problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
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values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
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2. If the real number $x$ satisfies $\log _{2} \log _{2} x=\log _{4} \log _{4} x$, then $x=$ $\qquad$ . | 2. $\sqrt{2}$.
Let $y=\log _{a} \log _{a} x$. Then $a^{a^{y}}=x$. Let $a=2$ and $a=4$. According to the problem,
$$
\begin{array}{l}
2^{2^{y}}=4^{4^{y}}=\left(2^{2}\right)^{\left(2^{2}\right) y}=2^{2^{2 y+1}} \\
\Rightarrow y=2 y+1 \Rightarrow y=-1 \\
\Rightarrow x=\sqrt{2} .
\end{array}
$$ | \sqrt{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,957 |
3. As shown in Figure 1, given a square $A B C D$ with a side length of $5, E$ and $F$ are two points outside the square, $B E=D F=4, A E=C F=$ 3. Then $E F=$ . $\qquad$ | 3. $7 \sqrt{2}$.
It is known that, $\angle A E B=\angle C F D=90^{\circ}$.
Extend $E A$ and $F D$ to intersect at point $G$.
Since $\angle A D G=90^{\circ}-\angle C D F=\angle D C F=\angle B A E$, $\angle D A G=90^{\circ}-\angle B A E=\angle A B E$,
we know that $\triangle A G D \cong \triangle B E A$.
Therefore, $E G... | 7 \sqrt{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,958 |
4. Try to determine the largest integer not exceeding $\frac{\sqrt{14}+2}{\sqrt{14}-2}$ | 4. 3 .
Notice that,
$$
\begin{array}{l}
\frac{\sqrt{14}+2}{\sqrt{14}-2}=\frac{(\sqrt{14}+2)^{2}}{(\sqrt{14}-2)(\sqrt{14}+2)} \\
=\frac{14+4+4 \sqrt{14}}{14-4}=\frac{18+4 \sqrt{14}}{10} .
\end{array}
$$
And $3<\sqrt{14}<4$, thus, $30<18+4 \sqrt{14}<34$.
Therefore, $3<\frac{18+4 \sqrt{14}}{10}<3.4$.
Hence, the largest ... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,959 |
5. As shown in Figure 2, in quadrilateral $ABCD$, it is known that $\angle ACB = \angle ADB = 90^{\circ}$, $AC$ intersects $BD$ at point $E$, $AC = BC$, $AD = 4$, $BD = 7$. Then $S_{\triangle AEB}=$ $\qquad$ | 5. $11 \frac{9}{11}$.
From the problem, we know that points $A$, $B$, $C$, and $D$ are concyclic, and
$$
\angle C B D = \angle C A D \text{, }
$$
$\triangle A B C$ is an isosceles right triangle, with $\angle C B A = 45^{\circ}$.
Rotate $\triangle B C D$ around point $C$ so that point $B$ coincides with point $A$, the... | 11 \frac{9}{11} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,960 |
6. On each face of a cube, write a positive integer, and at each vertex, write the product of the positive integers on the three faces meeting at that vertex. If the sum of the numbers written at the eight vertices is 2014, then the sum of the numbers written on the six faces is $\qquad$ .
| 6. 74 .
As shown in Figure 4, let the pairs of positive integers written on the opposite faces of the cube be $(a, b),(c, d),(e, f)$.
Then the sum of the numbers written at each vertex is
$$
\begin{array}{l}
a c f+a d f+a c e+a d e+b c f+b c e+b d f+b d e \\
=(a+b)(c+d)(e+f) \\
=2014=2 \times 19 \times 53 .
\end{arra... | 74 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,961 |
Example 7 As shown in Figure 7, $P$ is any point on the midline $M N$ of $\triangle A B C$, and the extensions of $B P$ and $C P$ intersect $A C$ and $A B$ at points $D$ and $E$ respectively. Prove: $\frac{A D}{D C}+\frac{A E}{E B}=1$. | Prove as shown in Figure 7, draw a line through point $A$ parallel to $BC$, intersecting the extensions of $CP$ and $BP$ at points $F$ and $G$ respectively.
Since $FG \parallel BC$, we have
$$
\begin{array}{l}
\frac{AD}{DC}=\frac{AG}{BC}, \frac{AE}{EB}=\frac{AF}{BC}. \\
\text{Therefore, } \frac{AD}{DC}+\frac{AE}{EB} \\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,962 |
7. Given $A, B$ are digits in the set $\{0,1, \cdots, 9\}$, $r$ is a two-digit integer $\overline{A B}$, $s$ is a two-digit integer $\overline{B A}$, $r, s \in\{00,01$, $\cdots, 99\}$. When $|r-s|=k^{2}$ ( $k$ is an integer), the number of ordered pairs $(A, B)$ is $\qquad$. | 7. 42 .
Notice,
$$
|(10 A+B)-(10 B+A)|=9|A-B|=k^{2} \text {. }
$$
Then $|A-B|$ is a perfect square.
When $|A-B|=0$, there are 10 integer pairs:
$$
(A, B)=(0,0),(1,1), \cdots,(9,9) \text {; }
$$
When $|A-B|=1$, there are 18 integer pairs:
$$
(A, B)=(0,1),(1,2), \cdots,(8,9) \text {, }
$$
and their reverse numbers;
W... | 42 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,963 |
8. Given that $P$ is a moving point on the incircle of the equilateral $\triangle A B C$ with side length $6 \sqrt{3}$. Then the minimum value of $B P+\frac{1}{2} P C$ is $\qquad$ | 8. $\frac{3 \sqrt{21}}{2}$.
As shown in Figure 5, let the incircle of $\triangle ABC$ be $\odot O$, and take point $M$ on $OC$ such that $OM = \frac{1}{4} OC$.
Given that the side length of the equilateral $\triangle ABC$ is $6 \sqrt{3}$, we know
$$
\begin{array}{l}
OC = 6, OP = 3, OM = \frac{3}{2} \\
\Rightarrow \fra... | \frac{3 \sqrt{21}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,964 |
2. A line $l$ with an inclination angle of $45^{\circ}$ is drawn through the left focus $F$ of an ellipse, intersecting the ellipse at points $A$ and $B$. If $|B F|=2|A F|$, then the eccentricity of the ellipse is ( ).
(A) $\frac{1}{3}$
(B) $\frac{\sqrt{2}}{3}$
(C) $\frac{1}{2}$
(D) $\frac{\sqrt{2}}{2}$ | 2. B.
As shown in Figure 1, let $A F=d$. Then $B F=2 d$.
Figure 1
By the second definition of an ellipse, we know $A A^{\prime}=\frac{d}{e}, B B^{\prime}=\frac{2 d}{e}$. Therefore, $B C=\frac{d}{e}$.
Also, $B C=A C=\frac{3 \sqrt{2}}{2} d$, so $e=\frac{\sqrt{2}}{3}$. | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 727,965 |
3. Given an arithmetic sequence $\left\{a_{n}\right\}$ with a common difference of $d$ satisfying $d>0$, and $a_{2}$ is the geometric mean of $a_{1}$ and $a_{4}$. Let $b_{n}=a_{2} \cdot\left(n \in \mathbf{Z}_{+}\right)$, for any positive integer $n$ we have
$$
\frac{1}{b_{1}}+\frac{1}{b_{2}}+\cdots+\frac{1}{b_{n}}<2 \t... | 3. A.
From the problem, we know
$$
\begin{array}{l}
a_{2}^{2}=a_{1} a_{4} \Rightarrow\left(a_{1}+d\right)^{2}=a_{1}\left(a_{1}+3 d\right) \\
\Rightarrow a_{1}=d .
\end{array}
$$
Thus, $a_{n}=n d$. Therefore, $b_{n}=a_{2^{n}}=2^{n} d$.
Then $\frac{1}{d}\left[\frac{1}{2}+\left(\frac{1}{2}\right)^{2}+\cdots+\left(\frac{... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,966 |
4. Given that $a, b$ are positive real numbers, let
$$
\begin{aligned}
P & =\sqrt{\frac{a^{2}+b^{2}}{2}}-\frac{a+b}{2}, Q=\frac{a+b}{2}-\sqrt{a b}, \\
R & =\sqrt{a b}-\frac{2 a b}{a+b} .
\end{aligned}
$$
Then the correct judgment is ( ).
(A) $P \geqslant Q \geqslant R$
(B) $Q \geqslant P \geqslant R$
(C) $Q \geqslant ... | 4. B.
Notice,
$$
\begin{array}{l}
P \leqslant Q \Leftrightarrow \sqrt{\frac{a^{2}+b^{2}}{2}}+\sqrt{a b} \leqslant a+b \\
\Leftrightarrow 2 \sqrt{\frac{a^{2}+b^{2}}{2}} \cdot a b \leqslant \frac{a^{2}+b^{2}}{2}+a b .
\end{array}
$$
Thus, the inequality holds.
$$
\begin{array}{l}
P \geqslant R \Leftrightarrow \sqrt{\f... | B | Inequalities | MCQ | Yes | Yes | cn_contest | false | 727,967 |
5. Given a sphere with a radius of 6. Then the maximum volume of a regular tetrahedron inscribed in the sphere is ( ).
(A) $32 \sqrt{3}$
(B) $54 \sqrt{3}$
(C) $64 \sqrt{3}$
(D) $72 \sqrt{3}$ | 5. C.
Let the angle between the height and a lateral edge of a regular tetrahedron be $\theta$. Then the length of the lateral edge $l=12 \cos \theta$, and the height $h=12 \cos ^{2} \theta$.
