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
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
Question 10 In quadrilateral $A B C D$, $\angle A B C=$ $\angle A D C=90^{\circ}, P$ is the intersection of diagonals $A C$ and $B D$, $B F$ $\perp A C$ at $F, D E \perp A C$ at $E$, a line through $P$ intersects $B F$ and $D E$ at points $H$ and $G$ respectively, $A H$ intersects $B E$ at $M$, $C G$ intersects $D F$ a... | Prove: As shown in Figure 10, the line $H M A$ intersects $\triangle B E F$ to yield
$$
\frac{B M}{M E} \cdot \frac{E A}{A F} \cdot \frac{F H}{H B}=1 \text {. }
$$
The line $G N C$ intersects $\triangle D E F$ to yield
$$
\begin{array}{l}
\frac{F N}{N D} \cdot \frac{D G}{G E} \cdot \frac{E C}{C F} \\
=1 .
\end{array}
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,633 |
1. The number of solutions to the system of equations $\left\{\begin{array}{l}|x|+y=12, \\ x+|y|=6\end{array}\right.$ is ( ) .
(A) 1
(B) 2
(C) 3
(D) 4 | -、1.A.
If $x \geqslant 0$, then $\left\{\begin{array}{l}x+y=12, \\ x+|y|=6 \text {. }\end{array}\right.$
Thus, $|y|-y=6$, which is obviously impossible.
If $x<0$, then $\left\{\begin{array}{l}-x+y=12, \\ x+|y|=6 \text {. }\end{array}\right.$ Thus, $|y|+y=18$.
Solving this gives $y=9$, and subsequently, $x=-3$.
Therefor... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,634 |
3. Given that $\triangle A B C$ is an acute triangle, $\odot O$ passes through points $B$ and $C$, and intersects sides $A B$ and $A C$ at points $D$ and $E$ respectively. If the radius of $\odot O$ is equal to the radius of the circumcircle of $\triangle A D E$, then $\odot O$ must pass through the ( ) of $\triangle A... | 3. B.
As shown in Figure 7, connect $B E$.
Since $\triangle A B C$ is an acute triangle, $\angle B A C$ and $\angle A B E$ are both acute angles.
Furthermore, the radius of $\odot O$ is equal to the radius of the circumcircle of $\triangle A D E$, and $D E$ is a common chord of the two circles, thus,
$$
\angle B A C=... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,636 |
4. Given three quadratic equations in $x$
$$
\begin{array}{l}
a x^{2}+b x+c=0, \\
b x^{2}+c x+a=0, \\
c x^{2}+a x+b=0
\end{array}
$$
have exactly one common real root. Then the value of $\frac{a^{2}}{b c}+\frac{b^{2}}{c a}+\frac{c^{2}}{a b}$ is ( ).
(A) 0
(B) 1
(C) 2
(D) 3 | 4.D.
Let $x_{0}$ be a common real root of them. Then
$$
(a+b+c)\left(x_{0}^{2}+x_{0}+1\right)=0 \text {. }
$$
Since $x_{0}^{2}+x_{0}+1=\left(x_{0}+\frac{1}{2}\right)^{2}+\frac{3}{4}>0$, we have
$$
\begin{array}{l}
a+b+c=0 . \\
\text { Therefore, } \frac{a^{2}}{b c}+\frac{b^{2}}{a a}+\frac{c^{2}}{a b}=\frac{a^{3}+b^{3... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,637 |
5. The number of integer solutions $(x, y)$ for the equation $x^{3}+6 x^{2}+5 x=y^{3}-y+2$ is ( ).
(A) 0
(B) 1
(C) 3
(D) infinitely many | 5.A.
The original equation can be rewritten as
$$
\begin{array}{l}
x(x+1)(x+2)+3\left(x^{2}+x\right) \\
=y(y-1)(y+1)+2 .
\end{array}
$$
Since the product of three consecutive integers is a multiple of 3, the left side of the equation is a multiple of 3, while the right side leaves a remainder of 2 when divided by 3, ... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,638 |
Example 8 When the algebraic expression
$$
\begin{array}{l}
\sqrt{9 x^{2}+4}+\sqrt{9 x^{2}-12 x y+4 y^{2}+1}+ \\
\sqrt{4 y^{2}-16 y+20}
\end{array}
$$
reaches its minimum value, find the values of $x$ and $y$. | Explanation: Although the ratio 7 is complex, the problem-solving approach is the same, the key is to transform the given expression. At this point, one can think of the distance formula between two points.
Transform the original algebraic expression into
$$
\begin{array}{c}
\sqrt{[0-(-2)]^{2}+(3 x-0)^{2}}+ \\
\sqrt{(1... | x=\frac{8}{15}, y=\frac{6}{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,639 |
6. As shown in Figure 1, in the right triangle $\triangle ABC$, $\angle ACB=90^{\circ}$, $CA=4$, $P$ is the midpoint of the semicircular arc $\overparen{AC}$, connect $BP$, the line segment $BP$ divides the figure $APCB$ into two parts. The absolute value of the difference in the areas of these two parts is $\qquad$. | Ni.6.4.
As shown in Figure 8, let $A C$ and $B P$ intersect at point $D$, and the point symmetric to $D$ with respect to the circle center $O$ is denoted as $E$. The line segment $B P$ divides the figure $A P C B$ into two parts, and the absolute value of the difference in the areas of these two parts is the area of $\... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,640 |
7. As shown in Figure 2, points $A$ and $C$ are both on the graph of the function $y=\frac{3 \sqrt{3}}{x}(x$ $>0)$, points $B$ and $D$ are both on the $x$-axis, and such that $\triangle O A B$ and $\triangle B C D$ are both equilateral triangles. Then the coordinates of point $D$ are $\qquad$. | 7. $(2 \sqrt{6}, 0)$.
As shown in Figure 9, draw perpendiculars from points $A$ and $C$ to the $x$-axis, with the feet of the perpendiculars being $E$ and $F$ respectively. Let $O E=a, B F=b$, then $A E = \sqrt{3} a, C F = \sqrt{3} b$.
Thus, $A(a, \sqrt{3} a)$,
$$
C(2 a+b, \sqrt{3} b) \text{. }
$$
Therefore, $\left\{... | (2 \sqrt{6}, 0) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,641 |
8. Given points $A(1,0)$ and $B(2,0)$. If the graph of the quadratic function $y=x^{2}+(a-3)x+3$ intersects the line segment $AB$ at only one point, then the range of values for $a$ is $\qquad$. | 8. $-1 \leqslant a<-\frac{1}{2}$ or $a=3-2 \sqrt{3}$.
In two cases:
(1) Since the graph of the quadratic function $y=x^{2}+(a-3) x+3$ intersects the line segment $A B$ at only one point, and $A(1,0), B(2,0)$, then
$$
\left[1^{2}+(a-3) \times 1+3\right] \times\left[2^{2}+(a-3) \times 2+3\right]<0 \text {. }
$$
Solving... | -1 \leqslant a<-\frac{1}{2} \text{ or } a=3-2 \sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,642 |
11. A. As shown in Figure 4, given points \( M(0,1) \), \( N(0,-1) \), and \( P \) is a moving point on the parabola \( y = \frac{1}{4} x^2 \).
(1) Determine the positional relationship between the circle with center \( P \) and radius \( PM \) and the line \( y = -1 \);
(2) Let the other intersection point of line \( ... | Three, 11.A. (1) Let $P\left(x_{0}, \frac{1}{4} x_{0}^{2}\right)$. Then
$$
P M=\sqrt{x_{0}^{2}+\left(\frac{1}{4} x_{0}^{2}-1\right)^{2}}=\frac{1}{4} x_{0}^{2}+1 \text {. }
$$
Since the distance from point $P$ to the line $y=-1$ is
$$
\frac{1}{4} x_{0}^{2}-(-1)=\frac{1}{4} x_{0}^{2}+1 \text {, }
$$
the circle with cen... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,645 |
12.A. Given that $a$ and $b$ are positive integers. Determine whether the equation $x^{2}-a b x+\frac{1}{2}(a+b)=0$ has two integer solutions. If it does, find them; if not, provide a proof. | 12.A. Let's assume $a \leqslant b$, and the two integer roots of the equation are $x_{1}, x_{2} \left(x_{1} \leqslant x_{2}\right)$. Then we have
$x_{1} + x_{2} = ab, x_{1} x_{2} = \frac{1}{2}(a + b)$.
Therefore, $x_{1} x_{2} - x_{1} - x_{2} = \frac{1}{2}a + \frac{1}{2}b - ab$.
Thus, $4\left(x_{1} - 1\right)\left(x_{2}... | x_{1} = 1, x_{2} = 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,646 |
12.B. The real numbers $a, b, c$ satisfy $a \leqslant b \leqslant c$, and $ab + bc + ca = 0, abc = 1$. Find the largest real number $k$ such that the inequality $|a+b| \geqslant k|c|$ always holds. | 12. B. When $a=b=-\sqrt[3]{2}, c=\frac{\sqrt[3]{2}}{2}$, the real numbers $a, b, c$ satisfy the given conditions, at this time, $k \leqslant 4$.
Below is the proof: The inequality $|a+b| \geqslant 4|c|$ holds for all real numbers $a, b, c$ that satisfy the given conditions.
From the given conditions, we know that $a, ... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,647 |
13.A. As shown in Figure 5, given that $AB$ is the diameter of the semicircle $\odot O$, and $P$ is any point on the diameter $AB$. With point $A$ as the center and $AP$ as the radius, draw $\odot A$, which intersects the semicircle $\odot O$ at point $C$; with point $B$ as the center and $BP$ as the radius, draw $\odo... | 13.A. As shown in Figure 12, connect $A C$, $A D$, $B C$, and $B D$, and draw perpendiculars from points $C$ and $D$ to $A B$, with the feet of the perpendiculars being $E$ and $F$. Then, $C E \parallel D F$.
Since $A B$ is the diameter of $\odot O$, we have $\angle A C B = \angle A D B = 90^{\circ}$.
