problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
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values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
6. If convex hexagons are classified according to the number of axes of symmetry, then hexagons can be divided into ( ) categories.
(A) 3
(B) 4
(C) 5
(D) 7 | 6.C.
It is easy to find that hexagons with 0, 1, and 6 axes of symmetry exist. Now let's discuss the cases of 2, 3, 4, and 5. As shown in Figure 3(a), when lines $a$ and $b$ are both axes of symmetry of the hexagon, the hexagon must be a regular hexagon. Therefore, the number of axes of symmetry of a hexagon cannot be... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,256 |
2. Given the equation in $x$, $x^{4}+k x^{2}+4=0$, has four distinct real roots. Then the range of values for $k$ is $\qquad$ . | $$
\left\{
\begin{array}{l}
k0, \\
y_{1}+y_{2}=-k>0, \\
y_{1} y_{2}=4>0 .
\end{array}\right.
$$
Solving yields $k<-4$. | k<-4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,258 |
3. A person statistically calculated the proportions of adults in Town A who have received primary education or below, secondary education, and higher education, with precision to $1 \%$ (using the round half up method). When he added up these three calculated percentages, the result was actually $101 \%$. If the stati... | 3. For example, $19650$, $9480$, $870$ (answers are not unique).
When rounding to $1 \%$ using the round half up method, the difference between the approximate value and the true value of a number is within $(-0.5 \%, 0.5 \%]$. The difference between the approximate value and the true value of the sum of three numbers... | 19650, 9480, 870 | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 716,259 |
4. Given that $AB$ is the diameter of $\odot O$, $C$ is a point on $\odot O$, and the tangent line through point $C$ intersects line $AB$ at point $D$. Let the radius of $\odot O$ be $r$. When $\triangle ACD$ is an isosceles triangle, its area is $\qquad$ . | 4. $\frac{\sqrt{3}}{4} r^{2}$ or $\frac{3 \sqrt{3}}{4} r^{2}$.
As shown in Figure 4(a), when point $D$ is on the extension of $BA$, since $\angle CAB$ is an acute angle, we have $AD = AC$. Therefore, $\angle OCA = \angle OAC = 2 \angle D = 2 \angle ACD = 60^{\circ}$. It is easy to see that $S_{\triangle ACD} = \frac{\... | \frac{\sqrt{3}}{4} r^{2} \text{ or } \frac{3 \sqrt{3}}{4} r^{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,260 |
One. (20 points) Two cars, A and B, start from the same point $A$ at the same time, driving in the same direction in a straight line. Each car can carry a maximum of $240 \mathrm{~L}$ of gasoline, and they cannot refuel during the journey. Each liter of gasoline allows a car to travel $12 \mathrm{~km}$. Both cars must ... | Let the car A travel $x \mathrm{~km}$ and car B travel $y \mathrm{~km}$, as far away from the starting point $A$ as possible. Then
$$
\left\{\begin{array}{l}
x+y \leqslant 240 \times 12 \times 2, \\
x-y \leqslant 240 \times 12 .
\end{array}\right.
$$
and $x=\frac{1}{2}(x+y)+\frac{1}{2}(x-y) \leqslant 4320$, which mean... | 4320 \mathrm{~km} | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 716,261 |
II. (25 points) If the quadratic trinomial $a x^{2} + b x + c$ (where $a$, $b$, and $c$ are integers, and $a b c \neq 0$) can be factored into the product of two linear terms with integer coefficients.
(1) Prove that $b^{2} - 4 a c$ is a perfect square.
(2) Considering the parity of $a$, $b$, and $c$, which cases exis... | (1) Let $a x^{2}+b x+c=(m x+n)(p x+q)$, where $m$, $n$, $p$, $q$ are all integers. Then $a=m p$, $b=m q+n p$, $c=n q$.
Therefore, $b^{2}-4 a c=(m q+n p)^{2}-4 m p n q=(m q-n p)^{2}$.
(2) The case where $a$, $b$, $c$ are all odd numbers does not exist.
Because when $a$ and $c$ are odd, $m$, $n$, $p$, $q$ are all odd, an... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 716,262 |
Three. (25 points) Given that $P(2,3)$ is a point on the graph of the inverse proportion function $y=\frac{k}{x}$.
(1) Find the equation of the line passing through point $P$ and having only one common point with the hyperbola $y=\frac{k}{x}$.
(2) $Q$ is a moving point on the branch of the hyperbola $y=\frac{k}{x}$ in ... | (1) From the given conditions, it is easy to obtain the equation of the hyperbola as $y=\frac{6}{x}$. Clearly, the lines $x=2$ and $y=3$ meet the requirements. Let the equation of the third line be $y=a(x-2)+3$.
Then $\left\{\begin{array}{l}y=a(x-2)+3, \\ y=\frac{6}{x}\end{array}\right.$, which has only one solution.
E... | A D / / B C | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,263 |
Example 7 Let $A M$ and $A N$ be the median and angle bisector of $\triangle A B C$, respectively. Draw a perpendicular line through point $N$ to $A N$, intersecting $A M$ and $A B$ at points $Q$ and $P$, respectively. Draw a perpendicular line through point $P$ to $A B$, intersecting $A N$ at point $O$. Prove that $O ... | Proof: As shown in Figure 7, let \( A(0, a) \), and define the line \( l_{1 B} \):
\[
y = k_{1} x + a,
\]
the line \( l_{B C} \):
\[
y = k_{2} x.
\]
Then the line \( l_{A C} \):
\[
y = -k_{1} x + a.
\]
Since \( O P \perp A B \), it is easy to get
the line \( l_{O P}: k_{1} x + k_{1}^{2} y + a = 0 \).
Let \( x = 0 \)... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,264 |
1. Given $x^{2}=y^{2}+4 y+20$. Then the number of all integer pairs satisfying the equation is ( ).
(A) 4
(B) 6
(C) 8
(D) infinitely many | - 1. B.
From the given equation, we have $x^{2}-\left(y^{2}+4 y+4\right)=16$, which simplifies to $(x+y+2)(x-y-2)=16$.
Since $x+y+2$ and $x-y-2$ have the same parity, we have the system of equations
$\left\{\begin{array}{l}x+y+2= \pm 2, \\ x-y-2= \pm 8\end{array}\right.$ or $\left\{\begin{array}{l}x+y+2= \pm 8, \\ x-y... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,265 |
2. The function $f(x)$ defined on the set of positive integers satisfies
$$
f(x)=\left\{\begin{array}{ll}
f(f(x-7)), & x>1005 ; \\
x+3, & x \leqslant 1005 .
\end{array}\right.
$$
Then the value of $f(2006)$ is ( ).
(A) 1005
(B) 1006
(C) 1007
(D) 1008 | 2.A.
Since $2006=1005+143 \times 7$, we have
$$
f(2006)=f_{2}(1999)=\cdots=f_{144}(1005) \text {. }
$$
Also, $f(1005)=1008$,
$$
\begin{array}{l}
f(1008)=f_{2}(1001)=f(1004)=1007, \\
f(1007)=f_{2}(1000)=f(1003)=1006, \\
f(1006)=f_{2}(999)=f(1002)=1005,
\end{array}
$$
Therefore, $f_{m}(1005)$ has a period of 4.
Since ... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,266 |
3. If $y=\sqrt{3} \sin \frac{x}{4} \cdot\left(1+\cos \frac{x}{2}\right)(0 \leqslant x \leqslant 2 \pi)$, then the maximum value of $y$ is ( ).
(A) $2 \sqrt{3}$
(B) $\frac{4}{3}$
(C) $\frac{4 \sqrt{3}}{9}$
(D) $\frac{\sqrt{3}}{2}$ | 3. B.
Since $0 \leqslant x \leqslant 2 \pi$, i.e., $0 \leqslant \frac{x}{4} \leqslant \frac{\pi}{2}$, hence
Given $1-\cos ^{2} \frac{x}{4}=\frac{1}{2} \cos ^{2} \frac{x}{4}$, i.e., $3 \cos ^{2} \frac{x}{4}=2$, and $\frac{x}{4} \in\left[0, \frac{\pi}{2}\right]$, then $\cos \frac{x}{4}=\frac{\sqrt{6}}{3}$.
Thus, $\frac... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,267 |
4. $B$ is the intersection point of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ with the positive $y$-axis, and point $P$ moves on the ellipse. Then the maximum value of $|B P|$ is ( ).
(A) $2 b$
(B) $\frac{a^{2}}{c}$
(C) $2 b$ or $\frac{b^{2}}{c}$
(D) $2 b$ or $\frac{a^{2}}{c}$ | 4.D.
Let $P(a \cos \alpha, b \sin \alpha)$. Since $B(0, b)$, we have
$$
\begin{array}{l}
|B P|^{2}=a^{2} \cos ^{2} \alpha+(b \sin \alpha-b)^{2} \\
=\left(b^{2}-a^{2}\right)\left(\sin \alpha-\frac{b^{2}}{b^{2}-a^{2}}\right)^{2}+\frac{a^{4}}{c^{2}} .
\end{array}
$$
When $\left|\frac{b^{2}}{b^{2}-a^{2}}\right| \leqslant... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,268 |
5. In the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, $E$ and $F$ are the midpoints of edges $A A_{1}$ and $B B_{1}$, respectively, and $G$ is a point on $B C$. If $D_{1} F \perp F G$, then the angle between line $C_{1} F$ and line $E G$ is $(\quad)$.
(A) $\frac{\pi}{3}$
(B) $\frac{\pi}{2}$
(C) $\frac{5 \pi}{6}$
(D) $\frac... | 5.B.
As shown in Figure 2, establish a coordinate system, and let the side length of the cube be 1. Then $E\left(1,0, \frac{1}{2}\right), F(1, 1, \frac{1}{2}), G(m, 1,0)$,
$$
\begin{array}{l}
D_{1}(0,0,1), C_{1}(0,1,1), \\
\quad E G=\left(m-1,1,-\frac{1}{2}\right) .