Thus, the side length of the equilateral triangle at the base is
$$
a=12 \sqrt{3} \sin \theta \cdot \cos \theta \text {. }
$$
... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 727,968 |
Example 8 In the hexagon $A B C D E F$,
\[
\begin{array}{l}
A B \parallel D E, B C \parallel E F, C D \parallel F A, \\
A B + D E = B C + E F, A_{1} D_{1} = B_{1} E_{1},
\end{array}
\]
$A_{1}, B_{1}, D_{1}, E_{1}$ are the midpoints of $A B, B C, D E, E F$ respectively.
Prove: $\angle C D E = \angle A F E$. ${ }^{[s]}$
... | Prove as shown in Figure 8, construct $\square A B P F$, connect $D P$, take the midpoint $M$ of $D P$, and connect $B_{1} M, E_{1} M$.
It is easy to see that $B_{1} M, E_{1} M$ are the midlines of trapezoids $B C D P$ and $D E F P$, respectively. Then
$$
\begin{array}{l}
B_{1} M=\frac{B P+C D}{2}=\frac{A F+C D}{2}, B... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,969 |
9. Let $a, b$ be real numbers, for any real number $x$ satisfying $0 \leqslant x \leqslant 1$ we have $|a x+b| \leqslant 1$. Then the maximum value of $|20 a+14 b|+|20 a-14 b|$ is . $\qquad$ | 9.80 .
Let $x=0$, we know $|b| \leqslant 1$;
Let $x=1$, we know $|a+b| \leqslant 1$.
Therefore, $|a|=|a+b-b| \leqslant|a+b|+|b| \leqslant 2$, when $a=2, b=-1$, the equality can be achieved.
If $|20 a| \geqslant|14 b|$, then
$$
\begin{array}{l}
|20 a+14 b|+|20 a-14 b| \\
=2|20 a|=40|a| \leqslant 80 ;
\end{array}
$$
If... | 80 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,970 |
11. Given that vectors $\boldsymbol{\alpha}, \boldsymbol{\beta}$ are two mutually perpendicular unit vectors in a plane, and
$$
(3 \alpha-\gamma) \cdot(4 \beta-\gamma)=0 .
$$
Then the maximum value of $|\boldsymbol{\gamma}|$ is . $\qquad$ | 11.5.
As shown in Figure 2, let
$$
\begin{array}{l}
\overrightarrow{O A}=3 \alpha, \\
\overrightarrow{O B}=4 \beta, \\
\overrightarrow{O C}=\gamma .
\end{array}
$$
From the given information, $\overrightarrow{A C} \perp \overrightarrow{B C}$. Therefore, point $C$ lies on the circle with $A B$ as its diameter, and thi... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,971 |
13. Given that $a$ is a constant, the function
$$
f(x)=\ln \frac{1-x}{1+x}-a x \text {. }
$$
(1) Find the interval where the function $f(x)$ is monotonically decreasing;
(2) If $a=-\frac{8}{3}$, find the extremum of $f(x)$. | Three, 13. (1) From the problem, we know that the domain of the function $f(x)$ is $(-1,1)$, and
$$
f(x)=\ln (1-x)-\ln (1+x)-a x .
$$
Notice that,
$$
f^{\prime}(x)=\frac{-1}{1-x}-\frac{1}{1+x}-a=\frac{-2}{1-x^{2}}-a \text {. }
$$
Since $-1\frac{a+2}{a}$
$\Rightarrow \sqrt{\frac{a+2}{a}}0$;
When $\frac{1}{2}<x<1$, $f^... | not found | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 727,972 |
14. Given that for all $x \in \mathbf{R}$, $3 \sin ^{2} x-\cos ^{2} x+4 a \cos x+a^{2} \leqslant 31$. Find the range of real numbers $a$.
untranslated text remains the same as requested. | 14. Let $f(x)=3 \sin ^{2} x-\cos ^{2} x+4 a \cos x+a^{2}-31$.
Then $f(x)=-4 \cos ^{2} x+4 a \cos x+a^{2}-28$.
Let $t=\cos x$. Then $t \in[-1,1]$.
Therefore, when $t \in[-1,1]$, we always have
$$
g(t)=-4 t^{2}+4 a t+a^{2}-28 \leqslant 0 \text {. }
$$
Note that the axis of symmetry of the quadratic function $g(t)$ is $... | [-4,4] | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 727,973 |
15. Given $k$ as a positive integer, the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=3, a_{n+1}=\left(3^{\frac{2}{2 k-1}}-1\right) S_{n}+3\left(n \in \mathbf{Z}_{+}\right) \text {, }
$$
where $S_{n}$ is the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$.
Let $b_{n}=\frac{1}{n} \log _{3} a_{1}... | 15. From the problem, we know
Also, $a_{n+1}=\left(3^{\frac{2}{2 k-1}}-1\right) S_{n}+3$,
$$
a_{n}=\left(3^{\frac{2}{2 k-1}}-1\right) S_{n-1}+3(n \geqslant 2) \text {, }
$$
Thus, $a_{n+1}-a_{n}=\left(3^{\frac{2}{2 k-1}}-1\right) a_{n}$
$\Rightarrow a_{n+1}=3^{\frac{2}{2 k-1}} a_{n}$
$$
\Rightarrow a_{n}=a_{2}\left(3^... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,974 |
1. Let the set
$$
A=\{x \mid a x+2=0\}, B=\{-1,2\},
$$
satisfy $A \subseteq B$. Then all possible values of the real number $a$ are $\qquad$ | $$
\text { -,1. }-1,0,2 \text {. }
$$
When $a=0$, $A=\varnothing \subseteq B$, satisfying the condition;
When $a \neq 0$, $-\frac{2}{a}=-1$ or 2, solving for $a$ gives $a=2$ or -1. | -1,0,2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,975 |
2. A bag contains 5 red balls, 6 black balls, and 7 white balls, all of the same size and shape. Now, 14 balls are drawn at random. The probability of drawing exactly 3 red balls is $\qquad$ . | 2. $\frac{13}{51}$.
When 14 balls are drawn from a bag of 18 balls, the number of ways to draw is $\mathrm{C}_{18}^{14}$, among which the number of ways to draw exactly three red balls is $\mathrm{C}_{5}^{3} \mathrm{C}_{13}^{11}$. Therefore, the required probability is
$$
P=\frac{\mathrm{C}_{5}^{3} \mathrm{C}_{13}^{11... | \frac{13}{51} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,976 |
2. As shown in Figure 10, with the three sides of $\triangle ABC$ as sides, construct squares $ABDE$, $CAFG$, and $BCHK$ outside the triangle. Prove: the segments $EF$, $GH$, and $KD$ can form a triangle, and the area of the triangle formed is three times the area of $\triangle ABC$. | As shown in Figure 10, translate $\triangle D B K$ to $\triangle E A P$.
Using $\square D E P K$, $\square A P K B$, $\square A P H C$, $\square P H G F$ and $S_{\triangle D B K}=S_{\triangle C C H}=S_{\triangle E A F}=S_{\triangle A B C}$ to prove. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,977 |
3. The value of the complex number $\left(\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}\right)^{6 n}\left(n \in \mathbf{Z}_{+}\right)$ is | 3. 1 .
$$
\begin{array}{l}
\left(\frac{1}{2}+\frac{\sqrt{3}}{2} i\right)^{6 n}=\left[\left(\frac{1}{2}+\frac{\sqrt{3}}{2} i\right)^{3}\right]^{2 n} \\
=\left(\frac{1}{8}+\frac{3}{4} \times \frac{\sqrt{3}}{2} i-\frac{3}{2} \times \frac{3}{4}-\frac{3 \sqrt{3}}{8} i\right) \\
=(-1)^{2 n}=1 .
\end{array}
$$ | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,978 |
4. Given $\left\{\begin{array}{l}1 \leqslant x+y \leqslant 3, \\ -1 \leqslant x-y \leqslant 1 \text {. }\end{array}\right.$ Then the maximum value of $2 x^{2}-3 y$ is $\qquad$ . | 4.5.
Let $z=2 x^{2}-3 y$. Then
$$
y=\frac{2}{3} x^{2}-\frac{z}{3} \text{. }
$$
The shaded area in Figure 2 is the plane region represented by
$$
\left\{\begin{array}{l}
1 \leqslant x+y \leqslant 3, \\
-1 \leqslant x-y \leqslant 1
\end{array}\right.
$$
When the parabola $y=\frac{2}{3} x^{2}-\frac{z}{3}$ passes throug... | 4.5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,979 |
5. Given a geometric sequence $\left\{a_{n}\right\}$ with all terms being positive, it satisfies $\left|a_{2}-a_{3}\right|=14, a_{1} a_{2} a_{3}=343$.
Then the general term formula of the sequence $\left\{a_{n}\right\}$ is $a_{n}=$ $\qquad$ | $5.7 \times 3^{n-2}$.
From $a_{2}^{3}=a_{1} a_{2} a_{3}=343=7^{3} \Rightarrow a_{2}=7$.
From $\left|a_{2}-a_{3}\right|=14 \Rightarrow a_{3}=21$ or -7 (discard).
Thus, the common ratio $q=3, a_{n}=7 \times 3^{n-2}$. | 7 \times 3^{n-2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,980 |
6. Given that $\alpha$ is an acute angle, vectors
$$
a=(\cos \alpha, \sin \alpha), b=(1,-1)
$$
satisfy $a \cdot b=\frac{2 \sqrt{2}}{3}$. Then $\sin \left(\alpha+\frac{5 \pi}{12}\right)=$ $\qquad$ | 6. $\frac{2+\sqrt{15}}{6}$.