In right triang... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,648 |
13. B. As shown in Figure 6, points $E$ and $F$ are on the extensions of sides $AD$ and $BC$ of quadrilateral $ABCD$, respectively, and satisfy $\frac{DE}{CF}=\frac{AD}{BC}$. If the extensions of $CD$ and $FE$ intersect at point $G$, and the circumcircles of $\triangle DEG$ and $\triangle CFG$ intersect at another poin... | 13. B. (1) As shown in Figure 13, connect $P E$, $P F$, and $P G$.
Since $\angle P D G = \angle P E G$,
then $\angle P D C = \angle P E F$.
Also, $\angle P C G = \angle P F G$,
thus $\triangle P D C \backsim \triangle P E F$.
Therefore, $\frac{P D}{P C} = \frac{P E}{P F}$, $\angle C P D = \angle F P E$.
Hence, $\trian... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,649 |
12. Given that $P$ is a point inside $\triangle A B C$ and satisfies $3 P A+4 P B+5 P C=0$. Then, the ratio of the areas of $\triangle P A B$, $\triangle P B C$, and $\triangle P C A$ is $\qquad$ | Establish the Cartesian coordinate system as shown in Figure 9. Let
$$
\begin{array}{l}
A(-2,0), \\
B(0,3 x), \\
C(1,2 y), \\
D(3,4) .
\end{array}
$$
Then \( \mid A B \mid \) is in the ratio 12.5:3:4.
Take points \( A^{\prime}, B^{\prime}, C^{\prime} \) on \( P A, P B, P C \) respectively, such that \( P A^{\prime}=3... | 5 : 3 : 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,650 |
14. A. (1) Do there exist positive integers $m, n$ such that $m(m+2)=n(n+1)$?
(2) Let $k(k \geqslant 3)$ be a given positive integer. Do there exist positive integers $m, n$ such that $m(m+k)=n(n+1)$? | 14.A. (1) The answer is negative.
If there exist positive integers $m, n$ such that $m(m+2)=n(n+1)$, then $(m+1)^{2}=n^{2}+n+1$. Clearly, $n>1$.
Thus, $n^{2}<n^{2}+n+1<(n+1)^{2}$, so the above equation cannot hold.
When $k \geqslant 4$, if $k=2 t$ ($t$ is an integer no less than 2), take $m=t^{2}-t, n=t^{2}-1$, then
$... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 718,651 |
11.B. Given the parabolas $C_{1}: y=-x^{2}-3 x+4$ and $C_{2}: y=x^{2}-3 x-4$ intersect at points $A$ and $B$, point $P$ is on parabola $C_{1}$ and lies between points $A$ and $B$; point $Q$ is on parabola $C_{2}$, also lying between points $A$ and $B$.
(1) Find the length of segment $A B$;
(2) When $P Q \parallel y$-ax... | 11. B. (1) Solve the system of equations $\left\{\begin{array}{l}y=-x^{2}-3 x+4, \\ y=x^{2}-3 x-4,\end{array}\right.$ to get
$$
\left\{\begin{array} { l }
{ x _ { 1 } = - 2 , } \\
{ y _ { 1 } = 6 , }
\end{array} \left\{\begin{array}{l}
x_{2}=2, \\
y_{2}=-6 .
\end{array}\right.\right.
$$
Therefore, $A(-2,6)$ and $B(2,... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,652 |
14.B. Proof: For any triangle, there must exist two sides, whose lengths $u$ and $v$ satisfy $1 \leqslant \frac{u}{v}<\frac{1+\sqrt{5}}{2}$. | 14. B. Let the three sides of an arbitrary $\triangle ABC$ be $a, b, c$ (assuming $a \geqslant b \geqslant c$). If the conclusion does not hold, i.e.,
$$
\frac{a}{b} \geqslant \frac{1+\sqrt{5}}{2}, \frac{b}{c} \geqslant \frac{1+\sqrt{5}}{2}.
$$
Then $a \geqslant \frac{1+\sqrt{5}}{2} b=b+\frac{\sqrt{5}-1}{2} b$
$$
\geq... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,653 |
1. The maximum value of the real number $k$ for which the inequality $\frac{1+\sin x}{2+\cos x} \geqslant k$ has solutions for $x$ is ( ).
(A) $-\frac{4}{3}$
(B) $-\frac{3}{4}$
(C) $\frac{3}{4}$
(D) $\frac{4}{3}$ | -.1.D.
This problem essentially seeks the upper limit of the range of $f(x)=\frac{1+\sin x}{2+\cos x}$.
View $f(x)$ as the slope of the line determined by point $A(\cos x, \sin x)$ and point $B(-2, -1)$. Point $A$ moves along the unit circle, and the slope takes its extreme values when $BA$ is tangent to the circle.
.... | D | Inequalities | MCQ | Yes | Yes | cn_contest | false | 718,654 |
2. Given the quadratic function $f(x)$ satisfies:
$$
\begin{array}{l}
f(1-x)=f(1+x),-4 \leqslant f(1) \leqslant-1, \\
-1 \leqslant f(2) \leqslant 5 .
\end{array}
$$
Then the range of $f(3)$ is ( ).
(A) $7 \leqslant f(3) \leqslant 26$
(B) $-4 \leqslant f(3) \leqslant 15$
(C) $-1 \leqslant f(3) \leqslant 32$
(D) $-\frac... | 2.C.
Let $f(x)=a x^{2}+b x+c$, then
$$
f(1)=a+b+c, f(2)=4 a+2 b+c, f(0)=c \text {. }
$$
Also, the axis of symmetry of $f(x)$ is $x=1$, so $f(2)=f(0)$. This gives $2 a+b=0$.
Thus, $f(1)=-a+c, f(2)=c$.
But $f(3)=9 a+3 b+c=3 a+c$, hence
$$
f(3)=-3 f(1)+4 f(2) \text {. }
$$
From the given ranges of $f(1)$ and $f(2)$, we... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,655 |
3. Given $x, y \in\left[-\frac{\pi}{4}, \frac{\pi}{4}\right], a \in \mathbf{R}$, and
$$
\left\{\begin{array}{l}
x^{3}+\sin x-2 a=0, \\
4 y^{3}+\sin y \cdot \cos y+a=0 .
\end{array}\right.
$$
Then the value of $\cos (x+2 y)$ is ( ).
(A) 1
(B) -1
(C) 0
(D) $\frac{\sqrt{2}}{2}$ | 3.A.
From the problem, we have $\left\{\begin{array}{l}x^{3}+\sin x=2 a, \\ (-2 y)^{3}+\sin (-2 y)=2 a \text {. }\end{array}\right.$ Let $f(t)=t^{3}+\sin t, t \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, then $f(t)$ is an increasing function on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. Therefore, from $f(x)=f(-... | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,656 |
4. Let $a$ and $b$ be two coprime natural numbers. Then the greatest common divisor of $a^{2}+$ $b^{2}$ and $a^{3}+b^{3}$ is ( ).
(A) 1
(B) 2
(C) 1 or 2
(D) Possibly greater than 2 | 4.C.
If we take $a=2, b=3$, then $a^{2}+b^{2}=13, a^{3}+b^{3}=35$. 13 and 35 are coprime.
If we take $a=3, b=5$, then $a^{2}+b^{2}$ and $a^{3}+b^{3}$ are both even, they have a common factor of 2.
Now assume $a^{2}+b^{2}$ and $a^{3}+b^{3}$ have a common factor of 4, then because
$$
a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}\... | C | Number Theory | MCQ | Yes | Yes | cn_contest | false | 718,657 |
5. On the $x O y$ plane, the coordinates of the vertices of a triangle are $\left(x_{i}, y_{i}\right)(i=1,2,3)$, where $x_{i}, y_{i}$ are integers and satisfy $1 \leqslant x_{i} \leqslant n, 1 \leqslant y_{i} \leqslant n$ ($n$ is an integer), and there are 516 such triangles. Then the value of $n$ is ( ).
(A) 3
(B) 4
(... | 5. B.
It is easy to calculate that when $n \geqslant 5$, the number of triangles all exceed 516. Therefore, we only need to consider the case where $n<5$: At this point, all sets of three points have $\mathrm{C}_{n}^{3}$ 2 sets, among which, the sets of three collinear points in the horizontal or vertical direction to... | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 718,658 |
6. In $\triangle A B C$, $A$ is a moving point, $B\left(-\frac{a}{2}, 0\right)$, $C\left(\frac{a}{2}, 0\right)(a>0)$ are fixed points, and the trajectory equation of the moving point $A$ is $\frac{16 x^{2}}{a^{2}}-\frac{16 y^{2}}{3 a^{2}}=1(y \neq 0)$ of the right branch. Then the three interior angles of $\triangle A ... | 6. A.
Write the trajectory equation as $\frac{x^{2}}{\left(\frac{a}{4}\right)^{2}}-\frac{y^{2}}{\left(\frac{\sqrt{3} a}{4}\right)^{2}}=1$.
From the definition of a hyperbola, we have
$$
|A B|-|A C|=2 \times \frac{a}{4}=\frac{a}{2}=\frac{1}{2}|B C| \text {. }
$$
By the Law of Sines, we know
$$
\begin{array}{l}
|A B|=2... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,659 |
7. Let non-zero distinct complex numbers $x, y$ satisfy $x^{2}+x y+$ $y^{2}=0$. Then the value of the expression
$$
\left[\frac{x y}{(x+y)(x-y)^{2}}\right]^{2000}\left(x^{2006}+y^{2006}\right)
$$
is $\qquad$ . | $=7 .-\frac{1}{3^{2000}}$.
From $x 、 y$ being non-zero and distinct, we get $\left(\frac{x}{y}\right)^{2}+\frac{x}{y}+1=0$, then $\frac{x}{y}=\omega$ is a complex cube root of unity. Therefore,
$$
\begin{array}{l}
\text { Original expression }=\left[\frac{\omega}{(1+\omega)(1-\omega)^{2}}\right]^{2006}\left(\omega^{200... | -\frac{1}{3^{2006}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,660 |
8. Given $a b=1$, and $\frac{1}{1-2^{x} a}+\frac{1}{1-2^{y+1} b}=1$, then the value of $x+y$ is $\qquad$. | 8. -1 .
Given that both sides of the equation are multiplied by $\left(1-2^{x} a\right)\left(1-2^{y+1} b\right)$, we get
$$
2-2^{x} a-2^{y+1} b=1-2^{x} a-2^{y+1} b+2^{x+y+1} a b \text {, }
$$
which simplifies to $1=2 \times 2^{x+y}$.
