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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class | __index_level_0__ int64 0 742k |
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7. The number of positive integers $n$ that make $\frac{n}{30-n}$ a positive integer is $(\quad)$.
(A) 4
(B) 5
(C) 6
(D) 7
(E) 8 | 7. D.
Notice,
$$
\frac{n}{30-n}=\frac{n-30+30}{30-n}=\frac{30}{30-n}-1
$$
is a positive integer.
Therefore, $(30-n) \mid 30$.
Hence, $30-n=1,2,3,5,6,10,15$.
Thus, $n=29,28,27,25,24,20,15$, a total of seven. | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 727,846 |
9. As shown in Figure 1, in the convex
quadrilateral $A B C D$,
$$
\begin{array}{l}
A B=3, B C=4, C D \\
=13, A D=12,
\end{array}
$$
$\angle A B C=90^{\circ}$. Then the area $S$ of quadrilateral $A B C D$ is ( ).
(A) 30
(B) 36
(C) 40
(D) 48
(E) 58.5 | 9. B.
As shown in Figure 1, connect $A C$. In the right triangle $\triangle A B C$,
$$
\angle A B C=90^{\circ}, A B=3, B C=4 \text {. }
$$
Therefore, $A C=\sqrt{A B^{2}+A C^{2}}=5$.
In $\triangle A C D$, $C D=13, A D=12$, and
$$
A C^{2}+A D^{2}=C D^{2} \text {, }
$$
Thus, $A C \perp A D$.
Then, $S=S_{\triangle A B C... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 727,847 |
12. A set $S$ consists of several triangles, each with three side lengths that are integers less than 5, and no two triangles are congruent or similar. The maximum number of elements in set $S$ is ( ).
(A) 8
(B) 9
(C) 10
(D) 11
(E) 12 | 12. B.
Take any triangle from the set $S$, and let its three sides be $a$, $b$, and $c$ (assuming $a \leqslant b \leqslant c$), and $a+b>c$. Moreover, no two triangles are congruent or similar. Enumeration yields:
$$
\begin{array}{l}
(1,1,1),(1,2,2),(1,3,3), \\
(1,4,4),(2,2,3),(2,3,3), \\
(2,3,4),(3,3,4),(3,4,4),
\end... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 727,848 |
13. Let real numbers $a, b$ satisfy $1<a<b$. If the three numbers $1, a, b$ and $1, \frac{1}{a}, \frac{1}{b}$ cannot form the sides of a triangle, then the minimum possible value of $b$ is ( ).
(A) $\frac{3+\sqrt{3}}{2}$
(B) $\frac{5}{2}$
(C) $\frac{3+\sqrt{5}}{2}$
(D) $\frac{3+\sqrt{6}}{2}$
(E) 3 | 13. C.
From the problem, we know
$$
\begin{array}{l}
\left\{\begin{array} { l }
{ a + 1 \leqslant b , } \\
{ \frac { 1 } { b } + \frac { 1 } { a } \leqslant 1 }
\end{array} \Rightarrow \left\{\begin{array}{l}
\frac{1}{a} \geqslant \frac{1}{b-1}, \\
\frac{1}{b} \leqslant 1-\frac{1}{a}
\end{array}\right.\right. \\
\Rig... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,849 |
Example 8 In isosceles $\triangle ABC$, with the midpoint $O$ of base $BC$ as the center, draw $\odot O$, which is tangent to the two legs $AB$ and $AC$ at points $E$ and $F$, respectively. $D$ is any point on the lower semicircular arc of $\odot O$. Draw a tangent line through point $D$, intersecting the extensions of... | 【Analysis and Proof】As shown in Figure 8, using the same method.
Let the line $N B$ intersect $F E$ at point $P_{1}$.
Since $O$ is the incenter of $\triangle A M N$, if we denote $\alpha=\frac{1}{2} \angle A, \beta=\frac{1}{2} \angle A M N, \gamma=\frac{1}{2} \angle A N M$,
then $\alpha+\beta+\gamma=\frac{\pi}{2}, \ang... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,850 |
14. The surface area of a rectangular box is 94 square inches, and the sum of the lengths of all its edges is 48 inches. Then the sum of the lengths of all its body diagonals is $(\quad)$.
(A) $8 \sqrt{3}$
(B) $10 \sqrt{2}$
(C) $16 \sqrt{3}$
(D) $20 \sqrt{2}$
(E) $40 \sqrt{2}$ | 14. D.
Let the length, width, and height of the box be $x$, $y$, and $z$ respectively. Then $2(xy + yz + zx) = 94$, $4(x + y + z) = 48$.
$$
\begin{array}{l}
\text { Hence } x^{2} + y^{2} + z^{2} \\
= (x + y + z)^{2} - 2(xy + yz + zx) \\
= 12^{2} - 94 = 50 \text {. }
\end{array}
$$
Therefore, the sum of the lengths of... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 727,851 |
15. Let $p=\sum_{k=1}^{6} k \ln k, \mathrm{e}^{p}$ be an integer, and $2^{n} \| \mathrm{e}^{p}$. Then $n=(\quad)$.
(A) 12
(B) 14
(C) 16
(D) 18
(E) 20 | 15. C.
Notice,
$$
p=\sum_{k=1}^{6} k \ln k=\ln \left(1^{1} \times 2^{2} \times 3^{3} \times 4^{4} \times 5^{5} \times 6^{6}\right) \text {. }
$$
Then $\mathrm{e}^{p}=\mathrm{e}^{\ln \left(1^{1} \times 2^{2} \times 3^{3} \times 4^{4} \times 5^{5} \times 6^{6}\right)}$
$$
\begin{array}{l}
=1^{1} \times 2^{2} \times 3^{... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,852 |
16. Let $P(x)$ be a cubic polynomial, and
$$
P(0)=k, P(1)=2 k, P(-1)=3 k \text {. }
$$
Then $P(2)+P(-2)=(\quad)$.
(A) 0
(B) $k$
(C) $6 k$
(D) $7 k$
(E) $14 k$ | $$
\begin{array}{l}
P(x)=a_{3} x^{3}+a_{2} x^{2}+a_{1} x+a_{0}\left(a_{3} \neq 0\right) . \\
\text { Then }\left\{\begin{array}{l}
P(0)=a_{0}=k, \\
P(1)=a_{3}+a_{2}+a_{1}+a_{0}, \\
P(-1)=-a_{3}+a_{2}-a_{1}+a_{0}
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
a_{0}=k, \\
a_{2}=\frac{3 k}{2}
\end{array}\right. ... | E | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,853 |
17. Let the parabola $C: y=x^{2}$, and the point $Q(20,14)$. A line passing through point $Q$ with a slope of $m$ (where $r<m<s, r, s \in \mathbf{R}$) does not intersect the parabola $C$. Then $r+s=(\quad)$.
(A) 1
(B) 26
(C) 40
(D) 52
(E) 80 | 17. E.
Let the equation of the line passing through point $Q$ with slope $m (r < m < s, r, s \in \mathbf{R})$ be
$$
y-14=m(x-20) \text {. }
$$
Since the line does not intersect the parabola $C$, we have
$$
x^{2}-m x+20 m-14=0
$$
has no real roots, i.e.,
$$
\Delta=m^{2}-4(20 m-14)=m^{2}-80 m+56<0 \text {. }
$$
There... | E | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,854 |
21. As shown in Figure 2, given that the side length of square $A B C D$ is 1, and rectangle $J K H G \cong$ rectangle $E B C F$. Then $B E=(\quad$.
(A) $\frac{\sqrt{6}-2}{2}$
(B) $\frac{1}{4}$
(C) $2-\sqrt{3}$
(D) $1-\frac{\sqrt{2}}{2}$
(E) $\frac{\sqrt{3}}{6}$ | 21. C.
Let $B E=x, \angle K J E=\theta$.
Since rectangle $J K H G \cong$ rectangle $E B C F$, we have,
$$
\begin{array}{l}
E J=x \cos \theta, A J=\sin \theta, \\
A G=\cos \theta, D G=x \sin \theta .
\end{array}
$$
Given that the side length of square $A B C D$ is 1, thus,
$$
\begin{array}{l}
\left\{\begin{array}{l}
x... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 727,856 |
23. Given that the number 2017 is a prime number, let $S=\sum_{k=0}^{62} \mathrm{C}_{2014}^{k}$. Then the remainder when $S$ is divided by 2017 is ().
(A) 32
(B) 684
(C) 1024
(D) 1576
(E) 2016 | 23. C.
Notice,
$$
\begin{array}{l}
\mathrm{C}_{2014}^{k}=\frac{2014 \times 2013 \times \cdots \times(2014-k+1)}{k!} \\
\equiv \frac{(-3) \times(-4) \times \cdots \times(-2-k)}{k!} \\
\equiv(-1)^{k} \mathrm{C}_{2+k}^{k} \equiv(-1)^{k} \mathrm{C}_{2+k}^{2}(\bmod 2017) . \\
\text { Therefore, } S=\sum_{k=0}^{62} \mathrm{... | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 727,857 |
24. Let pentagon $A B C D E$ be a cyclic pentagon, $A B=$ $C D=3, B C=D E=10, A E=14$, the sum of the lengths of all diagonals is $\frac{m}{n}\left(m, n \in \mathbf{Z}_{+},(m, n)=1\right)$. Then $m+n=$ ( ).
(A) 129
(B) 247
(C) 353
(D) 391
(E) 421 | 24. D.
As shown in Figure 3, from the given conditions, we know that quadrilaterals $ABCD$ and $BCDE$ are both isosceles trapezoids.
Thus,
$$
\begin{array}{c}
AC=BD=CE. \\
\text{Let } AC=BD=CE \\
=x, AD=y, BE=z.