\end{array}
$$
From $D_{1} F \perp F G$, we get $D_... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,269 |
6. Among the seven-digit numbers formed by the digits $0,1,2,3,4,5,6$, the number of permutations that do not contain "246" and "15" is ( ).
(A) 3606
(B) 3624
(C) 3642
(D) 4362 | 6.C.
Since the total number of permutations of the 7 digits $0,1, \cdots, 6$ into a seven-digit number with the first digit not being zero is $\mathrm{A}_{7}^{7}-\mathrm{A}_{6}^{6}=4320$, but among these, the number where “246” are together is $\mathrm{A}_{4}^{4} \mathrm{C}_{4}^{1}=96$, and the number where “15” are t... | 3642 | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 716,270 |
1. Given a sequence of non-negative numbers $\left\{a_{n}\right\}$ with the first term being $a$, the sum of the first $n$ terms is $S_{n}$, and $S_{n}=\left(\sqrt{S_{n-1}}+\sqrt{a}\right)^{2}$. If $b_{n}=$ $\frac{a_{n+1}}{a_{n}}+\frac{a_{n}}{a_{n+1}}$, the sum of the first $n$ terms of the sequence $\left\{b_{n}\right... | $$
=1 \cdot \frac{4 n^{2}+6 n}{2 n+1} \text {. }
$$
From $S_{n}=\left(\sqrt{S_{n-1}}+\sqrt{a}\right)^{2}$, we get
$$
\sqrt{S_{n}}=\sqrt{S_{n-1}}+\sqrt{a} \text {. }
$$
Also, $\sqrt{S_{1}}=\sqrt{a_{1}}=\sqrt{a}$, then $\left\{\sqrt{S_{n}} \mid\right.$ is an arithmetic sequence with the first term $\sqrt{a}$ and common... | \frac{4 n^{2}+6 n}{2 n+1} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,271 |
2. The number of non-negative integer solutions to the equation $x_{1}+x_{2}+\cdots+x_{99}+2 x_{100}=3$ is $\qquad$ . | 2. 166749 .
Classify by $x_{100}$:
(1) When $x_{100}=1$, there are 99 non-negative integer solutions;
(2) When $x_{100}=0$, there are
$$
\mathrm{C}_{99}^{1}+\mathrm{A}_{99}^{2}+\mathrm{C}_{99}^{3}=166 \text { 650 (solutions). }
$$
Combining (1) and (2), the total number of non-negative integer solutions is
$$
99+1666... | 166749 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,272 |
3. If $n \in \mathbf{R}$, and the equation $\sqrt{x^{2}-9}=-x-n$ has one solution, then the range of values for $n$ is $\qquad$ . | 3. $n \leqslant-3$ or $0<n \leqslant 3$.
$$
\text { Let } y=\sqrt{x^{2}-9} \text {. }
$$
Then $y \geqslant 0 . x^{2}-y^{2}=9$ is the part of the hyperbola above and on the $x$-axis (as shown in Figure 3).
$$
\text { Let } y=-x-n \text {. }
$$
This is a line.
From the original, we find one
point $1(-3,0)$ to
$t_{1}$ i... | n \leqslant -3 \text{ or } 0 < n \leqslant 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,273 |
4. Given the function
$$
y=f(x)=\log _{a+2}\left[a x^{2}+(a+2) x+a+2\right] \text {. }
$$
If $f(x)$ has a maximum or minimum value, then the range of values for $a$ is . $\qquad$ | $$
\begin{array}{l}
4 .(-2,-1) \cup(-1.0) \cup\left(\frac{2}{3},+\infty\right) \text {. } \\
\text { Let } a(x)=a x^{2}+(a+2) x+a+2 \text {. } \\
\end{array}
$$
Considering the possible values, we find that the solution is $a>\frac{2}{3}$.
Solving, we get $-2<a<0$. However, for $-1<a<0$, we have $1<a+2<2$, so $y=f(x)$... | (-2,-1) \cup(-1,0) \cup\left(\frac{2}{3},+\infty\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,274 |
For example, there are $4 n(n \geqslant 4)$ plates, and the total number of candies in the plates is no less than 4. Select any two plates and take 1 candy from each, then put these candies into another plate, which is called one operation. Can all the candies be concentrated into one plate after a finite number of ope... | Explanation: The goal of this problem is to concentrate all the candies in one plate through a finite number of operations. The solution can be designed as follows:
(1) First, try to reduce the number of plates with candies. As long as it can be reduced, find a way to concentrate the candies in two plates;
(2) Make the... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 716,275 |
Example 8 As shown in Figure 11, in quadrilateral $A B C D$,
$$
\begin{array}{l}
\angle C A B=30^{\circ}, \\
\angle A B D=26^{\circ}, \\
\angle A C D=13^{\circ}, \\
\angle D B C=51^{\circ} .
\end{array}
$$
Find the measure of $\angle A D B$. | Solution 1: Let $\angle A D B=x$, then,
$$
\angle A D C=x+43^{\circ} \text {. }
$$
Since $\angle C A B=30^{\circ}, \angle A B D=26^{\circ}$,
$$
\angle D B C=51^{\circ}, \angle A C D=13^{\circ} \text {, }
$$
then $\angle A C B=73^{\circ}, \angle B D C=43^{\circ}$.
Applying the trigonometric form of Ceva's Theorem to $... | 107^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,277 |
4. As shown in Figure 3, in $\triangle A B C$, a semicircle is drawn with side $B C$ as the diameter, intersecting sides $A B$ and $A C$ at points $D$ and $E$ respectively. If $D E = E C = 4$, and $B C - B D = \frac{16}{5}$, then $\sqrt{\frac{B D - A D}{B C}}=$ $\qquad$ . | 4. $\frac{3}{5}$.
As shown in Figure 7, connect $B E$. Since $B C$ is the diameter, we have
$$
\begin{array}{l}
\angle B E C=\angle B E A \\
=90^{\circ} .
\end{array}
$$
Also, $D E=E C$, thus
$$
\overparen{D E}=\overparen{E C}, \angle 1=\angle 2 \text {. }
$$
Therefore, Rt $\triangle A E B \cong \mathrm{Rt} \triangl... | \frac{3}{5} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,278 |
One, (20 points) Given that $a$ is an integer, the system of equations about $x, y$
$$
\left\{\begin{array}{l}
x+y=(a+2) x, \\
x y=\left(a^{2}+1\right) x-2 a^{3}+2
\end{array}\right.
$$
all solutions $(x, y)$ are integers. Try to find the value of $a$. | From equation (1), we get $y=(a+1) x$. Substituting into equation (2), we have
$$
(a+1) x^{2}-\left(a^{2}+1\right) x+2 a^{3}-2=0 \text {. }
$$
When $a=-1$, equation (3) can be simplified to $-2 x-4=0$.
Solving, we get $x=-2, y=0$, which meets the requirements.
When $a \neq-1$, we have
$$
x_{1}+x_{2}=\frac{a^{2}+1}{a+1... | -1,0,1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,279 |
II. (25 points) As shown in Figure 4, in quadrilateral $ABCD$, $AC$ bisects $\angle BAD$, and $CE \perp AB$ at point $E$.
(1) If $\angle ADC +$
$$
\begin{array}{l}
\angle ABC=180^{\circ}, \text { prove: } AD \\
+ AB = 2 AE ;
\end{array}
$$
(2) If $AD + AB = 2 AE$, prove: $CD = CB$. | II. (1) As shown in Figure 8, extend $AB$ to point $M$ such that $AE = ME$. Since $CE \perp AB$, $\triangle ACM$ is an isosceles triangle. Therefore, $AC = CM$, and $\angle 1 = \angle 3$.
Given that $\angle 1 = \angle 2$, it follows that $\angle 3 = \angle 2$. Also, $\angle ADC + \angle ABC = 180^\circ$, so $\angle ADC... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,280 |
Three, (25 points) Given positive integers $a, b, c, d$ satisfying $b<a<d<c$, and the sums of each pair are $26, 27, 41, 101, 115, 116$. Find the value of $(100a + b) - (100d - c)$. | Three, from $b<a<d<c$, we can get
$$
\begin{array}{l}
a+b<b+d<b+c<a+c<d+c, \\
b+d<a+d<a+c .
\end{array}
$$
Therefore, $a+d$ and $b+c$ are both between $b+d$ and $a+c$.
It is easy to know that $a+b=26, b+d=27, a+c=115, d+c=116$.
If $a+d=101$, then $b+c=41$.
Solving this gives $b=-24$ (discard).
Thus, $a+d=41, b+c=101$.... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,281 |
1. The range of the function $f(x)=\left(|\cot x|+\frac{1}{\sin x}\right)^{2}$ $\left(0<|x|<\frac{\pi}{2}\right)$ is ( ).
(A) $\mathbf{R}_{+}$
(B) $\mathbf{R}_{+} \backslash\{1\}$
(C) $\mathbf{R}_{+} \backslash\{\sqrt{2}\}$
(D) $\mathbf{R}_{+} \backslash\{1, \sqrt{2}\}$ | - 1.B.
When $0<x<\frac{\pi}{2}$,
$$
\begin{array}{l}
f(x)=\left(\cot x+\frac{1}{\sin x}\right)^{2}=\left(\frac{1+\cos x}{\sin x}\right)^{2} \\
=\left(\frac{2 \cos ^{2} \frac{x}{2}}{2 \sin \frac{x}{2} \cdot \cos \frac{x}{2}}\right)^{2}=\cot ^{2} \frac{x}{2} \in(1,+\infty) ;
\end{array}
$$
When $-\frac{\pi}{2}<x<0$,
$$... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,282 |
2. If the real number $a$ makes the inequality $\frac{1}{1+\sqrt{x}}$ $\geqslant a \sqrt{\frac{\sqrt{x}}{x-1}}$ always have non-zero real solutions for $x$, then the range of values for $a$ is ( ).