From the problem, we know
$$
\cos \alpha - \sin \alpha = \frac{2 \sqrt{2}}{3} \Rightarrow \cos \left(\alpha + \frac{\pi}{4}\right) = \frac{2}{3} > 0 \text{. }
$$
Since $\alpha$ is an acute angle, we have
$$
\sin \left(\alpha + \frac{\pi}{4}\right) = \frac{\sqrt{5}}{3} \text{. }
$$
Therefo... | \frac{2 + \sqrt{15}}{6} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,981 |
7. If the equation $x^{2}-x y-2 y^{2}+x+a=0$ represents two straight lines, then $a=$ | 7. $\frac{2}{9}$.
Notice that, $x^{2}-x y-2 y^{2}=(x+y)(x-2 y)$.
Let $x^{2}-x y-2 y^{2}+x+a$
$$
=(x+y+m)(x-2 y+2 m) \text {. }
$$
Expanding the right side of the above equation and comparing the coefficients on both sides, we get
$$
\left\{\begin{array} { l }
{ a = 2 m ^ { 2 } , } \\
{ 3 m = 1 }
\end{array} \Rightar... | \frac{2}{9} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,982 |
8. Given $(\sqrt{2}+1)^{21}=a+b \sqrt{2}$, where $a, b$ are positive integers. Then $(b, 27)=$ $\qquad$ | 8. 1 .
Notice,
$$
\begin{array}{l}
(\sqrt{2}+1)^{21} \\
=(\sqrt{2})^{21}+\mathrm{C}_{21}^{1}(\sqrt{2})^{20}+\mathrm{C}_{21}^{2}(\sqrt{2})^{19}+\cdots+ \\
\mathrm{C}_{21}^{20} \sqrt{2}+\mathrm{C}_{21}^{21}, \\
(\sqrt{2}-1)^{21} \\
=(\sqrt{2})^{21}-\mathrm{C}_{21}^{1}(\sqrt{2})^{20}+\mathrm{C}_{21}^{2}(\sqrt{2})^{19}-\c... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,983 |
9. (16 points) As shown in Figure 1, in rectangle $A B C D$, it is known that $A B=2, A D=4$. Points $E$ and $F$ are on $A D$ and $B C$ respectively, and $A E=1, B F=3$. Quadrilateral $A E F B$ is folded along $E F$, such that the projection of point $B$ on plane $C D E F$ is point $H$ on line $D E$. Find the size of t... | 9. As shown in Figure 3, draw $E R / / D C$, and draw $E S \perp$ plane $E F C D$. Establish a spatial rectangular coordinate system with $E R$, $E D$, and $E S$ as the $x$, $y$, and $z$ axes, respectively.
Notice that the projection of point $B$ on plane $C D E F$ is point $H$ on line $D E$. Let point $B(0, y, z)(y>0... | 135^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,984 |
10. (20 points) Through the right focus $F$ of the ellipse $C: \frac{x^{2}}{25}+\frac{y^{2}}{16}=1$, a line is drawn intersecting the ellipse $C$ at points $A$ and $B$. It is known that $|A B|=8$. Find the equation of the line $A B$.
untranslated part:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Note: The part "将上面的文本翻译成英文，... | 10. From the problem, we know the point $F(3,0)$. Let $l_{A B}: x=m y+3$.
From $\left\{\begin{array}{l}16 x^{2}+25 y^{2}=400, \\ x=m y+3,\end{array}\right.$ we have
$$
\left(16 m^{2}+25\right) y^{2}+96 m y-256=0 \text {. }
$$
Let points $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$. Then
$$
y_{1}+y_{2}=\fr... | 2 x \pm \sqrt{5} y-6=0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,985 |
11. (20 points) Given that for any positive integer $n$ we have $\left(1+\frac{1}{n}\right)^{n-a} \geqslant \mathrm{e}$.
Find the range of values for the real number $a$.
| 11. Taking the natural logarithm of both sides of the given inequality, we get
$$
\begin{array}{l}
(n-a) \ln \left(1+\frac{1}{n}\right) \geqslant 1 \\
\Rightarrow\left(1-\frac{a}{n}\right) \ln \left(1+\frac{1}{n}\right) \geqslant \frac{1}{n} \\
\Rightarrow\left(1-\frac{a}{n}\right) \ln \left(1+\frac{1}{n}\right)-\frac{... | a \leqslant -\frac{1}{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 727,986 |
1. A social club has $2 k+1$ members, each of whom is proficient in the same $k$ languages, and any two members can only communicate in one language. If there do not exist three members who can communicate pairwise in the same language, let $A$ be the number of subsets of three members, where each subset's three member... | 1. The maximum value of $A$ is $\frac{2 k(k-2)(2 k+1)}{3}$.
Consider the social club as a complete graph with $2 k+1$ vertices, where each language corresponds to a color, and the edge connecting two members who speak the same language is colored with the corresponding color.
Let the set of vertices be $V$, and the s... | \frac{2 k(k-2)(2 k+1)}{3} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,987 |
3. Let $P$ and $Q$ be points on the sides $BC$ and $AD$ of a convex quadrilateral $ABCD$, respectively, such that $\frac{AQ}{QD} = \frac{BP}{PC} = \lambda$. The line $PQ$ intersects the lines $AB$ and $CD$ at points $E$ and $F$, respectively. Prove:
$$
\lambda = \frac{AB}{CD} \Leftrightarrow \angle BEP = \angle PFC.
$$ | Given: As shown in Figure 11,
$P R // C G$.
We can prove
$$
\begin{array}{l}
\angle B E P = \angle P F C \\
\Leftrightarrow \angle B A R = \angle G A R \\
\Leftrightarrow \frac{B R}{R G} = \frac{A B}{A G} \\
\Leftrightarrow \lambda = \frac{A B}{C D} .
\end{array}
$$ | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,988 |
5. If non-congruent $\triangle A B C$ and $\triangle X Y Z$ are called a pair of "buddies," they must satisfy the following conditions:
(1) The areas of the two triangles are equal;
(2) Let the midpoints of $B C$ and $Y Z$ be $M$ and $W$, respectively. The sets $\{A B, A M, A C\}$ and $\{X Y, X W, X Z\}$ are the same t... | 5. The answer is affirmative.
First, we prove two lemmas.
Lemma 1 The following statement and its converse are both true.
If \( q, r, s \) are three distinct positive real numbers, and \( q, r, 2s \) are the side lengths of a certain triangle, then there exists a unique \(\triangle PQR\) such that \( PQ = q, PR = r, P... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,990 |
6. In the acute triangle $\triangle ABC$, the circle $\Gamma_{1}$ with diameter $AC$ intersects side $BC$ at point $F$ (different from point $C$), and the circle $\Gamma_{2}$ with diameter $BC$ intersects side $AC$ at point $E$ (different from point $C$). The ray $AF$ intersects circle $\Gamma_{2}$ at two points $K$ an... | 6. As shown in Figure 2, let $C D \perp A B$ at point $D$, and $H$ be the orthocenter of $\triangle A B C$. Then circles $\Gamma_{1}$ and $\Gamma_{2}$ both intersect $A B$ at point $D$.
By the power of a point theorem, we have
$$
L H \cdot H N=C H \cdot H D=K H \cdot H M .
$$
Therefore, points $K, L, M, N$ are concycl... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,991 |
7. Given that $n$ is a fixed positive integer, in an $n \times 2n$ rectangular grid, each cell contains 0 or 1, and each row has $n$ zeros and $n$ ones. For $1 \leqslant k \leqslant n$ and $1 \leqslant i \leqslant n$, define $a_{k, i}$ such that the $i$-th 0 in the $k$-th row is in the $a_{k, i}$-th column. For each in... | 7. First, give a bijection between the grid table $C \in \mathscr{F}$ and a "good" tiling of a regular hexagon $H$ with side length $n$, where the hexagon $H$ is tiled with parallelograms formed by two unit equilateral triangles, and such a tiling is called "good".
For a good tiling of a regular hexagon $H$, one side ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 727,992 |
8. Does there exist a ternary integer polynomial $P(x, y, z)$ that satisfies the following property: a positive integer $n$ is not a perfect square if and only if there is a ternary positive integer tuple $(x, y, z)$ such that
$$
P(x, y, z)=n ?
$$ | 8. Existence
Assume a ternary integer coefficient polynomial satisfies: for all integers $x, y, z$,
(1) $Q(x, y, z) \geqslant 0$;
(2) If $Q(x, y, z)=0$, then $x$ is not a perfect square;
(3) For every non-perfect square $x$, if $x$ is a positive integer, then there exist $y, z \in \mathbf{Z}_{+}$ such that $Q(x, y, z)... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 727,993 |
1.200 people stand in a circle, some of whom are honest people, and some are liars. Liars always tell lies, while honest people tell the truth depending on the situation. If both of his neighbors are honest people, he will definitely tell the truth; if at least one of his neighbors is a liar, he may sometimes tell the ... | 1. There can be at most 150 honest people.
Since liars do not say they are liars, the 100 people who say they are liars are all honest people, and what they say is false. This indicates that they are all adjacent to liars.
Since each liar is adjacent to at most two honest people, there must be at least 50 liars. This ... | 150 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 727,994 |
2. On a $13 \times 13$ chessboard, two opposite corner squares (Translator's note: such as the top-right and bottom-left corners) are removed. In the remaining part, some squares are colored black. Prove: It is possible to place no more than 47 kings on the black squares so that they can attack all the empty black squa... | 2. As shown in Figure 1, divide the remaining part of a $13 \times 13$ chessboard into 47 parts, such that a chess king placed on any square within each part can attack all other squares in that part.
Next, it is only necessary to place the kings based on the distribution of the black squares (at most one king per par... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 727,995 |
3. Given a convex quadrilateral $A B C D$ with two equal diagonals intersecting at point $O$, point $P$ is located inside $\triangle A O D$ such that $C D / / B P$, $A B / / C P$. Prove: Point $P$ lies on the bisector of $\angle A O D$. | 3. From $A B / / C P, C D / / B P$, we know
$S_{\triangle A P C}=S_{\triangle B P C}, S_{\triangle B P C}=S_{\triangle B P D}$.