\end{array}
$$
By Ptolemy's theorem, we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
x^{2}=10 y+9, \\
... | 391 | Geometry | MCQ | Yes | Yes | cn_contest | false | 727,858 |
25. The equation
$$
2 \cos 2 x \cdot\left(\cos 2 x-\cos \frac{2014 \pi^{2}}{x}\right)=\cos 4 x-1
$$
has the sum of all its positive real roots as ( ).
(A) $\pi$
(B) $810 \pi$
(C) $1008 \pi$
(D) $1080 \pi$
(E) $1800 \pi$ | 25. D.
From the problem, we have
$2 \cos 2 x \cdot\left(\cos 2 x-\cos \frac{2014 \pi^{2}}{x}\right)=2 \cos ^{2} 2 x-2$.
Let $a=\cos 2 x, b=\cos \frac{2014 \pi^{2}}{x}$. Then
$|a| \leqslant 1,|b| \leqslant 1$.
Thus, the original equation can be transformed into
$$
2 a^{2}-2 a b=2 a^{2}-2 \Rightarrow a b=1 \text {. }
$$... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,859 |
1. Solve the equation
$$
\sin \sin x=\sin (\cos x+1)
$$ | 1. Consider two cases.
(1) If $\sin x=\cos x+1+2 \pi l(l \in \mathbf{Z})$, then
$$
\begin{array}{l}
|2 \pi l|=|-1+\sin x-\cos x| \leqslant 1+\sqrt{2} \\
\Rightarrow l=0 .
\end{array}
$$
Notice that,
$$
\begin{array}{l}
\sin x-\cos x=1 \\
\Leftrightarrow \sin \left(x-\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2} \\
\Leftrigh... | x=\frac{\pi}{4}+(-1)^{n} \frac{\pi}{4}+\pi n(n \in \mathbf{Z}) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,860 |
1. As shown in Figure 9, given that $A B$ and $C D$ are two diameters of $\odot O$, $P$ is any point on the circumference, draw $P M \perp A B, P N \perp C D$, $A H \perp C D$. Prove:
$$
M N=A H .
$$ | First, prove that points $P$, $M$, $O$, and $N$ are concyclic and lie on the circle with diameter $OP$. Also, since the circumradii of $\triangle AOH$ and $\triangle PMN$ are equal, it follows that $AH=MN$. | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,861 |
2. Through a vertex of a trapezoid, draw two lines $l_{1}$ and $l_{2}$, where line $l_{1}$ is a diagonal of the trapezoid, and divides the line segment connecting the midpoints of the two bases in the ratio $3: 1$. Line $l_{2}$ bisects the area of the trapezoid. Find the ratio in which line $l_{2}$ divides the line seg... | 2. Let $M$ and $N$ be the midpoints of the bases $BC$ and $AD$, respectively, and $O$ be the intersection of $MN$ and $AC$.
From the problem, we know that the diagonal $AC$ divides $MN$ such that $\frac{MO}{ON}=\frac{1}{3}$.
Similarly, $\frac{BC}{AD}=\frac{MC}{AN}=\frac{1}{3}$.
Let $S$ represent the area of trapezoid $... | \frac{3}{4} \text{ or } 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,862 |
3. Solve the system of equations
$$
\left\{\begin{array}{l}
\frac{9-4^{y+1}-3^{x} \times 2^{y+2}-9^{x}}{4^{y+1}+3^{x} \times 2^{y+1}-3^{x+1}-9}=\frac{3^{x}+2^{y+1}-3}{3^{x}-1} \\
3^{x+1} \times 2^{y+1}=1
\end{array}\right.
$$ | 3. Let the variable substitution be $a=3^{x}, b=2^{y+1}(a>0, b>0)$.
Then, the second equation transforms to $a b=3$.
The first equation becomes
$$
\begin{array}{l}
\frac{9-a^{2}-2 a b-b^{2}}{b^{2}+a b-3 a-9}=\frac{a+b-3}{a-1} \\
\Rightarrow \frac{(3-a-b)(3+a+b)}{(b-3)(a+b+3)}=\frac{a+b-3}{a-1} \\
\Rightarrow \frac{3-a... | x=1, y=-1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,863 |
4. Find all values of $a$ such that for each of its values, the equation
$$
x^{3}-a x^{2}-\left(a^{3}-6 a^{2}+5 a+8\right) x-(a-3)^{3}=0
$$
has three distinct roots, which form a geometric progression, and find these roots. | 4. Let $x_{1}, x_{2}, x_{3}$ be the roots of the equation. By Vieta's formulas, we have
$$
\begin{array}{l}
x_{1}+x_{2}+x_{3}=a, \\
x_{1} x_{2}+x_{1} x_{3}+x_{2} x_{3}=-a^{3}+6 a^{2}-5 a-8, \\
x_{1} x_{2} x_{3}=(a-3)^{3} .
\end{array}
$$
Given that $x_{2}^{2}=x_{1} x_{3}$.
Substituting into (3) gives $x_{2}=a-3$.
Subs... | a=2,4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,864 |
3. Let non-negative real numbers $x, y, z$ satisfy $x+y+z=1$. Then
$$
t=\sqrt{9+x^{2}}+\sqrt{4+y^{2}}+\sqrt{1+z^{2}}
$$
the minimum value of $t$ is $\qquad$. | 3. $\sqrt{37}$.
First, $\sqrt{4+y^{2}}+\sqrt{1+z^{2}} \geqslant \sqrt{9+(y+z)^{2}}$.
Then $t \geqslant \sqrt{9+x^{2}}+\sqrt{9+(y+z)^{2}}$
$$
\geqslant 2 \sqrt{9+\frac{1}{4}}=\sqrt{37} \text {. }
$$
When and only when $z=\frac{1}{2} y=\frac{1}{3} x=\frac{1}{6}$, $t_{\text {min }}=\sqrt{37}$. | \sqrt{37} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,867 |
4. A square piece of paper with a side length of 2 (vertices are $A$, $B$, $C$, $D$, and the center is $O$) is folded into a regular quadrilateral pyramid $O-A B C D$. When the volume of the quadrilateral pyramid is maximized, the size of the dihedral angle $A-O B-C$ is $\qquad$ | 4. $\frac{2 \pi}{3}$.
As shown in Figure 2, draw $A E \perp O B$ at point $E$, and connect $C E$. Then $C E \perp O B$, and $\angle A E C$ is the plane angle of the dihedral angle to be found. Let $H$ be the center of the base. Then $O H \perp$ plane $A B C D$. Draw $H F \perp A B$, and connect $O F$.
By the theorem o... | \frac{2 \pi}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,868 |
5. A, B, C, and D each hold a soccer ball and simultaneously pass the ball once, with each person able to pass the ball to any of the other three. The probability that each person still has one ball after one pass is $\qquad$ | 5. $\frac{1}{9}$.
$$
P=\frac{3 \times 3}{3^{4}}=\frac{1}{9} \text {. }
$$ | \frac{1}{9} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,869 |
7. Given integers $a, b, c, d$. If the roots of the equation
$$
z^{4}+a z^{3}+b z^{2}+c z+d=0
$$
correspond to four points $A, B, C, D$ forming the vertices of a square in the complex plane, then the minimum value of the area of square $A B C D$ is $\qquad$ | 7. 2 .
Let the complex number corresponding to the center of the square be $m$. Then, after translating the origin of the complex plane to $m$, the vertices of the square are distributed on a circle, i.e., they are the solutions to the equation $(z-m)^{4}=n$ (where $n$ is a complex number).
From $z^{4}+a z^{3}+b z^{2... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,871 |
2. As shown in Figure 10, given that quadrilateral $ABCD$ is a cyclic quadrilateral, $AC$ is the diameter of the circle, $BD \perp AC$, $AC$ intersects $BD$ at point $E$, point $F$ is on the extension of $DA$, and $BF$ is connected. Point $G$ is on the extension of $BA$ such that $DG \parallel BF$, and $H$ is on the ex... | ```
Given: $C, B, H, G$ are concyclic
\[
\begin{array}{l}
\Rightarrow \angle B H C = \angle B G C = \angle A E F \\
\Rightarrow \angle B H F + \angle B E F = 180^{\circ} \\
\Rightarrow B, H, F, E \text{ are concyclic. }
\end{array}
\]
``` | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,872 |
8. Let the set $I=\{1,2, \cdots, 2 n+1\}\left(n \in \mathbf{Z}_{+}\right)$ have $k$ distinct subsets $A_{1}, A_{2}, \cdots, A_{k}$ that satisfy: for any $i, j(1 \leqslant i < j \leqslant k)$, we have $A_{i} \cup A_{j}=I$. Then the maximum value of $k$ is $\qquad$. | 8. $2 n+2$.
First, the $2 n+1$ subsets of $I$ with $2 n$ elements and the $2 n+2$ subsets of $I$ satisfy the requirement.
Second, if $k \geqslant 2 n+3$, then among $A_{1}, A_{2}, \cdots, A_{k}$, there must be one with the number of elements less than or equal to $2 n-1$, let it be $A_{1}$. Then the $2 n$-element sub... | 2 n+2 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,873 |
9. (16 points) Let $a_{n}=1+\frac{1}{2}+\cdots+\frac{1}{n}\left(n \in \mathbf{Z}_{+}\right)$. Find the smallest positive real number $\lambda$, such that for any $n \geqslant 2$, we have
$$
a_{n}^{2}<\lambda \sum_{k=1}^{n} \frac{a_{k}}{k} .
$$ | Sure, here is the translated text:
```
9. Notice that,
\[
\begin{array}{l}
a_{k}^{2}-a_{k-1}^{2}=\left(a_{k}-a_{k-1}\right)\left(a_{k}+a_{k-1}\right) \\
=\frac{1}{k}\left(2 a_{k}-\frac{1}{k}\right)(k \geqslant 2) .