Therefore, $x+y=-1$. | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,662 |
9. Let $\alpha, \beta, \gamma \in \left(0, \frac{\pi}{2}\right)$, and satisfy $\cos \alpha = \alpha, \cos (\sin \beta) = \beta, \sin (\cos \gamma) = \gamma$. Then the size relationship of $\alpha, \beta, \gamma$ is $\qquad$. | 9. $\gamma\cos \beta$,
$$
\gamma=\sin (\cos \gamma)<\cos \gamma,
$$
and $f(x)=x-\cos x$ is an increasing function on $\left[0, \frac{\pi}{2}\right]$, then we have
$$
\gamma<\alpha<\beta .
$$ | \gamma<\alpha<\beta | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,663 |
10. If the complex numbers $z_{1}$ and $z_{2}$ satisfy $\left|z_{1}\right|=\left|z_{2}\right|$, and $z_{1}-z_{2}=2-\mathrm{i}$, then the value of $\frac{z_{1} z_{2}}{\left|z_{1} z_{2}\right|}$ is $\qquad$. | 10. $-\frac{3}{5}+\frac{4}{5}$ i.
Notice that $\left|z_{1} z_{2}\right|=\left|z_{1}\right|^{2}=\left|z_{2}\right|^{2}$, then
$$
\begin{array}{l}
2-\mathrm{i}=z_{1}-z_{2}=\frac{1}{\left|z_{2} \overline{z_{2}}\right|} z_{1} z_{2} \overline{z_{2}}-\frac{1}{\left|z_{1} \overline{z_{1}}\right|} z_{2} z_{1} \overline{z_{1}}... | -\frac{3}{5}+\frac{4}{5} \mathrm{i} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,664 |
11. Let the sum of 10 consecutive positive integers not greater than 2006 form the set $S$, and the sum of 11 consecutive positive integers not greater than 2006 form the set $T$, then the number of elements in $S \cap T$ is $\qquad$ . | 11.181.
$S$ is the set of all integers ending in 5, starting from 55 to
$$
20015=\sum_{i=1}^{10}(1996+i)=10 \times 1996+55
$$
Similarly, $T$ is the set of integers starting from 66 and increasing by 11 each time, with the largest number being
$$
22011=\sum_{i=1}^{11}(1995+i)=11 \times 1995+66 \text {. }
$$
In $T$, one... | 181 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 718,665 |
12. Given $n$ vectors in the plane as $O P_{1}$, $O P_{2}$, $\cdots$, $O P_{n}$, the vector $\boldsymbol{O P}$ makes $\boldsymbol{P P}_{1}^{2}+\boldsymbol{P P}_{2}^{2}+\cdots$ $+P P_{n}^{2}$ minimal. Then the vector $\boldsymbol{O P}$ is $\qquad$ | $$
\begin{array}{l}
\text { 12. } \frac{1}{n} \sum_{k=1}^{n} O P_{k} \text {. } \\
\sum_{k=1}^{n} \boldsymbol{P P}_{\boldsymbol{k}}^{2}=\sum_{k=1}^{n}\left(\boldsymbol{O P} \boldsymbol{P}_{\boldsymbol{k}}-\boldsymbol{O P}\right)^{2} \\
=\sum_{k=1}^{n}\left(O P_{k}^{2}-2 O P_{k} \cdot O P+O P^{2}\right) \\
=\sum_{k=1}^{... | O P=\frac{1}{n} \sum_{k=1}^{n} O P_{k} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,666 |
13. Let $f$ be a function from the set of real numbers $\mathbf{R}$ to the set of real numbers $\mathbf{R}$, satisfying
$$
f(x+y)=f(x)+f(y)+2 x y .
$$
If the graph of $f(x)$ has a line of symmetry $x=k$ and in the interval | Three, 13. Let $x=0$, we get $f(0)=0$. From the given condition, we have
$$
0=f(0)=f(x-x)=f(x)+f(-x)-2 x^{2} \text {, }
$$
which means $f(x)+f(-x)=2 x^{2}$.
$$
\begin{array}{l}
\text { Also, } f(k+x)=f(k)+f(x)+2 k x, \\
f(k-x)=f(k)+f(-x)-2 k x,
\end{array}
$$
Subtracting the above two equations, and noting that $f(k+... | k \geqslant 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,667 |
14. Let $a, b$ be positive integers, two lines $l_{1}: y=$ $-\frac{b}{2 a} x+b$ and $l_{2}: y=\frac{b}{2 a} x$ intersect at $\left(x_{1}, y_{1}\right)$. For natural number $n(n \geqslant 2)$, the line passing through the points $(0, b)$ and $\left(x_{n-1}, 0\right)$ intersects the line $l_{2}$ at $\left(x_{n}, y_{n}\ri... | 14. The line $l_{1}$ passes through the points $(2 a, 0)$ and $(0, b)$. It is easy to see that the intersection point of $l_{1}$ and $l_{2}$ is
$$
\left(x_{1}, y_{1}\right)=\left(a, \frac{b}{2}\right).
$$
The line passing through the points $(0, b)$ and $\left(x_{n-1}, 0\right)$ has the equation
$$
\frac{x}{x_{n-1}}+\... | x_{n} = \frac{2a}{n+1}, \quad y_{n} = \frac{b}{n+1} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,668 |
15. Let $a>1, b>1$. Prove:
$$
\frac{a^{4}}{(b-1)^{2}}+\frac{b^{4}}{(a-1)^{2}} \geqslant 32 .
$$ | 15. Let $\left\{\begin{array}{l}x=a-1, \\ y=b-1,\end{array}\right.$ then $\left\{\begin{array}{l}a=x+1, \\ b=y+1 .\end{array}\right.$
The left side of the inequality
$$
\geqslant \frac{(2 \sqrt{x})^{4}}{y^{2}}+\frac{(2 \sqrt{y})^{4}}{x^{2}}=16\left(\frac{x^{2}}{y^{2}}+\frac{y^{2}}{x^{2}}\right) \geqslant 32 \text {. }... | 32 | Inequalities | proof | Yes | Yes | cn_contest | false | 718,669 |
$$
\begin{array}{l}
A=\left\{y \mid y=-x^{2}+1, x \in \mathbf{R}\right\}, \\
B=\{y \mid y=-x+1, x \in \mathbf{R}\},
\end{array}
$$
If the sets are defined as above, then $A \cap B$ equals ( ).
(A) $(0,1)$ or $(1,1)$
(B) $\{(0,1),(1,1)\}$
(C) $\{0,1\}$
(D) $(-\infty, 1]$ | $$
-1 . \mathrm{D} \text {. }
$$
From $A=(-\infty, 1], B=(-\infty,+\infty)$, we get $A \cap B=(-\infty, 1]$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,670 |
2. For a sequence $P=\left(p_{1}, p_{2}\right.$, $\cdots, p_{n}$ ), the "Cesàro sum" of $P$ (Cesàro is a mathematician) is defined as $\frac{S_{1}+S_{2}+\cdots+S_{n}}{n}$, where,
$$
S_{k}=p_{1}+p_{2}+\cdots+p_{k}(1 \leqslant k \leqslant n) .
$$
If the Cesàro sum of the sequence $\left(p_{1}, p_{2}, \cdots, p_{2006}\ri... | 2.A.
Let $S_{k}^{\prime}=1+p_{1}+p_{2}+\cdots+p_{k-1}=1+S_{k-1}$, then
$$
\begin{array}{l}
\frac{\sum_{k=1}^{2007} S_{k}^{\prime}}{2007}=\frac{2007+\sum_{k=1}^{2006} S_{k}}{2007} \\
=\frac{2007+2007 \times 2006}{2007}=2007 .
\end{array}
$$ | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,671 |
Example 11 Let $a, b, c, d, x, y, z, m$ all be positive real numbers, and satisfy
$$
a+x=b+y=c+z=d+m=1 .
$$
Prove: $a m+b x+c y+d z<2$. | Explanation: From the given conditions, we think of constructing a square $ABCD$ with side length 1, and taking points $E, F, G, H$ on sides $AB, BC, CD, DA$ respectively, such that
$$
A E = a, B F = b, C G = c,
$$
$D H = b$ (as shown in Figure 12). Then,
$$
\begin{array}{l}
E B = x, F C = y, \\
G D = z, H A = m .
\en... | a m + b x + c y + d z < 2 | Inequalities | proof | Yes | Yes | cn_contest | false | 718,672 |
3. As shown in Figure 1, for the cube $A B C D-$ $A_{1} B_{1} C_{1} D_{1}$, draw a line $l$ through vertex $A_{1}$ such that $l$ forms an angle of $60^{\circ}$ with both lines $A C$ and $B C_{1}$. How many such lines $l$ are there?
(A) 1
(B) 2
(C) 3
(D) More than 3 | 3.C.
Obviously, $A D_{1} // B C_{1}$. There are 3 lines passing through point $A$ that form a $60^{\circ}$ angle with both lines $A C$ and $A D_{1}$. Therefore, there are also 3 lines $l$ passing through point $A_{1}$ that satisfy the conditions. | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,673 |
4. Given that the graph of the function $y=\sin x+a \cos x$ is symmetric about the line $x=\frac{5 \pi}{3}$. Then the graph of the function $y=a \sin x+\cos x$ is symmetric about the line ( ) .
(A) $x=\frac{\pi}{3}$
(B) $x=\frac{2 \pi}{3}$
(C) $x=\frac{11 \pi}{6}$
(D) $x=\pi$ | 4.C.
Let $f(x)=\sin x+a \cos x, g(x)=a \sin x+\cos x$. According to the problem, we have $f(x)=f\left(\frac{10 \pi}{3}-x\right)$.
Also, $f(x)=g\left(\frac{\pi}{2}-x\right)$,
$f\left(\frac{10 \pi}{3}-x\right)=g\left(\frac{\pi}{2}-\frac{10 \pi}{3}+x\right)$,
thus $g\left(\frac{\pi}{2}-x\right)=g\left(\frac{\pi}{2}-\fra... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,674 |
5. If the function $f(x)=x^{3}-6 b x+3 b$ has a local minimum in $(0,1)$, then the range of the real number $b$ is ( ).
(A) $(0,1)$
(B) $(-\infty, 1)$
(C) $(0,+\infty)$
(D) $\left(0, \frac{1}{2}\right)$ | 5.D.