Therefore, $S_{\triangle A P C}=S_{\triangle B P D}$.
Since $A C=B D$, the heights from these two triangles to sides $A C$ and $B D$ are equal.
This indicates that the distances from point $P... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,996 |
4. Let \(0 \leqslant x, y, z \leqslant 1\). Prove:
\[
\frac{x^{2}}{1+x+x y z}+\frac{y^{2}}{1+y+x y z}+\frac{z^{2}}{1+z+x y z} \leqslant 1 \text {. }
\] | 4. Since $(1-y)(1-z) \geqslant 0$, therefore, $1+yz \geqslant y+z$.
$$
\begin{array}{l}
\text { then } 1+x+xyz=1+x(1+yz) \\
\geqslant 1+x(y+z)=1+xy+xz \\
\geqslant x^2+xy+xz.
\end{array}
$$
$$
\text { Hence } \frac{x^2}{1+x+xyz} \leqslant \frac{x^2}{x^2+xy+xz}=\frac{x}{x+y+z} \text {. }
$$
Similarly, $\frac{y^2}{1+y+x... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 727,997 |
6. In a convex quadrilateral $ABCD$, it is known that $AB=CD$. Points $K$ and $M$ are taken on sides $AB$ and $CD$ respectively, such that $AM=KC$, $BM=KD$. Prove that the angle between line $AB$ and $KM$ is equal to the angle between line $KM$ and $CD$. | 6. Proof 1: By the three sides being equal, we know that
$\triangle A B M \cong \triangle C D K$.
Thus, their corresponding altitudes $K K_{1}$ and $M M_{1}$ are equal.
If they are equal to $K M$, then they coincide with $K M$, and the conclusion of the problem is obviously true.
If they are less than $K M$, then
Rt $\... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,998 |
Example 1 Let $n$ be an odd positive integer, write the numbers $1,2,3, \cdots, 2 n$ on the blackboard, then take any two numbers $a, b$, erase these two numbers and write down $|a-b|$. Prove: The last number left is an odd number. | Proof Let $S$ be the sum of all numbers on the blackboard.
At the beginning, the sum
$S=1+2+\cdots+2 n=n(2 n+1)$ (odd).
Since each step reduces $S$ by $2 \min (a, b)$ (even), the parity of $S$ is an invariant. Throughout the simplification process, we always have $S \equiv 1(\bmod 2)$.
Therefore, the final result is an... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 727,999 |
Example 2 Find all integer triples $(x, y, z)$ such that $x^{3}+y^{3}+z^{3}-3 x y z=2003$.
untranslated text is retained in its original format and line breaks. | The left side of the equation is a familiar model, that is,
$$
\begin{array}{l}
x^{3}+y^{3}+z^{3}-3 x y z \\
=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-x z\right) .
\end{array}
$$
Notice that,
$$
\begin{array}{l}
(x-y)^{2}+(y-z)^{2}+(x-z)^{2} \\
=2\left(x^{2}+y^{2}+z^{2}-x y-y z-x z\right) .
\end{array}
$$
And $(x-y)^{2... | (668,668,667),(668,667,668),(667,668,668) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,000 |
3. Let integers $a, b, c, d$ not all be equal, starting from $(a, b, c, d)$ and repeatedly transforming $(a, b, c, d)$ into $(a-b, b-c, c-d, d-a)$. Prove: at least one number in the quadruple will eventually become arbitrarily large. | Let $p_{n}=\left(a_{n}, b_{n}, c_{n}, d_{n}\right)$ be the quadruple after $n$ iterations. Then
$$
\begin{aligned}
& a_{n}+b_{n}+c_{n}+d_{n}=0(n \geqslant 1) . \\
& \text { Also } a_{n+1}^{2}+b_{n+1}^{2}+c_{n+1}^{2}+d_{n+1}^{2} \\
& =2\left(a_{n}^{2}+b_{n}^{2}+c_{n}^{2}+d_{n}^{2}\right)+\left(a_{n}+c_{n}\right)^{2}+\le... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 728,001 |
8. A $3 \times 3$ grid with the following properties is called a "T-grid":
(1) Five cells are filled with 1, and four cells are filled with 0;
(2) Among the three rows, three columns, and two diagonals, at most one of these eight lines has three numbers that are pairwise equal.
Then the number of different T-grids is $... | 8. 68.
First, the number of all ways to fill a $3 \times 3$ grid with five 1s and four 0s is $\mathrm{C}_{9}^{4}=126$.
Next, consider the number of ways that do not satisfy property (2), i.e., methods that result in at least two lines of three equal numbers (hereafter referred to as good lines).
We will count these ... | 68 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,002 |
9. (16 points) Given the function
$$
f(x)=a x^{2}+b x+c(a>0),
$$
and $3 a+4 b+6 c=0$.
Prove: $f(x)$ must have a zero in the interval $(0,1)$. | $$
\begin{array}{l}
f(0)=c, f(1)=a+b+c=\frac{1}{4}(a-2 c), \\
f\left(\frac{2}{3}\right)=-\frac{1}{18} a<0$, then $f(0) f\left(\frac{2}{3}\right)<0$;
if $c \leqslant 0$, then $f(1) f\left(\frac{2}{3}\right)<0$.
Therefore, the conclusion holds.
$$ | proof | Algebra | proof | Yes | Yes | cn_contest | false | 728,003 |
10. (20 points) Given the hyperbola $x^{2}-y^{2}=2$ with its left and right foci at points $F_{1}$ and $F_{2}$, respectively, a tangent line is drawn through the fixed point $P(2,3)$ to the hyperbola $x^{2}-y^{2}=2$. The points of tangency are $A$ and $B$, with the x-coordinate of point $A$ being less than that of poin... | 10. (1) Let points $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)\left(x_{1}<x_{2}\right)$. Then the equations of the two tangent lines $l_{1}, l_{2}$ at points $A$ and $B$ are
$$
x_{1} x - y_{1} y = 2, \quad x_{2} x - y_{2} y = 2.
$$
Since point $P(2,3)$ lies on the lines $l_{1}$ and $l_{2}$, we have
$$
2 x_... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,004 |
11. (20 points) Given real numbers $x, y$ satisfy $3^{x}+3^{y}=9^{x}+9^{y}$.
Find the range of $U=27^{x}+27^{y}$. | 11. Let $a=3^{x}, b=3^{y}$. Then the given equation can be transformed into
$$
\begin{array}{l}
a+b=a^{2}+b^{2}(a, b>0) \\
\Rightarrow\left(a-\frac{1}{2}\right)^{2}+\left(b-\frac{1}{2}\right)^{2}=\left(\frac{\sqrt{2}}{2}\right)^{2}
\end{array}
$$
From the graph of equation (1) in the $a O b$ plane, we know that
$$
t=a... | (1,2] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,005 |
一、(40 points) As shown in Figure 1, given that $AB$ is the longest side of the convex quadrilateral $ABCD$, points $M$ and $N$ are on $AB$ and $BC$ respectively, and $AN$ and $CM$ both bisect the area of quadrilateral $ABCD$. Prove: line segment $MN$ bisects diagonal $BD$. | Given $S_{\text {quadrilateral } M A D C}=\frac{1}{2} S_{\text {quadrilateral } A B C D}=S_{\text {quadrilateral } A A D C}$, we know
$$
\begin{array}{l}
S_{\triangle M A C}=S_{\triangle N A C} \Rightarrow M N \parallel A C \\
\Rightarrow S_{\triangle G A C}=S_{\triangle M A C} \\
\Rightarrow S_{\text {quadrilateral } ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,006 |
II. (40 points) Given positive sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ satisfy for any positive integer $n$,
$$
a_{n+2}=a_{n}+a_{n+1}^{2}, \quad b_{n+2}=b_{n}^{2}+b_{n+1} \text {, }
$$
and $a_{1}>1, a_{2}>1, b_{1}>1, b_{2}>1$. Prove:
(1) For any positive integer $n(n \geqslant 2)$,
$$
a_{n+2}>a_{n}... | (1) From the sequence of positive numbers $\left\{a_{n}\right\}$, we know
$$
a_{n+2}=a_{n}+a_{n+1}^{2}>a_{n+1}^{2}=\left(a_{n-1}+a_{n}^{2}\right)^{2}>a_{n}^{4} \text {. }
$$
(2) Clearly, the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ are strictly increasing, hence each term is greater than 1.
Then $b_{... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 728,007 |
Three. (50 points) Let $I_{k}$ denote the decimal number consisting of $k$ ones (e.g., $I_{1}=1, I_{3}=111$), and define $\{n\}!=\prod_{k=1}^{n} I_{k}$.
(1) For any positive integers $m, n$, let
$$
f(m, n)=\frac{\{m+n\}!}{\{m\}!\cdot\{n\}!},
$$
write a recursive relation for $f(m, n)$ and prove it;
(2) Prove that for ... | (1) Supplement the definition $\{0\}!=1$.
For positive integers $m, n$ we have
$$
\begin{array}{l}
f(m, n)=\frac{\{m+n\}!}{\{m\}!\cdot\{n\}!} \\
=\frac{\prod_{k=1}^{m+n} I_{k}}{\left(\prod_{k=1}^{m} I_{k}\right)\left(\prod_{k=1}^{n} I_{k}\right)}=\frac{I_{m+n}^{m+n} \prod_{k=1}^{m-1} I_{k}}{\left(\prod_{k=1}^{m} I_{k}\... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 728,008 |
Four. (50 points) A country has 53 cities, and between any two cities, there is either a two-way road directly connecting them or no direct road. It is known that there are 312 roads between these 53 cities, and from any city, one can reach all other cities through the roads. Each city has at most 12 roads leading to o... | Four, it can definitely be reached.