\end{array}
\]
Then \(a_{n}^{2}-a_{1}^{2}=\sum_{k=2}^{n}\left(a_{k}^{2}-a_{k-1}^{2}\right)=2 \sum_{k=2}... | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 727,874 |
10. (20 points) Given that $AB$ is the major axis of the ellipse $\Gamma: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, and $CD$ is a chord of the ellipse $\Gamma$. The tangents at points $C$ and $D$ intersect at point $P$, the extension of $AD$ intersects the extension of $CB$ at point $E$, and the extension of $AC$ int... | 10. As shown in Figure 3, let the center of the ellipse be $O$. Connect $O P$, intersecting $C D$ at point $M$.
Let $C\left(x_{1}, y_{1}\right), D\left(x_{2}, y_{2}\right), P\left(x_{0}, y_{0}\right)$.
Then the equation of line $C D$ is $\frac{x_{0} x}{a^{2}}+\frac{y_{0} y}{b^{2}}=1$.
From $\left\{\begin{array}{l}\frac... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,875 |
11. (20 points) Natural numbers $a, b$ make $4a + 7b$ and $5a + 6b$ both multiples of 11, and $a + 21b \geq 792$. Find the minimum value of $T = 21a + b$. | 11. Let $4a + 7b = 11x$, $5a + 6b = 11y$. Then
$$
\begin{array}{l}
a = 7y - 6x \geqslant 0, \\
b = 5x - 4y \geqslant 0.
\end{array}
$$
Thus, $T = 11(13y - 11x)$.
Let $13y - 11x = r$.
From $a + 21b \geqslant 792$, we get
$$
9x - 7y \geqslant 72.
$$
Substituting equation (3) into equation (1) to eliminate $y$ gives $x ... | 44 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,876 |
One. (40 points) As shown in Figure 1, given that $\odot M$ and $\odot N$ intersect at points $B$ and $C$, points $A$ and $D$ are on $\odot M$ and $\odot N$ respectively, and $AC$ is tangent to $\odot N$, $CD$ is tangent to $\odot M$, the extension of $CB$ intersects the circumcircle of $\triangle ACD$ at point $E$. Pr... | As shown in Figure 4, connect $A B$, $B D$, and $A D$, and extend $A C$ to $F$. Since $A C$ and $C D$ are tangents to $\odot N$ and $\odot M$ respectively, we have
$$
\begin{array}{l}
\angle A C B = \angle B D C, \angle B C D = \angle B A C \\
\Rightarrow \triangle B C D \backsim \triangle B A C \\
\Rightarrow \frac{A ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,877 |
Sure, here is the translated text:
```
II. (40 points) Let $x, y, z$ be non-negative real numbers, and $x+y+z = 1$. Prove:
$$
\frac{\left(x^{2}+z^{2}\right) y}{x+z}+\frac{\left(y^{2}+z^{2}\right) x}{y+z}+\frac{\left(x^{2}+y^{2}\right) z}{x+y} \leqslant \frac{1}{2} \text {. }
$$
``` | Given the symmetry, without loss of generality, assume $x \leqslant y \leqslant z$. Then,
$$
x \geqslant \frac{1}{3} \Rightarrow \frac{1}{x} \leqslant 3 \text {. }
$$
Let the left side of the inequality be $S$.
Since $x+y+z=1$, we have,
$$
S=2 \sum y(x+z)-2 \sum \frac{x y z}{x+z},
$$
where “$\sum$” denotes the cyclic... | \frac{1}{2} | Inequalities | proof | Yes | Yes | cn_contest | false | 727,878 |
Three. (50 points) Given that $n$ is a positive integer, and the set $I=\{1,2, \cdots, n\}(n>3)$ has $k$ three-element subsets $A_{1}, A_{2}, \cdots, A_{k}$ satisfying: for any other three-element subset $B$ of $I$, there exists an integer $m$ and a subset $A_{i}(1 \leqslant i \leqslant k)$ such that
$$
\left\{y \mid y... | If $a, b, c \in \mathbf{Z}_{+}$, and $a+b+c=p$, then $(a, b, c)$ is called a "cyclic group" of length $p$, and it is agreed that $(a, b, c)$, $(b, c, a)$, and $(c, a, b)$ are the same cyclic group.
Consider the number of cyclic groups of length $n$.
The number of cyclic groups where two of $a, b, c$ are equal is $\left... | 1+\left[\frac{n^{2}-3 n}{6}\right] | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,879 |
Four. (50 points) Two people, A and B, take turns blowing up the same balloon. The balloon will burst if and only if the volume of gas inside the balloon \( V \) (unit: milliliters) exceeds 2014. A starts by blowing in 1 milliliter of gas, and it is agreed that the volume of gas blown in each time is either double or h... | (1) Let the maximum volume of the balloon be $A$ (the balloon bursts if and only if the volume of gas inside exceeds $A$). If player A has a winning strategy, then we denote $f(A)=1$; if player B has a winning strategy, then we denote $f(A)=2$.
When $A=3,4, \cdots, 9$, it is easy to verify that
$f(3)=2, f(4)=f(5)=1, f(... | proof | Logic and Puzzles | proof | Yes | Yes | cn_contest | false | 727,880 |
1.2014 Arrange chairs in a circle, with $n$ people sitting on the chairs, such that when one more person sits down, they will always sit next to one of the original $n$ people. Then the minimum value of $n$ is $\qquad$ | -、1. 672 .
From the problem, we know that after $n$ people sit down, there are at most two empty chairs between any two people.
If we can arrange for there to be exactly two empty chairs between any two people, then $n$ is minimized.
Thus, if we number the chairs where people are sitting, we easily get the arithmetic s... | 672 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,881 |
2. In $\triangle A B C$, it is known that
$$
|\overrightarrow{A B}|=2,|\overrightarrow{A C}|=3,|\overrightarrow{B C}|=4 \text {. }
$$
Let $O$ be the incenter of $\triangle A B C$, and $\overrightarrow{A O}=\lambda \overrightarrow{A B}+\mu \overrightarrow{B C}$. Then $\lambda+\mu=$ $\qquad$ | 2. $\frac{7}{9}$.
Let $A O$ intersect $B C$ at point $D$.
By the Angle Bisector Theorem, we have $\frac{B D}{D C}=\frac{A B}{A C}=\frac{2}{3}$.
Thus, $\overrightarrow{A D}=\frac{3}{5} \overrightarrow{A B}+\frac{2}{5} \overrightarrow{A C}$.
Also, $\frac{A O}{O D}=\frac{A B}{B D}=\frac{A C}{C D}=\frac{A B+A C}{B D+C D}=... | \frac{7}{9} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,882 |
Example 1 As shown in Figure 1, in $\triangle A B C$, it is given that $A B>A C, O$ and $I$ are the circumcenter and incenter of $\triangle A B C$ respectively, and $A B-A C=$ 2OI. Prove: OI //BC. ${ }^{\text {[1] }}$
(2013, National Junior High School Mathematics League) | Proof As shown in Figure 1, draw $O M \perp B C$ at point $M$, and $I N \perp B C$ at point $N$.
Let $B C=a, A C=b, A B=c$.
It is easy to see that $C M=\frac{1}{2} a, C N=\frac{1}{2}(a+b-c)$.
Therefore, $M N=C M-C N=\frac{1}{2}(c-b)=O I$.
Since $M N$ is a perpendicular segment between the parallel lines $O M$ and $I N$... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,883 |
Example 2 As shown in Figure 2, in $\triangle A B C$, it is known that $\angle A C B=90^{\circ}$, the altitude $C H$ from side $A B$ intersects the two angle bisectors $A M$ and $B N$ of $\triangle A B C$ at points $P$ and $Q$, respectively, and the midpoints of $P M$ and $Q N$ are $E$ and $F$, respectively. Prove: $E ... | Proof As shown in Figure 2, connect $C E$ and extend it to intersect $A B$ at point $I$, connect $C F$ and extend it to intersect $A B$ at point $G$.
$$
\begin{array}{l}
\text { From } \angle A B N=\angle C B N \text { and } C H \perp A B \text {, we get } \\
\angle C Q N=\angle B Q H=90^{\circ}-\angle A B N \\
=90^{\c... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,884 |
4. As shown in Figure 12, the diagonals $AC$ and $BD$ of convex quadrilateral $ABCD$ intersect at point $M$. A line through $M$ parallel to $AD$ intersects $AB$, $CD$, and $BC$ at points $E$, $F$, and $O$, respectively. $P$ is a point on the circle with center $O$ and radius $OM$. Prove: $\angle O P F = \angle O E P$. | ```
\begin{array}{l}
\text { Therefore, } \frac{O F}{O M}=\frac{K D}{K A}=\frac{O M}{O E} \\
\Rightarrow O P^{2}=O M^{2}=O E \cdot O F \\
\Rightarrow \triangle O F P \backsim \triangle O P E \\
\Rightarrow \angle O P F=\angle O E P .
\end{array}
``` | \angle O P F = \angle O E P | Geometry | proof | Yes | Yes | cn_contest | false | 727,885 |
7. On the blackboard, write $n$ consecutive positive integers in increasing order in a row, and below each number write a divisor of that number which is greater than 1 and less than the number itself. It is known that these divisors are also $n$ consecutive positive integers in increasing order. Prove: each of the ori... | 7. Notice that, the difference between each original number and the number written below it is equal, all equal to a certain positive integer $c$.