(A) $(-\infty,+\infty)$
(B) $(-\infty, \sqrt{2}]$
(C) $\left(-\infty, \frac{\sqrt{2}}{2}\right]$
(D) $(-\infty, \sqrt{2}-1... | 2. D.
If $a \leqslant 0$, then the inequality always has a non-zero solution.
If $a>0$, let $t=\sqrt{x}$. Then $x=t^{2}(t>1)$.
Thus, the original inequality becomes $\frac{1}{1+t} \geqslant \frac{a \sqrt{t}}{\sqrt{t^{2}-1}}$, which is
$$
a^{2} \leqslant \frac{t-1}{t(t+1)} \text {. }
$$
Let $s=t-1(s>0)$. From equation... | D | Inequalities | MCQ | Yes | Yes | cn_contest | false | 716,283 |
3. Let $a$ be an integer, and $\frac{a^{2}-a+3}{a^{3}-3 a^{2}+1}$ is also an integer. Then the number of all such $a$ is ( ).
(A) 2
(B) 3
(C) 4
(D) 5 | 3.B.
Obviously, when $a=0,1,3$, the condition is satisfied, and when $a=2$, it does not satisfy the condition.
If $a \geqslant 4$, then
$$
\begin{array}{l}
a^{3}-3 a^{2}+1=a^{2}(a-3)+1 \\
>a^{2}(a-3) \geqslant a^{2}>a^{2}-a+3 .
\end{array}
$$
If $a a_{1}^{2}+a_{1}+3 \text {. }
$$
Moreover, when $a_{1}=1$, the requir... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,284 |
4. The maximum value of the function $f(x)=(5-x)(x+1)^{\frac{2}{3}}(-1 \leqslant x \leqslant 5)$ is ( ).
(A) $\frac{658}{445} \sqrt[3]{90}$
(B) $\frac{748}{495} \sqrt[3]{90}$
(C) $\frac{36}{25} \sqrt[3]{90}$
(D) $\frac{22}{15} \sqrt[3]{90}$ | 4.C.
$$
\begin{array}{l}
f^{3}(x)=(5-x)^{3}(x+1)^{2}=\frac{1}{72}(10-2 x)^{3}(3 x+3)^{2} \\
\leqslant \frac{1}{72}\left[\frac{3(10-2 x)+2(3 x+3)}{5}\right]^{5}=\frac{1}{72} \times\left(\frac{36}{5}\right)^{5}
\end{array}
$$
When and only when $10-2 x=3 x+3$, i.e., $x=\frac{7}{5}$, the equality holds. Therefore, the ma... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,285 |
5. The number $10^{100^{100^{10}}}$ when divided by 77, the remainder is ( ).
(A) 45
(B) 56
(C) 67
(D) 68 | 5.C.
Let $x=10^{10^{10^{10}}}$. Note that $77=7 \times 11$, obviously,
$$
x=(11-1)^{10^{10^{10}}} \equiv(-1)^{10^{10^{10}}} \equiv 1(\bmod 11) \text {. }
$$
Since 7 and 10 are coprime, by Fermat's Little Theorem, we have $10^{6} \equiv 1(\bmod 7)$.
Next, find the remainder $r(0 \leqslant r \leqslant 5)$ when $10^{10^... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 716,286 |
6. In the Cartesian coordinate system $x O y$, it is known that points $A(3,5)$, $B(1,3)$, and $C(4,3)$. Now, $\triangle A B C$ is translated to the position of $\triangle A_{1} B_{1} C_{1}$, and the translation direction forms an angle of $30^{\circ}$ with the $x$-axis. If the area of the common part of $\triangle A B... | 6. A.
As shown in Figure 2, let $A_{1} B_{1}$ intersect $B C$ at point $K$, then
$$
\begin{array}{l}
\frac{C K}{C B}=\sqrt{\frac{S_{\text {common }}}{S_{\triangle U C}}}=\sqrt{m}, \\
\frac{B K}{C B}=1-\frac{C K}{C B}=1-\sqrt{m} .
\end{array}
$$
Let the line $l$ passing through points $C$ and $C_{1}$ intersect $A B$ a... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,287 |
Example 9 As shown in Figure 12,
in $\triangle A B C$, $A C$ $=B C, \angle A C B=$ $80^{\circ}$, take a point $M$ inside $\triangle A B C$, such that $\angle M A B=10^{\circ}$, $\angle M B A=30^{\circ}$. Find the degree measure of $\angle A M C$.
(1983, Former Yugoslavia Mathematical Olympiad) | Solution 1: Since $A C=B C, \angle A C B=80^{\circ}$, then $\angle C A B=\angle C B A=50^{\circ}$.
Since $\angle M A B=10^{\circ}, \angle M B A=30^{\circ}$, therefore,
$$
\angle C A M=40^{\circ}, \angle C B M=20^{\circ} \text {. }
$$
Let $\angle B C M=x$, then, $\angle A C M=80^{\circ}-x$.
Applying the trigonometric f... | 70^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,288 |
1. If $x, y \in R$, and $2^{x}=18^{y}=6^{x y}$, then $x+y$ $=$ . $\qquad$ | Ni.1.0 or 2.
If $x=0$ or $y=0$, then it must be that $x=y=0$. Therefore, $x+y=0$.
If $x \neq 0$ and $y \neq 0$, taking the logarithm with base 6 of $2^{x}=18^{y}=6^{\circ}$, we get
$$
x \log _{6} 2=y \log _{6} 18=x y .
$$
Then $y=\log _{6} 2, x=\log _{6} 18$.
Thus, $x+y=\log _{6} 18+\log _{6} 2=\log _{6} 36=2$.
In con... | 0 \text{ or } 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,289 |
2. Let $x_{1}=\frac{1}{4}, x_{2}=\frac{3}{16}, x_{n+2}=\frac{3}{16}-\frac{1}{4}\left(x_{1}\right.$ $\left.+x_{2}+\cdots+x_{n}\right)(n=1,2, \cdots)$. Then $x_{n}=$ | 2. $\frac{n+1}{2^{n+2}}$.
From the given, we have $x_{n+2}-x_{n+1}=-\frac{1}{4} x_{n}$, which means $x_{n+2}=x_{n+1}-\frac{1}{4} x_{n}$.
Let $a_{n}=2^{n} x_{n}$, then
$$
a_{1}=\frac{1}{2}, a_{2}=\frac{3}{4}, a_{n+2}=2 a_{n+1}-a_{n} \text {. }
$$
Then $a_{n+2}-a_{n+1}=a_{n+1}-a_{n}=\cdots=a_{2}-a_{1}=\frac{1}{4}$.
The... | \frac{n+1}{2^{n+2}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,290 |
3. Let the integer $k$ be such that the quadratic equation in $x$
$$
2 x^{2}+(11-5 k) x+3 k^{2}-13 k-2003=0
$$
has at least one integer root. Then all such integers $k$ are
$\qquad$ | 3. $-4030,-2012,2018,4036$.
Completing the square, we get
$$
(x-k+2)(2 x-3 k+7)=2017 \text {. }
$$
Since 2017 is a prime number, we have
$$
\left\{\begin{array} { l }
{ x - k + 2 = \pm 1 } \\
{ 2 x - 3 k + 7 = \pm 2 0 1 7 , }
\end{array} \text { or } \left\{\begin{array}{l}
x-k+2= \pm 2017, \\
2 x-3 k+7= \pm 1 .
\en... | -4030,-2012,2018,4036 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,291 |
4. As shown in Figure 1, let $A B C D-A_{1} B_{1} C_{1} D_{1}$ be a cube, and $M$ be the midpoint of edge $A A_{1}$. Then the size of the dihedral angle $C-M D_{1}-B_{1}$ is equal to $\qquad$ (express using radians or the arccosine function). | 4. $\arccos \frac{\sqrt{6}}{6}$.
As shown in Figure 3, let the dihedral angle $C$
$-M D_{1}-B_{1}$ be $\theta$, the dihedral angle $C-M D_{1}-$ $D$ be $\alpha$, and the dihedral angle $B_{1}-M D_{1}-A_{1}$ be $\beta$. Then $\theta+\alpha+\beta$ $=\pi$. Therefore,
$$
\theta=\pi-(\alpha+\beta) \text {. }
$$
Assume the ... | \arccos \frac{\sqrt{6}}{6} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,292 |
5. In the Cartesian coordinate system $x O y$, it is known that points $A(3,3)$, $B(-2,1)$, and $C(1,-2)$. If $T$ represents the set of all points inside and on the sides (including vertices) of $\triangle A B C$, then the range of the bivariate function $f(x, y)=\max \{2 x+y$, $\left.x^{2}-y\right\}($ where $(x, y) \i... | 5. $\left[\frac{129-12 \sqrt{139}}{25}, 9\right]$.
Divide $T$ into two parts and discuss them separately.
(1) When $2 x+y \geqslant x^{2}-y$, i.e., $y \geqslant \frac{x^{2}}{2}-x$, then $f(x, y)=2 x+y$.
Let the parabola $y=\frac{x^{2}}{2}-x$ intersect sides $A B$ and $A C$ at points $D$ and $E$, respectively. Point $(... | \left[\frac{129-12 \sqrt{139}}{25}, 9\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,293 |
6. In the Cartesian coordinate system $x O y$, it is known that points $A(6,4)、B(4,0)、C(0,3)$, and the line $l: y=k x$ bisects the area of $\triangle A B C$, $k=\frac{-b+\sqrt{c}}{a}(a、b、c$ are all positive integers, and $c$ has no square factors). Then $a+b+c=$ $\qquad$ . | 6.584.
As shown in Figure 5, take the midpoint $M\left(2, \frac{3}{2}\right)$ of side $BC$. Let the line $l$ intersect sides $BC$ and $CA$ at points $P$ and $Q$, respectively.