Considering these 53 cities as 53 vertices, if there is a two-way road directly connecting two cities, then connect an undirected edge between the corresponding vertices. Thus, construct the graph $G$. According to the problem, $G$ is a simple connected graph. Let $D$ be the diameter... | D \leqslant 12 | Combinatorics | proof | Yes | Yes | cn_contest | false | 728,009 |
As shown in Figure 1, given a square $A B C D$ with a side length of 1, the diagonals $A C$ and $B D$ intersect at point $O, E$ is a point on the extension of side $B C$, connect $A E$, which intersects $B D$ and $C D$ at points $P$ and $F$ respectively, connect $B F$ and extend it to intersect segment $D E$ at point $... | Given $P S / / A C$, we have
$\triangle D P S \backsim \triangle D O C \Rightarrow \frac{D S}{D C}=\frac{P S}{O C}$.
Similarly, $\frac{E S}{E O}=\frac{P S}{O A}$.
Since quadrilateral $A B C D$ is a square, we know $O A=O C$.
Thus, $\frac{D S}{D C}=\frac{E S}{E O} \Rightarrow D E / / A C$.
Also, $A D / / C E$, so quadri... | \frac{2 \sqrt{5}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,010 |
As shown in Figure 2, in the square $A B C D$, $P$ is a point on the $\frac{1}{4}$ circular arc with $C$ as the center and $C B$ as the radius, and $P B > P D$. Point $Q$ is on the line segment $P B$, such that $P Q = P D$, and $A Q$ intersects $D P$ at point $R$. Find the value of $\frac{A R}{R Q}$. | Solve as shown in Figure 3, draw a perpendicular from point $B$ to $DR$, with the foot of the perpendicular at $H$. Draw a line through point $R$ parallel to $PB$ intersecting $HB$ at point $S$, and connect $HA$ and $BD$.
Since quadrilateral $ABCD$ is a square,
$\Rightarrow \angle ABC=\angle ADC=90^{\circ}$
$\Rightarr... | \frac{\sqrt{2}}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 728,011 |
4. Let $x, y$ be positive integers. If for every positive integer $n$ we have
$$
\left(2^{n} y+1\right) \mid\left(x^{2^{n}}-1\right) \text {, }
$$
prove: $x=1$. | Notice,
$$
x^{2 n}-1=\left(x^{2}-1\right) \prod_{i=1}^{n-1}\left(x^{2 i}+1\right) \text {, }
$$
for $i \geqslant 1, x^{2 i}+1$ has all its odd prime factors congruent to 1 modulo 4.
Thus, every prime factor of $2^{n} y+1$ that is congruent to 3 modulo 4 must divide $x^{2}-1$.
It suffices to prove: there are infinitel... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 728,012 |
Given point $P$ inside $\triangle ABC$, the ray $AP$ intersects the circumcircle of $\triangle ABC$ at point $Q$. Through point $P$, draw $EF \parallel BC$, $DG \parallel AC$, and $NM \parallel AB$, where $E$, $F$, $D$, $G$, $N$, and $M$ are all on the sides of $\triangle ABC$. Prove:
$$
A P \cdot P Q = P E \cdot P F +... | Prove a lemma first.
Lemma In figure $4, \square A B C D$ has vertex $A$ on the circle, and rays $A B, A D, A C$ intersect the circle at points $B^{\prime}, D^{\prime}, C^{\prime}$.
Then $A B \cdot A B^{\prime}+A D \cdot A D^{\prime}=A C \cdot A C^{\prime}$.
Proof It is easy to see that $\triangle D^{\prime} C^{\prime... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,013 |
Given $\triangle ABC$ with its circumcenter, incenter, and orthocenter as $O$, $I$, and $H$ respectively, and semicircles $\odot O_{1}$, $\odot O_{2}$, $\odot O_{3}$ are constructed outwardly on $BC$, $CA$, $AB$ respectively. Lines through point $A$ parallel to $IB$ and $IC$ intersect the semicircles $\odot O_{3}$ and ... | Proof (1) Since $O_{1}, O_{2}, O_{3}$ are the midpoints of sides $BC, CA$, and $AB$ respectively, we have:
$$
\begin{array}{l}
\angle B O_{1} A_{1}=2 \angle O_{1} C A_{1}=2 \angle I B C \\
=\angle A B C=\angle O_{2} O_{1} C .
\end{array}
$$
Thus, $O_{2}, O_{1}, A_{1}$ are collinear.
Similarly, $B_{2}, O_{2}, O_{1}$ ar... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,014 |
Example 1: From the nine equally divided points on a circle, choose five points and color them red. Prove: There exist six different triangles $\triangle_{1}, \triangle_{2}, \cdots, \triangle_{6}$ with red points as vertices, satisfying
$$
\triangle_{1} \cong \triangle_{2}, \triangle_{3} \cong \triangle_{4}, \triangle_... | Prove as shown in Figure 1.
Notice,
(1) In the isosceles trapezoid $ABCD$ with bases $AD$ and $BC$, there are two pairs of congruent triangles:
$\triangle ABC \cong \triangle DCB, \triangle BAD \cong \triangle CDA$,
and each vertex of the trapezoid appears twice in one of the pairs of congruent triangles.
(2) If $M$ i... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,015 |
Example 2 As shown in Figure 3, two equilateral triangles overlap to form a six-pointed star. Now, the positive integers 1, 2, $\cdots, 12$ are to be filled in the 12 nodes of the figure, such that the sum of the four numbers on each straight line is equal.
(1) Try to find the minimum value of the sum of the numbers at... | (1) Solution: For any arrangement that satisfies the conditions, the number filled at point $a_i$ $(i=1,2, \cdots, 12)$ is still denoted as $a_i$. If the sum of the four numbers on each line is $s$, then
$6 s=2(1+2+\cdots+12) \Rightarrow s=26$.
Therefore, in $\triangle a_{1} a_{3} a_{5}$ and $\triangle a_{2} a_{4} a_{6... | 24 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,016 |
Example 3 Given irrational numbers $a, b$. Prove: The equation
$$
\left|x+a y+\frac{1}{3}\right|+\left|a x-y+\frac{a}{3}\right|=b
$$
has at most one integer solution $(x, y)$. | Proof for the case where $b \neq 0$.
Proof by contradiction.
Suppose there are two sets of integers $x, y$ and $x_{1}, y_{1}$ that both satisfy the equation. Then
$$
\begin{array}{l}
\left|x+a y+\frac{1}{3}\right|+\left|a x-y+\frac{a}{3}\right| \\
=\left|x_{1}+a y_{1}+\frac{1}{3}\right|+\left|a x_{1}-y_{1}+\frac{a}{3}\... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 728,017 |
Example 4 If the real number $a>\sqrt{5}$, prove: the rational number $\frac{p}{q}$ (in lowest terms) satisfying
$$
\left|\frac{\sqrt{5}-1}{2}-\frac{p}{q}\right|<\frac{1}{a q^{2}}
$$
is only finite in number. | Prove that for any reduced fraction $\frac{p}{q}$, let $\frac{\sqrt{5}-1}{2}-\frac{p}{q}=\frac{\alpha}{q^{2}}$, where $|\alpha|<\frac{\sqrt{5}}{2}$. If $\alpha>0$, then when $q$ is sufficiently large, $\left|\frac{\alpha}{\sqrt{5} q^{2}}-1\right|<1$. Hence, $\left|\frac{\alpha^{2}}{q^{2}}-\sqrt{5} \alpha\right|=|\sqrt{... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 728,018 |
Example 5 Let real numbers $a \geqslant b \geqslant c \geqslant d>0$. Find the minimum value of the function
$$
\begin{array}{l}
f(a, b, c, d) \\
=\left(1+\frac{c}{a+b}\right)\left(1+\frac{d}{b+c}\right)\left(1+\frac{a}{c+d}\right)\left(1+\frac{b}{d+a}\right)
\end{array}
$$ | When $a=b=c=d$, the value of $f$ is $\left(\frac{3}{2}\right)^{4}$.
It suffices to prove: For any positive numbers $a, b, c, d$ that satisfy the condition, we have
$$
f(a, b, c, d) \geqslant\left(\frac{3}{2}\right)^{4},
$$
which means we need to prove
$$
\begin{array}{l}
\frac{a+b+c}{3} \cdot \frac{b+c+d}{3} \cdot \fr... | \left(\frac{3}{2}\right)^{4} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 728,019 |
Example 2 Proof:
$$
y=\frac{a}{\sec ^{n} \alpha}-\frac{b}{\tan ^{n} \alpha} \leqslant\left(a^{\frac{2}{n+2}}+b^{\frac{2}{n+2}}\right)^{\frac{n+2}{2}},
$$
where, $\alpha \in\left(0, \frac{\pi}{2}\right), a>b>0, n \in \mathbf{Z}_{+}$. | Prove that from equation (4), taking $p=-\frac{n}{2}<0$, we have
$$
\begin{aligned}
y & =\frac{\left(\sec ^{2} \alpha\right)^{p}}{\left(a^{\frac{1}{1-p}}\right)^{p-1}}-\frac{\left(\tan ^{2} \alpha\right)^{p}}{\left(b^{\frac{1}{1-p}}\right)^{p-1}} \\
& \leqslant \frac{\left(\sec ^{2} \alpha-\tan ^{2} \alpha\right)^{p}}{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 728,022 |
Example 3: Each vertex of a regular pentagon is labeled with an integer, such that the sum of the five integers is positive. If among three consecutive vertices the numbers $x, y, z$ satisfy $y<0$, then the following operation can be performed: $(x, y, z) \rightarrow (x+y, -y, y+z)$. As long as there are still negative... | Notice that, during the operation process, the sum of the five numbers remains unchanged. Consider the sum of the squares of the differences of alternating two numbers.