From the problem, we know $f^{\prime}(0)0$, which means
$$
-6 b0 \Rightarrow 0<b<\frac{1}{2} \text {. }
$$ | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,675 |
6. Among 6 products, there are 4 genuine items and 2 defective items. Now, each time 1 item is taken out for inspection (the inspected item is not put back), until both defective items are found. What is the probability that both defective items are found exactly after 4 inspections?
(A) $\frac{1}{15}$
(B) $\frac{2}{15... | 6. C.
The required probability is
$$
\begin{array}{l}
\frac{2}{6} \times \frac{4}{5} \times \frac{3}{4} \times \frac{1}{3}+\frac{4}{6} \times \frac{2}{5} \times \frac{3}{4} \times \frac{1}{3}+ \\
\frac{4}{6} \times \frac{3}{5} \times \frac{2}{4} \times \frac{1}{3}=\frac{1}{5} .
\end{array}
$$ | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 718,676 |
7. If $\sin ^{2}\left(x+\frac{\pi}{12}\right)-\sin ^{2}\left(x-\frac{\pi}{12}\right)=-\frac{1}{4}$, and $x \in\left(\frac{\pi}{2}, \frac{3 \pi}{4}\right)$, then the value of $\tan x$ is . $\qquad$ | II, 7. $-2-\sqrt{3}$.
$$
\begin{array}{l}
\text { Given } \sin ^{2}\left(x+\frac{\pi}{12}\right)-\sin ^{2}\left(x-\frac{\pi}{12}\right)=-\frac{1}{4} \\
\Rightarrow \cos \left(2 x-\frac{\pi}{6}\right)-\cos \left(2 x+\frac{\pi}{6}\right)=-\frac{1}{2} \\
\Rightarrow 2 \sin 2 x \cdot \sin \frac{\pi}{6}=-\frac{1}{2} \\
\Rig... | -2-\sqrt{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,677 |
8. Given the function $f(x)=\log _{\frac{1}{2}} x$, let $x=\frac{a}{f(a)}$, $y=\frac{b}{f(b)}, z=\frac{c}{f(c)}(0<c<b<a<1)$. Then, the order of $x 、 y 、 z$ in terms of size is $\qquad$ . | $$
\begin{array}{l}
\text { 8. } x>y>z . \\
x-y=\frac{a \log \frac{1}{2} b-b \log _{\frac{1}{2}} a}{\log \frac{1}{2} a \cdot \log \frac{1}{2} b} \\
=\frac{(a \lg b-b \lg a)(-\lg 2)}{\lg a \cdot \lg b}=\frac{-\lg 2 \cdot \lg \frac{b^{a}}{a^{b}}}{\lg a \cdot \lg b} . \\
\text { From } 0<y$.
Similarly, $y>z$.
$$ | x>y>z | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,678 |
9. For the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, the right vertex is $A$, the upper vertex is $B$, and the left focus is $F$. If $\angle A B F=$ $90^{\circ}$, then the eccentricity of the ellipse is $\qquad$. | 9. $\frac{\sqrt{5}-1}{2}$.
Given $\angle A B F=90^{\circ}$, according to the projection theorem, we have $a c=b^{2}=a^{2}-c^{2}$,
i.e., $\left(\frac{c}{a}\right)^{2}+\left(\frac{c}{a}\right)-1=0$.
Thus, $e=\frac{c}{a}=\frac{-1 \pm \sqrt{5}}{2}$ (the negative value is discarded). | \frac{\sqrt{5}-1}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,679 |
10. As shown in Figure 2, in Pascal's Triangle, the numbers above the diagonal form the sequence: $1,3,6,10, \cdots$, let the sum of the first $n$ terms of this sequence be $S_{n}$. Then, as $n \rightarrow +\infty$, the limit of $\frac{n^{3}}{S(n)}$ is $\qquad$ | 10.6 .
It is known that $S_{n}=\mathrm{C}_{2}^{2}+\mathrm{C}_{3}^{2}+\cdots+\mathrm{C}_{n+1}^{2}=\mathrm{C}_{\mathrm{n}+2}^{3}$. Therefore, $\lim _{n \rightarrow+\infty} \frac{n^{3}}{S(n)}=\lim _{n \rightarrow+\infty} \frac{n^{3} \times 6}{(n+2)(n+1) n}=6$. | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 718,680 |
13. Prove: $1<\sqrt{2 \sqrt{3 \sqrt{4 \cdots \sqrt{n}}}}<3$, where $n$ is any positive integer. | Three, 13. It is easy to know that $\sqrt{2 \sqrt{3 \sqrt{4 \cdots \sqrt{n}}}}$ is a monotonically increasing sequence, so $1<\sqrt{2 \sqrt{3 \sqrt{4 \cdots \sqrt{n}}}}$.
We only need to prove that $\sqrt{2 \sqrt{3 \sqrt{4 \cdots \sqrt{n}}}}<3$.
When $n=2$, the proposition holds.
When $n \geqslant 3$,
$$
\begin{array}{... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,682 |
Example 1 Given 100 points on a plane, it is known that the distance between any two points does not exceed 1, and any three points form an obtuse triangle. Prove: These 100 points are covered by a circle with a radius of $\frac{1}{2}$. | In these 100 points, every two points have a distance. Among these finite numbers $\left(\frac{100 \times 99}{2}\right)$, there must be a maximum number (Property 1). Let's assume the distance between points $A$ and $B$ is the maximum (if there are multiple maximum distances, choose any one). Draw a circle with $A B$ a... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,683 |
Example 2 In space, 8 points are given, where no four points lie on the same plane, and 17 line segments are drawn between them. Prove: these line segments form at least one triangle.
untranslated text remains the same as requested, only the content has been translated. | Explanation: Each point is connected by several line segments (at most 7). Let's assume the point with the most connected line segments is $A$, which is connected by $n$ line segments.
If none of the 17 line segments form a triangle, then the $n$ points connected to $A$ are not connected to each other. For the remaini... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 718,684 |
3. Consider an infinitely large chessboard, where each square contains a positive integer. If each number in a square is the average of the four numbers in the squares directly above, below, to the left, and to the right, prove: all the numbers are equal.
| ( Hint: If these positive integers are not equal, let $a$ be the smallest among them, then there must be a $b$ on the left adjacent to $a$ and strictly greater than $a$, and it is also equal to the average of the 4 adjacent blocks, each of which is no less than $a$, and one is greater than $a$. Contradiction. ) | proof | Algebra | proof | Yes | Yes | cn_contest | false | 718,685 |
3. Given the equation $x^{2}+(b+2) x y+b y^{2}=0$ $(b \in \mathbf{R})$ represents two lines. Then the range of the angle between them is $\qquad$ . | 3. $\left[\arctan \frac{2 \sqrt{5}}{5}, \frac{\pi}{2}\right]$.
When $b=0$, the two lines represented are $x=0, x+2 y=0$, and their angle is $\arctan 2$.
$$
\begin{array}{l}
\text{When } b \neq 0, \text{ we have } b\left(\frac{y}{x}\right)^{2}+(b+2) \frac{y}{x}+1=0, \\
\Delta=(b+2)^{2}-4 b=b^{2}+4>0 .
\end{array}
$$
Le... | \left[\arctan \frac{2 \sqrt{5}}{5}, \frac{\pi}{2}\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,686 |
4. Given $a_{n}=\cos ^{n} \frac{\pi}{9}+\cos ^{n} \frac{5 \pi}{9}+\cos ^{n} \frac{7 \pi}{9}$ $\left(n \in \mathbf{N}_{+}\right)$. Then the recurrence relation of the sequence $\left\{a_{n}\right\}$ is $\qquad$ . | 4. $a_{n+3}=\frac{3}{4} a_{n+1}+\frac{1}{8} a_{n}, a_{1}=0, a_{2}=\frac{3}{2}, a_{3}=\frac{3}{8}$.
Let $x_{1}=\cos \frac{\pi}{9}, x_{2}=\cos \frac{5 \pi}{9}, x_{3}=\cos \frac{7 \pi}{9}$.
It is easy to see that $x_{1}+x_{2}+x_{3}=0$,
$$
x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}=-\frac{3}{4}, x_{1} x_{2} x_{3}=\frac{1}{8} \te... | a_{n+3}=\frac{3}{4} a_{n+1}+\frac{1}{8} a_{n} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,687 |
5. The coefficient of $x^{150}$ in the expansion of $\left(1+x+x^{2}+\cdots+x^{100}\right)^{3}$, after combining like terms, is $\qquad$ | 5.7651 .
By the polynomial multiplication rule, the problem can be transformed into finding the number of natural number solutions of the equation
$$
s+t+r=150
$$
that do not exceed 100.
Obviously, the number of natural number solutions of equation (1) is $\mathrm{C}_{152}^{2}$.
Next, we find the number of natural nu... | 7651 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,688 |
6. Spheres with radii $1,2,3$ are externally tangent to each other, and planes $\alpha$ and $\beta$ are tangent to all three spheres. Then the dihedral angle formed by plane $\alpha$ and plane $\beta$ is $\qquad$ . | $6.2 \arccos \frac{\sqrt{23}}{6}$.
According to the symmetry, the plane determined by the centers of the three spheres $\mathrm{O}_{1}, \mathrm{O}_{2}, \mathrm{O}_{3}$ must be the bisector plane of plane $\alpha$ and plane $\beta$.
Below, we find the area of $\triangle O_{1} O_{2} O_{3}$ and the area of the projection... | 2 \arccos \frac{\sqrt{23}}{6} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,689 |
Three. (20 points) Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=\frac{1}{4},\left(1-a_{n}\right) a_{n+1}=\frac{1}{4}$.