Since $AM$ and $PQ$ both bisect the area of $\triangle ABC$, we have $S_{\triangle PQC} = S_{\triangle MMC}$. Subtracting the common area $S_{... | 584 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,294 |
Three. (20 points) For the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with edge length $2$, $M$ is the midpoint of edge $A A_{1}$. A circle $\Gamma$ is constructed through vertices $A$, $B_{1}$, and $C$. Let $P$ be any point on circle $\Gamma$. Find the range of values for the segment $P M$.
---
Translation:
Three. (20 ... | Three, since $\triangle A B_{1} C$ is an equilateral triangle, let the body diagonal $B D_{1}$ intersect the plane of $\triangle A B_{1} C$ at point $O_{1}$, then $O_{1}$ is the center of circle $\Gamma$. It can be calculated that $B O_{1}=\frac{1}{3} B D_{1}, D_{1} O_{1}=\frac{2}{3} B D_{1}$.
As shown in Figure 6, wit... | \left[\frac{\sqrt{45-24 \sqrt{3}}}{3}, \frac{\sqrt{45+24 \sqrt{3}}}{3}\right] | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,295 |
Four, (20 points) Given a real number $a$, solve the inequality for $x$
$$
\sqrt{3 x-5 a}-\sqrt{x-a}>1 \text {. }
$$ | From the original inequality, we have
$$
\sqrt{3 x-5 a}>\sqrt{x-a}+1\left(x \geqslant a \text {, and } x \geqslant \frac{5}{3} a\right) \text {. }
$$
By squaring both sides, we get
$$
3 x-5 a>x-a+1+2 \sqrt{x-a}(x \geqslant a) \text {. }
$$
Thus, the original inequality is equivalent to
$$
2 x-4 a-1>2 \sqrt{x-a}(x \ge... | x>2 a+1+\sqrt{a+\frac{3}{4}} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 716,296 |
Five. (20 points) For each real number $a$, let the parabola $y=x^{2}+a x+a-2$ be denoted as $P_{a}$.
(1) Find the intersection of all $P_{a}$;
(2) Find the equation of the locus of the foci of all $P_{a}$;
(3) Find all lines $l$ such that $l$ intersects with all $P_{a}$;
(4) Find all $a$ such that there exists a downw... | (1) Let $P_{a_{1}}: y=x^{2}+a_{1} x+a_{1}-2$,
$$
P_{a_{2}}: y=x^{2}+a_{2} x+a_{2}-2\left(a_{1} \neq a_{2}\right) \text {. }
$$
Then $x^{2}+a_{1} x+a_{1}-2=x^{2}+a_{2} x+a_{2}-2$,
which simplifies to $\left(a_{1}-a_{2}\right) x=a_{2}-a_{1}$.
This equation has only one solution $x=-1$.
Thus, $y=1-a_{1}+a_{1}-2=-1$.
Ther... | not found | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,297 |
One, (50 points) Given that $P$ is the intersection point of the diagonals $AC$ and $BD$ of a convex quadrilateral $ABCD$, and $\angle BAC + \angle BDC = 180^{\circ}$. If the distance from point $A$ to line $BC$ is less than the distance from point $D$ to line $BC$, prove:
$$
\left(\frac{AC}{BD}\right)^{2} > \frac{AP \... | As shown in Figure 7, construct the point $A'$ symmetric to point $A$ with respect to $BC$. Given $\angle BAC + \angle BDC = 180^\circ$, we know that points $B, A', C, D$ are concyclic. Therefore,
$$
\begin{array}{l}
\frac{CD}{AB} = \frac{CD}{A'B} \\
= \frac{\sin \angle DBC}{\sin \angle BCA'} \\
= \frac{\sin \angle PBC... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,298 |
II. (50 points) Let $a_{0}=1, a_{n+1}=\frac{a_{n}}{1+a_{n}^{2}}(n=$ $0,1, \cdots)$. Prove:
(1) $a_{n} \leqslant \frac{3}{4} \times \frac{1}{\sqrt{n}}(n=1,2, \cdots)$;
(2) $a_{n} \leqslant \frac{n}{2}\left(\sum_{i=1}^{n} \sqrt{i}\right)^{-1}(n=1,2, \cdots)$. | (1) Let $b_{n}=\frac{1}{a_{n}}$, then $b_{0}=1, b_{n+1}=b_{n}+\frac{1}{b_{n}}$. We only need to prove $b_{n} \geqslant \frac{4}{3} \sqrt{n}(n \geqslant 1)$.
We will prove this by mathematical induction.
When $n=1$, the proposition is obviously true.
Assume that $b_{n} \geqslant \frac{4}{3} \sqrt{n}$.
Since the function... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 716,300 |
Three. (50 points) Let the natural number $n \geqslant 25$. Prove that all composite numbers not exceeding $n$ can be rearranged (not necessarily in the original order of magnitude) such that every three consecutive numbers have a common divisor greater than 1 (for example, when $n=25$, the arrangement $21,9,6,24,22,8,... | For $n \geqslant 25$, we construct a permutation below to meet the requirements.
Let $p_{1}4$, thus,
$4 p_{k} \leqslant 4 \sqrt{n}<\sqrt{n}, \sqrt{n}=n$,
and $p_{k}, p_{k-1}$ are both odd, so $p_{k-1} \leqslant p_{k}-2$, thus, $8 p_{k-1} \leqslant 8\left(p_{k}-2\right) \leqslant 8 \sqrt{n}-16 \leqslant n$ (which is eq... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 716,301 |
As 167 As shown in Figure 2, from a point $P$ outside $\odot O$, two secants $PAB$ and $PCD$ are drawn, intersecting $\odot O$ at points $A$, $B$, $C$, and $D$. Chords $AD$ and $BC$ intersect at point $Q$. A secant $PEF$ passing through point $Q$ intersects $\odot O$ at points $E$ and $F$. A line $DM$ parallel to $PF$ ... | Prove: Connect $D N$ ($N$ is the intersection of $M B$ and $E F$). Since $M D / / P F$, we have $\angle M D Q = \angle D Q F$:
Also, since $\angle M D A = \angle A B M$, we have
$\angle D Q F = \angle A B M$.
Thus, $A, B, N, Q$ are concyclic.
Therefore, $P Q \cdot P N = P A \cdot P B = P C \cdot P D$.
Hence, $C, D, N, ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,302 |
In the isosceles right triangle $\triangle ABC$ as shown in Figure 3, $D_{1}$ is any point on the leg $AC$, $D_{1} G \perp BC$ at point $G$, $B D_{2} \perp A G$ intersects $AC$ at point $D_{2}$, and perpendiculars are drawn from point $C$ to $B D_{1}$ and $B D_{2}$, intersecting the extensions of $B D_{1}$ and $B D_{2}... | Proof: As shown in Figure 4, extend $C E_{1}$ and $C E_{2}$ to intersect the extension of $B A$ at points $F_{1}$ and $F_{2}$. Since $B A \perp A C$ and $B E_{1} \perp C E_{1}$, points $B$, $A$, $E_{1}$, and $C$ are concyclic.
Therefore, $\angle A B E_{1} = \angle A C E_{1}$, which means $\angle A B D_{1} = \angle A C ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,303 |
Given $S=\{1,2, \cdots, n\}, A_{1}, A_{2}, \cdots, A_{10}$ are all subsets of $S$, and satisfy
(1) $\left|A_{i}\right|=3, i=1,2, \cdots, 10$;
(2) $A_{i} \cap A_{j} \neq \varnothing, 1 \leqslant i<j \leqslant 10$;
(3) Each element in $S$ belongs to at least one of $A_{1}, A_{2}, \cdots, A_{10}$, and at most to 5 subsets... | Solution: From (1) and (3), we know that each $A_{i}$ contains 3 elements, and the 10 subsets have a total of 30 elements. Each element belongs to at most 5 subsets, so $n \geqslant 6$.
When $n=6$, the following 10 subsets satisfy the problem's requirements:
$\{1,2,3\},\{1,3,4\},\{1,4,5\},\{1,5,6\},\{1,6,2\}$.
$\{2,4,5... | 6,7 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,304 |
168 cm tall. Try to find all functions $f: \mathbf{R} \rightarrow \mathbf{R}$ such that for all $x, y \in \mathbf{R}$, we have
$$
f(f(x+f(y))-1)=f(x)+f(x+y)-x .
$$ | First, we prove that $f$ is injective.
Let $y_{1}, y_{2} \in \mathbf{R}, f\left(y_{1}\right)=f\left(y_{2}\right)$.
We need to prove that $y_{1}=y_{2}$.
From $f\left(y_{1}\right)=f\left(y_{2}\right)$, we get
$f\left(x+y_{1}\right)=f\left(x+y_{2}\right)$ (for any $\left.x \in \mathbf{R}\right)$.
Thus, $f\left(x+y_{1}-y_{... | f(x)=x+1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,305 |
Second question Let positive numbers $a, b, c, x, y, z$ satisfy $c y + b z = a, a z + c x = b, b x + a y = c$.
Find the minimum value of the function $f(x, y, z) = \frac{x^{2}}{1+x} + \frac{y^{2}}{1+y} + \frac{z^{2}}{1+z}$. | Given the conditions, we have
$$
\begin{array}{l}
b(a z+c x-b)+c(b x+a y-c)- \\
a(c y+b z-a)=0,
\end{array}
$$
which simplifies to $2 b c x+a^{2}-b^{2}-c^{2}=0$.
Solving for $x$, we get $x=\frac{b^{2}+c^{2}-a^{2}}{2 b c}$.
Similarly, $y=\frac{a^{2}+c^{2}-b^{2}}{2 a c}$ and $z=\frac{a^{2}+b^{2}-c^{2}}{2 a b}$.