Let the original five numbers be $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$. Then
$$
\begin{array}{l}
f=\left(x_{1}-x_{3}\right)^{2}+\left(x_{2}-x_{4}\right)... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 728,023 |
Example 3 Let $a, b, c > 0, ab + bc + ca = \frac{1}{3}$. Prove:
$$
\frac{1}{a^{2}-bc+1}+\frac{1}{b^{2}-ca+1}+\frac{1}{c^{2}-ab+1} \leqslant 3 \text {. }
$$ | Proof Let $M=a+b+c$,
$$
N=a b+b c+c a=\frac{1}{3},
$$
where, “ $\sum$ ” denotes the cyclic symmetric sum.
Then the original inequality can be transformed into
$$
\begin{array}{l}
\sum \frac{1}{a^{2}-b c+N+2 N} \leqslant \frac{1}{N} \\
\Leftrightarrow \sum \frac{N}{a M+2 N} \leqslant 1 \\
\Leftrightarrow \sum\left(\fra... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 728,024 |
Example 4 Let $p, q, r$ be positive numbers, and satisfy $pqr=1$. Prove: for all $n \in \mathbf{Z}_{+}$, we have
$$
\frac{1}{p^{n}+q^{n}+1}+\frac{1}{q^{n}+r^{n}+1}+\frac{1}{r^{n}+p^{n}+1} \leqslant 1 .
$$ | Let's make the substitution $a=p^{n}, b=q^{n}, c=r^{n}$. Then $a b c=1$, and equation (5) $\Leftrightarrow \frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1} \leqslant 1$.
By the rearrangement inequality, we have
$$
\begin{array}{l}
a+b=a^{\frac{2}{3}} a^{\frac{1}{3}}+b^{\frac{2}{3}} b^{\frac{1}{3}} \geqslant a^{\frac{2}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 728,025 |
Example 5 If $m, r_{1}, r_{2}, \cdots, r_{k} \in \mathbf{R}_{+}, m \geqslant r_{1}+r_{2}+$ $\cdots+r_{k}+1, a_{i} \in \mathbf{R}_{+}, b_{i 1}, b_{i 2}, \cdots, b_{i k} \in \mathbf{R}_{+}, i=1,2$, $\cdots, n$, prove:
$$
\begin{array}{l}
\sum_{i=1}^{n} \frac{a_{i}^{m}}{b_{i 1}^{r_{1}} b_{i 2}^{r_{2}} \cdots b_{i k}^{r_{k... | Let \( c_{i}=\frac{a_{i}^{m}}{b_{i 1}^{\tau} b_{i 2}^{\tau_{2}} \cdots b_{i k}^{r}} \), \( \delta=m-\left(r_{1}+r_{2}+\cdots+r_{k}\right)-1 \).
Also, let \( \theta_{0}=\frac{1}{m} \), \( \theta_{j}=\frac{r_{j}}{m} \) (for \( j=1,2, \cdots, k \)), and \( \theta_{k+1}=\frac{\delta}{m} \).
Denote the left side of inequali... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 728,026 |
Example 6 Let $x_{1}, x_{2}, \cdots, x_{n}$ be positive numbers. Satisfy
$$
\begin{array}{l}
\sum_{i=1}^{n} x_{i}=\sum_{i=1}^{n} \frac{1}{x_{i}} . \\
\text { Prove: } \sum_{i=1}^{n} \frac{1}{n-1+x_{i}} \leqslant 1 .
\end{array}
$$ | $$
\begin{array}{l}
\text { Proof by contradiction. } \\
\text { Let } y_{i}=\frac{1}{n-1+x_{i}}. \\
\text { Then } x_{i}=\frac{1}{y_{i}}-(n-1) \text { for } 0<y_{i}<1, \text { and we can assume } \sum_{i=1}^{n} y_{i}=1+\frac{1}{k}n . \\
\text { However, } \sum_{i=1}^{n} x_{i}=\sum_{i=1}^{n}\left[\frac{1}{y_{i}}-(n-1)\... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 728,027 |
Example 7 Proof: For any positive real numbers $a$, $b$, $c$, we have $1<\frac{a}{\sqrt{a^{2}+b^{2}}}+\frac{b}{\sqrt{b^{2}+c^{2}}}+\frac{c}{\sqrt{c^{2}+a^{2}}} \leqslant \frac{3 \sqrt{2}}{2}$. | Prove that
$$
\begin{array}{c}
\frac{a}{\sqrt{a^{2}+b^{2}}}>\frac{a}{\sqrt{a^{2}+b^{2}+c^{2}}}, \\
\frac{b}{\sqrt{b^{2}+c^{2}}}>\frac{b}{\sqrt{a^{2}+b^{2}+c^{2}}}, \\
\frac{c}{\sqrt{c^{2}+a^{2}}}>\frac{c}{\sqrt{a^{2}+b^{2}+c^{2}}}.
\end{array}
$$
Adding the three inequalities, we get
$$
\sum \frac{a}{\sqrt{a^{2}+b^{2}... | \frac{3 \sqrt{2}}{2} | Inequalities | proof | Yes | Yes | cn_contest | false | 728,028 |
Example 8 Given $0 \leqslant a, b, c < 1$, and $ab + bc + ca = 1$. Prove: $\frac{a}{1-a^{2}} + \frac{b}{1-b^{2}} + \frac{c}{1-c^{2}} \geqslant \frac{3 \sqrt{3}}{2}$. | Given the equation, without loss of generality, let
$$
a=\cot A, b=\cot B, c=\cot C \text {, }
$$
where $\angle A, \angle B, \angle C$ are the three interior angles of a triangle, and from $0<a, b, c<1$, we know $\frac{\pi}{4}<\angle A, \angle B, \angle C<\frac{\pi}{2}$.
Thus,
$$
\frac{a}{1-a^{2}}+\frac{b}{1-b^{2}}+\f... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 728,029 |
Example 1 Solve the equation $\sqrt[4]{10+x}+\sqrt[4]{7-x}=3$.
| Solve by mean substitution
$$
\sqrt[4]{7-x}=\frac{3}{2}-d, \sqrt[4]{10+x}=\frac{3}{2}+d .
$$
Raising both sides of the two equations to the fourth power and then adding them yields
$$
16 d^{4}+216 d^{2}-55=0 \text {. }
$$
Solving for $d$ gives $d= \pm \frac{1}{2}$.
When $d=\frac{1}{2}$, $x_{1}=6$;
When $d=-\frac{1}{2... | x_1=6, x_2=-9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,030 |
Example 2 Solve the equation $\sqrt{x-\frac{1}{x}}+\sqrt{1-\frac{1}{x}}=x$. | Let $u=\sqrt{x-\frac{1}{x}}, v=\sqrt{1-\frac{1}{x}}$.
Then $u+v=x \neq 0, u^{2}-v^{2}=x-1$.
Dividing the two equations gives $u-v=1-\frac{1}{x}$.
Thus $2 u=x+1-\frac{1}{x}=u^{2}+1$
$\Rightarrow(u-1)^{2}=0 \Rightarrow u=\sqrt{x-\frac{1}{x}}=1$
$\Rightarrow x=\frac{1+\sqrt{5}}{2}$ (negative value discarded). | x=\frac{1+\sqrt{5}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,031 |
Example 3 Solve the equation
$$
x^{2}-x+1=\left(x^{2}+x+1\right)\left(x^{2}+2 x+4\right) \text {. }
$$ | Let $t=x^{2}+x+1>0$. Then $t-2 x=t(t+x+3)$.
Factoring, we get $(t+2)(t+x)=0$.
Since $t+2>0$, we have
$$
t=x^{2}+x+1=-x \text{. }
$$
Solving, we get $x=-1$. | x=-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,032 |
Example 4 Solve the equation
$$
x^{2}-x-1000 \sqrt{1+8000 x}=1000 .
$$ | Solution: Clearly, $x \neq 0$. The original equation can be transformed into
$$
\begin{array}{l}
x^{2}-x=1000(1+\sqrt{1+8000 x}) \\
=\frac{8 \times 10^{6} x}{\sqrt{1+8000 x}-1} .
\end{array}
$$
Then $x-1=\frac{8 \times 10^{6}}{\sqrt{1+8000 x}-1}$.
Let $a=10^{3}, t=\sqrt{1+8 a x}-1$.
Thus, $x=\frac{t^{2}+2 t}{8 a}$, an... | 2001 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,033 |
Example 4 Let $S_{n}$ denote the sum of the first $n$ prime numbers, for example, $S_{3}=$ $2+3+5=10$. Is there any pair of consecutive terms in the sequence $\left\{S_{n}\right\}$ that are both perfect squares? ${ }^{[3]}$ | Notice that, the sum of the first $n$ terms is an invariant.
Let the $n$-th prime number be denoted as $p_{n}$. Suppose there exists a positive integer $m$ $(m>1)$, such that
$$
S_{m-1}=k^{2}, S_{m}=l^{2}\left(k, l \in \mathbf{Z}_{+}\right) .
$$
Since $S_{2}=5, S_{3}=10$ are not perfect squares, we know $m>4$.