Let $p$ be a prime number less than $n$, and $p^{s}$ be the largest power of $p$ not exceeding $n$. Then, there exists a positive integer in the second row that is divisibl... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 727,886 |
8. It is known that 99 wise men are seated around a large round table, each wearing a hat of one of two different colors. Among them, 50 people's hats are of the same color, and the remaining 49 people's hats are of the other color. However, they do not know in advance which 50 people have the same color and which 49 p... | 8. Sure.
Suppose there are 50 white hats and 49 black hats. Those who see 50 white hats and 48 black hats are all wearing black hats. Thus, they can all write down the correct color, and there are 49 such people.
Next, let those who see 49 white hats and 49 black hats follow this strategy:
Observe which color of hat... | 74 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 727,887 |
5. In each rugby match, the winning team gets 2 points, the losing team gets 0 points, and if it's a draw, both teams get 1 point. It is known that 14 teams participated in a rugby tournament, and each pair of teams played one match. After the tournament, it was found that no two teams had the same total score in the m... | 5. Impossible.
First, the total points of the team ranking third from the bottom should be no less than 9 points.
In fact, the last three teams will play 3 matches among themselves, each match generating 2 points, totaling 6 points; these points are distributed among the three teams.
If their total points are less t... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,888 |
2. The left and right foci of a hyperbola are $F_{1}$ and $F_{2}$, respectively. A line passing through $F_{2}$ intersects the right branch of the hyperbola at points $A$ and $B$. If $\triangle F_{1} A B$ is an equilateral triangle, then the eccentricity of the hyperbola is $\qquad$ | 2. $\sqrt{3}$.
Let the equation of the hyperbola be
$$
\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>0, b>0) \text {, }
$$
with the semi-focal distance being $c$.
Then $c^{2}=a^{2}+b^{2}$.
By $\left|F_{1} A\right|-\left|F_{2} A\right|=\left|F_{1} B\right|-\left|F_{2} B\right|=2 a$,
$$
\left|F_{1} A\right|=\left|F_{2} A... | \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,891 |
3. Given positive real numbers $a, b, c$ satisfying $a+b+c=1$. Then the maximum value of $a+\sqrt{b}+\sqrt[3]{c}$ is $\qquad$ . | 3. $\frac{5}{4}+\frac{2 \sqrt{3}}{9}$.
By the AM-GM inequality, we have
$$
\begin{array}{l}
b+\frac{1}{4} \geqslant \sqrt{b}, \\
c+\frac{2}{3 \sqrt{3}}=c+\frac{1}{3 \sqrt{3}}+\frac{1}{3 \sqrt{3}} \geqslant 3 \sqrt[3]{c\left(\frac{1}{3 \sqrt{3}}\right)^{2}}=\sqrt[3]{c} .
\end{array}
$$
Thus, $a+\sqrt{b}+\sqrt[3]{c}$
$... | \frac{5}{4}+\frac{2 \sqrt{3}}{9} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,892 |
4. From nine classmates, select five to form a class committee, requiring that A and B either both be selected or both not be selected, and C and D not be selected at the same time. Then the number of selection methods that meet the requirements is $\qquad$ (answer with a number).
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。 | 4.41.
When A and B are both selected, there are $\mathrm{C}_{7}^{3}=35$ ways to choose the other members, and among these, the number of ways in which C and D are both selected is $\mathrm{C}_{5}^{1}=5$; When A and B are both not selected, there are $\mathrm{C}_{7}^{5}=21$ ways to choose all members, and among these, ... | null | Logic and Puzzles | other | Yes | Yes | cn_contest | false | 727,893 |
5. In isosceles $\triangle A B C$, it is known that $A C=B C=\sqrt{5}$, points $D, E, F$ are on sides $A B, B C, C A$ respectively, and $A D=D B=E F=1$. If $\overrightarrow{D E} \cdot \overrightarrow{D F} \leqslant \frac{25}{16}$, then the range of $\overrightarrow{E F} \cdot \overrightarrow{B A}$ is | 5. $\left[\frac{4}{3}, 2\right]$.
Let $D$ be the origin, and the rays $D B$ and $D C$ be the positive directions of the $x$ and $y$ axes, respectively, in a Cartesian coordinate system. Then
$$
A(-1,0), B(1,0), C(0,2) \text {. }
$$
Let the points $E\left(x_{1}, y_{1}\right), F\left(x_{2}, y_{2}\right)$, where,
$$
y_{... | \left[\frac{4}{3}, 2\right] | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,894 |
6. Each point on a circle is colored one of three colors: red, yellow, or blue, and all three colors appear. Now, $n$ points are chosen from the circle. If among these points, there always exist three points that form a triangle with vertices of the same color and an obtuse angle, then the minimum possible value of $n$... | 6.13.
First, we state that \( n \geqslant 13 \).
If \( n \geqslant 13 \), then by the pigeonhole principle, among these \( n \) points, there must be \(\left[\frac{13-1}{3}\right]+1=5\) points of the same color (let's assume they are red).
Take a diameter, whose endpoints do not belong to these five points.
By the pig... | 13 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,895 |
Example 1 Proof: When $n$ is a positive odd number, $60 \mid\left(6^{n}-3^{n}-2^{n}-1\right)$ | Prove that for $n$ being a positive odd number, we have
$$
(a-b)\left|\left(a^{n}-b^{n}\right),(a+b)\right|\left(a^{n}+b^{n}\right) \text {. }
$$
Then $3 \mid \left(6^{n}-3^{n}\right), 3 \mid \left(2^{n}+1\right), 4 \mid \left(6^{n}-2^{n}\right)$, $4 \mid \left(3^{n}+1\right), 5 \mid \left(6^{n}-1\right), 5 \mid \left... | 60 \mid \left(6^{n}-3^{n}-2^{n}-1\right) | Number Theory | proof | Yes | Yes | cn_contest | false | 727,896 |
7. Given $\alpha, \beta \in [0, \pi]$. Then
$$
[\sin \alpha + \sin (\alpha + \beta)] \sin \beta
$$
the maximum value is . $\qquad$ | 7. $\frac{8 \sqrt{3}}{9}$.
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
{[\sin \alpha+\sin (\alpha+\beta)]^{2}} \\
=(\sin \alpha+\sin \alpha \cdot \cos \beta+\cos \alpha \cdot \sin \beta)^{2} \\
=[\sin \alpha(1+\cos \beta)+\cos \alpha \cdot \sin \beta]^{2} \\
\leqslant\left(\sin ^{2} \alpha+\cos ^{2} ... | \frac{8 \sqrt{3}}{9} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,897 |
8. Let $\frac{1}{1-x-x^{2}-x^{3}}=\sum_{i=0}^{\infty} a_{n} x^{n}$. If $a_{n-1}=n^{2}$, then the set of values for $n$ is . $\qquad$ | 8. $\{1,9\}$.
Multiplying both sides of the given equation by $1-x-x^{2}-x^{3}$, we get
$$
1=a_{0}+\left(a_{1}-a_{0}\right) x+\left(a_{2}-a_{1}-a_{0}\right) x^{2}+\cdots \text {. }
$$
By comparing coefficients, we obtain
$$
\begin{array}{l}
a_{0}=1, a_{1}=1, a_{2}=2, \\
a_{n}=a_{n-1}+a_{n-2}+a_{n-3}(n \geqslant 3) .
... | \{1,9\} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,898 |
9. (16 points) Find the range of the function
$$
f(x)=2 x^{2}+x \sqrt{4 x^{2}-3}
$$ | 9. The domain of the function is easily obtained as
$$
\left(-\infty,-\frac{\sqrt{3}}{2}\right] \cup\left[\frac{\sqrt{3}}{2},+\infty\right) \text {. }
$$
(1) It is easy to see that the function
$$
f(x)=2 x^{2}+x \sqrt{4 x^{2}-3}
$$
is increasing on the interval $\left[\frac{\sqrt{3}}{2},+\infty\right)$.
Therefore, whe... | (1,+\infty) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,899 |
10. (20 points) As shown in Figure 1, in the Cartesian coordinate system $x O y$, a circle with center $P$ intersects the hyperbola $x y=1$ at points $A$, $B$, $C$, and $D$. Let the midpoints of segments $A B$ and $C D$ be $E$ and $F$, respectively. Prove: Quadrilateral $O E P F$ is a parallelogram. | 10. Let $A\left(t_{1}, \frac{1}{t_{1}}\right), B\left(t_{2}, \frac{1}{t_{2}}\right), C\left(t_{3}, \frac{1}{t_{3}}\right)$,
$$
D\left(t_{4}, \frac{1}{t_{4}}\right), P(a, b),
$$
The equation of the circle is $x^{2}+y^{2}-2 a x-2 b y+f=0$.
By combining the equation of the circle with $x y=1$, we get
$$
x^{4}-2 a x^{3}+f... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,900 |
11. (20 points) Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=0$, and for any $n \in \mathbf{Z}_{+}, a_{n+1}$ takes the value $a_{n}-1$ or $a_{n}+1$ with probability $\frac{1}{2}$ each.