(1) Find the general term of the sequence $\left\{a_{n}\right\}$;
(2) Prove that: $\frac{a_{2}}{a_{1}}+\frac{a_{3}}{a_{2}}+\cdots+\frac{a_{n+1}}{a_{n}}<n+\frac{3}{4}$. | (1) Let $b_{n}=a_{n}-\frac{1}{2}$, then
$$
\begin{array}{l}
\left(b_{n+1}+\frac{1}{2}\right)\left(\frac{1}{2}-b_{n}\right)=\frac{1}{4} \\
\Rightarrow \frac{1}{2} b_{n+1}-b_{n} b_{n+1}-\frac{1}{2} b_{n}=0 \\
\Rightarrow \frac{1}{b_{n}}-\frac{1}{b_{n+1}}=2 \\
\Rightarrow \frac{1}{b_{n}}=\frac{1}{b_{1}}+(n-1)(-2)=-4-2(n-1... | n+\frac{3}{4} | Algebra | proof | Yes | Yes | cn_contest | false | 718,690 |
Four. (20 points) Given the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>$ $b>0$ ). As shown in Figure 2, $F_{1}$ and $F_{2}$ are its left and right foci, and $P$ is any point on the ellipse $C$ (not an endpoint of the major axis). $PD$ is the angle bisector of $\angle F_{1} P F_{2}$ in the triangle $\triang... | Let $\angle F_{1} P F_{2}=\theta$, it is easy to know that $\angle M D N=180^{\circ}-\theta$,
$|D M|=|D N|=|P D| \sin \frac{\theta}{2}$.
From $S_{\triangle P F_{1} F_{2}}=S_{\triangle P F_{1} D}+S_{\triangle P F_{2} D}$, we get
$$
\begin{array}{l}
\frac{1}{2}\left|P F_{1}\right| \cdot\left|P F_{2}\right| \sin \theta \\... | \frac{b^{2} c^{2}}{a^{4}} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,691 |
$$
\begin{array}{l}
f(x)=x^{3}+a x^{2}+b x+c, \\
g(x)=3 x^{2}+2 a x+b,
\end{array}
$$
and $\triangle A B C$ satisfies the following conditions:
(1) $\angle A, \angle B, \angle C$ form an arithmetic sequence;
(2) $\tan B$ is a root of the equation $f(x)=0$;
(3) $\tan A, \tan C$ are two distinct real roots of the equati... | From condition (1), we get $\angle B=\frac{\pi}{3}, \angle A+\angle C=\frac{2 \pi}{3}$.
From condition (2), we get
$$
f(\tan B)=f(\sqrt{3})=3 \sqrt{3}+3 a+\sqrt{3} b+c=0 \text {. }
$$
From condition (3), we get $\Delta=4 a^{2}-12 b=4\left(a^{2}-3 b\right)>0$,
$\tan A+\tan C=-\frac{2 a}{3}, \tan A \cdot \tan C=\frac{b}... | \begin{array}{l}
f(x)=x^{3}+3 \sqrt{3} x^{2}-3 x-9 \sqrt{3}, \\
g(x)=3 x^{2}+6 \sqrt{3} x-3, \\
\angle A=\frac{\pi}{12}, \angle B=\frac{\pi}{3}, \angle C=\frac{7 \pi}{12}
\end{array} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,692 |
一、(50 points) As shown in Figure 3, $\odot O_{1}$ and $\odot O_{2}$ intersect at points $A$ and $B$. A chord $B C$ of $\odot O_{2}$ is drawn from point $B$, and $A C$ is connected to intersect $\odot O_{1}$ at point $D$. Prove that $B C = C D$ if and only if $\odot O_{2}$ passes through the center $O_{1}$ of $\odot O_{... | As shown in Figure 6, connect $AB$, $O_{1}O_{2}$, $AO_{1}$, $BO_{1}$, $AO_{2}$, $BO_{2}$, and $BD$, and extend $CB$ to intersect $\odot O_{1}$ at $E$, then connect $AE$.
Since $AB$ is the common chord of $\odot O_{1}$ and $\odot O_{2}$, and $O_{1}O_{2}$ is the line connecting the centers, then
$$
\angle AO_{1}O_{2} = ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,693 |
II. (50 points) In an isosceles right triangle $\triangle ABC$, $CA = CB = 1$, and $P$ is any point on the boundary of $\triangle ABC$. Find the maximum value of $PA \cdot PB + PB \cdot PC + PC \cdot PA$. | (1) If $P \in AB$, as shown in Figure 7, let $AP = t$, then $PB = \sqrt{2} - t$, and $PC = \sqrt{t^2 - \sqrt{2} t + 1} (t \in [0, \sqrt{2}])$.
Thus, $m = PA \cdot PB + PB \cdot PC + PC \cdot PA$
$= PA \cdot PB + PC(PA + PB)$
$= t(\sqrt{2} - t) + \sqrt{t^2 - \sqrt{2} t + 1} \cdot \sqrt{2}$.
Let $u = \sqrt{t^2 - \sqrt{2}... | \frac{3}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,694 |
Three. (50 points) Given sets $A_{1}, A_{2}, \cdots, A_{n}$ are different subsets of the set $\{1,2, \cdots, n\}$, satisfying the following conditions:
(i) $i \notin A_{i}$ and $\operatorname{Card}\left(A_{i}\right) \geqslant 3, i=1,2, \cdots, n$;
(ii) $i \in A_{j}$ if and only if $j \notin A_{i}(i \neq j$, $i, j=1,2, ... | (1) Let $A_{1}, A_{2}, \cdots, A_{n}$ be $n$ sets, and let $r_{i}$ be the number of sets that contain element $i$ for $i=1,2, \cdots, n$. Then,
$$
\sum_{i=1}^{n} \operatorname{Card}\left(A_{i}\right)=\sum_{i=1}^{n} r_{i} \text {. }
$$
Assume the $r_{1}$ sets that contain element 1 are
$$
A_{2}, A_{3}, \cdots, A_{r_{1}... | 7 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 718,695 |
4. Let the entire group of people be denoted as $S$. If any two people have the same number of friends within $S$, then they do not have any common friends in $S$. Prove: There is a person in $S$ who has exactly one friend in $S$ (assuming there is indeed someone who is a friend in $S$). | (Consider the person with the most friends, let's say he has $n$ friends $A_{1}, A_{2}, \cdots, A_{n}$. Then the number of friends $A_{1}, A_{2}, \cdots, A_{n}$ have in $S$ does not exceed $n$, and no two of them have the same number of friends. Therefore, their number of friends are $1,2, \cdots, n$.) | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 718,696 |
As shown in Figure 2, $P$ is a point inside $\triangle ABC$. $PD$, $PE$, and $PF$ are the perpendiculars from point $P$ to the sides $BC$, $CA$, and $AB$, with feet of the perpendiculars being $D$, $E$, and $F$, respectively. The circle passing through points $D$, $E$, and $F$ intersects $BC$, $CA$, and $AB$ at three a... | Proof: First, prove a lemma.
Lemma As shown in Figure 3, line $l$ intersects circle $\odot O$ at points $A$ and $B$. Draw perpendicular lines $m$ and $n$ to line $l$ through points $A$ and $B$ respectively, and a line through point $O$ intersects $m$ and $n$ at points $M$ and $N$. Then $M$ and $N$ are symmetric about p... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,697 |
Given that $\triangle ABC$ is an acute triangle inscribed in a circle, and the three chords $AA_1 \parallel BB_1 \parallel CC_1$. Points $A_1$ and $A_2$, $B_1$ and $B_2$, $C_1$ and $C_2$ are symmetric with respect to $BC$, $CA$, and $AB$ respectively. Prove:
$$
\triangle A_2 B_2 C_2 \cong \triangle A B C .
$$ | Proof: As shown in Figure 4, draw the altitudes from points $A$ and $B$ of $\triangle ABC$ to intersect the circumcircle at points $D$ and $E$, respectively, and let $H$ be the orthocenter. It is easy to prove that point $H$ and point $E$ are symmetric with respect to $AC$.
Connect $A B_{1}$, $A B_{2}$, $B A_{1}$, $B ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,698 |
In $\triangle A B C$, $\angle B>\angle A>\angle C$, and $\angle B, \angle A, \angle C$ form an arithmetic sequence. $I$ is the incenter of $\triangle A B C$, the extension of $B I$ intersects $A C$ at point $D$, and the extension of $C I$ intersects $A B$ at point $E$.
(1) Prove that $B C^{2}=B E \cdot B A+C D \cdot C ... | Proof: (1) As shown in Figure 5, since $\angle B, \angle A, \angle C$ form an arithmetic sequence, thus, $2 \angle A = \angle B + \angle C$.
Also, $\angle A + \angle B + \angle C = 180^{\circ}$, then
$$
\begin{array}{l}
\angle A = 60^{\circ}, \\
\frac{\angle B + \angle C}{2} = \angle A = 60^{\circ}.
\end{array}
$$
Si... | \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,699 |
Let $k$ be a real number. Try to find the range of the function
$$
f(x)=x+\frac{1}{x}-k \sqrt{x^{2}+\frac{1}{x^{2}}}(x>0)
$$ | Let $x+\frac{1}{x}=t$, then
$$
t \geqslant 2, f(x)=g(t)=t-k \sqrt{t^{2}-2}(t \geqslant 2) \text {. }
$$
Thus, we only need to find the range of $g(t)(t \geqslant 2)$.
Taking the derivative of $g(t)$, we get
$$
g^{\prime}(t)=1-\frac{k t}{\sqrt{t^{2}-2}}=1-\frac{k}{\sqrt{1-\frac{2}{t^{2}}}} \text {. }
$$
Since $\sqrt{1... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,700 |
5. Take any 21 points on a circle. Prove: Among all the arcs with these points as endpoints, there are no fewer than 100 arcs that do not exceed $120^{\circ}$. | (Assuming among all points, the chord with $A_{1}$ as an endpoint is the least, and denoting the chords with $A_{1}$ as an endpoint as $A_{1} A_{2}, A_{1} A_{3}, \cdots$, $A_{1} A_{n}$, a total of $n-1$ chords, and the chords with $A_{2}, A_{3}, \cdots, A_{n}$ as endpoints are no less than $n-1$ each. Therefore, these ... | 100 | Combinatorics | proof | Yes | Yes | cn_contest | false | 718,701 |
Example 1 Let $S$ be a subset of the set $\{1,2, \cdots, 50\}$ with the following property: the sum of any two distinct elements of $S$ cannot be divisible by 7. Then, what is the maximum number of elements that $S$ can have?