And $p_... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,034 |
Example 5 Solve the equation
$$
x \sqrt{4-y^{2}}-y \sqrt{9-z^{2}}-z \sqrt{9-x^{2}}=11 .
$$ | By the two-variable mean inequality, we have
$$
\begin{array}{l}
x \sqrt{4-y^{2}}-y \sqrt{9-z^{2}}-z \sqrt{9-x^{2}} \\
\leqslant \frac{1}{2}\left[x^{2}+4-y^{2}+(-y)^{2}+9-z^{2}+(-z)^{2}+9-x^{2}\right] \\
=11,
\end{array}
$$
Equality holds if and only if $\left\{\begin{array}{l}x=\sqrt{4-y^{2}}, \\ -y=\sqrt{9-z^{2}} \\... | x=\sqrt{2}, y=-\sqrt{2}, z=-\sqrt{7} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,035 |
Example 6 Solve the equation
$$
\sqrt{2-x}+\sqrt{x+3 y-5}+\sqrt{y+2}=\sqrt{12 y-3} .
$$ | By the three-variable mean inequality, we have
$$
\begin{array}{l}
\sqrt{2-x}+\sqrt{x+3 y-5}+\sqrt{y+2} \\
\leqslant 3 \sqrt{\frac{2-x+x+3 y-5+y+2}{3}} \\
=\sqrt{12 y-3} .
\end{array}
$$
The equality holds if and only if $2-x=x+3 y-5=y+2 \geqslant 0$.
Solving this, we get $x=-7, y=7$.
Upon verification, these values s... | x=-7, y=7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,036 |
Example 7 Solve the equation
$$
\sqrt{x^{2}+\sqrt{3} x+\frac{7}{4}}+\sqrt{x^{2}-3 \sqrt{3} x+\frac{31}{4}}=4 \text {. }
$$ | By the triangle inequality, we have
$$
\begin{array}{l}
\sqrt{\left(x+\frac{\sqrt{3}}{2}\right)^{2}+1^{2}}+\sqrt{\left(x-\frac{3 \sqrt{3}}{2}\right)^{2}+(-1)^{2}} \\
\geqslant \sqrt{\left[\left(x+\frac{\sqrt{3}}{2}\right)-\left(x-\frac{3 \sqrt{3}}{2}\right)\right]^{2}+[1-(-1)]^{2}} \\
=4 .
\end{array}
$$
Equality hold... | x=\frac{\sqrt{3}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,037 |
Example 8 Solve the equation
$$
x+\sqrt{x}+\sqrt{x+2}+\sqrt{x^{2}+2 x}=3 .
$$ | 【Analysis and Solution】Let the function be
$$
f(x)=x+\sqrt{x}+\sqrt{x+2}+\sqrt{x^{2}+2 x} \text {. }
$$
Then $f(x)$ is an increasing function on the interval $[0,+\infty)$.
Also, by observation, $f\left(\frac{1}{4}\right)=3$, thus, the original equation has a unique solution
$$
x=\frac{1}{4} \text {. }
$$ | x=\frac{1}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,038 |
Example 9 Solve the system of equations
$$
\left\{\begin{array}{l}
x^{3}=y^{2}+y+\frac{1}{3}, \\
y^{3}=z^{2}+z+\frac{1}{3} \\
z^{3}=x^{2}+x+\frac{1}{3} .
\end{array}\right.
$$ | 【Analysis and Solution】It is easy to know that $x, y, z > 0$.
Let the function $f(t) = t^2 + t + \frac{1}{3}$, which is monotonically increasing on the interval $(0, +\infty)$.
Then the system of equations becomes
$$
\left\{\begin{array}{l}
x^3 = f(y), \\
y^3 = f(z), \\
z^3 = f(x).
\end{array}\right.
$$
If $x > y$, th... | x = y = z = \frac{1}{\sqrt[3]{4} - 1} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,039 |
Question $1^{\prime}$ Let $x_{1}, x_{2}, \cdots, x_{n} \in \mathbf{R}_{+}$, and $\sum_{i=1}^{n} x_{i}=1$. Prove:
$$
\prod_{i=1}^{n}\left(1+\frac{1}{x_{i}}\right) \geqslant(n+1)^{n} .
$$ | Proof 1 uses the algebraic-geometric mean inequality. Note that,
$$
\begin{array}{l}
1+\frac{1}{x_{i}}=\frac{x_{i}+1}{x_{i}} \\
=\frac{x_{i}+x_{1}+x_{2}+\cdots+x_{i}+\cdots+x_{n}}{x_{i}} \\
\geqslant \frac{(n+1)^{n+1} \sqrt{x_{i} x_{1} x_{2} \cdots x_{i} \cdots x_{n}}}{x_{i}} . \\
\text { Therefore, } \prod_{i=1}^{n}\l... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 728,040 |
Question 2' Let $x_{1}, x_{2}, \cdots, x_{n} \in \mathbf{R}_{+}$, and $\sum_{i=1}^{n} x_{i}=1$. Prove:
$$
\prod_{i=1}^{n}\left(x_{i}+\frac{1}{x_{i}}\right) \geqslant\left(n+\frac{1}{n}\right)^{n} .
$$ | Prove that by Hölder's inequality,
$$
\prod_{i=1}^{n}\left(x_{i}+\frac{1}{x_{i}}\right) \geqslant\left(\sqrt[n]{\prod_{i=1}^{n} x_{i}}+\frac{1}{\sqrt[n]{\prod_{i=1}^{n} \frac{1}{x_{i}}}}\right)^{n} .
$$
By the Arithmetic Mean-Geometric Mean Inequality (AM-GM Inequality),
$$
\sqrt[n]{\prod_{i=1}^{n} x_{i}} \leqslant \f... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 728,041 |
Question 3' Let $x_{1}, x_{2}, \cdots, x_{n} \in \mathbf{R}_{+}$, and $\sum_{i=1}^{n} x_{i}=1$, $m>1$. Prove:
$$
\sum_{i=1}^{n}\left(x_{i}+\frac{1}{x_{i}}\right)^{m} \geqslant \frac{\left(n^{2}+1\right)^{m}}{n^{m-1}} .
$$ | $$
\begin{array}{l}
\sum_{i=1}^{n}\left(x_{i}+\frac{1}{x_{i}}\right)^{m}=\sum_{i=1}^{n} \frac{\left(x_{i}+\frac{1}{x_{i}}\right)^{m}}{1^{m-1}} \\
\geqslant \frac{\left[\sum_{i=1}^{n}\left(x_{i}+\frac{1}{x_{i}}\right)\right]^{m}}{n^{m-1}}=\frac{\left(1+\sum_{i=1}^{n} \frac{1^{2}}{x_{i}}\right)^{m}}{n^{m-1}} . \\
\text {... | \frac{\left(n^{2}+1\right)^{m}}{n^{m-1}} | Inequalities | proof | Yes | Yes | cn_contest | false | 728,042 |
1. Let $a_{0}<a_{1}<\cdots$ be an infinite sequence of positive integers. Prove: there exists a unique integer $n(n \geqslant 1)$, such that
$$
a_{n}<\frac{a_{0}+a_{1}+\cdots+a_{n}}{n} \leqslant a_{n+1} .
$$ | For $n=1,2, \cdots$, define
$$
d_{n}=\left(a_{0}+a_{1}+\cdots+a_{n}\right)-n a_{n} \text {. }
$$
Then the first inequality in the given problem is equivalent to
$$
d_{n}>0 \text {. }
$$
Notice that,
$$
\begin{array}{l}
n a_{n+1}-\left(a_{0}+a_{1}+\cdots+a_{n}\right) \\
=(n+1) a_{n+1}-\left(a_{0}+a_{1}+\cdots+a_{n}+a_... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 728,043 |
2. Let $n \geqslant 2$ be an integer. Consider an $n \times n$ chessboard composed of $n^{2}$ unit squares. If each row and each column contains exactly one “rook”, then a placement of $n$ rooks is called “peaceful”. Find the largest positive integer $k$, such that for any peaceful placement of $n$ rooks, there exists ... | 2. $[\sqrt{n-1}]$.
Let $l$ be a positive integer. Next, we prove two conclusions, thereby illustrating that the maximum value sought is $k_{\text {max }}=[\sqrt{n-1}]$.
(1) If $n>l^{2}$, then for any peaceful placement of $n$ rooks, there exists an empty $l \times l$ square.
(2) If $n \leqslant l^{2}$, then there exis... | [\sqrt{n-1}] | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 728,044 |
Example 5 Given positive integers $a>b>1$, and the equation
$$
\frac{a^{x}-1}{a-1}=\frac{b^{y}-1}{b-1}(x>1, y>1)
$$
has at least two different positive integer solutions $(x, y)$. Prove: the positive integers $a, b$ are coprime. | Prove that if $(a, b)=p$ (prime), $v_{p}(a)$ represents the exponent of the factor $p$ in $a$. Then
$$
v_{p}(a)=v_{p}(b) .
$$
Assume the equation has two roots $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)\left(x_{1}>y_{1}\right)$.
$$
\begin{array}{l}
\text { Then } \sum_{i=x_{2}}^{x_{1}-1} a^{i}=\sum_{i=y_{2}}... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 728,045 |
3. In the convex quadrilateral $A B C D$, it is known that $\angle A B C=$ $\angle C D A=90^{\circ}$, point $H$ is the foot of the perpendicular from $A$ to $B D$, and points $S, T$ are on sides $A B, A D$ respectively, such that $H$ is inside $\triangle S C T$, and
$$
\begin{array}{l}
\angle C H S-\angle C S B=90^{\ci... | 3. As shown in Figure 2, let the line passing through point $C$ and perpendicular to line $S C$ intersect $A B$ at point $Q$.
Then $\angle S Q C=90^{\circ}-\angle B S C=180^{\circ}-\angle S H C$.
Therefore, points $C, H, S, Q$ are concyclic.