(1) Let the value of $a_{2 n+1}$ be the random variable $X$. Find the probability distribution of $X$;
(2) Find the expec... | 11. (1) Let $d_{n}=a_{n+1}-a_{n}$. Then for any positive integer $n$, the probability that $d_{n}$ takes 1 or -1 is $\frac{1}{2}$, and
$$
a_{2 n+1}=a_{1}+\sum_{i=1}^{2 n}\left(a_{i+1}-a_{i}\right)=\sum_{i=1}^{2 n} d_{i} \text {. }
$$
Let $a_{2 n+1}=k$. Clearly, $|k| \leqslant 2 n$, and suppose at this point $d_{1}, d_... | \frac{n \mathrm{C}_{2 n}^{n}}{2^{2 n-1}} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,901 |
One, (40 points) As shown in Figure 2, in $\triangle ABC$, $AB > AC$, circle $\Gamma$ is the circumcircle of $\triangle ABC$, circle $\Gamma_{1}$ passes through point $B$ and is tangent to $AC$ at point $A$, circle $\Gamma_{2}$ passes through point $C$ and is tangent to $AB$ at point $A$, circles $\Gamma_{1}$ and $\Gam... | Given that points $A$, $D$, $C$, and $E$ are concyclic and by the tangent-secant angle theorem, we have
$$
\angle C E B = \angle C A D = \angle A B D \text{.}
$$
Thus, $A B \parallel P C$.
Similarly, $A C \parallel P B$.
Therefore, quadrilateral $A B P C$ is a parallelogram.
Let $A P$ intersect $B C$ at point $M$. The... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,902 |
II. (40 points) Let $c_{1}, c_{2}, \cdots$ be a sequence of positive integers, and for any positive integers $m, n$ satisfying $1 \leqslant m \leqslant \sum_{i=1}^{n} c_{i}$, there exist positive integers $a_{1}, a_{2}, \cdots, a_{n}$ such that $m=\sum_{i=1}^{n} \frac{c_{i}}{a_{i}}$. For each fixed $i\left(i \in \mathb... | Let $C_{1}=2, C_{i}=4 \times 3^{i-2}(i \geqslant 2)$.
First, prove that for each fixed $i, C_{i}$ is the maximum possible value of $c_{i}$.
Use mathematical induction on $i$ to prove $c_{i} \leqslant C_{i}$.
When $i=1$, if $c_{1}>1$, then let $(m, n)=\left(c_{1}-1,1\right)$, we know $a_{1}=\frac{c_{1}}{c_{1}-1}$ is an ... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,903 |
Three, (50 points) Prove: There exist infinitely many positive integers $n$, such that
$$
\left.\left([\sqrt{n}]+\left[\sqrt{\frac{n}{5}}\right]\right) \right\rvert\, n,
$$
where, $[x]$ denotes the greatest integer not exceeding the real number $x$. | Three, let $x=[\sqrt{n}], y=\left[\sqrt{\frac{n}{5}}\right]$. Then $x^{2} \leqslant n0 .
\end{array}
$
Therefore, $x^{2} \leqslant n \leqslant x^{2}+2 x$, which means there are $2 x+1$ integers $n$ such that
$$
x=[\sqrt{n}], y=\left[\sqrt{\frac{n}{5}}\right] \text {. }
$$
By $y \leqslant x$, we know that among the $2... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 727,904 |
Four. (50 points) Given that there are $n(n \geqslant 4)$ football teams participating in a round-robin tournament, each pair of teams plays one match. The winning team gets 3 points, the losing team gets 0 points, and in case of a draw, both teams get 1 point. After all the matches, it is found that the total scores o... | All teams play a total of $\mathrm{C}_{n}^{2}=\frac{n(n-1)}{2}$ matches.
Let the score of the last place be $k$. Then the total score is
$$
S=\sum_{i=0}^{n-1}(k+i)=\frac{n(2 k+n-1)}{2} .
$$
If a team has no draws, then each match score is either 0 or 3, so the total score is a multiple of 3.
Given that the total scor... | n-2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,905 |
Given $H$ is the orthocenter of $\triangle A B C$, and $\odot O$ intersects the sides $B C$, $C A$, and $A B$ of $\triangle A B C$ at points $D_{1}$ and $D_{2}$, $E_{1}$ and $E_{2}$, $F_{1}$ and $F_{2}$, respectively. Points $D$, $E$, and $F$ are the midpoints of $D_{1} D_{2}$, $E_{1} E_{2}$, and $F_{1} F_{2}$, respect... | Proof As shown in Figure 4, let the projections of the orthocenter $H$ of $\triangle ABC$ onto the three sides be $R, S, T$, and connect $EF, ET, FS, ST$.
By the given condition,
$$
E_{1}, E_{2}, F_{1}, F_{2} \text{ are concyclic}
$$
$$
\Rightarrow A E_{1} \cdot A E_{2}=A F_{1} \cdot A F_{2}.
$$
Notice that,
$$
\begin... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,906 |
For $I$ being the excenter of $\triangle ABC$ with respect to $\angle A$, prove:
$$
\frac{A I^{2}}{C A \cdot A B}-\frac{B I^{2}}{A B \cdot B C}-\frac{C I^{2}}{B C \cdot C A}=1 .
$$ | Proof As shown in Figure 5, let $\odot I$ be tangent to $B C$, $C A$, and $A B$ at points $D$, $E$, and $F$ respectively.
$$
\begin{array}{l}
\text { Let } \angle A I F=\alpha, \\
\angle B I D=\beta, \\
\angle C I E=\gamma .
\end{array}
$$
Then $\alpha=\beta+\gamma$.
Assume $I D=I E=I F=1$. Then $A I^{2}=\frac{1}{\cos... | 1 | Geometry | proof | Yes | Yes | cn_contest | false | 727,908 |
For the New Year's celebration in 2014, Hongqi Mall is hosting a shopping and voucher return event. For every 200 yuan spent, a customer can participate in one round of a lottery. The lottery rules are as follows: take out the 13 hearts cards $(A, 2, 3, \cdots, Q, K)$ from a deck of cards, shuffle them, and randomly dr... | Solve $\frac{365}{2014}$.
If the first $n$ draws in a round are all $K$ or $Q$, and do not contain three consecutive $Q$s, then at least $n$ coupons are won, with a probability of $\frac{a_{n}}{13^{n}}$, where $a_{n}$ is the number of sequences of length $n$ consisting only of $\mathrm{K}$ and $\mathrm{Q}$ and not cont... | \frac{365}{2014} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,909 |
Given positive real numbers $a, b, c$ satisfying $a^{3}+b^{3}+c^{3}=3$. Prove:
$$
\frac{a b+a}{a^{2}+a+3}+\frac{b c+b}{b^{2}+b+3}+\frac{c a+c}{c^{2}+c+3} \leqslant \frac{6}{5} .
$$ | To prove that “$\sum$” represents a cyclic symmetric sum.
Let $f(x)=x+\frac{3}{x}(0<x<\sqrt[3]{3})$.
Since $f^{\prime}(x)=1-\frac{3}{x^{2}}<0$, $f(x)$ is strictly monotonically
decreasing.
By the rearrangement inequality, we have
$$
\begin{array}{l}
\sum \frac{a b+a}{a^{2}+a+3}=\sum \frac{b+1}{1+a+\frac{3}{a}} \\
\leqs... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 727,910 |
Example 4 Find all positive integers $n$, such that
$$
323 \mid\left(20^{n}+16^{n}-3^{n}-1\right) \text {. }
$$ | Notice that, $323=17 \times 19$.
When $n$ is even, because
$$
\begin{array}{l}
20^{n}+16^{n}-3^{n}-1 \\
\equiv 1+3^{n}-3^{n}-1 \equiv 0(\bmod 19), \\
20^{n}+16^{n}-3^{n}-1 \\
\equiv 3^{n}+1-3^{n}-1 \equiv 0(\bmod 17),
\end{array}
$$
thus, equation (1) holds;
When $n$ is odd,
$$
\begin{array}{l}
20^{n}+16^{n}-3^{n}-1 \... | n \text{ is even} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,912 |
Example 5 Find the largest positive integer $x$, such that for every positive integer $y$, we have $x \mid\left(7^{y}+12 y-1\right)$.
| When $y=1$, $7^{y}+12 y-1=18$.
Let 18 I $\left(7^{y}+12 y-1\right)$. Notice that,
$$
\begin{array}{l}
7^{y+1}+12(y+1)-1 \\
=6 \times\left(7^{y}+2\right)+\left(7^{y}+12 y-1\right) .
\end{array}
$$
Since $7^{y}+2 \equiv 1+2 \equiv 0(\bmod 3)$, therefore,
$$
18 \mid\left[7^{y+1}+12(y+1)-1\right] \text {. }
$$
Thus, for ... | 18 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,913 |
Example 6 If \( n \) is a positive integer greater than 1, prove:
\[
n \nmid \left(2^{n}-1\right) .
\] | Prove that if $n$ is even, since $2^{n}-1$ is odd, then $n \nmid\left(2^{n}-1\right)$.
If $n$ is an odd prime, by Fermat's Little Theorem, $2^{n} \equiv 2(\bmod n)$.
Thus, $2^{n}-1 \equiv 1(\bmod n)$, which means $n \nmid\left(2^{n}-1\right)$. If $n$ is an odd composite number, let $p$ be the smallest prime factor of $... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 727,914 |
Example 7 Let $m$ be a positive integer. If $\left(2^{m+1}+1\right) \mid\left(3^{2^{m}}+1\right)$, prove: $2^{m+1}+1$ is a prime number.
(Adapted from the 2003 Korean Mathematical Olympiad) | Proof Let $q=2^{m+1}+1$. Then
$$
\begin{array}{l}
3^{2^{m}} \equiv-1(\bmod q) \\
\Rightarrow(3, q)=1 .
\end{array}
$$
Squaring both sides of equation (1) yields $3^{2^{m+1}} \equiv 1(\bmod q)$.
Let $k$ be the smallest positive integer such that $3^{k} \equiv 1(\bmod q)$. Then $k \mid 2^{m+1}$.
Thus, $k=2^{r}(r \leqsla... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 727,915 |
Example 9 Find the last two digits of $\left[(\sqrt{29}+\sqrt{21})^{2012}\right]$.