(43rd American High School Mathematics Examination) | For two different natural numbers $a$ and $b$, if $7 \times (a+b)$, then the sum of their remainders when divided by 7 is not 0. Therefore, the set $\{1,2, \cdots, 50\}$ can be divided into 7 subsets based on the remainders when divided by 7. Among them, each element in $K_{i}$ has a remainder of $i$ when divided by 7 ... | 23 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 718,702 |
Example 2 Let $X=\{1,2, \cdots, 2001\}$. Find the smallest positive integer $m$ that satisfies the following condition: for any $m$-element subset $W$ of $X$, there exist $u, v \in W$ (where $u$ and $v$ can be the same), such that $u+v$ is a power of 2.
(2001, China Mathematical Olympiad) | Explanation: When $u$ and $v$ are $2^{r} + a$ and $2^{r} - a$ respectively, $u + v = 2 \times 2^{r} = 2^{r+1}$ is a power of 2. Based on this, $X$ is divided into the following 5 subsets:
$$
\begin{array}{l}
2001 = 2^{10} + 977 \geqslant x \geqslant 2^{10} - 977 = 47, \\
46 = 2^{5} + 14 \geqslant x \geqslant 2^{5} - 14... | 999 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 718,703 |
Example 3 Let $M=\{1,2, \cdots, 1995\}, A$ be a subset of $M$ and satisfy the condition: when $x \in A$, $15 x \notin A$. Then the maximum number of elements in $A$ is $\qquad$
(1995, National High School Mathematics Competition) | Construct the subset $A$ as follows:
Notice that $1995 \div 15=133$, and let
$$
A_{1}=\{134,135, \cdots, 1995\} \text {, }
$$
then $\left|A_{1}\right|=1862$.
Also, $133=15 \times 8+13$, and let
$A_{2}=\{9,10, \cdots, 133\}$,
then $\left|A_{2}\right|=125$.
Let $A_{3}=\{1,2, \cdots, 8\}$, then $\left|A_{3}\right|=8$.
T... | 1870 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 718,704 |
Example 4 Let $n(n \geqslant 2)$ be an integer. $S$ is a subset of the set $\{1,2, \cdots, n\}$, where no number in $S$ divides another, and no two numbers are coprime. Find the maximum number of elements in $S$.
(2005, Balkan Mathematical Olympiad) | Construct a mapping $f$:
$$
S \rightarrow\left\{\left[\frac{n}{2}\right]+1,\left[\frac{n}{2}\right]+2, \cdots, n\right\} \text {, }
$$
$f$ transforms $x \in S$ into $2^{k} x \in\left(\frac{n}{2}, n\right]$ (where $k$ is a non-negative integer), and the elements in the set $\left\{\left[\frac{n}{2}\right]+1,\left[\frac{... | \left[\frac{n+2}{4}\right] | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 718,705 |
Example 5 Given that $A$ and $B$ are finite sets composed of different positive real numbers, $n$ ($n>1$) is a given positive integer, and both $A$ and $B$ have at least $n$ elements. If the sum of any $n$ different real numbers in $A$ belongs to $B$, and the product of any $n$ different real numbers in $B$ belongs to ... | Assume set $A$ contains $m$ elements, denoted as $a_{1}, a_{2}, \cdots, a_{m}$, and satisfies
$0n)$.
Let $S=a_{1}+a_{2}+\cdots+a_{n+1}$,
$P=\left(S-a_{1}\right)\left(S-a_{2}\right) \cdots\left(S-a_{n+1}\right)$.
Then $S-a_{k} \in B(k=1,2, \cdots, n+1)$.
Set $t_{i}=\frac{P}{S-a_{i}}(i=1,2, \cdots, n+1)$,
$$
\begin{array... | 2n | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 718,706 |
Example 6 Let $S=\{1,2, \cdots, 2005\}$. If any set of $n$ pairwise coprime numbers in $S$ contains at least one prime number, find the minimum value of $n$.
(2005, China Western Mathematical Olympiad) | Explanation: First, take a subset of $S$
$$
A_{0}=\left\{1,2^{2}, 3^{2}, 5^{2}, \cdots, 41^{2}, 43^{2}\right\},
$$
then $\left|A_{0}\right|=15, A_{0}$ contains any two numbers that are coprime, but there are no primes in it. This indicates that $n \geqslant 16$.
Second, it can be proven: For any $A \subseteq S, n=|A|... | 16 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 718,707 |
Example 7 For an integer $n(n \geqslant 4)$, find the smallest integer $f(n)$ such that for any positive integer $m$, in any $f(n)$-element subset of the set $\{m, m+1, \cdots, m+n-1\}$, there are at least 3 pairwise coprime elements.
(2004, National High School Mathematics Competition) | Given the condition "for any positive integer $m$", we can take $m=2$. Then, in the set $S=\{2,3, \cdots, n+1\}$, the numbers divisible by 2 are $\left[\frac{n+1}{2}\right]$, the numbers divisible by 3 are $\left[\frac{n+1}{3}\right]$, and the numbers divisible by both 2 and 3 are $\left[\frac{n+1}{6}\right]$. Therefor... | f(n)=\left[\frac{n+1}{2}\right]+\left[\frac{n+1}{3}\right]-\left[\frac{n+1}{6}\right]+1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 718,708 |
Example 3 Suppose there are $n(n \geqslant 7)$ circles, among which any 3 circles do not pairwise intersect (including being tangent). Prove: there must exist a circle that intersects with at most 5 other circles.
---
The translation maintains the original text's line breaks and format. | Consider the smallest circle among these $n$ circles (if there are multiple, choose any one of them) as $\odot O_{1}$. If $\odot O_{1}$ intersects with 6 (or more than 6) circles $\odot O_{2}, \odot O_{3}, \cdots, \odot O_{7}$, then connect $O_{1} O_{2}, O_{1} O_{3}, \cdots, O_{1} O_{7}$ (as shown in Figure 2). Thus, a... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 718,709 |
Example 8 Let sets $A$ and $B$ both consist of positive numbers, with $|A|=10, |B|=9$, and set $A$ satisfies the following condition: If $x, y, u, v \in A, x+y=u+v$, then \begin{aligned}\{x, y\} & =\{u, v\} . \text{ Let } \\ A & +B=\{a+b \mid a \in A, b \in B\} .\end{aligned}
Prove: $|A+B| \geqslant 50$ ($|X|$ denotes... | Explanation: Consider the general case.
Let $|A|=m,|B|=n$,
$A+B=\left\{s_{1}, s_{2}, \cdots, s_{k}\right\}$.
Suppose $s_{i}(1 \leqslant i \leqslant k)$ can be expressed in $f(i)$ ways as $a+b$, where $a \in A, b \in B$, i.e.,
$$
s_{i}=a_{i 1}+b_{i 1}=a_{i 2}+b_{i 2}=\cdots=a_{i(i)}+b_{i f(i)}.
$$
Then $f(1)+f(2)+\cdot... | 50 | Combinatorics | proof | Yes | Yes | cn_contest | false | 718,710 |
2. Let $S$ be a subset of $\{1,2, \cdots, 9\}$ such that the sum of any two distinct elements of $S$ is unique. How many elements can $S$ have at most?
(2002, Canadian Mathematical Olympiad) | (When $S=\{1,2,3,5,8\}$, $S$ meets the requirements of the problem. If $T \subseteq\{1,2, \cdots, 9\},|T| \geqslant 6$, then since the sum of any two different numbers in $T$ is between 3 and 17, at most 15 different sum numbers can be formed. And choosing any two numbers from $T$, there are at least $\mathrm{C}_{6}^{2... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 718,712 |
Example 11 Let $a \geqslant b \geqslant c>0, x \geqslant y \geqslant z>0$. Prove:
$$
\begin{array}{l}
\frac{a^{2} x^{2}}{(b y+c z)(b z+c y)}+\frac{b^{2} y^{2}}{(c z+a x)(c x+a z)}+ \\
\frac{c^{2} z^{2}}{(a x+b y)(a y+b x)} \geqslant \frac{3}{4} .
\end{array}
$$
(2000, Korean Mathematical Olympiad) | Proof: From the binary mean inequality, we have
$$
\begin{array}{l}
(b y+c z)(b z+c y) \\
\leqslant\left[\frac{(b y+c z)+(b z+c y)}{2}\right]^{2} \\
=\frac{(b+c)^{2}(y+z)^{2}}{4} .
\end{array}
$$
Similarly,
$$
\begin{array}{l}
(c z+a x)(c x+a z) \leqslant \frac{(c+a)^{2}(z+x)^{2}}{4}, \\
(a x+b y)(a y+b x) \leqslant \... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,715 |
Example 12 Let $a, b, c$ be positive real numbers. Prove that:
$$
\frac{(b+c-a)^{2}}{a^{2}+(b+c)^{2}}+\frac{(c+a-b)^{2}}{b^{2}+(c+a)^{2}}+\frac{(a+b-c)^{2}}{c^{2}+(a+b)^{2}} \geqslant \frac{3}{5} \text {. }
$$
(1997, Japan Mathematical Olympiad) | Proof: By replacing $a, b, c$ with $\frac{a}{a+b+c}, \frac{b}{a+b+c}, \frac{c}{a+b+c}$, the original inequality remains unchanged. Therefore, without loss of generality, we can assume $0 < a, b, c < 1$ and $a + b + c = 1$. Then,
$$
\begin{array}{l}
\frac{(b+c-a)^{2}}{a^{2}+(b+c)^{2}}=\frac{(1-2 a)^{2}}{a^{2}+(1-a)^{2}}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,716 |
Example 13 Given that $a, b, c$ are positive numbers, and $a^{4}+b^{4}+c^{4}=3$. Prove:
$$
\frac{1}{4-ab}+\frac{1}{4-bc}+\frac{1}{4-ca} \leqslant 1 .
$$
(2005, Moldova Mathematical Olympiad) | Proof: First, prove $\frac{2}{4-a b} \leqslant \frac{1}{4-a^{2}}+\frac{1}{4-b^{2}}$.
From the condition, we have $a^{4}0$.
Similarly, $4-b^{2}>0,4-a b>0$.
