$ respectively, $AE$ intersects the diagonal $BD$ at point $P$, and $AF$ intersects $BD$ and $BE$ at points $Q$ and $R$ respectively.
(1) Prove that $AB$, $PR$, and $EF$ are concurrent;
(2) Find the area of quadrilateral $PQRE$.
---... | Solution: (1) As shown in Figure 2, let the extension of $EF$ intersect $AB$ at point $T$. Since $F$ is the midpoint of $BC$,
we have $EF = FT, AB = 2CE = 2BT$.
Since $\triangle PED \sim \triangle PAB$, then
$$
\frac{PE}{PA} = \frac{DE}{AB} = \frac{1}{2}.
$$
Since $\frac{PE}{AP} \cdot \frac{AB}{\overline{BT}} \cdot \... | \frac{1}{8} | Geometry | proof | Yes | Yes | cn_contest | false | 714,330 |
Given the set $M=|A|$ where $A$ is a ten-digit positive integer with all distinct digits, and $11111|A|$. Find $|M|$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Solution: Since the digits of $A$ are all different, we have
$A \equiv 0+1+\cdots+9 \equiv 0(\bmod 9)$, i.e., $9 \mid A$.
Also, $11111 \mid A$, and since $(9,11111)=1$, it follows that 99 999. .
Let $A=99999 A_{0}, A_{0} \in \mathbf{Z}_{+}$.
Since $10^{9}\frac{10^{9}}{10^{5}}=10^{4}, 10^{5}+10, a_{i} \in\{0,1, \cdots, ... | 3456 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 714,331 |
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