(Adapted from the 25th IMO Preliminary Question) | Let $a_{n}=(\sqrt{29}+\sqrt{21})^{2 n}+(\sqrt{29}-\sqrt{21})^{2 n}$
$$
=(50+2 \sqrt{609})^{n}+(50-2 \sqrt{609})^{n} \text {. }
$$
Then the characteristic equation of the sequence $\left\{a_{n}\right\}$ is
$$
x^{2}-100 x+64=0 \text {. }
$$
Therefore, the sequence $\left\{a_{n}\right\}$ has the recurrence relation
$$
a... | 31 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,917 |
Example 3 As shown in Figure 3, given that $AB$ and $BC$ are the diameter and tangent of $\odot O$ respectively, $OC$ is parallel to chord $AD$. A perpendicular line $DE \perp AB$ is drawn through point $D$, intersecting at point $E$. Line $AC$ is connected, intersecting $DE$ at point $P$. Question: Are $EP$ and $PD$ e... | Extend $A D$ and $B C$, intersecting at point $F$.
From $O A=O B, O C \parallel A F$, we get $C B=C F$.
It is easy to prove that $D E \parallel B F$.
$$
\text { Then } \frac{E P}{B C}=\frac{A P}{A C}=\frac{P D}{C F} \Rightarrow E P=P D \text {. }
$$
[Summary] A set of lines passing through a point is called a linear pe... | EP=PD | Geometry | proof | Yes | Yes | cn_contest | false | 727,918 |
2. Find all integers $m, n$ such that $m n \mid \left(3^{m}+1\right), m n \mid \left(3^{n}+1\right)$.
(2005, Korean Mathematical Olympiad) | First prove that at least one of $m$ and $n$ is odd, then prove that at least one of them is 1, and find that the only $(m, n)$ satisfying the condition are $(1,1),(1,2),(2,1)$. | (1,1),(1,2),(2,1) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,920 |
3. Find all odd prime numbers $p$ such that
$$
p \mid \left(1^{p-1}+2^{p-1}+\cdots+2004^{p-1}\right) .
$$ | First, prove:
$$
\begin{array}{l}
1^{p-1}+2^{p-1}+\cdots+2004^{p-1} \\
\equiv 2004-\left[\frac{2004}{p}\right](\bmod p),
\end{array}
$$
where $[x]$ denotes the greatest integer not exceeding the real number $x$.
Let $2004=k p+r(0 \leqslant r \leqslant p-1, k \in \mathbf{N})$. Then
$$
\begin{array}{l}
k \equiv r(\bmod ... | p=2003 \text{ or } 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,921 |
Example 1 Find the range of the function $y=x^{2}+x \sqrt{x^{2}-1}$. ${ }^{[1]}$ (2013, Hubei Provincial Preliminary of the National High School Mathematics League) | It is known that the domain of the function is $\{x \mid x \geqslant 1$ or $x \leqslant-1\}$.
Let $x=\sec \theta\left(0 \leqslant \theta \leqslant \pi, \theta \neq \frac{\pi}{2}\right)$. Then $y=\sec ^{2} \theta+\sec \theta \cdot \sqrt{\sec ^{2} \theta-1}=\frac{1}{\cos ^{2} \theta}+\frac{|\tan \theta|}{\cos \theta}$.
(... | \left(\frac{1}{2},+\infty\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,922 |
Example 2 Let $0<x<\frac{9}{2}$. Find the range of the function
$$
y=\left[1+\frac{1}{\lg \left(\sqrt{x^{2}+10}+x\right)}\right]\left[1+\frac{1}{\lg \left(\sqrt{x^{2}+10}-x\right)}\right]
$$ | Given $0<x<\frac{9}{2}$, we know
$$
0<\lg \left(\sqrt{x^{2}+10}-x\right)<\lg \left(\sqrt{x^{2}+10}+x\right)<1 .
$$
Notice that,
$$
\lg \left(\sqrt{x^{2}+10}+x\right)+\lg \left(\sqrt{x^{2}+10}-x\right)=1 \text {. }
$$
We can set $\lg \left(\sqrt{x^{2}+10}+x\right)=\sin ^{2} \theta$,
$$
\lg \left(\sqrt{x^{2}+10}-x\righ... | (9,+\infty) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,923 |
Example 3 Given that $x, y, z$ are distinct real numbers. Prove:
$$
\left|\frac{1+x y}{x-y}+\frac{1+y z}{y-z}+\frac{1+z x}{z-x}\right| \geqslant \sqrt{3} .
$$ | Proof: Let $a=\frac{1+x y}{x-y}, b=\frac{1+y z}{y-z}, c=\frac{1+z x}{z-x}$,
$$
x=\tan \alpha, y=\tan \beta, z=\tan \gamma \text {. }
$$
It is easy to see that $x, y, z$ are distinct.
Thus, $\alpha, \beta, \gamma$ are also distinct real numbers.
$$
\begin{array}{l}
\Rightarrow\left\{\begin{array}{l}
\tan (\alpha-\beta)... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 727,924 |
Example 4 Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{n+1}=a_{n}^{3}-3 a_{n}$, and $a_{1}=\frac{3}{2}$. Find the general term formula of $\left\{a_{n}\right\}$. | Solve: From $a_{n+1}=a_{n}^{3}-3 a_{n} \Rightarrow \frac{a_{n+1}}{2}=4\left(\frac{a_{n}}{2}\right)^{3}-3 \times \frac{a_{n}}{2}$.
Let $b_{n}=\frac{a_{n}}{2}$. Then
$$
b_{n+1}=4 b_{n}^{3}-3 b_{n}, b_{1}=\frac{3}{4} \text {. }
$$
Using the triple angle formula $\cos 3 \theta=4 \cos ^{3} \theta-3 \cos \theta$.
If $b_{n}=... | a_{n}=2 \cos \left(3^{n-1} \arccos \frac{3}{4}\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,925 |
Example 5 Let $a_{1}=1$. When $n \geqslant 2$, $a_{n}=\frac{\sqrt{1+a_{n-1}^{2}}-1}{a_{n-1}}$. Prove: $a_{n}>\frac{\pi}{2^{n+1}}$. | Prove that obviously, $a_{n}>0$.
Let $a_{n}=\tan \theta_{n}\left(n=1,2, \cdots, \theta_{n} \in\left(0, \frac{\pi}{2}\right)\right)$.
Then when $n \geqslant 2$,
$$
\begin{array}{l}
\tan \theta_{n}=\frac{\sqrt{1+\tan ^{2} \theta_{n-1}}-1}{\tan \theta_{n-1}}=\tan \frac{\theta_{n-1}}{2} \\
\Rightarrow \theta_{n}=\frac{1}{2... | a_{n}=\tan \frac{\pi}{2^{n+1}}>\frac{\pi}{2^{n+1}} | Algebra | proof | Yes | Yes | cn_contest | false | 727,926 |
Example 6 Solve the system of equations in the real numbers
$$
\left\{\begin{array}{l}
x y+y z+z x=1, \\
5\left(x+\frac{1}{x}\right)=12\left(y+\frac{1}{y}\right)=13\left(z+\frac{1}{z}\right) .
\end{array}\right.
$$ | From equation (2), we know that $x, y, z$ have the same sign.
Therefore, we can first find the positive solutions.
In $\triangle ABC$, let the sides opposite to $\angle A, \angle B, \angle C$ be $a, b, c$ respectively.
From equation (1), we recall that in $\triangle ABC$, there is the identity
$$
\tan \frac{B}{2} \cdot... | \left(\frac{1}{5}, \frac{2}{3}, 1\right) \text{ and } \left(-\frac{1}{5}, -\frac{2}{3}, -1\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,927 |
Question: The sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ satisfy $a_{0}=1, b_{0}=0$, and
$$
\left\{\begin{array}{l}
a_{n+1}=7 a_{n}+6 b_{n}-3, \\
b_{n+1}=8 a_{n}+7 b_{n}-4
\end{array}(n=0,1, \cdots)\right. \text {. }
$$
Prove: $a_{n}$ is a perfect square.
$(2000$, National High School Mathematics Comp... | Prove that from equation (1) we get
$$
6 b_{n}=a_{n+1}-7 a_{n}+3 \text {. }
$$
Substituting $6 \times$ (2) and simplifying, we get
$$
a_{n+2}=14 a_{n+1}-a_{n}-6 \text {. }
$$
Let $a_{n}=c_{n}+\frac{1}{2}$. Then
$$
\begin{array}{l}
c_{n+2}=14 c_{n+1}-c_{n} \\
\Rightarrow c_{n}=\frac{1}{4}\left[(7+4 \sqrt{3})^{n}+(7-4 ... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 727,928 |
Example 4 As shown in Figure 4, given that $\triangle ABC$ is an acute triangle, and a circle $\odot O$ is drawn with $BC$ as its diameter. $AD$ is a tangent to $\odot O$, and from a point $E$ on $AB$, a perpendicular is drawn to $AB$ intersecting the extension of $AC$ at point $F$. If $\frac{AB}{AF}=\frac{AE}{AC}$, pr... | Let $\odot O$ intersect $AB$ at point $G$, and connect $CG$.
Since $BC$ is the diameter of $\odot O$, we have $CG \perp AB$.
Also, since $EF \perp AB$, then $CG \parallel EF$.
Therefore, $\frac{AG}{AC} = \frac{AE}{AF}$.
From the given conditions, we have $AE^2 = AG \cdot AB$.
Since $AD$ is tangent to $\odot O$ at point... | AD = AE | Geometry | proof | Yes | Yes | cn_contest | false | 727,929 |
Question 1: The number of five-digit numbers with no repeated digits (excluding 0) and where 4 and 5 are not adjacent is $\qquad$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Solve by the method of exclusion.