Thus,
$$
\frac{1}{4-a^{2}}+\frac{1}{4-b^{2}}-\frac{2}{4-a b}
=\frac{(4+a b)(a-b)^{2}}{\left(4-a^{2}\right)\left(4-b^{2}\right)(4-a b)} \geqslant 0 .
$$
Therefore, $... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,717 |
Example 14 Let $x, y, z \in \mathbf{R}_{+}$, and $x+y+z=1$. Prove:
$$
\frac{x y}{\sqrt{x y+y z}}+\frac{y z}{\sqrt{y z+z x}}+\frac{z x}{\sqrt{z x+x y}} \leqslant \frac{\sqrt{2}}{2} .
$$
(2006, China National Training Team Exam) | $$
\begin{array}{l}
\left(\frac{x y}{\sqrt{x y+y z}}+\frac{y z}{\sqrt{y z+z x}}+\frac{z x}{\sqrt{z x+x y}}\right)^{2} \\
=\left(\frac{x \sqrt{y}}{\sqrt{x+z}}+\frac{y \sqrt{z}}{\sqrt{y+x}}+\frac{z \sqrt{x}}{\sqrt{z+y}}\right)^{2} \\
=\left[\sqrt{x+y} \cdot \frac{x \sqrt{y}}{\sqrt{(x+y)(z+x)}}+\right. \\
\sqrt{y+z} \cdot... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,718 |
Example 15 Let positive real numbers $a, b, c$ satisfy $a+b+c = 1$. Prove:
$$
10\left(a^{3}+b^{3}+c^{3}\right)-9\left(a^{5}+b^{5}+c^{5}\right) \geqslant 1 \text {. }
$$
(2005, China Western Mathematical Olympiad) | Proof: Since $a+b+c=1$, then,
$$
\begin{array}{l}
10\left(a^{3}+b^{3}+c^{3}\right)-9\left(a^{5}+b^{5}+c^{5}\right) \geqslant 1 \text {. } \\
\Leftrightarrow 10\left(a^{3}+b^{3}+c^{3}\right)(a+b+c)^{2}- \\
9\left(a^{5}+b^{5}+c^{5}\right) \geqslant(a+b+c)^{5} \\
\Leftrightarrow 10\left(a^{3}+b^{3}+c^{3}\right)\left(a^{2}... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,719 |
Example 4 There are $n(n \geqslant 5)$ points in the plane. Color them with two colors, red and blue. Suppose no three points of the same color are collinear. Prove that there exists a triangle such that
(1) its three vertices are colored the same color;
(2) this triangle has at least one side that does not contain a p... | Explanation: Since the number of points $n \geqslant 5$, and all points are colored with only two colors, there must be at least three points of the same color. Therefore, there exists a triangle with three vertices of the same color.
Among these triangles with vertices of the same color, take the one with the smalles... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 718,720 |
Example 16 Let $a, b, c$ be positive real numbers, and $abc=1$. Prove:
$$
\frac{a}{a^{2}+2}+\frac{b}{b^{2}+2}+\frac{c}{c^{2}+2} \leqslant 1 .
$$
$(2005$, Baltic Way Mathematical Olympiad) | Proof: Notice
$$
\begin{array}{l}
\frac{a}{a^{2}+2}+\frac{b}{b^{2}+2}+\frac{c}{c^{2}+2} \\
=\frac{a}{a^{2}+1+1}+\frac{b}{b^{2}+1+1}+\frac{c}{c^{2}+1+1} \\
\leqslant \frac{a}{2 a+1}+\frac{b}{2 b+1}+\frac{c}{2 c+1} .
\end{array}
$$
Thus, it suffices to prove
$$
\begin{array}{l}
\frac{a}{2 a+1}+\frac{b}{2 b+1}+\frac{c}{2... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,721 |
Example 1 Arrange all positive divisors of 8128 that are less than itself in ascending order as $a_{1}, a_{2}, \cdots, a_{n}$. Prove that:
$$
\sum_{k=2}^{n} \frac{a_{k}}{k\left(a_{1}^{2}+a_{2}^{2}+\cdots+a_{k}^{2}\right)}<\frac{8127}{8128} .
$$ | Here, 8128 is used as a perfect number, and the divisors are arranged in increasing order. 8128 has 13 proper divisors: $1,2,2^{2}, 2^{3}$,
$$
\begin{array}{l}
2^{4}, 2^{5}, 2^{6}, 2^{7}-1,2^{8}-2,2^{9}-2^{2}, 2^{10}-2^{3}, 2^{11}- \\
2^{4}, 2^{12}-2^{5} \text {, and } \\
a_{1}+a_{2}+\cdots+a_{13}=8128 . \\
\text { The... | \frac{8127}{8128} | Number Theory | proof | Yes | Yes | cn_contest | false | 718,722 |
For positive real numbers $a_{1}, a_{2}, \cdots, a_{n}, b_{1}, b_{2}, \cdots, b_{n}$, we have
$$
\begin{array}{l}
\sum_{k=2}^{n} \frac{a_{k} b_{k}}{\left(a_{1}^{2}+a_{2}^{2}+\cdots+a_{k}^{2}\right)\left(b_{1}^{2}+b_{2}^{2}+\cdots+b_{k}^{2}\right)} \\
<\frac{1}{a_{1} b_{1}}-\frac{1}{a_{1} b_{1}+a_{2} b_{2}+\cdots+a_{n} ... | $$
\begin{array}{l}
\text { Prove: } \sum_{k=2}^{n} \frac{a_{k} b_{k}}{\left(a_{1}^{2}+a_{2}^{2}+\cdots+a_{k}^{2}\right)\left(b_{1}^{2}+b_{2}^{2}+\cdots+b_{k}^{2}\right)} \\
\leqslant \sum_{k=2}^{n} \frac{a_{k} b_{k}}{\left(a_{1} b_{1}+a_{2} b_{2}+\cdots+a_{k} b_{k}\right)^{2}} \\
<\sum_{k=2}^{n} \frac{a_{k} b_{k}}{\le... | \sum_{k=2}^{n} \frac{6}{(2 k+1)\left[2^{2}+3^{2}+\cdots+(k+1)^{2}\right]} < \frac{1}{2} - \frac{3}{n(n+1)(n+2)} | Inequalities | proof | Yes | Yes | cn_contest | false | 718,723 |
Example 4 Let the positive divisors of 8128 be denoted as $a_{1}, a_{2}, \cdots$, $a_{n+1}$, where $a_{1}=1, a_{n+1}=8128$. Prove:
$$
\sum_{k=2}^{n} \frac{a_{k}}{k\left(a_{1}^{2}+a_{2}^{2}+\cdots+a_{k}^{2}\right)}<\frac{1}{2} .
$$ | Prove: Since $8128=2^{6}\left(2^{7}-1\right)$ is a perfect number, we have
$$
\begin{array}{l}
\frac{1}{a_{2}}+\frac{1}{a_{3}}+\cdots+\frac{1}{a_{n+1}}=1 . \\
\text { Then } \sum_{k=2}^{n} \frac{a_{k}}{k\left(a_{1}^{2}+a_{2}^{2}+\cdots+a_{k}^{2}\right)} \\
<\sum_{k=2}^{n} \frac{a_{k}}{k a_{k}^{2}} \\
=\sum_{k=2}^{n} \f... | \frac{1}{2} | Number Theory | proof | Yes | Yes | cn_contest | false | 718,725 |
Example 5 Let $a$ be a perfect number, and let the positive divisors of $a$ be denoted as $a_{1}, a_{2}, \cdots, a_{n+1}$, where $a_{1}=1, a_{n+1}=a$. Prove:
$$
\sum_{k=2}^{n} \frac{a_{k}}{k\left(a_{1}^{2}+a_{2}^{2}+\cdots+a_{k}^{2}\right)}<\frac{a-1}{2 a} .
$$ | Proof: Knowing that $a$ is a perfect number,
$$
\begin{array}{l}
\frac{1}{a_{2}}+\frac{1}{a_{3}}+\cdots+\frac{1}{a_{n+1}}=1 . \\
\text { Hence } \sum_{k=2}^{n} \frac{a_{k}}{k\left(a_{1}^{2}+a_{2}^{2}+\cdots+a_{k}^{2}\right)} \\
<\sum_{k=2}^{n} \frac{a_{k}}{k a_{k}^{2}} \\
=\sum_{k=2}^{n} \frac{1}{k a_{k}} \\
<\frac{1}{... | \frac{a-1}{2a} | Number Theory | proof | Yes | Yes | cn_contest | false | 718,726 |
Question 1 Let $x, y, z > 0, x + y + z = 1$. Prove:
$$
\frac{x y}{\sqrt{x y + y z}} + \frac{y z}{\sqrt{y z + z x}} + \frac{z x}{\sqrt{z x + x y}} \leqslant \frac{\sqrt{2}}{2} \text{. }
$$
(2006, China National Training Team Exam (4)) | Proof 1: By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
\frac{x y}{\sqrt{x y+y z}}+\frac{y z}{\sqrt{y z+z x}}+\frac{z x}{\sqrt{z x+x y}} \\
=\frac{x \sqrt{y}}{\sqrt{x+z}}+\frac{y \sqrt{z}}{\sqrt{y+x}}+\frac{z \sqrt{x}}{\sqrt{z+y}} \\
=\sqrt{x+y} \cdot \frac{x \sqrt{y}}{\sqrt{(x+y)(x+z)}}+ \\
\sqrt{y+z} \... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,728 |
Given $E, F$ are the midpoints of sides $AB, AC$ of $\triangle ABC$, and $CM, BN$ are the altitudes on sides $AB, AC$, respectively. Connecting $EF$ and $MN$ intersect at point $P$. Also, let $O, H$ be the circumcenter and orthocenter of $\triangle ABC$, respectively, and connect $AP, OH$. Prove: $AP \perp OH$. ${ }^{[... | Proof: As shown in Figure 1, connect $A O$ and $A H$. Let the midpoints of segments $A O$ and $A H$ be $O_{1}$ and $H_{1}$, respectively. Then $O H \parallel O_{1} H_{1}$. Therefore, it is sufficient to prove that $A P \perp O_{1} H_{1}$.
Since $\angle A M H = \angle A N H = 90^{\circ}$, we have
$$
H_{1} M = H_{1} N = ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,729 |
Question 3 Let $x, y, z$ be real numbers, $0 < x, y, z < \pi$. Prove that:
$$
\begin{array}{l}
\sin x \sin y \sin z < \sin 2 x+\sin 2 y+\sin 2 z .