Thus, $f... | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,311 |
For each positive integer $n$, define the function
$$
f(n)=\left\{\begin{array}{cl}
0, & \text { when } n \text { is a perfect square; } \\
{\left[\frac{1}{\{\sqrt{n}\}}\right],} & \text { when } n \text { is not a perfect square. }
\end{array}\right.
$$
where $[x]$ denotes the greatest integer not exceeding $x$, and ... | Solution: If $k$ is not a perfect square, then there exists $a \in \mathbf{N}$, such that $a^{2}<k<(a+1)^{2}$, then $a<\sqrt{k}<a+1$, i.e.,
$$
\{\sqrt{k}\}=\sqrt{k}-a \text {. }
$$
Therefore, $\left[\frac{1}{\{\sqrt{k}\}}\right]=\left[\frac{1}{\sqrt{k}-a}\right]=\left[\frac{\sqrt{k}+a}{k-a^{2}}\right]$.
$$
\begin{alig... | 768 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 716,312 |
$6 \rightarrow$ In a regular $n$-sided polygon, the difference between any two adjacent interior angles is $18^{\circ}$, find the maximum value of $n$, | Hedgehog, this problem can be solved in three steps:
(1) Prove that $n$ is even
(2) Prove that $n<40$, and $n \leqslant 38$;
(3) Prove that $n=38$, i.e., the maximum value of $n$ is 38. | 38 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 716,313 |
Example 1 Given $a, b, c \in \mathbf{R}_{+}$. Prove:
$$
\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geqslant \frac{3}{2} \text {. }
$$
(26th Moscow Mathematical Olympiad) | Proof: $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}$
$$
\begin{array}{l}
=\frac{a^{2}}{a(b+c)}+\frac{b^{2}}{b(c+a)}+\frac{c^{2}}{c(a+b)} \\
\geqslant \frac{(a+b+c)^{2}}{2(a b+b c+c a)} \\
\geqslant \frac{3(a b+b c+c a)}{2(a b+b c+c a)}=\frac{3}{2} .
\end{array}
$$ | \frac{3}{2} | Inequalities | proof | Yes | Yes | cn_contest | false | 716,314 |
Example 2 Let $a, b, c \in \mathbf{R}_{+}$, and $abc=1$. Then $\frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)} \geqslant \frac{3}{2}$.
(26th IMO) | $\begin{array}{l}\text { Prove: } \frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)} \\ =\frac{a^{2} b^{2} c^{2}}{a^{3}(b+c)}+\frac{a^{2} b^{2} c^{2}}{b^{3}(c+a)}+\frac{a^{2} b^{2} c^{2}}{c^{3}(a+b)} \\ =\frac{b^{2} c^{2}}{a(b+c)}+\frac{a^{2} c^{2}}{b(c+a)}+\frac{a^{2} b^{2}}{c(a+b)} \\ \geqslant \frac{(a b... | \frac{3}{2} | Inequalities | proof | Yes | Yes | cn_contest | false | 716,315 |
Example 3 Proof: For any numbers $a>1, b>1$, the inequality $\frac{a^{2}}{b-1}+\frac{b^{2}}{a-1} \geqslant 8$ holds.
(26th CIS Mathematical Olympiad) | Proof: Given $a>1, b>1$, we know $a-1>0, b-1>0$.
Then $\frac{a^{2}}{b-1}+\frac{b^{2}}{a-1} \geqslant \frac{(a+b)^{2}}{a+b-2}$.
Since $(a+b-4)^{2} \geqslant 0$, we have $(a+b)^{2} \geqslant 8(a+b-2)$. Therefore, $\frac{(a+b)^{2}}{a+b-2} \geqslant 8$.
Thus, the original inequality holds. | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 716,316 |
Example 4 If $\alpha$ is an acute angle, prove:
$$
\left(1+\frac{1}{\sin \alpha}\right)\left(1+\frac{1}{\cos \alpha}\right)>5 \text {. }
$$
(1963, Hungarian Mathematical Olympiad) | $\begin{array}{l}\text { Prove: }\left(1+\frac{1}{\sin \alpha}\right)\left(1+\frac{1}{\cos \alpha}\right) \\ =1+\frac{1}{\sin \alpha}+\frac{1}{\cos \alpha}+\frac{1}{\sin \alpha \cdot \cos \alpha} \\ \geqslant 1+\frac{4}{\sin \alpha+\cos \alpha}+\frac{2}{\sin 2 \alpha} \\ =1+\frac{4}{\sqrt{2} \sin \left(\alpha+\frac{\pi... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 716,317 |
Example 5 Let $x, y \in (0, +\infty)$, and $\frac{19}{x} + \frac{98}{y} = 1$. Find the minimum value of $x + y$.
(1998, Hunan Province High School Mathematics Competition) | Solution: Since $x, y \in \mathbf{R}_{+}$, we have
$$
1=\frac{19}{x}+\frac{98}{y} \geqslant \frac{(\sqrt{19}+\sqrt{98})^{2}}{x+y} \text {, }
$$
which means $x+y \geqslant(\sqrt{19}+\sqrt{98})^{2}=117+14 \sqrt{38}$.
Equality holds if and only if $\frac{\sqrt{19}}{x}=\frac{\sqrt{98}}{y}$, i.e.,
$$
x=\sqrt{19}(\sqrt{19}+... | 117+14\sqrt{38} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,318 |
Example 6 Given $x, y, z \in \mathbf{R}_{+}$. Prove:
$$
\frac{x}{2 x+y+z}+\frac{y}{x+2 y+z}+\frac{z}{x+y+2 z} \leqslant \frac{3}{4} \text {. }
$$ | Prove: Let $a=y+z, b=z+x, c=x+y$. Then, from Example 1, we know
$$
\begin{aligned}
& \frac{y+z}{(z+x)+(x+y)}+\frac{z+x}{(x+y)+(y+z)}+\frac{x+y}{(y+z)+(z+x)} \\
= & \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geqslant \frac{3}{2} .
\end{aligned}
$$
Therefore, $\frac{x}{2 x+y+z}+\frac{y}{x+2 y+z}+\frac{z}{x+y+2 z}$
$$
\b... | \frac{3}{4} | Inequalities | proof | Yes | Yes | cn_contest | false | 716,319 |
Question 1 Let $a_{1}, a_{2}, \cdots$ be a sequence of integers, with infinitely many positive integers and infinitely many negative integers. Assume that for each positive integer $n$, the numbers $a_{1}, a_{2}, \cdots, a_{n}$ have distinct remainders when divided by $n$. Prove: In the sequence $a_{1}, a_{2}$, ..., ea... | Proof: Subtracting an integer from each term of the sequence does not change the conditions and conclusions of the problem, so we may assume \(a_{1}=0\). In this case, for each positive integer \(k\), we must have \(\left|a_{k}\right|<k\) (if \(\left|a_{k}\right| \geqslant k\), take \(n=\left|a_{k}\right|\), then \(a_{... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 716,320 |
Question 2 If $x, y, z$ are positive real numbers, and $x y z \geqslant 1$, prove:
$$
\frac{x^{5}-x^{2}}{x^{5}+y^{2}+z^{2}}+\frac{y^{5}-y^{2}}{x^{2}+y^{5}+z^{2}}+\frac{z^{5}-z^{2}}{x^{2}+y^{2}+z^{5}} \geqslant 0 .
$$ | Prove: The original inequality is equivalent to
$$
\sum \frac{x^{5}}{x^{5}+y^{2}+z^{2}} \geqslant \sum \frac{x^{2}}{x^{5}+y^{2}+z^{2}}.
$$
By $\frac{a^{2}}{b} \geqslant 2 a-b\left(a, b \in \mathbf{R}_{+}\right)$, we get
$$
\begin{array}{l}
x^{5}+y^{2}+z^{2}=x\left(x^{4}+\frac{y^{2}}{x}+\frac{z^{2}}{x}\right) \\
\geqsl... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 716,321 |
一、 $A B$ is a chord of $\odot O$, with its midpoint being $M$. Draw a non-diameter chord $C D$ through point $M$. Draw two tangents to $\odot O$ from points $C$ and $D$, intersecting line $A B$ at points $P$ and $Q$ respectively. Prove: $P A=Q B$. | As shown in Figure 1, connect $O M, O P, O Q, O C, O D$. Since $P C$ is a tangent to $\odot O$ and $M$ is the midpoint of chord $A B$, then
$$
\begin{array}{l}
\angle P C O=\angle P M O \\
=90^{\circ} .
\end{array}
$$
Therefore, points $P, C, M, O$ are concyclic.
Similarly, points $Q, D, O, M$ are concyclic.
Thus, $\a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,322 |
Define a function $f(x)$ on $\mathbf{R}$ that satisfies:
(1) $f(0)=0$;
(2) For any $x, y \in(-\infty,-1) \cup(1, +\infty)$, we have $f\left(\frac{1}{x}\right)+f\left(\frac{1}{y}\right)=f\left(\frac{x+y}{1+x y}\right)$;
(3) When $x \in(-1,0)$, we have $f(x)>0$.
Prove:
$$
f\left(\frac{1}{19}\right)+f\left(\frac{1}{29}\ri... | In (2), let $y=-x$, we get
$$
f\left(\frac{1}{x}\right)+f\left(-\frac{1}{x}\right)=f(0)=0,
$$
i.e., $f\left(-\frac{1}{x}\right)=-f\left(\frac{1}{x}\right)$.
Thus, $f(x)$ is an odd function.
Suppose $-10 \Leftrightarrow x_{1}-x_{2}>x_{1} x_{2}-1 \\
\Leftrightarrow-10$, then $f\left(x_{1}\right)>f\left(x_{2}\right)$.
Th... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 716,323 |
Example 1 As shown in Figure 4, in $\triangle ABC$, $O$ is the circumcenter, the three altitudes $AD, BE, CF$ intersect at point $H$, line $ED$ and $AB$ intersect at point $M$, $FD$ and $AC$ intersect at point $N$. Prove:
(1) $OB \perp DF, OC \perp DE$;
(2) $OH \perp MN$.
untranslated text:
(1) $O B \perp D F, O C \p... | Proof: (1) Since point $O$ is the circumcenter of $\triangle ABC$, then $\angle BOC=2\angle BAC$.