The number of five-digit numbers without repeated digits and not containing 0 is $\mathrm{A}_{9}^{5}=15120$. Among them, the number of five-digit numbers without repeated digits, not containing 0, and with 4 and 5 adjacent is $\mathrm{C}_{7}^{3} \mathrm{~A}_{4}^{4} \mathrm{~A}_{2}^{2}=... | null | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,930 |
Question 2 In the tetrahedron $P-ABC$, the base is an equilateral triangle with side length 3, $PA=3, PB=4, PC=5$. Then the volume $V$ of the tetrahedron $P-ABC$ is $\qquad$ | 【Analysis】A tetrahedron has four faces, and each face can serve as the base, so the method is not unique.
Solution As shown in Figure 1, it is easy to know that $\triangle P B C$ is a right triangle, and we take the midpoint $O$ of the hypotenuse $P C$.
Since $A P=A B=A C$, point $A$'s projection on plane $P B C$ is ... | \sqrt{11} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 727,931 |
Question 3 Given $x, y \in(0,1)$, and $3x+7y$, $5x+y$ are both integers. Then there are $\qquad$ pairs of $(x, y)$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 【Analysis】Let $\left\{\begin{array}{l}3 x+7 y=u, \\ 5 x+y=v .\end{array}\right.$
We can first determine the integer values of $u$ and $v$ based on $x, y \in(0,1)$.
Solution Let $\left\{\begin{array}{l}3 x+7 y=u, \\ 5 x+y=v,\end{array}\right.$ where the possible values of $u$ and $v$ are $u \in\{1,2, \cdots, 9\}, v \in\... | 31 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 727,932 |
Question 4 Given that $x, y, z$ are all positive numbers, and
$$
x y z=\frac{1}{2}, x^{2}+y^{2}+z^{2} \leqslant 2 \text{. }
$$
Then the maximum value of $x^{4}+y^{4}+z^{4}$ is | 【Analysis】This problem discusses the maximum and minimum values of a ternary function. The idea is to eliminate variables and transform it into a unary function to solve.
Solution Without loss of generality, assume $x^{2} \leqslant y^{2} \leqslant z^{2}$. Then $x^{2} \leqslant \frac{2}{3}$.
Since $x^{2}+y^{2}+z^{2} \le... | \frac{3}{2} | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 727,933 |
Question 5 Let $m, n (m<n)$ be positive integers. Prove: For any $n$ consecutive positive integers, there always exist two different positive integers whose product is a multiple of $mn$.
保留源文本的换行和格式，直接输出翻译结果。 | 【Analysis】In any sequence of $n$ consecutive positive integers, there must be a multiple of $n$ and a multiple of $m$, denoted as $x, y$, respectively. We then discuss two cases: $x \neq y$ and $x=y$.
Proof: In any sequence of $n$ consecutive positive integers, there must be a multiple of $n$ and a multiple of $m$, de... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 727,934 |
Question 6 Let $P$ be a moving point on the curve $2 x^{2}-5 x y+2 y^{2}=1$. Find the minimum distance from point $P$ to the origin. | By the AM-GM inequality, we have $-2 x y \leqslant x^{2}+y^{2}$.
$$
\begin{array}{l}
\text { Therefore, } 1=2 x^{2}-5 x y+2 y^{2}=2 x^{2}+\frac{5}{2}(-2 x y)+2 y^{2} \\
\leqslant 2 x^{2}+\frac{5}{2}\left(x^{2}+y^{2}\right)+2 y^{2}=\frac{9}{2}\left(x^{2}+y^{2}\right) \\
\Rightarrow x^{2}+y^{2} \geqslant \frac{2}{9} \Rig... | \frac{\sqrt{2}}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,935 |
7 A function $f(x)$ defined on $(0,+\infty)$ satisfies: for any $x, y$,
$$
f(x+y)=f(x y) \text {. }
$$
Prove: $f(x)$ is a constant on $(0,+\infty)$. | 【Analysis】First, we obtain that $T=1$ is a period of $f(x)$. To prove $f(a)=f(b)$, we can consider assigning values to $x$ and $y$ in the given conditions cleverly.
Proof It is easy to see that $T=1$ is a period of $f(x)$.
For any $a, b>0$, since there are infinitely many positive integers, we can always find a positiv... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 727,936 |
Question 8 Let $x, y \in \left(0, \frac{\pi}{2}\right)$, and $\tan x \cdot \tan y \geqslant 3$. Prove: $\frac{\cos x+\cos y}{2} \geqslant \sqrt{\frac{\cos x \cdot \cos y}{\cos (x-y)}}$. | 【Analysis】First convert the trigonometric inequality into an algebraic inequality, then consider comparing the sizes by taking the difference.
Proof Using $\cos x=\sqrt{\frac{1}{1+\tan ^{2} x}}$, the problem can be transformed into:
Given $a b \geqslant 3$. Prove:
$$
\frac{1}{\sqrt{1+a^{2}}}+\frac{1}{\sqrt{1+b^{2}}} \... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 727,937 |
Question 9 Let $n$ be a positive integer no less than 2, and $a_{1}, a_{2}, \cdots$, $a_{n} \in(0,1]$. Prove:
$$
\left(\sum_{i=1}^{n} a_{i}\right)\left(\sum_{i=1}^{n} \frac{a_{i}}{a_{i}+1}\right) \geqslant n\left(\prod_{i=1}^{n} a_{i}\right)\left(\sum_{i=1}^{n} \frac{1}{a_{i}+1}\right) .
$$ | 【Analysis】First, determine the condition for equality: $a_{1}=a_{2}=\cdots=a_{n}=1$, then consider simplifying and unifying the expression.
Proof Note that,
$$
\begin{array}{l}
n\left(\prod_{i=1}^{n} a_{i}\right)\left(\sum_{i=1}^{n} \frac{1}{a_{i}+1}\right)=n\left(\prod_{i=1}^{n} a_{i}\right)\left(n-\sum_{i=1}^{n} \fra... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 727,938 |
Question 10 Given $a_{1} \in(0,1), a_{n+1}=a_{n}+\frac{a_{n}^{2}}{n^{2}}$. Prove: There exists $M>0$ such that for any positive integer $n$ we have $a_{n}<M$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 【Analysis】Let $b_{n}=\frac{a_{n}}{n}$.
First, study $b_{n}$, then estimate the upper bound of $a_{n}$.
Proof: Let $b_{n}=\frac{a_{n}}{n}$.
From $a_{n+1}=a_{n}+\frac{a_{n}^{2}}{n^{2}} \Rightarrow(n+1) b_{n+1}=n b_{n}+b_{n}^{2}$
$$
\begin{array}{l}
\Rightarrow \frac{1}{b_{n+1}}=\frac{n+1}{b_{n}\left(b_{n}+n\right)}=\frac... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,939 |
Example 5 As shown in Figure 5, let $P$ be a point inside $\square A B C D$, and $\angle P B A=\angle P D A$. Prove: $\angle P A B=\angle P C B$.
保留源文本的换行和格式，直接输出翻译结果。 | Prove that in Figure 5, draw $PE \parallel BC$, and $PE = BC$. It is easy to prove that quadrilateral $PEBC$ and quadrilateral $AEPD$ are both parallelograms.
$$
\begin{array}{l}
\Rightarrow BE \parallel PC, AE \parallel DP \\
\Rightarrow \angle ABP = \angle ADP = \angle AEP \\
\Rightarrow A, E, B, P \text{ are concycl... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,940 |
Example 1: Do there exist four positive real numbers such that their pairwise products are $2, 3, 5, 6, 10, 16$?
(2011, Peking University and other schools' independent admission examination) | Assume there exist four positive real numbers $x_{1}, x_{2}, x_{3}, x_{4}$ that satisfy the conditions.
Since the pairwise products are all distinct, these four numbers are also distinct.
Without loss of generality, let $x_{1}<x_{2}<x_{3}<x_{4}$. Then
$$
\begin{array}{l}
x_{1} x_{2}=2, x_{1} x_{3}=3, \text { so } \fra... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,941 |
Example 2 What is the maximum number of positive integers such that the sum of any three of them is a prime number? Prove your conclusion.
(2013, Peking University and other universities' independent enrollment examination) | From the question, we know that at least three positive integers are needed to talk about "the sum of any three numbers."
Take the numbers $1, 3, 7$, the sum of these three numbers is 11 (a prime number), which satisfies the condition.
Take the numbers 1, 3, 7, 9, the sums of any three numbers are 11, 13, 17, 19 (all... | 4 | Number Theory | proof | Yes | Yes | cn_contest | false | 727,942 |
Example 3 Does there exist $0<x<\frac{\pi}{2}$, such that $\sin x$, $\cos x$, $\tan x$, $\cot x$ form an arithmetic sequence?
(2010, Peking University and other universities' independent admission examination) | Assume there exists such an $x$. Then
$$
\begin{array}{l}
\cos x - \sin x = \cot x - \tan x \\
= \frac{(\cos x - \sin x)(\cos x + \sin x)}{\sin x \cdot \cos x}.
\end{array}
$$
Thus, $\cos x - \sin x = 0$ or $1 = \frac{\cos x + \sin x}{\sin x \cdot \cos x}$. If $\cos x - \sin x = 0$, then $x = \frac{\pi}{4}$. At this p... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,943 |
Example 4 Let $a_{1}, a_{2}, \cdots, a_{13}$ be an arithmetic sequence,
$$
M=\left\{a_{i}+a_{j}+a_{k} \mid 1 \leqslant i<j<k \leqslant 13\right\} .
$$
Question: Can $0$, $\frac{7}{2}$, and $\frac{16}{3}$ all be in the set $M$? Prove your conclusion.