Since $S Q$ is the diameter of this circle, the circumcenter $K$ of $\triangle... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,046 |
4. Let points $P, Q$ be on the side $BC$ of acute $\triangle ABC$, satisfying
$$
\angle PAB = \angle BCA \text{, and } \angle CAQ = \angle ABC \text{, }
$$
Points $M, N$ lie on lines $AP, AQ$ respectively, such that $P$ is the midpoint of $AM$, and $Q$ is the midpoint of $AN$. Prove: The intersection of lines $BM$ and ... | 4. As shown in Figure 4, let the intersection of line $B M$ and $C N$ be point $S$. Denote $\angle Q A C=\angle A B C=\beta, \angle P A B=\angle A C B=\gamma$. Therefore, $\triangle A B P \backsim \triangle C A Q$. Hence, $\frac{B P}{P M}=\frac{B P}{P A}=\frac{A Q}{Q C}=\frac{N Q}{Q C}$.
Furthermore, since $\angle B P ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 728,047 |
5. For every positive integer $n$, the Cape Town Bank issues a coin with a value of $\frac{1}{n}$. Given a finite number of such coins (not necessarily distinct) with a total value not exceeding $99+\frac{1}{2}$, prove that they can be divided into at most 100 groups, such that the sum of the values of the coins in eac... | 5. Prove a general conclusion:
For any positive integer $N$, given a finite number of such coins with a total value not exceeding $N-\frac{1}{2}$, it is always possible to divide them into at most $N$ groups such that the sum of the values of the coins in each group is at most 1.
If the sum of the values of some coins ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 728,048 |
6. A family of lines in the plane is said to be "in general position" if no two lines are parallel and no three lines are concurrent. A family of lines in general position divides the plane into several regions, and the regions with finite area are called "bounded regions" of this family of lines. Prove: For sufficient... | 6. The case $c=1$.
If a point is the intersection of a red line and a blue line, then it is called a red point. The intersection of two blue lines is called a "blue point". For a red line $l$, there exists a unique finite region $A$ whose only red edge is a part of the red line $l$. Denote its vertices in clockwise or... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 728,049 |
1. Given $n$ is a positive integer, $a_{1}, a_{2}, \cdots, a_{n-1}$ are arbitrary real numbers, define the sequences $u_{0}, u_{1}, \cdots, u_{n}$ and $v_{0}, v_{1}, \cdots, v_{n}$ satisfying
$$
\begin{array}{l}
u_{0}=u_{1}=v_{0}=v_{1}=1, \\
u_{k+1}=u_{k}+a_{k} u_{k-1}, v_{k+1}=v_{k}+a_{n-k} v_{k-1},
\end{array}
$$
wh... | 1. Prove by mathematical induction on $k$
$$
u_{k}=\sum_{\substack{02}} a_{i_{1}} a_{i_{2}} \cdots a_{i_{i}},
$$
where the trivial sum is defined as 1 (this case corresponds to $t=0$, when the sequence is empty, and its product is 1).
For $k=0,1$, the sum on the right-hand side of (1) only includes the product of the... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 728,050 |
2. Prove: For any set of 2000 distinct real numbers, there exist real numbers $a, b(a>b)$ and $c, d(c>d)$, and $a \neq c$ or $b \neq d$, such that
$$
\left|\frac{a-b}{c-d}-1\right|<\frac{1}{100000} .
$$ | 2. For any set $S$ consisting of $n(n=2000)$ distinct real numbers, let the absolute values of the differences between any two different numbers in these $n$ numbers be $D_{1}, D_{2}, \cdots, D_{m}$, and $D_{1} \leqslant D_{2} \leqslant \cdots \leqslant D_{m}$.
Then $m=\frac{n(n-1)}{2}$.
By rescaling the scale, we can ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 728,051 |
4. Let $n$ be a positive integer, and consider the sequence of positive integers $a_{1}, a_{2}, \cdots, a_{n}$ with period $n$. Extend this sequence to an infinite sequence, i.e., for all positive integers $i$, we have $a_{n+i}=a_{i}$. If $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n} \leqslant a_{1}+n$, and f... | 4. First prove
$$
a_{i} \leqslant n+i-1(i=1,2, \cdots, n) .
$$
Assume there exists $i$ such that $a_{i}>n+i-1$. Consider the smallest $i$ for which this inequality holds.
By $a_{n} \geqslant a_{n-1} \geqslant \cdots \geqslant a_{i} \geqslant n+i$, and
$$
a_{a_{i}} \leqslant n+i-1 \text {, }
$$
then $a_{i} \neq i, i+1... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 728,052 |
5. Find all functions $f: \mathbf{N} \rightarrow \mathbf{N}$, such that for all $n \in \mathbf{N}$, we have
$$
f(f(f(n)))=f(n+1)+1 .
$$ | 5. For $n \in \mathbf{N}$, there are two functions satisfying the conditions:
$f(n)=n+1$,
$$
f(n)=\left\{\begin{array}{ll}
n+1, & n \equiv 0 \text { or } 2(\bmod 4) ; \\
n+5, & n \equiv 1(\bmod 4) ; \\
n-3, & n \equiv 3(\bmod 4) .
\end{array}\right.
$$
Let $h^{0}(x)=x$,
$h^{k}(x)=\underbrace{h(\cdots h}_{k \uparrow}(x... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,053 |
6. Let $m \neq 0$ be an integer. Find all real-coefficient polynomials $P(x)$ such that for any real number $x$, we have
$$
\begin{array}{l}
\left(x^{3}-m x^{2}+1\right) P(x+1)+\left(x^{3}+m x^{2}+1\right) P(x-1) \\
=2\left(x^{3}-m x+1\right) P(x) .
\end{array}
$$ | 6. Let $P(x)=a_{n} x^{n}+\cdots+a_{1} x+a_{0}\left(a_{n} \neq 0\right)$.
Comparing the coefficients of $x^{n+1}$ on both sides of equation (1), we get
$a_{n}(n-2 m)(n-1)=0$.
Thus, $n=1$ or $2 m$.
If $n=1$, it is easy to see that $P(x)=x$ is a solution to equation (1), and $P(x)=1$ is not a solution to equation (1).
Si... | P(x)=t x(t \in \mathbf{R}) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 728,054 |
1. If $a+b=\sqrt{5}$, then
$$
\frac{a^{4}+a^{2} b^{2}+b^{4}}{a^{2}+a b+b^{2}}+3 a b=(\quad) \text {. }
$$
(A) 5
(B) $\frac{3 \sqrt{5}}{2}$
(C) $2 \sqrt{5}$
(D) $\frac{5 \sqrt{5}}{2}$ | $\begin{array}{l}\text {-1. A. } \\ \frac{a^{4}+a^{2} b^{2}+b^{4}}{a^{2}+a b+b^{2}}+3 a b \\ =\frac{\left(a^{2}+a b+b^{2}\right)\left(a^{2}-a b+b^{2}\right)}{a^{2}+a b+b^{2}}+3 a b \\ =\left(a^{2}-a b+b^{2}\right)+3 a b \\ =(a+b)^{2}=5 .\end{array}$ | 5 | Algebra | MCQ | Yes | Yes | cn_contest | false | 728,055 |
3. Among the numbers $29998$, $29999$, $30000$, $30001$, the one that can be expressed as the sum of the products of three consecutive natural numbers taken two at a time is ( ).
(A) 30001
(B) 30000
(C) 29999
(D) 29998 | 3. C.
Notice that, for three consecutive natural numbers $n-1, n, n+1$, the sum of their pairwise products has the form
$$
n(n-1)+n(n+1)+(n-1)(n+1)=3 n^{2}-1 \text {. }
$$
The above expression leaves a remainder of 2 when divided by 3, while 30001 and 29998 leave a remainder of 1 when divided by 3, and 30000 is divis... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 728,057 |
4. Given $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$ are two points on the graph of the inverse proportion function $y=\frac{1}{x}$ in the first quadrant of the Cartesian coordinate system $x O y$, satisfying $y_{1}+y_{2}=\frac{7}{2}, x_{2}-x_{1}=\frac{5}{3}$. Then $S_{\triangle A O B}=$ ( ).
(A) $2 \frac{... | 4. B.
As shown in Figure 2, add auxiliary lines, and let $O B$ intersect $A C$ at point $P$.
$$
\begin{array}{l}
\text { Given } x_{1} y_{1}=x_{2} y_{2}=1 \\
\Rightarrow S_{\triangle A O C}=S_{\triangle B O D} \\
\Rightarrow S_{\triangle A O B}=S_{\text {quadrilateral } A O D B}-S_{\triangle B O D} \\
\quad=S_{\text {... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 728,058 |
5. There are 2015 integers, and by taking any 2014 of them and adding them together, their sum can exactly take on the 2014 different integer values $1, 2, \cdots, 2014$. Then the sum of these 2015 integers is ( ).
(A) 1004
(B) 1005
(C) 1006
(D) 1008 | 5. D.
Let 2015 integers be $x_{1}, x_{2}, \cdots, x_{2015}$, and denote $x_{1}+x_{2}+\cdots+x_{2015}=M$.
Assume without loss of generality that $M-x_{i}=i(i=1,2, \cdots, 2014)$, $M-x_{2015}=A$.
Then $2014 M=1+2+\cdots+2014+A$.
Thus, the remainder when $A$ is divided by 2014 is 1007.
Therefore, $A=1007, M=1008$.
When $... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 728,059 |
1. Among the natural numbers from $1 \sim 10000$, the integers that are neither perfect squares nor perfect cubes are $\qquad$ in number. | In the natural numbers from $1 \sim 10000$, there are 100 perfect squares.
$$
\begin{array}{l}
\text { Because } 22^{3}=10648>10000, \\
21^{3}=9261<10000,
\end{array}
$$
Therefore, there are 21 perfect cubes.
Next, consider the number of natural numbers from $1 \sim 10000$ that are both perfect squares and perfect cub... | 9883 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 728,060 |
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