Therefore, $\angle OBC=\frac{1}{2}\left(180^{\circ}-\angle BOC\right)$
$$
=90^{\circ}-\angle BAC \text{. }
$$
Since $A, F, D, C$ are concyclic, thus,
$\angle BDF=\angle BAC$.
Let $OB \cap FD=P$. Then,
$$
\angle BPD=180^{\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,324 |
Three, in an arithmetic sequence of positive numbers $a_{1}, a_{2}, \cdots, a_{3 n}(n \geqslant 2)$ with a common difference of $d(d>0)$, any $n+2$ numbers are taken. Prove: there must exist two numbers $a_{i} 、 a_{j}(i \neq j)$, satisfying the inequality $1<\frac{\left|a_{i}-a_{j}\right|}{n d}<2$. | Three, in the $n+2$ numbers taken out, let $a_{l}$ be the maximum, then $a_{l} \leqslant a_{3 n}$. Add $a_{3 n}-a_{l}$ to each number, this processing does not change the absolute value of any two number differences. Thus, it can always be assumed that the $n+2$ numbers taken out include $a_{3 n}$, and let $a_{l}=a_{3 ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 716,325 |
Given that the digits of an $n$-digit number can only take elements from the set $\{1,2,3,4,5\}$, let the number of $n$-digit numbers containing the digit 5 and having no 3 before the 5 be $f(n)$. Find $f(n)$.
（Supplied by Jiang Ximing） | For the number of $(n+1)$-digit numbers $f(n+1)$ that satisfy the conditions, we can divide it into the following two scenarios:
Scenario 1: When the unit digit is not 5, then the first $n$ digits must contain the digit 5, making it a valid $n$-digit number, the count of which is $f(n)$. For each such $n$-digit number... | f(n)=n \cdot 4^{n-1} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,326 |
Five, if three positive real numbers $x, y, z$ satisfy
$$
\begin{array}{l}
x^{2}+x y+y^{2}=\frac{25}{4}, y^{2}+y z+z^{2}=36, \\
z^{2}+z x+x^{2}=\frac{169}{4} .
\end{array}
$$
Find the value of $x y+y z+z x$. | Five, it is known that the three equations can be transformed into
$$
\begin{array}{l}
x^{2}+y^{2}-2 x y \cos 120^{\circ}=\left(\frac{5}{2}\right)^{2}, \\
y^{2}+z^{2}-2 y z \cos 120^{\circ}=\left(\frac{12}{2}\right)^{2}, \\
z^{2}+x^{2}-2 z x \cos 120^{\circ}=\left(\frac{13}{2}\right)^{2} .
\end{array}
$$
Construct a r... | 10 \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,327 |
Six, let $0 \leqslant \alpha, \beta, \gamma \leqslant \frac{\pi}{2}, \cos ^{2} \alpha+\cos ^{2} \beta+$ $\cos ^{2} \gamma=1$. Prove:
$$
\begin{aligned}
2 \leqslant & \left(1+\cos ^{2} \alpha\right)^{2} \sin ^{4} \alpha+\left(1+\cos ^{2} \beta\right)^{2} \sin ^{4} \beta+ \\
& \left(1+\cos ^{2} \gamma\right)^{2} \sin ^{4... | Six, let $a=\cos ^{2} \alpha, b=\cos ^{2} \beta, c=\cos ^{2} \gamma$, then $0 \leqslant a, b, c \leqslant 1$, and $a+b+c=1$.
Thus, the original inequality is equivalent to
$$
\begin{array}{l}
0 \leqslant a^{4}+b^{4}+c^{4}-2\left(a^{2}+b^{2}+c^{2}\right)+1 \\
\leqslant a b+b c+c a+a b c .
\end{array}
$$
Let $a b+b c+c ... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 716,328 |
1. If $S=\{1,2,3,4,5\}, M=\{1,3,4\}$, $N=\{2,4,5\}$, then, $\left(\mathrm{C}_{S} M\right) \cap\left(\mathrm{C}_{S} N\right)$ equals ( ).
(A) $\varnothing$
(B) $\{1,3\}$
(C) $\{4\}$
(D) $\{2,5\}$ | - 1. A.
It is easy to know that $\mathrm{C}_{S} M=\{2,5\}, \mathrm{C}_{S} N=\{1,3\}$, so,
$$
\left(\mathcal{C}_{S} M\right) \cap\left(\mathcal{C}_{S} N\right)=\varnothing .
$$ | A | Logic and Puzzles | MCQ | Yes | Yes | cn_contest | false | 716,329 |
3. If $c$ and $d$ are two non-zero planar vectors that are not collinear, then among the four pairs of $\boldsymbol{a}, \boldsymbol{b}$ given below, the pair that is not collinear is $(\quad$.
(A) $a=-2(c+d), b=2(c+d)$
(B) $a=c-d, b=-2 c+2 d$
(C) $a=4 c-\frac{2}{5} d, b=c-\frac{1}{10} d$
(D) $a=c+d, b=2 c-2 d$ | 3. D.
From $a=-2(c+d)=-b$, we know that $a$ and $b$ are collinear, thus eliminating option (A);
$$
a=c-d=-\frac{1}{2}(-2 c+2 d)=-\frac{1}{2} b \text{, so } a \text{ and } b
$$
are collinear, eliminating option (B);
$$
a=4 c-\frac{2}{5} d=4\left(c-\frac{1}{10} d\right)=4 b \text{, so } a \text{ and } b \text{ are coll... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,331 |
4. For a function $f(x)$ defined on the interval $[a, b]$, if there exists a constant $c$, such that for any $x_{1} \in[a, b]$ there is a unique $x_{2} \in[a, b]$, satisfying $\frac{f\left(x_{1}\right)+f\left(x_{2}\right)}{2}=c$, then the function $f(x)$ is said to have a "mean" of $c$ on $[a, b]$. Then, the mean of th... | 4. C.
Let for any $x_{1} \in [10,100]$, there exists a unique $x_{2} \in$ $[10,100]$, such that $x_{1} x_{2}=10^{3}$. Therefore,
$$
\begin{array}{l}
\frac{f\left(x_{1}\right)+f\left(x_{2}\right)}{2}=\frac{\lg x_{1}+\lg x_{2}}{2} \\
=\frac{\lg x_{1} x_{2}}{2}=\frac{\lg 10^{3}}{2}=\frac{3}{2} .
\end{array}
$$ | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,332 |
5. In a triangle, the three sides forming an arithmetic sequence is the ratio of the three sides being $3: 4: 5$ ( ).
(A) a sufficient but not necessary condition
(B) a necessary but not sufficient condition
(C) a sufficient and necessary condition
(D) neither a sufficient nor a necessary condition | 5. B.
The three sides of a triangle are $4, 5, 6$, forming an arithmetic sequence, and their ratio is not $3: 4: 5$; conversely, if the ratio of the three sides of a triangle is $3: 4: 5$, let the two side lengths be $3k, 4k, 5k (k>0)$, then they form an arithmetic sequence with a common difference of $k$. Therefore, ... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,333 |
1. In the Cartesian coordinate system, a point whose both horizontal and vertical coordinates are integers is called an integer point, such as $(-1,7)$ is an integer point. If the line $l$ passes through the points $A\left(\frac{1}{2}, \frac{1}{3}\right)$ and $B\left(\frac{1}{4}, \frac{1}{5}\right)$, then the integer p... | Ni.1. $(-2,-1)$.
Suppose the equation of line $l$ is $y=k x+b$. According to the problem, we have
$$
\left\{\begin{array}{l}
\frac{1}{3}=\frac{1}{2} k+b, \\
\frac{1}{5}=\frac{1}{4} k+b .
\end{array}\right.
$$
Solving these equations, we get $k=\frac{8}{15}, b=\frac{1}{15}$.
Thus, the equation of line $l$ is $y=\frac{8... | (-2,-1) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,334 |
Example 2 Quadrilateral $ABCD$ is inscribed in $\odot O$, its sides $AB$ and $DC$ intersect at point $P$, and $AD$ and $BC$ intersect at point $Q$. Two tangents $QE$ and $QF$ are drawn from point $Q$ to $\odot O$, with points of tangency at $E$ and $F$ respectively. Prove that points $P$, $E$, and $F$ are collinear. | Proof: As shown in Figure 5, connect $A E$, $C E$, $D E$, and $D F$.
Since $Q E$ and $Q F$ are both tangents to $\odot O$, we have,
$$
\begin{array}{l}
\angle A E F = \angle A D F \\
= 180^{\circ} - \angle Q D F, \\
\angle F E D = \angle Q F D.
\end{array}
$$
$$
\begin{array}{l}
\text{Also, } \angle P D A = 180^{\circ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,335 |
3.2005 real numbers $x_{1}, x_{2}, \cdots, x_{2000}$ satisfy
$$
\begin{array}{l}
\left|x_{1}-x_{2}\right|+\left|x_{2}-x_{3}\right|+\cdots+ \\
\left|x_{2004}-x_{2000}\right|+\left|x_{2000}-x_{1}\right|=1 .
\end{array}
$$
Then $\left|x_{1}\right|+\left|x_{2}\right|+\cdots+\left|x_{2000}\right|$ has a minimum value of $\... | 3. $\frac{1}{2}$.
For any positive integers $i, j$, we always have
$$
\left|x_{i}-x_{j}\right| \leqslant\left|x_{i}\right|+\left|x_{j}\right| \text { . }
$$
$$
\begin{array}{c}
\text { Then } 1=\left|x_{1}-x_{2}\right|+\left|x_{2}-x_{3}\right|+\cdots+ \\
\left|x_{2} \cos -x_{2} \cos \right|+\left|x_{2000}-x_{1}\right|... | \frac{1}{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 716,337 |
4. Given $x \in \mathbf{R}$. Then the function
$$
f(x)=\max \left\{\sin x, \cos x, \frac{\sin x+\cos x}{\sqrt{2}}\right\}
$$
The sum of the maximum and minimum values is $\qquad$ . | 4. $1-\frac{\sqrt{2}}{2}$.