(2014, Peking University and other schools' independent admission exa... | Let $a_{n}=a_{1}+(n-1) d$.
Then the elements in set $M$ can be represented as
$$
m_{k}=3 a_{1}+k d(k=3,4, \cdots, 33) \text {. }
$$
Assume $0, \frac{7}{2}, \frac{16}{3}$ can simultaneously be in set $M$, let
$$
m_{k_{1}}=0, m_{k_{2}}=\frac{7}{2}, m_{k_{3}}=\frac{16}{3} \text {. }
$$
Then $m_{k_{2}}-m_{k_{1}}=\frac{7}... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 727,944 |
Example 5 Prove: $\tan 3^{\circ}$ is an irrational number.
$(2014$, Peking University and other schools' independent recruitment examination) | Notice,
$$
\tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \cdot \tan \beta} .
$$
According to the closure of rational numbers under the four arithmetic operations, if $\tan \alpha$ and $\tan \beta$ are rational numbers, then $\tan (\alpha+\beta)$ is also a rational number.
Assume $\tan 3^{\circ}$ is a... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 727,945 |
Example 6 Given $m n$ real numbers, arranged in an $m \times n$ matrix, denoted as $\left\{a_{i j}\right\}_{m \times n}$, such that each row in the matrix is non-decreasing from left to right, i.e., for any $i=1,2, \cdots, m$, when $j_{1}<j_{2}$, we have $a_{i j 1} \leqslant a_{i j j_{1}}$.
Now, rearrange the numbers ... | To prove that in the number array $\left\{a_{i j}^{\prime}\right\}_{m \times n}$, the $n$ numbers in each row are increasing from left to right, it is sufficient to prove: for any $i=1,2, \cdots, m$, we have $a_{i j}^{\prime} \leqslant a_{i(j+1)}^{\prime}(j=1,2, \cdots, n-1)$.
Assume there exists a pair of numbers $a_... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 727,946 |
Define the sequence of positive integers $\left\{a_{n}\right\}$ as
$$
a_{1}=1, a_{2}=2, a_{k+2}=2 a_{k+1}+a_{k}(k \geqslant 1) \text {. }
$$
(1) Find all positive real numbers $\beta$ such that there are infinitely many pairs of positive integers $(p, q)$ satisfying
$$
\left|\frac{p}{q}-\sqrt{2}\right|<\frac{\beta}{q^{... | First, we provide two lemmas.
Lemma 1 The positive integer solutions of the Pell equation $\left|p^{2}-2 q^{2}\right|=1$ are
$$
p=b_{n}, q=a_{n}\left(n \in \mathbf{Z}_{+}\right),
$$
where $b_{1}=1, b_{2}=3, b_{k+2}=2 b_{k+1}+b_{k}(k \geqslant 1)$.
Lemma 2 The positive integer solutions of the Pell equation $\left|p^{... | \left(70-49 \sqrt{2}, \frac{\sqrt{2}}{2}\right) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,947 |
1. Let $P$ be the set of prime numbers, and $H$ be the set of composite numbers. Define $I(n)=\left\{\begin{array}{ll}1, & n \in P ; \\ 0, & n \in H .\end{array}\right.$
Below are three propositions:
(1) For any $x 、 y \in P$, we have $I(x+y)=0$;
(2) For any $x 、 y \in H$, we have $I(x+y)=0$;
(3) For $x \in P, y \in H$... | $-1 . A$
If $x=2, y=5$, then $x+y=7$, so
$$
I(x+y)=I(7)=1 \neq 0,
$$
thus proposition (1) is not true;
If $x=4, y=9$, then $x+y=13$, so
$$
I(x+y)=I(13)=1 \neq 0,
$$
thus proposition (2) is not true;
If $x=3, y=8$, then $x+y=11$, so
$$
I(x+y)=I(11)=1 \neq 0,
$$
thus proposition (3) is not true.
In summary, there are ... | A | Number Theory | MCQ | Yes | Yes | cn_contest | false | 727,949 |
2. In the Cartesian coordinate system $x O y$, the minimum distance from a moving point $P$ on the graph of the function $y=\frac{1}{2 x}$ to the origin $O$ is ( ).
(A) $\frac{1}{2}$
(B) $\frac{\sqrt{2}}{2}$
(C) 1
(D) $\sqrt{2}$ | 2. C.
According to the symmetry, we only need to consider the case where $x>0$.
Let the moving point be $P(x, y)$. Then $x y=\frac{1}{2}$.
Also, $x^{2}+y^{2}=(x-y)^{2}+2 x y=(x-y)^{2}+1$, so
when $x=y=\frac{\sqrt{2}}{2}$, $x^{2}+y^{2}$ reaches its minimum value of 1, at which point, $O P$ $=\sqrt{x^{2}+y^{2}}$ has its... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,950 |
Example 6 As shown in Figure 6, in $\square A B C D$, $A E \perp B C, A F \perp$ $C D, H$ is the orthocenter of $\triangle A E F$. Prove:
$$
A C^{2}=A H^{2}+E F^{2} \text {. }
$$ | Prove: As shown in Figure 6, connect $H E, H F, G F, G C$.
Since $H$ is the orthocenter of $\triangle A E F$,
$H E \perp A F, H F \perp A E$.
Combining with the given $A E \perp B C, A F \perp C D$, we get
$H E \parallel F C, H F \parallel C E$.
Therefore, quadrilateral $C E H F$ is a parallelogram.
Draw $G E \parallel... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 727,951 |
3. In $\triangle A B C$, it is known that $A B=A C, D$ is a point on $A C$, and $A D=B D=B C$. Then $\cos \angle B A C=(\quad)$.
(A) $\frac{\sqrt{5}-1}{2}$
(B) $\frac{\sqrt{5}+1}{2}$
(C) $\frac{\sqrt{5}+1}{4}$
(D) $\frac{\sqrt{5}-1}{4}$ | 3. C.
As shown in Figure 3, let $A B=A C=1$, $A D=B D=B C=a$. Then
$$
C D=1-a \text {. }
$$
It is easy to see that,
$$
\begin{array}{l}
\triangle A B C \backsim \triangle B C D \\
\Rightarrow \frac{A B}{B C}=\frac{B C}{C D} \Rightarrow \frac{1}{a}=\frac{a}{1-a} \\
\Rightarrow a^{2}=1-a \\
\Rightarrow a=\frac{\sqrt{5}... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 727,952 |
4. A sequence of numbers
$$
1,1,2,3,5,8,13,21,34,55,89,144,233, \cdots
$$
is constructed such that the first and second numbers are both 1, and from the third number onwards, each number is the sum of the two preceding numbers. What is the remainder when the 2014th number in this sequence is divided by 7? $(\quad)$.
(... | 4. D.
Let the sequence of numbers be $a_{1}, a_{2}, \cdots, a_{n}, \cdots$, with the rule $a_{1}=a_{2}=1, a_{n}=a_{n-1}+a_{n-2}(n \geqslant 3)$.
Consider the sequence of remainders when these numbers are divided by 7:
$$
1,1,2,3,5,1,6,0,6,6,5,4,2,6,1,0,1,1 \cdots \text {. }
$$
Observing that the 17th and 18th numbers... | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 727,953 |
5. The function $f(x)$ defined on $\mathbf{R}$. If for any real numbers $x, y$, the equation
$$
f(x-y)=f(x)+f(y)-2xy,
$$
is satisfied, then $f(201)=(\quad)$.
(A) 40400
(B) 40401
(C) 40402
(D) 40403 | 5. B.
In the given equation, let $x=y$, we get
$$
\begin{array}{l}
f(0)=f(x)+f(x)-2 x^{2} \\
\Rightarrow f(x)=x^{2}+\frac{a}{2}(f(0)=a) .
\end{array}
$$
Taking $x=y=0$, we get
$$
a=2 a \Rightarrow a=0 \text {. }
$$
Upon verification, $f(x)=x^{2}$ satisfies the equation.
Therefore, $f(201)=40401$. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,954 |
6. Given the function $f(x)=\frac{x^{2}+3 x+1}{x^{2}-x+1}$. Then the range of $f(x)$ is ( ).
(A) $\left[-1, \frac{5}{3}\right]$
(B) $\left[-\frac{1}{3}, 5\right]$
(C) $\left[-\frac{5}{3}, 1\right]$
(D) $\left[-5, \frac{1}{3}\right]$ | 6. B.
Notice that, $x^{2}-x+1>0$.
Therefore, the domain of $f(x)$ is all real numbers.
Let $y=\frac{x^{2}+3 x+1}{x^{2}-x+1}$. Then
$$
\begin{array}{l}
y\left(x^{2}-x+1\right)=x^{2}+3 x+1 \\
\Rightarrow(y-1) x^{2}-(y+3) x+(y-1)=0 .
\end{array}
$$
If the coefficient of the quadratic term $y-1=0$, then $x=0$.
If the coe... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 727,955 |
1. Below is the inscription by the famous Chinese mathematician, Academician Wang Yuan:
数棈妻好
If different Chinese characters represent different digits from $0 \sim 9$, and assuming “数学竞赛好” represents the largest five-digit number that is a perfect square, formed by different digits. Then this five-digit number is $\q... | Notice that, $300^{2}=90000,310^{2}=96100$,
$$
320^{2}=102400>100000 \text {. }
$$
Thus, this five-digit number with distinct digits might be among the nine square numbers between $310^{2} \sim$ $320^{2}$.
$$
\begin{array}{l}
\text { Also, } 311^{2}=96721,312^{2}=97344,313^{2}=97969, \\
314^{2}=98596,315^{2}=99225,316... | 96721 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 727,956 |
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