\end{array}
$$
(1990, China National Training Team Test Question) | Prove: After rearranging the inequality to be proven, it is equivalent to proving:
$$
\begin{array}{l}
\sin x(\cos x-\cos y)+\sin y(\cos y-\cos z)+ \\
\sin z \cdot \cos z<\frac{\pi}{4} .
\end{array}
$$
Since $0<x<y<\frac{\pi}{2}$, we have
$$
\sin x<\sin y, \cos y<\cos x.
$$
Then $\sin x(\cos x-\cos y)$
$$
<\frac{1}{2... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 718,730 |
Example 5 Proof: The indeterminate equation $x^{3}+2 y^{3}=4 z^{3}$ has no positive integer solutions $(x, y, z)$. | Explanation: Using proof by contradiction.
Assume the equation has a positive integer solution. Let $\left(x_{1}, y_{1}, z_{1}\right)$ be the solution with the smallest $x$ among all positive integer solutions.
Since $x_{1}^{3}+2 y_{1}^{3}=4 z_{1}^{3}$, $x_{1}^{3}$ is even.
Therefore, $x_{1}$ is even.
Let $x_{1}=2 x_{2... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 718,731 |
Question As shown in Figure 1, given quadrilateral $A B C D$, a moving line $l$ through point $A$ intersects the rays $B C$ and $D C$ at points $X$ and $Y$ respectively. In $\triangle A B X$, the excenter opposite to $\angle B A X$ is $K$, and in $\triangle A D Y$, the excenter opposite to $\angle D A Y$ is $L$. Prove ... | Proof: Let $\angle B A X=2 \alpha, \angle D A Y=2 \beta$, then we have
$$
\begin{array}{l}
\angle K A B=\angle K A X=\alpha, \\
\angle L A D=\angle I A Y=\beta .
\end{array}
$$
Let the points on the extensions of line segments $A B$ and $A D$ be $B^{\prime}$ and $D^{\prime}$, respectively, then
$$
\angle L D D^{\prime... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 718,732 |
1. Given $x, y, z$ satisfy $\frac{2}{x}=\frac{3}{y-z}=\frac{5}{z+x}$. Then the value of $\frac{5 x-y}{y+2 z}$ is ( ).
(A) 1
(B) $\frac{1}{3}$
(C) $-\frac{1}{3}$
(D) $\frac{1}{2}$ | -、1.B.
From $\frac{2}{x}=\frac{3}{y-z}=\frac{5}{z+x}$, we get $y=3x, z=\frac{3}{2}x$. Therefore, $\frac{5x-y}{y+2z}=\frac{5x-3x}{3x+3x}=\frac{1}{3}$. Note: This problem can also be solved using the method of special values. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,733 |
2. When $x$ takes the values $\frac{1}{2007}, \frac{1}{2006}, \cdots, \frac{1}{2}, 1$, $2, \cdots, 2006, 2007$, calculate the value of the algebraic expression $\frac{1-x^{2}}{1+x^{2}}$, and add up the results obtained. The sum is equal to ( ).
(A) -1
(B) 1
(C) 0
(D) 2007 | 2.C.
Notice that
$$
\frac{1-\left(\frac{1}{n}\right)^{2}}{1+\left(\frac{1}{n}\right)^{2}}+\frac{1-n^{2}}{1+n^{2}}=\frac{n^{2}-1}{n^{2}+1}+\frac{1-n^{2}}{1+n^{2}}=0,
$$
that is, when $x$ takes the values $\frac{1}{n}$ and $n\left(n \in \mathbf{N}_{+}\right)$, the sum of the values of the algebraic expressions calculat... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,734 |
3. Let $a$, $b$, and $c$ be the lengths of the sides of $\triangle ABC$, and the quadratic function $y=\left(a-\frac{b}{2}\right) x^{2}-c x-a-\frac{b}{2}$ has a minimum value of $-\frac{8}{5} b$ at $x=1$. Then $\triangle ABC$ is ( ).
(A) isosceles triangle
(B) acute triangle
(C) obtuse triangle
(D) right triangle | 3. D.
From the problem, we have $\left\{\begin{array}{l}-\frac{-c}{2\left(a-\frac{b}{2}\right)}=1, \\ a-\frac{b}{2}-c-a-\frac{b}{2}=-\frac{8}{5} b .\end{array}\right.$
Therefore, $c=\frac{3}{5} b, a=\frac{4}{5} b$.
Thus, $a^{2}+c^{2}=b^{2}$.
Hence, $\triangle A B C$ is a right triangle. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 718,735 |
4. Given that in the acute triangle $\triangle A B C$, the distance from vertex $A$ to the orthocenter $H$ is equal to the radius of its circumcircle. Then the degree of $\angle A$ is ( ).
(A) $30^{\circ}$
(B) $45^{\circ}$
(C) $60^{\circ}$
(D) $75^{\circ}$ | 4.C.
In the acute triangle $\triangle ABC$, the orthocenter is inside the triangle. As shown in Figure 2, let the circumcenter of $\triangle ABC$ be $O$, and $D$ be the midpoint of $BC$. The extension of $BO$ intersects $\odot O$ at point $E$. Connecting $CE$, $AE$, and $CH$, we know that $CE \parallel AH$, $AE \paral... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,736 |
5. Let $K$ be any point inside $\triangle ABC$, and the centroids of $\triangle KAB$, $\triangle KBC$, and $\triangle KCA$ be $D$, $E$, and $F$ respectively. Then the ratio $S_{\triangle DEF}: S_{\triangle ABC}$ is ( ).
(A) $\frac{1}{9}$
(B) $\frac{2}{9}$
(C) $\frac{4}{9}$
(D) $\frac{2}{3}$ | 5.A.
Extend $K D$, $K E$, $K F$ to intersect the sides $A B$, $B C$, and $C A$ of $\triangle A B C$ at points $M$, $N$, and $P$ respectively. Since $D$, $E$, and $F$ are the centroids of $\triangle K A B$, $\triangle K B C$, and $\triangle K C A$ respectively, it is easy to see that $M$, $N$, and $P$ are the midpoints... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 718,737 |
6. A bag contains 5 red balls, 6 black balls, and 7 white balls. Now, 15 balls are drawn from the bag. The probability that exactly 3 of the drawn balls are red is ( ).
(A) $\frac{6}{65}$
(B) $\frac{65}{408}$
(C) $\frac{13}{816}$
(D) $\frac{13}{4896}$ | 6.B.
There are a total of $5+6+7=18$ balls in the bag. Drawing 15 balls is equivalent to leaving 3 balls in the bag. Considering that the balls are indistinguishable in order, the number of different drawing results is $\frac{18 \times 17 \times 16}{3 \times 2 \times 1}=816$.
If exactly 3 red balls are drawn, we firs... | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 718,738 |
1. Let $x=\frac{1}{\sqrt{2}-1}, a$ be the fractional part of $x$, and $b$ be the fractional part of $-x$. Then $a^{3}+b^{3}+3 a b=$ $\qquad$ . | ニ、1. 1 .
Since $x=\frac{1}{\sqrt{2}-1}=\sqrt{2}+1$, and $2<\sqrt{2}+1<3$, therefore,
$$
\begin{array}{l}
a=x-2=\sqrt{2}-1 . \\
\text { Also }-x=-\sqrt{2}-1, \text { and }-3<-\sqrt{2}-1<-2 \text {, so, } \\
b=-x-(-3)=2-\sqrt{2} .
\end{array}
$$
Then $a+b=1$.
Thus $a^{3}+b^{3}+3 a b=(a+b)\left(a^{2}-a b+b^{2}\right)+3 a... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,739 |
2. For all natural numbers $n$ not less than 2, the two roots of the quadratic equation in $x$, $x^{2}-(n+2) x-2 n^{2}=0$, are denoted as $a_{n} 、 b_{n}(n \geqslant 2)$. Then
$$
\begin{array}{c}
\frac{1}{\left(a_{2}-2\right)\left(b_{2}-2\right)}+\frac{1}{\left(a_{3}-2\right)\left(b_{3}-2\right)}+ \\
\cdots+\frac{1}{\le... | 2. $-\frac{1003}{4016}$.
By the relationship between roots and coefficients, we have
$$
\begin{array}{l}
a_{n}+b_{n}=n+2, a_{n} b_{n}=-2 n^{2} . \\
\text { By }\left(a_{n}-2\right)\left(b_{n}-2\right)=a_{n} b_{n}-2\left(a_{n}+b_{n}\right)+4 \\
=-2 n^{2}-2(n+2)+4=-2 n(n+1),
\end{array}
$$
we have
$$
\begin{array}{l}
\... | -\frac{1003}{4016} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 718,740 |
3. Given a right trapezoid $A B C D$ with side lengths $A B=2, B C=C D=10, A D=6$, a circle is drawn through points $B$ and $D$, intersecting the extension of $B A$ at point $E$ and the extension of $C B$ at point $F$. Then the value of $B E-B F$ is $\qquad$ | 3.4 .
As shown in Figure 3, extend $C D$ to intersect $\odot O$ at point $G$. Let the midpoints of $B E$ and $D G$ be $M$ and $N$, respectively. It is easy to see that $A M = D N$. Since $B C = C D = 10$, by the secant theorem, it is easy to prove that
$$
\begin{aligned}
B F = & D G. \text{ Therefore, } \\
& B E - B F... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 718,741 |
Example 6 Let positive integers $a, b, k$ satisfy $\frac{a^{2}+b^{2}}{a b-1}=k$. Prove: $k=5$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | For every $k$ that satisfies the above equation, let $a_{0}, b_{0}$ satisfy $\frac{a_{0}^{2}+b_{0}^{2}}{a_{0} b_{0}-1}=k$, and $a_{0}+b_{0}$ is the smallest pair. Without loss of generality, assume $a_{0} \geqslant b_{0}$.
(1) If $a_{0}=b_{0}$, then
$$
k=\frac{a_{0}^{2}+b_{0}^{2}}{a_{0} b_{0}-1}=2+\frac{2}{a_{0}^{2}-1}... | 5 | Number Theory | proof | Yes | Yes | cn_contest | false | 718,742 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.