Notice that
$$
\begin{array}{l}
f(x)=\max \left\{\sin x, \cos x, \frac{\sin x+\cos x}{\sqrt{2}}\right\} \\
=\max \left\{\sin x, \cos x, \sin \left(x+\frac{\pi}{4}\right)\right\},
\end{array}
$$
Obviously, the maximum value of $f(x)$ is 1. The minimum value of $\max \{\sin x, \cos x\}$ can b... | 1-\frac{\sqrt{2}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,338 |
5. The six faces of a unit cube are painted in 6 different colors, and a different number of roosters are drawn on each face. The colors and the number of roosters on each face correspond as shown in Table 1:
Table 1
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline Color on the face & Red & Yellow & Blue & Green & Purple & Gree... | 5.17.
Since the 4 aforementioned unit cubes are identical, it can be observed that the colors of the 4 adjacent faces to the red face are blue, yellow, cyan, and purple, so the color of the face opposite the red face is green. The colors of the 4 adjacent faces to the yellow face are cyan, blue, red, and green, thus t... | 17 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 716,339 |
Three, (10 points) Given that $a$, $b$, and $c$ are real numbers, the quadratic function $f(x) = a x^{2} + b x + c$ satisfies $f\left(\frac{a-b-c}{2 a}\right) = 0$. Prove that at least one of -1 and 1 is a root of $f(x) = 0$. | (1) From $f\left(\frac{a-b-c}{2 a}\right)=0$, we know that the quadratic function $f(x) = a x^{2} + b x + c$ has a zero.
If the quadratic function $f(x) = a x^{2} + b x + c$ has only one unique zero, then this zero is the vertex of the parabola. Therefore, we have $\frac{a-b-c}{2 a} = -\frac{b}{2 a}$, solving this give... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 716,340 |
Four. (15 points) For a cyclic convex quadrilateral $ABCD$, let the area be denoted as $S$, and the sides be $AB=a$, $BC=b$, $CD=c$, $DA=d$. Prove:
(1) $S=\sqrt{(p-a)(p-b)(p-c)(p-d)}$, where $p=\frac{a+b+c+d}{2}$;
(2) If the quadrilateral $ABCD$ has both an inscribed circle and a circumscribed circle, then $S=\sqrt{abc... | (1) Connect $A C$. Since quadrilateral $A B C D$ is a cyclic quadrilateral, we have
$$
\angle B+\angle D=180^{\circ} \text {. }
$$
Thus, $\cos B=-\cos D, \sin B=\sin D$. It is easy to see that,
$$
\begin{array}{l}
S=S_{\triangle B B C}+S_{\triangle A D C} \\
=\frac{1}{2}(a b \sin B+c d \sin D),
\end{array}
$$
which m... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,341 |
Five. (15 points) Let $p$ prime numbers $a_{1}, a_{2}, \cdots, a_{p}$ form an arithmetic sequence with a common difference of $d(d>0)$, and $a_{1}>p$. Prove:
(1) When $p$ is a prime number, $p \mid d$;
(2) When $p=15$, $d>30000$. | (1) Since $a_{1}>p, d>0$, therefore, $a_{1}, a_{2}, \cdots, a_{p}$ are all prime numbers greater than $p$. Thus, each $a_{i}(i=1,2, \cdots, p)$ cannot be divisible by $p$.
When $a_{1}, a_{2}, \cdots, a_{p}$ are divided by $p$, they can only take $p-1$ different remainders. According to the pigeonhole principle, at lea... | d \geqslant 30030 | Number Theory | proof | Yes | Yes | cn_contest | false | 716,342 |
1. Let $f(x)=\lg \frac{1-x}{1+x},|x|<1$. Then $f\left(\frac{x^{3}+3 x}{1+3 x^{2}}\right)$ equals ( ).
(A) $f^{2}(x)$
(B) $f^{3}(x)$
(C) $2 f(x)$
(D) $3 f(x)$ | \begin{array}{l}\text {-.1.D. } \\ f\left(\frac{x^{3}+3 x}{1+3 x^{2}}\right)=\lg \left(\frac{1-\frac{x^{3}+3 x}{1+3 x^{2}}}{1+\frac{x^{3}+3 x}{1+3 x^{2}}}\right)=\lg \left(\frac{1-x}{1+x}\right)^{3} \\ =3 f(x) .\end{array} | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,343 |
3. In $\triangle A B C$, $B C=a, A C=b, A B=c$. The necessary and sufficient condition for the equation $\sin ^{2} \frac{A}{2}+\sin ^{2} \frac{B}{2}+\sin ^{2} \frac{C}{2}=\cos ^{2} \frac{B}{2}$ to hold is ( ).
(A) $a+b=2 c$
(B) $b+c=2 a$
(C) $c+a=2 b$
(D) $a c=b^{2}$ | 3.C.
From the given, we have
$$
\frac{1-\cos A}{2}+\frac{1-\cos B}{2}+\frac{1-\cos C}{2}=\frac{1+\cos B}{2},
$$
it follows that $2 \sin \frac{B}{2}=\cos \frac{A-C}{2}$, hence $2 \sin B=\sin A+\sin C$. Therefore, $a+c=2 b$. The converse is also true. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,345 |
Example 3 In a right-angled $\triangle ABC$ inscribed in $\odot O$, tangents to $\odot O$ are drawn through points $B$ and $C$, and they intersect the tangent to $\odot O$ through point $A$ at points $M$ and $N$, respectively. $AD$ is the altitude from $A$ to side $BC$. Prove that $AD$ bisects $\angle MDN$.
(1988, All-... | Prove: As shown in Figure 6, let
\[
\angle M D A=\alpha, \angle N D A=\beta.
\]
We only need to prove that \(\alpha=\beta\).
Since \(M N, M B,\) and \(N C\) are all tangents to \(\odot O\),
\[
\begin{array}{l}
\angle M A B=\angle M B A=\angle A C B, \\
\angle N A C=\angle N C A=\angle A B C .
\end{array}
\]
Applying t... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,346 |
4. The vertex of the parabola is at the origin, the axis of symmetry is the $x$-axis, and the focus is on the line $3 x-4 y=12$. Then the equation of the parabola is $(\quad)$.
(A) $y^{2}=-12 x$
(B) $y^{2}=12 x$
(C) $y^{2}=-16 x$
(D) $y^{2}=16 x$ | 4. D.
From the vertex at the origin and the axis of symmetry being the $x$-axis, we know the equation of the parabola is $y^{2}=2 p x$. In $3 x-4 y=12$, setting $y=0$, we get the focus at $(4,0)$, so $p=8$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 716,347 |
5. Let $k<3, k \neq 0$. Then the conic sections
$$
\frac{x^{2}}{3-k}-\frac{y^{2}}{k}=1 \text { and } \frac{x^{2}}{5}+\frac{y^{2}}{2}=1
$$
must have ( ).
(A) different vertices
(B) different directrices
(C) the same foci
(D) the same eccentricity | 5.C.
When $00$, and $3-k>-k$. Then $\frac{x^{2}}{3-k}+\frac{y^{2}}{-k}=1$ represents an ellipse with foci on the $x$-axis. $a^{2}=3-k, b^{2}=-k$. Therefore, $a^{2}-b^{2}=3=c^{2}$, which means the ellipse has the same foci as the given ellipse. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,348 |
6. Connecting the diagonals of the regular pentagon $A_{1} A_{2} A_{3} A_{4} A_{5}$ intersects to form a regular pentagon $B_{1} B_{2} B_{3} B_{4} B_{5}$, and connecting the diagonals of the regular pentagon $B_{1} B_{2} B_{3} B_{4} B_{5}$ again intersects to form a regular pentagon $C_{1} C_{2} C_{3} C_{4} C_{5}$ (as ... | 6.C.
For any point $P$, the number of isosceles triangles with $P$ as the "vertex" (the common point of the two equal sides) is denoted as $[P]$, then
$$
\begin{array}{l}
{\left[A_{1}\right]=6\left(\triangle A_{1} A_{2} A_{5}, \triangle A_{1} B_{3} B_{4}, \triangle A_{1} B_{2} B_{5},\right.} \\
\left.\triangle A_{1} A... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 716,349 |
7. Let $f(x)$ satisfy the equation $f(x)-2 f\left(\frac{1}{x}\right)=x$. Then the range of $f(x)$ is $\qquad$ . | $=7 .\left(-\infty,-\frac{2 \sqrt{2}}{3}\right] \cup\left[\frac{2 \sqrt{2}}{3},+\infty\right)$.
From $f(x)-2 f\left(\frac{1}{x}\right)=x$, by replacing $x$ with $\frac{1}{x}$ we have
$$
f\left(\frac{1}{x}\right)-2 f(x)=\frac{1}{x} \text {. }
$$
Eliminating $f\left(\frac{1}{x}\right)$ from the two equations, we get $f(... | \left(-\infty,-\frac{2 \sqrt{2}}{3}\right] \cup\left[\frac{2 \sqrt{2}}{3},+\infty\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,350 |
8. If for any $\alpha, \beta$ satisfying $\alpha \pm \beta \neq k \cdot 360^{\circ}$, we have
$$
\begin{array}{l}
\frac{\sin \left(\alpha+30^{\circ}\right)+\sin \left(\beta-30^{\circ}\right)}{\cos \alpha-\cos \beta} \\
=m \cot \frac{\beta-\alpha}{2}+n,
\end{array}
$$
then the array $(m, n)=$ $\qquad$ | $$
\begin{array}{l}
\text { 8. }(m, n)=\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right) . \\
\text { The left side of equation (1) }=\frac{2 \sin \frac{\alpha+\beta}{2} \cdot \cos \frac{\alpha-\beta+60^{\circ}}{2}}{-2 \sin \frac{\alpha+\beta}{2} \cdot \sin \frac{\alpha-\beta}{2}} \\
=\frac{\sqrt{3}}{2} \cot \frac{\beta-\al... | (m, n)=\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 716,351 |
9. In the arithmetic sequences $3,10,17, \cdots, 2005$ and $3,8, 13, \cdots, 2003$, the number of terms that have the same value is $\qquad$. | 9.58.
Subtract 3 from each term of the two sequences, transforming them into $0,7,14, \cdots$, 2002 and $0,5,10, \cdots, 2000$. The first sequence represents multiples of 7 not exceeding 2002, and the second sequence can be seen as multiples of 5 within the same range. Therefore, the common terms are multiples of 35. ... | 58 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 716,352 |
10. If for all positive numbers $x, y$, the inequality $\sqrt{x}+\sqrt{y}$ $\leqslant a \sqrt{x+y}$ holds, then the minimum value of $a$ is $\qquad$ | 10. $\sqrt{2}$.
From $\left(\frac{\sqrt{x}}{\sqrt{x+y}}\right)^{2}+\left(\frac{\sqrt{y}}{\sqrt{x+y}}\right)^{2}=1$, we have $\frac{\sqrt{x}}{\sqrt{x+y}}+\frac{\sqrt{y}}{\sqrt{x+y}} \leqslant \sqrt{2}$.
When $x=y$, the equality holds, hence the minimum value of $a$ is $\sqrt{2}$. | \sqrt{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 716,353 |
11. If $2^{6}+2^{9}+2^{n}$ is a perfect square, then the positive integer $n=$ $\qquad$ . | 11. 10 .
$$
2^{6}+2^{9}=2^{6}\left(1+2^{3}\right)=2^{6} \times 3^{2}=24^{2} \text {. }
$$
Let $24^{2}+2^{n}=a^{2}$, then
$$
(a+24)(a-24)=2^{n} \text {. }
$$
Thus, $a+24=2^{r}, a-24=2^{t}$,
$$
2^{r}-2^{t}=48=2^{4} \times 3,2^{t}\left(2^{r-t}-1\right)=2^{4} \times 3 \text {. }
$$
Then $t=4, r-t=2$.
So $r=6, n=t+r=10$. | 10 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 716,354 |
12. Using weights labeled $1 \mathrm{~g}, 2 \mathrm{~g}, 3 \mathrm{~g}, 15 \mathrm{~g}, 40 \mathrm{~g}$, each one of them, to weigh objects on a balance scale without graduations. If weights can be placed on both ends of the balance, then the maximum number of different gram weights (positive integer weights) that can ... | 12.55.
Using $1 \mathrm{~g}, 2 \mathrm{~g}, 3 \mathrm{~g}$ these three weights, all integer grams in the interval $A=$ $[1,6]$ can be measured; after adding a $15 \mathrm{~g}$ weight, the measurement range is expanded to the interval $B=[15-6,15+6]=[9,21]$. After adding a $40 \mathrm{~g}$ weight, the measurement range... | 55 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,355 |
13. As shown in Figure 2, in the right triangle $\triangle ABC$, $E$ and $F$ are arbitrary points on the legs $AB$ and $AC$, respectively. From point $A$, perpendiculars are drawn to $BC$, $CE$, $EF$, and $FB$, with the feet of the perpendiculars being $M$, $N$, $P$, and $Q$, respectively. Prove that $M$, $N$, $P$, and... | Three, 13. Since points $A, E, N, P$ are concyclic, we have
$$
\angle C N P=\angle E A P=\angle A F P.
$$
Since points $A, N, M, C$ are concyclic, we have $\angle C N M=\angle C A M$.
Also, since points $A, B, M, Q$ are concyclic, we have $\angle M Q B=\angle M A B$.
Since points $A, P, Q, F$ are concyclic, we get $\a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,356 |
Example 4 In a convex pentagon $A B C D E$, $\angle A E D = \angle A B C = 90^{\circ}, \angle B A C = \angle E A D, B D \cap C E = F$. Prove that $A F \perp B E$.
(23rd IMO Preliminary Problem) | Proof: As shown in Figure 7, draw $AH \perp BE$ at $H$. Therefore, it is sufficient to prove that $AH, BD, CE$ are concurrent.
Since $\triangle ABC \sim \triangle AED$, we have
$$
\frac{AB}{AE} = \frac{AC}{AD} = \frac{BC}{DE}.
$$
Also, $\angle BAC = \angle EAD$, thus
$$
\angle BAD = \angle CAE.
$$
Therefore, $S_{\tr... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,357 |
14. The three sides of $\triangle ABC$ are of lengths $a$, $b$, and $c$. Prove:
$$
\frac{\left|a^{2}-b^{2}\right|}{c}+\frac{\left|b^{2}-c^{2}\right|}{a} \geqslant \frac{\left|c^{2}-a^{2}\right|}{b} .
$$ | 14. Since $a=2 R \sin A, b=2 R \sin B, c=2 R \sin C$, it suffices to prove
$$
\begin{array}{l}
\frac{\left|\sin ^{2} A-\sin ^{2} B\right|}{\sin C}+\frac{\left|\sin ^{2} B-\sin ^{2} C\right|}{\sin A} \\
\geqslant \frac{\left|\sin ^{2} C-\sin ^{2} A\right|}{\sin B}
\end{array}
$$
Notice that
$$
\begin{array}{l}
\sin ^{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 716,358 |
15. Find the smallest positive integer $n$, such that among any $n$ consecutive positive integers, there is at least one number whose sum of digits is a multiple of 7. | 15. First, we can point out 12 consecutive positive integers, for example,
$$
994,995, \cdots, 999,1000,1001, \cdots, 1005 \text {, }
$$
where the sum of the digits of any number is not a multiple of 7. Therefore, $n \geqslant 13$.
Next, we prove that in any 13 consecutive positive integers, there must be one number ... | 13 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 716,359 |
6.2. In the forest, oak trees and spruce trees grow. The owner cut down one third of the oaks and one sixth of the spruces. The environmental organization “Green Avengers” claimed that half of the trees in the forest were cut down. Prove: The claim contains incorrect elements. | 6.2. Because the forest contains oak and spruce trees, and the number of oaks that have been cut is less than half of the total number of oaks, and the number of spruces that have been cut is less than half of the total number of spruces, therefore, the number of trees that have been cut is less than half. | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 716,361 |
6.3. A five-digit decimal number $A$ has all its digits as 2 or 3, and a five-digit decimal number $B$ has all its digits as 3 or 4. Can the digits of the product $A B$ all be 2? Explain your reasoning. | 6.3. The product $A B$ lies between
$22222 \times 33333$ and $33333 \times 44444$,
which is between
740725926 and 1481451852.
Therefore, the first digit of $A B$ cannot be 2. | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 716,362 |
6.4. Multiplying a positive integer by 2 and then rearranging its digits in any order (but 0 cannot be placed at the beginning) is called an operation. Prove: It is impossible to obtain 74 from 1 after several such operations. | 6.4. Solution 1: If it is possible to get from 1 to 74, then by rearranging the digits and dividing even numbers by 2, it should also be possible to get from 74 to 1. First, rearrange the digits of 74 to get 47, which is an odd number and cannot be divided by 2, so the operation ends here. Next, divide 74 by 2 to get 3... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 716,363 |
7.1. Try to arrange four 1s, three 2s, and three 3s on a circle so that the sum of any three consecutive numbers is not a multiple of 3. | 7.1. It can be arranged as $1,1,2,1,2,2,3,3,1,3$. | 1,1,2,1,2,2,3,3,1,3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 716,364 |
7.3. On an island, there live 100 people, some of whom always lie, while the rest always tell the truth. Each resident of the island worships one of three gods: the Sun God, the Moon God, and the Earth God. Each resident was asked three questions:
(1) Do you worship the Sun God?
(2) Do you worship the Moon God?
(3) Do ... | 7.3. People who always tell the truth are called "honest people", and those who always lie are called "liars". Each honest person will only answer "yes" to one question, while each liar will answer "yes" to two questions. Let the number of honest people be $x$, and the number of liars be $y$. Therefore, $x+2y=130$. Sin... | 30 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 716,366 |
7.4. If a mushroom is parasitized by more than 11 worms, it is called "bad"; if a worm eats no more than $\frac{1}{5}$ of the mushroom it parasitizes, the worm is called "thin." It is known that $\frac{1}{4}$ of the mushrooms in the forest are bad. Prove: no less than $\frac{1}{3}$ of the worms are thin. | The total number of mushrooms is $4 k$. If a worm eats more than $\frac{1}{5}$ of the mushroom it is parasitizing, it is called "fat." Clearly, on 1 mushroom, there can be at most 4 fat worms, so the number of fat worms does not exceed $16 k$. Since at least 12 worms parasitize each bad mushroom, at least 8 of them are... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 716,367 |
Example 5 As shown in Figure 8, in $\triangle A B C$, $\angle B A C=40^{\circ}$, $\angle A B C=60^{\circ}$. Points $D$ and $E$ are taken on sides $A C$ and $A B$ respectively, such that $\angle C B D=40^{\circ}$, $\angle B C E=70^{\circ}$, and $B D \cap C E=F$. Prove that $A F \perp B C$. | Proof 1: Since $\angle A B C=60^{\circ}, \angle C B D=40^{\circ}$, $\angle B A C=40^{\circ}, \angle B C E=70^{\circ}$, then
$\angle A B D=20^{\circ}, \angle A C B=80^{\circ}, \angle A C E=10^{\circ}$.
Let $\angle B A F=x$, thus, $\angle F A C=40^{\circ}-x$.
Applying the trigonometric form of Ceva's Theorem to $\triangl... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 716,368 |
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